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Students get through the MP Board Class 11th Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure which are most likely to be asked in the exam.

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

Chemical Bonding and Molecular Structure Class 11 Important Questions Very Short Answer Type

Question 1.
What is a chemical bond?
Answer:
The attraction force between two atoms which binds them to form a molecule is called chemical bond.

Question 2.
What is electronic theory of valency? Write its main postulates. ‘
Answer:
Electronic theory of valency and its main postulates are as follows :

• Valency of an element depends on the number of electrons present in the valence shell of its atom.
• Atoms of all elements possess the ability to achieve stable electronic configuration of inert gas.
• Electrons present in the valence shell of any element are called valence electrons :
  When electron is lost from the valence shell, then the remaining vacant orbital is known as Kernel.
• If the atom is unstable, it tries to achieve stability. For this it exchange or shares electrons.

On this basis there are three types of bonds :

1. Ionic bond,
2. Covalent bond,
3. Co-ordinate bond.

Question 3.
Write the favourable factors for the formation of ionic bond.
Answer:
The favourable factors for the formation of ionic bond are :

• Ionisation enthalpy of element forming cation should below.
• High negative electron gain enthalpy (electron affinity) of element forming anion.
• High lattice enthalpy of ionic compound formed.

Question 4.
Write Lewis structures of the following molecules and ions :
H₂S, SiCl₄, BeF₂, CO₃^-2 HCOOH
Answer:

Question 5.
Write Lewis dot symbol of atoms of the following elements :
Mg, Br, Na, O.
Answer:

Question 6.
Use Lewis symbols to show electron transfer between the following atoms to form cations and anions
(a) K and S
(b) Ca and O
(c) Al and N.
Answer:
(a)
(b)
(c)

Question 7.
What is Lewis symbol? Write Lewis symbol of each of the following elements: Na, Ca, B, Br,Xe, C, N, O
Answer:
Gilburt Newton Lewis utilized the Lewis symbol to represent the valence electrons present in an atom. In this method, electrons present in the valence shell of an atom are represented by dots around the symbol of the element equal to the number of electrons present in the valence shell.

Question 8.
What do you mean by lone pair of electron?
Answer:
Electron pair of an atom which do not take part in bonding is known as lone pair of electron. For example, N in NH₃ has one lone pair of electron.

Question 9.
Which molecule show trigonal bipyramidal geometry?
Answer:
Dipole moment is defined as, the product of the magnitude of charge on any one of the atoms and distance between them. It is represented by Greek letter υ(mu).
Mathematically, dipole moment is expressed as
υ = 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 υ is of the order 10^-18 esu cm and this unit of υ is known as Debye (D).
Thus, 1D = 1×10^-18 esu cm.

Question 10.
Is He₂ molecule possible? Justify.
Answer:
He₂ molecule is not possible.
₂He → 1s²
There are 2 electrons in Is orbital of He atom. It is a complete and stable orbital and cannot accept extra electron. Its bond order is zero, therefore he cannot form molecule.

Question 11.
What is Bond energy?
Answer:
Bond energy of a diatomic molecule, is the energy released when t gm molecule of two neutral atoms are bonded in gaseous state.
A + B → A – B + Energy
In other words, the energy required to break the covalent bond formed between 1 gm molecule of two atoms in gaseous state is called bond energy.
Its unit is Kilo calori per mole or Kilo joule per mole.

Question 12.
Discuss the geometery of BrF₅^–.
Answer:
In BrF₅^–, there are seven electrons in the valence shell of Br (₃₅Br – 2, 8, 18, 7) atom, out of which five electrons form bonds with five fluorine (F) atoms and remaining two electrons, lie as lone pairs. Thus, total 6 pairs (5 bond pair and 1 lone pair) are present. For the minimum repulsion between lone pairs and bond pair its geometry is square-pyramidal

Question 13.
What is the significance of plus and minus sign shown in representing the orbitals?
Answer:
Atomic orbitals are represented in the form of wave functions. In the orbital (+) sign represents positive wave function and (-) sign represents negative wave function. By the combination of two wave functions of similar sign bonding molecular orbital is formed whereas by the combination of two wave functions of opposite signs anti-bonding molecular orbital is formed.

Question 14.
What is the total number of sigma (σ) and pi (π) bonds in the following molecules :
(a) C₂H₂,
(b) C₂H₄
Answer:
(a) 
(b)

Question 15.
What is hydrogen bond? Write its types and effects.
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. This bond is represented by dotted line (……………).
It is of two types :

1. Inter-molecular hydrogen bond.
2. Intra-molecular hydrogen bond.

Question 16.
At normal temperature water is a liquid, whereas hydrogen sulphide is a gas. Why?
Answer:
H₂O and H₂S are the hydride of oxygen and sulphur lying in the same group but H2O is liquid and H2S is gas, because oxygen is highly electrbnegative so H₂O is polar in nature due to this biter-molecular hydrogen Bonding occurs in it. Thus, because of association H₂O exists in liquid form and its boiling point is high.

In H₂S, due to larger size, electronegativity of S is less, therefore formation of hydrogen bond between H2S molecule is not possible. Their molecules possess only weak van der Waals’ force between them. Thus, H₂S is a gas at normal temperature.

Question 17.
Viscosity of glycerol is higher than ethanol. Why?
Answer:
Ethanol molecule contain one hydroxyl group whereas each glycerol contain three OH group which form three hydrogen bond. That is glycerol has more tendency to form hydrogen bond than ethanol. Therefore, viscosity of glycerol is higher.

Question 18.
Justify the cause of high boiling point and high viscosity of Sulphuric acid.
Answer:
H₂SO₄ molecule, due to high electronegativity, possess the tendency to form H- bond. Due to H- bond, H₂SO₄ molecules get associated. Due to this intermolecular hydrogen bond, it does not vaporise easily, therefore, their boiling point and viscosity is high.

Question 19.
What is the cause of solubility of ethyl alcohol in water?
Answer:
Ethyl alcohol easily forms intermolecular hydrogen bond with water molecules. Due to the nature of formation of hydrogen bond, ethyl alcohol is easily soluble in water.

Question 20.
What are the main conditions for the formation of hydrogen bond by a molecule?
Answer:
Following are the conditions for the formation of hydrogen bond by a molecule:
(i) The atom joined by covalent bond to hydrogen atom in a molecule should have very high electronegativity.
(ii) Atomic radius of the atom joined to hydrogen bond should below.

Question 21.
What is meant by Lattice energy and Hydration energy?
Answer:
Lattice energy: Lattice energy of an ionic solid, is the energy released during the formation of one mole solid crystal by the component gaseous ion situated at infinity from each other or in other words, energy released during the formation of a crystal is known as Lattice energy.

Hydration energy: The energy released on dissolving an ion in water or the energy released by the combination of ion with water is called hydration energy.

Question 22.
What conditions are necessary for the overlapping of atomic orbitals?
Answer:
The following conditions are necessary for the overlapping of atomic orbitals :

• Energy of both the atomic orbitals should be nearly same.
• Nuclear axis of both the atoms should by symmetrical.
• For combination, there should be a definite limit of overlapping i.e, limit of overlapping should be maximum.

Question 23.
What are sigma and pi bond? When are they formed?
Answer:
Sigma bond: The bond formed by overlapping of two orbitals along their axis is called a sigma bond. The line joining the two nuclei of the combining atoms is called the intemuclear axis or bond axis.
Example: This type of overlapping takes place between s-s orbital, s-p orbital and p_x – p_x orbitals.

Question 24.
Why is σ-bond stronger than π-bond?
Answer:
Strength of bond depends upon the extent of overlapping. In σ-bond formation, the extent of overlapping is more because overlapping takes place along the inter-nuclear axis while in case of π -bond formation extent of overlapping is less due to sidewise overlapping, hence σ-bond is stronger than π-bond.

Question 25.
Write the differences between atom and ion.
Answer:
Differences between Atom and Ion

Question 26.
Differentiate Sigma (σ) and Pi (π) bond.
Answer:
Differences between Sigma (σ) and Pi (π) bond

Question 27.
HCl is a covalent compound, still its aqueous solution gets ionized.
Answer:
In HCl due to high electronegativity of chlorine than hydrogen the shared electron pair gets displaced towards Cl and a partial negative charge is developed, on Cl and partial positive charge is developed on H. As a result, when HCl is dissolved in water, then due to partial polarizability HCl gets ionized.

Question 28.
Among electro-valent anti covalent compounds, whose boiling point is higher and why?
Answer:
Crystals of electro-valent compounds are formed by the combination of ions and strong electrostatic force is present between these ions and high energy is required to separate these ions and vaporize them. Therefore, boiling point of electro-valent compounds is higher than covalent compounds because covalent compounds are formed by the combination of atoms and only weak van der Waal’s force is present between them.

Question 29.
BaSO₄ is insoluble in water. Why?
Answer:
Solubility of an ionic compound in water depend on Lattice energy and hydration energy. If lattice energy of a compound is more than its hydration energy, then the ionic compound is insoluble in water. Magnitude of Lattice energy of BaSO₄ is higher than its hydration energy, therefore, BaSO₄ is insoluble in water.

Question 30.
Write the resonance structures of SO₃, NO₂, and NO₃^–.
Answer:
(SO₃)

(NO₂)

(NO₃^–)

Question 31.
Sodium is a strong electropositive element. Why?
Answer:
Atomic number of Sodium is 11 and its electronic configuration is 2,8,1. Due to large size, its ionization energy is very less, so it easily loses electron and forms a positive ions. Therefore, is a strongly electropositive element.

Question 32.
Electrovalent bond is formed between which type of compounds?
Answer:
Ionic bond is formed between strongly electropositive and strongly electrovalent elements because strongly electropositive element easily lose electron and forms cation and electronegative element easily gains electron and forms anion.

Strong electrostatic force of attraction is present between cation and anion. This strong electrostatic force is called electro valent or ionic bond.
Example : Na₂,₈,₁+ Cl₂,₈₇ → Na⁺₂₈+Cl^–₂,₈,₈.

Question 33.
HF molecule is more polar than HI molecule. Why?
Answer:
Electronegativity of Fluorine is more than Iodine, therefore, displacement of shared electrons in HF is more than in HI, as a result division of charge is more in HF, then HI therefore, HF is more polar than HI.

Question 34.
C-Cl bond is polar, but CCl₄ is non-polar. Given reason.
Answer:
In C-Cl bond, electronegativity of Chlorine is more than in C, due to which shared electrons gets displaced more towards Chlorine, by which partial positive charge is developed on C and partial negative charge is developed on Cl and the bond represents polar nature.
Whereas, CCl₄ has symmetrical structure due to which dipole moment of C-Cl bonds cancel each other therefore, CCl₄ molecule is non-polar.

Question 35.
What do you mean by resonance?
Answer:
When properties of a molecule are not explained by one structure and two or more than two structures are assigned to express its characteristics, it is said that molecule is resonance hybrid of these structures and this property is known as resonance.
Different resonating structures are exhibited by using sign (↔) in between these structures.

Question 36.
What are the main conditions of Resonance?
Answer:
Conditions for resonance are as follows :

• Enthalpy of formation of all resonating forms is nearly same.
• In every formula, arrangement of atoms should be same.
  
• Number of unpaired electrons in all the resonating forms should be equal.
  

Question 37.
What do you understand by Resonance energy?
Answer:
There is a difference in actual energy of a molecule and the energy calculated by its formula. This difference between both these energies is called resonance energy. In other words, difference between resonance hybrid and most stable resonance structure is called resonance energy. Higher the value of resonance energy, more is the stability of the molecule.

Question 38.
Calculate the bond order of N₂.
Answer:
Electronic configuration of N₂( 14 electron):

Bond order = \(\frac{1}{2}\) [N_b -N_a] = \(\frac{1}{2}\)[10-4] = 3

Question 39.
On the basis of molecular orbital theory explain that Be₂ molecule has no existence.
Answer:
Electronic configuration of ₄Be = 1s²,2s²
Electronic configuration of ₄Be molecule (4 + 4 = 8e^–)
=
Bond order = \(\frac{1}{2}\)[N_b-N_a] = \(\frac{1}{2}\)[4-4] = 0
Thus, Be₂ molecule has no existence.

Question 40.
What is hybridization?
Answer:
Hybridization: The hybridization may precisely be defined as the concept of mixing or merging of orbitals of an atom having equal energy to produce entirely new orbitals (equal number to the mixing orbitals) which have same energy contents, identical shapes and symmetrically disposed in space.

Question 41.
Write the differences between s-orbital and p-orbital.
Answer:
Differences between s-orbital and p-orbital

Question 42.
Boiling point of water is higher than HF whereas electronegativity of Fluorine is more than oxygen. Explain.
Answer:
In a molecule of HF one Fluorine is linked to one H, due to which each HF molecule can form two hydrogen bonds whereas in H₂O molecule, oxygen atom is linked to two H^– atom due to which each water molecule can form four hydrogen bond. This way, tendency to form H-bond in water is more than in HF. Thus, boiling point of water is higher than HF.

Question 43.
Write the properties of Co-ordinate compounds.
Answer:
Properties of Co-ordinate compounds :

• Co-ordinate compounds generally soluble in water and soluble organic solvents.
• Their melting and boiling points are higher than covalent compounds and lower than electrovalent compounds.
• They do not ionize on dissolving in water or in other solvents.
• These bonds like covalent bonds are directional thus these compounds also show stereoisomerism.

Question 44.
Write the type of geometrical shapes and value of bond angle of the following molecules:
(i) C₂H₂
(ii) BCl₃
(iii) NH₃
(iv) PCl₅.
Answer:
Types of geometrical shapes and value of Bond angle :

Question 45.
Draw the resonating structure of the following:
(i)
(ii) CO₂.
Answer:
(i) Resonating structure of :

(ii) Resonating structures of CO₂:

Chemical Bonding and Molecular Structure Class 11 Important Questions Short Answer Type 

Question 1.
On what factors does formation of ionic compound depend?
Answer:
Formation of ionic bond depends on the following factors :

• Electronegativity: There should be very high difference in electronegativity be¬tween the two combining atoms. Higher the difference, higher will be the ionic property of the bond.
• Ionisation energy : Ionisation energy of the atom donating electron should be very low, due to which the atom can easily lose electron and form cation.
• Electron affinity: Electron affinity of the atom gaining electron should be high, due to which the atom can easily gain electron.
• Lattice energy: Higher the lattice energy of the compound higher will be its ionic nature.

Question 2.
Write the characteristics of hybridization.
Answer:
Characteristics of hybridization :

• The orbitals taking part in hybridization belongs to same atom.
• The orbitals which participate in hybridization must have a small difference of energies.
• Both half filled and completely filled orbitals can take part in hybridization.
• Only orbitals participate in hybridization not electrons.

• Number of hybrid orbitals formed is equal to the number of atomic orbitals hybridized.
• Hybrid orbitals have equivalent energies and identical shape. They may have different orientation.
• Hybridization is a concept, which is used to justify the experimental facts.

Question 3.
How a covalent bond is formed? Explain by giving examples.
Answer:
Covalent bond: The bond formed by mutual sharing of electrons between the atoms is covalent bond. The compound obtained as a result of such bonding is called cova¬lent compound. When one-one electron is shared between two atoms, single bond is formed which is shown by line (-).

When two-two electrons are shared between two atoms results in the formation of double bond and is expressed with (=) line. Similarly, sharing of three electrons between two atoms forms triple bond expressed with (≡) line.

Example: Formation of Cl₂: Electronic configuration of chlorine is 2, 8, 7. Its valence shell contains 7 electrons. It needs one electron to complete its octet. One chlorine atom shares its electron with other chlorine atoms to complete their octet.
In this way, in one chlorine molecule (Cl₂) one single bond is formed between chlo¬rine atoms.

Question 4.
What is the difference between polar and non-polar covalent bonds?
Answer:
Polar covalent bond: The molecules formed of unlike atoms which differ in their electronegativity value contain polar bonds e.g., HCl, NH₃, HF, H₂O etc. as their bonds have ionic nature also.
Example: H^δ+ — Cl^δ- is a polar molecule.

