Category: Class 11

MP Board Class 11th Chemistry Important Questions Chapter 10 s – Block Elements

s – Block Elements Important Questions

s – Block Elements Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Plaster of Paris is :
(a) (CaSO₄)₂.H₂O
(b) CaSO₄.2H₂O
(c) CaSO₄.H₂O
(d) CaSO₄
Answer:
(a) (CaSO₄)₂.H₂O

Question 2.
Lowest melting point compound:
(a) LiCl
(b)NaCl
(c) KCl
(d) RbCl
Answer:
(a) LiCl

Question 3.
Active constituent of Bleaching powder:
(a) CaOCl₂
(b) Ca(OCl)Cl
(c) Ca(O₂Cl₂)
(d) CaCl₂O₂
Answer:
(b) Ca(OCl)Cl

Question 4.
Hydration energy of Mg²⁺ ion will be more than:
(a) Al³⁺
(b) Na⁺
(c) Be²⁺
(d) Mg³⁺
Answer:
(b) Na⁺

Question 5.
Which magnetic property is present in alkaline earth metals:
(a) Diamagnetic
(b) Paramagnetic
(c) Ferro – magnetic
(d) Anti – magnetic
Answer:
(a) Diamagnetic

Question 6.
Solubility of which sulphate is the least:
(a) BaSO₄
(b) MgSO₄
(c) SrSO₄
(d) CaSO₄
Answer:
(a) BaSO₄

Question 7.
Compounds of which element are mainly covalent:
(a) Ba
(b) Sr
(c) Ca
(d) Be
Answer:
(d) Be

Question 8.
Important ore of magnesium is:
(a) Malachite
(b) Kaesiterite
(c) Camallite
(d) Galena.
Answer:
(c) Camallite

Question 2.
Fill in the blanks:

1. Radius of Ca²⁺ ion is less than K⁺ because of ……………………………
2. On heating Rb(ICl₂) decomposes to form ………………………….. and ………………………….
3. Be(OH)₂ is soluble in both acid and base because it is of ……………………… nature.
4. Lithium resembles ………………………… element of group 2.
5. Sodium metal gives blue colour in liquid ammonial this is because of ………………………………
6. Potassium forms three oxides ………………………. and …………………………..

Answer:

1. High positive charge
2. RbCl, ICI
3. Amphoteric
4. Mg
5. e^–(NH₃)₂
6. K₂O, K₂O₂, KO₂

Question 3.
Answer in one word/sentence:

1. Sodium metals are kept in kerosene oil. Why?
2. What is lithopone?
3. Which element is present in teeth and bones?
4. What is Caustic soda?
5. What is Sorrel cement?
6. Write formula of Plaster of Paris?

Answer:

1. Highly reactive
2. BaSO₄ and ZnS
3. Calcium
4. NaOH
5. Mixture of MgCl₂ and MgO
6. (CaSO₄)2H₂O

s – Block Elements Very Short Answer Type Questions

Question 1.
Arrange the alkali metals in the increasing order of their reactivity?
Answer:
The reactivity of alkali metals increases from top to bottom.
Be < Mg < Ca < Sr < Ba < Ra.

Question 2.
Write the name of two ores of lithium with formula?
Answer:
Ores of lithium are:

1. Spodumene LiAl(SiO₃)₂
2. Lepidolite Li₂Al₂(SiO₃)₃.F(OH)₂.

Question 3.
Write the name of two ores of magnesium?
Answer:

1. Camolite
2. Dolomite.

Question 4.
Why alkali metals are kept in kerosene oil?
Answer:
Due to high reactivity.

Question 5.
What is the formula of global salt?
Answer:
Na₂SO₄.10H₂O.

Question 6.
What is Lithophone?
Answer:
BaSO₄ + ZnS.

Question 7.
What is the formula of hydrolite?
Answer:
CaH₂.

Question 8.
Carnolite is ore of which metal?
Answer:
Magnesium (Mg).

Question 9.
Which alkali metal show radioactivity?
Answer:
Radium.

Question 10.
Which compound is useful in purification of air in aircrafts?
Answer:
Potassium superoxide.

Question 11.
Name one mineral in which Ca and Mg both are present?
Answer:
Name: Dolomite, Formula: MgCO₃.CaCO₃.

Question 12.
Which flux is used for the removal of acidic impurities in metallic processes?
Answer:

1. Limestone CaCO₃
2. Magnesite MgCO₃.

Question 13.
Formula of Nitrolium is?
Answer:
CaCN₂ and C.

Question 14.
Which forms curtain of smoke?
Answer:
SiCl₄.

Question 15.
What is used to join the fractured bone and make statues?
Answer:
Plaster of paris.

Question 16.
What is the order of stability of carbonates of alkali metals?
Answer:
BeCO₃ < MgCO₃ < CaCO₃ < SrCO₃.

Question 17.
Write the configuration of alkali metals?
Answer:
ns^1-2.

Question 18.
What is the formation of Apsum and Gypsum salt?
Answer:
Magnesium sulphate (MgSO₄.2H₂O) and Calcium sulphate (CaSO₄.2H₂O).

s – Block Elements Short Answer Type Questions – I

Question 1.
Why alkali metal do not found in free state?
Answer:
Alkali metals are highly reactive and electropositive because the value of ionization energy is very low due to larger size thus alkali metals easily donate electron and form positive ion. (MPBoardSolutions.com) These positive ion easily combine with oxygen, moisture and other electronegative element present in nature and form ionic compound.

Question 2.
Explain, why sodium is less reactive than potassium?
Answer:
The ionization enthalpy of sodium is higher than that of potassium. Therefore, sodium lose electron less readily as compared to potassium. Hence, sodium is less reactive than potassium.

Question 3.
Potassium carbonate cannot be prepared by Solvay process. Why?
Answer:
Potassium carbonate cannot be prepared by Solvay process because potassium bicarbonate being more soluble than sodium bicarbonate and does not get precipitated when CO₂ is passed through a concentrated solution of KCl with NH₃.

Question 4.
Why is Li₂CO₃ decomposed at a lower temperature whereas Na₂CO₃ at higher temperature?
Answer:
Lithium is less electropositive than sodium and therefore, carbonates of lithium is less stable than that of sodium. Li₂CO₃ is not so stable to heat and therefore, decomposes at lower temperature. (MPBoardSolutions.com) This is because lithium being veiy small in size polarizes a large. CO₃^2- ion leading to the formation of Li₂O and CO₂. On the other hand, Na₂CO₃ is very stable and decomposes at higher.

Question 5.
I – A and II – A group is called s – block elements. Why?
Answer:
I – A and II – A group elements have their electronic configuration of ns¹ and ns². So the last electron fill in 5 – block so the I – A and II – A group is called 5 – block elements.

Question 6.
What is Soral cement and write its uses?
Answer:
When MgCl₂ solution reacts with MgO and product is formed as MgCl₂.2MgO. nH₂O. It is a white paste and called as Soral cement.
Uses:

1. It is used to fill the teeth cavity.
2. They use in porcelain in point.

Question 7.
Find the oxidation state (O.S.) of Na in Na₂O₂?
Answer:
Let the O.S. of Na in Na₂O₂ = x
One peroxide bond (Na – O – O) is present in Na₂O₂ where the O.S. of O = -1

So, in Na₂O₂ the O.S. of Na = +1.

Question 8.
How calcium sulphate is prepared? Write its uses?
Answer:
In the lab, calcium sulphate is formed by Ca oxide, carbonate, chloride. They react with dil. H₂SO₄ to form calcium sulphate.
CaO + H₂SO₄ → CaSO₄ + H₂
Ca(OH)₂ + H₂SO₄ → CaSO₄ + 2H₂O
CaCl₂ + H₂SO₄ → CaSO₄ + 2HCl
CaCO₃ + H₂SO₄ → CaSO₄ + H₂O + CO₂

Question 9.
What is Gypsum? How plaster of Paris is formed by gypsum?
Answer:
Calcium sulphate CaSO₄.2H₂O is called gypsum. Gypsum is heated at 120° – 130°C and 3 parts of water molecule are released and plaster of Paris is formed.

The plaster of Paris again absorb the water molecule and again get converted to calcium sulphate (Gypsum).

Question 10.
Why during the formation of quick lime the temperature of furnance is not kept more than 1000°C? Explain with equation?
Answer:
During the formation of quick, lime the temperature is maintained at 1000°C because at high temperature the clay present as impurity in the limestone combines with lime producing fusible silicate which fill the pores of lime. Due to this reason slaking of lime become very difficult.
1000°C.
CaO + SiO₂ \(\underrightarrow { 1000^{ \circ }C } \) CaSiO_3.

Question 11.
Why sodium kept in kerosene oil?
Answer:
Alkali metals are very reactive specially against electronegative elements like oxygen, moisture and CO₂. These are oxidized quickly from oxide and hydroxide due to their reactivity. Sodium is kept in kerosene oil to prevent oxidation on exposure to air.
4Na + O₂ → 2Na₂O
2Na + H₂O → Na₂O + H₂
Na₂O + H₂O → 2NaOH.

Question 12.
Write the formula of lime water. What happens when CO₂ passes through it?
Answer:
The formula of lime water is Ca(OH)₂. Due to the flow of the CO₂ gas in Ca(OH)₂ it becomes milky due to formation of a white precipitate of calcium carbonate. (MPBoardSolutions.com) On passing excess of CO₂, milkyness disappears and calcium bicarbonate is formed which is soluble in water.
Ca(OH)₂ + CO₂ → CaCO₃ + H₂O
CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂.

Question 13.
Why K₂CO₃ is not prepared by Solvay method?
Answer:
Potassium carbonate cannot be prepared by Solvay method because the potassium salt analogous to NaHCO₃ is KHCO₃ which is much soluble and hence, cannot be obtained by crystallization.

Question 14.
Which metal is used in photochemical cell and why?
Answer:
In photochemical cell potassium and caesium metals are used, as their ionization energy is very low.

Question 15.
Why the extraction of sodium from sodium chloride cannot be done by general reducing agents?
Answer:
Sodium is a strong reducing agent. In electrochemical series, it is present at top position. Due to new availability of strong reducing agent than sodium, it cannot be reduced by normal reducing agents. It can only be reduced by electrolytes.

Question 16.
Why Li and Be have the tendency to form covalent compounds? Explain?
Answer:
Due to small size of Li and Be atoms and their high ionization energy, the electrons of the valence shell are firmly bound to their nuclei. The polarizing power of their ions is also high due to high charge density, hence Li and Be form covalent compounds.

Question 17.
Write the Lewis structure of O^–₂ ion and write the O.N. of each O atom. In this ion what is the average oxidation state of O?
Answer:
Lewis structure of O^–₂ ion =
Oxygen without charge have 6 electrons. So, its O.S. = 0, but oxygen with -1 oxidation state have 7 electrons so its O.S. = —1.
Oxidation state of each oxygen atom = \(\frac{-1}{2}\)
O₂^– = 2x = -1
x = \(\frac{-1}{2}\)

Question 18.
In Solvay’s process, can be obtain sodium carbonate by direct reaction of ammonium carbonate and sodium chloride?
Answer:
No, because the reaction between ammonium carbonate and NaCl.
(NH₄)₂CO₃ + 2NaCl ⇄ Na₂CO₃ + 2NH₄Cl
As the products obtained are highly soluble the equilibrium will not shift to,yards forward direction. This is the reason why NaCO₃ cannot be prepared by reaction of (NH₄)₂CO₃ and NaCl in Solvay’s process.

Question 19.
AH compounds of alkali metals are easily soluble in water but lithium compounds are more soluble in organic solvents? Explain?
Answer:
The size of LT ion is small and thus, it has high polarizing power. This brings covalent character in lithium compounds. Due to covalent character, Li compounds are soluble in organic solvents.

Question 20.
Why are potassium and caesium rather than lithium used n photoelectric cells?
Answer:
Potassium and caesium have much lower ionization enthalpy than that of lithium. Therefore, these metals on exposure to light, easily emit elec arm but lithium does nN. Therefore, K and Cs rather than Li are used in photoelectric cells.

Question 21.
Beryllium chloride (BeCl₂) produces smoke when kept in air. Why?
Answer:
Normally air contains moisture, thus beryllium halide hydrolyses in water and releases HCl due to which it produces smoke in air.
BeCl₂ + 2H₄O → Be(OH)₂ + 2HCl.

Question 22.
On moving downward in the first group the hardness of the elements increases. Why?
Answer:
In the first group on moving downwards along with the increase in size cf the elements their density also increases and the force of attraction between its atoms increases due to which there is an increase in their hardness.

s – Block Elements Short Answer Type Questions – II

Question 1.
Why s – block and p – block elements are called representative elements?
Answer:
The elements present in the s and p-block are called normal or representative elements.

1. The elements of group 1 and 2 constitute the s – block of the periodic table. These are known as s – block elements because the last electron in them enters the s – orbital of the valence shell.
2. The elements of p group of the periodic table these are known as p – block elements because the last electron in them enters the p-orbital of the valence shell.

Question 2.
Why alkaline metals are strong or reducing agents?
Answer:
Due to less ionization energy of alkali metals. They have tendency to loose electron (Get oxidized) and form positive ion (M⁺) also alkali metals have negative value of standard reduction potential. (MPBoardSolutions.com) Therefore, alkali metals are good reducing agent.

Question 3.
Why are BeSO₄ and MgSO₄ readily soluble in water while CaSO₄, SrSO₂ and BaSO₄ are insoluble?
Answer:
The hydration energy of SO₄ of alkaline earth metals decreases down the group. The high hydration energy of Be²⁺ and Mg²⁺ diminishes the lattice energy due to this their SO₂ are soluble in water. But in case of Ca²⁺, Sr²⁺ and Ba²⁺ the hydration energy is low due to this the SO₄ are insoluble in water.

Question 4.
Why the reducing power of lithium is high in solution?
Answer:
Electrode potential is a measure of the tendency of an element to loose electron in aqueous solution. It may depend on the following three factors:

With the small size of its ion, Li has the highest hydration enthalpy. However, ionization enthalpy of Li is highest among alkali metals but hydration (MPBoardSolutions.com) enthalpy predominates over I.P. Therefore, Li is the strongest reducing agent in aqueous solution.

Question 5.
Explain, why can alkali and alkaline earth metals not be obtained by chemical reduction method?
Answer:
The s – block elements themselves are good reducing agents therefore, reducing agent better than s – block elements are not available. Therefore, the chlorides, oxides etc. of s – block elements cannot be reduced to obtain metal.

Question 6.
Explain why alkaline metals form M⁺ cation not M⁺² type of cation?
Answer:
There is one electron in the valence shell of alkali metals. The ionization energy is very low due to bigger size. So they can easily donate electron and can form M⁺ cation. (MPBoardSolutions.com) In the M⁺ state the electronic configuration becomes similar to nobel gases and stable so in this state they become unreactive and the ionization energy become high. So they did not form M²⁺ ion.

Question 7.
In alkali metals, which metal is strongest reducing agent. Why?
Answer:
The reducing character increases from sodium to caesium. However, lithium is the strongest reducing agent among all the alkali metals. Inspite of its highest I.P. This is because of extensive hydration of Li⁺ ions and large amount of energy released during hydration more than compensates the higher I.P. value of Lithium.

Question 8.
Why alkali metals are not found in free state in nature? Or, Why alkali metals always form ionic compounds?
Answer:
Due to bigger size of alkali metals, the ionization energy is very low. So they can easily donate electron and form positive ion. So due to positively charge and highly reactive nature they combine with the (MPBoardSolutions.com) electronegative elements in the nature like moisture, CO₂ etc. and form ionic compounds. Therefore, they cannot be found in free state in nature.

Question 9.
Why alkali metal give flame test?
Answer:
The ionization energy of alkali metal is very low. When these elements or their compounds are heated in Bunsen flame electron present in valency shell absorb energy and easily goes to higher energy level. When these excited electron comes to ground state they emit energy in the form of radiation in visible region and give characteristic colour to flame.

Question 10.
Why Be and Mg do not give flame test?
Answer:
Beryllium and Magnesium atoms in comparison to other alkaline earth metals are comparatively smaller and their ionization energy are very high. (MPBoardSolutions.com) Hence, the energy of the flame is not sufficient to excite their electrons to higher energy levels. These elements therefore do not give any colour in Bunsen flame temperature.

Question 11.
When an alkali metal dissolves in liquid ammonia, the solution acquires different colours? Explain the reason for this type colour change?
Answer:
All alkali metals dissolve in liquid ammonia giving highly conducting deep blue solutions.
M + (x + y )NH₃ M⁺ (NH₃) → x + e^– (NH₃ )_y
When ordinary light falls on these ammoniated electrons. They get excited and jump to higher energy levels by absorbing energy corresponding to red region of the visible light. (MPBoardSolutions.com) As a result transmitted light is blue which imparts blue colour to the solution. However, when the concentration increases the ammoniated metal ion may get bound by free electrons and colour becomes copper bronze.

Question 12.
Why Be and Mg do not give flame test but other metals give? Why?
Answer:
Except beryllium and magnesium all the alkaline earth metals impart characteristic colours to Bunsen flame. Due to small size of Be and Mg atom, the energy required to excite the valency electrons is very high which is not obtained in Bunsen flame. That is why Be and Mg do not impart any colour to flame.

Question 13.
Give two uses of each:

1. Caustic soda
2. Sodium carbonate
3. Quick lime.

Answer:
1. Uses of Caustic soda:

• In soap and paper industry.
• For mercerization of cotton thread.

2. Uses of Sodium carbonate:

• Washing soda is used for washing clothes.
• Removes permanent hardness of water.

3. Uses of Quick lime:

• For making statues, floor, buildings in the form of marble.
• For manufacture of lime, cement, glass and washing soda.

Question 14.
Alkali metals form blue coloured solution when dissolved in ammonia, which is a strong electrolyte. Give reason with equation?
Answer:
Alkali metals dissolve in liquor ammonia to form deep blue coloured solution of high electrical conductivity.
M + (x+y) NH₃ → [M(NH₃)_x] + [e(NH₃)_y]^–
The blue colour of the solution is due to ammoniated electrons positive ion and electrons are responsible for the conductivity.

Question 15.
Na is alkaline or Na₂O? Clarify the statement?
Answer:
Monoxide of all alkali metals are alkaline and form strongly alkaline solution in water.

Na₂O is alkaline because it reacts with water to form NaOH. Na also react with water to form NaOH but it first forms Na₂O and then NaOH, thus Na₂O us alkaline not Na.
4Na + 2H₂O → 2Na₂O + 2H₂
Na₂O + H₂O → 2NaOH

Question 16.
Compare Alkali metals and Alkaline earth metals on the basis of following points:

1. Reaction of heat on carbonate
2. Reaction with nitrogen
3. Solubility of sulphates in water.

Answer:
Comparison of Alkali metals and Alkaline earth metals:

Question 17.
Hydroxides and Carbonates of (Sodium and Potassium) Alkali metals are completely soluble in water whereas that of (Magnesium and Calcium) Alkaline earth metals are partially soluble?
Answer:
All alkali metal carbonates are soluble in water because the value of their Lattice energy is less than their hydration energy.
Lattice energy < Hydration energy (Compound soluble)
Alkaline earth metal carbonates are insoluble because the value of their Lattice energy is more than their hydration energy.
Lattice energy < Hydration energy (Compound insoluble)
∆H_solution = ∆H_{Hydrogen energy} + ∆H_{Lattice energy}

Question 18.
Why are lithium salts commonly hydrated and those of the other alkali ions usually anhydrous?
Answer:
Because of smallest size among alkali metals, Li⁺ can polarize water molecule more easily than the other alkali metal ions and hence get attached to lithium salts as water of crystallization.

Question 19.
Why is LiF almost soluble in water whereas LiCI soluble not only in water but also in acetone?
Answer:
To make compound water soluble, its lattice energy should be low and hydration energy should be high. Lattice energy of LiCl is less than that of LiF. The difference in these two energies for LiCl and LiF is 31 and 14 kJ/mol respectively. (MPBoardSolutions.com) The difference is larger for LiCl, which is soluble in water. Moreover, lattice energy of LiF is much higher than that of LiCl. It makes LiF sparingly soluble in water. Moreover, LiCl is largely covalent, so it is soluble in organic solvent such as acetone.

Question 20.
If the alkali metals are kept open in the air, after sometime the metallic brightness is lost. Why?
Answer:
On exposure to moist air alkali metals soon get covered with a thick crust of their oxide, hydroxides and carbonate hence, their surface get tarnished. For this reason these metals are stored under kerosene and which prevents them from coming in contact with air and moisture.
4M + O₂ → 2M₂O
M₂O + H₂O → 2MOH
2MOH + CO₂ → M₂CO₃ + H₂O
Here, M is any alkali metal.

Question 21.
Among LiCl and RbCl which will ionize more. Why?
Answer:
Among LiCI and RbCl, RbCl is more reactive because LiCl is slightly covalent in nature due to which it is soluble in organic solvent like pyridine and alcohol.

Question 22.
Among alkali metals and alkaline earth metals whose carbonates are soluble in water? Or, The carbonates of alkali metals are souble in water but that of alkaline metals are insoluble?
Answer:
Carbonates of alkali metals are soluble in water because its hydration energy is greater than lattice energy. Whereas alkaline earth metals are smaller in size and have high density, so its hydration energy is lower than lattice energy. So the carbonates of alkaline earth metals are insoluble in water.

Question 23.
Give the differences between the 8eCl₂ and other chlorides of alkaline earth metals?
Answer:
1. Anhydrous halide are deliquescent and they absorb moisture or water forming hydrated salt.
Example: MgCl₂.6H₂O, CaCl₂.6H₂O, BaCl₂.2H₂O, etc.

2. BeCl₂ fumes on hydrolysis in moist air.
BeCl₂ + 2H₂O → Be(OH)₂ + 2HCl

3. BeCl₂ has different structure in solid and in vapour state. In solid state, it exists as polymeric chain in which each Be atom is surrounded by four chlorine atom. Two of the chlorine atom are covalently bonded while, other two are bonded through co – ordinate bond. In vapour state, at a temperature above 1200 K it has linear monomeric structure with zero dipole moment. Below 1200 K it exist as a dimer.

4. Except BeCl₂ and MgCl₂ other metal chloride give characteristic colour to the flame.

5. Anhydrous CaCl₂ has strong affinity for water therefore, it is used as dehydrating agent.

Question 24.
Why solubility of BaSO₄ is less than CaSO₄?
Answer:
The solubility of sulphates decreases from top to bottom as lattice energy is same. But as we move top to bottom the atomic size increases and hydration energy decreases so solubility decreases.

Question 25.
Why ionization energy of Be is more than B?

These orbitals of Be is full – filled so this element is stable but orbitals of B is not full – filled (Completely), so this element is not stable.

Question 26.
How sodium carbonate is formed by soda process? Write its principle?
Answer:
Solvay method or Ammonia soda process: It has replaced Le – Blanck process and is most commonly employed.
Principle:
In this process, concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammoniacal sodium chloride. On passing CO₂ gas to this solution, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate.

Question 27.
How will you prepare

1. Sodium bicarbonate
2. Sodium hydroxide
3. Sodium silicate from sodium carbonate?

Answer:
1. In sodium carbonate (Aqueous solution) pass CO₂ gas and sodium bicarbonate is formed (White precipitate).
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

2. Sodium carbonate is boiled with limewater then sodium hydroxide is formed.
Na₂CO₃ + Ca(OH)₂ → 2Na0H + CaCO₃

3. Sodium carbonate is treated with silica to form sodium silicate.

Question 28.
What is Baking soda? Write its method of preparation, properties and uses?
Answer:
Sodium Bicarbonate (NaHCO₃):
It is also known as sodium hydrogen carbonate or baking soda.

Method of preparation:
Sodium bicarbonate is obtained as an intermediate in the manufacture of sodium carbonate by Solvay’s process. It can be also obtained by passing carbon dioxide through aqueous solution of sodium carbonate.
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

Physical property:
It is a white, crystalline solid, slightly soluble in water.

Chemical properties:
1. Effect of heat:
When heated to 373 K it decomposes into Na₂CO₃ and CO₂ is set free.

2. Action of water:
Sodium bicarbonate hydrolyses when dissolved in water. Its aqueous solution is alkaline.

Uses:

1. As a source of carbon dioxide in fire extinguisher.
2. It acts as mild antiseptic for skin infection.
3. As a component of baking powder.
4. As a digestive powder to remove acidity of stomach.

Question 29.
Describe the Le – Blanck method of preparation of sodium carbonate with chemical equation? How Solvay method is better than this?
Answer:
Le – BIanck method:
Sodium chloride when heated with cone. H₂SO₄ produces sulphates and HCl gas. Mixture of sodium sulphate, calcium carbonate and coke on heating provide sodium carbonate and calcium sulphide. This mixture is called blanck ash. (MPBoardSolutions.com) Now in this mixture water is added and filtered. Gas is removed as residue while sodium carbonate being soluble goes into filtrate. Evaporation of filtrate provide solid sodium carbonate.
2NaCl + H₂SO₄ → Na₂SO₄ + 2HCl
Na₂SO₄ + 4C → Na₂S + 4CO
Na₂S + CaCO₃ → Na₂CO₃ + CaS

Advantages of Solvay method:

1. It is cheap
2. Pure Na₂CO₃ is formed
3. No harmful smoke is obtained
4. In the middle of reaction NaHCO₃ is formed which is a useful compound.

s – Block Elements Long Answer Type Questions – I

Question 1.
Describe the manufacture of caustic soda by Nelson cell on the following point: Labelled diagram, Chemical reactions.

Answer:
It consists of perforated steel tube Graphite anode lined inside with asbestos act as cathode. It is fitted with NaCl solution (brine) and is suspended in the steel tank. Here asbestos lining separates cathode from anode.

One passing current electrolysis of NaCl takes place. (MPBoardSolutions.com) Chlorine is liberated at anode and escapes out. Sodium ions passed through asbestos and is liberated at cathode they reacts with steam for caustic soda.


At anode:

At cathode:
2Na⁺ + 2e^– → 2Na (Reduction)
2Na + 2H²O → 2NaOH + H₂

Question 2.
Ions of an element of group 1 participate in the transmission of nerve signals and transport of sugars and amino acids into ceils. This element imparts yellow colour to the flame in flame test and forms an oxide and a peroxide with oxygen. Identify the element and write chemical reaction to show the formation of its peroxide. Why does the element impart colour to the flame?
Answer:
The element is sodium. This imparts yellow colour to the flame. Na⁺ ions participate in the transmission of nerve signals and transport of sugars and amino acids to cell. It forms a monoxide (Na₂O) and a peroxide (Na₂O₂).

The reason for flame colouration is, ionisation energy of Na is low. (MPBoardSolutions.com) Therefore, when Na metal or its salts is heated in Bunsen flame, its valence shell electron is excited to higher energy levels by absorption of energy. When the excited electron returns to the ground state, it emits extra energy in the yellow region of electromagnetic spectrum. Therefore, Na imparts yellow colour to the flame.

Question 3.
Write the diagonal relationship between Be and Al?
Answer:
Diagonal relationship between Beryllium and Aluminium:
Beryllium resembles aluminium of group 3 in the following properties:

1. Both beryllium and aluminium form covalent compounds.
2. Both metals are passive towards reaction with concentrated HN03, because they are covered with a layer of oxide on their surface.
3. Both the metals are of weak electropositive nature.
4. Both do not form hydride quickly.
5. Carbides of both the metals react with water to form methane.

Be₂C + 2H₂O → 2BeO + CH₄
Al₄C₃ + 6H₂O → 2AlCl₃ + 3CH₄

6. Oxides of both are soluble in water and form hydroxides which are amphoteric in nature and react with acid and base to form salt.

BeO + 2HCl → BeCl₂ + H₂O
Al₂O₃ + 6HCl → 2AlCl₃ + 3H₂O
BeO + 2NaOH → Na₂BeO₂ + H₂O
Al₂O₃ + 2NaOH → 2NaAlO₂ + H₂O

7. Both Be and Al do not give colour to the flame.

Question 4.
When water is added to compound (A) of calcium, solution of compound (B) is formed. When carbon dioxide is passed into the solution, it turns milky due to the formation of compound (C). (MPBoardSolutions.com) If excess of carbon dioxide is passed into the solution milkiness disappears due to the formation of compound (D). Identify the compounds A, B, C and D. Explain why the milkyness disappears in the last step?
Answer:
The compound (A) is quick lime, CaO. This combines with water and forms calcium hydroxides Ca(OH)₂.

When CO₂ is passed through the solution having Ca(OH)₂, the solution turns milky due to formation of calcium carbonate, CaCO₃.

When excess of CO₂ is passed into milky solution the milkiness disappears due to formation of calcium bicarbonate Ca(HCO₃)₂ which is soluble.

Question 5.
How is lithium different from other members of its group?
Answer:
Anomalous behaviour of Lithium:
Properties of lithium are different than other members of the group due to the following reasons:

1. Size of lithium atom and ion is very small.
2. Due to small size, polarizing power of lithium ion is high, due to w hich compounds possess covalent character.
3. In comparison to other alkali metals its electropositive nature is less and ionization energy is high.

