Category: Class 12

MP Board Class 12th Biology Solutions Chapter 1 Reproduction in Organisms

Reproduction in Organisms NCERT Textbook Questions and Answers

Question 1.
Why is reproduction essential (neccessary) for organisms?
Answer:
An organism gives rise to young ones by reproduction. The offspring grow, mature and in turn produce new offspring. Thus, there is a cycle of birth, growth and death. Reproduction enables the continuity of the species, generation after generation. So, therefore reproduction is essential.

Question 2.
Which is a better mode of reproduction sexual or asexual? Why?
Answer:
Sexual mode of reproduction is better because it is biparental reproduction and introduces variation among offsprings and their parants due to crossing over and recombination during gamete formation by meiosis.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
In sexual reproduction, the offspring is morphologically and genetically identical to the parent and to each other. Hence, it is called clone.

Question 4.
Offsprings formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?
Answer:
Offspring formed due to sexual reproduction have better chance of survival because:

• The offspring consists its hybrid characters which may adapt better with the different environment.
• Genetic variations are introduced among the offspring’s, which increases the biological tolerance.
• Sexual reproduction occurs in adverse condition in lower plant kingdom, so sexual spores survive in adverse condition.
• Sexual reproduction may not always show better chances of survival because the offspring may be inferior to the parents.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Answer:
The progenies have similar genetic make up and are exact copies of their parents in asexual reproduction but the progenies have different genetic make up and different from each other and dissimilar to the parent in sexual reproduction.

Variation is absent in asexual reproduction but it is a common phenomenon of sexual reproduction. In asexual reproduction, variation may occur due to mutation whereas varia¬tion occurs due to mutation, crossing over and recombination in sexual reproduction.

Question 6.
Distinguish between asexual and sexual reproduction. Why is vegetative reproduction aslo considered as a type of asexual reproduction?
Answer:
Differences between Asexual and Sexual reproduction:

S.No.	Asexual reproduction	Sexual reproduction
1.	In this type of reproduction only one parent is required.	In this type of reproduction two parents of different sexes are required.
2.	Whole body or a single cell acts as reproductive unit.	Reproductive units are called as gametes which are produced by specific tissues.
3.	Offsprings remain pure, i.e., alike their parents.	Offsprings differ from their parents.
4.	It occurs by mitosis cell division.	Gametes are formed by meiosis cell division and zygote develops by mitosis cell division
5.	Variation does not occur.	Variation occurs.

Vegetative reproduction is considered as a type of asexual reproduction because :

• It is uniparental reproduction.
• There is no involvement of gametes or sex cells.
• Cell division and no reductional division takes place.
• Vegetative propagules are somatic cells.

Question 7.
What is vegetative propagation ? Give two suitable examples.
Answer:
In plants, the vegetative propagules (runner, rhizome, sucker, etc.) are capable of producing new offsprings by the process called vegetative propagation. As the formation of these vegetative propagules does not involve both the parents, the process involved is asexual.
Examples:

• Adventitious buds in the notches along the leaf margins of bryophyllum grow to form new plants.
• Potato tuber having buds when grown, develops into a new plant.

Question 8.
Define:
(a) Juvenile phase
(b) Reproductive phase
(c) Senescent phase.
Answer:
(a) Juvenile phase : It is the pre-reproductive in which all organisms require a certain growth and maturity in the life before reproducing sexually.
(b) Reproductive phase: Reproductive phase is the phase in the life cycle, where an organism possess all the capacity and potential to reproduce sexually. It is the end of juvenile phase or vegetative phase.
(c) Senescent phase : It is the post reproductive phase in the life cycle where an organism slowly losses the rate of metabolism, reproductive potential and show deterioration of the physiological activity of the body.

Question 9.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Higher organisms have resorted to sexual reproduction to :

• Get over the unfavourable condition
• Restore high gene pool in a population
• Restore vigour and vitality of the race and .
• Get proper parental care ‘
• Introduce variation to enable better adaptive capacity.

Question 10.
Explain, why meiosis and gametogenesis are always interlinked.
Answer:
Gametogenesis (formation of male and female gametes) is associated with reduction in chromosome number thus, the gamete formed contains half chromosome set of the parental cell. So, gametogenesis is interlinked with meiosis because in meiosis reduction of chromosome number from diploid set (2n) to haploid set (n) takes place.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).
Answer:

Question 12.
Define external fertilization mention its disadvantages.
Answer:
The fusion of compatible gametes (Male and female) outside the body of an organism is called external fertilization, e.g., in Frog.
Disadvantages of external fertilization:

• It requires a medium for fusion of gametes.
• The young ones are often exposed to the predators.
• After fertilization, offsprings are produce large in number but no parental care is provided

Question 13.
Differentiate between a Zoospore and Zygote.
Answer:
Differences between Monera and Protista:

S.No.	Zoospore	Zygote
1.	These are endogenously, asexually	Zygote is a diploid cell formed by fusion
3.	Zoospores takes part in dispersal.	Zygote does not have significant role in dispersal.
3.	Zoospores takes part in dispersal.	Zygote does not have significant role in dispersal.

Question 14.
Differentiate between Gametogenesis and Embryogenesis.
Answer:
Differences between Gametogenesis and Embryogenesis:

S.No.	Gametogenesis	Embryogenesis
1.	It is the formation of gametes from meiocytes.	It is the formation of embryo from zygote cell.
2	This is a pre-fertilization event.	This is a post fertilization event.
3.	It occurs inside reproductive organs.	It occurs outside or inside the female body.
4.	It produces haploid gamete.	It gives rise to diploid embryo.
5.	The cell division during gametogenesis is meiotic in diploid organism.	The cell division during embryogenesis is mitotic in diploid organisms.

Question 15.
Describe the post-fertilization changes in a flower.
Answer:
The post-fertilization changes that take place in a flower are as follows :

• The formation of zygote which later develops into an embryo and primary endosperm cell which develops into endosperm takes place.
• While the sepals, petals and stamens are shed, the pistil remains intact.
• The fertilized ovule develops into seeds.
• The ovary matures into a fruit that later develops a thick protective wall, called pericarp.
• Seeds after dispersal germinate under favourable conditions which later develop into a new plant.

Question 16.
What is a bisexual flower ? Collect five bisexual flowers from your neigh bourhood and with the help of your teacher find our their common and scientific names.
Answer:

S.No.	Common Name	Scientific Name
1.	China rose	Hibiscus rosa sinensis
2	Chandni	Ervatamia divaricata
3.	Makoy	Solatium nigrum
4.	Sunflower	Helianthus annuus
5.	Mustard	Brassica compestris.

Question 17.
Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that bears unisexual flowers ?
Answer:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow colored petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure. Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Answer:
In viviparous animals, the young one develops inside the body of the female organism. As a result of this, the young one gets better production and nourishment for proper development. In case of oviparous animals, they lay eggs and the young ones develop inside the calcareous shell, outside the body of the female. So, the young ones are not effectively protected and nourished and are vulnerable, to predators so, they are at a greater risk as compared to the offsprings of the viviparous animals.

Reproduction in Organisms Other Important Questions and Answers

Reproduction in Organisms Objective Type Questions

1. Choose the Correct Answers:

Question 1.
Callose wall is found: (CBSE PMT 2007)
(a) In male gamete
(b) In ovum
(c) In pollen grain
Answer:
(d) In megaspore mother cell.

Question 2.
Name the plant in which new plant arise from the notches of leaves through vegetative propagation: (AFMC2012)
(a) Asparagus
(b) Chrysanthemum
(c) Agave
(d) Bryophyllum.
Answer:
(d) Bryophyllum.

Question 3.
Which type of vegetative propagation is occurs in banana: (AMU 2012)
(a) From tuber
(b) From rhizome
(c) From bulb
(d) From stolen.
Answer:
(b) From rhizome

Question 4.
Which of the following is viviparous:
(a) Tortoise
(b) Bony fish
(c) Hummingbird
(d) Whale
Answer:
(d) Whale

Question 5.
Which of the following is characteristics of the ‘clone’:
(a) Similar to ancestors in genetic character
(b) A group of plants developed in vegetative propagation
(c) Plants developed from same parents
(d) All of the above.
Answer:
(d) All of the above.

Question 6.
Organism which give birth to young ones:
(a) Viviparous
(b) Amphibians
(c) Oviparous
(d) Coelomate.
Answer:
(a) Viviparous

Question 7.
Set of chromosome found in zygote:
(a) X
(b) 2X
(c) 3X
(d) 4X.
Answer:
(b) 2X

Question 8.
The plant which can be obtained same as parent plant:
(a) Through seeds
(b) Through fruits
(c) By cutting the stem
(d) By hybridization.
Answer:
(c) By cutting the stem

Question 9.
Vegetative propagation in potato is:
(a) By rhizome
(b) By bulb
(c) By stem
(d) By tuber.
Answer:
(d) By tuber.

Question 10.
Which of the following is oviparous animal:
(a) Bat
(b) Whale
(c) Penguin
(d) Amoeba.
Answer:
(c) Penguin

2. Fill in the Blanks:

1. Reproductive organs in plants are developed in ……………….. phase.
2. ……………….. nucleus are present in pollen tube of angiosperms.
3. The ……………….. are not form in Rose and Banana, therefore, reproduction occurs through ………………..
4. Reproductive organs of animals are differentiate in ……………….. stage.
5. Eichomia propagate through ………………. in water.
6. Rate of multiplication is greater ……………….. reproduction than ……………….. reproduction.

Answer:

1. Adolescence
2. Three
3. Seeds, Vegetative propagation
4. Embryo
5. Offset
6. Sexual, Asexual.

3. Match the Following

Answer:

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

Answer in One Word/Sentence:

1. What we called uniparental structure formed from asexual reproduction which is genetically and morphologically similar.
2. Write the name of an algae reproduce by means of zoospores.
3. Animal reproduce asexually.
4. What we called spores having flagella.
5. In Bryophyllum what kind of bud reproduce vegetatively ?

Answer:

1. Clone
2. Chlamydomonas
3. Spongilla
4. Zoo-spores
5. Adventitious bud.

Reproduction in Organisms Very Short Answer Type Questions

Question 1.
What is meant by life span ?
Answer:
The period from birth to the natural death of an organism is called life span.

Question 2.
Give definition of fertilization.
Answer:
Fertilization is a process in which male and female gamete fused to form the zygote.