Non-polar covalent bond: The covalent bond that have no ionic character are called non-polar bonds. The molecules which are formed of like atoms contain non-polar bonds. e.g., H₂, N₂, O₂ etc. are a non-polar molecule.
H:H

Question 5.
Write characteristics of covalent compounds.
Answer:
Following characteristics of covalent compounds :

• Generally these compounds are in gaseous or liquid state. Solids are of high molecular weight.
• Crystal lattice consist of neutral molecules or atoms.
• Boiling points and melting points are low.
• These are bad conductor of electricity.
• Reactions of these compounds are slow being molecular reactions.
• These compounds are soluble in non-polar solvents like benzene, pyridine, ether etc.
• Covalent bonds are rigid and directional. So many types of isomerism is found.

Question 6.
Differentiate Electrovalent, Ionic compound and Covalent compound.
Answer:
Differences between Ionic and Covalent compounds

Question 7.
Ice is lighter than water. Why? Explain. Or, Density of ice Is less than water. Why?
Answer:
In ice each oxygen atom is tetrahedrally surrounded by four hydrogen atoms in which two hydrogen atoms are linked to oxygen atom by covalent bond and other two hydrogen atoms are linked by hydrogen bond.
The molecules of H₂O are not packed closely. This gives rise to open cage-like structure for ice having a
larger volume for thé given mass of water.

Thus density of ice is less than water. Ice is actually hydrogen-bonded crystal. Three-dimensional structure of protein and H ucleic acids like biologically important substances is due to hydrogen bond. Energy of hydrogen bond is between 35 kJ mol^-1 and 8 kJ mol^-1. Thus, hydrogen bond is stronger than Van der Wall’s force and weaker than covalent bond.

Question 8.
Explain co-ordinate bond with example.
Answer:
Coordinate bond is also formed by the sharing of electrons between two atoms but the shared electron pair is donated by one atom only whose outer orbit is already stable the other atom gains that electron pair and both the atoms achieve stable configuration.

Atom which donates electron is called donor atom and the atom which accept electron is called acceptor. Coordinate bond is represented by an arrow (→) which starts from the atom which donates electron pair and points to the atom which accepts them. Example: In the combination of ammonia and Boron trifluoride N atom donates its electron pair to B
atom forming a co-ordinate bond.

Question 9.
Is there any change in the hybridization of B and N atoms as a result of the following reaction:
BF₃ + NH₃ → F₃B.NH₃
Answer:
In BF₃ there are 3 bond pairs and 0 lone pair due to which Boron is sp² hybridized and in NH₃ due to the presence of 3 bond pairs and 1 lone pair Nitrogen is sp³ hybridized. After the reaction hybridization of Boron changes to sp³ but hybridization of nitrogen remains same because N shares its lone pair of electron with electron-deficient B.

Question 10.
Describe the change in hybridization (if any) of the A1 atom in the following reaction:
AICI3 + Cl^– → AICI₃^–
Solution:
Electronic configuration of Al is as follows :
In ground state configuration ₁₃Al – 1s², 2s²2p⁶,3s²3p_x¹ .
In excited state – 1s², 2s²2p⁶,3s²3p_x¹3p_y¹
In the formation of AlCl₃, Al is sp²-hybridized and its geometeiy is trigonal planar. On the other side in the formation of AlCL₄^– due to participation of vacant 3pz orbital, A1 becomes sp³-hydrized and its geometry is tetrahedral.

Question 11.
What is metallic bond? Explain with example.
Answer:
To explain the metallic bond, there are two theories of the bond present in metals:
1. Independent (Free) electron theory: This theory was proposed in 1990 by Drud and modified later in 1916 by Lorenz. According to this theory, metallic solid is that group of cations which are orderly immersed in a sea of mobile electrons and the force which bonds these cations in the electron sea is called metallic bond.
Example: Each atom of Mg metal donates 2 electron to form lattice of Mg2+ ion and electron sea is formed by valence electrons.

2. Valence bond theory: According to this theory “Bond in metals is of covalent nature and resonance is found in it.”
Example: Each Li atom is surrounded by eight other Li atoms and it forms covalent bond with each Li turn by turn. This way, various resonance structures are obtained.

Question 12.
What do you understand by intermolecular and intramolecular hydrogen bonding?
Answer:
(i) 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.

(ii) 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.

Question 13.
Which out of NH₃ and NF₃ has higher dipole moment and why?
Answer:
Shape of both NH₃ and NF₃ is pyramidal, but dipole moment of NH₃ is higher than NF₃. In NH₃, dipole of lone pair electron is in the direction of three N-H bond, where as dipole of lone pair in NH₃ is in opposite direction of three N-F bond due to which magnitude of dipole moment of NF₃ decreases.

Question 14.
Explain why BeH₂ molecule has zero dipole moment although the Be-H bonds are polar?
Answer:
BeH₂ molecule is linear. Two similar bond dipoles in opposite directions cancel the effect of each other. That is why, dipole moment of BeH₂ molecule is zero.

Question 15.
What are the main postulates of atomic orbital overlapping of covalent bond?
Answer:
According to this theory:

• A covalent bond is formed by the overlapping of atomic orbital of one atom with the orbital of the other atom.
• At the formation of bond, electrons are related to the nuclei of both the atoms.
• Spin of the bonded electron should be in opposite direction.
• Strength of bond depends on the extent of overlapping.
• On the basis of overlapping bonds are of two types : a bond and n bond.
• Valency of an element depends on the number of unpaired electrons present in its n valence shell.
• Overlapping takes place only in those orbitals which takes part in bond formation.
• Multiple bonds are formed by the overlapping of orbitals containing more than one unpaired electrons.

Question 16.
Explain the cause and arrangement of the following compounds in the increasing order of their ionic behaviour : N-H, F-H, C-H and O-H
Answer:
Higher the electronegativity difference between two bonded atoms in a molecule higher is the ionic behaviour.

Thus, correct order of ionic behaviour is as follows : C-H < N-H < O-H < F-H.

Question 17.
Justify: Why is PCl₅ trigonal bipyramidal and IF₅ is square pyramidal?
Answer:
PCl₅: In it, Phosphorus (Z = 15) is the central atom whose electronic configuration in ground state and excited state is as follows :

In PCl₅, Phosphorus is sp³d hybridized, thus its geometry is trigonal bipyramidal.
IF₅: In it, central atom is iodine (Z = 53), whose ground state and excited state elec¬tronic configuration is as follows :

In IF₅, Iodine is sp³d² hybridized, thus its geometery is square pyramidal, distortion is due to lone pair electron.

Question 18.
What are sigma (σ) and pi (π) bonds? When are they formed? Explain.
Answer:
Sigma (σ) bond: Sigma bond is formed by axial or end to end overlapping of orbitals. In sigma bonds, electron density is maximum at intemuclear axis due to maximum overlapping so sigma bond is stronger than π-bond. Formation of sigma bonds occur on s- s, s-p and p-p overlapping e.g.,
(i) s-s overlap :

(ii) s-p overlap:

(iii) p-p overlap :

Pi (π) Bond: Bond formed by the lateral or side wise overlapping of two p-orbitals is called π-bond. It is a weak bond.

Question 19.
Tell the hybridization and geometry of the following compounds on the basis of VSEPR theory: SnCl₂, SiCl₄, SF₆, H₂S, HF, NCl₃, PCl₅.
Answer:
Hybridization and geometry of the given compounds :

Question 20.
Which hybrid orbitals are used by carbon atoms in the following molecules: (a) CH₃-CH₃ (b) CH₃-CH = CH₂
(c) CH₃CH₂-OH
(d) CH₃-CHO
(e) CH₃COOH.
Answer:
(a)
(b)
(c)
(d)
(e)

Question 21.
What do you understand by bond pairs and lone pairs of electrons ? Illustrate by giving one example of each type.
Answer:
Covalent bond is formed by the mutual sharing of electrons. Electron pairs shared between bonded atoms are called bond pair. Electrons which do not participate in bond formation are called lone pair.
Example: In ammonia (NH₃), three bond pairs and one lone pair electron are present.
In water (H₂O), two bond pairs and two lone pairs electrons are present.
In Methane only four bond pair electrons are present.

Question 22.
What is s-s overlapping? Explain with example.
Answer:
When half filled s-orbital of an atom overlaps with half-filled s-orbital of the other atom σ bond is formed, then it is known as s-s overlapping. This bond is symmetrical along the nuclear axis.

Example: Formation of H₂ molecule: Atomic number of H-atom is 1. Its normal electronic configuration is 1.s¹. During the formation of H₂ molecule half-filled orbital of H- atom, overlaps with half-filled orbital of other H-atom to form a bond.

Question 23.
What is s-p overlapping? Describe with example.
Answer:
s-p overlapping: When half filled s-orbital of any atom overlap with p-orbital (half-filled) of other atoms, this type of overlapping is called s-p overlapping. Molecular orbital thus formed is called s-p molecular orbital. Bond formed by this overlapping is a (sigma) bond. 5-orbital is non-directional or spherical so it can be overlap with P_x, p_y or p_z

Example: Formation of HCI: Electronic configuration of hydrogen and chlorine atoms are:
₁H = 1s¹
₁₇Cl= 1s²,2s²2p⁶,3s² 3p²_x3p_y² 3p_z¹
Thus, 1s-orbital of hydrogen atom overlap with 3p² orbital of chlorine atom and form a s-p molecular orbital and HCl molecule is formed.

Question 24.
Write Lewis sü .cture of the following compound and express formal charge on each atom.
(a) HNO₃,
(b) NO₂,
(c) H₂SO₄.
Answer:
(a) HNO₃:

Formal charge on H =1-0 – \(\frac{1}{2}\) – × 2 =0
Formal charge on N =5-0 – \(\frac{1}{2}\) × 8 = 1
Formal charge on O(1) =6-4- \(\frac{1}{2}\) × 4 = 0
Formal charge onO(2) =6-4- \(\frac{1}{2}\)×4=0
Formal charge on 0 (3) = 6-6- \(\frac{1}{2}\) ×2 = -1.

(b)NO₂:

Formal charge on 0 (1) =6-4- \(\frac{1}{2}\) × 4 = O
FormalchargeonN =5-1- \(\frac{1}{2}\)×6=+1
Formal charge on 0(2) =6-6-\(\frac{1}{2}\) × 2= -1

(c) H₂SO₄:

Formal charge on H (1) and H (2) =1-0 – \(\frac{1}{2}\) × 2 =0
Formal charge on 0(1) and 0 (3) =6-4- \(\frac{1}{2}\) × 4 =0
Formal chargeonO(2) and 0(4) =6-6- \(\frac{1}{2}\)× 2=-1
Formal charge on S =6 – 0- \(\frac{1}{2}\) × 8 = +2.

Question 25.
Justify the cause of difference in the properties of the two aiiotropes diamond and graphite.
Answer:
Diamond and graphite, both are the allotropes of carbon but due the difference in arrangement of C, their properties are different. In diamond, C-atom is in sp³ hybrid state and each C-atom is linked with other four carbons and forms tetrahedral structure. This way, it forms a three-dimensional structure, therefore diamond is hard and its melting point is high.

In graphite, each C-atom is in sp² hybrid state i.e. each C-atom is surrounded by three Other carbon and fourth valency of each C-atom is unsaturated. There are various layers in graphite which are joined by weak van der Waals’ attractive force. Therefore, graphite is soft and due to the presence of free electrons, it is a good conductor of electricity.

Question 26.
Explain the hybridized structure of acetylene molecule with figure.
Answer:
Explanation of acetylene by sp hybridization: In this molecule, carbon is the central atom. Thus, both the carbon atom assumes sp hybridization as shown below :

One 2s and one 2p -orbital of each carbon atom hybridizes while other two 2p-orbitals are left unhybridized. The two sp hybrid orbitals formed by each carbon atom are oriented linearly while unhybridized 2p_z orbitals are oriented perpendicularly to hybrid orbitals. During the formation of acetylene molecule one sp hybrid orbital of each carbon atom overlaps forming carbon-carbon sp-sp sigma bonds.

Remaining two sp hybrid orbitals of both the C-atoms overlaps with half-filled 1s orbital of H-atom to form two carbon-hydrogen sp-s sigma bonds. Each unhybridized orbital overlaps sidewise forming two Pi (π) bonds. Thus, sp hybrid linear acetylene molecule containing two π-bonds and three sigma bonds is formed. In this molecule, C≡C bond length is 120 pm while C-H bond length is 100 pm. H-C-C bond angle is 180°.

Question 27.
Considering X-axis as the internuclear axis which out of the following will not form a sigma bond and why :
(a) 1s and 1s
(b) Is and 2p_x
(c) 2p_y and 2p_y
(d) 1s and 2s.
Answer:
Only (c) will not form carbon, because on taking X-axis on internuclear axis, 2p_y and 2p_y undergo lateral overlapping in between the axis, as a result of which π bond is formed.

Question 28.
Apart from tetrahedral geometry, another possible geometry for CH₄ is square planar with the four H-atoms at the corners of the square and C at its centre. Explain why CH₄ is not square planar?
Answer:
Electronic configuration of C is as follows:
In ground state ₆C = 1s²,2s²,2p_x¹, 2p_y¹,2p_z⁰
In excited state ₆C =
In CH₄ molecule, C is sp³-hybridized due to which its geometry is tetrahedral. For square planar geometry dsp² hybridization is required. But due to absence of rf-orbitals in carbon, it is not possible. Along with this, according to VSEPR theory, four bond pairs of carbon atom are situated in tetrahedral geometry.
In tetrahedral structure, bond angle is 109°28′ and in square planar geometry is 90°. Thus, repulsion of bonded electrons in tetrahedral geometry is less as compared to square planar geometry.

Question 29.
Justify on the basis of hybridization, that structure of BeCl₂ molecule is linear.
Answer:
In BeCl₂, Be is in sp-hybrid state.
sp-hybridization: When one s-orbital and one orbital of p-subshell redistribute their energy to form two hybrid orbitals, it is known-as sp-hybridization. Shape of BeCl₂ molecule : In BeCl₂, B is the central atom, its normal oxidation state is 1s²,2s²,2p⁰.

In excited state one electron of 2s orbital gets excited and goes to vacant 2p subshell. One orbital of 2s and one orbital of 2p redistribute their energy to form two hybrid orbitals which arrange themselves making an angle of 180°. Both these hybrid orbitals overlap with two chlorine atoms to form two a bonds. Therefore structure of BeCl₂ is linear.

Question 30.
What is sp3 hybridization? Justify with example.
Answer:
sp³-hybridization: When s-orbital and three orbitals of. p-subshell mutually redistribute their energy and form four hybrid orbitals, then it is known as sp³-hybridization.
Due to repulsion, all these four orbitals form an angle of 109°28′ with one another i. e., these four orbitals directed towards the four comers of a tetrahedron.

Example: CH₄: Atomic number of carbon is 6 and its normal electronic configuration is 1s², 2s², 2p².
In excited state, one electron of 2s gets excited and moves to 2p_z orbital. Thus, in excited state C has four unpaired electrons. 2s orbital of carbon and all the three orbitals of 2p subshell redistribute their energy and form four hybrid orbitals which overlaps with four hydrogen atoms to form a bond. Due to sp³-hybridization, structure of methane molecule is tetrahedral and bond angle is 109°28′. ‘

Question 31.
Explain the structure of HNO₃ and H₂SO₄.
Answer:
Structure of HNO₃ :

Structure of H₂SO₄

Question 32.
Write the conditions for the formation of molecular orbitals by Linear Combination of Atomic Orbitals.
Answer:
The following conditions are required for the formation of molecular orbitals by Linear Combination of Atomic Orbitals: ‘
(i) The combining atomic orbitals should possess same or nearly similar energy. For example:1s-orbital can combine with 1s-orbital, not with 2s because energy of 2s-orbital is more than 1s-orbital. This way, combination is possible when the combining atoms are different (Intemuclear diatomic molecule).

(ii) The combining atomic orbitals must have the same symmetry about the molecular
axis.
(iii) Combining atomic orbitals should undergo maximum overlapping. Higher the overlapping between the orbitals, higher the electron density between their nuclei.