Lithium differ from alkali metals of group I due to following properties:

1. 1. Lithium is comparatively harder than the other alkali metals.
   2. Melting and boiling point of lithium is comparatively high.
   3. Lithium reacts with oxygen to form only normal oxide (Li₂G), whereas other metals form peroxide (M₂O₂) and superoxide (MO₂).
   4. Lithium hydride (LiH) is more stable as compared to hydrides of other metals.
   5. Lithium hydroxide (LiOH) is a weak base and is partially soluble in water, but hydroxides of other metals of the group are more soluble in water.
   6. Lithium forms nitride (Li₃N) with nitrogen, whereas other metals of the group do not form nitride.
   7. Lithium nitrate when heated decomposes to nitrogen dioxide and oxygen.

4LiNO₃ \(\underrightarrow { \Delta } \) 2Li₂O + 4NO₂ + O₂

Whereas sodium nitrate and potassium nitrate when heated strongly form corresponding nitrate with the release of oxygen.

2NaNO₃ \(\underrightarrow { \Delta } \) 2NaNO₂ + O₂

Question 6.
What happens when:

1. Magnesium is burnt in air –
2. Quick lime is heated with silica –
3. Chlorine reacts with slaked lime –
4. Calcium nitrate is heated?

Answer:

s – Block Elements Long Answer Type Questions – II

Question 1.
How is sodium carbonate manufactured by Solvay process?
Answer:
Principle:
In this process, first concentrated solution of sodium chloride (Brine) is saturated with NH₃ to form ammanica sodium chloride.(MPBoardSolutions.com) On passing CO₂ gas to this, ammonium bicarbonate is formed which reacts with sodium chloride and forms sodium bicarbonate. Precipitate of sodium bicarbonate is filter ed which on calcination gives sodium carbonate.
NH₃ + CO₂ + H₂O → NH₄HCO₃
NH₄CO₃ + 2NaCl → NaHCO₃ + NH₄Cl
Na₂HCO₃ → Na₂CO₃ + H₂O + CO₂

CO₂ formed is used agiain for carbonation NH₄Cl formed on treatment with slaked lime gives NH₃ which is used for saturation of brine.
2NH₄CL + Ca(OH)₂ → CaCl₂ + 2H₂O + 2NH₃

The various steps and reactions involved are given below:

1. Saturating tank:
In this tank brine (NaCl) is saturated with ammonia. NaCl solution is introduced from the top while ammonia is introduced from bottom so that ammonium brine is formed and which collects at bottom. Calcium and magnesium salts are present as impurity in sodium chloride solution. They get precipitated as hydroxides by ammonium hydroxide.

2. Filter:
Ammonical brine is filtered to removed the precipitated hydroxides of calcium and magnesium.

3. Cooler:
The filtrate is cooled by passing through condenser. Cooling is necessary because when ammonia dissolves in brine a lot of heat is produced.

4. Carbonating tower:
The cold solution of brine saturated with ammonia is introduced into the carbonating tank from top. Carbonating tower is fitted with a number of compound diaphragms each made of a horizontal iron plate. Following reactions take place here:
2NH₃ + CO₂ + H₂O → (NH₄)₂CO₃
(NH₄)₂CO₃ + 2NaCl → Na₂CO₃ + 2NH₄Cl
Na₂CO₃ + H₂O + CO₂ → 2NaHCO₃

5. Vacuum filter:
The precipitated sodium bicarbonate along with solution of traces of ammonium carbonate and ammonium chloride is passed through rotatory vacuum filter at the bottom of tower where sodium bicarbonate separates leaving behind mother liquor containing ammonium chloride.

6. Limekiln:
Here limestone is burnt to produce quick lime and carbon dioxide.
CaCO₃ \({ \underrightarrow { \Delta } }\) CaO + CO₂
Formed CaO reacts with water to form slaked lime.
CaO + H₂O → Ca(OH)₂

7. Ammonia recovery tower:
The mother liquor obtained from rotatory filter pump containing ammonium chloride is introduced into the ammonia recovery tank from the top part while the milk of lime is introduced from the top of lower part and steam is introduced into the tower from the bottom, ammonium chloride reacts with milk of lime to produce ammonia.
2NH₄Cl + Ca(OH)₂ → 2NH₃ + CaCl₂ + 2H₂O
Sodium bicarbonate obtained from rotatory filters is ignited in a specially constructed cylindrical vessels when sodium carbonate is formed and CO₂ evolved is collected and used again.
2NaHCO₃ → Na₂CO₃ + H₂O + CO₂

Question 2.
What happens when:

1. Reaction of NaOH with Zn or A1
2. Reaction of NaOH with S
3. Reaction of NaOH with halogen
4. Reaction of NaOH with metal oxides
5. Reaction of NaOH with metal salts.

Answer:
1. Zn + 2NaOH → Na₂ZnO₂ + H₂
2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂

2. 4S + 6NaOH → Na₂S₂O₃ + 2Na₂S + 3H₂O Hypo
8S + 2Na₂S → 2Na₂S₅

3. 2NaOH + Cl₂ → NaCl + NaCIO + H₂O
(Cold) Sodium hypochlorate
6NaOH + 3Cl₂ → 5NaCl + NaClO₃ + 3H₂O
(Warm)

4. ZnO + 2NaOH → Na₂ZnO₂ + H₂O

5. CuSO₄ + 2NaOH → Na₂SO₄ + Cu(OH)₂
3NaOH + FeCl₃ → Fe(OH)₃ + 3NaCl
2NaOH + 2 AgNO₃ → Ag₂O + 2NaNO₃ + H₂O.

Question 3.
How Lithium show similarity with Magnesium?
Answer:
In the periodic table some elements differ from elements of the same period but show diagonal similarity with elements of next period.
Example:
Elements of second period show similarities with elements of third period. Like Lithium (second) period with magnesium (third period), Beryllium with A1 and Boron shows similarity with Si.

The relationship between Li and Mg:

1. The Li atomic radius is 1.34 Å and atomic radius of Mg is 1.36 Å .
2. The polar capacity of Li and Mg is equal.
3. Li and Mg is harder elements.
4. The Li and Mg is electronegativity is equal to (1.0 and 1.2).
5. The melting and boiling point of Li and Mg is very high.
6. Li and Mg reacts with N₂ and they form nitrite.
7. Li and Mg reacts with O₂ and they form monoxide.
8. Li and Mg reacts with water to release of hydrogen gas.
9. Li and Mg carbonate is heated and release CO₂ gas.
10. LiOH and Mg(OH)₂ is the weak basicity.

Question 4.
Write the formation of calcium oxide and write its properties and uses?
Answer:
Manufacture:
Commercially, calcium oxide is manufactured by heating limestone.

This process is exothermic and reversible. In order to get good yield of lime, carbon dioxide is to be removed from time to time. (MPBoardSolutions.com) The temperature of the reaction should not exceed 900°C because at higher temperature lime and clay reacts to form fusible silicate.

Furnace used for the manufacture of lime contain two fire boxes, one each on each side. Lime stone is added from the top. It decomposes on reach¬ing down. CO₂ produced is collected and stored in cylinders in liquid state. Lime farmed gets collected at the bottom of the furnace.
Physical properties:

1. It is a white solid.
2. Its melting point is high i.e., 2870K.
3. It is highly stable and it does not decompose in oxy-hydrogen flame but produces a brilliant white light called lime light.

Chemical properties:
1. It reacts with moist air absorbing CO₂ to form Ca(OH)₂ and CaCO₃
CaO + H₂O → Ca(OH)₂ (Slaked lime)
CaO + CO₂ → CaCO₃ (Lime stone)

2. When heated with ammonium salt, ammonia gas evolved.
2NH₄Cl + CaO → CaCl₂ + H₂O + 2NH₃

3. When heated with coke at high temperature, calcium carbide is produced.

4. It forms calcium chloride when chlorine gas is passed over hot and dry calcium oxide.
2CaO + 2Cl₂ → 2CaCl₂ + O₂

5. Calcium oxide is a strong basic oxide. It reacts with acid to form salt and water while, with acidic oxide it forms respective salt.
CaO + 2HCl → CaCl₂ + H₂O
CaO + H₂SO₄ → CaSO₄ + H₂O
CaO + CO₂ → CaCO₃ (Calcium carbonate)
CaO + SiO₂ → CaSiO₃ (Calcium silicate)
CaO + SO₂ → CaSO₃ (Calcium sulphite)
3CaO + P₂O₅ → Ca₃(PO₄)₂ (Calcium phosphate)

6. It forms slaked lime when dissolved in water. It is an exothermic reaction.
CaO + H₂O → Ca(OH)₂ + 15,000 cal.

Uses:

1. For making basic lining in furnaces.
2. For drying alcohol and gases.
3. Lime water is used as laboratory reagent and in medicine.
4. For purification of coal gas and in paper industry.
5. As flux in metallurgical process.
6. For producing lime light.
7. For manufacture of ammonia, sodalime, calcium carbide, cement and glass.

Question 5.
Differentiate between Alkali metals and Alkaline earth metals?
Answer:
Differences between Alkali metals and Alkaline earth metals:
Alkali metals:

1. They show +1 oxidation state.
2. Their hydroxides are strong bases.
3. Their carabonate, sulphate and phosphate compounds of alkaline earth plate are soluble in water.
4. Their ionisation energy is comparatively less.
5. They are melleable, ductile and lustrous.

Alkaline earth metals:

1. These show +2 oxidation state.
2. Their hydroxides are comparatively weaker bases.
3. These compounds of alkaline earth metals are insoluble in water.
4. Their ionization energy is high.
5. These are comparatively less.

Question 6.
Explain the Castner – Kellner cell with diagram to obtained sodium hydroxide?
Answer:
Castner – Kellner Cell:
It consists of rectangular iron tank which is fitted with three slate portion which do not touch the bottom but fit into grooves at bottom. The bottom of cell is covered with mercury and this is brought into circulation with the help of an eccentric wheel. (MPBoardSolutions.com) The outer compartmentA is filled with brine and inner Graphite compartment B with caustic soda solution. Two slot graphite electrode project from the ceiling of the outer compartment of the vessel and act anode. The iron cathode consisting of several rods is filled in middle compartment.

There are exit pipes fitted in compartment and for removal of hydrogen in central compartment. On passing current sodium chloride (brine) solution is electrolyzed in the two outer compartment. Chlorine is liberated at anode. (MPBoardSolutions.com) Sodium amalgam formed is pushed by the help of eccentric wheel into the central compartment. In the central compartment B when caustic soda is electrolyzed OH^– move to the mercury layer which act as anode and after discharge react with the sodium of Na – Hg to form caustic soda with the release of H₂ gas. Simultaneously an equivalent amount of sodium liberated at cathode reacts with water to produce more caustic soda.
2Na – Hg + 2H²O → 2NaOH + 2Hg + H₂↑

The mercury obtained can again be used in the cell. Caustic soda solution is evaporated to obtain solid caustic soda which is fused and cast into flakes or sticks.
Cell reactions are as follows:

1. In outer compartments:
Ionization:
2 NaCl ⇄ 2Na⁺ + 2Cl^–; (Ionization)

At cathode:
2Na⁺ + 2e^– → 2Na (Reduction)
2Na + xHg → Hg_xNa₂ (Sodium amalgam)

At anode:
2Cl^– → 2Cl + 2e^–; (Oxidation)
2Cl → Cl^2-

2. In central compartment:
Ionization:
NaOH ⇄ Na⁺ + OH^–; (Ionization)

At cathode:
2Na – Hg + 2H₂0 2NaOH + 2Hg + H₂
Na⁺ + e^– → Na, (Reduction)
2Na +2H₂O 2NaOH + H_2(g)

At anode:
2OH^– → 20H + 2e^–, (Oxidation)
Hg_xNa₂ + 20H → 2NaOH + xHg

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 9 Hydrogen

Hydrogen Important Questions

Hydrogen Very Short Answer Type Questions

Question 1.
In which compound the oxidation state of hydrogen is negative?
Answer:
CaH₂.

Question 2.
The bleaching property of H₂O₂ is due to oxidation or reduction?
Answer:
H₂O₂ easily gives oxygen therefore, it is used as bleaching agent. This is due to oxidation property it is used in bleaching.
H₂O₂ → H₂O + [O].

Question 3.
Who discovered hydrogen?
Answer:
Henry Cavendish.

Question 4.
What is used as moderator in atomic reactors?
Answer:
Heavy water (D₂O).

Question 5.
H₂O₂ reduces Cl₂ in which compound?
Answer:
In HCl.

Question 6.
The bleaching property of H₂O₂ depends upon?
Answer:
Oxidation.

Question 7.
Which oxide forms H₂O₂ with dii. HCI?
Answer:
Na₂O₂ and BaC₂.

Question 8.
Why the vapourisation of ethanol takes place faster than water?
Answer:
Due to weak hydrogen bonding.

Question 9.
Which type of compounds formed lattice carbides?
Answer:
d and f – block elements.

Question 10.
What is the use of lattice hydrides?
Answer:
For storage of H₂ and to catalyze the hydrogenation reaction.

Question 11.
Which acts as propellant in rockets?
Answer:
H₂O₂ (Hydrogen peroxide).

Question 12.
What is known as the absorbtion of hydrogen by palladium?
Answer:
Absorption.

Question 13.
Which chemical compound is called calgon?
Answer:
Sodium hexa metaphosphate.

Question 14.
What is main difference between ortho and meta hydrogen?
Answer:
Nuclear Spin.

Question 15.
What is prepared by the combustion of kerosene?
Answer:
Oil gas.

Question 16.
What is used as trace to study the processes occuring in organisms?
Answer:
Heavy water.

Question 17.
What is the radioactive isotope of hydrogen?
Answer:
Tritium.

Question 18.
What is formed by the reaction of calcium phosphate with water?
Answer:
Phosphene.

Question 19.
What is the bond angle in H – O – O in H₂O₂?
Answer:
97°.

Question 20.
Which hydrides are not in simple proportion?
Answer:
Interfacial hydrides.

Hydrogen Short Answer Type Questions – I

Question 1.
What is the reason for the temporary and permanent hardness of water? Explain?
Answer:
The temporary hardness of water is due to calcium and magnesium carbonate and the permanent hardness is due to presence of calcium chloride, magnesium chloride, calcium sulphate and magnesium sulphate.

Question 2.
Write an expression to show the amphoteric nature of water?
Answer:
Water has amphoteric nature. It behaves both as acid and base. With strong acid it behaves as base and with strong bases it behaves as acid.

Question 3.
How the impure hydrogen is purified?
Answer:
When hydrogen gas is passed on platinum black or palladium metal, the hydrogen gas is adsorbed this is called adsorption. The impure hydrogen gets purified.

Question 4.
Write the disadvantages of hard water?
Answer:

1. Hard water cannot be used in laboratory and for injection in medical.
2. Much quantity of soap is wasted if clothes are cleaned with hard water due to the formation of insoluble calcium and magnesium soap.
3. Hard water is unuseful in dyeing and printing.
4. Cooking with hard water take longer time and spoil the taste and quality of food.

Question 5.
Explain the Lane’s method to prepare hydrogen?
Answer:
By passing alternate currents of steam and water gas over red hot iron. The oxidation and reduction processes are alternatively carried out.

Oxidizing stage:
Superheated steam is passed over red hot iron at about 1023 K and 1073K, hydrogen gas is produced and magnetic oxides of iron (Fe₃O₄) is left.
3Fe + 4H₂O → Fe₃O₄ + 4H₂↑.

Question 6.
Write the method of removal of permanent as well as temporary hardness of water?
Answer:
By adding washing soda (Na2C03) when permanent hard water is treated with calculated quantity of sodium carbonate solution calcium and magnesium salts present in water get precipitated as insoluble carbonate. Soft water is then decanted off.
CaCl₂ + Na₂CO₃ → CaCO₃ + 2NaCl
MgSO₄ + Na₂CO₃ → MgCO₃ + Na₂SO₄
MgCl₂ + Na₂CO₃ → MgCO₃ + 2NaCl
CaS0₄ + Na₂CO₃ → Na₂SO₄ + CaCO₃.

Question 7.
How the strength of H₂O₂ is expressed?
Answer:
Strength of Hydrogen Peroxide Solution:
The strength of hydrogen peroxide solution is expressed in terms of the volume of oxygen obtained from it. For example, 10 volume, 20 volume etc. (MPBoardSolutions.com) The strength is equal to the volume of oxygen produced at NTP which is obtained by heating one unit of that solution. The ‘10 volume’ solution of hydrogen peroxide will give 10 ml oxygen at NTP when 1 ml of the sample is heated.

Question 8.
Write the uses of hydrogen?
Answer:
Uses of hydrogen:

1. As a reducing agent.
2. On account of its lighter nature, it was previously used in filling of balloons and aeroplanes but due to its combustible nature, a mixture of inert gas helium (85%) and hydrogen (15%) is used now.
3. In the manufacture of methyl alcohol, ammonia, synthetic petrol and fertilizers.
4. For preparing vegetable ghee: Hydrogenation of vegetable oil gives vegetable ghee.
5. Oxyhydrogen flame: Hygrogen produces high temperature when bums with oxygen which is used for welding and cutting purposes.
6. Hydrogen is used as a rocket propellent.

Question 9.
What is the difference between the terms “hydrolysis” and “hydration”?
Answer:
Hydrolysis:
Hydrolysis is the interaction of H⁺ and OH^– ions of water with the anion and cation of salt respectively to form acid and base.
Example:

Hydration:
Hydration is the interaction of water with the salts to form co – ordinated or hydrated ions or hydrated salts.
Example:
1. Water molecules are co-ordinated to metal ion in a complex.
[Cr(H₂O)₆]Cl₃.

2. Water occupying interstitial sites in the crystal lattice.
BaCl₂.2H₂O.

Question 10.
What is oil gas? How it is prepared?
Answer:
Coal gas is a mixture of many gases. When a thin layer of kerosene is dropped on red hot retort, then big molecules break and converts into methane, ethylene, acetylene. It is used in burners.

Question 11.
What is coal gas? Write its constituent?
Answer:
Coal gas is mixture of many gases where hydrogen gas is main element.
H₂ = 43.55%, N₂ = 2.12%, CH₄ = 25.35%, CO₂ = 0.3%, CO = 4.11%, O₂ = 0 – 1.5%.

Question 12.
When sodium reacts with water, which gas is released? Write name and formula?
Answer:
The more reactive metals (Na, K, Ca) react with water and metal hydroxide are formed and H₂ gas is released.
2Na + 2H – OH → 2NaOH + H₂↑
Ca + 2H – OH → Ca(OH)₂ + H₂↑
2K + 2H – OH → 2K—OH + H₂↑.

Question 13.
Write the reason for hardness of water and how many types of hardness are there?
Answer:
Water which does not readily produce lather with soap solution and cause of hardness of water. It is due to the presence of bicarbonate, sulphates and chlorides of calcium and magnesium in it. These salts get dissolved in it as it passes through the rock or ground.
Types of hardness:

1. Temporary hardness: It is due to the presence of bicarbonates of calcium, magnesium ion dissolved in water.
2. Permanent hardness: It is due to the presence of chlorides and sulphates of calcium, magnesium and iron in water.

Question 14.
What are hard and soft water?
Answer:
Hard water:
Water which does not readily produce lather with soap solution is known as hard water.
Soft water:
Water which readily produce lather with soap solution is known as soft water.

Question 15.
What is distilled water? Give its one use?
Answer:
The water which is formed with the help of distillation, this water is called distilled water. Its main use is in medicines and to prepare reagents in laboratory.

Question 16.
Write the uses of heavy water?
Answer:
Deuterium oxide is known as heavy water, its chemical formula is D20. It contains two deuterium atom in place of normal hydrogen atom.
Uses of heavy water:
1. Heavy water is used in nuclear reactors to slow down the speed of neutrons to carry out many important reactions. It is also used in the preparation of deuterium D₂. It is employed as a tracer in the study of reactions occurring in living organisms.

2. D₂O is used in atomic reactors where it serves two purposes:

• It acts as a coolant and
• Act as a moderator.

3. The water of some rivers like the Ganga remains clear and pure even when kept for years. It is due to presence of heavy water in it.

Question 17.
Explain the oxidising and reducing nature of H₂O₂?
Answer:
1. H₂O₂ liberates oxygen atom easily so it is strong oxidising agent but when it reacts with other oxidising agent it easily extracts oxygen atom from them hence it possesses reducing property too.
Oxidising nature:

1. PbS + 4H₂O₂ → PbSO₄ + 4H₂O
2. NaNO₂ + H₂O₂ → NaNO₃ + H₂O
3. Na₂SO₃ + H₂O₂ → Na₂SO₄ + H₂O

Reducing nature:

1. H₂O₂ + O₃ → H₂O + 2O₂
2. H₂O₂ + Na₂O₂ → Na₂O + H₂O + O₂
3. Ag₂O + H₂O₂ → 2Ag + H₂O + O₂.

Question 18.
Why the solution H₂O₂ did not concentrated on heating? How it can be concentrated?
Answer:
Hydrogen peroxide obtained by any method is dilute. It cannot be concentrated by boiling because it decomposes at a temperature below its boiling point. Therefore it is concentrated by following step:
1. Evaporation:
Evaporation of dilute solution of H₂O₂ on water bath at 70°C gives 45 – 50% H₂O₂ solution.

2. Vacuum evaporation:
The evaporation is further carried on, in a vacuum desiccator over concentrated sulphuric acid. In this way 66% solution of hydrogen peroxide is obtained.

3. Distillation under reduced pressure:
66% solution of hydrogen peroxide on distillation under reduced pressure, yields hydrogen peroxide of 99% concentration.

4. Crystallization:
H₂O₂ solution obtained in the above step is placed in a freezing mixture of solid CO₂ and ether. Crystals of H₂O₂ formed are separated and melted to obtain pure H₂O₂.

Question 19.
For the manufacture of hydrogen peroxide from peroxides, phosphoric acid is more useful than sulphuric acid why?
Answer:
H₂SO₄ acts as catalyst in the decomposition of H₂O₂. So for the manufacture of H₂O₂ from peroxides instead of H₂SO₄, weak acids like H₃PO₄, H₂CO₃ etc. are more useful.
3BaO₂ + 2H₃PO₄ → Ba₃(PO₄)₂ + 3H₂O₂
insoluble.

Question 20.
An ionic crystal of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride? Write its reaction with Al₂Cl₆?
Answer:
The hydride is LiH. Due to Li it behaves as covalent compound. It is highly stable.
8LiH + Al₂Cl₆ → 2LiAlH₄ + 6LiCl

Question 21.
Why hard water does not give fast lather with soap?
Answer:
In hard water bicarbonate, chlorides and sulphates of Ca and Mg are present. Soaps are sodium salt of higher fatty acids e,g., sodium stearate (C₁₇H₃₅COONa). (MPBoardSolutions.com) Salts of calcium and magnesium react and form precipitates of calcium and magnesium stearate.
2C₁₇ H₃₅ COONa + M²⁺ → (CH₁₇ H₃₅ COO)₂ M + 2Na⁺
Until all the salts are precipitated, no lather is formed with soap and this soap is wasted.

Question 22.
Write four uses of H₂O₂?
Answer:

1. It has antiseptic properties and hence used cleaning teeth, ears, wounds, etc.
2. It is used for bleaching hair, silk, wool, feathers, ivory, etc.
3. As oxidizing agent in laboratory.
4. In the preservation of milk, wine and some other drinks.
5. As a rocket fuel (as propellant) by producing oxygen.

Question 23.
Write the method of production of hydrogen from water gas?
Answer:
Process of preparation of hydrogen from water gas is known as Bosch process. Bosch process : In this process, hydrogen is obtained by separating it from water gas. Water gas is produced by passing steam over red hot coke.

Water gas is mixed with steam and heated at a temperature of 450 °C in presence of Fe₂O₃ as catalyst and chromic oxide as promoter when carbon monoxide gets oxidized to carbon dioxide.

The mixture of hydrogen and carbon dioxide is passed through water under 25 atmospheric pressure. Carbon dioxide gets dissolved leaving behind hydrogen. (MPBoardSolutions.com) Hydrogen so obtained still contains some carbon monoxide. It is removed by passing the gas through ammoniacal solution of cuprous chloride. The hydrogen so obtained contains nitrogen as impurity.

Question 24.
What is conducting water? Write their uses?
Answer:
Kohlrausch distilled the water 42 times at low pressure in an aperture made up of quartz. This water is called conducting water. This is used for
conductivity.

Question 25.
Write the redox reaction between fluorine and water?
Answer:
Fluorine is strong oxidising agent. It oxidizes H₂O into O₂ and O₃.
2F_2(g) + 2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4F^–_(aq) 3F_2(g) + 3H₂O_(l) → O_3(g) + 6H⁺_(aq) + 6F^–_(aq).

Question 26.
Explain why HCI is a gas and HF is a liquid?
Answer:
F is smaller and more electronegative than Cl, so it forms stronger H – bonds as compared to Cl. As a consequence, more energy is needed to break the H – bonds in HF than HCI and hence the boiling point of HF is higher than that of HCI. That’s why HF is liquid and HCI is a gas.

Question 27.
Write an equation for the manufacturing of D₂O₂?
Answer:
On reaction of D₂SO₄ with BaO₂, D₂O₂ is prepared.
BaO₂ + D₂SO₄ → BaSO₄ + D₂O₂

Question 28.
Why H₂O₂ cannot be kept for a longer time?
Answer:
Since hydrogen peroxide decomposes on storage, therefore H₂O₂ is stored in brown bottle to avoid the effect of light. This is because the light cause the decomposition of H₂O₂. (MPBoardSolutions.com) A small quantity of acetanilide is also added which retard the decomposition of H₂O₂. Here acetanilide acts as an inhibitor.

Question 29.
Why H₂O₂ is called Antichlor?
Answer:
In neutral medium H202 reduce halogen to acid, metal oxides to metals and ozone to 02.
Cl₂ + H₂O₂ → 2HCl + O₂
Br₂ + H₂O₂ → 2HBr + O₂.
Due to its ability to reduce chlorine it acts as an Antichlor in bleaching by destroying the unreacted chlorine.

Question 30.
When the temporary hard water is boiled with lime water, it becomes soft Why?
Answer:
This method is called Clark method. When temporary hard water is treated with calculated quantity of lime, bicarbonate present in water change to insoluble carbonates which settle down. Soft water is then decanted off.
Ca(HCO₃)₂ + Ca(OH)₂ → 2CaCO₃ + 2H₂O
Mg(HCO₃)₂ + Ca(OH)₂ CaCO₃ + MgCO₃ + 2H₂O

Question 31.
Do you expect the carbon hydrides of the type (C_n H_2n+2) to act as Lewis acid or base? Justify your answer?
Answer:
C_nH_2n+2 such as CH₄, C₂H₆ etc. neither act as Lewis acid nor Lewis base. It is because octet of all the carbon atoms are completed.

Question 32.
Explain the effect of high enthalpy of H – H bond on reactivity of dihydrogen?
Answer:
Due to high enthalpy of dissociation of H – H bond. Hydrogen is unreactive at room temperature. But at high temperature and in presence of catalyst it forms hydrides with metals and non – metals.

Question 33.
In which compound the oxidation number of hydrogen is negative?
Answer:
When hydrogen reacts with higher metal or highly reactive metals like Na, K, Ca etc then electrovalent hydrides are formed. In this hydride the hydrogen has negative oxidation state.

Hydrogen Short Answer Type Questions – II

Question 1.
What is deuterium? Write its two uses?
Answer:
Deuterium is obtained by electrolysis of heavy water. It is collected at cathode and shows reaction similar to hydrogen like its form heavy ammonia (ND₃) with nitrogen heavy water (D₂O) with oxygen.

Uses:

1. Chemical and Biological reaction.
2. It is used in artificial transmutation as target particle.

Question 2.
Give reason:

1. Lakes freeze from top towards bottom.
2. Ice floats on water.

Answer:

1. Water has maximum density at 4°C. In severe cold, the surface of the lake almost freezes but below the surface, there is water at a temperature about 4°C. This property is extremely helpful for animals living under lake water.
2. Ice floats on liquid water as its density is less than liquid water.

Question 3.
If same mass of liquid water and a piece of ice is taken, then why is the density of ice less than that of liquid water?
Answer:
The mass per unit volume (i.emass/volume) is called density. Since water expands on freezing, therefore, volume of ice for the same mass of water is more than liquid water. In other words, density of ice is lower than liquid water and hence ice floats on water.