Question 3.
What is clone ?
Answer:
Morphologically and genetically similar individuals who are produced by single parent is called clone.

Question 4.
Give the name of an organism in which asexual reproduction occurs through conidia.
Answer:
Penicillum.

Question 5.
In which plant vegetative propagation occurs through rhizome ?
Answer:
In Ginger, Termeric.

Question 6.
Give the name of an organism in which transverse binary fission occurs.
Answer:Paramoecium.

Question 7.
What is aging ?
Answer:
When humans are not capable for reproduction is known as aging.

Question 8.
Give two examples of Hermaphrodite plants.
Answer:
Cucurbita and Coconut are hermaphrodite plant.

MP Board Class 12th Biology Solutions

MP Board Class 12th Chemistry Solutions Chapter 4 Chemical Kinetics

Chemical Kinetics NCERT Intext Exercises

Question 1.
For the reaction R → P, the concentration of reactant changes from 0.03M to 0.02M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:
For the reaction R → P

Question 2.
In a reaction, 2A → Products, the concentration of A decreases from 0.5 mol L^-1 to 0.4 mol L^-1 in 10 minutes. Calculate the rate during this interval ?
Answer:

Question 3.
For a reaction, A + B → Product; the rate law is given by, r = k [A]^1/2 [B]². What is the order of the reaction ?
Answer:
Order of reaction = sum of powers of concentration of reactants.
Order of reaction = \(\frac { 1 }{ 2 } \) + 2 = \(\frac { 5 }{ 2 } \) = 2.5.

Question 4.
The conversion of molecules A; toy follows second order kinetics. If concetration of JC is increased to three times how will it affect the rate of formation of y².
Answer:
For the reaction x → y
Rate of reaction (r) = k[x]² …(1)
If concentration of x is increased three times, now
Rate of reaction (r)¹ = K[3x]² = k[9x²]
Dividing equation (2) by eq. (1)
\(\frac{r^{1}}{r}=\frac{k\left[9 x^{2}\right]}{k\left[x^{2}\right]}=9\)
Thus, rate of reactions will become 9 times.

Question 5.
A first order reaction has a rate constant 1.15 × 10^-3 s^-1. How long will 5g of this reactant take to reduce to 3g ?
Solution:
According to question,

Applying first order kinetic equation

Question 6.
Time required to decompose SO₂Cl₂ to half of its initial amount is 60 minutes. If the decomposition is a first order reaction, calculate the rate constant of the reaction.
Solution:
For a first order reaction,

Question 7.
What is the effect of rate constant on temperature ? How can this effect of temperature be measured quantitatively ?
Answer:
Rate of reaction increases with increase in temperature and with 10°C rise in temperature its value becomes two times.
Arrhenius equation expresses the effect of temperature on constant
\(\mathrm{K}=\mathrm{A}_{e}^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Where A is frequency factor and Efl is activation energy of the reaction.

Question 8.
The rate of the chemical reaction doubles for ah increase of 10K in absolute temperature from 298K. Calculate Ea.
Solution:
Given : k₂ = 2k₁, T₁ = 298K, T₂ = 308K, R = 8.314 JK^-1 mol^-1
We know that

Question 9.
The activation energy for the reaction
2HI_(g) → H_2(g) + I_2(g)
is 209.5kJ mol^-1 at 581K. Calculate the fraction of molecules of reactants having energy equal to or greater than activation energy ?
Solution:
According to question, E_a = 209.5kJ mol^-1 = 209.5 × 103 J mol^-1, T = 581K Fraction of molecules with energy equal to or greater than activation energy is given by
[R = 8.314J]

Chemical Kinetics NCERT TextBook Exercises

Question 1.
From the rate expression for the following reactions, determine their order of reaction and the dimensions of the rate constants.

Solutions:

Question 2.
For the reaction :
2A + B → A2B
the rate= k [A][B]² with k = 2.0 × 10^-6 mol^-2 L² s^-1. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. Calculate the rate of reaction after [A] is reduced to 0.06 mol L^-1
Solution:
Given, Rate = k [A][B]²
Initial rate = 2 × 10-6 × [0.1] × [0.2]²
= 8 × 10^-9 mol L^-1 s^-1.
Decrease in concentration of A
i.e. ∆[A] = 0.1 – 0.06 = 0.04 mol L^-1
Decrease in concentration of B
i.e. ∆[B] = \(\frac { 1 }{ 2 } \) × 0.04 = 002 mol L^-1
Remaining concentration of B
= 0.2 – 0.02 = 0.18M
Rate = k[A] [B]²
= 2 × 10-6 × 0.06 × [0.18]²
= 3.89 × 10^-9 mol L^-1 s^-1.

Question 3.
The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol^-1 L s^-1 ?
Solution:
2NH₃ → N₂ + 3H₂

Question 4.
The decomposition of dimethyl ether leads to the formation of CH₄, H₂ and CO and the reaction rate is given by Rate = k [CH₃OCH₃]^3/2. The rate of reaction is followed by increase in pressure in a closed vessel, so the rate can also be expressed in terms of the partial pressure of dimethyl ether, i.e.,
Rate = k(P_CH3OCH3)^3/2
If the pressure is measured in bar and time in minutes, then what are the units of rate and rate constants ?
Solution:
The rate law of reaction is
Rate = k [P_CH3OCH3]^3/2
Rate = Pressure change/Time change
Unit of rate = bar min^-1

Question 5.
Mention the factors that affect the rate of a chemical reaction.
Answer:
Rate of reaction depends upon the following factors:

1. Concentration of reactants
2. Temperature
3. Catalyst
4. Nature of reactant etc.

Question 6.
A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of the reactant is :
(i) doubled
(ii) reduced to half ?
Solution:
Let the reaction A → B is a second order reaction, for A

(i) When the concentration of [A] is doubled
A = 2 a
The new rate of reaction

The rate of reaction will become four times when concentration is doubled.

(ii) When concentration of [A] is 1/2.
A = \(\frac { a }{ 2 } \)
So new rate of reaction

Therefore, the rate of the reaction could be reduced to l/4^th.

Question 7.
What is the effect of temperature on the rate constant of a reaction ? How can this temperature effect on rate constant be represented quantitatively ?
Answer:
Rate constant of the reaction increases with temperature, it is independent whether the reaction is exothermic or endothermic. Arrhenius equation shows the dependence of rate constant on temperature.

Question 8.
In a pseudo first order hydrolysis of ester in water, the following results were obtained:

(i) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester.
Solution:

Where [R]0 = 0.55 M is initial concentration of ester (t = 0) and [R] is the concentration of ester at time ‘t’

Question 9.
A reaction is first order in A and second order in B.
(i) Write the differential rate equation.
(ii) How is the rate affected on increasing the concentration of B three times?
(iii) How is the rate affected when the concentrations of both A and B are doubled ?
Solution:

Let [A] = x and [B] = y
Rate (r₁) = kxy² ….(i)

(ii) When [B] = 3y
r₂ = k x (3y)² ….(ii)
On dividing equation (ii) by (i),
r₂ = 9r₁
i. e. rate increases 9 times.

(iii) When both [A] and [B] are doubled.
r₃ = k (2x) (2y)² ….(iii)
Dividing equation (iii) by (i),
r₃ = 8r₁
i.e. rate increases 8 times.

Question 10.
In a reaction between A and B, the initial rate of reaction (r₀) was measured for different initial concentrations of A and B as given below:

What is the order of the reaction with respect to A and B ?
Solution:
Assuming that the order of reaction w.r.t. A is x and w.r.t. B is y.
Rate = k [A]^x [B]^y
Rate₁ = 1(0.20)^x (0.30)^y = 5.07 × 10^-5 ….(i)
Rate₂ = k(0.20)^x (0.10)^y = 5.07 × 10^-5 …..(ii)

Order of reaction w.r.t A = 1.5
Order of reaction w.r.t B = 0.

Question 11.
The following results have been obtained during the kinetic studies of the reaction :
2A + B → C + D

Determine the rate law and the rate constant for the reaction.
Solution:
Comparing experiments I and IV and substituting the values.
As concentration of ‘B’ remains constant therefore we get order w.r.t. ‘A’.
(Rate)₁ = k (0.1)^x (0.1)^y = 6.0 × 10^-3 ….(i)
(Rate)₂ = k (0.4)^x (0.1)^y = 2.40 × 10^-2 ….(ii)
Dividing equation (ii) by (i),

Similarly by comparing experiments II and III, we get the order w.r.t. ‘B’.
(Rate)₂ = k (0.3)^x (0.2)^y = 7.2 × 10^-2 ….(iii)
(Rate)₃ = k (0.3)^x (0.4)^y = 2.88 × 10^-1 …(iv)
Dividing equation (iv) by (iii),

Question 12.
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:

Solution:
Given that the reaction between A and B is first order w.r.t. A and zero order w.r.t. B.
∴ Rate = it [A]¹ [B]0 but [B]⁰ = 1.
Rate = k [A]
From experiment I,
2 × 10^-2 = k(0.1) ⇒ k = 0.2 min^-1
From experiment II,
4 × 10^-2 = 0.2 [A] ⇒ [A] = 0.2 min L^-1
From experiment III,
Rate = (0.2) (0.4) = 0.08 mol L^-1 min^-1
From experiment IV,
2 × 10^-2 = 0.2[A] ⇒ [A] = 0.1 mol L^-1

Question 13.
Calculate the half-life of a first order reaction from their rate constants given below:
(i) 200 s^-1
(ii) 2 min^-1
(iii) 4 years^-1.
Solution:

Question 14.
The half-life for radioactive decay of ¹⁴C is 5730 years. An archaeological artifact containing wood had only 80% of the ¹⁴C found in a living tree. Estimate the age of the sample.
Solution:
Radioactive decay follows first order kinetics,

Question 15.
The experimental data for decomposition of N₂O₅
[2N₂O_s → 4NO₂ + O₂]
In gas phase at 318K are given below:

(i) Plot [N₂O₅] against t.
(ii) Find the half-life period for the reaction.
(iii) Draw a graph between log [N₂O₅] and t.
(iv) What is the rate law?
(v) Calculate the half-life period from k and compare it with (ii).
Solution:
(i) The plot of [N₂O₅] versus time is shown below :

(ii) Initial concentration of N₂O₅ = 1.63 × 10^-2 M
Half of initial concentration = 1.63 × 10^-2 \(\frac { 1 }{ 2 } \) × = 0.815 × 10^-2 M
Time corresponding to half of initial cone. (t_1/2) from the plot = 1440 sec.