Question 33.
Explain the formation of N₂ molecule on the basis of orbital theory or overlapping.
Answer:
Atomic number of N-atom is 7, it normal electronic configuration is 1s²,2s²,2p³. It has three electrons in the p-subshell of its valence shell and to complete its octet, it requires three more electrons. p_x orbital of N-atom, overlaps with p_xorbital of another N-atom and form a bond and p² orbital of both nitrogen form two n bonds by lateral overlapping. This way three bonds are formed between both N-atoms in N₂ molecule.

Question 34.
Write the main points of Molecular Orbital Theory.
Answer:
Main points of Molecular Orbital Theory :

• Molecular orbitals are formed by the combination of atomic orbitals of atoms.
• It is supposed that molecular orbitals are formed by the linear combination of atomic orbitals.

Molecular orbitals are of two types :

1. Bonding molecular orbitals: These are formed by addition of wave functions.
2. Antibonding molecular orbitals: These are formed by the difference of wave functions.

• Molecular orbitals are polycentric, thus electrons are effected by two or more than two nuclei.
• Atomic orbitals of inner shells (apart from valence shell) also combine to form molecular orbitals. These are normally called nonbonding molecular orbitals. Molecular orbitals are formed by overlapping.
  Filling of electrons follow Aufbau’s principle, Hund’s Rule and Pauli’s principle.

Question 35.
Write the hybridization, geometry and bond angle present in some specific compounds.
Answer:
Hybridization, geometry and bond angle in compounds :

Chemical Bonding and Molecular Structure Class 11 Important Questions Long Answer Type

Question 1.
State important applications of dipole moment.
Answer:
Important applications of dipole moment are as follows :

• By the help of dipole moment, polar and non-polar nature of a molecule can be determined. Higher the magnitude of dipole moment (υ = q × d), higher is its polarity. For non-polar molecules, magnitude of dipole moment is zero. For example: CO₂.
• By its help, percent ionic character can be determined as follows :
  Percent ionic character = \(\frac{\mu \text { observed }}{\mu \text { ionic }}\) × 100
• Dipole moment of Symmetric molecules is zero, though they contain two or more polar bonds. Thus, used in measuring the symmetry of molecules.
• By its help, cis and trans isomers can be differentiated. Normally, dipole moment of cis isomer is higher than that of trans isomer.
• It is also used in the differentiation of ortho, meta and para isomers. Dipole mo¬ment of para isomer is zero. Dipole moment of ortho isomer is more than meta isomer.
• Information regarding shape of molecule is obtained.

Question 2.
What is valence shell electron pair repulsion theory? State its limitations. Or, Explain valence shell Electron pair repulsion theory with example.
Answer:
Geometry of covalent molecules can be justified on the basis of VSEPR theory. According to this theory:
Geometry of a molecule depends on the number of bonded and lone pair electron or non-bonding electron in valence shell of central atom.

If only bonded electron pairs are present in the valence shell of central atom of a molecule, then the geometry of the molecule is symmetrical and depends on the type of hybridization of valence shell contain 2,3,4,5,6 and 7 bond electron pair the geometry of the molecule is linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral and pen¬tagonal bipyramidal.

If valence shell of central atom of molecule contains both bond and lone pair electrons, then the geometry of the molecule is not symmetrical because due to the presence of electron pair, these is more repulsion between bond and lone pair electrons which destroy the geometry. Order of repulsion in various electron pair is as follows :
Lone pair – Lone pair > Lone pair – Bond pair > Bond pair – Bond pair.

Due to increase in the number of lone electron pairs on central atom, magnitude of bond angle between bonded electron pair decreases.

If valence shell of central atom is fully filled with electrons then repulsion in bonded electron pair is more than the repulsion by bonded electron pair in incomplete va¬lence shell.

Limitations :

• On the basis of this theory, geomtry of complex salts of transition metals cannot be explained.
• By the help of this theory, geometry of such molecules which have highly displaced π-electrons cannot be explained. Shape of highly polar compound like Li₂O cannot be discussed.

Question 3.
Compare the relative stability of the following species and indicate their magnetic properties:
O₂,O₂⁺,O₂^– (Superoxide), O₂^–(Peroxide).
Answer:
Electronic configuration of species is as follows :
(a) O₂(16c^–) →
Bond order = \(\frac{1}{2}\) (N_b – N_a ) = \(\frac{1}{2}\) (10 – 6) = 2
Thus, 0 = 0
Due to the presence of two unpaired electrons, O₂ is paramagnetic.

(b) O₂⁺ ion is formed by the removal of one electron from O₂ molecule
O₂ → O₂⁺ + e^–
O₂⁺ (15 e^–) →
Bond order = \(\frac{1}{2}\) (N_b-N_a)= \(\frac{1}{2}\) (10-5) = 2.5
InO₂⁺ one electron is unpaired, thus it is paramagnetic.

(c) O₂^– (Superoxide) ion : O₂^– is formed by the addition of one electron in O₂ molecule.
O₂ + e → O₂
O₂^– (17e^–) →
Bond order = \(\frac{1}{2}\)(N_a – N_b) = 1(10-7) = 1.5
Due to the presence of one unpaired electron, O₂^– is paramagnetic.

(d) O₂^2- (Peroxide) ion : O₂^2- ion is formed by the addition of two electrons in O₂ molecule.
O₂+2e^– → O₂^2-
O₂^2- (18e^–) →
Bond order =\(\frac{1}{2}\) (N_b-N_a) = \(\frac{1}{2}\)(10-8) =1
Due to presence of paired electrons O₂^2- is diamagnetic.

Decreasing order of stability : We know that, Bond order « Stability.
Thus, decreasing order of stability is as follows :
O₂⁺ > O₂ > O₂^– > O₂^2-.

Question 4.
What is hydrogen bond? Write its types. How does it affect the properties of compounds?
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. This bond is represented by dotted line (….). Hydrogen bond exists between H-atom of one molecule and electronegative (O, N, F) atom of another molecule:
Example : HF:
It is a weak bond.
Types of hydrogen bond : 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. This bond is represented by dotted line (……………).
It is of two types :

1. Inter-molecular hydrogen bond.
2. Intra-molecular hydrogen bond.

Effects of hydrogen bond: Hydrogen bond affects the physical and chemical properties like boiling point, solubility etc. of compounds. Due to intermolecular hydrogen bond, molecules get associated, therefore their melting and boiling points are high whereas due to intramolecular hydrogen bond melting point, boiling point and solubility etc. of compound becomes low.

Example:
1. Physical state: In H₂O, due to hydrogen bond, molecules of water are in associated state, therefore at normal temperature H₂O is a liquid whereas due to low
electronegativity of S, hydrogen bond in H₂S is not possible. Their molecules possess weak van der Waals’ force, therefore it is a gas at normal temperature.
2. Solubility: Alcohol and carboxylic acid easily form intermolecular hydrogen bond, therefore they are soluble in water.

Question 5.
Describe the hybridisation in case of PCl₅. Why are the axial bonds longer as compared to equatorial bonds ?
Answer:
Electronic configuration of phosphorus is as follows :
₁₅ P = 1s²,2s² 2p⁶,3s²,3p³

In the formation of PCl₅ molecule, one s, three p and one rf-orbital are available for hybridisation. These orbitals combine to form five sp³d hybrid orbitals. These five hybrid orbitals are oriented towards the five comers of a trigonal bipyramid.

Again, in PCl₅ three P – Cl bonds are situated on one plane and form an angle of 120°. These are known as equatorial bonds, remaining two P – Cl bonds lie above and below the equatorial bonds making an angle of 90°. These bonds are called axial bonds. Since axial bond feel more repulsion than equatorial bonds, thus they are comparatively longer than the equatorial bonds.

Question 6.
What do you mean by resonance? Explain with example.
Answer:
When properties of a molecule are hot explained by one structure and two or more than two structures are assigned to express its characteristics, it is said that molecule is resonance hybrid of these structures and this property is known as resonance.

Different resonating structures are exhibited by using sign (↔) in between these structures.
Resonance structures do not have any physical significance. Actually resonance theory is an imagination theory. As a result of resonance, energy of the compound decreases and stability increases. Energy of resonance hybrid is always less than the energy of various resonating structures.
Example: Carbon dioxide is normally represented by the following formula :
0 = C=0

On the basis of this structure bond length between O and C should be 1.22Å, but actual bond length is 1.15Å, which is in between the double and triple bond. Energy of formation of CO₂ should be 300 kcal/mol but actual bond energy is 310 kcal/mol. Thus, there can be two formulae of CO₂:

Other example of Resonance : Carbonate ion :

Conditions for resonance are as follows :

• Enthalpy of formation of all resonating forms is nearly same.
• In every formula, arrangement of atoms should be same.
  
• Number of unpaired electrons in all the resonating forms should be equal.

Question 7.
Explain the electrostatic attractive and repulsive force on the basis of formation of covalent bond. Or, What is Quantum theory of covalency for hydrogen molecule? Explain.
Answer:
During the formation of covalent bond, when two atoms come close to one another then two types of forces act between them:

1. Attraction by nuclei of both the atoms to the electrons of the molecule.
2. Repulsive force between the nuclei and the electrons of both the atoms.

When the resultant force of acting attractive force and repulsive force is attraction the energy of the system decrease and formation of bond is possible in this situation. But, if the resultant force of acting attractive force and repulsive force is repulsion, there is increase in energy and bond formation is not possible.

Example: Formation of H₂ molecule :
If both the H-atoms are situated far from each other, then there is no attractive or repulsive force between them but when both the H-atoms are brought close to each other then the ahead to situations can be possible :
(i) When two H-atoms whose spin of electron is parallel to one another come closer then their potential energy continuously increases because strong repulsive force is present between both the electrons. Therefore this bond will be unstable.

(ii) If H-atoms of opposite spin come closer, strong attractive force between them leads to decrease in energy. Due to opposite spin electron density between both the nuclei increases which binds both the nuclei by strong electrostatic force. This bond is called covalent bond. If both the H-atoms are brought closer than the minimum distance, then repulsive force between nucleus-nucleus increases, by which energy increases continuously and bond again becomes unstable.

Question 8.
Structure of two molecules are given below :

(i) Of the above in which compound intermolecular hydrogen bond and in which intramolecular hydrogen bond is present?
(ii) Melting point of a compound apart from other concepts is based on hydrogen bond. On this basis justify which of the above compound will have higher melting point?
(iii) Solubility of a compound depends upon its ability to form hydrogen bond. Which of the above compound will easily form hydrogen bond and will be more soluble in water?
Answer:
(i) Compound will form intramolecular hydrogen bond, when hydrogen atom in a molecule is present between two more electronegative atoms (O, F, N etc.) then intramolecular hydrogen bond is formed. In ortho nitrophenol (compound I), hydrogen atom is situated between two oxygen atom.

In compound (II) intermolecular hydrogen bond is formed. In p-nitro phenol vacant space is present between -NO₂ and -OH group. Thus, hydrogen atom of one molecule forms intermolecular hydrogen bond with oxygen atom of other molecules.

(ii) Melting point of compound(II) will be higher because its various molecules are associated with hydrogen bonds.
(iii) Due to intramolecular H-bond compound (I) cannot form H-bond with water, due to which it is less soluble in water. Whereas molecules of compound (II) can easily form H-bond with water due to which it is more soluble in water.

Question 9.
Explain the structure of the following molecules on the basis of VSEPR theory:
(a) CH₄,(b) H₂O,(c) NH₃.
Answer:
(a) Structure of CH₄: In methane the central carbon atom is surrounded by four electrons and it forms four similar C – H bonds. This molecule is of tetrahedral geometry due to sp3-hybridization. On the basis of maximum distance the paired electrons are at the four tips of the tetrahedron. Bond angle is 109°28′.

(b) Structure of H₂O: In water molecule, central oxygen atom is also sp³-hybridized and structure is tetrahedral but oxygen atom has two lone pair of electrons. Thus, lp-lp and lp-bp repulsion exist. Due to this, bond angle reduces to 104°5′ in place of 109°28′ and shape becomes V shaped.

(c) Structure of NH₃: In ammonia, central N atom is surrounded by four electron pairs, so shape of molecule is tetrahedral. According to VSEPR theory, the four groups around the central atom of ammonia should be tetrahedrally arranged at bond angle of 109°. But, the measured bond angle is 107°. This is explained on the basis of repulsive effect of the lone pair of electrons on bonding electrons. In ammonia molecule there is a lone pair of electrons on the N atom. Thus, the Shape of NH₃ molecule is distorted and it looks like pyramidal and it is polar in nature.

Question 10.
Justify the following with reason :
(i) Covalent bonds are directional whereas ionic bond are non-directional.
(ii) Structure of water molecule is angular whereas carbon dioxide molecule is linear.
(iii) Ethyne molecule is linear.
Answer:
(i) Covalent bond, is formed by the overlapping of atomic orbitals. Direction of overlapping states the direction of bond. In ionic compounds, electrostatic area of ion is non-directional. On the basis of size of each cation is surrounded by anion in any direction. Similarly, anion is surrounded by cation. That is why, covalent bond is directional and ionic bond is non-directional.

(ii) In H₂ oxygen is sp-hybridized with two lone pairs, sp³ hybrid orbitais are at the apices of the four tetrahedron. Two apices are surrounded by hydrogen atoms and remaining two apices are surrounded by lone pairs. Thus, due to lone pair, lone pair repulsive force bond angle decreases from 109-5° to 104-5° due to which shape of the molecule becomes V-shaped or bent or angular.
In CO₂ molecule, C-atom is sp-hybridized. Two hybrid orbitals are situated in opposite direction to each other forming an angle of 180°.

This is the reason that H₂O molecule is angular whereas structure of CO₂ is linear.
(iii) In Ethyne, both carbon atoms are sp -hybridized and the two orbitals (2px and 2py) are unhybridized. These two sp-hybrid orbitals lie in opposite directions making an angle of 180° due to which ethyne is linear.

Question 11.
Write the main postulates of valence bond theory and state the limitations of this theory.
Answer:
Valence bond theory was proposed by Hitler and London which was modified by Pauling and Slater. The main postulates of this theory are as follows :

• According to this theory, covalent bond is formed by the partial overlapping of half filled orbitals between the atoms.
• Atomic orbitals which participate in overlapping possess electrons with opposite spin.
• Strength of bond depends upon the extent of overlapping.
• Directional strong bond is formed between two orbitals of similar stability, similar energy and symmetry.
• As a result of overlapping and pairing of electrons, energy is released and the system achieves a state of lower energy.
• Of both complete bond orbitals of each orbital is obtained. Therefore it is considered to be the property of both the atomic orbitals.

Limitations of Valence bond Theory :

• According to this theory, O₂ molecule does not contain any unpaired electron. Thus, nature of O₂ should be diamagnetic but O₂ molecule is paramagnetic.
• Nothing is discussed about the formation of co-ordinate bond.
• This theory is unsuccessful in explaining the formation of ions like H₂⁺ .
• It is unable to give any information regarding resonance.

Question 12.
Discuss the structure of PCl₅ and SF₆. Axial bonds in PCl₅ is longer than equatorial bonds. Whereas in SF₆ bond length of both axial and equatorial bonds are same discuss.
Answer:
Formation of PCl₅:Electronic configuration of phosphorus is as follows :
₁₅ P = 1s²,2s² 2p⁶,3s²,3p³

In the formation of PCl₅ molecule, one s, three p and one rf-orbital are available for hybridisation. These orbitals combine to form five sp³d hybrid orbitals. These five hybrid orbitals are oriented towards the five comers of a trigonal bipyramid.

Again, in PCl₅ three P – Cl bonds are situated on one plane and form an angle of 120°. These are known as equatorial bonds, remaining two P – Cl bonds lie above and below the equatorial bonds making an angle of 90°. These bonds are called axial bonds. Since axial bond feel more repulsion than equatorial bonds, thus they are comparatively longer than the equatorial bonds.

Formation of SF₆.
Electronic configuration of S in ground state
Electronic configuration of S in excited state
In SF₆, sulphur is sp³d² hybridized by which towards 6 corners octahedron the orbitals are orientated. These six sp³d² hybrid orbitals overlap with p-orbitals of six fluorine atoms to form six similar S-F bonds.

Question 13.
What is bond order? What informations are given by bond order?
Answer:
Bond order is defined as half of the difference between the number of electrons in bonding molecular orbitals and anti-bonding molecular orbitals.
Bond order = \(\frac{1}{2}\) [No. of electron in bonding molecular orbitals – No. of electron in anti-bonding molecular orbital]
= \(\frac{1}{2}\) [N_b – N_a].