Question 4.
How is the detection of hydrogen peroxide is performed?
Answer:
Test for hydrogen peroxide:

1. When few drops of H₂O₂ is added to acidified potassium dichromate solution containing ether, blue colour solution is formed which confirms the presence of H₂O₂.
2. Titanium dioxide is mixed with hot and cone. H₂SO₄ and then cooled followed by addition of few drops of hydrogen peroxide, yellow – orange pertitanic acid is produced.
3. On adding few drops of H₂O₂ to acidified glycol solution, blue colour is obtained.
4. Addition of few drops of H₂O₂ to a solution of FeSO₄ and starch and potassium iodide solution gives blue colour.
5. Addition of few drops of H₂O₂ to a mixture of aniline and potassium chlorate in dilute sulphuric acid give violet solution.

Question 5.
On the basis of electronic configuration justify the position of hydrogen in periodic table?
Answer:
Hydrogen is the first and the lightest element of the periodic table. It is not a metal but a non – metallic element. But on the basis of electronic configuration it is kept in first group of 5 – block. It is found in atomic form only at high temperature. (MPBoardSolutions.com) In elemental form it is found as diatomic molecule i.e., as H₂ and is also called as dihydrogen. One proton and one electron is found in hydrogen atom. Hydrogen forms large number of compounds and is an element of high industrial importance.

Hydrogen is the first element of periodic table with one proton in the nucleus and one electron in the first shell (K – shell). It is not possible to assign a to hydrogen in Mendeleev’s and modem periodic table because it shows similarities and dissimilarities with alkali metals (IA group), halogens (VIIA group) and carbon group (IV A group). This makes position of hydrogen very controversial. Because of this it is also known as Rogue element.

Question 6.
H₂O₂ is used for shining the old oil paintings?
Answer:
Old oil paintings contain basic lead oxide, H₂S present in atmosphere change lead oxide into lead sulphide. Thus, white paintings become black.

When the paintings are washed with hydrogen peroxide black lead sulphide oxidised into lead sulphate and painting shine again.

Question 7.
Draw the atomic structures of all isotopes of hydrogen?
Answer:
Atoms of an element having same atomic number but different atomic mass is known as isotope. Hydrogen has three isotopes :

1. Protium or ordinary hydrogen \(_{ 1 }^{ 1 }\)H
2. Deuterium or heavy hydrogen \(_{ 1 }^{ 1 }\)H
3. Tritium or radioactive hydrogen \(_{ 1 }^{ 3 }\)H

Structural diagram:

Isotopic effect:
The difference in properties mass is called isotopic effect.

Question 8.
On the basis of structure and chemical reactions? Explain, what are the characteristics of electron deficient hydrides?
Answer:
Electron deficient hydrides do not have sufficient number of electrons to form normal covalent bonds. They generally exist in polymeric forms such as B₂H₆, B₄H₁₀, (AlH₃)_n etc.
Due to deficiency of electrons, these hydrides act as Lewis acids and thus, form complex entities with Lewis bases such as NH₃, H^– ions.
B₂H₆ + 2NH₃ → BH₂(NH₃)₂]⁺(BH₄)^–
B₂H₆ + 2NaH → 2Na⁺(BH₄)^–

Question 9.
Describe ortho and para hydrogen. How will you obtain para hydrogen?
Answer:
Hydrogen atom consists of one proton in the nucleus and one electron in the extra nuclear part. Both electron and nucleus spin about their own axis. (MPBoardSolutions.com) When two atoms of hydrogen combine to form a hydrogen molecule, the spins of electron should be in opposite direction but the spins of nuclei may either be in the same direction or in opposite directions. When nuclear spins are in same direction, it is called ortho hydrogen. When nuclear spins are in opposite directions, it is called para hydrogen.

Preparation of para hydrogen:
Para hydrogen can be obtained from ordinary hydrogen by keeping it in a quartz vessel with active charcoal at 20K for about 3 to 4 hours.

Question 10.
Why, hard water does not used in boiler?
Answer:
1. Hard water is not used in boiler because compound of calcium and magnesium are deposited on internal side of boiler. This coating is bad conductor of heat so wastes of fuel take place. Due to this coating boiler have to the heated very much for heating water.

2. The boiler may burst if sudden cracks appear in boiler scale. Due to these sudden cracks, the water that comes in contact with the heated boiler is at once converted into steam and the sudden increase in pressure may cause the boiler to burst.

3. Hydrogen is volatile gas therefore, it is risk for explosion.

Question 11.
How the production of dihydrogen obtained from coal gasification can be increased?
Answer:
The gasification of producing syn gas from coal or coke is called coal gasification.

This is called water gas shift reaction. The CO₂ thus produced is removed by scrub-bing with sodium arsenite solution.

Question 12.
Expiain the structure of hydrogen peroxide?
Answer:
Dipole moment of H₂O₂ is 2.1 D, it indicates that structure of H₂O₂ is non – planar. By X – rays and other physical methods it is known that structure of H₂O₂ is like open book. (MPBoardSolutions.com) In gaseous state planes form an angle of 111.5°. In the axis, there are two oxygen atoms and one hydrogen in each plane.

The H – O and O – O bond lengths are 0.95Å and 147Å respectively. There is a slight change in the crystalline state due to hydrogen bonding. The angle between the planes is 90.2°, angle O – O – H = 101.9°, bond length H – O = 0.99Å and O – O is 1.46Å.

According to modem and latest belief probably both forms occur in equilibrium in aqueous solutions involving ionisation which explains the feeble acidic nature of hydrogen peroxide.

Question 13.
What is water gas? Write its constituents and uses?
Answer:
Bosch processes:
When steam is passed over red hot coke, water gas is produced.

Water gas is mixed with steam and passed over Fe₂O₃ oxidized to CO₂.

This mixture of CO₂ and H₂ is passed in water at 25 atmospheric pressure CO₂ gas is absorbed and H₂ gas remains.
Constitution:
H₂ = 49%, N₂ = 4%, CO = 44%, CO₂ = 2.7%, CH₄ = 0.3%
Uses:

1. As fuel in furnaces
2. For preparation of carbonated water gas
3. For manufacturing of ammonia.

Question 14.
What do you understand by the term “non – stoichiometric” hydrides? Do you expect this type of the hydrides to be formed by alkali metals? Justify your answer?
Answer:
The hydrides in which the ratio of the metal and hydrogen is found to be fractional are called non – stoichiometric hydrides. It is also observed that this fractional ratio is not fixed but varies with the temperature and pressure. (MPBoardSolutions.com) Such hydrides are formed by d and f – block elements. Usually all the holes are not occupied i.e., some holes always remain vacant and thus, these compounds are non – stoichiometric. Alkali metals do not form non – stoichiometric hydrides as each sodium atom loses its valency electron which is accepted by hydrogen atom to form H^– ion. In this way, an ionic compound, Na⁺ H^–, is formed.

Question 15.
What is Calgon? How hardness of water is removed by using it?
Answer:
Calgon is a trade name for the complex salt (NaPO₃)6 as Na₂ [Na₄ (PO₃)₆]. When calgon is added to hard water are rendered ineffective due to the formation of water soluble complex ion. This is known as sequstration.
Na₂[Na₄(PO₃)₆] + 2CaCl₂ → Na₂[Ca₂(PO₃)₆] + 4NaCl
Na₂ [Na₄(PO₃)₆] + 2MgSO₄ → Na₂[Mg₂ (PO₃)₆] + 2Na₂SO₄
The complex calcium and magnesium ion do not form any precipitate with soap. Therefore, they readily form lather with soap solution. Modem detergent contains sodium hexa – meta phosphate to remove the hardness of water due to Ca²⁺ and Mg²⁺ ion hence wastage of soap is checked.

Question 16.
Explain that H₂O₂ acts as both oxidizing and reducing agent. Why?
Answer:
H₂O₂ liberates oxygen atom easily, so it is strong oxidizing agent, but when it reacts with other oxidizing agent, it easily extracts oxygen atom from them hence, it possesses reducing property too.
Oxidizing nature:
1. 2KI + H₂O₂ → 2KOH + I₂

Reducing nature:
(a) In basic medium, it converts ferricyanide to ferrocyanide.
2[Fe(CN₆)]^3- + 20H^– + H₂O₂ → 2[Fe(CN)₆]^4- + 2H₂O + O₂

(b) In acidic medium; it reduces permanganate to manganese (II) salt.
2MnO₄^– + 6H⁺ + 5H₂O₂ → Mn²⁺ + 8H₂O + 5O₂

Question 17.
Consider a reaction of water with F₂ and suggest, in terms of oxidauon and reduction, which species will oxidized/reduced?
Answer:
2F_2(g) + 2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4F^–_(aq)
3F_2(g) + 3H₂O_(l) → O_2(g) + 6H⁺_(aq) + 6F^–_(aq)
In these reactions, water acts as a reducing agent as it gets oxidized to either O₂ or O₃ while fluorine acts as an oxidizing agent as it is reduced to F^– ion.

Question 18.
How does H₂O₂ behave as bleaching agent?
Answer:
Hydrogen peroxide acts as a bleaching agent due to the release of nascent oxygen.
H₂O₂ → H₂O + [0]
The nascent oxygen combines with colouring matter which in turn gets oxidized. The bleaching action of H₂O₂ is due to the oxidation of colouring matter by nascent oxygen. It is used for bleaching materials like jury, silk, wool, feathers etc.

Question 19.
Why the density of water is maximum at 4°C?
Answer:
Maximum density of water:
Melting of ice decreases hydrogen bonds because cage – like structure is broken. With increase in temperature from 0°C to 4°C with the cleavage of H – bonds, water molecules starts moving closer to each other as a result volume decreases while density increases. (MPBoardSolutions.com) When the temperature becomes greater than 4°C, kinetic energy of H₂O molecule increases resulting in expansion of water. As a result of this volume increases while density decreases. That is at 4°C, density of water is maximum.

Question 20.
What do you understand by

1. Electron deficient
2. Electron precise and
3. Electron rich compounds of hydrogen? Provide justification with suitable examples?

Answer:
1. Electron deficient hydride:
Electron deficient hydrides are those which do not have sufficient number of electrons to form normal covalent bond.
Examples: BH₃, AlH₃
To makeup their deficiency they generally exist in polymeric form such as B₂H₆, Al₂H₆.

2. Electron precise hydride:
Electron precise hydrides are those which have suffi-cient number of electrons required for forming covalent bond.
Examples:
CH₄, SiH₄ etc.

3. Electron rich hydride:
Electron rich hydrides are those which have excesses electron as required to form normal covalent bond. The excesses electron present as lone pair.
Examples: NH₃, PH₃ etc.

Question 21.
Tell the group of hydrides which include H₂O, B₂H₆ and NaH?
Answer:

• H₂O: Covalent or molecular hydride (Electron rich hydride)
• B₂H₆: Covalent or molecular hydride (Electron deficient hydride)
• NaH: Ionic or salt hydride.

Hydrogen Long Answer Type Questions

Question 1.
Explain Permutit process of removal of hardness of water?
Answer:
Permutit process or Zeolite process:
The water is passed over such substances which readily replace the Ca²⁺ and Mg²⁺ ions of hard water by sodium ions. The water does not become hard in presence of sodium salts. Ofcourse too much of sodium salts also make it hard just as in sea water. (MPBoardSolutions.com) The water thus obtained is soft. These products are called as zeolites. Zeolite is a technical name given to certain hydrated silicates of aluminium and sodium. It is also known as permutit. It is obtained by fusing sodium carbonate with alumina and silica.

Na₂CO₃ + Al₂O₃ + 2SiO₂ → Na₂Al₂Si₂O₈ + CO₂
Na₂Al₂Si₂O₈ is known as sodium zeolite and Al₂Si₂O₈ is zeolite and can be represented by Z. Sodium zeolite reacts with chlorides, sulphates of Ca and Mg as given below and hardness is removed.
CaCl₂ + Na₂ → CaZ + 2NaCl
MgSO₄ + Na₂ → MgZ + Na₂SO₂
Reactions have been given by representing sodium zeolite as Na₂Z.
After sufficient use permutit gets exhausted and cannot be used further. It is regenerated and made usable by passing 10% solution of sodium chloride.
CaZ + 2NaCl → CaCl₂ + Na₂Z
MgZ + 2NaCl → MgCl₂ + Na₂Z

Question 2.
What happens when:

1. Calcium hydride reacts with water
2. H₂O₂ is added in acidic KMnO₄ solution
3. Reaction of potassium ferricyanide with H202
4. Reaction of H₂ with N₂ at high temperature and pressure in presence of Fe?

Answer:

1. On reaction of calcium hydride with water, H₂ is formed
   CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂
2. The pink colour disappear of KMnO₄
   2KMnO₄ + 3H₂SO₄ + 5H₂O₂ → K₂SO₄ + 2MnSO₄ + 8H₂O + 5O₂
3. On adding H₂O₂ in potassium ferricyanide, it reduces to potassium ferrocyanide
   2K₃[Fe(CN)₆] + 2KOH + H₂O₂ → 2K₄[Fe(CN)₆] + 2H₂O + O₂
4. Ammonia is formed when hydrogen reacts with N₂ 200 atm. pressure

Question 3.
Explain the processes of formation of hydrogen in laboratory?
Answer:
Preparation of hydrogen (laboratory method):
Hydrogen is prepared in laboratory by action of dil. H₂SO₄ on granulated Zn. Zn is taken into woulf bottle fitted with thistle funnel and a outlet tube.
Hydrogen formed by this method is collected over water by displacement method.
Zn + H₂SO₄ → ZnSO₄ + H₂↑
Purification of hydrogen : Hydrogen prepared by this method contains following impurities:

1. Arsine (AsH₃) and phosphene (PH₃)
2. H₂S
3. SO₂, CO₂ and oxides of nitrogen (NO₂)
4. Water vapours.

In order to remove these impurities, gas is passed in a series of U – tubes containing different solutions:

1. Passed in tube filled with AgNO₃ solution to absorb AsH₃ and PH₃.
2. Passed in the tube containing lead nitrate solution to remove H₂S.
3. Passed in the tube filled with cone. KOH solution to remove SO₂, CO₂ and NO₂.

Question 4.
Rohan heard that instructions were given to the laboratory attendent to store a particular chemical Le., keep it in the dark room, add some urea in it, and keep it away from dust This chemical acts as an oxidizing as well as a reducing agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents?

1. Write the name of this compound.
2. Explain, why such precautions are taken for storing this chemical?

Answer:

1. H₂O₂ (Hydrogen peroxide).
2. The following precautions must be taken while storing hydrogen peroxide:
3. It must be kept in wax lined coloured bottles because the rough glass surface, light and dust particles are responsible for its decomposition.
4. A small amount of negative catalyst such as urea, glycerol, phosphoric acid etc. is generally added which retards its decomposition.

Question 5.
Calculate the strength of 5 volume of H₂O₂ solution?
Answer:
5 volumes ₂O₂ solution means that 1 volume of this solution will decompose to give 5 volumes of oxygen at S.T.P.

5 L oxygen is obtained from \(\frac{68×5}{22.4}\) g H₂O₂ = 15.17 g H₂O₂
i.e. 15g H₂O₂ is present in 1L solution. Thus, strength of the solution = 15 g/L.

Question 6.
Complete the following reactions:

1. PbS_(s) + H₂O_2(aq) →
2. MnO₄^–_(aq) + H₂O_2(aq) →
3. CaO_(s) + H₂O_(g) →
4. AlCl_3(g) + HzO_(l) →
5. Ca₃N_2(s) + H₂O_(l) →

above

1. Hydrolysis
2. Redox
3. Solvation process.

Answer:

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 8 Redox Reactions

Redox Reactions Important Questions

Redox Reactions Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Metals used in Daniel cell are:
(a) N and Cu
(b) Zn and Ag
(c) Ag and Cu
(d) Zn and Cu
Answer:
(d) Zn and Cu

Question 2.
Which of the following is strongly reducing:
(a) r
(b) cr
(c) Bf
(d) r
Answer:
(d) r

Question 3.
Oxidation in electrolyte occurs:
(a) At anode
(b) At cathode
(c) At both the electrodes
(d) None of these
Answer:
(a) At anode

Question 4.
Which of the following statement is correct? Galvanic cell converts:
(a) Chemical energy to electrical energy
(b) Electrical energy to chemical energy
(c) Electrical state of metal to combined state
(d) Electrolytes to ions
Answer:
(b) Electrical energy to chemical energy

Question 5.
Oxidation number of chlorine in HOCl is:
(a) – 1
(b) 0
(c) +I
(d) +2
Answer:
(c) +I

Question 6.
Oxidation number of Mn in K₂MnO₄ is:
(a) +2
(b) +6
(c) +7
(d) 0
Answer:
(b) +6

Question 7.
Oxidation number of Cr in K₂Cr₂O₇ is:
(a) -6
(b) +6
(c) +2
(d) -2.
Answer:
(b) +6

Question 8.
Oxidation number of Ni in Ni(CO)₄ is:
(a) 0
(b) +2
(c) +1
(d) – 1.
Answer:
(a) 0

Question 9.
Unit of cell constant is:
(a) ohm^-1cm^-1
(b) ohm cm
(c) cm
(d) cm^-1
Answer:
(d) cm^-1

Question 10.
Oxidation state of nitrogen is maximum in:
(a) N₃H
(b) NH₂OH
(c) N₂H₄
(d) NH₃
Answer:
(a) N₃H

Question 11.
Oxidation state of oxygen is zero in:
(a) CO
(b) O₃
(c) SO₂
(d) H₂O₂
Answer:
(b) O₃

Question 12.
Oxidation state of Fe in K₄[Fe(CN)₆] is:
(a) +2
(b) +6
(c) +3
(d) +4
Answer:
(a) +2

Question 13.
Oxidation state of S In H₂S₂O₈ is:
(a) +2
(b) +4
(c) +6
(d) +7
Answer:
(c) +6

Question 14.
Oxidation state of Mn in KMnO₄ is:
(a) +4
(b) +6
(c) +7
(d) +5
Answer:
(c) +7

Question 15.
oxidation state of chlorine is in:
(a) HCl
(b) HClO₄
(c) ICI
(d) Cl₂O
Answer:
(d) Cl₂O

Question 16.
Oxidation state of Cl in ClO₃^– ion is:
(a) +4
(b) +5
(c) +3
(d) +2
Answer:
(b) +5

Question 17.
Halogen which is reduced most easily is:
(a) I₂
(b) Br₂
(c) Cl₂
(d) F₂
Answer:
(a) I₂

Question 18.
Oxidation state of C in CCl₄ is:
(a) +4
(b) -4
(c) +6
(d) – 6
Answer:
(a) +4

Question 19.
Oxidation state of S in S0₄^-2 is:
(a) +6
(b) -6
(c) +5
(d) – 5
Answer:
(a) +6

Question 20.
Oxidizing agent are:
(a) Electron acceptor
(b) Electron donor
(c) Proton acceptor
(d) Neutron acceptor
Answer:
(a) Electron acceptor

Question 2.
Fill in the blanks:

1. Process of loss of electrons is called ……………………..
2. Process of gain of electrons is called …………………….
3. The deterioration of metals in presence of atmospheric gases and moisture is called …………………………
4. A device in which electric energy gets converted to chemical energy is called …………………….. cell
5. In electrochemical series the ability to reduce ………………………… on moving down
6. Most strong reducing element is ………………………….
7. Most strong oxidizing element is ………………………..
8. Theory of dissociation of electrolyte was proposed by ……………………….

Answer:

1. Oxidation
2. Reduction
3. Corrosion
4. Electrochemical cell
5. Decreases
6. Lithium
7. Fluorine
8. Arrhenius

Question 3.
Answer in one word/sentence:

1. What is balanced redox reaction?
2. Why does iron displaced copper from its salt solution?
3. What is the oxidation state of chlorine in Cl₂O?
4. Why does an electrolyte dissociate into ions when dissolved in water?
5. A saturated solution of KNO₃ is used for the formation of salt bridge. Why?
6. In SnCl₂ + 2 FeCl₃ → SnCl₄ + 2 FeCl₂, reaction which element is oxidizing agent?
7. What is oxidation state of Xe in XeO₃?

Answer:

1. Redox reaction in which amount of oxidation and reduction is to the same extent.
2. Because standard eletrode potential value of iron is less then the standard electrode potential of copper.
3. +1
4. Because the electrostate attractive force between the ions break due to water.
5. Because the speed of K⁺ and NO₃ and is almost same.
6. FeCl₃
7. +6

Question 4.
Match the following:

Answer:

1. (c)
2. (d)
3. (a)
4. (b)
5. (e)

Redox Reactions Very Short Answer Type Questions

Question 1.
Why iron displaces copper from its salt solution?
Answer:
Because the standard electrode potential of iron is less than copper.

Question 2.
What is the oxidation number of Cl in Cl₂O?
Answer:
+1.

Question 3.
When an electrolyte is dissolved in water, it dissociates into ions? Why?
Answer:
Because due to water the electrostatic attraction force between the ions breaks.

Question 4.
Why the saturated solution of KNO₃ is used to prepare salt bridge?
Answer:
Because the speed of K⁺ and NO₃^– ions is similar.

Question 5.
In SnCl₂ + 2FeCl₃ → SnCl₄ + 2FeCl₂, which is oxidizing agent?
Answer:
FeCl₃.

Question 6.
What is the oxidation number of Xe in XeO₃?
Answer:
+6.

Question 7.
Which metal is used in Daniel cell?
Answer:
Lithium.

Question 8.
What is the oxidation state of Mn in K₂MnO₄?
Answer:
+6.

Question 9.
Which is the most reducing element?
Answer:
Lithium.

Question 10.
Who maintains the neutrality among the two half cells?
Answer:
Salt bridge.

Question 11.
What is the standard electrode potential of SHE?
Answer:
0.00V (Zero).

Question 12.
Which is the best conducting metal?
Answer:
Silver (Ag).

Question 13.
The reaction 3ClO^–_aq → Cl₃^– + 2Cl^– is an example of?
Answer:
Disproportionation reaction.

Question 14.
Which halogen reduces very easily?
Answer:
F₂.

Question 15.
Which reaction is called electron acceptance reaction?
Answer:
Reduction.

Question 16.
Where oxidation occurs in electrolysis?
Answer:
At anode.

Redox Reactions Short Answer Type Questions – I

Question 1.
What is oxidation process? Explain redox reaction with example?
Answer:
Oxidation state of an element is defined as “The residual charge left on its atom”. When all the other atoms are removed from the molecule as ion.

2Mg + O₂ → 2MgO [Addition of oxygen]
2FeCl₂ → 2FeCl₃ [Addition of electronegativity element]
H₂S + Cl₂ → S + 2HCl [Removal of hydrogen]
2Ki + H₂O₂ → 2KOH + I₂ [Removal of electropositive element]

Question 2.
What is reduction process? Explain with example?
Answer:
The ability of any substance to accept hydrogen atom or +ve electrolytic element or to release oxygen or – ve element is called reduction process.
CuO + H₂ → Cu + H₂O [Remove of oxygen]
2FeCl₃ + H₂ S → 2FeCl₂ + 2HCl + S [Removal of electromagnetive element]
Cl₂ + H₂S → 2 HCl + S [Addition of H]
S + Fe → Fes [Addition of electropositive element]

Question 3.
AgF₂ is unstable compound, but if it is formed then it will acts as strong oxidising agent Why?
Answer:
In AgF₂, Ag has oxidation state +2, it is highly unstable, it very easily accept electron and forms oxidation state +1. Because Ag⁺ has stable configuration so it is stable.
Ag2⁺ + e^– → Ag⁺
This is the reason why AgF₂ behaves as a strong oxidizing agent.

Question 4.
By the structure of S₄O₆^2-clarify that the oxidation state of S is +5?
Answer:
The oxidation state of two central sulphur atom is 0. As the electron pair remains in the centre cf the bond. So the remaining sulphur atom has oxidation state 4 – 5.

⇒ 6(-2) + 2x = -2
⇒ -12 + 2x = -2
⇒ 2x = -2 + 12 = 10
⇒ x = \(\frac{10}{2}\) = 5.

Question 5.
What is electrochemical series?
Answer:
Electrochemical series:
When different elements and ions are arranged in increasing order of their standard electrode potential, a series is obtained which is called electrochemical series as activity series.

Question 6.
What is salt bridge? Write its two uses?
Answer:
1. It connects the solution in two half ceil and completes the cell circuit

2. The salt bridge which contains the solution of an electrolyte say K⁺NO₃^– maintains the electrical neutrality by diffusion of ion through it. As positive ion begins to be formed in oxidation half cell negative ion from the salt bridge diffuse out into this half cell. Similarly, as the positive Ion begin to be consumed in the reduction.

Hair cell positive:
On from the salt bridge diffuse out into this half cell. As a result the solution in the two half cell remain electrically neutral and the current continous to flow.

Question 7.
On the basis of electron transfer explain oxidation and reduction reactions?
Answer:
Reactions involving exchange of electrons are called redox reaction. Reaction given below
2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄
It is an example of redox reaction in which FeCl₃ is oxidizing agent while SnCl₂ is a reducing agent.

According to electronic concept during oxidation substance loses one or more electron and its positive oxidation number increases while negative oxidation number decreases. (MPBoardSolutions.com) On the other hand, during reduction substance gains one or more electron and is positive oxidation number decreases while negative oxidation number increases.

Question 8.
What is electrochemical equivalent?
Answer:
According to Faraday’s first law:
The mass of any substance deposited at any electrode is directly proportional to the quantity of electricity passed.

Thus, if W gm substance is deposited on passing Q coulomb electricity, then W ∝ Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. (MPBoardSolutions.com) So that, W = Z × Q = Z × I × r

Thus, if, Q = 1 coulomb, 1=1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second”.

As one Faraday (96500 C) deposits one gram equivalent of the substance, hence, electrochemical equivalent can be calculated from the equivalent mass.
\(\frac { Equivalentmassofthesubstance\quad }{ 96500 } \)

Question 9.
What happens when Zn is kept in CuS0₄ solution? Explain with equation?
Answer:
The reaction occurring in the beaker may be written as

Cancelling the common S0₄²⁺ ions
Zn_(s) + Cu²⁺_(eq) → Zn²⁺_(eq) + Cu_(s)
In this case Zinc metal loses electron and gets oxidised to Zn⁺⁺ ions which goes into the solution. Due to this, weight of zinc plate gradually decreases. On the other hand, electrons lost by Zinc atom are taken up by Cu²⁺ ions of CuS0₄ solution.

Question 10.
What do you mean by standard electrode potential?
Answer:
Standard electrode potential: Standard electrode potential (E°) of a half cell (electrode) is the potential difference which is produced when one electrode is dipped in the solution of molar concentration of its ion at 298 K. (MPBoardSolutions.com) If electrode is gaseous, the pressure of gas must be one atmosphere.
In IUPAC system reduction potentials are known as standard electrode potential.

Question 11.
The reaction Cl_2(g) + 2OH^–_(aq) → ClO^–_(aq) + Cl^–_(aq) + H₂O_(l) represents the process of bleaching. Identify and name species that bleaches the substances due to its oxidizing action?
Answer:

In this reaction, O. N. of Cl increases from 0 in Cl₂ to +1 in ClO^– and decreases to -1 in Cl^–. Therefore, Cl₂ is both oxidized to ClO^– and reduced to Cl^– (Disproportionation reaction). Since, Cl^– ion cannot act as oxidizing agent (Because it cannot decrease its O.N. lower than -1) therefore, Cl₂ bleaches substance due to oxidizing action of ClO^– (Hypochlorite) ion.

Question 12.
Define oxidizing agent and reducing agent?
Answer:
Oxidizing agent:
A substance which brings oxidation is called oxidising agent. On the basis of electronic theory, an oxidizing agent may be defined as an electron acceptor. i.e. (MPBoardSolutions.com) in the reaction between Zn atom and Cu²⁺ ion in aqueous solution of Cu²⁺ salt, Cu²⁺ ion is the oxidizing agent as it brings about the oxidation of Zn atom by gaining the electron lost by it.
Zn + Cu²⁺ → Zn² + Cu
(Gains e^–)
(∴ Oxidising agent)
Thus, oxidizing agent or oxidant undergoes gain of electrons during the reaction.
Reducing agent:
A substance which brings about reduction is called a reducing agent. On the basis of electronic theory, a reducing agent may be defined as an electron donor, e.g., in reaction between Zn atom (MPBoardSolutions.com) Cu²⁺ ion in an aqueous salt solution of Cu²⁺, Zn atom is the reducing agent as it brings about reduction of Cu²⁺ ion by losing the electrons.
Zn + Cu²⁺ → Zn²⁺ + Cu
(Loses e^–)
(∴Reducing agent)

Question 13.
If iron rod is dipped in CuS0₄ solution then copper is displaced from its solution by iron but when copper rod is dipped in FeS04 solution. Why iron is not displaced by copper? Explain?
Answer:
Fe occupies higher position in electrochemical series than Cu. So, Fe is more active than Cu. That is why Fe displaces Cu from CuS0₄ solution, but Cu is not able to displace Fe from FeS0₄ solution.

Question 14.
What is oxidation number?
Answer:
Oxidation number of an element is defined as, “The residual charge left on its atom when all the other atoms are removed from the molecule as ions.” Atoms can have positive, negative or zero oxidation state depending upon their state of combination. (MPBoardSolutions.com) In fact, oxidation number is the charge assigned to the atom in a species according to some rules. It is also known as oxidation state.