(iii) The plot of log [N₂O₅] Versus time :

(iv) Since the graph between log [N₂O₅] vs time is a straight line, the reaction is of 1st order.
Rate = k [N₂O₅]

Question 16.
The rate constant for a first order reaction is 60s^-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16^th value ?
Solution:

Question 17.
During nuclear explosion, one of the products is ⁹⁰Sr with half-life of 28.1 years. If µg of ⁹⁰Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically.
Solution:

Question 18.
For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Solution:

Question 19.
A first order reaction takes 40 min for 30% decomposition. Calculate t_1/2.
Solution:
30% decomposition means that x = 30% of a = 0.30 a
As reaction is of first order

Question 20.
For the decomposition of azoisopropane to hexane and nitrogen at 543 K, the following data are obtained :

Calculate the rate constant.
Solution:

Question 21.
The following data were obtained during the first order thermal decomposition of SO₂Cl₂ at a constant volume.
SO₂Cl₂(g) → SO₂(g) + cl₂(g)

Calculate the rate of the reaction when total pressure is 0.65 atm.
Solution:

For the first order reaction in terms of pressure

Question 22.
The rate constant for the decomposition of N₂O₅ at various temperatures is given below:

Draw a graph between In k and 1/T and calculate the values of A and E_a. Predict the rate constant at 30°C and 50°C.
Solution:
To draw the plot of log k versus 1/T, we can rewrite the given data as follows:

Draw the graph as shown in the ahead figure :


Compare it with y = mx + c
log A = Value of intercept on y axis i.e„ on logk axis
[y₂ – y₁ = -1 – (-7.2)] = (-1 + 7.2) = 6.2
log A = 6.2
A = Antilog 6.2
= 1.585 × 10⁶ s^-1
The value of rate constant k can be found from graph as follows :

We can also calculate the value of k from the following formula :

Question 23.
The rate constant for the decomposition of hydrocarbons is 2.418 × 10^-s s^-1 at 546 K. If the energy of activation is 179-9 kj/mol, what will be the value of pre-exponential factor.
Solution:
According to the question, Rate constant (k) = 2.418 × 10^-5 s^-1, Temperature (T) = 546K, Activation energy (E_a) = 179.9 kJ mol^-1

Question 24.
Consider a certain reaction A → Products, with k = 2.0 × 10^-2 s^-1. Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L^-1.
Solution:
Unit of k is s^-1.

Question 25.
Sucrose decomposes in acid solution into glucose and fructose according to the first order rate law, with t_1/2 = 3.00 hours. What fraction of sample of sucrose remains after 8 hours?
Solution:

Let initial concentration of sucrose (a) = 1M
Concentration after 8hrs = (a – x) = (1 – x). Where ‘x’ is the amount of sucrose decomposed.

Question 26.
The decomposition of hydrocarbon follows the equation k = (4.5 × 10¹¹ s^-1) e^– 28000 K/T Calculate E_a.
Solution:
k = (4.5 × 10¹¹ s^-1) e^-28000K/T
Comparing the equation with Arrhenius equation
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)

Question 27.
The rate constant for the first order decomposition of H₂O₂ is given by the following equation:
log k = 14.34 – 1.25 × 10⁴ K/T
Calculate Ea for this reaction and at what temperature will its half-period be 256 minutes?
Solution:
(i) We know that
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Taking logarithm on both sides of equation (i), we get

Question 28.
The decomposition of A into product has value of k as 4.5 × 10³ s^-1 at 10°C and energy of activation 60 kj mol^-1. At what temperature would k be 1.5 × 10⁴ s^-1 ?
Solution:
Given k₁ = 4.5 × 10³ s^-1, k₂ = 1.5 × 10⁴ s^-1, T₁ =10°C = 10 + 273 = 283K, E_a = 60kJ mol^-1 = 60000J mol^-1
Applying Arrhenius equation and substituting the values, we get

Question 29.
The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308 K. If the value of A is 4 × 10¹⁰ s^-1. Calculate k at 318 K and Ea.
Solution:
For 10% completion of the reaction

Question 30.
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.
Solution:

Chemical Kinetics Other Important Questions and Answers

Chemical Kinetics Objective Type Questions

Question 1.
Choose the correct answer :

Question 1.
For most of the reactions, the value of temperature coefficient lies in between:
(a) 1 and 3
(b) 2 and 3
(c) 1 and 4
(d) 2 and 4.

Question 2.
For first order reaction value of t_1/2 is :

Question 3.
A first order reaction gets completed to 75% in 32 minutes. How much time would have been required for 50% completion :
(a) 24 minute
(b) 16 minute
(c) 8 minute
(d) 4 minute.

Question 4.
For reaction H_2(g) + Br_2(g) → 2HBr_(g) the experimental values indicate that:
Rate of reaction = k [H₂] [Br₂]^1/2 Molecularity and order of reaction is :
(a) 2, \(\frac { 3 }{ 2 } \)
(b) \(\frac { 3 }{ 2 } \),\(\frac { 3 }{ 2 } \)
(c) 1,1
(d) 1,\(\frac { 1 }{ 2 } \)

Question 5.
The reaction 2FeCl₃ + SnCl₂ → 2FeCl₂ + SnCl₄ is an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) None of these.

Question 6.
In the reaction 2A + B → A₂B, the rate of consumption of reactant A is:
(a) Half of the consumption rate of B
(b) Equal to the consumption rate of B
(c) Twice to the consumption rate of B
(d) Equal to the rate of formation of A₂B.

Question 7.
Hydrolysis of sucrose to glucose and fructose is :
C₁₂H₂₂O₁₁ + H₂O > C₆H₁₂O₆ + C₆H₁₂O₆
sucrose  glucose  fructose
an example of:
(a) First order reaction
(b) Second order reaction
(c) Third order reaction
(d) Zero order reaction.

Question 8.
At a given temperature the rate of reaction becomes slow when :
(a) Energy of activation becomes high
(b) Energy of activation becomes low
(c) Entropy changes
(d) Initial concentration of reactants remains constant.

Question 9.
Plants prepare starch in the process of:
(a) Flash photolysis
(b) Photolysis
(c) Photosynthesis
(d) None of these.

Question 10.
For first order reaction the specific reaction constant depends upon:
(a) Concentration of reactants
(b) Concentration of products
(c) Time
(d) Temperature.

Question 11.
In the reaction between A and B to form C, A represents first order and B rep¬resents second order. Rate equation will be written as :
(a) Rate = k [A]² [B]
(b) Rate = k [A] [B]²
(c) Rate = k [A]^1/2 [B]
(d) Rate = k [A] [B]^1/2.

Question 12.
Molecularity of reaction of Inversion of sugar :
(a) 3
(b) 2
(c) 1
(d) 0.

Question 13.
Minimum energy required for molecules to react is called :
(a) Potential energy
(b) Kinetic energy
(c) Nuclear energy
(d) Activation energy.

Question 14.
2A + B → A₂B, if concentration A is doubled and concentration of B is being half, the rate of reaction will:
(a) Increase 4 times
(b) Decrease 2 times
(c) Increase 2 times
(d) Remains unchanged.

Question 15.
Half life for second order reaction :
(a) Is proportional to initial concentration
(b) Does not depend upon initial concentration
(c) Inversely proportional to initial concentration
(d) Inversely proportional to square root of intial concentration.

Question 16.
The reaction 2H₂O₂ → 2H₂O + O₂, r = k[H₂O₂] is:
(a) Zero order reaction
(b) First order reaction
(c) Second order reaction
(d) Third order reaction.

Question 17.
Thermal decomposition of a compound is first order reaction. If a sample of compound decomposes 50% in 120 minutes, how much time it will take for 90% decomposition:
(a) About 240 minutes
(b) About 480 minutes
(c) About 450 minutes
(d) About 400 minutes.

Question 18.
Arrhenius equation is :

Question 19.
When temperature is raised, rate of reaction increases. This is due to :
(a) Decrease in activation energy
(b) Increase in number of collisions
(c) Decrease in number of active molecules
(d) Decrease in number of collisions.

Question 20.
Unit of rate constant of second order reaction is :
(a) mol^-1 litre^-1 second^-1
(b) mol litre^-1 second^-1
(c) mol litre second
(d) mol^-1 litre second^-1.

Question 21.
Reaction

(a) Bimolecular and second order
(b) Unimolecular and first order
(c) Bimolecular and first order
(d) Bimolecular and zero order.

Question 22.
Unit of rate constant of first order reaction is :
(a) mol L^-1 S^-1
(b) mol^-1 LS^-1
(c) S^-1
(d) mol L^-1 S.

Question 23.
Time required for the completion of 90% reaction of first order is :
(a) 1-1 times that of half life
(b) 2-2 times that of half life
(c) 3-3 times that of half life
(d) 4-4 times that of half life.

Question 24.
The rate of chemical reaction depends upon :
(a) Active mass
(b) Atomic mass
(c) Equivalent weight
(d) Molecular mass.

Question 25.
Unit of reaction rate is :
(a) mol litre^-1 second^-1
(b) mol^-1 litre second^-1
(c) mol^-1 litre^-1 second
(d) mol litre second.

Answer:
1. (b), 2. (a), 3. (b), 4. (a), 5. (d), 6(b), 7. (a), 8. (a), 9. (c), 10. (d), 11. (b), 12. (c), 13. (d), 14. (d), 15. (c), 16. (b), 17. (d), 18. (d), 19. (b), 20. (d), 21. (c), 22. (c), 23. (c), 24. (a), 25. (a).

Question 2.
Fill in the blanks :

1. The rate of a reaction does not depends on the concentration of the reacting species, then the reaction is of …………………
2. Half life period of a radioactive element is 140 days. On taking 1 gm element initially the amount left after 560 days will be …………………
3. Fast reactions are completed in less than ………………… seconds.
4. In the mechanism of a reaction, the slowest step is called …………………
5. Study of fast reactions is done by …………………
6. Unit of rate constant for third order reaction is …………………
7. Difference in the minimum and maximum energy state of reactants is called …………………
8. Reactions which take place by the absorption of radiations are called …………………
9. The total number of molecules which participate in a reaction is called …………………
10. ………………… gives the idea about the mechanism of the reaction.
11. Hydrolysis of Ethyl acetate in acidic medium is an example of ………………… order reaction.
12. Molecularity is always a …………………
13. For the excitation of one mole reactant ………………… photons are required.
14. Rate of change in concentration of reactant at a specific instant of time is known as ………………… rate.
15. Rate of reaction is ………………… to the concentration of reactant.
16. cm-1 is the unit of …………………
Answer:
1. Zero
2. \(\frac { 1 }{ 16 } \) gm
3. 10^-9
4. Rate determining step
5. Flash photolysis
6. mol^-2 Litre² second^-1
7. Activation energy
8. Photochemical reactions
9. Molecularity
10. Order of reaction
11. Pseudo unimolecular
12. whole number
13. one mole
14. Instantaneous
15. Proportional
16. cell constant.