Bond order in a molecule gives following informations :

• If bond order has positive value it indicates a stable molecule and if bond order has negative value or zero the molecule is unstable and is not formed.
• It tells about the number of covalent bonds in a molecule. Bond order of a molecule is equal to the number of covalent bond between the atoms in the molecule.
• Bond dissociation energy of a molecule is directly proportional to the bond order of the molecule. Greater the bond order, greater is the bond dissociation energy.
• Bond length is inversely proportional to the bond order of the molecule, That is greater the bond order, shorter will be the bond length.
• Paramagnetic molecules have fractional value of bond order while when bond order is a whole number, the molecule may or may not be paramagnetic.

Question 14.
Calculate bond order of O₂, O₂^–, O₂² and O₂⁺.
Answer:
1. O₂ [16] –
Bond order = \(\frac{1}{2}\) [10 – 6] = 2
2. O₂^–[17] –
Bond order = \(\frac{1}{2}\) [10 – 7] = 1.5
3. O₂^2- [18] –
Bond order = \(\frac{1}{2}\) [10-8] = 1
4. O₂⁺ [15] –
Bond order = \(\frac{1}{2}\) [10 — 5]
= 2.5.

Question 15.
Represent N₂ molecular by figure on the basis of molecular energy level.
Answer:
Nitrogen molecule (N₂): Each nitrogen atom contains seven electrons. Thus, total 14 electrons are filled in seven orbitals of increasing energy.

Molecular orbital structure of N₂ molecule will be as follows :
N₂:
This structure shows that it contains 10 bonding and 4 anti-bonding electrons.
Bondorder =\(\frac{1}{2}\) (N_b-N_a) = \(\frac{1}{2}\) (10-4) = 3.
Value of bond order is more. Hence value of bond energy should also be high. Experimental value of bond energy is 945 KJ mol^-1 which proves the presence of paired electrons in nitrogen molecule. Thus, it is a diamagnetic molecule.

Question 16.
Represent by molecular energy level diagram that diatomic Neon molecule has no existence in nature.
Answer:
Electronic configuration of ₁₀Ne = 1s²,2s²,2p_x²2p_y²2p_z2.
Bondorder =\(\frac{1}{2}\) (N_b -N_a) = \(\frac{1}{2}\) (10-10) = 0
Bond order zero means, no bond is formed between two Ne atoms. Thus, Ne2 (di¬atomic molecule) has no existence in nature.

Question 17.
Draw molecular orbital diagram for O₂ molecule.
Answer:
Oxygen molecule O₂: Each oxygen atom has eight electrons. When two oxygen atom combines, molecular orbitals are formed. These molecular orbitals have following configuration.

From the following configuration we have N_b = 10, N_a = 6
∴ Bond order =\(\frac{1}{2}\) [N_b – N_a] = [10-6] = 2
Hence, there is a double bond in oxygen molecule. Due to the presence of two unpaired electrons it is paramagnetic.
Its molecular orbital diagram is :

Chemical Bonding and Molecular Structure Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Which type of bond is formed between similar atoms :
(a) Ionic
(b) Covalent
(c) Co-ordinate
(d) Metallic.
Answer:
(b) Covalent

Question 2.
An atom form an anion :
(a) By loosing one or more electron
(b) By gaining one or two electrons
(c) By sharing electron pair
(d) None of these.
Answer:
(c) By sharing electron pair

Question 3.
Maximum number of hydrogen bonds formed by one water molecule in ice is :
(a) 4
(b) 3
(c) 2
(d) 1.
Answer:
(b) 3

Question 4.
Molecule with highest dipole moment is :
(a) CH₄
(b) CHCl₃
(c) CHI₃
(d) CCl₄.
Answer:
(b) CHCl₃

Question 5.
Shape of ethylene molecule is :
(a) Tetrahedral
(b) Pyramidal
(c) Planar
(d) Linear.
Answer:
(b) Pyramidal

Question 6.
Ammonia molecule is formed due to hybridization of:
(a) dsp²
(b) sp³
(c) sp³d
(d) d²sp.
Answer:
(b) sp³

Question 7.
The high boiling point of water is due to :
(a) Weak dissociation of water molecules
(b) Hydrogen bonding among water molecules
(c) Its high specific heat
(d) Its high dielectric constant.
Answer:
(b) Hydrogen bonding among water molecules

Question 8.
Hydrogen bond is present in :
(a) HF
(b) HCl
(c) HBr –
(d) HI.
Answer:
(a) HF

Question 9.
Which of the following molecule has dipole moment zero :
(a) H₂O
(b) CO₂
(c) SO₂
(d) NO₂.
Answer:
(b) CO₂

Question 10.
In which compound, sp² hybrid orbitals are used :
(a) BCl₃
(b) CH₄
(c) NH₃
(d) BeH₂
Answer:
(a) BCl₃

Question 11.
Which of the following molecules has linear structure :
(a) CO₂
(b) H₂O
(c) SO₂
(d) H₂O₂.
Answer:
(a) CO₂

Question 12.
Hybridization present in PCl₅ molecule is :
(a) sp²d²
(b) sp³d
(c) spd³
(d) sp²d³.
Answer:
(b) sp³d

Question 13.
Number of electrons involved in the formation of bond in O₂ :
(a) 2
(b) 4
(c) 6
(d) 10.
Answer:
(b) 4

Question 14.
In which molecule bond angle of central atom is maximum:
(a) NH₃
(b) NH₄
(c) PCl₃
(d) SCl₂.
Answer:
(b) NH₄

Question 15.
H-bond is present in :
(a) CH₄
(b) NaCl
(c) H₂O
(d) CHCl₃.
Answer:
(c) H₂O

Question 16.
VSEPR (Valence Shell Electron Pair Repulsion) Theory was proposed by :
(a) Hund and Mullican
(b) Heisenberg
(c) Sizwick and Powel
(d) Heitler and London.
Answer:
(c) Sizwick and Powel

2. Fill in the blanks:

1. In CH₄ ………………………… hybridization is found.
Answer:
sp³

2. Full name of LCAO is ………………………… .
Answer:
Linear combination of atomic orbital

3. Boiling point of NH₃ is ………………………… than boiling point of PH3.
Answer:
Higher

4. ………………………… bond is formed by s-s overlapping.
Answer:
σ

5. H-O-H bond angle in water molecule is ………………………… .
Answer:
104.5°

6. ………………………… bond is formed due to p-p overlapping.
Answer:
π

7. σ bond is comparatively ………………………… than π -bond.
Answer:
Stronger

8. Geometry of molecule due to sp3 hybridization is ………………………… .
Answer:
Tetrahedral

9. With the formation of chemical bonds, there is a ………………………… in energy.
Answer:
Decrease

10. Diamond is a ………………………… of electricity but graphite is ………………………… of electricity.
Answer:
Bad conductor, good conductor

11. ………………………… hybridization is found in diambnd and graphite.
Answer:
sp² and sp³

12. Hybridization in water molecule is ………………………… .
Answer:
sp³.

3. Match the following:

Answer:
1. (c) Linear
2. (d) Trigonal pyramidal
3. (e) Three-dimensional (planar tetrahedral)
4. (b) Angular
5. (a) Square planar.

4. Answer in one word/sentence:

1. What type of bond is present in nitrogen molecule?
Answer:
Triple covalent bond

2. Which type of bond is directional?
Answer:
Covalent bond

3: What is the shape of sulphur tetrafluoride?
Answer:
Square pyramidal

4. What type of chemical bond is found in sodium chloride?
Answer:
Electrovalent

5. Sugar is a covalent compound, yet it is soluble in water.
Answer:
Due to hydrogen bond

6. Bond angle of water molecule is.
Answer:
104.5°

7. Dipole moment of BF₃ molecule is.
Answer:
Zero

8. Write the name of linear molecule.
Answer:
CO₂

9. Which molecule show trigonal bipyramidal geometry?
Answer:
PCl₅.

Students get through the MP Board Class 11th Chemistry Important Questions Chapter 3 Classification of Elements and Periodicity in Properties which are most likely to be asked in the exam.

MP Board Class 11th Chapter 3 Classification of Elements and Periodicity in Properties

Classification of Elements and Periodicity in Properties Class 11 Important Questions Very Short Answer Type

Question 1.
What is Mendeleev’s periodic law?
Answer:
The law on the basis of which Mendeleev classified the elements, is called Mendeleev’s periodic law. According to this, “physical and chemical properties of elements are periodic functions of their atomic weights.” i.e.,When elements are arranged in the increasing order of their atomic masses they show periodicity in their properties after a definite interval.

Question 2.
Define modern periodic law.
Answer:
Modern periodic law : Physical and chemical properties of elements are periodic functions of their atomic numbers. On the basis of this law when elements are arranged in the increasing order of their atomic number they show a periodicity in properties after a definite interval.

Question 3.
Discuss the law on the basis of which Mendeleev classified the elements. What modifications are done in this law ?
Answer:
The law on the basis of which Mendeleev classified the elements is known as Mendeleev’s Periodic law. According to this law “The physical and chemical properties of elements are a periodic function of their atomic mass”. Now this law is modified and is known as Modem Periodic law. According to this law “The physical and chemical properties of elements are a periodic function of their atomic number.”

Question 4.
What is Dobereiner’s law of Triad ?
Answer:
This law was proposed by Dobereiner. According to this law, “Of the three similar atoms, the atomic mass of the middle element is the average mean of the other two elements.
Li₇, Na₂₃, K₃₉,i.e \(\frac{39+7}{2}\) = 23

Question 5.
On the basis of quantum numbers prove that there should be 32 elements in the sixth period of the periodic table.
Answer:
In long form of periodic table each period starts with the filling of a new energy shell (n). For the sixth period n = 6. In this period, electrons enter in 6s,4f, 5d and 6p. Total 16(1+7 + 5 + 3) orbitals are there in these subshells. According to Pauli’s exclusion prin¬ciple, each orbital can have maximum two electrons. Thus, in 16 orbitals these can be 32 electrons. Therefore, sixth period should have 32 elements.

Question 6.
What is Newland octave rule?
Answer:
Newland arrange the element in order of their atomic masses he found that the eighth element is a kind of repetition of first element, just like eighth note of music. This arrangement of element is known as Newland law of octave or Newland octave rule.

Question 7.
State the period and group where Z = 114 will be situated?
Answer:
Maximum value of n will be represented by the electronic configuration of the element.
₁₁₄Z=86[Rn]7s²,5f¹⁴,6d¹⁰,7f₂
Period = 7,
(∵ n = 7)
Group =14(10 + 2 + 2).

Question 8.
Write the atomic number of the element which is situated in the 3rd period and 17th group.
Answer:
Outermost configuration of elements situated in 17th group is ns² np². In the third-period valence shell configuration of the element will be 3s² 3p⁵. In the 3rd period, atomic number starts from atomic number Z = 11 and ends in Z = 18. Thus, the given atomic number =10 + 2 + 5=17.

Question 9.
Name of which element is proposed by the following :
(i) By Lawrence Berkeley Laboratory,
(ii) By Sea borg’s group.
Answer:
(i) Lawrencium (Z = 103) and Berkelium (Z = 97)
(ii) Seaborgium (Z = 106).

Question 10.
Why are the physical and chemical properties of the elements of a group same?
Answer:
The outer electronic configuration of the elements of a group is same due to which they show similar physical and chemical properties.

Question 11.
Why are cations smaller than their parent atoms and anions larger than their parent atoms? Explain.
Answer:
Cations are always smaller than their parent atoms because due to the removal of one or two electrons, magnitude of their effective nuclear charge increases. This results in greater attractive force of the nucleus on the valence electrons due to which ionic radius decreases.

Alternatively, anions are always larger than their parent atoms because due to the gain of one or two electrons magnitude of effective nuclear charge decreases by which attractive force of nucleus with the valence electron decreases. This results in decrease in ionic radius.

Question 12.
How many periods and groups are in Mendeleev’s periodic table? What is the number of elements in each period?
Answer:
There are 7 periods and 9 groups in Mendeleev’s Periodic table. First period consists of 2 elements, second and third period consist of 8-8 elements, fourth and fifth period contain 18-18 elements and sixth period consist of 32 elements. Seventh period is incomplete. In it, elements from atomic number 90 to 103 called Actinides are also included.

Question 13.
What changes occur on moving from left to right in every period?
Answer:
In every period, on moving from left to right ionization potential and electron affinity increases but metallic nature, basicity of oxides and atomic radius decreases.

Question 14.
What changes occur on moving from top to bottom in a group?
Answer:
On moving from top to bottom in a group atomic radius, ionic radius increases but ionization energy, electronegativity, electron affinity and melting point decreases.

Question 15.
Compare Modern periodic table with Mendeleev’s periodic table (any 3 points).
Answer:
Comparison between Modern periodic table and Mendeleev’s periodic table

Question 16.
What is diagonal relationship? Give an example.
Answer:
Certain elements of the 2nd period show resemblance with some elements of the 3rd period. Thus Li resembles Mg, Be resembles Al, B resembles Si in certain respects.

Question 17.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
Answer:
Decrease in ionization enthalpy down the group in the elements of main groups is due to the following two reasons :
(i) On moving down the group atomic size increases, due to increase in the formation of new energy shell due to which value of effective nuclear charge of the nucleus for the valence electron gradually decreases.

(ii) On moving down the group, screening effect increases, due to which ionization enthalpy decreases (Because attractive force between the nucleus and electron decreases).

Question 18.
What is the main difference between electron gain enthalpy and electronegativity?
Answer:
Electron gain enthalpy is the nature of formation of a gaseous anion by gaining an electron by an isolated gaseous atom whereas electronegativity is the nature of an atom of an element to attract the shared pair of electrons present in a covalent bond towards itself. Like electron gain enthalpy, electronegativity also has no measuring unit.

Question 19.
In all nitrogen compounds, Electronegativity of N in Pauling scale is 3.0. What reaction will you give on this statement?
Answer:
Electronegativity of N in all nitrogen compounds in Pauling scale is 3.0. This statement is incorrect because electronegativity of any element is not definite, the effect of the other element with which it is bonded is also applied on its value. Its value increases with the oxidation state of the element and with the increase in percent 5-character of the hybrid orbital.

Question 20.
What is covalent radius?
Answer:
Covalent radius: In homoatomic molecules containing same type of atoms bonded together by single covalent bond, covalent radius may be defined as, “one half of the distance between the centres of the nuclei of two adjacent similar atoms joined together by a single covalent bond”.

Question 21.
What is van der Waals’ radius ?
Answer:
van der Waals’ radius: It is defined as, “one half of the internuclear distance between two adjacent atoms belonging to two nearest neighbouring molecules of the substance in the solid-state”.

Question 22.
What do you understand by normal elements?
Answer:
Such elements which represent their group are called typical elements. Their only one outer shell is incomplete and all inner shells are completely filled according to the rules. Elements of second and third period are called typical elements.
Example: C, N, O etc.

Question 23.
First ionisation enthalpy of two isotopes of an element will be same or different ? What do you believe ? Justify your answer.
Answer:
First ionisation enthalpy of two isotopes of an element should be same because its value depends on electronic configuration and effective nuclear charge. Electronic configuration of the isotopes of an element is same due to which nuclear charge is also similar.

Question 24.
Write general outer electronic configuration of s-, p-, d- and f-block elements.
Answer:
s- block elements : ns^1-2 (n = 2-7)
p- block elements : ns²np^1-6 (n = 2-7)
d- block elements : (n – l)d^1-10ns^0-2 (n = 3-7)
f-block elements : (n-2)f ^1-14 (n-1)d,10ns²(n =6-7)

Question 25.
Ionization potential of Nitrogen is higher than oxygen. Explain.
Answer:
Electronic configuration of nitrogen and oxygen is 7N = 1.s², 2s² 2p³ (half filled 2p -orbitals) 8^O = 1.s², 2s² 2p⁴
Thus in case of nitrogen, 2p subshell is half-filled and removal of electron from half-filled orbitals or full-filled orbitals require more energy than that of other incomplete orbitals. So, ionization energy of nitrogen is greater than oxygen because of half filled p-subshell.