Question 15.
In electrochemical series, the activity of metals increases or decreases in which sequence?
Answer:
Species present in the right of electrochemical series are reducing agent. Smaller the value of E°, stronger is the reducing agent. The activity of metals decrease on going down the electrochemical series.

Question 16.
What is E° value be + ve shows in Galvanic cell?
Answer:
The positive value of E° shows:

1. Oxidation occurs at anode and
2. The electrode where reduction occurs can be taken as cathode.

Question 17.
What is electrode potential? Its value depends on what factors?
Answer:
The electrical potential difference setup between the metal and its solution is known as electrode potential.
The electrode potential depends upon:

1. Concentration of ions in the solution
2. Temperature.

Question 18.
On the basis of standard electrode potential given below, arrange the metals in their increasing reducing power?
K⁺/ K = – 2.93V, Ag⁺/ Ag = 0.80V, Hg²⁺/ Hg = 0.79V, Mg²⁺/ Mg = -2.37V, Cr³⁺/ Cr = -0.74V?
Answer:
Less the value of E°, the reducing power increases of the metal. So the sequence is Ag < Hg < Cr < Mg < K.

Question 19.
Among MgO, ZnO, CuO and CaO which oxide will reduce by hydrogen?
Answer:
Mg, Zn and Ca are more reactive than H in electrochemical series. So they will not reduced by H. Cu is present below H, so the reactivity of Cu is less than H. So CuO will easily get reduced by H.

Question 20.
The electrode potential (E°) of Ag, Ba, Mg and Au are +0.80, -2.90, -2.37 and +1.42 volt. Among these metals which will displace H from acids and which will not?
Answer:
Those metals which have E° value negative they are more reactive than hydrogen and can easily displace H from weak acids. So, Ba and Mg can displace H from acid solution.

Question 21.
Can be put CuSO₄ solution in silver container? Why?
Answer:
In electrochemical series Ag comes below Cu, so reactivity of Ag is less than Cu. There Ag will not displace Cu from CuSO₄ solution. CuSO₄ can be kept in Ag container.

Redox Reactions Short Answer Type Questions – II

Question 1.
Fluorine reacts with ice as follows:
H₂O_(s) → HF_(g) + HOF_(g)
Justify that, this reaction is a redox reaction.
Answer:

Since, fluorine can undergo oxidation as well as reduction, it is an example of redox reaction.

Question 2.
In acidic medium MnO₄^2- shows disproportionation reaction but MnO₄^– does not show why?
Answer:
In oxidation state of Mn in MnO₄^-2 is +6. It can increase oxidation state (+7) and decrease (+4, +3, +2, 0 etc.). So in acidic medium it shows disproportionation reaction.

In MnO₄^–, Mn has maximum oxidation state (+7). It can only decrease its oxidation state. Due to which it cannot show disproportionation reaction.

Question 3.
In the following reaction whose oxidation and whose reduction occur:
PbS + 4H₂O₂ → PbSO₄ + 4H₂O
Answer:
In this reaction PbS is oxidized as PbSO₄ and H₂O₂ is reduced to form H₂O.

Question 4.
Differentiate between oxidation number and valency?
Answer:
Differences between Oxidation number and Valency:
Oxidation:

1. It is the residual charge left on the atom of an element when all other atoms are removed from molecule or ion.
2. It refers to charge and can be positive, negative or zero, e.g., oxidation number of C in CO₂ is +4 and that of oxygen is -2.
3. It may have fractional value.
4. Elements like C, N, O have variable oxidation number.

Valency:

1. It is the combining capacity of the element expressed as number of H – atom or double the number of oxygen atoms which combine with one atom of the element.
2. It refers to number only and does not carry any sign, e.g., in CO₂ valency of carbon is 4 and that of oxygen is 2.
3. It is always a whole number.
4. Elements like C, N, O exhibit constant valency.

Question 5.
Why fluorine doesn’t show disproportionation reaction?
Answer:
Like chlorine, bromine and iodine also undergo similar disproportionation reactions but fluorine does not. The reason for this anomalous behaviour is that fluorine being the strongest oxidizing agent does not show positive oxidation states. It, therefore, reacts in a different way forming oxygen difluoride (OF2).
2F_2(g) + 2OH^–_(g) → 2F^–_(aq) + OF_2(g) + H₂O_(l)

Question 6.
Show that galvanic cell where following reaction occurs:
Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq) + 2 Ag_(s)
Now explain :

1. Which electrode is -ve?
2. What is the motive of electricity in cell?
3. What is the reaction occurring on each electrode?

Answer:
Galvanic cell Zn_{(s) ∥} Zn²⁺_(aq)Ag(aq) ∥ Ag⁺_(aq) ∥ Ag_(s)

1. Zn electrode is negatively charged. Zn is oxidized into Zn⁺².
2. Electricity flows from Ag electrode to Zn electrode and electrons flow from Zn to Ag electrode.
3. Reactions on electrode:

Cathode — Zn_(s) → Zn⁺²_(s) + 2e^–
Anode — 2Ag⁺_(aq) + 2e^– → 2Ag_(s)

Question 7.
Write the definition of electrochemical cell? Write the chemical reactions of Daniel cell?
Answer:
An electrochemical cell (Galvanic cell) is a device in which the redox reaction takes place indirectly and the decrease in free energy appears in the form of electrical work i.e. chemical energy is converted into electrical energy.

Question 8.
Write the factors affecting electrode potential?
Answer:
The factors affecting electrode potential are:

1. Tendency of metal to donate electrons: More the tendency to release electron more will be electrode potential.
2. Temperature: With increasing temperature electrode potential increases.
3. Concentration of solution: More the concentration of solution more will be electrode potential.

Question 9.
Determine the oxidation number of following:

1. Cr in K₂Cr₂O₇
2. Mn in KMnO₄
3. S in Na₂S₄O₆.

Solution:
1. Cr in K₂Cr₂O₇
2 (+1) + 2x + 7 (-2) = 0
or 2 + 2x – 14 = 0
or 2x = 12
or x = +6.

2. Mn in KMnO₄
1 (+1) + 1x + 4(-2) = 0
or 1 + x – 8 = 0
or x = +7.

3. S in Na₂S₄O₆
2 (+1) + 4x + 6 (-2) = 0
or 2 + 4x – 12 = 0
or 4x = 10
or x = 2.5.

Question 10.
Why does the following reactions occur:
XeO_6(aq)^4- + 2F^–_(aq) + 6H⁺_(aq) → XeO_3(g) + F_2(g) + 3H₂O_(l)?
What conclusion about the compound Na₄XeO₆ (of which XeO₆^4- is a part) K₂MnF₆ can be drawn from the reaction?
Answer:

O.N. of Xe decreases from +8 to +6. This shows that XeO₆^4- is an oxidizing agent. It oxidises F^– to F₂. This reaction shows that Na₂XeO₆ (or XeO₆^–) is a stronger oxidizing agent than F₂.

Question 11.
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water?
Answer:

Multiplying Cl^– by 2 (As 2 atoms are in Cl₂)
Cl_2(aq) + SO_2(aq) + H₂O_(l) → 2Cl^–_(aq) + SO₄^2-_(aq)
On adding 4H⁺ in left side and multiplying in H₂O by 2,
Cl_2(aq) + SO_2(aq) + 2H₂O_(l) → 2Cl^–_(aq) + SO₄^2- + 4H⁺
It is a balanced disproportionation reaction.

Question 12.
Differentiate between electrochemical cell and electrolytic cell?
Answer:
Differences between electrochemical cell and electrolytic cell:
Electrochemical (Galvanic) cell:

1. In Galvanic cell, chemical energy is changed into electrical energy.
2. Reaction in Galvanic cell is spontaneous.
3. Oxidation and reduction occurs in different half cell.
4. Anode is negative while cathode is positive terminal.
5. The electrons move from anode to cathode in the external circuit.

Electrolytic cell:

1. In electrolytic cell, electrical energy is changed into chemical energy.
2. Reaction in electrolytic cell is non – spontaneous.
3. Oxidation and reduction occurs in same container.
4. Anode is positive while cathode is negative terminal.
5. The electrons are supplied by the external circuit which enter through the cathode and came out through anode.

Question 13.
What is Nernst equation? Derive relation between E and E°?
Answer:
In Standard cell, the concentration of the solution used is 1M and the temperature is maintained at 25°C (For 298K). But the Galvanic cells are often constructed under non standard condition. In such cases the cell voltage (or EMF) is calculated by Nernst equation as follows:
E = E° – \(\frac{RT}{nF}\) log_e \(\frac { M_{ (s) } }{ M^{ n+ }_{ (aq) } } \)
or E = E° + \(\frac{RT}{nF}\) log_e \(\frac { M^{ n+ }_{ (aq) } }{ M_{ (s) } } \)
In changing log_e in log₁₀
or E = E° + \(\frac{2.303}{nF}\) log₁₀ \(\frac { M^{ n+ }_{ (aq) } }{ M_{ (s) } } \) [M_(s) = 1]
or E = E° + \(\frac{2.303RT}{nF}\) log₁₀[Mⁿ⁺_(aq)]
or E = E° + \(\frac{0.0591}{nF}\) log₁₀ [Mⁿ⁺_(aq)].

Question 14.
Consider the reactions:
2S₂O₃^2-_(aq) + I_2(s) → S₄O₆^2-_(aq) + 2I^–_(aq)
S₂O₃^2-_(aq) → 2SO₄^2-_(aq) + 4Br^–_(aq) + 10H⁺_(aq)
Why does the same reductant, thiosulphate reacts differently with iodine and bromine?
Answer:
Br₂ is stronger oxidizing agent than I₂, it oxidises SO₃^2- to SO₄^2- i. e. from +2 state to +6 state of sulphur. However, I₂ being weaker oxidizing agent, oxidises S₂O₃^2- to S₄O₆^2- ion i.e. from + 2 to + 2.5 state of sulphur.

Redox Reactions Long Answer Type Questions – I

Question 1.
What is standard hydrogen electrode? How it is prepared?
Answer:
Standard hydrogen electrode: This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HCl at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to atmospheric increase its surface. (MPBoardSolutions.com) The hydrogen electrode thus Pressure constructed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the 1 M H+solution circumstances works both as anode or cathode.

Cell reaction of standard hydrogen electrode platinum electrode (SHE) when it acts as anode is
H_2(g) → 2H⁺ + 2e^–
It is represented as
H_2(g)(1 atm) Pt | H₃O⁺_(aq)(1.0 M)
When it acts as cathode, the cell reaction is
2H⁺ + 2e^– → H_2(aq)
and it is represented as
H₃O⁺_(aq) (1.0 M) | H_2(g) (1 atm) Pt
Standard hydrogen electrode (SHE) is arbitarily assigned a potential of zero.

Question 2.
On the basis of electrode potential explain which reaction is feasible:
E°_Cu2+/Cu = 0.34 V, E°_Zn2+/Zn = -0.76 V, E°_Mg2+/Mg = 2.37 V, E°_Fe2+/Fe = -0.74V

1. Cu + Zn²⁺ → Cu²⁺ + Zn
2. Mg + Fe²⁺ → Mg²⁺ + Fe.

Solution:
1. Cu + Zn²⁺ → Cu²⁺ + Zn
E°_Cu²⁺/Cu = 0.34 V and E°_Zn2+/Zn = – 0.76 V
In given cell, Cu oxidised to Cu²⁺. So Cu²⁺/Cu will act as anode and Zn²⁺/Zn acts as cathode.
E°_cell = E°_cathode – E°_anode
E°_cell = – 0.74 – (-2.37) = +1.63V
E°_cell = +Ve, shows that the reaction is feasible.

2. Mg + Fe²⁺ → Mg²⁺ + Fe
E°_Mg²⁺_/Mg = 2.37 V and E°_Cu²⁺/Cu = – 0.74 V
Mg oxidised to Mg²⁺ and acts as anode Mg²⁺/Mg.
Fe²⁺ reduced to Fe and acts as cathode Fe²⁺/Fe.
E°_cell = E°_cathode – E°_anode
E°_cell = – 0.74 – (-2.37) = +1.63 V

Question 3.
While sulphur dioxide and hydrogen peroxide can act as oxidizing as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants, why?
Answer:
In sulphur dioxide (SO₂) and hydrogen peroxide (H₂O₂), the oxidation states of sulphur and oxygen are +4 and -1 respectively. Since, they can increase as well as decrease when there compounds take part in chemical reaction, they can act as oxidizing as well as reducing agents. For example,
Increase in O.N.

In ozone (O₃), the oxidation state of oxygen is zero while in nitric acid (HNO₃), the oxidation state of nitrogen is +5. Since, both of them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidizing agents and not as reducing agents.

In ozone (O₃), the oxidation state of oxygen is zero while in nitric acid (HNO₃), the oxidation state of nitrogen is + 5. Since both them can undergo decrease in oxidation state and not an increase in its value, they can act only as oxidizing agents and not as reducing agents.

Question 4.
How do you count for the following observation though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant, why? Write a balanced redox equation for the reaction?
Answer:
Toluene can be oxidized to benzoic acid in acidic, alkaline and neutral medium by using potassium permanganate.

1. In acidic medium:
[MnO^–_4(aq) + 8H⁺_(aq) + 5e^– → Mn²⁺_(aq) + 4H₂O_(l)] × 6

In the manufacture of benzoic acid alcoholic KMnO₄ is more useful.
Alcohol if used as solvent will help in the formation of a homogeneous mixture between toluene (non – polar) and KMnO₄ (ionic). Actually alcohol has non – polar alkyl group as well as polar OH^– group.

Question 5.
How the redox reactions are balanced by ion – electron method?
Answer:
The redox reactions are balanced by ion – electron method are as follows:
1. Indicate oxidation number of each element and identify the elements which undergoes change in oxidation number.

2. Indicate increase and decrease in oxidation number per atom. Multiply the increase or decrease in oxidation number with number of atoms undergoing the change.

3. Multiply the formula with suitable integer to equalise the increase and decrease in oxidation number.

4. Balance O atoms:
K₂Cr₂O₇ + 6HCl + xHCl → 2KCl + 2CrCl₃ + yCl₂ + H₂O + 6H₂O

5. Balance H atoms:
K₂Cr₂O₇ + 6HCl + 8 HCl → 2KCl + 2CrCl₃ + 3Cl₂ + 7H₂O
Here x = 8 and y = 3 to balance Cl atoms
K₂Cr₂^2-O₇ + 14 HCl → 2KCl + 2rCl₃ + 3Cl₂ + 7H₂O.
Example 2. Cr₂O₇^2- + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O.
Solution:
Step 1. (Cr₂O₇)^2- + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O.

Step 2.

Step 3.
Multiply Fe by 6 to equalise increase or decrease in O.N.
Cr₂O₇^2- + 6Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O

Step 4.
Balance atoms other than O and H
Cr₂O₇^2- + 6Fe²⁺ + H⁺ → 2Cr³⁺ + 6Fe³⁺ + H₂O

Step 5.
Balance O atoms
Cr₂O₇^2- + 6Fe²⁺ + H⁺ → 2Cr³⁺ + 6Fe³⁺ + H₂O + 6H₂O

Step 6.
Balance H atoms
Cr₂O₇^2- + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O.

Question 6.
On the basis of electrode potential values given below? Determine reaction between reactants will occur or not?

Answer:
For feasibility of any reaction E° = +ve.
E°_cell = E°_cathode – E°_anode
(a) Fe³⁺_(aq) and I^–_(aq):
2Fe³⁺ + 2I^–_(aq) → 2Fe²⁺_(aq) + 2I_2g
Half cell reaction (oxidation)
2I^–_(aq) → I_2(g) + 2e^– E° = – 0.54 V
Reducion half cell,
2Fe³⁺_(aq) + 2e^– → 2Fe²⁺_(aq) E° = +0.77 V
Complete reaction,
2Fe³⁺_(aq) + 2I^–_(aq) → 2Fe²⁺_(aq) + I_2(g) E°_cell = +0.23 V
E° = +ve, reaction is feasible.

(b) Ag⁺_(aq) and Cu_(s)²⁺:
2Ag_(aq) + Cu_(s) → Cu²⁺_(aq) + 2Ag_(aq)
Oxidation half reaction,
Cu_(s) → Cu²⁺_(aq) + 2e^– E° = – 0.34 V
Reduction half reaction,
2Ag⁺_(aq) + 2e^– → 2Ag_(s) E° = – 0.80 V
Complete reaction,
Cu_(s) + 2Ag⁺_(aq) → Cu²⁺_(aq) + 2Ag_(s) E°_cell = +0.46 V

Question 7.
Give the main characteristics of electrochemical series?
Answer:
The main characteristics of electrochemical series are:

1. Tendency to loose electron becomes greater and the corresponding element or ion behaves as strong reducing agent if it has greater negative value of E°.
2. The substances above hydrogen are weak oxidising agents and substances below it are strong oxidising agents.
3. If the reducing agent of electrode pair is strong reducing agent, its oxidising agent is weak and when reducing agent of electrode pair is weak, its oxidising agent is strong.
4. In series the reducing efficiency decreases from above to downwards.
5. The more the positive value of E°(MPBoardSolutions.com) the stronger is its oxidising properties. In the series the oxidising power of cation decreases as we move upwards from downwards.
6. In series the electropositive character of metals decreases from upwards to downwards.
7. Along the series the electronegative character of non – metal increases from upward to downwards.
8. The metals which come prior to hydrogen displaces hydrogen from acids,
9. The metals which occupy higher position in the series displaces metals which come later in series from their salt solution.

Question 8.
Balance the equation by oxidation number method and identify the oxidising agent and reducing agent?
Cl₂O_7(g) → ClO₂^–_(aq) + O_2(g).
Answer:
Oxidation number method:

Balance the increase and decrease in O.N. multiply H₂O₂ by 4,
Cl₂O₇ + 4H₂O₂ → ClO₂^– + O₂
Balance atom other,
Cl₂O_7(g) + 4H₂O_2(aq) → 2ClO₂^–_(aq) + 4O_2(g)
Balance ‘H’,
Cl₂O_7(g) + 4H₂O_2(aq) + 2OH^–_(aq) → 2ClO₂^–_(aq) + 4O_2(g) + 5H₂O.

Question 9.
Whenever a reaction between an oxidizing agent and a reducing agent is carried out compound of lower oxidation state is formed if the reducing agent is ¡n excess and compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement by giving three illustrations?
Answer:
1. Let us consider the reaction between carbon and oxygen.
Reducing agent + Oxidising agent

2. Let us consider the reaction between white phosphorous (P₄) and Cl_2(g)

3. Let us consider the reaction between sulphur and oxygen.
Reducing agent + Oxidising agent

Redox Reactions Long Answer Type Questions – II

Question 1.
Balance the following reaction by ion – electron method:
Cr₂O₇^2- + Fe²⁺ + H⁺ → Cr⁺³ + Fe⁺³.
Solution:
Cr₂O₇^2- + Fe²⁺ → Cr⁺³ + Fe⁺³ + H₂O

1. Write the oxidation number of each atom in skeleton equation.

Species undergo change is Cr and Fe.

2. Dividing the equation into two half reaction.
Cr₂O₇^2- → Cr⁺³ (Reduction half)
Fe⁺² → Fe⁺³ (Oxidation half)

3. Balancing reduction half:
(a) Balancing Cr atom,
Cr₂O₇^2- → 2Cr⁺³

(b) Adding e to make up the difference in O.N.

(Since each Cr atom gain 3e^– ∴ for 2 Cr atom 6e^–)

(iv) Balancing oxidation half:
(a) Balancing Fe atom (Balanced)
(b) Adding e^– to make up the difference in O.N.
Fe⁺² → Fe⁺³ + e^–

(v) Balancing ‘O’ atom,
Cr₂O₇^2- + 6e^– → 2Cr⁺³ + 7H₂O
Fe⁺² → Fe⁺³ + e^–

(vi) Balancing H by adding H⁺ ions to the side which is deficient on H atoms.
Cr₂O₇^2- + 6e^– + 14H⁺ → 2Cr⁺³ + 7H₂O
Fe²⁺ → Fe⁺³ + e^–

(vii) Equalising e^–,

Question 2.
To determine the oxidising power/reducing power in a solution which method is used? Explain with example?
Answer:
Determination of standard electrode potential ; To determine electrode potential of an electrode, a cell is setup using this electrode and standard hydrogen electrode. The EMF of the cell is measured with the help of a voltameter (or more accurately by potentiometer). (MPBoardSolutions.com) The electrode potential of hydrogen electrode is taken as zero. Therefore, the E.M.F. of such a cell will directly give the value for electrode potential for the given electrode at standard condition (298 K, 1M, 1 atm pressure).

1. If the deflection in galvanometer is towards metal electrode (experimental electrode) then it is anode (-ve terminal) of the cell. It is allotted negative value of standard reduction electrode potential.
For example, in the determination of E° value of zinc, zinc acts as anode and has got negative value of E°.

E°_cell = E°_cathode – E°_anode
∴ 0.76 V = 0.00V – E°_zn⁺²/Zn = -0.76V

Question 3.
What is oxidation number? Write its main rules of determination?
Answer:
Oxidation number of an element is defined as, “The residual charge left on its atom when all the other atoms are removed from the molecule as ions”. Atoms can have positive, negative or zero oxidation state depending upon their state of combination. (MPBoardSolutions.com) In fact, oxidation number is the charge assigned to the atom in a species according to some rules. It is also known as oxidation state.

Rules for assigning oxidation number:
The assignment of oxidation numbers is arbitrary and is usually governed by the following conclusions:

1. The oxidation number of an element in free state i. e., elementary state is regarded as zero. For example,
Oxidation state of hydrogen in H₂ = 0
Oxidation state of helium in He = 0
Oxidation state of sulphur in S₈ = 0
Oxidation state of sodium in Na = 0
Oxidation state of magnesium in Mg = 0

2. In a compound, the more electronegative elements are assigned negative (-) oxidation state and less electronegative elements are assigned positive (+) values. For example, in HCl as chlorine is more electronegative than hydrogen its oxidation state will have negative value while that of hydrogen will be positive.

3. In the formula of a compound, the sum of the negative oxidation states is equal to the sum of the positive oxidation states.

4. The sum of oxidation number of all the atoms in a neutral molecule is taken as zero.

5. Hydrogen has an oxidation state +1 in compounds like H₂S, H₂O, HCl, etc. Exceptionally it has the oxidation number -1 in metallic hydrides, such as NaH, CaH₂, etc.

6. Oxygen is usually assigned oxidation number -2, except in H₂O₂ and in oxide of fluorine [F₂O], in which the oxidation number -1 and +2 respectively. In all, oxidation number of oxygen is -1.

7. Fluorine being the most electronegative element is assigned oxidation number -1 in all its compounds. Other halogens also show -1 oxidation number.

8. The number of monoatomic ion in an ionic compound is equal to its electric charge. Thus, the elements of group IA of the periodic table (Li, Na, K, Rb, Cs) all have oxidation number + 1, while the alkaline earth metals of group IIA (Ca, Sr, Ba) have oxidation number + 2.

Question 4.
Balance the following reaction by ion – electron method?
Answer:
Ion – electron method:
Oxidation half equation is:

The oxidation number is balanced by adding 2 electrons as:
H₂O_2(aq) → O_2(g) + 2e^–
The charge is balanced by adding 2OH^– ions as:
H₂O_2(aq) + 2OH^–_(aq) → O_2(g) + 2e^–
The oxygen atoms are balanced by adding 2H₂O as:
H₂O_2(aq) + 20H^–_(aq) → O_2(g) + 2H₂O_(s) + 2e^– ……………… (1)
The reduction half equation is:

The Cl atoms are balanced as:
Cl₂O_7(g) → 2ClO₂^–_(aq)
The oxidation number is balanced by adding 8 electrons as:
Cl₂O_7(g) + 8e^– → 2ClO₂^–_(aq)
The charge is balanced by adding 6OH^– as:
Cl₂O_7(g) + 8e^– → 2ClO₂^–_(aq) + 6OH^–_(aq) ………………………… (2)
The balancing equation can be obtained by multiplying equation (1) with 4 and adding equation (2) to it as:
Cl₂O_7(g) + 4H₂O_2(aq) + 2OH^–_(aq) → 2ClO₂^–_(aq) + 4O_2(g) + 5H₂O_(l)

Question 5.
Write the main uses of electrochemical series?
Answer:
Electrochemical series:
When different elements and ions are arranged in increasing order of their standard electrode potentials, a series is obtained which is called electrochemical series or activity series.
The uses of electrochemical series are:
1. Electropositive character of metal:
Metal which loses electron easily from its outermost shell shows high electropositive character while the atom which loses electron with difficulty shows least electropositive character. In electrochemical series, Li shows maximum electropositive character while fluorine shows least electropositive character.

2. Comparison of reactivity of metals:
Metal having negative value of E° loses electron easily. Hence, greater the negative value of E° more is the reactivity and reducing character of a metal.

3. Knowledge of oxidizing agent and reducing agent:
Substance which gains electron and shows maximum value of standard electrode potential E° is strongest oxidizing agent. On the basis of this F is strongest oxidizing agent while Li is weakest oxidizing agent.

4. Displacement of elements:
Metal having greater tendency to form ion displaces others. Therefore, elements getting priority in the series displace ions of the elements following them. For example. When Cu turnings are added to AgNO₃ solution, Ag gets precipitated.
2AgNO_3(aq) + Cu_(s) → Cu(NO₃)_2(aq) + 2Ag_(s)

5. Electroplating:
Process of depositing layer of gold or silver on copper, brass, iron etc. is called electroplating. It makes articles lustrous and attractive.

6. Metallurgy:
More reactive metal displaces less reactive metal from their aqueous salt solution.

7. Corrosion of metals:
Destruction of metals due to action of chemicals, air and moisture is called corrosion.
e.g., iron rust in moist air. To prevent this a layer of more reactive metal like Sn or Zn is deposited on iron. Coating iron with zinc is called galvanization.

Question 6.
What information did you get from the following reaction:
(CN)_2(g) + 20H^–_(aq) → CN^–_(aq) + CNO^–_(aq) + H₂O_(l)
Answer:
(CN)_2(g) + 2OH^–_(aq) → CNO^–_(aq) + H₂O_(l).
1. Let the O.N. of C in (CN)₂ = x
2x + 2 (-3) = 0
or x = +3.

2. Let the O.N. of C in CN^– = x
x + (-3) = -1
or x = +2.

3. Let the O.N. of C in CNO^– = x
x + (-3) + (-2) = -1 or
x = +4.

Following informations obtained from above equation:

1. It is a disproportionation reaction.
2. The reaction occurs in basic medium.
3. O.N. of N in (CN)₂ is -3 and that in CN^– is -2 and in CNO^– it is -5.
4. Cyanogen (CN)₂ gets simultaneously reduced to CN^– ion as well as oxidized to cyanate, CNO^– ion.

Question 7.
The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction?
Answer:
The reaction is:
Mn³⁺_(aq) → Mn²⁺_(aq) + MnO_2(s) + H⁺_(aq)
Reduction half cell:
Mn³⁺_(aq) + e^– → Mn²⁺ …………. (1)
Oxidation half cell:
Mn³⁺_(aq) → Mn⁺⁴ + O_2(s) + e^–
To balance the equation added 4H⁺ in right and in left 2H₂O is added:
Mn³⁺_(aq) + 2H₂O_(l) → MnO_2(s) + e^– + 4H⁺_(aq) ……………….. (2)
Adding eqns. (1) and (2),
2Mn³⁺_(aq) + 2H₂O_(l) → Mn²⁺ + MnO_2(s) + 4H⁺_(aq)
This is last equation balanced (disproportionation reaction).