Question 3.
Match the following :


Answers:

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

Question 4.
Answer in one word / sentence :

1. What is the relation between threshold energy and activation energy ?
2. What is the expression of rate constant for first order reaction ?
3. Write alternative form of Arrhenius equation.
4. If reaction A + B → C is a zero order reaction. Write rate law expression for it.
5. If rate equation for a reaction is Rate = k [NO₂]² [Cl₂] then state the order of reaction with respect to Cl₂ and order with respect to NO₂ and total order of reaction.
6. Write rate equations on the basis of concentration of reactants and products for the following reactions.
2NO₂ → 2NO + O₂
7. What is rate determining step ?
8. Write an example of zero order reaction.
9. What is Quantum Efficiency ?
10. Write Arrhenius Equation.
11. Write the expression of half life period for first order reaction.
12. What is the effect of surface area of reactant on the rate of reaction ?
13. What is instantaneous rate ?
14. What is half-life period of a reaction ?
15. What is the unit of k for zero order reaction ?
16. Explain Threshold energy.
17. What are fast reactions ?
18. For a zero order reaction tm is proportional to what ?
Answer:
1. Activation energy = Threshold energy – Energy of molecules in normal state
2.

3.

4.

5. Order with respect to cl₂ = l, Order with respect to NO₂ = 2, Thus, Total order = 1 + 2 = 3
6.

7. The slowest step is the rate determining step
8.

9. Quantum Efficiency

10.

11.

12. Larger surface area increases the rate of reaction
13. Rate of change in concentration of reactant or product at a specific instant of time is known as instantaneous rate
14. Half-life period of a reaction is the time in which concentration of reactant is reduced to one half of the initial concentration
15. Litre mol^-1 sec^-1
16. The minimum amount of energy which the reactant molecules should possess for effective collision
17. Reactions which gets completed in 10-9 second or even lesser time interval are called fast reactions
18. Initial concentration of reactant [A]

Chemical Kinetics Very Short Answer Type Questions

Question 1.
What is first order reaction ?
Answer:
If the rate of reaction depends on the first power of concentration of reactant then the order of reaction is said to be first order reaction.

Question 2.
What is Energy Barrier ?
Answer:
The minimum energy achieved by the reactant only after which it can be converted to product is known as energy barrier. Reactant molecules cannot form activated complex till they reach this height (activation energy) and cannot be converted to form product.

Question 3.
Explain the rate determining step.
Answer:
Some chemical reactions complete in one or more steps. Rate of reaction is determined by the slowest step which is known as rate determining step.

Question 4.
Write the characteristics of photochemical reactions. Any three.
Answer:
Characteristics :

1. Absorption of magnetic radiations is necessary for these reactions.
2. These reactions are unaffected by temperature but intensity of radiations affects them.
3. Light is necessary for activating the reactants.

Question 5.
What are the applications of Arrhenius equation ? (Any two)
Answer:

1. In calculating activation energy.
2. In determining the rate constant at a temperature by the help of rate constant at another temperature of the reaction.

Question 6.
What is photochemical reaction ? Give an example.
Answer:
Such reactions which are induced by light or other electromagnetic radiations are known as photochemical reactions. Wavelength of these radiations is from 2000Å to 8000Å.

Question 7.
What is specific reaction rate ?
Answer:
Specific reaction rate of a reaction at a given temperature is equal to that rate of a reaction when concentration of each reactant is unity.

Rate of reaction = k where k = Specific reaction rate constant.

Question 8.
When is the average rate of the reaction equal to its instantaneous rate ?
Answer:
When value of time interval is nearly zero or when time by infinite form is minute then the average rate of reaction is comparable to its instantaneous rate.

Question 9.
What are pseudo unimolecular reactions ?
Answer:
There are some reactions in which molecularity is more than one i.e. two or more molecules are present, but in the chemical reaction concentration of only one reactant molecule is changed and it is only responsible for rate of reaction. Thus, order is one. These reactions are called pseudo unimolecular reactions.

Above reaction is bimolecular but concentration of water does not affect the rate of reaction, thus, rate of reaction is proportional to the concentration of sucrose only.
Rate = k[C₁₂H₂₂O₁₁]
Thus, inversion of sucrose is a first order reaction. It is known as pseudo unimoleciilar reaction.

Question 10.
What is temperature coefficient ?
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10 C, velocity of reaction is increased up to 2-3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10 C, the ratio of these two constant is called temperature coefficient, i.e.,

Thus, at various time the ratio of rate of reaction which differ by 10°C is known as temperature coefficient.

Chemical Kinetics Short Answer Type Questions

Question 1.
Write down the expression representing the rate of reaction.
Answer:
Rate of reaction is the rate of change in concentration of reactant or concentration of product in unit time interval.

Unit of rate of reaction depend on the units of concentration and time. If concentration is represented in mole per litre and time in second, then unit of rate of reaction is mol per litre per second. If time is expressed in minute then unit of reaction rate is mole per litre per minute.

Question 2.
What is the meaning of instantaneous rate of reaction ?
Answer:
The rate of reaction does not remain constant during the whole time interval because rate of reaction depends upon the concentration of reactants. As the concentration of reactants decreases with time, the rate of reaction also decreases with time.

In order to express the reaction rate as accurately as possible, the instantaneous rate of reaction is expressed. For this the time interval (∆t) is taken as small as possible.

Where, d [A] is a change in concentration of reactant A.

Question 3.
What do you know about molecularity of any reaction ?
Answer:
Molecularity of Reaction : ‘Number of moles of reactant participating in elementary step of chemical reaction is called molecularity of the reaction.’

There are many reactions which proceed through the number of steps and each step being independent is called elementary step. The rate of reaction is determined by slowest step and it is known as rate determining step. Here molecularity can be defined as “Number of molecules/atoms or ions participating in rate determining step is called molecularity.”

Example : (i) Unimolecular reaction :
O₃ → O₂ + O
(ii) Bimolecular reaction:
NO + O₃ → NO₂ + O₂

Question 4.
Differentiate molecularity and order of reaction.
Answer:
Differences between Molecularity and Order of reaction :

Molecularity:

1. It is the total number of molecules which participate in the reaction.
2. It is a theoretical concept.
3. It is always a whole number.
4. Its value is never zero.
5. It does not provide any information about mechanism of reaction.

Order of reaction

1. It is the number of molecules which participate in reaction and whose concentration is changed.
2. Order of reaction is determined by experimentally.
3. Fractional values are also possible.
4. Zero value is possible.
5. It provides information about mechanisms of reaction.

Question 5.
Write any four factors which affects rate of a chemical reaction.
Answer:
The rate of reaction depends upon the following factors :
(i) Concentration of reactants: At constant temperature, the rate of a reaction increases by increasing the concentration of the reactants.

(ii) Temperature of the system : If the concentration of the reactants are constant then the rate of reaction increases by increasing temperature. For 10 degree rise of temperature, the reaction rate becomes double or triple.

(iii) Presence of catalyst: Positive catalyst increases the rate of reaction and negative catalyst decreases the reaction rate.

(iv) Nature of the reactants: Nature of reactants also affect the reaction rate. In any chemical reaction some old bonds are broken and new bonds are formed. Thus, in case of more simple molecules the lesser is the number of bond breaking and the rate of reaction increase whereas in case of complex molecules more bonds are broken and rate decreases.

(v) Exposure to radiations: The rate of some reactions increases due to some special radiations.

Question 6.
Write characteristics of rate constant
Answer:

1. At fixed temperature, the value of k is constant.
2. For a particular reaction, k is independent of concentration but depends on temperature.
3. The value of k is different for different reactions.
4. It is a measure of intrinsic rate of reaction i. e., larger the value of k, faster will be the reaction and viceversa.

Question 7.
What do you understand by order of reaction ? Give example.
Answer:
Order of reaction : The order of a reaction is defined as the sum of all the powers to which concentration terms in the rate law are raised to express the observed rate of the reaction. Suppose there is a general reaction,
aA + bB + cC → product
For which the rate law is
Rate = – \(\frac { dx }{ dt } \) = k[A]^p [B]^q [C]^r
Then the order of the reaction n = p + q + r, where, p, q and r are the orders with respect to individual reactants and overall order is the sum of the exponents i.e.,p + q + r.
When n = 1 the reaction is of first order, if n = 2 the reaction is of second order and so on.
For example : Decomposition of ammonium nitrite occurs as follows :
NH₄NO₂ → N₂ + 2H₂O
Rate of reaction = – \(\frac { dx }{ dt } \) = k[NH₄NO₂]
Thus, order of this reaction will be 1.

Question 8.
What do you mean by zero order reaction ? Give one example,
Answer:
In some reactions, rate does not depends, upon the concentration of reactants. This type of reactions are called zero order reaction.
Example : In contact of Au or Pt, the ammonia molecule dissociate and rate of this reaction is independent of concentration of ammonia.

Question 9.
How does rate of any reaction depend on temperature ? Explain.
Answer:
Generally, rate of any reaction increases due to increase in temperature. On increasing temperature by 10°C, velocity of reaction is increased up to 2-3 times. If velocity constant of a reaction is and it is increased to increasing temperature by 10°C, the ratio of these two constant is called temperature coefficient, i.e.,
\(\frac{k_{t+10}}{k_{t}} \approx 2 \text { to } 3\)
Arrhenius provide the following relation for showing the effect of temperature on velocity constant.
i.e., \(k=\mathbf{A} \cdot e^{-\mathrm{E}_{\alpha} / \mathrm{RT}}\)

Question 10.
What is activation energy ?
Answer:
According to Arrhenius, any chemical reaction is only possible when reacting . molecules are activated with minimum energy which is called threshold energy. Kinetic energy of most of the molecules are less than this minimum energy. The excess energy which is required to activate reactant molecules, is called activation energy.