Question 26.
Ionisation potential of inert gases are very high. Why?
Answer:
Electronic configuration of inert gases is most stable so it is very difficult to pull an electron. So in each period, ionisation potential of inert gases are very high.

Question 27.
Halogens have very high electron affinity. Why?
Answer:
Halogens are one electron short to attain the stable electronic configuration of inert gases. So these elements have very high tendency to accept electron. Thus the electron affinity of halogens is very high.

Question 28.
Among F and Cl, whose electron affinity is higher and why?
Answer:
Electron affinity of F is less than Cl because due to small size of F repulsion of its 2p electrons is more. Therefore added electron does not provide extra stability to the atom

Question 29.
Electron affinity of Noble Gases is zero. Why?
Answer:
Electronic configuration of Noble gases is stable, because their valence shells are completely filled and all the electrons are paired, due to which there is no affinity for the incoming electron. Therefore, electron affinity of noble gases is zero.

Question 30.
Differentiate between Atomic radius and Ionic radius.
Answer:
Differences between Atomic and Ionic radius

Question 31.
Value of second ionization potential of an element is more than the value of first ionisation potential. Why?
Answer:
By the removal of an electron from an atom, the remaining atom becomes posi¬tively charged. Ionic radius of cation is smaller than atomic radius, due to which more energy is required to remove the second electron from the outermost shell. Therefore, value of second ionisation potential is higher than first ionization potential.

Question 32.
Ionic radius of cation is smaller than atomic radius. Why? Or, Size of cation is smaller than the corresponding atom. Why?
Answer:
A cation is formed by the loss of one or more electrons from an atom. In this process, number of protons in the nucleus and nuclear charge present in the nucleus remains the same, but number of electrons present in the outermost shell decreases. As a result nuclear attractive force per electron increases and outer shell gets strongly attracted by the nucleus. Therefore, size of cation is smaller than the corresponding atom.

Question 33.
Ionic radius of anion is larger than the atomic radius. Why? Or, Size of anion is bigger than the corresponding atom. Why?
Answer:
An anion is formed by the gain of one or more electrons by an atom. In this process, number of protons in the nucleus and nuclear charge present in the nucleus remains the same, but number of electrons present in outermost shell increases. As a result, nuclear attractive force per electron decreases due to which size of anion becomes larger than the corresponding atom.

Question 34.
Electron affinity of Be and N is almost zero. Why?
Answer:
2s sub-shell of Be is more stable as it is fully filled and it does not allow any extra electron to enter into the s sub-shell. Similarly, p sub-shell of N is more stable due to half filled configuration. It does not gain any extra electron. Therefore, electron affinity of Be and N is almost zero.

Question 35.
Mg²⁺ion is smaller than O^2- while electronic configuration of both are same. Why?
Answer:
The nuclear charge on Mg²⁺ and O^2- is +12 and +8 respectively but in both number of electrons is 10. Their electronic configuration is 1s²,2s² 2p⁶. In Mg²⁺ there is attraction between 10 electrons and 12 protons and in O^{2 –}ion there is force of attraction
between 10 electrons and 8 protons and force of Mg⁺² is greater than oxygen, so Mg⁺² is smaller than O^2- ion.

Question 36.
First ionization potential of A1 is less than Mg. Why?
Ans. Electronic configuration of Al and Mg is as follows :
Al₁₃: 1s², 2s²2p⁶a, 3s²3p¹
Mg₁₂ : 1s², 2s²2p⁶, 3s²,
Half filled or fully filled orbitals are more stable, p sub-shell of Al is incomplete whereas s sub-shell of Mg is completely filled. Thus, sub-shell of Al is less stable than sub shell of Mg. Therefore to remove an electron from Al atom less energy is required as compared to Mg. Therefore, ionization energy of Al is less than Mg.

Question 37.
What do you understand by screening effect ?
Answer:
The obstruction caused in nuclear attraction by electrons present in between the outermost shell and nucleus is called screening effect. Due to screening effect, outermost electrons are loosely bound.
Due to increase in screening effect, atomic radius or atomic size increases.

Question 38.
Discuss the periodicity in ionization energy on moving from left to right of a period in the periodic table.
Answer:
In a periodic table on moving from left to right, nuclear charge increases and ” atomic radii decreases. Due to this it is difficult to pull out any electron and requires more energy. Thus, in a period ionization energy increases on moving from left to right side.

Question 39.
Magnitude of Electron affinity increases on moving from left to right in a period. Explain.
Answer:
On moving from left to right in a period, nuclear charge increases and atomic radius decreases, due to which attraction of the nucleus for the added electron increases. Therefore, in any period, on moving from left to right electron affinity increases.

Question 40.
Explain the periodicity of metallic character in the periodic table.
Answer:
Elemeiits which lose electrons are electropositive elements and which gain electrons are called electronegative elements. Electropositive elements represent metallic nature whereas electronegative elements represent non-metallic nature. On moving from top to bottom in a group metallic nature increases but in any period on moving from left to right metallic nature decreases.

Question 41.
Of the following pair N and O, whose electron gain enthalpy is more?
Answer:
Electron gain enthalpy of O is more than N because the valence shell of N -1s²,2s²2p³ is half filled, thus it is stable. Thus, tendency to gain an extra e is more in O than in N. Apart from this size of O is smaller than N, thus due to high nuclear charge e^– is added easily.

Classification of Elements and Periodicity in Properties Class 11 Important Questions Short Answer Type 

Question 1.
What do you think? Second electron gain enthalpy like first electron gain enthalpy will be positive, negative or less negative Justify your answer.
Answer:
O_(g)+ e ^–→ O^–;
Δ_eg H = -141 kJ mol ^-1
O,sup>-_(g) + e^– → O^2-;
Δ_eg H = +780 kJ mol^-1

When oxygen atom gains one electron, to form O^– ion energy is released. Thus, first electron gain enthalpy of oxygen is negative.
But, when O^– ion gains one more electron to form O^-2 ion, it feels strong electrostatic repulsion. Thus, for the addition of second electron, energy is to be absorbed. Therefore, second electron gain enthalpy of oxygen is positive.

Question 2.
Electron gain enthalpy of fluorine is less negative as compared to chlorine. Give reason.
Answer:
Electron gain enthalpy (Electron affinity) of fluorine is less negative than chlo¬rine because when one electron enters F, then it enters into a small size (n = 2) energy level where it feels the repulsion of the other electrons present in this level. In Cl electron enters into a larger (n = 3) energy level where it feels very less repulsion force of other electrons present in it.

Question 3.
Long form of the periodic table is more superior than Mendeleev’s table. Why?
Answer:
Long form of periodic table is more superior than Mendeleev’s table because of following reasons:

• Long form of periodic table is based on atomic number which is more fundamental property of an atom.
• Position of every element is decided on the basis of its electronic configuration.
• It represents the similarities in chemical properties, dissimilarities and gradual change in properties of various elements more clearly.
• Here classification of elements is more scientific and reasonable.

Question 4.
Write the main characteristics of Modern periodic table.
Answer:
The main characteristics of Modem periodic table are :

• This periodic table is based on the electronic configuration of elements.
• Metals and non-metals are placed separately.
• Transition elements are separated from normal elements.
• On the basis of filling of electrons in elements, all the elements are divided into s,p, d and f block.
• Rare earth elements are placed in a suitable place outside, below the main periodic table.

Question 5.
Atomic number of an element is 17. Locate its position in the periodic table giving its electronic configuration.
Answer:
Electronic configuration of element of atomic number 17 =1s²,2s² 2p⁶,3s² 3p⁵

• Electronic configuration of outermost shell is 3s² 3p⁵. So it is element of p-block.
• Group: If a shell is incomplete, then the number of electrons represents the group of that element. Therefore, it is an element of 7th group.
• Sub-group: If last electron enters into s or p sub-shell, then it is the element of sub-group A.
• Period: Number of shells represents the period.This element is the member of third period.

1. Group = 7
2. Sub-group = A
3. Period = 3

Question 6.
Electronic configuration of an element is 1s²,2s² 2p⁶,3s² 3p⁶. In which group this element will be placed and why? Explain.
Answer:
Electronic configuration of outermost orbit of this element is 3s² 3p⁶. It is well known that electronic configuration of outermost orbit ns² np b is of noble gases (except He). Therefore, this element is also a noble gas which belongs to group 18 in modem periodic table.

Question 7.
All transition elements are d -block elements but all d -block elements are not transition elements. Explain.
Answer:
Elements in which last electron enters into rf-block orbital are called if-block elements or transition elements. General outer electronic configuration of these elements is (n – 1)d^1-10, ns^{0 – 2}Electronic configuration of Zn, Cd and Hg is (n-l)d^1-10, ns² but they do not show the characteristics of transition elements.

The d-orbitals of these elements is completely filled in their ground state or in common oxidation state. Therefore, they could not be considered as transition elements. Thus, on the basis of properties all transition elements are d-block elements and on the basis of electronic configuration, all d-block elements are not transition elements.

Question 8.
What do you understand by iso-electronic species? Write the name of such a species which is iso-electronic with the following atoms and ions :
(i) F^–
(ii) Ar
(iii) Mg²⁺
(iv) Rb⁺.
Answer:
In iso-electronic species number of electrons are same but nuclear charge is different.
In these species, with the increase in nuclear charge, size decreases.
(i) No. of electrons in F^– = 9 + 1 = 10
(ii) No. of electrons in Ar = 18
(iii) No. of electrons in Mg²⁺ = 12-2=10
(iv) No. of electrons in Rb⁺ = 37-1 =36
N^3-, O^2-, Ne, Na⁺ and Al³⁺ species are iso-electronic species of F and Mg²⁺, P^3-, s², Cl^–, K⁺ and Ca²⁺ species are iso-electronic species of Ar. Similarly, Br^– Kr and Sr²⁺ species are iso-electronic species of Rb⁺.

Question 9.
Among the following pairs whose ionization energy is low and why?
(i) Cl or F
(ii) Cl or S
(iii) K or Ar
(iv) Kr or Xe.
Answer:
(i) In Cl and F, ionization energy of chlorine is less because size of Cl is bigger than F.

(ii) In Cl and S, ionization energy of S is less because both have same number of shells but nuclear charge of S is less than Cl. Therefore, size of S is bigger than Cl.

(iii) In K and Ar, ionization energy of K is less. There is only 1 electron in the valence shell of K which it can lose easily and form K⁺ ion whereas due to complete octet in Ar, it is stable and more energy is required to remove an electron.

(iv) In Kr and Xe, ionization energy of Xe is less because size of Xe is bigger than Kr.

Question 10.
On the basis of electronic configuration, how many types of elements are there? Give two examples of each. Or, Explain the various types of elements on the basis of Electronic configuration.
Answer:
Types of elements on the basis of electronic configuration :
(i) Noble Gas: They are placed in 18th group. Their outermost shell is stable whose configuration is ns² or ns²p⁶. Example He, Ne.

(ii) Representative elements or s and p-block element: Elements of group 1,2 and 13 to 17 are called normal or Representative elements. Their outermost shell is incomplete and general electronic configuration of outermost shell is ns¹ to ns²p⁵.
It includes s-and p-block. (s-block element Na, Mg and p-block element Cl, Br)

(iii) Transition element or d-block elements: It includes elements of group 3 to 12. Their outer two shells are incomplete. Elements of group 3-12 which are situated in the middle of the periodic table are called Transition elements. General Electronic configuration of outermost shell is (n – 1) d^1-10 ns^1-2
Example: Cr, Cu.

(iv) Inner Transition Elements or f-block Element: Their outer three shells are incomplete. These are known as /-block or rare earth elements. It includs Lanthanides and Actinides. These are mainly situated at the bottom of the periodic table. General electronic configuration of its outermost shell is (n – 1)f^1-14 (n-1) d^0-1 ns²
Example: Lanthanide: Cerium (Ce), Praseodymium (Pr)
Actinide: Thorium (Th), Uranium (U).

Question 11.
What are s-block elements? Write their main properties.
Answer:
In their electronic configuration, last electron enters into s sub-shell and only outermost shell is incomplete, s sub-shell of outermost shell contains 1 or 2 electrons and their general electronic configuration is ns^{1 or 2}.

Main properties of s-block :

• In the outermost orbit of element, there are 1 or 2 electrons.
• The electronic configuration of its outer orbit is nslor2.
• The oxidation state of it is +1 or +2.
• Except He and H all elements are metals. In 1st group the elements are alkaline elements and in II nd group, the elements are alkaline earth elements.
• These form ionic compounds.
• These are strong reducing agents.
• The oxides of these are alkaline.
• These react with water and acid and give out hydrogen.

Question 12.
What are the p-block elements? What are their main characteristics?
Answer:
In the electronic configuration the last electron enters in top sub-shell, it is known as p -block element. In the outermost shell, there are 2 electrons in s sub-shell and 1 to 6 electrons in p sub-shell configuration of their valence shell is ns² np^1-6

General characteristic of p-block elements :

• These elements have 2 electrons in s sub-shell and 1 to 6 electrons in p sub-shell in outermost orbit. Only in zero group elements, outer shell is completed. ‘
• Definite positive or negative oxidation numbers are represented by the elements. Some of the elements show variable valencies.
• These elements form simple ions as well as complex ions of CO₃^2-, NO₃^–.
• These are generally non-metals and metalloids. Some of the elements are heavy elements, e.g., Pb, Bi, etc.
• Oxides are acidic in nature. Some oxides are amphoteric, e.g., PbO, SnO etc.
• These elements form covalent compounds with each other but with the elements of 5-block form electrovalent compounds.

Question 13.
Electronic configuration ofthe element is 1s²,2s² 2p⁶,3s² 3p⁶,4s²and its mass number is 40. State the total number of neutrons in the nucleus and also deduce its position and name of the element in the periodic table.
Answer:
Number of electrons in the element = 20
Number of protons = 20
Mass number = Number of protons + Number of neutrons
40 = 20 + Number of neutrons
∴ Number of neutrons = 20
Atomic number of that element is 20 and element is Calcium (Ca) which belongs to period 4 and group 2.

Question 14.
What is meant by Electronegativity of an element? How is it different from electron affinity? How does Electronegativity change in the periodic table?
Answer:
Electronegativity: The property of attracting the shared pair of electrons between two atoms in a compound towards itself is called electronegativity.

Electron affinity: The energy released due to entrance of an electron in outermost shell of an atom in gaseous state is called electron affinity. value of first electron affinity of an element is negative because this process is exothermic, whereas in second eleciron affinity energy is absorbed because negatively charged ion repels the incoming electron.

Difference: Electron affinity is the property of binding the extra electron whereas electro-negativity is the capacity of attracting electron by an atom of a molecule.

Gradation: On moving from left to right, due to decrease in atomic size and increase in nuclear charge electro-negativity increases. On moving from top to bottom in a group, with the increase in atomic size electronegativity decreases

Question 15.
Write uses of Mendeleev’s periodic table.
Answer:
Following are the importance or uses of Mendeleev’s periodic table :
(i) Helpful in the study of Elements: In the study of properties of elements in the periodic table.

(ii) Elements of same group possess similar properties, so by studying the general properties of the group, properties of all the elements can be known.

(iii) In finding of atomic mass: From the position of elements in the periodic table its valency can be known. If the equivalent mass of that element is known, its atomic mass can be known.
Atomic mass = Equivalent mass × Valency

(iv) Helpful in discovery of new elements: At the time of preparation of periodic table, Mendeleev had left many places for new elements but the properties of these elements were categorized by him. After discovery of these new elements, it is seen that the properties are exactly same as predicted by Mendeleev.

(v) Helpful in research: Elements and their compounds are studied comparatively with the help of periodic table. So, periodic table is helpful in research.

Question 16.
Where are Transition elements placed in the periodic table? Or, What are transition elements ? Explain four main properties of transition elements.
Answer:
Transition elements: The elements which are in between s-block and p-block are called as transition elements or elements in which d-orbitals are partially filled are called transition elements. In this, last electron goes to d-orbital of orbit inner to outer so called d- block elements. Boiling point, melting point and densities of these elements are high.
Example: Cr, Mn, Fe, Cu, etc.