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 7 Equilibrium

Equilibrium Important Questions

Equilibrium Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Which of the following reaction have equal K_c and K_p:
(a) N_2(g) + 3H_2(g) ⇄ 2NH₃
(b) 2H₂S_(g) + 3O_2(g) ⇄ 2SO_2(g) + 2H₂O_(g)
(c) Br₂ + Cl_2(g) ⇄ 2BrCl_(g)
(d) P_4(g) + 6Cl_2(g) ⇄ 4PCl_3(g)
Answer:
(c) Br₂ + Cl_2(g) ⇄ 2BrCl_(g)

Question 2.
For the reaction N_2(g) + 3H_2(g) ⇄ 2NH_3(g), ∆H = 92 kJ, the concentration of NH₃ at the equilibrium increase by temperature:
(a) Increase
(b) No change
(c) Decrease
(d) None of the above
Answer:
(c) Decrease

Question 3.
In one litre container equilibrium mixture of reaction 2H₂S_(g) ⇄ 2H₂S_(g) + S₂S_(g) is filled 6.5 mol H₂S. 0.1 mol H₂ and 0.4 mol S₂ are present in it. Equilibrium constant of this reaction will be:
(a) 0.004 mol litre^-1
(b) 0.080 mol litre^-1
(c) 0.016 mol litre^-1
(d) 0.160 mol litre^-1
Answer:
(c) 0.016 mol litre^-1

Question 4.
Favourable condition for exothermic reaction of ammonia synthesis N₂S_(g) + 3H₂S_(g) ⇄ 2NH_3(g) are:
(a) High temperature and high pressure
(b) High temperature and low pressure
(c) Low temperature and high pressure
(d) Low temperature and pressure
Answer:
(c) Low temperature and high pressure

Question 5.
Oxidation of SO₂ by O₂ in SO₃ is an exothermic reaction manufacture of SO₃ will be maximum if:
(a) Temperature increased and pressure decreased
(b) Temperature decreased and pressure increased
(c) Temperature and pressure both increased
(d) Temperature and pressure both decreased.
Answer:
(b) Temperature decreased and pressure increased

Question 6.
At 440°C HI was heated in a closed vessel till equilibrium is attained. It is dissociated 22%. Equilibrium constant dissociation will be:
(a) 0.282
(b) 0.0796
(c) 6.0199
(d) 1.99
Answer:
(c) 6.0199

Question 7.
K_p and K_c can be expressed as:
(a) K_c = Kp (RT)^∆n
(b) K_p = K_c (RT)Q^∆n
(c) K_p = K_c(RT)^∆n
(d) K_c = K_p (RT)^∆n
Answer:
(c) K_p = K_c(RT)^∆n

Question 8.
The equilibrium constant of the reaction H_2(g) + I_2(g) ⇄ 2HI_2(g), is 64. If the volume of the container is reduced to one – fourth of its original volume. The value of the equilibrium constant will be:
(a) 16
(b) 32
(c) 64
(d) 128
Answer:
(c) 64

Question 9.
What would happen to a reversible reaction at equilibrium when an inert gas is added while the presence remain unchanged:
(a) More of the product will be formed
(b) Less of the product will be formed
(c) More of the reactant will be formed
(d) It remains unchanged
Answer:
(d) It remains unchanged.

Question 10.
SO_2(g) + \(\frac{1}{2}\) O_2(g) ⇄ SO_3(g), K₁
2SO_3(g) ⇄ 2SO_2(g) + O_2(g), K₂ which of the following is correct:
(a) K₂ = K₁²
(b) K₂ = K₁^-2
(C) K₂ = K₁
(d) K₂ = K₁^-1
Answer:
(b) K₂ = K₁^-2

Question 11.
Which reaction is not affected by change in pressure:
(a) N_2(g) + O_2(g) ⇄ 2NO_(g)
(b) 2O_3(g) ⇄ 3O_2(g)
(c) 2NO_2(g) ⇄ N₂O_4(g)
(d) 2SO_2(g) + O_2(g) ⇄ 2SO_3(g)
Answer:
(a) N_2(g) + O_2(g) ⇄ 2NO_(g)

Question 12.
For N₂ + 3H₂ ⇄ 2NH₃ + heat:
(a) K_p = K_c
(b) K_p = K_cRT
(c) K_p = K_c (RT)^-2
(d) K_p = K_c(RT)^-1
Answer:
(c) K_p = K_c (RT)^-2

Question 13.
Sodium sulphate dissolve in water with the release of heat. Imagine a saturated solution of sodium sulphate. If temperature is increased then according to Le – Chatelier principle:
(a) Mass of solid will be dissolved
(b) Some solid will be precipitated in solution
(c) Solution will be more saturated
(d) Concentration of solution will be unchangeble.
Answer:
(b) Some solid will be precipitated in solution

Question 14.
For reaction, PCl_3(g) + Cl_2(g) ⇄ PCl_5(g) value of K_c at 250° is 26 value of K_p at this temperature will be:
(a) 0.61
(b) 0.57
(c) 0.83
(d) 0.46
Answer:
(a) 0.61

Question 15.
According to Le – Chatelier principle when heat is given on solid – liquid equilibrium then:
(a) Amount of solid decreases
(b) Amount of liquid decreases
(c) Temperature increases
(d) Temperature decreases
Answer:
(a) Amount of solid decreases

Question 2.
Fill in the blanks:

1. For an endothermic process, PCl₅ ⇄ PCl₃+ Cl₂; ∆H = + Q cal. Hence, in this reaction, temperature should be kept ………………………. and pressure ………………………. (If reaction is to be carried out in forward direction)
2. Ice ⇄ Water – Q cal. High temperature in this reaction favours …………………………. direction while increase in pressure favour reaction in ………………………… direction.
3. According to Ostwald’s dilution law, mathematical relation between degree of dissociation and dissociation constant is expressed by ……………………….. Degree of dissociation of weak electrolyte is inversely proportional to its …………………………..
4. Relation between solubility and solubility product for the reaction AB ⇄ A⁺ + B^– is expressed as ……………………….
5. For the maximum yield of SO₃ in the reaction 2SO₂ + O₂ → 2SO₃, ……………………… temperature and pressure is required.
6. Mixture of acetic acid and sodium acetate is an example of ……………………….. solution
7. Mixture of ammonium hydroxide and ammonium chloride is an example of ……………………… solution
8. Degree of dissociation of a weak electrolyte is inversely proportional to ………………………… of concentration.
9. Ostwald dilution law is not applicable for ……………………..
10. Heneiy’s law is related to solubility of in …………………………. solution
11. Relation between K_p and K_c at constant temperature for reaction is ………………………

Answer:

1. High, low
2. Forward, forward
3. α = \(\sqrt { \frac { K }{ C } } \), square root of concentration
4. K_sp = [A⁺] [B^–]
5. Low temperature, high pressure
6. Acidic buffer
7. Basic buffer
8. Square root
9. Strong electrolyte
10. Gas
11. K_p = K_c × RT^∆n.

Question 3.
Answer in one word/sentence:

1. What is the value of K_p and K_c for the given reaction:
   • PCl₅ ⇄ PCl₃ + Cl₂
   • H₂ + I₂ ⇄ 2HI
2. For the unit of K_c, concentration is expressed in?
3. Manufacture of nitrogen peroxide is exothermic. What should be temperature and pressure for maximum yield?
4. Ammonia gas dissolves in water giving NH₄OH. How is water reacting in it?
5. When NH₄Cl is added to NH₄OH solution, dissociation of NH₄OH decreases, why?
6. pH of water at 298K?
7. What is hydrogen ion concentration in pure water?
8. pH of water at 25°C is 7. If water is heated to 50°C, what change in its pH will occur?
9. Write conjugate base of H₂PO₄^– and HCO₃^–?
10. Name one ion which behaves as both Bronsted acid and base?

Answer:

1. • K_p > K_c
   • K_p = K_c
2. mol/litre
3. Low temperature and high pressure
4. Like acid
5. Common ion effect
6. 7
7. 1 × 10^-7
8. pH value decreases,
9. HPO₄^2- CO₃^2-
10. HCO₃^–

Question 4.
Match the following:
[I]

Answer:

1. (d)
2. (e)
3. (b)
4. (a)
5. (c)
6. (f)

[II]

Answer:

1. (d)
2. (c)
3. (a)
4. (e)
5. (b)

Equilibrium Very Short Answer Type Questions

Question 1.
For unit of Kc the concentration is expressed as?
Answer:
mol/litre.

Question 2.
Ammonia when dissolve in water give NH₄OH, here the water acts as?
Answer:
Like acid.

Question 3.
When NH₄Cl is mixed with NH₄OH solution, then the ionisation of NH₄OH decreases. What is the reason?
Answer:
Due to Common ion effect.

Question 4.
The pH of water at 25°C pH = 7. If water is heated upto 50°C, then what will be the change in pH?
Answer:
The value of pH decreases.

Question 5.
Write the conjugate base of H₂PO₄^– and HC0₃^–?
Answer:
HPO₄^2- and CO₃^2-.

Question 6.
Write the name of an ion which acts as both Bronsted acid and base?
Answer:
HCO₃^–.

Question 7.
Give an example of a salt formed by weak acid and weak base?
Answer:
Ammonium acetate.

Question 8.
What is the pH value of human blood?
Answer:
7.4.

Question 9.
What will happen when HCl gas is passed through NaCl?
Answer:
NaCl will precipitate.

Question 10.
Among conjugate bases CN^– and F^–, which is stronger base?
Answer:
CN^– is stronger base.

Question 11.
What is the change in free energy (∆G) of reversible process at equilibrium?
Answer:
Zero (0).

Question 12.
What is solubility product?
Answer:
The product of the concentration of ions in the saturated solution of a sparingly soluble salt is a constant at a given temperature and is called solubility’ product.

Question 13.
What is buffer solution?
Answer:
The solution in which on adding acid or base in iess quantity, the pH change is negligible, is called buffer solution.

Question 14.
In the exothermic reaction of ammonia formation N_2(g) + 3H_2(g) ⇄ 2NH_3(g), when be the formation of ammonia will be more?
Answer:
At low temperature and high pressure the formation of anunorua is more.

Question 15.
Give an example of acidic buffer?
Answer:
Mixture of acetic acid and sodium acetate.

Question 16.
What is the condition for precipitation?
Answer:
For this ionic product should he more than solubility product.

Question 17.
What is the nature of aqueous solution of KCN?
Answer:
It is of basic nature.

Question 18.
What is the nature of aqueous solution of CH₃COONH₄?
Answer:
It is of neutral nature.

Question 19.
What will be the relation between K_c and K_p for the reaction
PCl₅ ⇄ PCl₃ + Cl₂
Answer:
K_p > K_c.

Question 20.
What is Lewis concept?
Answer:
Acid is an electron pair acceptor and base is an electron pair donor.

Question 21.
What is the definition of acid and base according to Bronsted and Lowry concept?
Answer:
Acid is that which furnishes proton and base accepts the proton.

Question 22.
What is pH?
Answer:
The pH value of a solution is the numerical value of the negative power to which 10 should be raised in order to express the Hydrogen ion concentration of the solution i.e. [H⁺] = 10^-pH or pH = – log [H⁺].

Question 23.
The Ostwald dilution law applied on which electrolyte?
Answer:
On weak electrolytes.

Equilibrium Short Answer Type Questions – I

Question 1.
Explain the effects of catalyst at equilibrium?
Answer:
Effect of catalyst:
Le – Chatelier’s principle ignores the presence of a catalyst since the catalyst cannot displaces the equilibrium and simply reduce the time required for attaining the equilibrium when a catalyst is added to a reversible reaction in equilibrium. (MPBoardSolutions.com) It increases the speed of both forward and backward reaction i.e. r_f and r_b to the same extent. However the addition of a catalyst reduces the time required for a reaction to attain the equilibrium.

Question 2.
Explain Ostwald’s law of dilution?
Answer:
Ostwald’s time dilution law:
Ostwald gave law for weak elecrolytes. By applying law of mass action Ostwald gave a law for expressing the dissociation of weak electrolyte. It states that:
“The degree of dissociation of weak electrolyte is directly proportional to the square root of its dilution.”
α = \(\sqrt { KV } \) = \(\sqrt { \frac { K }{ C } } \)
Where, α = Degree of dissociation, K = Dissociation constant, V = Volume in litre in which one gram equivalent is dissolved, C = No. of gram equivalent in one litre.

Question 3.
What is the effect of pressure on chemical equilibrium?
Answer:
On increasing the pressure on chemical equilibrium, the equilibrium shifts in that direction where the volume decreases i.e., the number of moles decreases.
Example: On combining of SO₂ and O₂, SO₃ is formed and 45.2 kcal energy is liberated.
2SO_2(g) + O_2(g) ⇄ 2SO_3(g); ∆H = -45.2 kcal
In this reaction, 2 moles of SO₂ react with 1 mole of O₂ to form 2 moles of SO₃. So, on increasing the pressure the reaction shifts toward forward direction.

Question 4.
What do you mean by buffer solution?
Answer:
Buffer solution: Solution in which,

1. pH value is definite.
2. pH is not changed on dilution or on keeping for sometime.
3. On adding acid or base in less quantity, pH change is negligible.

Such solutions are called buffer solutions.
Or
Buffer solutions are solutions which retain their pH constant or unaltered.

Question 5.
What are acidic buffer and basic buffer?
Answer:
1. Acidic buffers:
Acidic buffers are formed by mixing an equimolar quantities of weak acid and its salt with a strong base.
Example: Acetic acid and sodium acetate, boric acid and borax, citric acid and so-dium citrate, etc.

2. Basic buffers:
Basic buffers contain equimolar quantities of weak base and its salt with strong acid.
Example: Ammonium hydroxide and ammonium chloride.

Question 6.
On the basis ofthe equation pH = – log[H⁺], the pH of the 10^-8mol dm^-3 HCl solution should be pH = 8. But the observed value is less than 7. Explain the reason?
Answer:
The 10^-8moldm^-3HCl shows that the solution is very dilute. So, we cannot neglect the H₃O⁺ ions formed from water in the solution. So, the total [H₃O⁺ ] = (10 ^-8 + 10^-7) M comes to be near 7 or less than 7 i.e., the pH is less than 7. (As the solution is acidic).

Question 7.
Ammonia is a Lewis base. Why?
Answer:
According to Lewis base, a base, is a substance (molecule or ion) which can donate an electron pair to form a co – ordinate. In the structure of ammonia, nitrogen is present in one lone pair of electron. So, ammonia donate a lone pair electron.
For example:

Question 8.
Write the uses of Buffer solution?
Answer:
Uses of Buffer solution:

1. To keep the pH of the reaction constant during the determination of velocity of chemical reactions.
2. The pH should be between pH – 5 to 6 – 8 in the formation of alcohol by fermentation.
3. Preparation of sugar, paper and electroplating occurs at definite pH.

Question 9.
What is the effect of pressure and temperature on solubility of gases in liquids?
Answer:
1. Effect of Pressure:
On increasing the pressure the solubility of gases increases as the molecules of gases enters in the intermoiecular spaces of solvent.

2. Effect of Temperature:
On increasing the temperature the solubility of gases decreases as the kinetic energy of gas molecules increases.

Question 10.
Write the name of the factors affecting the chemical equilibrium?
Answer:
The factors affecting the chemical equilibrium are:

1. Temperature
2. Pressure
3. Change in concentration
4. Catalyst.

Question 11.
Why the solubility of C0₂ decreases on Increasing the temperature?
Answer:
CO_2(g) + aq ⇄ CO_2(eq)
The solubility of CO₂ in water is an exothermic reaction. So, according to Le – Chatelier’s principle, on increasing temperature the reaction proceeds towards backward direction. That is why on increasing temperature the solubility of CO₂ decreases.

Question 12.
What do you understand by ionization of water?
Answer:
Auto ionisation occurs in water molecules. The ionic equilibrium of water can be shown by following equation:
H₂O + H₂O ⇄ H₃OH⁺ + OH^–
Equilibrium constant K = \(\frac { [H_{ 3 }O^{ + }][OH^{ – }] }{ [H_{ 2 }O]^{ 2 } } \)
⇒ K_H2O = [H₃O⁺][OH^–] [∵ K_H2O = K_w]
K_w = 1 × 10^-14
Here K_w is a constant, called ionic product of water.

Question 13.
What is concentration Quotient?
Answer:
The ratio of concentration of products and reactants is called concentration quotient. It is denoted by Q. For any reversible reaction the concentration quotient is equal to the equilibrium constant K_c.
K_c = \(\frac { [C]^{ c }[D]^{ d } }{ [A]^{ d }[B]^{ b } } \)
and Q = \(\frac { [C_{ C }]^{ c }[C_{ D }]^{ d } }{ [C_{ A }]^{ d }[C_{ B }]^{ b } } \)
At equilibrium Q = K_c.

Question 14.
What do you understand by Lewis acid and Lewis base? Explain with example?
Answer:
Lewis acid:
Acid is a substance (an atom, molecule or ion) which can accept a pair of electrons to complete its octet.
Example: BF₃, AlCl₃, Br⁺, NO₂⁺ etc.

Lewis base:
Lewis base is a substance (an atom, molecule or ion) which have complete octet of the central metal atom and have a lone pair of electron to donate in a chemical reaction to form co – ordinate bond.
Example:

Question 15.
At 310 K, ionic product of water is 2.7 × 10^-14. at this temperature determine the pH of neutral water?
Solution:
K_w = [H₃O⁺].[OH^–] = 2.7 × 10^-14 (At 310 K)
For reaction H₂O + H₂O ⇄ [H₃O⁺] [OH^–]
[H₃O⁺] = [OH^–]
So, [H₃O⁺] = \(\sqrt { 2.7\times 10^{ -14 } } \) = 1.643 × 10^-7M
∴_pH = -log [H₃O⁺] = -log 1.643 × 10^-7
= 7 + (- 0.2156) = 6.7844.

Question 16.
Determine the concentration quotient for following reactions:

1. CrO₄^-2_(eq) + Pb_(eq) ⇄ PbCrO_4(s)
   
2. CaC0_3(s) ⇄ CaO_(s) + CO_2(g)
   
3. NH_3(eq) + H₂O_(l) ⇄ NH₄⁺_(eq) + OH^–_(eq)
4. H₂O_(l) ⇄ H₂O_(g)

Answer:

Question 17.
What is the importance of solubility product in precipitation of soap?
Answer:
Soap is produced on alkaline hydrolysis of oil and fats. Soaps are actually sodium or potassium salts of higher fatty acids. (MPBoardSolutions.com) Soap is obtained in the form of concentrated liquid. Concentrated salt solution is added for its precipitation. In presence of increased concentration of common sodium ion, ionic product of [Na⁺] [C_n H_2n+1 COO] exceeds its solubility product (K_sp) and it gets precipitated. It is separated by filtration.

Equilibrium Short Answer Type Questions – II

Question 1.
By comparing the values of Kc and how will you find out the following state of the reaction:

1. Resultant reaction proceeds towards forward direction.
2. Resultant reaction proceeds towards backward direction.
3. No change in the reaction.

Answer:

1. If Q_c < K_c; reaction proceeds towards the direction by products (Forward reaction).
2. If Q_c > K_c; reaction proceeds towards reactant side (Backward reaction).
3. If Q_c = K_c; at equilibrium, the reaction mixture remains as before. So no change in the reaction.

Question 2.
The aqueous solution of sodium carbonate is basic in nature, why?
Answer:
Na₂CO₃ ⇄ 2Na⁺ + CO₃^-2
2H – OH ⇄ = 2H⁺ + 2OH^–
Na₂CO₃ + 2H – OH ⇄ 2NaOH + H₂CO₃
Neither solid Na₂CO₃ nor water alone has any action on litmus paper. However aqueous solution of Na₂CO₃ turned red litmus blue. The problem is successfully explained by the Arrhenius theory of ionization in terms of hydrolysis, when a salt is dissolved in water, it undergoes ionization. (MPBoardSolutions.com) The ions of salt interact with opposite ion of water to form acidic basic or neutral solution. This process is called hydrolysis may be defined as the interaction of ion of a salt with oppositely charged ion of water to give acidic or basic solution. Consider the hydrolysis of a general salt (AB).
Dissociation of salt
AB +Water → A⁺ + B^–
Dissociation of water

Question 3.
What is the effect on equilibrium when gases are dissolved in liquids? Explain with example?
Answer:
When the soda water bottle is opened, dissolved CO₂ gas in it comes out rapidly. This is an example of equilibrium of any gas in equilibrium between gas and its liquid. At finite pressure of the gas there exists an equilibrium between soluble and insoluble molecules of the gas.
CO_2(g) ⇄ CO_2(eq)
Henry’s law: The solubility of gas in a given solvent is directly proportional to the pressure to which the gas is subjected, provided the temperature remains the same.
Thus, m ∝ P
or m = KP
Where, K is a constant of proportionality and known as Henry’s constant.
Example: You have seen that when a soda water bottle is opened, the carbon dioxide dissolved in it freezes out rapidly. This can be explained on the basis of Henry’s law. (MPBoardSolutions.com) Carbonated beverages are bottled under pressure to ensure high concentration of carbon dioxide. When the bottle is opened the pressure above the solution falls and the excess carbon dioxide comes out.

Question 4.
What is common ion effect? Explain?
Answer:
Common ion effect:
When, in a solution of weak electrolyte, a solution of strong electrolyte containing the same ion is added, then the ionisation of weak electrolyte decreases. This effect is called common ion effect.
Example: During precipitation of radicals of group II taken care that the radicals of group IV does not get precipitated. (MPBoardSolutions.com) On the basis of this, H₂S gas is passed in presence of HCl in group II. Due to common ion effect dissociation of H₂S is suppressed.

This reduces concentration of sulphide ion and hence only radicals of group II get precipitated as sulphide. For the precipitation of radicals of group IV, H₂S is passed in presence of NH₄OH. H⁺ ion of H₂S combines with OH^– ions of NH₄OH to produce water. This increases dissociation of H₂S. In this way, increase in sulphide ion concentration helps in the precipitation of positive sulphides of group IV.

Question 5.
What do you understand by conjugate acid and conjugate base?
Answer:
The relative strengths of conjugate pairs can be found out, if we know whether forward reaction is favoured or backward reaction is favoured.
For example:

In the above reaction, the reaction proceeds almost to completion. We must therefore conclude that HCl is a stronger acid than H₃O⁺ i.e., HCl has stronger tendency to donate proton than H₃O⁺. Similarly H₂O is stronger base than H₂O has stronger tendency to accept proton than Cl^–. Thus, the strong acid (HCl) has a weak conjugate base Cl^–.
Consider another reaction

For the reaction, backward reaction is strongly favoured. This shows that H₃O⁺ is a stronger acid than CH₃COOH and CH₃COO^– is a stronger base than H₂O. Thus, we again observe that the strong acid (H₃O⁺) has the weak conjugate base (H₂O). Thus, we conclude that, A stronger acid has a weak conjugate base and vice – versa.

Question 6.
Explain the solubility product by giving definition?
Answer:
Solubility product:
The product of the concentration of ions in the saturated solution of a sparingly soluble salt as AgCl is a constant at a given temperature and is called solubility product. (MPBoardSolutions.com) If any sparingly soluble electrolyte at any temperature forms saturated solution, then the equilibrium is

According to law of mass action from (b) and (c),
\(\frac { [A^{ + }][B^{ + }] }{ [AB] } \) = K
∴ [A⁺ ][B⁺ ] = K[AB]
When solution is saturated [AB] will be constant and value of K[AB] will also be a constant and is written as KJ which is the solubility product.

Question 7.
Explain the acid – base concept of Bronsted – Lowry by giving example?
Answer:
Bronsted – Lowry concept of acid and base (1923):
Scientist Bronsted and Lowry gave the theory for acids and bases which is equally applied to aqueous and non – aqueous solutions of acids and bases. According to it, “Acid is that which furnishes proton and base accepts the proton.”

Neutralization reaction:
A reaction in which a proton is transferred from acid to a base or a reaction between H⁺ and OH^– ion to form H₂O molecule.
Relation between acid and base:

That means, every acid has conjugate base and every base has conjugate acid.

According to the definition given by Bronsted and Lowry, acid and base may be in molecular, cationic or anionic states.

Question 8.
Calculate the value of equilibrium constant for following reaction:
PCl₅ ⇄ PCl₃ + Cl₂
Answer:
Let ‘a’ mole of PCl₅, initiate the reaction and x moles dissociate at equilibrium state. If the volume of container is V litre then,

According to law of mass of action
K_c = \(\frac { [PCl_{ 3 }][Cl_{ 2 }] }{ [PCl_{ 5 }] } \)
On putting the values,

Question 9.
With the help of Le – Chatelier’s principle at equilibrium for reaction 2SO₃ + O₂ ⇄ 2SO₃;
∆H = – 188.2 kJ, determine the restrictions for the formation of sulphur trioxide?
Answer:
Le – Chatelier proposed a law. It states “If as equilibrium m £ chemical system is disturbed by changing temperature, pressure or concentration of the system, the equilibrium shift is such a way so that the effect of change gets minimised.”
On the basis of this principle, changes in chemical equilibria and physical equilibria can be explained.
Formation of SO₃ by SO₂ and O₂:
2SO₂ + O₂ ⇄ 2SO₃; ∆H = – 45.2 kcal

1. Effect of concentration:
On increasing concentration of SO₂ and O₂ more SO₃ will be formed.

2. Effect of pressure:
On increasing pressure, the equilibrium shifts on that side in which number of moles are less or towards less volume. In this reaction, 2 moles of SO₂ and 1 mole of O₂ (Total 3 moies) change into 2 moles of SO₃, So, on increasing pressure, more quantity of SO₃ will be formed.

3. Effect of temperature:
Formation of SO₃ is an exothermic reaction, so on increasing temperature, more SO₃ will dissociate and on decreasing temperature, more SO₃ will be formed.

Question 10.
What is the importance of solubility product m the purifkation of salt?
Answer:
Purification of common salt; Common salt contains impurities in it. For removal of these HCl gas is passed through the saturated solution of salt. HCl ionizes tc greater extent being a strong electrolyte.
HCl ⇄ H⁺ + Cl^– (More ionized)
NaCl ⇄ NaCl ⇄ Na⁺ + Cl^–

Equilibrium Long Answer Type Questions – I

Question 1.
Explain the buffering action of alkaline or basic buffer? Give its importance?
Answer:
Bask buffer:
Basic buffer contains equimolar quantities of weak base and sodium citrate etc.
Buffer action of basic buffer: Consider a basic buffer of NH₄OH and NH₄Cl. This buffer solution contains a large amount of NH₄⁺ ions, Cl^– ions and excess of undissociated NH₄OH molecules along with a small amount of OH^– ions.
NH₄Cl → NH₄⁺ + Cl^–
NH₄OH ⇄ NH₄⁺ + OH^–
On adding a drop of HCl, H₃O⁺ ions produced combines with OH^– ions of the buffer to form weakly ionised H₂O molecules. As a result of this ionisation of NH₄OH increases to restore the concentration of OH^– ions. Sc, pH of solution remains unchanged.

When a few drops of NaOH is added, it provides OH^– ions. The additional OH^– ions combines with NH₄⁺ ions to form weakly ionised NH₄OH resulting an increase in ionisation of NH₄OH to restore the concentration of NH₄⁺ ion. Thus, pH remains unchanged.

Importance of Buffer solution:

1. Buffer solutions are very important in laboratories, industries, botanical or zoological, physiology. Human blood is an example of buffer solution which has pH 7.34.
2. In the extraction of phosphate the buffer solution of CH₃COONa and CH₃COOH is used.
3. In industries, for the manufacturing of sugar and paper and electroplating occurs at fixed pH.

Question 2.
What is pH? Explain?
Or, What is meant by pH? What is its relationship with hydrogen ion concentration?
Answer:
To express the acidity or basicity of a solution Sorensen in 1909 established a scale known as the pH scale.
The pH value of a solution is the numerical value of the negative power to which 10 should be raised in order to express the hydrogen ion concentration of the solution, i.e;
[H⁺] = 10^-pH …………… (1)
On taking log of eqn.(1),
or log₁₀[H⁺] = log₁₀ 10^-pH
log₁₀[H⁺] = -pH log₁₀ 10
or pH = -log [H⁺], (∵ log₁₀ 10 = 1)
or pH = log₁₀\(\frac { 1 }{ [H^{ + }] } \)
This is the required relationship between pH and H⁺ ion concentration.
Thus, the pH value of a solution is the logarithm of its hydrogen ion concentration to base 10 with negative sign.
pH scale expresses acidic and basic nature of solution in terms of numbers from 0 to 14. Acidic solutions have pH value less than 7 while basic solutions have pH value greater than 7. pH value of neutral solution is 7.
Uses of pH measurement:

1. In industries:
   • In the manufacturing of alcohol by fermentation, pH is maintained between 5 and 6.8.
   • Manufacturing of paper, sugar and electroplating is being done at fixed pH.
2. In the study of velocities of chemical reactions, for maintaining pH, buffer solutions are used.
3. In qualitative analysis for removal of phosphate to maintain pH, buffer solutions of CH₃COONa and CH₃COOH are used.
4. pH of human blood is 7.34. Medium of stomach is acidic while in the intestine it is basic medium.

Question 3.
Write the characteristics of equilibrium constant?
Answer:
Characteristics of equilibrium constant:
1. The value of equilibrium constant for a reaction is constant at a given temperature and it changes with change in temperature.

2. The value of molar equilibrium constant does not depends upon the initial molar concentration of reactants and products but it depends upon their concentration at equilibrium state.

3. If the reaction is reversed the value of equilibrium constant is inversed. For example,
Then for, H_2(g) + I_2(g) ⇄ 2HI, K_c = 50
2HI_(g) ⇄ H_2(g) + I_2(g)
∴ K’ = \(\frac { 1 }{ K } \) or K’_c = \(\frac{1}{50}\) = 0.02

4. If the equation having equilibrium constant is divided by 2, the equilibrium constant for the new equation is the square root of K (i.e., \(\sqrt { K } \))
For example:

5. If the equation (Having equilibrium constant K) is multiplied by 2, the equilibrium constant for the new equation is square of K (i. e., K²)
K’ = K²

Question 4.
Give only the definition of:

1. Salt hydrolysis
2. Solubility product
3. Common ion effect
4. Buffer solution

Answer:
1. Salt hydrolysis:
The interaction of cation/anion or both with water making the solution acidic or basic is called salt hydrolysis.