Activation energy can be determined by the use of Arrhenius equation.

Question 11.
Write four differences between Rate of reaction and Rate constant.
Answer:
Differences between Rate of reaction and Rate constant:

Rate of reaction:

1. It is expressed in terms of consumption of reactants or formation of product per unit time.
2. It depends on concentration of reactant at particular moment.
3. It generally decreases with the progress of reaction.
4. Its unit is mol L^-1 cm^-1.

Rate constant:

1. It is proportionality constant in differential form in rate law or rate equation.
2. It is independent of concentration of reactant.
3. It does not depend on the progress of reaction.
4. It changes according to order of reaction.

Question 12.
Prove that half-life period of zero order reaction is proportional to initial concentration of reactant.
Answer:
If rate of reaction does not depend on the concentration of reactants, then it is known as zero order reaction.

For a zero order reaction the relation between rate constant and concentration is expressed by the following equation :

Thus, half-life period of zero order reaction is proportional to initial concentration of reactant.

Question 13.
Prove that half-life period is independent of the initial concentration for the first order reaction.
Answer:
Half-life period for the reaction is that period in which the initial concentration of reactant is reduced to half.
For first order reaction :

Thus, the half-life period is independent of the initial concentration for the first order reaction.

Question 14.
What is Integrated rate law method ? Write a note.
Answer:
Integrated rate equation for first order reaction is :

By this equation, order of reaction can be calculated. By knowing the initial concen¬tration (a) of reactant, concentration at a definite time (a – x) can be known.

This way, by substituting the values of t and (a – x), value of k is calculated. If value of k comes to be constant, the reaction is of first order. For, first order reaction, on plotting a graph between concentration log)0 (a – x) and t, a straight line is obtained. For other orders of reaction suitable equations are used and orders are determined.

Question 15.
A first order reaction is 90% complete in 40 minutes. Calculate its half- life period. (log 2 = 0.3010).
Solution:
Let initial cone, of reactant (a) = 100, t = 40 minutes
90% reaction gets completed in 40 minutes i.e.,

Question 16.
Show that time required for completing 99.9% of a first order reaction is 10 times of its half-life period.
Solution:
If initial concentration of reactant is a, then t = ? for x = 0.999 a
We know that for a first order reaction

Question 17.
Write unit of rate constant k for the zero order, first order and second order reaction
Solution:

Question 18.
(1) 2N₂O_s 4NO₂ + O₂
(2) H²₂ + I₂ → 2HI
Reaction (1) is first order reaction and (2) is second order reaction. Why ?
Answer:
(1) In reaction 2N₂O₅ → 4NO₂ + O₂, when graph is plotted between rate of reaction and then again graph is plotted between rate of reaction and [N₂O₅]², it is shown that in the first graph straight line is obtained i.e.
rate ∝ [N₂O⁵]
or rate = k[N₂O⁵]
Therefore, 2N₂O₅ → 4NO₂ + O₂ is first order reaction.

(2) In reaction H₂ + I₂ → 2HI when graph is plotted between rate of reaction and (H₂) (I₂), it is seen that straight line is obtained.
Therefore rate ∝ [H₂][I₂]
Hence, this reaction is of 2^nd order reaction.

Chemical Kinetics Long Answer Type Questions

Question 1.
Determine the expression for zero order reaction.
Answer:
Reactions in which rate of reaction does not depend on the concentration of the reactants are called zero order reactions. Consider a zero order reaction
A → B
Where, A and B are concentration of reactants and products. Since in this type of reaction, rate of reaction does not depend on the concentration of reactant therefore, rate of change in concentration of reactant remains constant.
Rate of reaction = Constant.
Let initial cone, of reactants be a moles /It and after time ‘t’ x moles are converted into product. Then cone, of A after time t will be (a – x) mol /It.

Thus, eqn. (7) is the velocity equation for zero order reaction.

Question 2.
Write a note on Arrhenius equation.
Answer:
Arrhenius equation : Arrhenius gave a relation between rate constant of reaction and temperature which is known as Arrhenius equation i.e.,
\(k=\mathbf{A} e^{-\Delta \mathbf{E}_{a} / \mathbf{R} \mathbf{T}}\)
Where, k = Rate constant of reaction, A = Frequency factor, Ea = Activation energy, R = Gas constant, T = Absolute temperature.

Taking logarithm both sides of above equation we get

E_a and A can be determined by measuring rate constants of the reaction at two different temperature. Let k₁ and k₂ are the rate constant for the reaction at two temperatures T₁ and T₂ respectively. Then from eqn. (1)

Application : Activation energy AE_a may be calculated easily from this equation.

Question 3.
Write the Arrhenius equation in the form of equation of straight line. What will be the slope of a graph using this equation ? Calculate the activation energy for the decomposition in a decomposition reaction in which the value of slope obtained is – 9920 when log k is plotted against \(\frac { 1 }{ T } \).
Answer:
Arrhenius equation : Arrhenius gave a relation between rate constant of reac¬tion and temperature which is known as Arrhenius equation, i.e.,
\(k=\mathrm{A} e^{-\mathrm{E}_{a} / \mathrm{RT}}\)
Where, k = Rate constant of reaction, A = Frequency factor, Ea = Activation energy, R = Gas constant and T = Absolute temperature.
Taking logarithm both sides of above equation, we get

This is a straight line equation. If log₁₀ A and \(\frac { 1 }{ T } \) is plotted at different temperatures, a straight line is obtained whose slope = \(\frac{-\mathrm{E}_{a}}{2 \cdot 303 \mathrm{R}}\) From the graph, value of slope = \(\frac{\mathrm{E}_{a}}{2 \cdot 303 \mathrm{R}}\) can be calculated and activation energy (E_a) can be calculated.

Alternatively, E_a and A can be determined by measuring rate constant of the reaction at two different temperatures. Let k₁ and k₂ are the rate constants for the reaction at two temperatures T₁ and T₂ respectively. Then, from eqn. (1),

Calculation of activation energy :
E_a = -2.303 × slope × R = -2.303 × (-9920) × 1.986
∴ E_a = 45334 cal/gm/mol.

Question 4.
Derive an expression for velocity constant of first order reaction.
Answer:
The reaction in which velocity of reaction depends upon the concentration of one mole, are called first order reaction.
Let this reaction is
A → Product
Suppose intial concentration of A is a gram mole and after t second x mole consumed and remaining concentration is (a – x) gram mole.
So after t time, the rate of reaction will be proportional to (a – x)

Putting the value of c from eqn. (4), into eqn. (3).
– ln(a – x) = kt – In a
or In a – ln(a – x) = kt

It is the desired expression for first order of reaction.

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 3 Electrochemistry

Electrochemistry NCERT Intext Exercises

Question 1.
How would you determine the standard electrode potential of the system Mg²⁺ | Mg ?
Answer:
\(\mathrm{E}_{\mathrm{Mg}^{2+} / \mathrm{Mg}}^{\circ}\) can be determined by forming a cell with standard hydrogen electrode (SHE).
Mg_(s) | Mg²⁺ 1 M || H⁺_g | H₂ (1 atm) Pt.

Question 2.
Can you store copper sulphate solutions in a zinc pot ?
Answer:
No, it is not possible. The E° values of the copper and zinc electrodes are as follows :
Zn2+(aq) + 2e– → Zn(s) ; E° = – 0·76 V
Cu2+(aq) + 2e– → Cu(s) ; E° = + 0·34 V
This shows that zinc is a stronger reducing agent than copper. It will lose electrons to Cu2+ ions and redox reaction will immediately set in.
Zn(s) + Cu2+ (aq) → Zn2+(aq) + Cu(s)
Thus, copper sulphate solution cannot be stored in zinc pot.

Question 3.
Consult the table of standard electrode potentials and suggest three sub-stances that can oxidise ferrous ions under suitable conditions.
Answer:
Only those species can oxidise ferrous ions (Fe²⁺) whose standard reduction potential is more positive than 0.77V. Thus, suitable oxidising agents will be F₂, Cl₂, Br₂.

Question 4.
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Solution:
For hydrogen electrode, H+ + e– → 1/2H2
Applying Nernst equation,

Question 5.
Calculate the emf of the cell in which the following reaction takes place :
Ni(s) + 2Ag⁺ (0.002 M) → Ni²⁺ (0.160 M) + 2Ag_(s) Given that E^∘_cell = 1.05 V.
Solution:

Question 6.
The cell in which the following reactions occurs :
2Fe³⁺(aq) + 2I^–(aq) → 2Fe²⁺(aq) + I₂(s) has E^∘_cell = 0-236 V at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Solution:
The two half reactions are :
2Fe³⁺ + 2e^– → 2Fe²⁺ and 2I^– → I₂ + 2e^–
For the above reaction, n = 2

Question 7.
Why does the conductivity of a solution decreases with dilution ?
Answer:
Because number of ions per cm³ decreases.

Question 8.
Suggest a way to determine the \(\wedge_{m}^{0}\) value of water.
Answer:
Conductance of weak electrolytes can be determined by kohlraush’s law. Thus, molar conductance of water of infinite dilution can be determined by the molar conductances of NaOH, HCl and NaCl at infinite dilution.

Question 9.
The molar conductivity of 0.025 mol L^-1 methanoic acid is 46.1 S cm² mol^-1 Calculate its degree of dissociation and dissociation constant Given \(\lambda_{\left(\mathbf{H}^{+}\right)}^{\circ}\) = 349.6 S cm² mol^-1 and \(\lambda_{\left(\mathrm{H} \mathrm{COO}^{-}\right)}^{\circ}\) = 54.6 S cm² mol.
Solution:

Question 10.
If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?
Solution:
Q (coulomb) = 1 (ampere) × t (sec)
Q = 0.5 ampere × 2 × 60 × 60
= 3600 C
A flow of IF, i.e. 96500 C is equivalent to flow of 1 mole of electrons
i. e., = 6.023 × 10²³ electrons
3600 C is equivalent to flow of electrons

Question 11.
Suggest a list of metals that are extracted electrolytically.
Answer:
The metals like Na, K, Mg, Al, Ca etc. i.e., reactive metals can be extracted electrolytically.

Question 12.
What is the quantity of electricity in coulombs needed to reduce 1 mol of Cr₂O^-2₇ ? Consider the reaction :

Answer:
1 mole Cr₂O^-2₇ requires 6 moles electrons for reduction.
∴ Required charge = 6F
= 6 × 96500 coulomb.
= 579000 coulomb.