Properties :

• In these two outer orbits are incomplete,
• Electronic configuration is (n – 1) d^1-10 ns^{1 or 2},
• It has metallic properties,
• These elements show variable oxidation state,
• These form coloured ions,
• These form complex salt,
• These are good catalysts,
• These form complex compounds,
• These are generally diamagnetic and
• These also form with non-metal compounds.

Question 17.
What are Inner transition elements? Write general characteristic of inner transition elements.
Answer:
Three outer shells of these elements are incomplete. In their electronic configuration, the last electron enters into the sub-shell of the third inner (anti penultimate) shell. Therefore, they are known as d-block elements. Both the series of Lanthanides and Actinides elements are the elements which join group 3 and group 4, therefore they are known as inner transition elements and their general electronic configuration is ns² f^1-14 (n-1)d^0-1 ns²
Lanthanides: In their electronic configuration, last electron enters 4/sub-shell.
Actinides: In their electronic configuration, last electron enters 5/sub-shell.

General characteristic of Inner Transition Elements :

• Electron enters in/sub shell of (n – 2) shell.
• They exhibit variable oxidation state.
• They all show metallic nature.
• Their ions are generally coloured.
• They form complex ions.
• Form electrovalent compounds.

Question 18.
Define modern periodic law. Discuss briefly the periodic table on the basis of this law.
Answer:
On the basis of modem periodic table, Bohr gave a systematic arranged periodic table which is called long form of periodic table. The physical and chemical properties of elements are periodic function of their atomic number.

Periods: In modern periodic table, there are also seven horizontal rows which are known as periods.

• First period is very short period which contains only two elements.
• Second and third periods are short period which contain 8-8 elements.
• Fourth and fifth periods have 18-18 elements.
• In sixth period total 32 elements are present. From atomic number 58 to 71 are 14 lanthanides which are placed separately below the periodic table.
• Seventh period is incomplete which contains 21 elements. From atomic number 90 to 103, total 14 actinides are also included which are placed separately.
• Period number shows the principal quantum number of outer shell.
• Outer orbit of each period start with ns and ends with ns , np configuration.
• Except first period, all period start with alkali metal and finishes at inert gases.
• 14 lanthanides and 14 actinides in two separate series are placed out-side of the periodic table.

Groups: There is 18 vertical columns in the modem periodic table which are called groups. These groups are expressed by numbers 1, 2, 3,…………….18.
1. All the elements of same group have similar electronic configuration in outermost shell.
2. On the basis of electronic configuration elements are classified into 4 blocks :

• s-block elements,
• p-block elements,
• d-block elements,
• f – block elements.

Question 19.
Compare modern periodic table with Mendeleev’s periodic table (Any 3 points).
Answer:
Comparison between Modern periodic table and Mendeleev’s periodic table

Question 20.
How does atomic size change with the increase in atomic number?
Answer:
In group: In a group on moving below, atomic size increases because electron enters in new orbit (shell). Along with increase in number of shells, effective nuclear charge also increases but increase in number of shells dominates over nuclear charge.

In period: In a period on moving from left to right atomic size decreases because electrons enter same shell and effective nuclear charge increases, which reduce the size.

Question 21.
What changes occur in the reducing and oxidizing properties of elements in periodic table ?
Answer:
Reducing property: Elements which loose electrons to form positive ions are called reducing agents. Reducing nature of any element depend on the nature of an element to loose its electron. Due to larger size of alkali metals, their ionization potential is low.

Therefore, they can easily loose electrons, thus alkali metals are strong reducing agents. On moving from left to right in a period reducing power decreases and on moving from top to bottom, reducing power increases.

Oxidizing property: Elements which gain electron act like oxidizing agents. Oxidiz¬ing nature of an element depends on its tendency to gain electron. Halogens can easily accept electrons therefore, they are strong oxidizing agents. On moving from left to right in a period. Oxidizing power increases and on moving from top to bottom in a group oxidizing power decreases.

Question 22.
Explain Ionization enthalpy (energy) and Electron affinity.
Answer:
Ionization enthalpy: The amount of energy which is required to separate the outermost electron from atom is called ionic energy or ionic potential i.e., to convert atom into positive ion the necessary energy required is called as ionization energy. Its unit is kJ mol^-1.

So, M_(g) + Energy → M(g)⁺e ^– ( ΔH = + ve)

Electron affinity: The electron affinity of an atom is the energy released, when an electron is added to a neutral atom. It is expressed in kcal/mol. Electronegativity is a relative number on an arbitrary scale while electron affinity is expressed in the units of energy (kcal/ mol).

This way in the addition of an electron to neutral isolated atom energy is released and electron affinity is positive, but in an anion energy is required to add an electron because the anion opposes the entrance of an electron. Thus, electron affinity for charged ions is negative.

Question 23.
Explain Atomic radius and Ionic radius.
Answer:
Atomic radius: Atomic size is a very important property of atom, because it is related to various other physical and chemical properties. For the study of atomic size, atom is considered to be spherical. Thus, its radius determines its size.
“Normally the distance between the centre of nucleus, to the outermost shell of an atom is called Atomic radius.”

Ionic radius: The distance of the affected region from the nucleus to the electron cloud is called ionic radius or in an ionic crystal half the distance of the nuclei between two ions is called ionic radius of those ions.

Classification of Elements and Periodicity in Properties Class 11 Important Questions Long Answer Type

Question 1.
What is the significance of the terms: ‘Isolated gaseous atom’ and ‘ground state’ while defining the Ionization enthalpy and Electron gain enthalpy.
Answer:
Ionization enthalpy is the minimum energy required to remove the outermost electron from an isolated gaseous atom (X).
X_(g) → X⁺_(g) + e^–

This force by which electron is attracted by the nucleus, by other atoms present in the molecule or is also affected by other neighbouring atoms. Thus, ionization enthalpy is always determined in gaseous state because in gaseous state intermolecular space is maximum and intermolecular attractive force is minimum. Again, ionization enthalpy is measured at low pressure, because it is not possible to isolate an atom but by lowering the pressure intermolecular force can be reduced. That is why, in the definition of Ionization enthalpy, isolated gaseous atom is added in ground state.
Electron affinity is the energy released when an electron is gained by an isolated gas¬eous atom (X) in ground state to form an anion.
X⁺_(g) + e^– →X_(g)

Most stable state of an atom is ground state. If isolated gaseous atom is in excited state then on gaining an electron comparatively less energy is released. Thus, electron gain en¬thalpy of gaseous atoms is always measured in ground state. Thus, ‘ground state’ and isolated gaseous atom should always be included in the definition of electron gain enthalpy.

Question 2.
In the following pairs of elements which element has more negative electron affinity :
(i) O or F,
(ii) F or Cl.
Answer:
(i) Oxygen and Fluorine both are situated in second period. On moving from left to right in a period electron gain enthalpy goes on becoming more negative. On moving from oxygen to fluorine, along with the increase in atomic number, value of effective nuclear charge increases and size of atom goes on decreasing, due to which attractive force by the nucleus for the incoming electron increases. That is why, electron gain enthalpy of Fluorine is comparatively more negative than the electron gain enthalpy of oxygen. Also Fluorine gains one electron and achieves stable configuration.

Thus, electron gain enthalpy of Fluorine (-328 kJ mol^-1) is more negative than oxygen (-141 kJ mol^-1).
(ii) On moving down in a group, negativity of electron gain enthalpy gradually decreases. But, electron gain enthalpy of Chlorine (-349 kJ mol^-1) is more negative than the electron gain enthalpy of Fluorine (-328 kJ mol^-1). This is due to small size of Fluorine compared to 3p-orbital (Cl), electron-electron repulsion in 2p-orbital (F) is more. Thus, the incoming electron feels more repulsion in Fluorine as compared to Chlorine. That is why, electron gain enthalpy of Chlorine is more negative than Fluorine.

Question 3.
What are the main defects of Mendeleev’s periodic table? How can they be removed by modern periodic table?
Answer:
Defects of Mendeleev’s periodic table :

• Position of hydrogen is indefinite. ,
• Heavier elements were placed before lighter elements.
• Elements with dissimilar properties were kept in the same group.
• Element of similar properties were kept separately in different groups. ‘
• Position of eighth group elements is not suitable.
• Isotopes were not given place.
• No position was given to inert gases.

In modem periodic table, elements are arranged in the increasing order of atomic number by which various defects of Mendeleev’s table were removed.

1. Elements of higher atomic mass were placed before elements of lower atomic mass : Keeping the elements of higher atomic mass before the elements of lower atomic mass. In modem periodic table elements are kept in the increasing order of atomic number due to which the defects on the basis of atomic mass are removed.
2. Position of isotopes: Isotopes of an element possess same atomic number, therefore there is no necessity of giving them separate position in the periodic table.
3. Elements with different properties have been placed in different positions: In modem periodic tables, elements with different properties are kept separate. Like elements of sub-group IA are placed in group I and elements of sub-group IB Cu, Ag, Au are placed in group II.
4. Position of inert gases: In modem periodic table, inert gases are placed between strongly electronegative elements (group 17) and strongly electropositive elements (group I) which acts as a bridge.

Question 4.
What is ionization energy? Explain the factors affecting ionization energy.
Answer:
Ionization energy: The amount of energy which is required to separate the outermost electron from atom is called ionic energy or ionic potential i.e., to convert atom into positive ion the necessary energy required is called as ionization energy. Its unit is kJ mol-1.
So, M_(g) + Energy → M⁺_(g) + e^– (ΔH = +ve)
M_(g) – Electron → M⁺_(g) + Ionisation energy.
Unit: Unit of ionization energy is kJ/mol

Factors affecting ionization energy :
(i) Size of atom or ion : Greater the size of atom or ion, weaker are the forces of attraction and lower is the value of ionization energy.
(ii) Nuclear charge: Greater the nuclear charge, more is the attraction for electrons and hence, greater is the value of ionization energy.

(iii) Shielding effect: In multielectron atom with increase in atomic number, shielding effect increases due to which valence shell electron feels lesser attraction and hence, value of ionization energy is lower.

(iv) Penetration effect: Closer the orbitals to the nucleus, more is the penetration of it for the nucleus. That is it experience greater attraction. Thus, value of ionization energy is higher, s electron of the same shell is strongly held as compared to p, d,f. Thus, order of ionization energy is s>p> d>f.

(v) Half filled and fully filled orbitals: Completely filled and half-filled orbitals are more stable than any other arrangement. Thus, value of ionization energy is higher for it.

(vi) Configuration of outermost shell: Elements which have 8 electrons in their outermost shell (except He) are more stable and their ionization energy is very high.

(vii) Periodicity:
(a) Periodicity in group: On moving from top to bottom in a group, atomic radius increases due to which less energy is required to remove the electron from the valence shell. Thus, on moving from top to bottom in a group ionization energy decreases.
(b) Periodicity in period: On moving from left to right in a period, due to increase in nuclear charge and decrease in atomic radius, more energy is required to remove an electron from the valence shell. Thus, on moving from left to right in a period value of ionization energy increases.

Question 5.
What do you understand by Atomic radius and Ionic radius?
Answer:
Atomic radius: Atomic radius means size of atom. It can be measured by X-rays or other spectroscopic methods. For non-metals it is also known as covalent radius and for metal elements it is known as metallic radius.

Covalent radius: Half the distance between the nucleus of two atoms bonded by a covalent bond in a molecule is called covalent radius. For example, for Chlorine molecule bond distance is 198pm, thus half of this, 99pm is the covalent radius or atomic radius of Chlorine atom.

Ionic radius: Ionic radius means size of ion (cation or anion). It can be measured by measuring the distance between the cation and anion in an ionic crystal. A cation is always smaller than its parent atom because due to removal of one or more electron, effective nuclear charge increases. Anion is always larger than the present atom because due to gain of electron value of effective nuclear charge decreases. For example, Ionic radius of Na+ ion (95pm) is less than atomic radius of Na atom (186pm) whereas ionic radius of F^– (136pm) is more than atomic radius of F (72pm).

Question 6.
What do you understand by electron affinity? Explain the factors affecting it
Answer:
Electron affinity: The electron affinity of an atom is the energy released, when an electron is added to a neutral atom. It is expressed in kcal/mol. Electronegativity is a relative number on an arbitrary scale while electron affinity is expressed in the units of energy (kcal/mol).

A_(g)+e^– → A^–_(g) + E₁ (Exothermic)
A^–(g) + e^– → A^2- – E₂ (Exothermic)
A^2-(g) + e^– → A^3-(g) – E₃(Exothermic)

This way in the addition of an electron to neutral isolated atom energy is released and electron affinity is positive, but in an anion energy is required to add an electron because the anion opposes the intrance of an electron. Thus, electron affinity for charged ions is negative.

Factors affecting electron gain enthalpy: Some important factors affecting electron gain enthalpy are:
(i) Atomic size: As the atomic size increases, the distance between the nucleus and the incoming electron increases. This results in lesser attraction. Consequently, electron gain enthalpy becomes less negative.

(ii) Effective nuclear charge: Greater the nuclear charge, greater will be attraction for the incoming electron and as a result, Δ_eg H becomes more negative.

(iii) Outermost electronic configuration: Atoms having stable electronic configuration have lesser tendency to accept the electron. Due to this the value of electron gain enthalpy becomes less iiegative.

(iv) Half filled and fully filled orbitals: Half filled and fully filled orbitals are more stable and does not easily accept an electron in this state. Therefore, electron affinity of such elements is zero.

(v) Periodicity :
(a) Periodicity in period: In a period, on moving from left to right, value of electron affinity generally increases because nuclear charge increases and atomic size decreases. As a result, attraction of nucleus for the accepted electron increases.
(b) Periodicity in group: On moving from top to bottom in a group; due to increase
in atomic size attractive force between the nucleus and incoming electron decreases, due to which electron affinity decreases.

Question 7.
In periodic table on moving left to right, what change in following proper-ties occurs:
(i) Atomic radii,
(ii) Electronegativity,
(iii) Ionization potential,
(iv) Metallic character,
(v) Valency.
Answer:
(i) Atomic radii: On moving from left to right in the period, the atomic radii decreases because electron enter in same shell and due to increase in nuclear charge atomic size shrinks. For example, in 3rd period, Na atom has largest size while Cl atom is shortest.

(ii) Electronegativity: In a period on moving towards right side electronegativity increases due to decrease in atomic size and increase in nuclear charge.

(iii) Ionisation potential (I.P.): When we move in a period from left to right ionisation potential of atoms also increases. It is due .to increase in nuclear charge and decrease in atomic size.

(iv) Metallic character: Metallic character decreases on moving from left to right in a period.
(v) Valency: On moving left to right in a period, valency of element with respect to oxygen increases from 1 to 7 while in respect of hydrogen or chlorine it increases from 1 to 4 and then reduces up to 1.

Question 8.
The first (Δ_iH₁) and the. second (Δ_iH₂) ionization enthalpies (in kJ mol^-1) and (Δ_egH) electron gain enthalpy (in kJ mol^-1) of a few elements are given below :

Which of the above elements is likely to be :

• The least reactive element
• The most reactive metal
• The most reactive non-metal
• The least reactive non-metal
• The metal which can form stable binary halide of the formula MX₂ (X = halogen).

Question 9.
Which metal is alkali and which forms stable covalent compound.
Answer:
(i) The least reactive element is (V) because its first ionization enthalpy is maximum and electron gain enthalpy is positive. Electron gain enthalpy of inert gas element is positive. Values given for element (V) corresponds to He.

(ii) Most reactive metal is element (II). Its first ionization enthalpy is least and electron gain enthalpy is less negative. The values given for element (II) corresponds to K.

(iii) Most reactive non-metal is (III) because its first ionization enthalpy is highest and electron gain enthalpy is maximum negative. Values given for element (III) corresponding to F.

(iv) Least reactive non-metal is (IV) because its first ionization enthalpy is not high but electron gain enthalpy is more negative. Values given for element (IV) corresponds to I.

(v) Metal which forms stable binary halide (MX₂) is (VI). Element (VI) is alkaline earth metal because its first ionization enthalpy is less but is more than Alkali metals (II). Secondly, difference between first and second enthalpies is very less. Values given for element (VI) corresponds to Mg.

(vi) First ionization enthalpy of element (I) is low but second is high. Its electron gain enthalpy is less negative. Thus element (I) is alkali metal. Values given for element (I) corresponds to Li. Li predominantly forms stable covalent compound of formula MX.