2. Solubility product:
At a particular temperature the product of concentration of ions in a saturated solution of a sparingly soluble electrolyte is known as solubility product. (K_sp).

3. Common ion effect:
The dissociation of weak electrolyte (Weak acid or weak base) is suppressed by the addition of a strong electrolyte having a common ion is called common ion effect.

4. Buffer solution:
Buffer solutions are solutions which retain their pH constant or unaltered after addition of less quantity of acid or base.

Question 5.
Before precipitation of IIIrd group radicals as hydroxides on adding NH₄OH, why it is necessary to add NH₄Cl first?
Answer:
Precipitation of hydroxide of group ill:
Radicals of group III i e.. Al³⁺, Fe³⁺, Cr³⁺ gets precipitated as hydroxides. Radicals cf succeeding groups such as Zn²⁺, Ni²⁺, Mg²⁺, etc. are also precipitated as hydroxide. (MPBoardSolutions.com) Solubility product of hydroxide of group III is lesser than that of other as given below:
Table: Solubility Product of Hydroxide at 18°C

Due to the difference in the solubility product of the hydroxides, concentration of OH^– ion is kept low such that only radicals of group III gets precipitated.
Hence, NH₄OH is added before adding NH₄Cl which remains completely ionized. Due to common (NH₄⁺) ion dissociation of weak electrolyte NH₄OH is suppressed.
NH₄Cl ⇄ NH₄⁺ + Cl^– (More ionised)
NH₄OH ⇄ NH₄⁺ + OH^– (Less ionised)
This decreases comcentration of OH^– ion and only hydroxides with low solubility product gets precipated.

Equilibrium Long Answer Type Questions – II

Question 1.
Derive a relation between equilibrium constant K_p and K_c, Or, Prove that K_p = K_cRT^∆n?
Answer:
In year 1867 Guldberg and Waage put forward a relationship between the rate of reaction and the molar concentration of reactants. The reUucrjhip is.0 as law of mass action. (MPBoardSolutions.com) It states that “At constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentration of the reaction each raised to a power equal to its corresponding stoichiometric coefficient that appears in the balanced chemical equation.”
Example: For solids and liquids: Consider a hypothetical reversible reaction in the state of equilibrium.
A + B ⇄ X + Y
Applying law of mass action: Rate of forward reaction
R_f ∝ [A] [B] or R_f = K_f[A][B] ……………… (1)
Where, K_f is rate constant or velocity constant for the forward reaction. Similarly, rate of backward reaction,
R_b ∝ [X] [Y] or R_b = K_b [X][Y] ………………… (2)
Where, K_b rate constant or velocity constant for the backward reaction.
At equlibrium,
Rate of forward reaction = Rate of backward reaction.
K_f[A][B] = K_b[X][Y]

Its value remains constant at particular temperature. The equation (3) is called law of chemical equilibrium. For a general type of the reaction,
aA + bB ⇄ xX + yY

For gaseous reactants and products:
In the expression for K_c, the concentrations of the various species are generally expressed in terms of moles/litre. (MPBoardSolutions.com) However, in case of gaseous reactions the concentrations of gases may also expressed in terms of their partial pressures. Therefore, for a gaseous reaction:
aA + bB ⇄ xX + yY
The law of chemical equilibrium may be expressed as:

Relationship between K_pc:
Let us consider the following general reaction,
aA + bB ⇄ xX + yY
In which all the substances A, B, X and Y are present in gaseous state. For this reaction, K_p and K_c is written as follows:

Since, for an ideal gas,
PV = nRT
∴ P = \(\frac{n}{V}\) RT = CRT ………………… (3)
Where, the term C (equal to \(\frac{n}{V}\)) represents the molar concentration of the gas. On the basis of equation (3), we have
P_A = C_ART = [A] RT
P_B = C_BRT = [B] RT
P_X = C_X RT = [X] RT
P_Y = C_Y RT = [Y] RT
Substituting this value in equation (2), we get

Proved.

Question 2.
Derive Ostwald’s dilution law of ionization of weak electrolytes? Write its limitations?
Or, Deduce the relationship between degree of ionization and ionization constant?
Answer:
Ostwald’s dilution law: Weak electrolytes are partially ionised. The ions produced due to ionisation of weak electrolyte exist in dynamic equilibrium with the undissociated molecules.
The fraction of the total number of molecules of electrolyte dissolved, which ionises at equilibrium is called degree of ionisation or degree of dissociation. It is denoted by α.
Consider ‘C’ mol per litre be the initial concentration of weak electrolyte AB dissolved in water. Let α be its degree of ionisation. (MPBoardSolutions.com) Thus, the molar concentration of different species before ionisation and at equilibrium be as given below:

Under normal conditions, value of a is very small for weak electrolyte and it can be neglected in comparison to 1.
∴K = Cα²
or α² = \(\frac{K}{C}\)
∴α = \(\sqrt { \frac { K }{ C } } \)
If V is the volume of the solution in litres containing 1 mole of electolyte, then C = \(\frac{1}{V}\). Thus,
α = \(\sqrt { KV } \)
The above equation helps us to conclude that:
The degree of ionisation is inversely proportional to the square root of the molar concentration or directly propotional to the square root of the volume of the solution containing one mole of electrolyte. This is called Ostwald’s dilution law.

Limitations:
This law is not applicable for strong electrolyte and concentrated solutions because attractive force also acts between the solutions of such electrolytes.

Question 3.
Derive a relationship between pH and pOH value? Or, Prove that pH + pOH = 14.
Answer:
The water ionizes as:
H₂O + H₂O ⇄ H₃O⁺ + OH^–
K = \(\frac { [H_{ 3 }O^{ + }][OH^{ – }] }{ [H_{ 2 }O]^{ 2 } } \)
K [H₂O ]² = [H₃O⁺] [OH^–]
[∵K [H₂O ]² = K_w]
K_w = [H₃O⁺][OH^–] ……………. (1)
K_w is a constant, it is called product solubility.
At 298K, Temperature K_w = 1 × 10^-14.
On putting the value in equation (1)
10^-14 = [H₃O⁺][OH^–]
Taking log on both sides,
-14 log₁₀10 = log₁₀[H₃O⁺] + log₁₀[OH^–]
-14 = log₁₀[H₃O⁺] + log₁₀[OH^–], [∵log₁₀10 = 1]
0r 14 = [-log₁₀[H₃O⁺]] + [-log₁₀[OH^–]]
[∵log₁₀ [H₃O⁺] = pH]
[∵log₁₀ [OH^–] = pOH]
14 = pH + pOH

Question 4.
For the determination of pH value of buffer solution derive Henderson – Hazel equation?
Answer:
Henderson’s equation:
pH of a buffer solution can be calculated with the help of Henderson’s equation. For this consider a buffer of weak acid HA and its salt.
HA ⇄ H⁺ + A^–
K_a = \(\frac { [H^{ + }][A^{ – }] }{ [HA] } \)
K_a is dissociation constant of acid.
or [H⁺] = K_a \(\frac { [HA] }{ [A^{ – }] } \)
Salt is completely ionised while due to presence of excess A^– from the salt, the dissociation of weak acid will be depressed more due to common ion effect.
or [H⁺] = K_a \(\frac { [Acid] }{ [Salt] } \) [A^–] ≈ [Salt]
Taking log value,

This equation is called Henderson equation.
In the same way, for basic buffer solution:

pH of a buffer solution does not change woth dilution because on dilution the ration of conc. of acid or base does not change.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 6 Thermodynamics

Thermodynamics Important Questions

Thermodynamics Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
In adiabatic expansion of an ideal gas always:
(a) Increase in temperature
(b) ∆H = 0
(c) q = 0
(d) W = 0
Answer:
(c) q = 0

Question 2.
For a reversible process, free energy change at equilibrium:
(a) More than zero
(b) Less than zero
(c) Equal to zero
(d) None of these.
Answer:
(c) Equal to zero

Question 3.
In isothermal expansion of an ideal gas:
(a) 9 = 0
(b) AE = 0
(c) W = 0
(d) dV = Q
Answer:
(b) AE = 0

Question 4.
Hess’s law is an application of the following:
(a) First law of Thermodynamics
(b) Second law of Thermodynamics
(c) Entropy change
(d) Free energy change.
Answer:
(a) First law of Thermodynamics

Question 5.
When the value of heat of neutralization of an acid with a base is 13.7 kcal, then:
(a) Acid and base both are weak
(b) Acid and base both are strong
(c) Acid is strong and base is weak
(d) Acid is weak and base is strong
Answer:
(b) Acid and base both are strong

Question 2.
Fill in the blanks:

1. Enthalpy is an …………………….. property.
2. Nicely closed thermos flask is an example of an ……………………….
3. Value of heat of combustion (∆H) is always ………………………..
4. Extensive property depend on the ………………………. of matter.
5. Value of heat of neutralization is always …………………………. kilocalorie.

Answer:

1. Extensive
2. Isolation
3. Negative
4. Amount
5. – 13.7

Question 3.
Answer in one word/sentence:

1. Tell the value of heat of neutralization of strong acid and strong base?
2. Give two examples of state function?
3. What is entropy?
4. Among NaCl, H₂O_(s) and NH_3(g) value of whose entropy is higher?
5. Expect the system, what is the remaining part of the universe called?
6. System in which changes occur spontaneously and by which entropy of the system increases, what are they called?
7. What is the type of heat of combustion?

Answer:

1. – 57 kJ
2. Enthalpy, Entropy
3. Measure of disorder
4. NH₃
5. Surroundings
6. Spontaneous process
7. Exothermic.

Thermodynamics Very Short Answer Type Questions

Question 1.
Give two examples of state function?
Answer:
Enthalpy and Entropy.

Question 2.
Whose entropy is greater among NaCl, H₂O_(s) and NH_3(g)?
Answer:
NH_3(g).

Question 3.
Who gave the equation ∆G = ∆H – T∆S?
Answer:
Gibbs – Helmholtz.

Question 4.
Write the equation of first law of thermodynamics?
Answer:
∆E = q + W.

Question 5.
What is the value of entropy when ice melts?
Answer:
Entropy increases.

Question 6.
What is Closed system?
Answer:
System which can exchange energy only and not matter with the surroundings.

Question 7.
The value of which enthalpy is always negative?
Answer:
Enthalpy of combustion.

Question 8.
Heat of neutralization of strong acid and strong base is equal to?
Answer:
-13.7 kcal or -57.1 kJ.

Question 9.
What is Adiabatic process?
Answer:
A process in which no exchange of heat between system and surroundings occur is known as adiabatic process.

Question 10.
What is Enthalpy?
Answer:
Heat change at constant pressure is known as enthalpy.

Question 11.
The process in which pressure remains constant is called?
Answer:
Isobaric process.

Question 12.
In Exothermic reaction the value of ∆H is?
Answer:
Negative.

Question 13.
Unit of specific heat capacity is?
Answer:
joule per kelvin per gm.

Question 14.
The relation between ∆G, ∆S and ∆H is given by?
Answer:
∆G = ∆H – T∆S.

Question 15.
Which type of property is heat capacity?
Answer:
Extensive property.

Question 16.
What type of properties are temperature, pressure and surface tension?
Answer:
Intensive property.

Question 17.
What is the unit of molar heat capacity?
Answer:
joule kelvin^-1 mol^-1.

Question 18.
For which process dq = 0?
Answer:
Adiabatic process.

Question 19.
The efficiency of any fuel is measured by which value?
Answer:
Calorific value.

Question 20.
What is entropy?
Answer:
The measurement of degree of disorder or randomness of the molecule of the system.

Question 21.
Write the relation between standard free energy change ∆G° and equilibrium constant (K)?
Answer:
∆G° = -RTlnK.

Question 22.
What is the value of ∆G for Spontaneous process?
Answer:
∆G < 0.

Question 23.
“The electricity obtained from an electrochemical ceil is equivalent to decrease in free energy”. Write expression for this sentence?
Answer:
∆G° = -nE°F.

Thermodynamics Short Answer Type Questions – I

Question 1.
What is System?
Answer:
System:
A specified portion of the universe which is selected for experimental or theoretical investigations is called the system. (MPBoardSolutions.com) In the system, the effects of certain properties such as pressure, temperature, etc. are observed. A system is said to be homogeneous if it consists of only one phase. On the other hand, it is heterogeneous if it consists of more than one phase.

Question 2.
What is process and what are its kinds?
Answer:
The operation which brings about change in the state of a system is called a thermodynamics process.
Thermodynamics process may be further classified as follows:

1. Isothermal process
2. Adiabatic process
3. Isobaric process
4. Isochoric process
5. Reversible process
6. Irreversible process
7. Cyclic process.

Question 3.
Explain Exothermic reaction with example?
Or, Explain why the value of enthalpy change negative for exothermic reaction?
Answer:
The reactions in which heat is emitted are called exothermic reactions. In such reactions total heat content of products is less than that of reactants. Hence, ∆H is negative.
Example: 2NO_(g) → N_2(g) + O_2(g); ∆H = – 180.5 kJ/mol.

Question 4.
Explain extensive and intensive properties?
Answer:
1. Intensive properties:
The properties of the system which are independent of the amount of matter present in it, are called intensive properties.
Example: Temperature, viscosity, surface tension, refractive index, specific heat, density, etc.

2. Extensive properties:
The properties of the system which depend upon the amount of matter present in it, are called extensive properties.
Example: Mass, volume, energy, etc.

Question 5.
State and Explain Zeroth law of Thermodynamics?
Answer:
According to this law, “Two bodies which are separately in thermal equilibrium with a third body are also in thermal equilibrium with each other”.

Question 6.
Explain Enthalpy of neutralization with example?
Answer:
The enthalpy of neutralization is defined as:
“Change in enthalpy when one gram equivalent of an acid is neutralized with one gram equivalent of base in dilute solution at constant temperature.
Example: NaOH_(aq) + HCl_(aq) ⇄  NaCl_(aq) + H₂O_(l); ∆H =- 57.1 kJ

Question 7.
The enthalpy of neutralization of weak acid and weak base is less than enthalpy of neutralisation of strong acid and base. Why?
Answer:
If either acid or base weak then its ionisation in solution remains incomplete. As a result a part of energy liberated during combination of H+ and OH” ion is used up for the ionisation of weak acid and weak base. Therefore, the value of enthalpy of neutralisation of weak acid with strong base or vice – versa is numerically less than – 57.1 kJ.

Question 8.
What is Bond enthalpy and bond dissociation energy?
Answer:
It is a well known fact that during the formation of a chemical bond, energy is required. Therefore the breaking of a bond energy is to be supplied. Thus, the energy required to break a particular bond in a gaseous molecule is called bond dissociation energy.
Example: 2HCl_(g) → H_2(g) + Cl_2(g)

Question 9.
What is the first law of thermodynamics? Write its mathematical form?
Or, Write the first law of thermodynamics and derive the mathematical expression of it?
Answer:
First law of thermodynamics is the law of conservation of energy .The common statement of this law is:
“Energy can neither be created nor be destroyed but it can be converted from one form to another form.” Let internal energy of the system is E₁ and q calorie heat is supplied to the system. E₂ is the energy of the final stage and work done is W. Therefore,
E₂ – E₁ = q + W
or ∆E = q + W.

Question 10.
Define the term Entropy?
Answer:
A change that brings about disorder or randomness is more likely to occur than one that brings about order. To account for the randomness or disorder of a system a state function called entropy was introduced. It is (MPBoardSolutions.com) defined as the measure of degree of disorder or randomness of the molecule of the system. Entropy is represented by symbol S. It is easier to define entropy change than entropy of a system.

Question 11.
Whose entropy is more: Water vapour, water or ice, why?
Answer:
Entropy is the measure of randomness. In solid state the molecules are completely arranged, therefore its entropy is minimum and in gas the molecules move randomly in all directions, so the value of entropy is more. So
S_(ice) < S_(water) < S_{(water vapour)}.

Question 12.
Among NaCl, H₂O and NH₃ whose entropy will be maximum and why?
Answer:
Entropy is the measure of randomness. In solid state the molecules are regularly arranged, so entropy is minimum whereas in gas the molecules move randomly in all directions, so the entropy is maximum. So, in the above example, NaCl is a solid, H₂O is liquid and NH₃ is a gas. So, the entropy of NH₃ is maximum and entropy of NaCl is minimum.

Question 13.
Prove that: P∆V = ∆ntRT?
Answer:
From ideal gas equation
P V = nRT
If the volume of gas at initial state is V₁ and the number of moles of gas is n₁ then,
PV₁ = ∆n₁RT …………….. (1)
If at final state, the volume of gas is V₂ and number of moles of gas is n₂, then,
PV₂ = ∆n₂RT …………. (2)
From eqn. (i) and (2),
P(V₂ – V₁) = (n₂ – n₁)RT
or P∆V = ∆nRT.

Question 14.
What is relation between ∆H and ∆U?
Answer:
If H be the enthalpy of any system and U is the internal energy, then the relation will be
H = U + PV
For enthalpy change, ∆H = ∆U + P∆V
We know that P∆V = ∆nRT
On putting the value,
∆H = ∆U + ∆nRT.

Question 15.
What do you mean by specific heat capacity?
Answer:
It is the heat required to raise the temperature of 1 gram of substance by 1°C. It is denoted by C_s.
C_s = \(\frac{C}{m}\)
Where, C = Heat capacity, C_s = Specific heat capacity, m = Mass of substance.
Its S.I. unit is joule K^-1g^-1.

Question 16.
What do you mean by molar heat capacity?
Answer:
It is the heat required to raise the temperature of 1 mole of substance by 1°C.
Molar heat capacity = \(\frac{C}{n}\) \(\frac{q}{∆T × n}\)
Where C = Heat capacity (Absorbed heat), ∆T = Increase in temperature, n = Number of moles.
Its S.I. unit is joule K^-1 mol^-1.

Question 17.
Write Hess’s law?
Answer:
In 1840, G.H. Hess formulated a law known as Hess’s law. According to law, “The enthalpy change in a physical or chemical change is same whether the process is carried out in one step or in several steps.”

Question 18.
What is Adiabatic process?
Answer:
A process in which no exchange of heat between system and surroundings occur is known as adiabatic process. This process mainly occurs in isolated system. For this type of process dq = 0.

Question 19.
What is standard enthalpy of reaction?
Answer:
The enthalpy change takes place at standard state i.e., at 298 K temperature (25°C), 1 atm pressure (760 mm) condition is called standard enthalpy of reaction. It is denoted by ∆H°_r or ∆_rH°.

Question 20.
What is enthalpy of solution? Explain with example?
Answer:
Enthalpy change taking place during the dissolution of one mole of a substance in excess of a solvent such that further addition of solvent does not produce any heat change is known as enthalpy of solution.
Example: KCl_(s) + aq → KCl_(aq)

Question 21.
What is enthalpy of hydration? Explain with example?
Answer:
Enthalpy change taking place when one mole of anhydrous salt combines with the required number of moles of water to form hydrated salt is called enthalpy of hydration.
Example: CuSO_4(s) + 5H₂O_(l) → CuS0₄.5H₂O ∆H = – 78.2kJ.

Thermodynamics Short Answer Type Questions – II

Question 1.
What is law of energy conservation? Write its mathematical expression. Or, What is the first law of thermodynamics? Deduce its mathematical expression?
Answer:
This law was first expressed by Meyer and Helmholt. According to this law, “Energy cannot be created or destroyed although one form and vice – versa”.
Or
Whenever a certain quantity of energy in one form disappear, an equivalent amount of energy in another form reappear.

Mathematical form:
Let us suppose that system has internal energy equal to U₁. If it absorbs heat energy 17 from the surroundings then internal energy will increase and becomes U₁ + q. If the work is done on the system then its final internal energy will become
U₂ = U₁ + q + w
U₂ – U₁ = q + w,
∆U – q + w, (∴ U₂ – U₁ = ∆U
If the work done on the system is w, then
∆U = q + w
If work done by the system
∆U = q – w

Question 2.
What is Enthalpy of fusion? Explain with example?
Answer:
Enthalpy change taking place during the conversion of 1 mole of a solid substance into liquid at its melting point is known as enthalpy of fusion.
Example: Enthalpy of fusion of ice at 273 K is 6.0 kJ.
H₂O_(s) → H₂O_(l);
i.e 6.0 kJ of energy is absorbed for the conversion of 1 mole of ice into water. ∆H = +6.0 KJ

Question 3.
For an isolated system if ∆U = 0, what will be ∆S?
Answer:
For an isolated system, AU = 0 and for a spontaneous process, the change in entropy should be positive. For example: For a closed container, which is an isolated system, two gases A and B are diffused (MPBoardSolutions.com) Both gases A and B are separated by a kinetic separator. When the separator is removed, the gases starts fusing with each other and randomness increases in the system. For this process, ∆S > 0 and ∆U = 0.
Again ∆S = \(\frac { q_{ rev } }{ T } \) = \(\frac { \Delta H }{ T } \) = \(\frac { \Delta U+P\Delta V }{ T } \) = \(\frac { P\Delta V }{ T } \) [∴∆U = 0]
So, T∆S or ∆S > 0.

Question 4.
On the basis of the following equations, write a note on thermodynamic stability of NO_(g):
\(\frac{1}{2}\) N_2(g) + \(\frac{1}{2}\) O_2(g) → NO_(g); ∆_rH° = 90kJ mol^-1
NO_(g) + \(\frac{1}{2}\) O_2(g) → NO_2(g); ∆_rH° = 94kJ mol^-1
Answer:
NO_(g) is unstable, as the preparation of NO is an endothermic reaction. But preparation of NO_2(g) occurs because it is an exothermic reaction (energy evolved). So unstable NO_(g) is converted into NO_2(g).

Question 5.
At equilibrium state, whose value will be zero ArG or ArG°?
Answer:
∆_rG = ∆_rG° + RTlnK
At equilibrium, 0 = ∆_rG° + RTlnK
or ∆_rG° = – RTlnK
∆_rG° = 0 (When K = 1)
For other values K, ∆_rG° will be zero.

Question 6.
What is the meaning of calorific value of any fuel? Explain with example?
Answer:
The heat or energy evolved in joule or calorie by the (combustion) burning of 1 gram food or fuel is known as calorific value of fuel.
C₂H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H_2(g); ∆H = -2840 kJ
In this process from 1 mole glucose or 180 gm = 2480 kJ energy obtained
So, energy obtained from 1 gm glucose = \(\frac{2480}{180}\) = 15.78 kJ/gm.
So, calorific value of glucose is 15.78 kJ/gm.

Question 7.
Explain heat of vapourization and heat of reaction?
Answer:
Heat or Enthalpy of vapourization : Enthalpy change taking place during the conversion of 1 mole of liquid into vapours at its boiling point and 1 atm pressure is called enthalpy of vapourization.
Example:
H₂O ⇄ H₂O_(g); ∆H = +40.7 kJ

Heat of reaction:
Enthalpy change taking place when number of moles of reactants as represented by the chemical equation have completely reacted is known as enthalpy of reaction. It is denoted by ∆H_f.
Example: C_(s) + O_2(g) → CO_2(g); ∆H_f = – 393.5 kJ/mol

Question 8.
Explain enthalpy of fusion and enthalpy of sublimation with example?
Answer:
Enthalpy of fusion:
Enthalpy change taking place during the conversion of I mole of a solid substance into liquid at its melting point is known as enthalpy of fusion.
H₂O_(s) ⇄ H₂O_(l); ∆H = +6.01 kJ
Enthalpy of Sublimation:
Enthalpy change taking place when one mole of solid changes into vapours without passing into intermediate liquid state at a temperature below its melting point is known as enthalpy of sublimation.
I_2(s) ⇄  I_2(g). ∆H = +62.4 kJ

Question 9.
How the change in entropy takes place in vapourisation process?
Answer:
Entropy of vapourisation:
Change in entropy when one mole of a liquid at its boiling point changes to the vapour state at the same temperature.
∆S_vapour = S_(vapour) – ∆S_(liquid) = \(\frac { \Delta H_{ vap } }{ T_{ b } } \)
Here, ∆_vap H is the latent heat of vapourisation (enthalpy of vapourisation) and T_b is the boiling point when liquid is converted into vapour, entropy of the system increases. Thus, ∆S_vap is +ve.

Question 10.
What is Internal Energy? Is its absolute value can be determined?
Answer:
Physical and chemical process occurs by some energy change. This energy may appear in the form of light, heat and work. (MPBoardSolutions.com) This evolution and absorption of energy clearly shows that every’ substance or system is associated with some definite amount of inherent energy’. The actual value of this inherent energy depends upon:

1. Chemical nature of substance
2. Conditions, like temperature, pressure and volume, and
3. Composition of the substance.

Thus, “The energy stored within a substance is called internal energy or intrinsic energy”.
Actually, internal energy is the sum of various forms of energy such as; electronic energy E_e, nuclear energy E_n, chemical bond energy E_c, potential energy E_p and kinetic energy E_k . Kinetic energy is the sum of translational energy (E_t), vibrational energy (E_v) and rotational energy (E_r).
It is represented by the symbol ‘U’,
U = E_e + E_n + E_c + E_p + E_k
It may be noted that, absolute value of internal energy cannot be determined, because it is not possible to determine the exact values for the constituent energies, such as: translational, vibrational, rotational energies etc.

Question 11.
Expansion of any gas in vacuum is called free expansion. 1 L of an ideal gas expands isothermally upto 5 L, then determine the change in internal energy and work done?
Solution:
Work done, W = – P_(external) (V₂ – V₁)
When P_(external) = 0,
So, W = – 0 (5 – 1) = 0
For isothermal expansion,
∆U = 0
So, ∆T = 0.

Question 12.
An ideal gas filled in a cylinder (according to figure.) is compressed in a single step with external pressure P_(external) Then what will be the work done on the gas? Explain with graph?
Solution:
Let the initial volume of the gas is V_i and pressure of cylinder is P. On compressing the gas by pressure P the final volume of gas is V_f.
So, change in volume ∆V = (V_f – V_i)
If W is the work done by the piston on the system
W = P_(external) (- ∆V)
W = P_(external) (V_f – V_i)
This can be shown in the figure by (P – V) graph, The work done is equal to ABV_fV_i. The positive sign shows that the work is done on the system.

Thermodynamics Long Answer Type Questions – I

Question 1.
What is heat capacity? Deduce the expression C_p – C_v = R?
Answer:
Heat Capacity: It is equal to the amount of heat required to raise the temperature of the system through 1°C. Its unit is JK^-1.
Relationship between C^p and C^v:
At constant volume: q_v = C_v∆T = ∆U
At constant pressure: q_p = C_p∆T = ∆H
∆H and ∆U are related to each other as
∆H = ∆U + ∆n_gRT
or ∆H = ∆U + ∆n_g(PV)
For 1 mole of ideal gas PV = RT
∴ ∆H = ∆U + ∆(RT)
or ∆H = ∆U + R∆T
on putting the value of ∆H and ∆U
C_p∆T = C_v∆T = C_v∆T + R∆T
Dividing whole equation by ∆T,
C_p = C_v + R
C_p – C_v = R
Thus, value of C_p is always more than C_v and the difference between them is about 2 calories or 8.314 joule.
This relationship is known as Meyer’s relationship.
The ratio C_p/C_v:
The ratio of molar heat capacities at constant pressure (C_p) to that at constant volume (C_v) is represented by γ. Value of γ gives information about the atomicity of the gas. Thus,
For monoatomic gases γ = 1.67
For diatomic gases γ = 1.40
For triatomic gases γ = 1.30

Question 2.
Explain Enthalpy of combustion? Write its uses also?
Answer:
Enthalpy of combustion:
The enthalpy change taking place when one mole of substance is completely oxidised or burnt in presence of excess of oxygen is known as enthalpy of combustion. For example, the enthalpy of combustion of methane is 890.4 kJ.
CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g) ∆Hc = – 890.4 kJ

Carbon on the other hand is oxidised to carbon monoxide and carbon dioxide.
C_(s) + \(\frac{1}{2}\) O_2(g) → CO_(g); ∆H = – 110.5kJ
and C_(s) + O_2(g) → CO_2(g); ∆H_c = – 393.5 kJ

In this case enthalpy of combustion of carbon is – 393.5 kJ and not – 110.5 kJ as formation of CO is a result of incomplete combustion of carbon.
Uses of Enthalpy of combustion:

1. To determine the calorific value of fuel.
2. To determine the enthalpy of reaction of compounds.
3. Determination of structure of compounds.
4. To calculate the calorific value of food.