Question 13.
Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Answer:
Lead storage battery : It is a secondary cell i.e., a cell which is rechargeable because the products of cell reaction sticks to the electrode. It is also called as lead storage cells. It consists of six cells connected in series. Each cell consist of spongy lead anode and a grid of lead packed with lead dioxide (PbO₂) acts as cathode. An aqueous solution of H₂SO₄ (38% by mass) acts as electrolyte. The reactions which takes place at electrodes can be represented as:

Concentration of H₂SO₄ decreases as sulphate ions are consumed to form PbSO₄ during the working of the cell. As a result of this the density of solution also decreases.

Recharging the cell / battery: Lead storage battery can be recharged by connecting it to an external source of direct current. This reverses the flow of electron with the deposition of Pb on the anode and PbO₂ on the cathode. That is, during recharging operation the cell behaves as electrolytic cell. Following reaction occurs during recharging.

Question 14.
Suggest two materials other than hydrogen that can be used as fuels in fuel cells.
Answer:
Methane, Methanol.

Question 15.
Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Answer:
Formation of carbonic acid takes place on the surface of iron

In presence of H⁺ ion, oxidation of iron takes place

The electrons are used at other spot where reduction takes place.

Overall reaction,

Electrochemistry NCERT TextBook Exercises

Question 1.
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
Answer:
A metal with lesser standard potential (more reactive) can displace the other metal from solution of its salts.

Question 2.
Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74V. Arrange these metals in their increasing order of reducing power.
Answer:
Less the electrode potential, more will be the reducing power.
Ag < Hg < Cr < Mg < K.

Question 3.
Depict the galvanic cell in which the reaction Zn_(s) + 2Ag⁺_(aq) → Zn²⁺_(aq)+ 2Ag_(s) takes place. Further show :
(i) Which of the electrode is negatively char-ged ?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Answer:

(i) Anode i.e., Zn electrode is negatively charged.
(ii) Electrons are current carriers.

Question 4.
Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

Calculate the ∆_rG°, and equilibrium constant of the reactions.
Solution:

Question 5.
Write the Nernst equation and emf of the following cells at 298 K :

Solution:

Question 6.
In the button cells widely used in watches and other devices the following reaction takes place:

Solution:

Question 7.
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer:
Conductivity : Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section.

Molar conductivity : Molar conductivity of a solution at a dilution (V) is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are one cm apart and area of cross-section of the electrodes is so large that the whole of the solution is contained between them. It is usually represented by Λm.

Variation with concentration : The conductivity of a solution (Both for strong and weak electrolytes) decreases with decrease in concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increase in the decrease in concentration of the electrolyte i.e., on dilution. This is due to the decrease in the number of ions per unit volume of the solution on dilution. The molar conductivity of a solution increases with decrease in concentration of the electrolyte. This is because both number of ions as well as mobility of ions increases with dilution. When concentration approaches zero, the molar conductivity is known as limiting molar conductivity.

Question 8.
The conductivity of 0.20 M solution of KCI at 298 K is 0.0248 Scm^-1. Calculate its molar conductivity.
Solution:

Question 9.
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500Ω What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^-3 S cm^-1.
Solution:

Question 10.
The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below :
Concentration/M 0.001 0.010 0.020 0.050 0.100 10² × kSm^-1 1.237 11.85 23.15 55.53 106.74 for all concentrations and draw a plot between Λ_m and C^1/2 Find the value calculate Λ_m and C^1/2 Find the value calculate Λ_m of Λ⁰_m.
Answer:

Λ⁰_m = intercept on Λ_m axis = 124-0 S cm² mol^-1 (on extrapolation to zero concentration)

Question 11.
Conductivity of 0.00241 M acetic acid is 7.896 × 10^-5 S cm^-1. Calculate its molar conductivity and if Λ⁰_m for acetic acid is 390.5 S cm² mol^-1, what is its dissociation constant ?
Solution:

Question 12.
How much charge is required for the following reductions :
(i) 1 mol of Al³⁺ to Al.
(ii) 1 mol of Cu²⁺ to Cu.
(iii) 1 mol of MnO^–₄ to Mn²⁺.
Solution:
(i) The electrode reaction is :

(ii) The electrode reaction is :

(iii) The electrode reaction is :

Question 13.
How much electricity in terms of Faraday is required to produce (i) 20.0 g of Ca from molten CaCl₂.
(ii) 40.0 g of Al from molten Al₂O₃.
Solution:

Question 14.
How much electricity is required in coulomb for the oxidation of (i) 1 mole of H₂O to O₂
(ii) 1 mole of FeO to Fe₂O₃.
Answer:
(i) 2H₂O → O₂ + 4H⁺ + 4e^–
2mole – 4F
lmole – 2F
Electricity required for one mole H₂O = 2F = 2 × 96500 coulomb.

(ii) 4FeO + O₂ → 2Fe₂O₃
or Fe²⁺ → Fe³⁺ + e^–
1 mole – IF
1 × 96500 coulomb.

Question 15.
A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode ?
Answer:
According to Faraday’s first law:
W = ZIr [z = \(\frac { M }{ nF } \)]

Question 16.
Three electrolytic cells A, B, C containing solutions of ZnSO₄, AgNO₃ and CuSO₄, respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow ? What mass of copper and zinc were deposited ?
Solution:

Question 17.
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible :
(i) Fe³⁺_(aq) and I^–_(aq)
(ii) Ag⁺_(aq) and Cu_(s)
(iii) Fe³⁺_(aq) and Br^–_(aq)
(iv) Ag_(s) and Fe³⁺_(aq)
(v) Br_2(aq) and Fe²⁺_(aq)
Answer:
A reaction is feasible if EMF of the cell is +ve.
Cathode : At which reduction occurs.
Anode : At which oxidation occurs.

Question 18.
Predict the products of electrolysis in each of the following :
(a) An aqueous solution of AgNO₃ with silver electrodes.
(b) An aqueous solution of AgNO₃ with platinum electrodes.
(c) A dilute solution of H₂SO₄ with platinum electrodes.
(d) An aqueous solution of CuCl₂ with platinum electrodes.
Answer:
(a) At cathode : Possible reactions are :
(i) Reduction of Ag⁺ E°_Ag = 0.80V
(ii) Reduction of H₂O E°_H2O = -0.83 V
∴ Ag⁺ ion reduced because its E°_Ag > E°_H2O
At anode : Possible reactions are :
(i) Oxidation of H₂O E° = 1.23 V
(ii) Oxidation of Ag E° = 0.80V
Since E°_Ag < E°H₂O ∴ Ag gets oxidised.

(b) Possible reactions at cathode :
(i) Reduction of Ag⁺ E°_g = 0.80V
(ii) Reduction of water E°_H2O = -0.8 V .
Since E°_Ag > E°H₂O ∴Ag⁺ gets reduced.
Possible reactions at anode :
Oxidation of water E°_Ag = 1.23
∴ O₂ gas is liberated at anode.

(c) Possible reaction at cathode :
(i) Reduction of H⁺ E°_H+/H2 0.0
(ii) Reduction of H₂O E°H₂O = -0.87
Since E°_H⁺/H2 > E°H₂O ∴ H₂ gas is liberated at cathode.
Possible reaction at anode :
Oxidation of water E°_H2O = 1.23
∴ O₂ gas is liberated at anode.

(d) Possible reaction at cathode :
(i) Reduction of water E° H₂O = -0.87V
(ii) Reduction of Cu²⁺ E°Cu²⁺/Cu = 0.34V
Since E°_cu²⁺/cu > E°_H2O ∴ Cu²⁺ gets reduced at cathode.
Possible reaction at anode :
(i) Oxidation of water E°H₂O = 1.23V
(ii) Oxidation of Cl- E° Cl^–/Cl₂ = 1.36V
Cl₂ gas is liberated at anode because oxidation of water is kinetically slow.

Electrochemistry Other Important Questions and Answers

Electrochemistry Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
If specific conductance = K, R = Resistance, l = distance between the electrodes A = Cross-sectional area of a conductor C_m = mol L^-1, C_eq = g.eq L^-1 then specific resistance of electrolyte will be equal to :

Question 2.
K is equal to :

Question 3.
Molar conductivity Λ_m is equal to :

Question 4.
Cell constant is :

Question 5.
For electrolyte (v₊ = v_– = 1) like NaCI:

Question 6.
Which among the following is not a conductor of electricity :
(a) NaCl_(aq)
(b) NaCl_(s)
(c) NaCl_(mol)
(d) Ag.

Question 7.
By which of the following the resistance of the solution is multiplied to obtain cell constant:
(a) Specific conductance (K^–)
(b) Molar conductance (Λ_m)
(c) Equivalent conductance (Λ_eq)
(d) None of these.

Question 8.
Increase in equivalent conductance of the solution of an electrolyte by dilution is due to:
(a) Increase in ionic attraction
(b) Increase in molecular attraction
(c) Increase in association of electrolyte
(d) Increase in ionization of electrolyte.

Question 9.
If the specific conductance and observed conductance of an electrolyte is same then its cell constant will be :
(a) 1
(b) 0
(c) 10
(d) 1000

Question 10.
Unit of cell constant is :
(a) ohm^-1 cm^-1
(b) cm
(c) ohm cm
(d) cm^-1.

Question 11.
Unit of specific conductance is :
(a) ohm^-1
(b) ohm^-1 cm^-1
(c) ohm^-2 cm^-1 equivalent^-1
(d) ohm^-1 cm^-2.

Question 12.
If the concentration of any solution is C gram equivalent/ litre and specific resistance is A, its equivalent conductance will be :

Question 13.
The coating of layer of zinc on iron to prevent it from corrosion is called :
(a) Galvanization
(b) Cathodic protection
(c) Electrolysis
(d) Photoelectrolysis.

Question 14.
A saturated solution of KNO₃ is used for salt bridge because :
(a) Speed of K⁺ is more than NO^–₃
(b) Speed of NO^–₃ is more than K⁺
(c) Speed of both is nearly same
(d) Solubility of KNO₃ is high in water.

Question 15.
Acts as dipolar in dry cells :
(a) NH₄Cl
(b) Na₂CO₃
(c) pbSO₄
(d) MnO₂

Question 16.
Reduction is known as :
(a) Electronation
(b) De-electronation
(c) Protonation
(d) De-protonation.