Classification of Elements and Periodicity in Properties Class 11 Important Questions Objective Type

1. Choose the correct answer:

Question 1.
Which of the following electronic structures is of a metal:
(a) 2, 8, 8
(b) 2, 7
(c) 3, 8, 2
(d) 2,8,4.
Answer:
(c) 3, 8, 2

Question 2.
Which of the following has the highest ionization potential:
(a) Na
(b) Mg
(c) C
(d) F.
Answer:
(d) F.

Question 3.
Which of the following is most electronegative :
(a) Oxygen
(b) Chlorine
(c) Nitrogen
(d) Fluorine.
Answer:
(d) Fluorine.

Question 4.
With increase in atomic mass in a period in periodic table there is :
(a) Increase in property of electropositivity
(b) Increase in property of electronegativity
(c) Increase in chemical activity
(d) Decrease in chemical activity.
Answer:
(b) Increase in property of electronegativity

Question 5.
Of the following, whose ionic radii is largest:
(a) K⁺
(b) O²⁺
(c) B²⁺
(d) CT.
Answer:
(a) K⁺

Question 6.
Which of the following is the electronic configuration in last orbit of a noble gas:
(a) s²p³
(b) s²p⁴
(c) s²p⁵
(d) s²p⁶.
Answer:
(d) s²p⁶.

Question 7.
Those elements in which s and p orbitals are complete :
(a) Normal elements
(b) Transition elements
(c) Inert gas
(d) Halogens.
Answer:
(c) Inert gas

Question 8.
Which of the following has bigger size :
(a) Br
(b) F
(c) I
(d) I.
Answer:
(d) I.

Question 9.
Which of the following is of smallest size :
(a) Al
(b) Al⁺
(c) Al²⁺
(d) Al³⁺.
Answer:
(d) Al³⁺.

Question 10.
Which of the following is a strong reducing agent:
(a) F^–
(b) Cr^–
(c) Br^–
(d) I^–.
Answer:
(d) I^–.

Question 11.
The element having highest value of first ionization potential:
(a) Boron
(b) Carbon
(c) Nitrogen
(d) Oxygen.
Answer:
(c) Nitrogen

Question 12.
Which of the following has highest value of ionization potential:
(a) Ca
(b) Ba
(c) Sr
(d) Mg
Answer:
(d) Mg

Question 13.
An element having atomic number 20, will be placed in periodic table in the group:
(a) 4
(b) 3
(c) 2
(d) 1.
Answer:
(c) 2

Question 14.
Chromium [atomic number = 24] is a :
(a) Transition element
(b) Inert gas element
(c) Normal element
(d) Inner transition element.
Answer:
(a) Transition element

Question 15.
Ionization potential of elements in a group of periodic table:
(a) Increases with atomic size
(b) Decreases with increase in atomic size
(c) Remains constant
(d) Changes irregularly.
Answer:
(b) Decreases with increase in atomic size

Question 16.
Metallic character on moving from left to right in a periodic table :
(a) Increases
(b) Decreases
(c) Remains constant
(d) First decreases then increases.
Answer:
(b) Decreases

2. Fill in the blanks:

1. Electron gain enthalpy of noble gases is …………….. .
Answer:
zero

2. Electron gain enthalpy of Be and Mg in second group is ……………… .
Answer:
zero

3. On moving from left to right in a period ionization energy …………….. .
Answer:
increases

4. Second ionization potential of any element is always …………….. than the first ionization potential.
Answer:
higher

5. Most electronegative element is placed in …………….. group of the periodic table.
Answer:
VII A

6. Similarity in properties of Li and Mg, and Be and A1 is due to …………….. relationship.
Answer:
diagonal

7. In modem peiodic table, there are total …………….. columns and …………….. periods.
Answer:
18,7

8. With the increase in nuclear charge, atomic radius in a period …………….. .
Answer:
decreases

9. Electron affinity of chlorine is …………….. than fluorine.
Answer:
more

10. Newland stated that if elements are arranged in increasing order of elements then every …………….. element which show properties similar to the first element.
Answer:
eighth.

3. Match the following:

Answer:
1. (d) Neon
2. (c) Fluorine
3. (b) Potassium
4. (a) Carbon.

4. Answer in one word/sentence:

1. Other name of Eka silicon.
Answer:
Germanium

2. Number of electrons in N^3- and O^2- is same. Thus they are called ……… .
Answer:
Isoelectronic

3. Which is the most stable oxidation state of aluminium?
Answer:
+3

4. In the periodic table, the most electronegative element is.
Answer:
Fluorine

5. Atomic number of an element is 16. In which group of the periodic table this element is present?
Answer:
Sixth

6. How many elements are present in the first period of the periodic table?
Answer:
Two.

These MP Board Class 11th Chemistry Notes for Chapter 7 Equilibrium help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 7 Equilibrium

→ Equilibrium : Dynamic state in which velocities of forward reaction and backward reaction are equal.

→ Physical equilibrium : Equilibrium related to physical reactions.

→ Reversible reaction : The reaction which proceeds in forward and backward directions.

→ Chemical equilibrium : The state of a revesible reaction at which velocities of forward and backward reactions are same.

→ Law of mass action: The rate of reaction of any substance is proportional to its active mass at constant temperature. The rate of chemical reaction is proportional to the product of active masses of reactants.

→ Equilibrium constant K_c = \(\frac{[\text { Product }]}{\text { [Reactant }]}\)

→ Relation between K_c and K_p : K_p = K_c (RT)_Δn

→ Le-Chatelier’s principle : At equilibrium, when temperature, pressure or concentration of any system- is changed, the equilibrium* * shifts in the direction, so that effect of change is minimized.

→ Degree of ionisation: Ratio of number of molecules dissociated and total number of molecules.

→ Weak electrolyte : Electrolyte whose ionisation is less.

→ Ostwald’s dilution law: Degree of dissociations of weak electrolyte is proportional to square root of dilution.
K_α = \(\) K_α = Cα² or α = \(\sqrt{\frac{\mathrm{K}_{a}}{\mathrm{C}}}\)or α = \(\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{V}}\)
(1 – α) = 1 (for weak electrolyte) ,

→ Arrhenius concept of acid-base: The substances producing H+ ions in aqueous solution are acids and the substances which produce OH- ions in aqueous solution are bases.

→ Bronsted-Lowry concept of acid-base : Proton donors are acids while proton acceptors are bases.

→ Lewis theory: Bases are the species which can donate lone pair of electrons while electron
pair acceptors are acids.

→ Ionic product of water : K_w = [H⁺][OH^–] = [H₃O^{+][OH– ] = 1 x 10-14}

→ pH value : Negative logarithm value of molar concentration of H⁺ ion or H₃O⁺ (in aqueous solution) as base 10 is called pH value.
pH = -log₁₀[H+]

→ Solubility product: At constant temperature; product of molar concentration of ions in a saturated, solution of solute.

→ Henry’s law: The mass of gas dissolved in definite quantity of solvent at constant temperature is proportional to pressure of gas.
m x P or m = KP m = Mass of gas per unit volume
P = Gas pressure.

→ Salt formed by strong base and weak acid : CH₃COONa, Na₂CO₃, Na₃PO₄, K₂CO₃ pH = 7 + \(\frac { 1 }{ 2 }\) [pK_a + log C]

→ Salt formed by strong acid and weak base : CaCl₂, NH₄NO₃, NH₄Cl, (NH₄)₂SO₄, AlCl₃ PH = \(\frac { 1 }{ 2 }\) [pK_w – pK_b – log C]

→ Salt formed by weak acid and weak base : CH₃COONH₄, (NH₄)₂CO₃, Ca₃(PO₄)₂

→ Lewis base : Species which possess lone electron pair.

→ Lewis acid : Species whose central atom does not possess octet.
Like AlCl₃, BF₃, FeCl₃, Ag+, H+, CO₂, SO₂, NH₄, BCl₃ etc.

→ Water is an amphoteric solution.

These MP Board Class 11th Chemistry Notes for Chapter 8 Redox Reactions help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 8 Redox Reactions

→ Redox reaction : When both oxidation and reduction reactions occur simultaneously.

→ Oxidising agent: Substance which accept electron and get reduced.

→ Reducing agent: Substance which donate electron and get oxidised.

→ Electrode potential: Potential developed on electrodes by oxidation or reduction. Its value is different for different electrode.

→ Electromotive force (EMF): Potential difference of both electrodes in the cell.

→ Nernst equation : Equation which express relation of electrode potential with ionic concentration and temperature.
E = E°+ \(\frac{\mathrm{RT}}{n \mathrm{~F}}\)ln[Mⁿ⁺(aq)]

→ Oxidation: Addition of oxygen/ electronegative element in a substance or removal of hydrogen/ electropositive element.

→ Reduction : Removal of oxygen/electronegative elements from a substance or addition of hydrogen/electropositive element.

→ Reducing agent: Species which is oxidized.

→ Oxidising agent: Species which gets reduced is called oxidizing agent.

→ Redox reactions are classified into four groups :
1. Addition , 2. Decomposition, 3. Displacement, 4. Disproportionation reaction.

→ Disproportionation reaction: Such redox reaction in which a single species undergo oxidation and reduction. Example:

In such reactions in active species a single element obtains at least three oxidation state.

→ Increase in oxidation state leads to oxidation and decrease in oxidation state leads to reduction.

→ Oxidation number of an element can never be more than its group number.

→ Electrochemical cell: It is such a combination in which oxidation and reduction indirectly as half equations and in the reaction, chemical energy gets converted to electrical energy.

→ Electrolytic cell: It is such an arrangement in which electrical energy is supplied from an external source to produce chemical reaction.

→ Electrode potential (E) : Potential difference between metal electrode and metal ion in which the electrode is dipped is called electrode potential.

→ Standard Electrode Potential (E°): If Pressure is 1 atm and concentration of electrolyte is 1 M then electrode potential is known as standard electrode potential (E°). Lower the value of E°, stronger reducing is the electrolyte. Higher the E° value, stronger is the oxidizing nature. The table of orderly arrangement of elements on the basic of their E° values is called Electro chemical series.

→ Electromotive force (EMF) or cell potential: The potential difference between two half cells of a cell is called cell potential
E°_cell = E°_cathode – E°_anode

→ If E°_cell is positive, the reaction proceeds in forward direction. If Ecell is negative the reaction proceeds in backward direction.

→ Redox couple : The pair formed in presence of oxidized and reduced forms of substances participating in oxidation and reduction half reactions is called redox pair.

These MP Board Class 11th Chemistry Notes for Chapter 9 Hydrogen help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 2 Structure of Atom

→ Hydrogen : Represents similarity with both alkali metals of I-group and halogens of VH-group.

→ Hydrogen has three isotopes : (1) Protium (₁H¹), (2) Deuterium ((₁H²), (3) Tritium (₁H³).

→ Ortho hydrogen : Nuclei of both hydrogen atoms of hydrogen molecule spin in the same direction.

→ Para hydrogen : Nuclei of both hydrogen atoms of hydrogen molecule spin in the opposite direction.

→ Water : Exist in three states on earth : (1) Solid, (2) Liquid, (3) Vapour.

→ Structure of water molecule : Angular V-shape, H – O – H bond angle is 104°27′. Density of ice less than water and maximum density of water is at 4°C.

→ Hardness of water: Hardness of water is of two types : (1) Temporary hardness, (2) Permanent hardness.

→ Temporary hardness : Due to soluble bicarbonates of Ca and Mg.

→ Permanent hardness : Due to soluble chlorides and sulphates of Ca and Mg.

→ Hard water : Produce less lather slowly with soap.

→ Heavy water : D₂0, as a neutron moderator.

→ Ionic hydrides are also known as Saline hydrides.

→ Tritium (₁H³) is radioactive in nature. It emits low energy β-particles (t_1/2 = 12-33).

→ Hydrogen is prepared in the laboratory by the action of dilute HCl on granulated zinc.

→ Bond enthalpy of H – H bond in H₂ is 435.88 kJ mol^-1, which is maximum in single bond.

→ Hydrogen combines with elements to form hydrides. Hydrides are of three series :
(i) Ionic or Saline hydrides, (ii) Covalent or Molecular hydrides, (iii) Metallic or Interstitial hydrides.

→ Metallic hydrides are used in the purification of dihydrogen and in storage of hydrogen and as an energy source.

→ Dihydrogen is used in the manufacture of ghee from oil, in reduction in metal extraction, as a rocket fuel.

→ Dihydrogen is the basis of hydrogen economy, transport and storage of hydrogen as a liquid or gas H2 in the form of energy. Main advantage of hydrogen economy is the observation of H₂ in the form or energy. Presently, it is used in the manufacture of electrical energy and in fuel cells,

→ High heat of evaporation and heat capacity of water is responsible for maintaining the body temperature of organisms and climatic conditions.

→ Due to the formation of hydrogen bond with polar molecules, covalent compounds like alcohol, carbohydrates are also soluble in water.

→ Due to bent structure molecules of water are extremely polar.

→ Hydrogen bond is maximum in ice and minimum in water vapour.

→ Water which forms lather with soap is called soft water. Water which does not give lather is called hard water.

→ Temporary hardness of water is removed simply by boiling.

→ Treatment of permanent hardness of water is done by washing soda method, calgon process, ion exchange method or synthetic resin process.

→ 30% solution of H₂O₂ is known as perhydrol which is used as an insecticide and antiseptic.

→ 20% volume H₂O₂ means by the decomposition of 1 ml of it at NTP 20 ml oxygen is obtained.

→ H₂O₂ behaves as an oxidizing, reducing and bleaching agent. Its bleaching property is due to oxidation.

→ H₂O₂ is stored in plastic or glass container sealed with wax in the dark because it decomposes by the effect of light. Presence of surface of metal and minute quantity of base catalysis its decomposition.

These MP Board Class 11th Chemistry Notes for Chapter 10 The s-Block Elements help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 2 Structure of Atom

→ General electronic configuration of j-block elements is ns^1-2.

→ IA group elements are called Alkaline earth metals. Its general electronic configuration is ns¹.

→ Order of basicity of oxides : Li₂O < Na₂O, K₂O < Rb₂O < Cs₂O.

→ Order of stability of carbonate : Li₂CO₃ < Na₂CO₃< K₂CO₃ < Rb₂CO₃ < Cs₂CO₃.

→ s-block elements are metals and strongly electropositive.

→ 2nd group elements are known as Alkaline earth metals. Their general electronic configuration is ns².
Oxidation number of Alkali metals is +1 and oxidation number of Alkaline earth metals is +2.

→ s-block elements are strongly reducing.

→ Li represents diagonal relationship with Mg and Be with Al.

→ Carbonates of Alkali metals are soluble in water whereas Alkaline earth metals are insoluble in water.

→ Alkali metals combine with halogens to form halides, these are ionic crystalline solids. Sodium is a highly reactive metal, it is stored in kerosene oil.

→ Alkali metals are paramagnetic but their compounds are diamagnetic.

→ Magnesium sulphate is known as Epsom salt.

→ Calcium sulphate is known as Gypsum salt.

→ Order of stability of metal hydroxides is as follows :
CsOH > RbOH > KOH > NaOH > LiOH
Ba(OH)₂ > Sr(OH)₂ > Ca(OH)₂ > Mg(OH)₂

→ Order of stability of IInd group metal carbonates is as follows :
BeCO₃ < MgCO₃ < CaCO₃ < SrCO₃

→ Order of stability of halides is as follows : .
MF > MCI > MBr > MI

→ Order of stability of metal sulphates is as follows :
BeSO₄ > MgSO₄ > CaSO₄ > SrSO₄ > BaSO₄

→ Density of Alkali metals decreases from Li to Cs. Though weight of K < weight of Na. → Hydration energy of Alkali metal decreases on moving down in the group. → Its value increases from left to right in a period. → Heat of hydration of IInd group > Heat of hydration of Ist group.

→ Order of conductivity of ions of Alkali metals in aqueous solution is as follows :
Li⁺ < Na⁺ < K⁺ < Rb⁺ < Cs⁺ < Fr⁺

→ Although ionization potential of Li is highest, still due to very high hydration energy, it is a strong reducing agent.

→ Compounds of Alkaline earth metals are less ionic than the compounds of Alkali metals.