Question 3.
Prove that qr = ∆H_p?
Or, Prove that at constant pressure and constant temperature the heat of reaction is equal to the change in enthalpy of the system?
Answer:
Suppose enthalpy, internal energy and volume of a system in initial state are H₁, U₁ and V₁ respectively and after gaining heat these values becomes H₂, U₂ and V₂ respectively, then according to definition.
H₁ = U₁ + PV₁, (in initial state) ……………… (1)
H₂ = U₂ + PV₂, (in final state) …………….. (2)
Subtracting eqn. (1) from eqn eqn. (2),
H₂ – H₁ = U₂ – U₁ + P(V₂ – V₁)
or ∆H = ∆U + P∆V
Where ∆H is enthalpy change, ∆U is change in internal energy and ∆V is change in volume. Therefore at constant pressure enthalpy change is equal to sum of internal energy change and expansion type of mechanical work.
According to first law of thermodynamics,
∆U = q – P∆V
q = ∆U + P∆V
= (U₂ – U₁) + P(V₂ – V₁)
= (U₂ + PV₂) – (U₁ + PV₁)
= H₂ – H₁
= ∆H
∴ ∆H = q_p
Thus, enthalpy change represents the heat change occurring at constant temperature and pressure. It is noteworthy that though q is path dependent, q_p is not because ∆H is a state function.

Question 4.
Differentiate between Reversible and Irreversible processes?
Answer:
Differences between Reversible and Irreversible process:
Reversible process:

1. It is carried out infinitesimally slowly i.e., the difference between driving force and the opposing force is very very small.
2. It is an ideal process requiring infinite time for completion.
3. Equilibrium is not disturbed at any stage during the process.
4. Work obtained is maximum.
5. It is an imaginary process and cannot be realised in actual practice.

Irreversible process:

1. This process is carried out rapidly /.e.,the difference between driving force and the opposing force and the opposing force is quite large.
2. It is a spontaneous process requiring finite time for completion.
3. Equilibrium may occur only after the completion of the process.
4. Work obtained is not maximum.
5. It is a natural process which occurs in
6. particular direction under given set of conditions.

Question 5.
What are the factors affecting enthalpy of reaction?
Answer:
Factors on which enthalpy of a reaction (∆H) depends: Enthalpy of a reaction i.e., ∆H depends upon the following factors:
1. Physical state of reactants and products:
Enthalpy of reaction is affected by the physical state of reactants and products because latent heat of substance is also involved. For example, value of enthalpy of reaction for the formation of liquid water and water vapour is different.
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 286 kJ
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 249 kJ

2. Quantities of the reactants involved:
Enthalpy of reaction is affected by the physical state of reactants and products because latent heat of substance is also involved. For example, value of enthalpy of reaction for the formation of liquid water and water vapour is different.
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 286 kJ
H_2(g) + \(\frac{1}{2}\) O_2(g) → H₂O_(l); ∆H = – 249 kJ

3. Allotropic modifications:
Allotropes of an element may have different enthalpies. For example, enthalpy change during combustion of graphite and diamond is – 393.5 kJ/mol^-1 and – 395.4 kJ/mol^-1 respectively.
C_(graphite) + O_2(g) → CO_2(g); ∆H = – 393.5 kJ
C_(diamond) + O_2(g) → CO_2(g); ∆H = – 395.4 kJ

4. Temperature:
Value of enthalpy of reaction is dependent on the temperature at which the reaction is carried out. For example, at 25°C enthalpy of formation of HCl_(g) is 184.6 kJ while at 75°C it is 184.4 kJ.
H_2(g) + Cl_2(g) → 2Hl_(g); ∆H = 184.6 kJ at 25°C
H_2(g) + Cl_2(g) → 2HCl_(g); ∆H = 184.4 kJ at 75°C

Question 6.
Prove that at constant volume q_v = ∆U?
Answer:
When a reaction occurs at constant volume, then no work is done by the system. So W = 0.
So, ∆U = q + W
On putting the value, ∆U = q
So, at constant volume the energy is absorbed which increases the internal energy of the system. So,
∆U = q + W = q + p∆V, (W = P∆V)
Since, the reaction occurs at constant volume. So, ∆V = 0.
On putting vallue, ∆U = q_v

Question 7.
From following data determine the heat of reaction of CH₄ or enthalpy, ∆H:
C_(s) + O_2(g) → CO₂; ∆H = – 97k cal ………… (1)
2H_2(g) + O_2(g) → 2H₂O_(g) ∆H = – 136k cal …………. (2)
CH₄ + 2O_2(g) → CO_2(g) + 2H₂O_(g); ∆H = – 212k cal ………………… (3)
Solution:
To determine
C_(s) + 2H_2(g) → cH_4(g); ∆H = ?
On adding eqn.(1) and (2),

Thermodynamics Long Answer Type Questions – II

Question 1.
What is Hess’s Law of constant heat summation? Explain with an example?
Answer:
In 1840, G.H. Hess gave an important law of constant heat summation according to which, “The enthalpy change in a particular reaction is always constant and does not depend on the path in which reaction takes place”.
Or
“The enthalpy change in a physical or chemical process is the same whether the process is carried out in one or in several steps.”
This law is based on the law of conservation of energy. Suppose that the conversion of substance A to (MPBoardSolutions.com) substance Z takes place in a single step by first method and through several steps in second method. In single step by first method and through several steps in second method. In single step:
A → Z + Q₁

Where, Q₁ is the energy in several steps:
A → B + q₁
B → C + q₂
C → Z + q₃
Total energy evolved in several steps = q₁ + q₂ + q₃
= Q₂ calories (suppose)
Acoording to Hess’s law,
Q₁ = Q₂
Suppose Hess’s law is incorrect and Q₂ > Q₁. In this stage if we convert A to Z by several steps and then Z directly to A, then heat equal to (Q₂ – Q₁) is produced. By repeating this cyclic process several times, an (MPBoardSolutions.com) unlimited amount of heat (energy) may be produced in an isolated system. But this is against the law of conservation of energy.
Practically also, Hess’s law is proved to be true.
Example: Carbon can be directly burnt to produce C0₂ or in the second method it is first converted to carbon monoxide and then oxidized to carbon dioxide.

Energy evolved by both the methods is nearly same. Different of 0.3 kcal is due to experimental error.
∴ ∆H = ∆H₁ + ∆H₂

Question 2.
Prove that ∆H = ∆U + P∆V?
Or, Explain the relationship between ∆H and ∆U?
Answer:
Relation between AH and AU: In case of solids and liquids, the difference between ∆H and ∆U is not significant but in gases it is significant. Let us consider a reaction involving gases. Let the process be isothermal and carried out at constant pressure (P). If V_A is the total volume of the gaseous reactants and V_B be the total volume of gaseous products, also n_A be the number of moles of gaseous reactants and n_B be the number of moles of gaseous products. Then
PV_A = n_ART ……………………. (1)
and PV_B = n_BRT ………………….. (2)
Substarcting eqn. (1) from eqn. (2) we get
PV_B – PV_A = (n_B – n_A)RT
or P(V_B – V_A) = (n_B – n_A)RT
P∆V = ∆nRT
For gaseous reactants PV_R = n_RRT …………………… (3)
and for gaseous products PV_p = n_pRT …………………… (4)
Substracting eqn. (3) from eqn. (4), we get
P(V_p – V_R) = (n_p – n_R) RT
or P∆V = ∆n_gRT ……………………. (5)
But enthalpy change ∆H = ∆U + P∆V …………………….. (6)
Substracting the value of P∆V from eqn. (5) into eqn. (6), we get
∆H = ∆U + ∆n_gRT ………………. (7)
Thus, using above eqn. (7) ∆H can be converted into ∆U or vice – versa.eqn. (7) can be wriien as
q_p = q_v + ∆n_gRT ………………….. (8)
Because ∆H = q_p and ∆U = q_v.
Conditions:

1. If the number of moles of products is greater than that of reactants than ∆n will be +ve and ∆H is greater than ∆U. So,
∆H = ∆U + ∆nRT

2. If the number of moles of reactants is greater than the number of moles of products then ∆n will be – ve and the value of ∆H is less than ∆U.
∆H = ∆U – ∆nRT

3. If the No. of moles of reactants is equal to No. of moles of products then ∆n = 0, then in this condition ∆H = ∆U.

Question 3.
Deduce an expression for PV work done?
Answer:
Let us, consider a cylinder, fitted with a weightless, frictionless piston having a cross – sectional area A, filled with 1 mole of an ideal gas. The total volume of gas is V_i and pressure inside the cylinder is P_in.
Suppose, external pressure on the gas is P_ex which is slightly greater than the internal pressure of the gas. (MPBoardSolutions.com) Due to this difference in pressure the gas is compressed till the pressure inside becomes equal to P_ex Suppose, the change is achieved in one single step and the final volume of the gas is V_f The gas is compressed and suppose the piston moves a distance l. Work done during compression is
W = Force × Displacement = F × l

or F = P × ABV
∴W = P × A × large
or W = – P∆V, [∴ A × l = ∆V]

The negative sign in the expression is required to obtain conventional sign W. In compression, V_f < V_i and therefore, (V_f – V_i) or ∆V is – ve.
Hence, W will come out to be +ve from the above expression. In expansion type work V_f > V_i and value of ∆V is positive, therefore, work done will be negative.

This expression is useful for all types of PVwork and for irreversible flow.
Now, we will calculate, the work done during expansion of ideal gas in a reversible manner and in isothermal condition.

Question 4.
What is free energy? Derive its mathematical form and write Gibbs – Helmholtz equation?
Answer:
The free energy of a system is defined as the maximum amount of energy of the system which can be converted into useful work.
Free energy is related to enthalpy (H), entropy (S) and absolute temperature (T) as,
G = H – TS
Since, we know H = E + PV
∴ G = E + PV – TS
Change in free energy may be given as,
∆G = ∆E + ∆(PV) – ∆(TS)
If the process is carried out at constant temperature and pressure, then
∆(TS) = T∆S and ∆(PV) = P∆V
∆G_Tp = ∆E + P∆V – T∆S or
∆G = ∆H – T∆S
The above equation is called Gibbs – Helmholtz equation and it helps to predict the spontaneity of a process.
Free energy change and spontaneity:
For a system which is not isolated with surroundings
∆S_total = ∆S_system + ∆S_surrounding …………………… (1)
when reaction takes place at constant temperature and constant pressure, heat is supplied to surrounding.
∆S_surrounding = \(\frac { -q_{ p } }{ T } \) = \(\frac { -\Delta H }{ T } \), (q_p = ∆H at constant pressure) ………………………. (2)
From eqns. (1) and (2),
∆S_total = ∆S_system – \(\frac { -\Delta H }{ T } \) …………….. (3)
Since, all the quantities on the right – hand side are system properties, the subscript ‘system’ is not used in equations.
Multiplying both sides by T, we get
T∆S_total = T∆S – ∆H, (Where, ∆S = ∆S_system)
or -T∆Sc = ∆H – T∆S …………………… (4)
For Gibbs free energy (G),
G = H – TS
∆G = ∆H – T∆S – S∆T ………………… (5)
or ∆G = ∆H – T∆S_total
For the process taking place at constant temperature and constant pressure, eqn. (5) will be
So that, ∆G = ∆H – T∆S ……………………… (6)
Comparing eqns. (4) and (6),
∆G = – T∆S_total ………………….. (7)
So that, ∆G = – ve(for sontaneous chnages)
We know that, for spontaneous chemical change ∆S_total is positive. Eqn. (7) shows that the spontaneity of a change can be predicted on the basis of the value of ∆G.
Three special cases may be considered according to eqn. (7):

1. If AG is negative, the change is spontaneous.
2. If AG is zero, the system is in equilibrium.
3. If AG is positive, the change is non – spontaneous.

Conditions for spontaneity of a process (conditions for ∆G to be – ve):

1. If AH is negative and AS is positive, AG would certainly be negative and the process will be spontaneous.
2. If AH is negative and AS is also negative, then AG would be negative if AH is greater than TAS in magnitude.

Question 5.
Explain the determination of internal energy ∆U by Bomb calorimeter under the following heads:

1. Labelled diagram of the apparatus,
2. Explanation of the process
3. Calculations.

Answer:
Experimental determination of change in internal energy:
The change in internal energy in a chemical reaction is determined with the help of an apparatus called bomb calorimeter. (MPBoardSolutions.com) It is made up of steel so that it can bear high pressure developed during the chemical reaction taking place in the calorimeter. The inner side of the steel vessel is coated with some non – oxidizable metal like Pt or Au. It is also fitted with a pressure tight screw – cap. The two electrodes are connected to each other through a platinum wire dipped in a platinum cup.

A small known mass of the substance under investigation is taken in the platinum cup. The bomb is filled with excess of oxygen under a pressure of 20 – 25 atm and sealed. Now it is kept in an insulated water – bath which contains a known amount of water. The water – bath is also provided with a thermometer and mechanical stirrer.

The initial temperature of water is noted and the reaction (i.e., combustion of the sample) is started by passing an electric current through the Pt wire. (MPBoardSolutions.com) The heat evolved during the chemical reaction raises the temperature of water which is recorded by the thermometer. When rise in temperature and the heat capacity of the calorimeter are known, the amount of heat evolved in the chemical reaction can be calculated. This will be equal to the change in internal energy (∆E) of the reaction.

Calculation: Let W = Mass of calorimeter in gm, w = Water equivalent of calorimeter, bomb stirrer, etc. t°C = Rise in temperature, x = Mass of compound ignited in gm and m = Molecular mass of the compound.
Heat produced by x gm compound = (W + w) t calories
∴ Heat of combustion of the compound at constant volume,
∆U or ∆E = \(-\frac { m }{ x } \) (W + w) t calorie/mol.
Heat is evolved so negative sign is used.
Using the equation ∆H = ∆U + ∆ng RT heat of combustion ∆H at constant pressure can be caluculated.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 5 States of Matter

States of Matter Important Questions

States of Matter Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
At the time of opening the bottle of ammonia, it can be recognized from a distance because:
(a) It is very reactive
(b) It diffuses very fast
(c) It possess a pungent smell
(d) It is lighter than air
Answer:
(b) It diffuses very fast

Question 2.
Who established the relationship between density and rate of diffusion of a gas:
(a) Boyle
(b) Charles
(c) Graham
(d) Avogadro
Answer:
(c) Graham

Question 3.
The value of in Calorie (approx):
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(b) 2

Question 4.
Value of gas constant R is:
(a) 8314 × 10⁷ ergs/degree/mol
(b) 83.14 × 10⁶ ergs/degree/mol
(c) 83.14 × 10⁵ ergs/degree/mol
(d) 8.314 × 10⁷ ergs/degree/mol
Answer:
(d) 8.314 × 10⁷ ergs/degree/mol

Question 5.
Absolute temperature is:
(a) 0° C
(b) – 100° C
(c) – 273° C
(d) – 373°C
Answer:
(c) – 273° C

Question 6.
To get general gas equation which two laws are combined:
(a) Charle’s law and Dalton’s law
(b) Graham’s law and Dalton’s law
(c) Boyle’s law and Charle’s law
(d) Avogadro’s law and Dalton’s law
Answer:
(c) Boyle’s law and Charle’s law

Question 7.
Which is correct in the following:
(a) r.m.s. velocity = 0.9212 × average velocity
(b) average velocity = 0.9212 × r.m.s. velocity
(c) r.m.s. velocity = 0.9013 × average velocity
(d) average velocity = 0.9013 × r.m.s. velocity
Answer:
(b) average velocity = 0.9212 × r.m.s. velocity

Question 8.
At constant volume the pressure of monoatomic gas depends on:
(a) Thickness of the wall of the vessel
(b) Absolute temperature
(c) Atomic number of the element
(d) Number of valence electron.
Answer:
(b) Absolute temperature

Question 9.
Behaviour of real gases near to that of ideal gases if:
(a) Temperature is low
(b) Pressure is high
(c) Low pressure and high temperature
(d) Gas is monoatomic
Answer:
(c) Low pressure and high temperature

Question 10.
At mountains of high altitude, water boils at lower temperature because of:
(a) Low atmospheric pressure
(b) High atmospheric pressure
(c) Hydrogen bonding in water is more strong at height
(d) Water vapour is lighter than liquid
Answer:
(a) Low atmospheric pressure

Question 11.
If molecular masses of two gases A and B are 16 and 64 respectively ratio of rates of diffusion A and B will be:
(a) 1:4
(b) 4:1
(c) 2:1
(d) 1:2
Answer:
(c) 2:1

Question 12.
Gases deviate from ideal behavior at high pressure because:
(a) At pressure number of colloision of molecule increases
(b) Attraction between molecules increases at high pressure
(c) Size of molecule decreases at high pressure.
(d) Molecule become steady at high pressure.
Answer:
(b) Attraction between molecules increases at high pressure

Question 13.
Distance on which magnitude of energy is minimum is called:
(a) Atomic radius
(b) Lattice radius
(c) Critical distance
(d) Molecular distance
Answer:
(a) Atomic radius

Question 14.
Diffusion rate of methane is twice than that of gas x, molecular mass of gas x will be:
(a) 64
(b) 32
(c) 40
(d) 8.0
Answer:
(a) 64

Question 15.
Part of van der Waals’ equation which illustrate the inter molecule force of real gases:
(a) (V – b)
(b) RT
(c) [P+\(\frac{a}{VL}\)]
(d) (RT)^-1
Answer:
(c) [P+\(\frac{a}{VL}\)]

Question 16.
Temperature which is same in both Celsius scale and Fahrenheit scale:
(a) 0°C
(b) 32°F
(c) – 40°C
(d) 40°C
Answer:
(c) – 40°C

Question 2.
Fill in the blanks:

1. With increase in temperature viscocity ………………………..
2. S.I. unit of surface tension is ………………………….
3. ………………………… is more heavy among dry air and moist air
4. AT N.T.P., volume of gas equal to Avogrado’s numbers ………………………….
5. Unit of vander Waals’ constant ‘b’ is ………………………..
6. …………………………. has minimum value among average velocity, root mean square velocity and most probable velocity
7. Process of diffusion of air from the puncture of an autommobile tube is known as ……………………….
8. …………………………… is obtained if a graph is plotted between volume and absolute temperature
9. Apparatus used for measuring gas pressure is …………………………
10. S.I. unit of viscosity is ………………………….
11. Absolute temperature of ideal gas is ………………………… to the average kinetic theory of the molecule.
12. A liquid which is permanently super cooled is called ………………………..
13. According to kinetic theory, the average kinetic energy of a gas is directly proportional to its …………………………. temperature.
14. The kinetic energy of one mole of a gas is equal to …………………………..
15. Root mean square velocity is …………………………..

Answer:

1. Decreases
2. Nm^-1
3. Dry air
4. 22.41
5. Litre/mol
6. Most probable velocity
7. Effiission
8. Straight line
9. Monometer
10. Nm^-2S
11. Proportional
12. Glass
13. Absolute
14. \(\frac{3}{2}\) RT
15. \(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \)

Question 3.
Answer in one word/sentence:

1. Vaporization of ethanol is faster as compared to water?
2. The resistance produced in the flow of a liquid is called?
3. Unit of surface tension is?
4. Molecular mass of two gases A and B are 36 and 64 respectively. What will be the ratio of diffusion of the two gases?
5. Write S.I. unit of pressure?
6. If rate of diffusion of oxygen is r, then tell the rate of diffusion of hydrogen?
7. Compressibility factor of ideal gases is?
8. Vander Waals’ equation is?
9. Adiabatic expression of ideal gas is?
10. Number of electron in outer most orbit of crypton is?

Answer:

1. Value of molar vaporization enthalpy of water is high
2. Viscosity
3. Dyne per cm
4. 4:3
5. pascal
6. r₁ = 4r [\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { m_{ 2 } }{ m_{ 1 } } } \)]
7. 1.0
8. [P + \(\frac { an^{ 2 } }{ V^{ 2 } } \)] [V – nb] = nRT
9. q = 0
10. 8 electron

Question 4.
Match the following:
[I]

Answer:

1. (f)
2. (a)
3. (b)
4. (a)
5. (d)
6. (c)

[II]

Answer:

1. (d)
2. (c)
3. (e)
4. (a)
5. (b)

States of Matter Very Short Answer Type Questions

Question 1.
The resistance produced in the flow of a liquid is called?
Answer:
Viscosity.

Question 2.
Unit of surface tension is?
Answer:
Dyne per cm.

Question 3.
Write S.I. unit of pressure?
Answer:
Pascal.

Question 4.
Name the scientist who developed relation between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 5.
What is the Value of gas constant in S.I. unit?
Answer:
8.314 JK^-1 mol^-1

Question 6.
1 Pascal is equal to?
Answer:
1 Nm².

Question 7.
Who established the relationship between density and rate of diffusion of gas?
Answer:
Graham’s.

Question 8.
Absolute temperature is?
Answer:
~273°C.

Question 9.
The value of R in calorie (approx)?
Answer:
2.

Question 10.
Behaviour of real gases is near to that of ideal gases of?
Answer:
Low pressure and high temperature.

Question 11.
At mountains or high altitude, water boils at lower temperature because of?
Answer:
Low atmospheric pressure.

Question 12.
To get general gas equation. Which two laws are combined?
Answer:
Boyle’s law and Charle’s law.

Question 13.
At constant volume the pressure of gas depends upon?
Answer:
On absolute temperature.

Question 14.
Part of van der Waals’ equation which illustrate the inter – molecular forces of real gases?
Answer:
P + \(\frac { a }{ V^{ 2 } } \).

Question 15.
Temperature which is same in both Celsius scale and fahrenheit scale?
Answer:
– 40°C.

Question 16.
What is the formula for kinetic theory of gases?
Answer:
PV = \(\frac{1}{3}\) mnv².

Question 17.
The average kinetic energy of gases is proportional to?
Answer:
Absolute temperature.

Question 18.
What is the kinetic energy of 1 mole of gas?
Answer:
\(\frac{3}{2}\) RT.

Question 19.
What is the formula of Root mean square velocity?
Answer:
\(\sqrt { \frac { 3PV }{ M } } \) or \(\sqrt { \frac { 3RT }{ M } } \).

Question 20.
Which type of Crystals are diamond and ice?
Answer:
Diamond – Covalent crystal
Ice – Ionic crystal.

Question 21.
What is Critical temperature?
Answer:
The temperature above which the gas cannot be liquefied.

Question 22.
Poise is the unit of which basic property?
Answer:
Viscosity (1 poise = dynes/cm²s).

Question 23.
What is the unit of a (volume correction), b (pressure correction) in van der Waals’ equation?
Answer:
a (Volume correction) = atm L² mol^-2
b (Pressure correction) = L mol^-1

States of Matter Short Answer Type Question – I

Question 1.
Explain Anisotropic and Isotropic?
Answer:
Anisotropic:
The crystalline solid exhibits different physical properties in the , three direction. In this way, crystalline solids are called anisotropic.

Isotropic:
The amorphous solid exhibits same physical properties in all directions. Due to this property they are called isotropic.

Question 2.
Define the term absolute zero? Write its value in centigrade scale?
Answer:
The lowest possible temperature at which all the gases are supposed to occupy zero volume is called absolute zero. The actual value of absolute zero is – 273.15°C. It is , related to temperature in centigrade scale by this relation.
t°C = t + 273 K.

Question 3.
What is unit cell?
Answer:
Smallest unit is a crystal which is formed by systematic arrangements of constituent particles as atom, molecule or ions, is called unit cell. The unit cell generates the whole lattice translation.

Question 4.
What is crystal lattice?
Answer:
Geometry or shape of any crystal in which unit cells are arranged systematically and three dimensionally is called crystal lattice.

Question 5.
How does volume of balloon used for weather study, change with height?
Answer:
At height, atmospheric pressure decreases. The volume of gas inside the balloon increases with decrease in pressure. A stage comes when due to decrease in atmospheric pressure in larger extent, volume increases and balloon bursts.

Question 6.
In winter season, a layer of ice is formed in the lake but the fishes and other organisms present in the lake remain alive. Why?
Answer:
The maximum density of water is at 4°C but below 4°C temperature the density decreases. When the temperature of lake decreases then the water present on the surface become denser and goes downward. This occurs upto the level when the temperature rises to t 4°C. (MPBoardSolutions.com) If the temperature of the upper layer is less than 4°C, the water remains on the upper surface and converts into ice slowly and the water below the surface remains as such due to high density and remains as liquid. That is why, the fishes and micro – organisms remains alive.

Question 7.
Why is the density of hot gas is less in comparison to cold gas?
Answer:
According to Charle’s law, volume of any gas of definite mass is directly proportional to absolute temperature. On increasing the temperature volume of gas also increases, but increase in volume results decrease in density.

Question 8.
Why one feel sluggish, breathlessness and headache at high altitude?
Answer:
At high altitude, the pressure is less and the corresponding volume of air is more, Thus, air becomes less dense at high altitude and the oxygen in air becomes insufficient for normal breathing. This causes what is known as altitude sickness.

Question 9.
What is Critical temperature?
Answer:
The temperature to which gas must be cooled before it can be liquefied by com-pression is known as critical temperature and is represented by Tc.
Example: Critical temperature of CO₂ gas is 31.1°C or 304.1K.

Question 10.
What is critical pressure and critical volume?
Answer:
Critical pressure:
The minimum pressure required to liquefy the gas at its critical temperature is known as critical pressure and denoted by P_c.
Example: Critical pressure of CO₂ is 72.8 atm.
Critical Volume: The volume occupied by 1 mole of gas at the critical temperature and critical pressure is known as critical volume and denoted by V_c.
Example: Critical volume of C0₂ gas is 94 cm³/mol.

Question 11.
Why are tyres of automobile inflated to lesser pressure in summer than in winter?
Answer:
As the automobiles move the temp, of tyre increases due to friction against the road. Consequently the air inside the tyre expand thereby the pressure exerted by air against the wall of tyre also increases. (MPBoardSolutions.com) In summer, there is increase in temperature hence increase in pressure is much more. This may leads bursting at tyre. In order to check bursting of tyre, the tyre are inflated with looser amount of air than in winter.

Question 12.
What would be the S.I. unit for the quantity PV²T²/n?
Answer:
\(\frac { PV^{ 2 }T^{ 2 } }{ n } \) = \(\frac { (Nm^{ -2 })(m^{ 3 })^{ 2 }(K)^{ 2 } }{ mol } \) = Nm⁴K²mol^-1.

Question 13.
Explain on the basis of Charle’s law that minimum possible temperature is – 273°C
Answer:
According to Charle’s law:
V_t = V₀ [1 + \(\frac { t }{ 273 } \)]
At t = – 273°C V_t = V₀ [ 1 – \(\frac { 273}{ 273 } \)] = 0
Therefore at – 273°C, the volume of gas becomes 0 and below this temperature the volume becomes – ve which is meaningless.

Question 14.
Why are liquid drops spherical?
Answer:
Small drops are spherical in shape:
Surface tension tries to decrease the surface area of a given liquid for a given volume. Therefore, drops of liquid are spherical because for a given volume sphere has minimum volume.

Question 15.
What is Root Mean Square velocity and Average velocity?
Answer:
1. Root Mean Square velocity:
It is defined as root of, mean of, square of, velocity of large no. of molecules of same gas. It is denoted by V.
V = \(\sqrt { \frac { v_{ 1 }^{ 2 }+v_{ 2 }^{ 2 }+v_{ 3 }^{ 2 }…..v_{ n }^{ 2 } }{ n } } \)
2. Average velocity:
It is defined as average of, velocity of all the molecules present in gas. It is denoted by V_a.
V_a = \(\frac { v_{ 1 }+v_{ 2 }+v_{ 3 }…..v_{ n } }{ n } \)

Question 16.
Explain the difference between Evaporation and Boiling?
Answer:
Differences between Evaporation and Boiling:
Evaporation:

1. Evaporation decreases spontaneously and occur at all temperatures.
2. Evaporation is a process of the liquid surface.
3. Evaporation is a slow process.

Boiling:

1. Boling takes place only when the vapour pressure of this liquid becomes equal to atmospheric pressure.
2. Boling is a process of the entire liquid and occurs in the form of bubbles inside the liquid.
3. Boling is a fast process.

Question 17.
What do you mean by compressibility factor of gases?
Answer:
The ratio of observed volume and caluculated volume of a gas at a given temperature and pressure is known as compressibility factor. It is denoted by Z.
Thus,

or Z = \(\frac { PV }{ nRT } \)
For ideal gases, PV = nRrt
∴For ideal gas, Z = 1.

Question 18.
What is the effect of pressure on melting of ice?
Answer:
By increasing pressure there occur tremendous increase in kinetic energy of molecules, due to this at low temperature, the molecules move freely, hence on increasing pressure the ice below its melting point converted into liquid.

Question 19.
Mountaineers carry oxygen cylinders with them at the lance of climbing mountains. Why?
Answer:
Atmospheric pressure is relatively low at heights. Quantity of oxygen is low in mountain and climbers feel difficulties in breathing. Therefore, they carry oxygen cylinders along with them.