Question 17.
What is the path of electric current in a Daniel cell when Zn and Cu electrodes are connected:
(a) From Cu to Zn inside the cell
(b) From Cu to Zn outside the cell
(c) From Zn to Cu inside the cell
(d) From Zn to Cu outside the cell.

Question 18.
In a cell containing Zn electrode and normal hydrogen electrode (NHE), Zn acts like:
(a) Anode
(b) Cathode
(c) Neither cathode nor anode
(d) Both anode and cathode.

Question 19.
If salt bridge is removed from half cells then voltage :
(a) Reduces and becomes zero
(b) Increases
(c) Immediately increases
(d) Does not change.

Question 20.
When lead storage battery is discharged :
(a) SO₂ is released
(b) Pb is manufactured
(c) PbSO₄ is used
(d) H₂SO₄ is used.

Question 21.
Process of Rusting of iron is :
(a) Oxidation
(b) Reduction
(c) Corrosion
(d) Polymerisation.

Question 22.
Value of standard potential of hydrogen electrode is :
(a) Positive
(b) Negative
(c) Zero
(d) No definite value.

Answers:
1. (b), 2. (c), 3. (d), 4. (b), 5. (a), 6 (b), 7. (a), 8. (d), 9. (a), 10. (d), 11. (b), 12. (a), 13. (a), 14. (c), 15. (d), 16. (a), 17. (d), 18. (a), 19. (a), 20. (d), 21. (c), 22. (c).

Question 2.
Fill in the blanks :

1. Acetic acid is a ………………… electrolyte.
2. Conductance of electrolyte ………………… with the increase in temperature.
3. On increasing dilution the value of specific conductance of a solution …………………
4. ………………… decreases with increase in size of ion.
5. Unit of specific resistance is …………………
6. Primary cells cannot be ………………… again.
7. Device which converts chemical energy into electrical energy is known as ………………… cell.
8. Amount of electric current which produces one gram equivalent of a substance is known as …………………
9. In metallic conduction ………………… property remains unchanged.
10. Reciprocal of resistance is known as …………………
11. Conductance of 1 cm cube of a conductor is called …………………
12. 1 Faraday is equal to ………………… coulomb.
13. Rusting of iron is an example of …………………
14. Potential of standard hydrogen is assumed to be …………………

Answers:

1. Weak
2. Increases
3. Decreases
4. Conductance
5. ohm cm
6. Charged
7. Electrochemical
8. Faraday
9. Chemical property
10. Conductance
11. Specific conductance
12. 96500 coulomb
13. Corrosion process
14. 0.0 volt
15. Electrochemical cell.

Question 3.
Match the following :

I.

Answers:

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

II.


Answer:

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

Question 4.
Answer in one word / sentence :

1. Give two examples of strong electrolyte.
2. Give two examples of weak electrolytes.
3. What is the effect of temperature on electrolytic conductivity ?
4. Write the formula of Kohlrausch’s law.
5. Write the formula of equivalent conductance.
6. Write the formula of molar conductance.
7. Device which converts electrical energy into chemical energy.
8. What is the potential of both the electrodes of the cell due to which electric current flows in the cell ?
9. What is the potential produced due to redox reaction between a metal electrode and its ions called ?
10. What is the unit of potential difference ?
11. Write the formula of cell constant.
12. Cell which can be recharged are known as.
13. State the unit of Equivalent conductance.
14. What is the chemical composition of rust ?
15. Write the relation between Electromotive force and Equilibrium constant of a cell.
16. What is the name of the reaction in which oxidation and reduction occur simultaneously ?
Answers:
1. Strong electrolyte : HCl, NaOH, NaCl
2. Weak electrolyte : CH₃COOH, H₂CO₃,
3. Conductivity increases
4.

5. Equivalent conductance

6.

7. Electrolytic cell
8. Electromotive force
9. Electrode potential
10. Volt
11. Cell constant = \(\frac { l }{ A } \)
12. Secondary cell
13. Ohm cm² gm eq^-1
14. Fe₂O₃ . xH₂O
15.

16. Redox reaction.

Electrochemistry Very Short Answer Type Questions

Question 1.
Write the definition of Electrochemical cell.
Answer:
System in which chemical energy is converted to electrical energy by oxidation reduction is known as electrochemical cell or voltaic cell.

Question 2.
What is an Electrolytic cell ?
Ans.
The container or system in which electrical energy is passed by which chemical reaction takes place thus, electrical energy is converted to chemical energy is known as electrolytic cell.

Question 3.
What is Electrode potential ?
Answer:
The potential difference developed between the electrodes and electrolyte of an Electrolytic cell is known as Electrode potential.

Question 4.
What is a strong electrolyte ? Write two examples.
Answer:
Electrolyte which completely dissociate in aqueous solution are known as strong electrolyte.
Example : NaCl, KCl, NH₄Cl etc.

Question 5.
What is a weak Electrolyte ? Write two examples.
Answer:
Electrolyte which dissociate partially in aqueous solution are known as weak electrolyte.
Example : NH₄OH, CH₃COOH, HCN etc.

Question 6.
What is meant by standard electrode potential ?
Answer:
Standard electrode potential (E°) of a half cell is the potential difference when one electrode is dipped in molar solution of its ion at 298 K. If electrode is gaseous the pressure of gas must be one atmosphere.

In IUPAC system, reduction potential are known as standard electrode potential.

Question 7.
Write Ohm’s law.
Answer:
According to Ohm’s law “It states that potential difference across the conductor is directly proportional to the current (I) flowing through it” i.e.,
Mathematically, it can be written as :
I α V
V = IR (R = Resistance, unit = ohm, Ω)

Question 8.
What is cell constant ?
Answer:
For a conductivity cell, the ratio of distance between two electrodes (l) and area of cross-section of electrode (A) is called as cell constant.
∴ Cell constant = \(\frac { l }{ A } \) or x = \(\frac { l }{ A } \)
Unit of cell constant = cm^-1.

Question 9.
What is galvanization ? Explain.
Answer:
Iron is coated with the layer of zinc to protect it from rusting. This process is known as galvanization. The galvanized iron articles keep their lustre due to the coating of invisible protective layer of basic zinc carbonate, ZnCO₃.Zn(OH)₂.

Question 10.
What is Electrochemical Equivalent ?
Answer:
Electrochemical equivalent of a substance is that mass of substance released or deposited on an electrode when a current of one ampere is passed for one second.

Electrochemistry Short Answer Type Questions

Question 1.
What is salt bridge ? Write its two functions.
Answer:
‘U’ shaped tube filled with KCl or KNO₃ in Agar-Agar solution or gelatin, is known as salt bridge. It connects the two half cell.
Functions : (i) It allows the flow of current by completing the circuit.
(ii) It maintains the electrical neutrality.

Question 2.
Derive relation between standard electromotive force and equilibrium constant
Answer:
Relation between standard electromotive force and equilibrium constant can be derived using van’t Hoff isochore. For any given reaction equilibrium constant IQ is equal to the ratio of rate constant of forward reaction and rate constant of backward reaction.

Value of equilibrium constant Kc can be calculated using standard free energy change (∆G°) because

Question 3.
What do you understand by oxidation-reduction reactions ?
Answer:
Oxidation-Reduction reactions: Chemical reactions in which valency of elements changes are known as oxidation-reduction reactions. In this process both oxidation and reduction reactions occur simultaneously, in which one of the substance is oxidized and the other substance is reduced. Like

In this reaction FeCl₃ is reduced to FeCl₂ and SnCl₂ is oxidized to SnCl₄. In the reaction valency of Fe decreases and valency of Sn increases.

Question 4.
Write difference between Metallic conduction and Electrolytic conduction.
Answer:
Differences between Metallic conduction and Electrolytic conduction :

Metallic conduction:

1. Metallic conduction takes place by movement of electrons.
2. There is no chemical change.
3. There is no transfer of matter.
4. In metallic conduction conductivity decreases with increase in temperature.

Electrolytic conduction

1. Electrolytic conduction takes place by movement of ions.
2. Due to chemical change decomposition of electrolyte takes place.
3. Transfer of matter takes place as ions.
4. In electrolytic conduction conductivity increases with increase in temperature.

Question 5.
What are the difference between emf (Cell potential) and potential difference
Answer:
Difference between EMF and Potential difference :

EMF / Cell potential:

1. It is the potential difference between the two terminals of the cell when no current is flowing in the circuit, i.e., in an open circuit.
2. It is the maximum voltage which can be obtained from a cell.
3. It can be measured by potentiaometrie method.
4. Work performed by electromotive force is the maximum work done by a cell.
5. It is responsible for continuous flow of current in electric circuit.

Potential difference:

1. It is the difference of the electrodes potentials of the two electrodes when the cell is sending current through the circuit.
2. It is the less than the maximum voltage as it is the difference of electrode potential.
3. It can be measured by simple voltmeter also.
4. Work performed by potential difference is less than the maximum work done by a cell.
5. It is not responsible for the continuous flow of current in circuit.

Question 6.
What is specific conductance ? Give its unit.
Answer:
Specific conductivity: The reciprocal of resistivity is called specific conductiv¬ity. It is defined as the conductance between the opposite faces of one centimeter cube of a conductor. It is denoted by K (kappa).


The specific conductivity of a solution at a given dilution is the conductance of one cm cube of the solution. It is represented by K (kappa).
Note : The specific conductivity of a solu¬tion of electrolyte depends upon the dilution or molar concentration of the solution.

Question 7.
What is resistivity of any solution ?
Answer:
Resistivity : When current flow in the solution through two electrodes the resistance is proportional to length and inversely proportional to cross-sectional area A.

The constant p (rho) is called resistivity or specific resistance.
Unit: If l is expressed in cm, A in cm2 and R in ohm, the unit of resistivity will be

or Resistivity of any solution is the resistance of 1 cm cube.

Question 8.
Differentiate between Electrochemical cell (Galvanic cell) and Electrolytic cell.
Answer:
Differences between Electrochemical and Electrolytic cells :

Electrochemical cell or Galvanic cell:

1. It is a device to convert chemical energy into electrical energy.
2. It consists of two electrodes in different compartments joined by a salt bridge.
3. Redox reactions occurring in the cell are spontaneous.
4. Free energy decreases with operation of cell, i.e., ∆G < 0.
5. Useful work is obtained from the cell.
6. Anode works as negative and cathode as positive electrodes.
7. Electrons released by oxidation process at anode go into external circuit and pass to cathode.
8. To set-up this cell, a salt bridge/porous pot is used.