→ Solution of Alkali metals in liquid ammonia, is a conductor of electricity. Blue colour of the solution is due to ammoniated electrons. In concentrated solution, blue colour changes to bronze colour.

→ Alkali metals form normal oxides, peroxides and superoxides. Their order of stability is Normal oxide > Peroxide > Superoxide

→ Oxides of Alkali metals react with water to form alkali metal hydroxides. Due to high hydration energy these are soluble in water and large amount of energy is released.

→ Na₂CO₃ is known as soda ash and Na₂CO₃.10H₂O is known as washing soda or salt soda. It is manufactured by Solvay process.

→ 8% NaCl (according to mass) aqueous solution is known as brine.

→ NaHCO₃ is known as Baking soda.

→ Some difference in properties of Alkaline earth metals with Alkali metals is due to their small atomic or ionic radius and high positive charge.

→ Plaster of Paris (POP) [CaSO₄. \(\frac { 1 }{ 2 }\) H₂O] is obtained on heating Gypsum(CaSO₄.2H₂O) to 120°C.

→ Normal composition of Portland cement is 70% CaO, 18% SiO₂, 5% Al₂O₃, 3% Fe₂O₃ and 2% MgO. To slow down its setting process some amount of gypsum is mixed in it.

These MP Board Class 11th Chemistry Notes for Chapter 11 The p-Block Elements help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 11 The p-Block Elements

→ General electronic configuration of group-13 (Boron family) is ns²np¹.

→ Boron is a non-metal whereas other members of this group are metals.

→ Aluminium exists in maximum amount in earth’s crust.

→ Boron represents abnormal behaviour than other members of the group.

→ Boiling and melting point of Boron is higher than the other members of the group.

→ Oxides and hydroxides of Boron represent acidic behaviour whereas A1 represents amphoteric nature and other oxides are basic.

→ Boron forms covalent compounds whereas other members of the group show ionic behaviour.

→ Except Boron, other members of the group exhibit +1 and +3 oxidation state.

→ Due to inert pair effect, on moving down the group, stability of +3 oxidation state decreases but stability of+1 oxidation state increases.

→ Boron forms more than one type of hydrides. Mainly there are two series of hydrides whose general formula is B_nH_n+4 and B„H_n+6.

→ Halides of Boron act as strong Lewis acids.

→ Aluminium chloride acts as a dimer.

→ Boron is the hardest element after diamond.

→ Borazine is known as inorganic benzene.

→ In diborane, a two electron-three centred bond is present which is known as banana bond.

→ Aluminium sulphate forms a double salt with a monovalent metal sulphate which are known
as Alum. .

→ Group-14 of p-block is known as carbon family because the first element is carbon and its general electronic configuration is ns²np².

→ Of the five elements of this group C and Si are non-metals, Ge is metalloid and Sn and Pb are metals.

→ Diamond: It is a purest crystalline allotrope, very hard, high melting point and bad conductor of electricity and in it carbon is in sp³ hybridized state.

→ Graphite : Good conductor of electricity, soft, crystalline allotrope of carbon with lubricant property. In this sp² hybridized carbon forms a hexagonal ring.

→ Crystalline forms of carbon : (1) Charcoal, (2) Coke, (3) Gas carbon.

→ ¹⁴₆C of carbon is a radio-active.isotopes.

→ All members of carbon family exhibit +2 and +4 oxidation state. On moving down the group

→ Cause of stability of +2 oxidation state inert pair effect.

→ Silicon exhibits diagonal relationship with Boron.

→ Carbon forms three types of series : (1) Alkane, (2) Alkene, (3) Alkyne.

→ Carbon establishes covalent bond with metals to form organo-metallic compounds.

→ Three crystalline forms of carbon are : (1) Diamond, (2) Graphite, (3) Fullerene.

→ Fullerene is the third crystalline form of carbon whose formula is C₆₀, C₈₄, C₇₀ or C₆₀ is known as Buckminster Fullerene or Bucky ball.

→ Silica (SiO₂) is hard, which forms three-dimensional crystalline network. It exists in the form of quartz or sand.

→ Basic unit of Silicate is SiO₄^4- ion which has tetrahedral structure.

→ Orthosilicate (SiO₄^4-), Pyrosilicate (S₄O_4- ), Cyclic silicate (SiO₃)_n²ⁿ , Chain silicate (SiO₃)_n²ⁿ and Sheet silicate (Si₂O₅)_n²ⁿ are basic units.

These MP Board Class 11th Chemistry Notes for Chapter 12 Organic Chemistry: Some Basic Principles and Techniques help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 12 Organic Chemistry: Some Basic Principles and Techniques

→ Isomerism : Two or more compounds which are different in their properties but have same molecular formula, are called isomers and this phenomenon is called isomerism.

→  Types:

1. Chain isomerism : When difference exists in chain of carbon atoms.
2. Position isomerism : When functional group is attached in different positions in same chain.
3. Functional isomerism : When functional groups are different.
4. Metamerism : Functional group is same but alkyl groups attached with it are different.
5. Tautomerism : Special type of functional group isomerism in which both the isomers of a compound are in dynamic equilibrium.
6. Ring chain isomerism : Structure of compound is open chain or cyclic.
7. Optical: Formation of enantiomers of different optical properties by chiral molecules, (viii) Geometrical: Formation of cis-trans isomers due to restricted rotation in compounds of C = C double bond.
8. Conformation : Formation of various spatial arrangements due to free rotation around C—C single bond.

→ Alkane : Saturated hydrocarbon, paraffin (less reactive) general formula C„H2)1+2, regular tetrahedral structure, H—C—H bond angle 109°28′ and sp3 hybridization in C-atom.

→ Conformation in ethane : Two forms (i) Eclipsed : In which H-atoms attached to both carbon are in front of each other.
(ii) Staggered : In which H-atoms of both carbon are in between each other. Low energy and more stable.

→ Conformation of cyclohexane: Two forms (i) Chair or staggered and (ii) Boat or eclipsed.

→ Chirality : Such compounds whose mirror images are non-superimposed.

→ Asymmetric carbon atom : Such a carbon atom to which four different groups are linked.

→ Optical isomers: Compounds which rotate the plane polarized to opposite direction and are mirror image of each other.

→ d – or (+) form : Which rotates the plane polarized light to the right.
Organic compounds of carbon : Compounds of carbon and hydrogen or their derivatives which have carbon-carbon bond.

→ Alkyl radicals : Part of molecule after ignoring one hydrogen atom in paraffin hydrocarbon.

→ Functional group : Atom or group of atoms determining the general chemical properties of the organic compound.

→ Homologous series : Series of compounds having common functional group and composed by the difference of —CH₂

→ Nomenclature of organic compounds : IUPAC system.

→ l-(or) (-) form : Which rotates the plane polarised light to the left.

→ Racemic mixture : Equimolar proportion of d- and l-form is optically inactive.

→ Inductive effect is pennanent displacement of electrons along the chain of C-atoms.

→ Electromeric effect is temporary effect and occurs at the requirement of attacking species.

→ Resonance effect results in generation of centres of high and low electron density in a mol-ecule.

→ Free radical: Obtained by heterolytic cleavage. No charge but contain unpaired electron.

→ Carbocation : Positively charged carbon atom or group.

→ Carbanion : When negative charge is on carbon atom.

→ Meso form : Molecule with one or more chiral centre which is optically inactive due to internal arrangement.

→ Alkene : Carbon atoms of double bond is sp² hybridized. One cr and 7t-bond is in between carbon. These can show chain, position and geometrical isomerism. General formula C_nH_2n

→ Alkyne : Carbon atoms of triple bonds are sp hybridized and are linear molecules with bond angle 180°. General formula C_nH_2n-2. These show chain and position isomerism.

→ Arene : Aromatic hydrocarbons which contain benzene ring. Each carbon atom in benzene is sp² hybridized in which each bond angle is 120°.

→ Isomerism in Arenes : They have position isomerism. Disubstituted benzene occur in three isomeric forms (ortho, meta and para).

→ Crystallisation:
(a) Simple crystallisation : When hot saturated solution of any substance in a suitable sol-vent is cooled slowly, solid crystalline pure substance is obtained.
(b) Fractional crystallisation : Substances of different solubilities crystallise one by one from the solution which contains two or more than two substances as solute. Substance with least solubility crystallise first.

→ Sublimation : On heating, solid substance directly changes into gaseous state without chang-ing into liquid state and vice-versa.

→ Distillation : The process in which a liquid changes into vapour state and again it condenses back to liquid state. ’

→ Fractional distillation : Mixture of liquids which differ sufficiently in their melting points distil at different temperature. The liquid with low melting point distil first.

→ Steam distillation : Liquids having boiling point less than 100°C and immiscible with water, distil with steam. When condensed, different layers can be separated by separating funnel.

These MP Board Class 11th Chemistry Notes for Chapter 13 Hydrocarbons help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 13 Hydrocarbons

Main methods of preparation of alkanes :

→ Main methods of preparation of alkenes :

→ Main methods of preparation of alkynes:

→ Physical properties of alkanes :

1. Boiling point increases with increase in molecular mass, b.p. of branched chain compounds are lower than that of straight chain compounds.
2. Melting points also increases with increasing molecular mass. Compounds having odd number of carbon atoms have lower value of m.p.
3. Insoluble in water or polar solvents but soluble in non-polar solvents like ether, benzene, etc.
4. Generally, density increases with increasing molecular mass.

→ Chemical properties of hydrocarbons;

• Combustion: Exothermic reactions, give CO₂ and water vapours.
• Alkenes and alkynes decolorize alcoholic KMnO₄ solution.
• Alkyl benzene oxidize into benzoic acid.
• Unsaturated hydrocarbons give addition reactions (H₂, Cl₂, HCI, etc.).
• Addition of HX in unsaturated hydrocarbons is according to MarkownikofFs rule but opposite in presence of peroxide.
• Alkenes, alkynes and arenes ozonolyze and form carbonyl compounds.
• 
• 
• Polymerization: (a) Ethylene → polyethylene, (b) Propylene → polypropylene, (c) Isoprene → polyisoprene, (d) Vinyl chloride → polyvinyl chloride (PVC), (e) Styrene → polystyrene, (f) Tetrafluoroethene → teflon (PTFE).

(x) Substitution reactions: (a) In alkanes substitution of all hydrogen atoms by halogen atoms take place in presence of sunlight.
(b) Benzene gives halogenation, nitration, sulphonation, alkylation and Friedel-Crafts reaction.
(c) Acetylene and other alkynes give substitution reactions with silver(I) and Cu(I) ions in ammoniacal medium and white and red coloured precipitates of acetylides are formed.

→ Acetylene and others : Alkynes undergo substitution reactions in ammoniacal medium with Silver(I) and Copper(II) ions and form white and red precipitate of acetylides. The ≡ C—H group is of acidic nature.

→ Methods of preparation of Benzene :

→ Alkanes undergo free radical substitution reactions.

→ Alkenes and Alkynes undergo electrophilic addition reaction.

→ Main chemical reaction of benzene is electrophilic substitution reaction.

→ Conformation: Due to free rotation around C – C (single bond), formation of different special arrangements.

→ Alkanes : Saturated hydrocarbons or paraffins, general formula C_nH_2n+2tetrahedral structure, sp³ hybridization, H – C – H bond angle is 109° 28′.

→ Alkenes : Unsaturated hydrocarbons whose general fonnula is C_nH_2n.  Carbon-carbon double bond is present.

→ Alkynes : Unsaturated hydrocarbons whose general formula is C_nH_{2n – 2}. Carbon-carbon triple bond is present.

→ Chirality : Compounds not superimposable on their mirror images.

→ Unsymmetrical carbon atom : Such carbon atom to which four different atoms or groups are linked.

→ Optical isomers : Mirror image isomers which rotate the plane of polarised light in opposite direction equally.

→ Racemic mixture : Mixture of equal amount of d and f forms which is optically inactive.

→ Meso form : Symmetrical molecule with more than one chiral centre which is optically inactive.

→ Markownikoff’s rule : The negative part of polar molecule adds to that carbon atom of an unsymmetrical double bond which contains less number of hydrogen atoms.

→ Kharasch effect: Addition reaction on unsymmetrical alkene in presence of peroxide occurs against Markownikoff’s rule.

→ Huckel’s rule : Compound behaves as Aromatic compound if it contains (4n + 2) π-electrons, mete directing groups : — CHO, – COOH, – NO₂, – C ≡ N,- SO₃H, — COR, — COOR. ortho and para directing groups: – OH, – NH₂, – X, – OR, – R, – NHR, – NHCOCH₃, – CH₃, – C₂H₅.

→ meta directing groups makes the ring inactive towards electrophilic substitution reactions.

→ ortho and para directing groups make the ring active towards electrophilic substitution reagents.

→ Dienes : These are unsaturated hydrocarbons. They contain carbon-carbon double bond.

→ Hydrocarbons are compounds of carbon and hydrogen only. They are mainly obtained from coal and petroleum which are main sources of energy.

→ Liquefied Petroleum Gas (LPG) and Compressed Natural Gas (CNG) which is a domestic fuel and are main source of energy for automobiles. These are also obtained from petroleum.

→ Classification of hydrocarbons on the basis of structure :

These MP Board Class 11th Chemistry Notes for Chapter 14 Environmental Chemistry help students to get a brief overview of all the concepts.

MP Board Class 11th Chemistry Notes Chapter 14 Environmental Chemistry

→ Substances produced by natural causes or by human activities which cause harmful effect on the environment are called pollutant.

→ Such substances which when present in the environment pollute it, they are called contaminant.

→ Air pollutants are maximum in Troposphere.

→ Ozone layer situated in stratosphere prevent us from ultraviolet rays.

→ Oxides of nitrogen and sulphur cause acid rain.

→ CFC is the main cause of depletion of ozone layer.

→ The heat radiations (IR rays) coming from the sun are reflected from the earth’s surface and absorbed by gases like CO₂. This increases the temperature and is called green house effect.

→ Along with CO₂ CFC also cause green house effect.

→ 111 effect of green house effect is Globa! warming.

→ Pollutants are of two types :

• Degradable pollutants : Pollutants which can easily break down by micro-organisms. Example : Discarded vegetables, cow dung.
• Non-degradable pollutants : Pollutants which do not break down by micro-organisms. Example : DDT, plastic, heavy metals, chemicals etc.

→ Tropospheric pollution : It occurs due to presence of undesirable solid or gaseous particles in air. Gaseous air pollutants are oxides of S, N and C, H₂S, hydrocarbons, ozone and other oxidants. Particulate pollutants are dust, mist, fumes, smoke, smog etc.

→ Vaporized material formed by minute carbon particles, obtained by combustion of a substance is called smoke.

→ Thin layer of fog, smoke and dust is called Smog. It is harmful mixture of smoke and fog.

→ Smog is of two types : (i) Classical smog, (ii) Photochemical smog.

→ Fine solid particles present in air whose size range from 01 µ to 25 µ are called dust.

→ In Green house system allow low wavelength solar radiation to enter but does not allow higher wavelength infrared radiations to go out.

→ Main green house gases are CO₂, CH₄, N₂O and CFCs.

→ Main components of Acid rain are H₂SO₄ and HNO₃. When pH of acid rain becomes less than 5-6, it is known as Acid rain.

→ Main source of soil pollution are : Detergent, poisonous metals, insecticides, acid, base, salt, insects, pharmaceutical waste.

→ Bio-chemical oxygen demand (BOD). Amount of oxygen required by bacteria to break down the organic matter (at 206° for 5 days) present in a certain volume of water sample is called B.O.D.

BOD = \(\frac{\text { Number of milligrams of } \mathrm{O}_{2} \text { required }}{\text { Volume of sample (in litres) }}=\frac{\text { Parts of } \mathrm{O}_{2}}{\text { Million part of sample }}\)

BOD for clean water is less than 5 ppm, whereas BOD of highly polluted water is 17 ppm or can be more than this.

→ Eutrophication : The process in which nutrient enriched water bodies support a dense plant population which kills animal life by depriving it of oxygen and results in subsequent loss of biodiversity is known as Eutrophication.

→ Green chemistry: It is a way of utilizing the existing knowledge and principles of chemistry to reduce the adverse production of pollutants.