Question 20.
Define viscosity of liquid. Explain the effect of temperature on viscosity?
Answer:
Resistance in flow of any liquid is called viscosity. Such resistance is produced due to internal friction of different layers of liquid. (MPBoardSolutions.com) When temperature is increased, the cohesive force, which opposes liquid flow, decreases and molecular velocity increases. Due to this, viscosity decreases.

Question 21.
On same temperature when ether and water pour on different hands, then ether seems to be more colder than water. Why?
Answer:
In ether the intermolecular attractive forces between the molecules is less in comparison to water, so ether evaporates more quickly than water and the energy required for evaporation is absorbed from hand, that is why ether seems to be colder.

Question 22.
What is surface tension? Write its S.L unit?
Answer:
It is an important property of a liquid which is related with interatomic attraction force. The molecules present inside the liquid is attracted equally by molecules present in all direction. (MPBoardSolutions.com)
But molecule present on the surface of liquid is attracted by molecules at bottom and in sides, as a result the molecules at surface are pulled downward and nature of surface is to lessen the area. Due to compactness, the surface of liquid behaves as a stretched membrane. This effect is called surface tension.

“Surface tension is a measure of work which is necessary to increase the unit cross-section of liquid.”
Its S.I. unit is Joule/metre² or Newton metre.

Question 23.
The compressibility factor Z of a gas is as follows:

1. What will be the value of Z for an ideal gas?
2. What will be the effect on Z above Boyle’s temperature for real gas?

Answer:

1. For ideal gas, compressibility factor Z = 1.
2. Above Boyle’s temperature, real gases show positive deviation. So, Z > 1.

Question 24.
What is Ideal gas? Write its characteristics?
Answer:
Ideal Gas:
The gases which obey Ideal gas equation under all conditions of temperature and pressure is called ideal gas.
Characteristics:

1. At constant temperature, product of pressure and volume of ideal gas are always constant. Therefore, horizontal line should be obtained in a graph. If graph is plotted between PV and P.
2. If an ideal gas is called at constant pressure, then its volume requestly decreased and become at – 273°C.
3. There is no force of attraction between gas molecules.
4. The compressibility factor of ideal gas is equal to one.

Question 25.
What is Real gas? What are its properties?
Answer:
Gas which does not obey Boyle’s law, Charle’s law and Ideal gas equation strictly is called Real gas.
The gases which does not follow ideal gas equation behaviour under all condtions of temperature and pressure called real gas.
Properties:

1. They do not follow gas law at low temperature and high pressure.
2. At – 273°C their volume is not zero because most of the gases converted into liquid on cooling.
3. The attractive force between gas molecule is negligible.
4. The compressibility factor of real gas is not equal to zero.

States of Matter Short Answer Type Questions – II

Question 1.
State and explain Boyle’s Law?
Answer:
According to this law: “At constant temperature, the volume of a known amount of gas is inversely proportional to the pressure.”
P ∝ \(\frac{1}{V}\) (at constant temperature)
⇒P = Constant × \(\frac{1}{V}\)
⇒ PV = Constant.
Thus, “at constant temperature, product of volume of a given mass of gas and pressure remain constant.”
At initial condition, P₁V₁ = K …………….. (1)
At final condition, P₂V₂ = K ………………. (2)
From eqn. (1) and eqn. (2).
P₁V₁ = P₂V₂

Question 2.
What is the nature of gas constant R?
Answer:
We know that,
PV = nRT
R = \(\frac{PV}{nT}\)

Question 3.
Explain concept of absolute zero from Charle’s law?
Answer:
Charle’s law: According to this law “At a constant pressure the volume of certain mass of a gas increases or decreases by of its previous volume for every 1°C changes in temperature (increases or decreases).”
Suppose V₀ is the volume of certain mass of gas at 0°C then,
Volume of the gas at 1°C temperature = V₀ [1 + \(\frac{1}{273}\)]
Volume of the gas at t°C temperature = V₀ [1 + \(\frac{t}{273}\)]
Volume of the gas at – 1°C temperature = V₀ [1 – \(\frac{t}{273}\)]
Volume of gas at – 273°C = V₀ [1 – \(\frac{273}{273}\)] = 0
Thus decrease of temperature results in the decrease in volume of the gas and ultimately the volume should become zero at – 273°C. (MPBoardSolutions.com) This lowest possible temperature at which all the gases are suppossed to occupy zero volume is called absolute zero.

Question 4.
What is Gay Lussac’s law?
Answer:
Gay Lussac law:
According to this law, “At constant volume the pressure of a given mass of gas is directly proportional to its absolute temperature.”
P ∝ T (Mass and volume are constant)
⇒ P = K × T
⇒ \(\frac{P}{T}\) = K
Suppose at initial condition,
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = K
At final condition,
\(\frac { P_{ 2 } }{ T_{ 2 } } \) = K
From eqn. (1) and (2),
\(\frac { P_{ 1 } }{ T_{ 1 } } \) = \(\frac { P_{ 2 } }{ T_{ 2 } } \)

Question 5.
State and explain Avogadro’s law?
Answer:
According to this law, “Equal volume of all gases under identical conditions of temperature and pressure contain equal number of molecules”.
V ∝N (at constant temperature and pressure) …………. (1)
At constant temperature and pressure number of moles of a gas n is directly proportional to number of molecules N.
Hence, N ∝n
⇒\(\frac{V}{n}\) = Constant
Suppose at initial condition volume of gas is V₁ and no. of mole of gas is n₁ hence
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = Constant …………… (2)
Similarly at final condition no. of moles and volume of gas is n₂ and V₂, hence
\(\frac { V_{ 2 } }{ n_{ 2 } } \) = Constant …………… (3)
From eqns. (2) and (3)
\(\frac { V_{ 1 } }{ n_{ 1 } } \) = \(\frac { V_{ 2 } }{ n_{ 2 } } \)

Question 6.
Write the applications of Graham’s law of diffusion?
Answer:
1. To determine the density and molecular weight of a gas:
If the time of diffusion and density of a gas is known and the time of diffusion of other gas is known, then the density and molecular weight of other gas can be calculated.

2. Marsh gas indicator:
The persons working in the mines get aware by the leakage of the poisonous gases by this indicator.

3. In separation of gases:
The gases can be separated easily due to difference in the rate of diffusion of gases.

4. Smell:
Bad smell and poisonous gases get separated due to diffusion in air.

Question 7.
Derive Charle’s law on the basis of Kinetic gas theory?
Answer:
According to Kinetic gas theory, kinetic energy of gases is directly proportional to absolute temperature.
K.E ∝T
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT
⇒PV = \(\frac{2}{3}\) KT,
⇒V = \(\frac{2}{3}\) \(\frac{K}{P}\).T
At constant pressure \(\frac{2}{3}\) \(\frac{K}{P}\) = constant
V = constant ∝ T
V ∝ T

Question 8.
How are rates of diffusion of different gases compared?
Answer:
Let two gases are A and B, equal volume V of both gases diffuse in times t_A and t_B respectively.
Rate of diffusion of gas A,
r _A = \(\frac { V }{ t_{ A } } \)
Rate of diffusion of gas B,
r _B = \(\frac { V }{ t_{ B } } \)
∴\(\frac { r_{ A } }{ r_{ B } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \)
From equation (3) and equation (4),
\(\frac { t_{ B } }{ t_{ A } } \) = \(\sqrt { \frac { d_{ B } }{ d_{ A } } } \).

Question 9.
The ratio between the rate of diffusion of an unknown gas (x) s&d CO₂ Is 40:45 Find out the molecular mass of unknown gas (x).
Solution:
According to Graham’s law of diffusion,
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \)
Given, r₁:r₂ = 40:45 M₂(CO₂) = 44, M₁ = ?
⇒\(\frac{40}{45}\) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
⇒ \(\frac { (40)^{ 2 } }{ (45)^{ 2 } } \) = \(\sqrt { \frac { 44 }{ M_{ 1 } } } \)
M₁ = \(\frac { 44\times 45\times 45 }{ 40\times 40 } \) = 55.68.

Question 10.
If relative density of chlorine is 36, diffusion of 25 volume of hydrogen takes 40 sec. under the condition how much time will be taken for the diffusion of 30 volume of chlorine?
Solution:
Hydrogen d₁ = 1, r₁ = \(\frac{25}{40}\), Chlorine d₂ = 36, r₂ = \(\frac{30}{t}\)

States of Matter Long Answer Type Questions

Question 1.
State and explain Graham’s law of diffusion?
Answer:
Graham’s Law of Diffusion:
The rate of diffusion of gas under similar condition of temperature and pressure is inversely proportional to the square roots of their density.”
Thus, Rate of diffusion ∝\(\frac { 1 }{ \sqrt { density } } \)
⇒r ∝\(\frac { 1 }{ \sqrt { d } } \)
If the rate of diffusion of two gases are r₁ and r₂ and their density are d₁ and d₂ respectively.
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 1 } } } \)
r₁ = K \(\frac { 1 }{ \sqrt { d_{ 2 } } } \)
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \)
∵ M. Mass = 2 × Vapour density = \(\frac { Molecular\quad mass }{ 2 } \) = Vapour density
\(\frac { r_{ 1 } }{ r_{ 2 } } \) = \(\sqrt { \frac { d_{ 2 } }{ d_{ 1 } } } \) = \(\sqrt { \frac { M_{ 2 } }{ M_{ 1 } } } \).

Question 2.
Write difference between Real and Ideal gas?
Answer:
Differences between Ideal gas and Real gas:
Ideal gas:

1. Ideal gas obeys die equation, PV = RT at all temperature and pressure.
2. There is no ideal gas, they are hypothetical.
3. Total volume of gas molecules is supposed to be negligible in comparison to total volume.
4. There is no attraction force between gas molecules.

Real gas:

1. Real gas obeys PV = RT only at low pressure and high temperature.
2. All existing gases are real gases and show deviation from ideal gas behaviour, positive or negative.
3. Volume of gas molecules are not negligible in comparison to total volume.
4. Gas molecules attract each other.Therefore, total pressure is less than ideal gas.

Question 3.
What are the main differences between Crystalline and Amorphous solids?
Answer:
Differences between Crystalline and Amorphous solids:
Crystalline:

1. In it the constituent particles are arranged in regular manner.
2. Melting point of crystalline solid is fixed.
3. They are an anisotropic i.e. some of their physical properties are different in different directions.
4. They are rigid and their shape is not distorted by mild distorting tone.
5. They have a definite heat of fusion.
6. They are true solid in real meaning.

Amorphous solids:

1. In it the constituent particles are arranged in irregular manner.
2. Melting point of amorphous solid is not fixed.
3. They are Isotropic i.e. their physical properties are same in all directions.
4. They are not very rigid, they can be distorted easily.
5. They do not have definite heat of fusion.
6. They are super cooled liquid.

Question 4.
Using state equation clarify that the density of a gas at given temperature is proportional to pressure of gas?
Answer:
PV = nRT
PV = \(\frac{m}{M}\) RT,
Or P = \(\frac{mRT}{VM}\),

Or P = \(\frac{dRT}{M}\)
Or d = \(\frac{PM}{RT}\)
if T = known constant
∴d ∝P

Question 5.
Derive Ideal gas equation on the basis of kinetic gas equation?
Answer:
According to Kinetic gas theory, “ The average kinetic energy of gas molecules is directly proportional to absolute temperature.”
Average Kinetic energy = \(\frac{1}{2}\) mnv²
\(\frac{1}{2}\) mnv² ∝ T
⇒\(\frac{1}{2}\) mnv² = KT
⇒\(\frac{3}{2}\) × \(\frac{1}{3}\) mnv² = KT
⇒\(\frac{1}{3}\) mnv² = \(\frac{2}{3}\) KT,
⇒\(\frac{PV}{T}\) = R
⇒ PV = RT.

Question 6.
Explanation of Dalton’s law on the basis of Kinetic gas theory?
Answer:
If the volume of container V litre and no. of moles of gas A is n₁ ar.d mass of each particles in m₁ R.M.S. velocity is V₁ then,
P_A = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \)
Kinetic gas equation PV = \(\frac{1}{3}\) mnv²
For B gas number of moles = n₂
Mass = m₂
R.M.S. Velocity = v₂
P_B = \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
If both the gases are kept in same container total pressure of mixture
P = \(\frac{1}{3}\) \(\frac { m_{ 1 }n_{ 1 }v_{ 1 }^{ 2 } }{ V } \) + \(\frac{1}{3}\) \(\frac { m_{ 2 }n_{ 2 }v_{ 2 }^{ 2 } }{ V } \)
⇒ P = P_A + P_B
so in general, for more than two gases
P = P_A + P_B + P_C + …………………..
It is Daltons law of partial pressure.

MP Board Class 11 Chemistry Important Questions

MP Board Class 11th Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure Important Questions

Chemical Bonding And Molecular Structure Very Short Answer Type Questions

Question 1.
What type of bond is present generally in same atoms?
Answer:
Covalent bond.

Question 2.
Which hybridization is present in ammonia (NH₃)?
Answer:
sp³.

Question 3.
What is the reason for high boiling point of water?
Answer:
Presence of H – bond between molecules of water.

Question 4.
What is the dipole moment of C0₂?
Answer:
Zero.

Question 5.
What type of bonds are directional?
Answer:
Covalent bonds.

Question 6.
What is the bond angle in water?
Answer:
104°5′.

Question 7.
What is the structure of [Ni(CN)₄]^2-?
Answer:
Square planner.

Question 8.
Which bond is present in s – s overlapping?
Answer:
cr (Sigma) bond.

Question 9.
What is the structure of diamond?
Answer:
Crystal lattice (Tetrahedral).

Question 10.
Which bond is present in sidewise overlapping of p – p orbitals?
Answer:
n – bond.

Question 11.
Which type of hybridization found in PCl₅?
Answer:
sjy’d hybridization.

Question 12.
What is full form of LCAO?
Answer:
Linear Combination of Atomic Orbitals.

Question 13.
What is the structure of NH₃?
Answer:
Trigonal bipyramidal.

Question 14.
What is dipole moment (µ) of a linear covalent molecule?
Answer:
Zero.

Question 15.
What is the unit of electron gain enthalpy?
Answer:
eV per atom or kJ per mole.

Question 16.
What is the formula of bond order?
Answer:
Bond order = \(\frac{1}{2}\) (N_b – N_a).

Question 17.
What is the symbol of superoxide and peroxide?
Answer:
Superoxide – O^–₂ and peroxide – O^2-₂.

Question 18.
What do you mean by bond order?
Answer:
The number of electrons present between two atoms of molecules or ions.

Question 19.
More polarizing power and more polarisability increases which property of molecule?
Answer:
Covalent property.

Question 20.
What is determined by Born – Haber cycle?
Answer:
Lattice energy.

Question 21.
What is full form of VSEPR?
Answer:
Valence Shell Electron Pair Repulsion.

Chemical Bonding And Molecular Structure Short Answer Type Questions – I

Question 1.
What is the electronic theory of covalency? Write its main postulates?
Answer:
The electronic theory of valency and its main postulates are as follows:

1. The covalency of any element depends upon the no.of electrons present in its valence shell.
2. All the elements have the tendency to acquire the nobel gas configuration.
3. The electrons present in valence shell are called valence electrons and when the electron comes out than the vacant space is called kernel.
4. If an element is unstable, than it works for its stability for this it gives and takes the electron. On this basis the bonds are of three types:
   1. Ionic bond
   2. Covalent bond and
   3. Co – ordinate bond.

Question 2.
What do you mean by lone pair of electron?
Answer:
The electron pair which present in the valence orbital of the element and does not take part bond formation is called lone pair electron.
Example: In the H₂O atom the lone pairs of electrons on an atom:

Bond pairs = 2, Lone pair electrons = 2.

Question 3.
What do you mean by dipole moment?
Answer:
Dipole moment is defined as, the product ofthe magnitude of charge on any one of the atoms and distance between them. It is represented by Greek letter µ(mu).
Mathematically, dipole moment is expressed as µ = e × d
Where, e is charge on any one of the atoms and d is distance between the atoms.
As e is of the order of 10^-10 esu while d is of the order of 10^-8 cm µ is of the order 10¹⁸ esu cm and this unit of p is known as Debye (D). Thus,
1D = 1 × 10^-18esu cm

Question 4.
Is He₂ molecule is possible? Clearify it.
Answer:
He₂ molecule is not possible.
2He → 1s²
Two electrons are present in Is orbital of He atom. It is complete and stable orbital and so it cannot accept an extra electron. The bond order is zero. So the formation of He₂ molecule is impossible.

Question 5.
What is the total number of σ (sigma) and π (pi) bond are present in following molecules:

1. C₂H₂
2. C₂H₄

Answer:

1. C₂H₂
2. C₂H₄

Question 6.
What do you mean by hydrogen bond?
Answer:
Hydrogen bond is defined as the electrostatic force of attraction which exists between the covalently bonded hydrogen atom of one molecule and the electronegative atom of the other molecule. Hydrogen bond is of two types:

1. Intermolecular hydrogen bonding
2. Intra – molecular hydrogen bonding.

Question 7.
At normal temperature H20 is in liquid state but H2S is in gaseous state. Why?
Answer:
Hydrogen bonding affects the physical state of the molecule. For example, H₂0 and H₂S are the hydrides of group 16 elements, but H₂O is liquid whereas H₂S is gas at room temperature. Actually, H₂O fulfils the (MPBoardSolutions.com) condition of hydrogen bonding therefore, in H₂O hydrogen bonding is found and they get associated, grow in molecular size and exist in liquid state. The H₂S molecule, the magnitude of hydrogen bonding is negligible, therefore, molecules remains separated from each other and acquires gaseous state at room temperature.

Question 8.
Why, viscosity of glycerol is more than ethanol?
Answer:
In ethanol molecule, one hydroxyl gap is present whereas in every glycerol, three OH – groups are present which form hydrogen bond. Therefore, in glycerol hydrogen bond formation is more in comparison to ethanol. That is why, the viscosity of glycerol is more than ethanol.

Question 9.
Why σ – bond is stronger than π – bond?
Answer:
The strength of any bond depends upon the overlapping limit, σ – bonds are formed by axial overlapping whereas σ – bonds are formed by sidewise overlapping. So, the extent of overlapping in σ – bond is more than π – bond. So σ – bond is more stronger than π – bond.

Question 10.
Why HF molecule is more polar than HI?
Answer:
The electronegativity of F is more than I. So, the displacement of electron in covalency in HF is more than HI, resultant the deviation of charges in HF is greater than HI. So, HF is more polar than HI.

Question 11.
C – Cl bond is polar but CCl₄ is non – polar, why? Give the reason?
Answer:
In C – Cl bond the electronegativity of Cl is more than C, due to this the electrons in covalent bond shifts towards Cl, due to which partial +ve charge appears on C and partial -ve charge develops once and this bond become polar in nature. Whereas the structure of CCl₄ is symmetric, due to which the dipole moment of C – Cl bond cancelled each other. So, CCl₄ molecule is non – polar.

Question 12.
What do you mean by resonance?
Answer:
Some compounds cannot be represented by a single definite structure, rather more than one structure is required by none of them is able to explain all the known properties of the compound alone. Thus, the various structures written for a compound to explain the known properties of it completely are called resonating structure. This phenomenon is called resonance.

Question 13.
What are the conditions for resonance?
Answer:
The conditions are as follows:

1. The heat of formation of each resonating structure should be same.
2. The arrangement of atoms in each formula should be same.
3. The number of unpaired electron in each structures should be same.

Question 14.
What is resonance energy ?
Answer:
The difference between the actual energy of the resonance hybrid and the most stable one of the resonating structures is called resonance energy.

Question 15.
Determine the bond order in N₂?
Answer:
N₂ (14 electrons): The electronic configuration
σ1s², σ^*1s², σ2s², σ^*2s², (π2p_x² = π2p_y²), σ2p^z2
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (10 – 4) = 3.

Question 16.
On the basis of molecular orbital theory explain that the Be₂ atom is not formed?
Answer:
₄Be: Electronic configuration = 1s² 2s²
Be₂ (molecule) (4 + 4 = 8e^–)
Electronic configuration = σ1s², σ^*1s², σ2s², σ^*2s²
Bond order = \(\frac{1}{2}\) (N_b – N_a) = \(\frac{1}{2}\) × (4 – 4) = 0.
So, the Be₂ molecule does not formed.

Question 17.
Write the definition of hybridization?
Answer:
The process of intermixing of atomic orbitals of nearly equal energy and proper symmetry giving rise to equal number of new orbitals of same energy is called hybridization and the orbitals so formed hybridized orbital.

Question 18.
What do you understand by lattice energy and solvation energy?
Answer:
Lattice energy:
Once the gaseous ions are formed, the ions of opposite charges come close together and pack up three – dimensionally in a definite geometric pattern to form ionic crystal (Crystal lattice). Since, the packing of ions of opposite charges takes place as a result of attractive force between them, the process is accompanied (MPBoardSolutions.com) with the release of energy referred to as lattice enthalpy. Lattice enthalpy may be defined as; the amount of energy released when one mole of ionic solid is formed by the close packing of its constituents. It is denoted by ∆_LH and negative in nature.

Solvation energy:
The energy released when an ion get soluble in water is called solvation energy.

Question 19.
Whose boiling points are high electrovalent compounds or covalent compounds, why?
Answer:
High boiling points: The boiling points of ionic solids are very high. This is due to strong electrostatic force of attraction between the oppositely charged ions. To change the physical state of ionic compounds, high temperature is required.

Question 20.
Why BaS0₄ is insoluble in water?
Answer:
The solubility of any ionic compound depends upon the lattice energy and solvation energy. If the lattice energy of any compound is more than solvation energy, than ionic compound is insoluble in water. The solvation energy of BaS04 is less than lattice energy. So it is insoluble in water.

Question 21.
If Be – H bond is polar, the dipole moment of Be – H₂ is zero. Why?
Answer:
BeH₂ is linear. The bond moment present in opposite direction cancelled each other. That is why Be the dipole moment of BeH₂ is zero.

Chemical Bonding And Molecular Structure Short Answer Type Questions – II

Question 1.
Why ice is lighter than water? Explain?
Or, Density of ice is less than water. Why?
Answer:
Density of ice is less than water:
In ice each oxygen atom is tetrahedrally surrounded by four hydrogen atoms in which two hydrogen atoms are linked to oxgyen atom by covalent bond and other two hydrogen atoms are linked by hydrogen bond.

The molecules of H₂O are not packed closely. This H gives rise to open cage like structure for ice having a larger volume for the given mass of water. Thus, density of ice is less than water. Ice is actually hydrogen bonded crystal. (MPBoardSolutions.com) Three dimensional structure of protein and nucleic acids like biologically important substances is due to hydrogen bond. Energy of hydrogen bond is between 3.5 kJmol^-1 and 8 kJmol^-1. Thus, hydrogen bond is stronger than van der Waals force and weaker than covalent bond.

Question 2.
Write the rules of hybridization?
Answer:
Conditions for hybridization: Following are the conditions for hybridization:

1. The orbitals of one and same atom participate in hybridization. Only the orbitals and not electrons get hybridized.
2. The energy difference between the hybridizing orbitals should be small.
3. Promotion of electron is not essential prior to hybridization.
4. It is not necessary that only the half-filled orbitals may participate in hybridiza¬tion. In some cases, even the filled orbitals may participate in hybridization.

Question 3.
Explain covalent bond with example?
Answer:
Lewis – Langmuir theory:
Lewis and Langmuir suggested that, atoms may combine by sharing of electrons in their outermost shell to complete their respective octet. The shared electrons becomes the property of both the atoms. This types of linkage is known as covalent linkage or covalent bond. Thus,

1. The force which binds atoms of same or different elements by mutual sharing of electrons is called a covalent bond.

2. This type of bond is formed between two similar non-metalic elements (A, A) or (B, B) or dissimilar atoms (A and B).
Example:
Chlorine molecule:
Both the chlorine atoms (Z = 17) contain 7 electrons in their valence shells and short in one electron each. They share one electron pair in which an electron is contributed by both as shown below:

Question 4.
Differentiate between atom and ion?
Answer:
Differences between Atom and Ion:
Atom:

1. Atoms are electroneutral.
2. Not present in free state.
3. Takes part in chemical reaction.
4. No. of e^– and proton is same.

Ion:

1. Ions are charged.
2. Ions present in free state.
3. Not take part in chemical reaction.
4. No. of e^– is more than proton.

Question 5.
Differentiate between Sigma (σ) and Pi (π) bond?
Answer:
Differences between Sigma (σ) and Pi (π) bond:
Sigma (σ) bond:

1. This bond is formed by end to end or head on overlapping of orbitals along the inter nuclear axis.
2. This is formed by overlapping of s – s, s – p or p – p orbitals.
3. Overlapping is large, hence it is strong bond.
4. Free rotation about σ – bond is possible.
5. Electron cloud is symmetrical about inter nuclear axis.

Pi (π) bond:

1. This bond is formed by the sidewise overlapping of orbitals.
2. This is formed by overlapping of p – p orbitals only.
3. Overlapping is small, hence it is weak bond.
4. Free rotation about a π – bond is not possible.
5. Electron cloud of π – bond is unsymmetrical.

Question 6.
Write difference between s – and p – orbitals?
Answer:
Differences between s – and p – orbitals:
s – orbital:

1. They are oval – symmetrical.
2. Non – directional
3. 1 = 0 and m = 0

p – orbitals:

1. They are dumbelled shape and lines symmetry of axis.
2. Directional
3. l = 1 and m = -1, 0, +1

Question 7.
Explain inter molecular and intramolecular hydrogen bonds with example?
Answer:
Types of hydrogen bond:
1. Intermolecular hydrogen bonding:
When these atoms (hydrogen and electronegative atom) are of different molecules, it is called intermolecular hydrogen bonding as in H₂O, HF, C₂H₅OH etc.

Hydrogen bond is represented by dotted lines. Many molecules of HF associates and form (HF)_n. On the same way molecules of water and alcohols are linked with hydrogen bonds.

2. Intramolecular hydrogen bonding:
If these atoms (hydrogen and electronegative atoms) are present in same molecule, this type of hydrogen bonding is called intramolecular hydrogen bonding.
e.g., o – nitrophenol

Question 8.
Among NH₃ and NF₃ whose dipole moment is high. Why?
Answer:
Both these molecules have pyramidal shape with one lone pair of electron on nitrogen atom. As fluorine is more electronegative than hydrogen therefore N – F bond should be more polar than N – H bonds. Consequently, the resultant dipole moment of NF₃ should be much larger than that of NH₃. However, the dipole moment of NH₃ (µ = 1.47D) is larger than that of NF₃ (µ = 0.24D).

The anomalous behaviour can be explained due to the presence of lone pair on nitrogen. In case of NH₃, the orbital dipole due to lone pair of electron and the bond moments of three N – H bonds are in same direction. Therefore, it adds on the resultant dipole moment of the N – H bonds. On the other hand in case of NF₃, the orbital dipole moment is in the opposite direction to resultant dipole moment of three N – F bonds. Thus, the lone pair moment cancels the resultant N – F bond moments as shown in figure. Consequently, the dipole moment of NF₃ is low.

Question 9.
Is according to following equation, is the hybridization changes in B and N:
BF₃ + NH₃ → F₃B.NH₃.
Answer:
In BF₃ three bonded pair and zero lone pair electrons are present. Due to this B is sp² hybridized and in NH₃ three bonded pair and one lone pair of electron is present. So, N is sp₃ hybridized. After reaction the hybridization of B becomes sp₃ but the hybridization of N remains same as N gives its lone pair to B atom.

Question 10.
Explain the change in hybridization in A1 atom in following reaction:
AlCl₃ + cl^– → AlCl^–₄
Answer:
The electronic configuration of Al is:
At ground state = ₁₃Al = ls² 2s² 2p⁶ 3s² 3p_x¹
At excited state = 1s², 2s², 2p⁶ 3s² 3p_x¹ _y¹
In the formation of AlCl₃, Al is sp² hybridized and its geometry is trigonal bipyramidal. Whereas in the formation of AlCl₄^–, due to inclusion of 3p_z orbital. Al is sp³ hybridized and its geometry is tetrahedral.

Question 11.
What are the postulates of orbital overlap concept of covalent bond?
Answer:
According to this concept:
1. The covalent bond is formed due to partial overlap of the two half-filled atomic orbitals of the valence shells of the combining atoms. Partial overlap means that a part of the electron cloud of each of the two half – filled domic orbitals becomes common. As a result the probability of finding electrons in the region of overlap is much more at the other places. This reduces the intemuclear repulsion and hence decreases the energy.

2. The orbitals undergoing overlap must have electrons with opposite spins.

3. Greater the extent of overlapping, stronger is the bond formed.

4. Larger the size of the orbitals, less effective is the overlapping and thus weaker is the bond formed.

Question 12.
Explain that the geometry of PCl₅ is trigonal bipyramidal and that of IF₅ is pyramidal?
Answer:
PCl₅: P is central metal atom.

As P in PCl₅ in sp³d hybridized so its geometry is square pyramidal.
IF₅: Central Metal atom is I (Z = 53)

In IF₇ , I is sp³d²