Electrolytic cell:

1. It is a device to convert electrical energy into chemical energy.
2. Both the electrodes are in same solution.
3. Redox reactions occurring in the cell are non-spontaneous.
4. Free energy increases with operation of cell, i.e., ∆G > 0.
5. Work is done on the system.
6. Anode is positive and cathode is negative.
7. Electrons enter into cathode electrode from external source and leave the cell at anode.
8. No salt bridge is used in this cell.

Question 9.
What is equivalent conductance ?
Answer:
Equivalent conductance:
“Conductance of total ion produced by one gram equivalent of electrolyte in the solution is called equivalent conductance.” It is denoted by Λ_eq.

Question 10.
What is molar conductance ?
Answer:
Molar Conductivity : The molar conductivity of a solution at definite concentration (or dilution) and temperature is the conductivity of that volume which contains one mole of the solute and is placed between two parallel electrodes 1 cm apart and having sufficient area to hold whole of the solution. It is denoted by Λ_m.
Mathematically,
Λ_m = K × V …(1)
Where V is the volume in ml in which one gram mole of substance is dissolved.
If M is molarity or m moles are dissolved in 1000 ml.

Question 11.
Define cell constant Develop a relation between specific conductance and cell constant.
Answer:
Cell constant: In any conductive cell, the distance between two electrodes and surface area of electrode A are constant. The ratio of l and a is called cell constant i.e.

Unit of cell constant is cm^–1 and it is generally expressed by x.
Relation between specific conductance and cell constant : For a conductor, the resistance R is directly proportional to length R and inversely proportional to area of cross – section of electrolyte.

Question 12.
What are the factors which influence the electrical conductance of electrolytes ?
Answer:
Factor which influence electrical conductivity of electrolytes :

The main factor which influence the electrical conductivity are following:

1. Temperature: It influence following interactions.
(a) Interionic attractions: It depends upon the solute-solute interactions. Which is found between the ions of solute.
(b) Solvation of ions: It depends upon solute-solvent interactions. It is relation between ions of solute and solvent molecules.
(c) Viscosity of solvent: It depends upon solvent-solvent interactions. Solvent molecules are related with each other.
With increase in temperature all these three effects decrease and average kinetic energy of ions increases. Thus, with increase of temperature, resistance of solution decreases and hence conductance increases.

2. Nature of electrolyte : The conductance of solution depends upon the nature of electrolyte. On the basis of conductance measurement electrolytes are classified as strong electrolyte and weak electrolyte. Strong electrolytes have high value of conductance even at higher concentration also.

3. Dilution or concentration : It is main factor which influence electrical conductance. Effect of dilution or concentration can be studied indivisually in equivalent conductance, specific conductance and molar conductance. But for a general concept of electrical conductance of solution as the concentration is lowered or dilution increases, electrical conductance of whole solution increases.

Question 13.
On what factors does the various conductivities of an electrolytic solution depend ?
Answer:
Conductivities of electrolytic solution depend on the following factors :

1. Dilution : On increasing dilution, value of specific conductance of a solution decreases, value of equivalent conductance and molar conductance increases.
2. Nature of solvent: A solvent with high dielectric constant has high conductivity and with low dielectric constant has low conductivity.
3. Number of ions present in solution: Conductivity of strong electrolytes is higher than the conductivity of weak electrolytes.
4. Size of ion : In aqueous solution, small ions are heavily hydrated due to which their conductivity decreases.
5. Effect of Temperature: With the increase in temperature conductivity increases.

Question 14.
With the increase in dilution how do specific conductance, Equivalent conductance and molar conductance change ?
Answer:
With the increase in dilution, specific conductance decreases. This is because by the increase in dilution number of ions present in 1 cm cube of solution decreases.
But, Equivalent conductance Λ_eq = K × V
and Molar conductance Λ_m = K × V
Equivalent conductance and molar conductance are the product of specific conductance and dilution. By the increase in dilution (or decrease in concentration) magnitude of K decreases but that of V increases.
Increase in magnitude of V is comparatively much more than the decrease in magnitude of K. Thus, by the combined effect of both, by the increase in dilution Λ_eq and Λ_m increases.

Question 15.
What is an Electrolytic cell and how does it work ?
Answer:
Electrolytic cells : In these cells electric current is supplied through an external source, as a result of which chemical reactions take place which is called electrolysis like : Electrolysis of water, NaCl, Al₂O₃ etc. For example in Solvay trough cell electrode is immersed in sodium chloride solution and electric current is passed due to which NaCl electrolyses.

At mercury cathode sodium is released and at anode chlorine is released. Sodium forms amalgam with mercury and is taken out of the cell.

Question 16.
What is meant by electromotive force of an electrochemical cell ?
Answer:
The difference in electrode potentials of the two electrodes of an electro- chemical cell is known as electromotive force or cell potential. It is expressed in volt.

Due to difference in potential electric current flows from an electrode of lower potential to an electrode of higher potential. EMF of the cell can be expressed in terms of reduction potential as :

Cell potential = Standard electrode potential – Standard electrode potential
of R.H.S. electrode of L.H.S electrode

EMF of a cell is measured by connecting the voltmeter between the two electrodes of a cell. EMF of a cell depend on the concentration of solutions of both half cells and nature of the two electrodes. For example, In Daniel cell, concentration of CuSO₄ and ZnSO₄ solutions in the two half cells is 1M and at 298 K EMF of the cell is 1.10 volt.

Electrochemistry Long Answer Type Questions

Question 1.
What is standard hydrogen electrode ? How is it prepared ?
Answer:
Standard hydrogen electrode : This consists of gas at 1 atmospheric pressure bubbling over a platinum electrode immersed in 1 M HC1 at 25°C (298 K) as shown in figure. The platinum electrode is coated with platinum black to increase its surface. The hydrogen electrode thus con¬structed forms a half cell which on coupling with any other half cell begins to work on the principle of oxidation or reduction. Electrode depending upon the circumstances works both as anode or cathode.

Standard hydrogen electrode (SHE) is arbitrarily assigned a potential of zero.

Question 2.
Derive Nernst Equation for single electrode potential.
Answer:
Value of standard electrode potential given in electrochemical series is applicable only when the concentration of electrolyte is 1M and temperature is 298 K. But in electrochemical cells the concentration of electrolyte is not definite and electrode potential depends on concentration and temperature. In such condition single electrode potential can be expressed by Nernst equation.
For a reduction half reaction, Nernst equation can be expressed as follows :

Where E = Reduction electrode potential
E° = Standard electrode potential (Mⁿ⁺ concentration 1M and at 298 K)
R = Gas constant = 8.31 JK^-1 mol^-1 = Temperature (in kelvin) = 298 K
n = Valency of metal ion, F = 1 Faraday (96,500 coulomb)

Equation (2) is Nernst equation for single electrode potential.

Question 3.
Write the Faraday’s laws of electrolysis.
Answer:
Faraday’s first law of electrolysis : The law states that, “The mass of any substance deposited or liberated at any electrode is directly proportional to the quantity of electricity passed.”
Thus, if W gm of the substance is deposited on passing Q coulomb of electricity, then
W α Q or W = ZQ
Where, Z is a constant of proportionality and is called electrochemical equivalent of the substance deposited. If a current of I ampere is passed for t second, then Q = I × t. So that,
W = Z × Q = Z × I × t
Thus, if Q = 1 coulomb, I = 1 ampere and t = 1 second, then W = Z. Hence, electrochemical equivalent of a substance may be defined as, “The mass of the substance deposited when a current of one ampere is passed for one second.”

As one faraday (96500 C) deposits one gram equivalent of the substance, hence electrochemical equivalent can be calculated from the equivalent mass.
Faraday’s second law of electrolysis : It states that, “When the same quantity of electricity is passed through solutions of different electrolytes connected in series, the weight of the substances produced at the electrodes are directly proportional to their equivalent mass.”

For example, for CuSO₄ solution and AgNO₃ solution connected in series, if the same quantity of electricity is passed, then

Question 4.
What is rusting of iron ? Describe Electrochemical theory of rusting.
Answer:
Corrosion: Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric condition are called corrosion. Rusting of iron is an example of corrosion, chemically it is Fe₂O₃xH₂O.

Electrochemical theory of rusting :
Anode reaction : On one spot of iron sheet, oxidation takes place and this spot behaves as an anode.

The electrons which are released at this spot travel through the metal and reach another spot on the metal which acts as cathode. These electrons cause the reduction of oxygen in the presence of hydrogen ions (H⁺). H⁺ ions are formed due to decomposition of carbonic acid formed by dissolution of CO₂ in H₂O.

Question 5.
What is corrosion ? Write three factors affecting it and any three methods to prevent it.
Answer:
Corrosion : Process by which the layers of undesirable compounds are formed on the surface of a metal on its exposure to atmospheric conditions are called corrosion.

Factors affecting corrosion :

1. Nature of metal: Reactive metal corrodes readily.
2. Impurities in metal: Impure metal corrodes quickly to a greater extent.
3. Environment: If environment around metal contains oxygens, carbon dioxide, moisture, salts and acidic gases like CO₂, SO₂, SO₃ etc. then corrosion occurs quickly.

Methods to prevent corrosion :

1. Barrier protection : In this method, the surface of the metal is coated with paint, oil or grease. Due to this the surface of the metal remain unexposed to atmospheric conditions and hence corrosion is prevented. Surface of the metal can also be protected by :
(i) Coating metal surface with non-corroding metals like nickel and chromium is called electroplating.
(ii) Dipping iron article in molten metal like zinc. This process is called galvanization.

2. Sacrificial protection : The rusting of iron can be prevented by covering the iron with more electropositive metals like zinc. Zinc metal has more tendency to get oxidized as compared to iron. So, iron articles will not be harmed till the layer of zinc present on its surface hence zinc metal is called the sacrificial metal.

3. Antirust solution : The alkaline antirust solution are employed to prevent rusting, alkaline solution prevent the availability of H⁺ ions.

In this method, iron articles are dipped in alkaline sodium phosphate or chromate solution. Due to this an insoluble sticking film of iron phosphate is formed on the surface which prevents rusting.

Question 6.
Describe dry cell with labelled diagram.
Answer:
Dry cell: It is a primary cell based on Leclanche cell invented by G. Leclanche in 1868. In a primary cell, the electrode reactions cannot be reversed by an external source of electrical energy. In this cell, the cell reaction takes place only once i.e., this cell is not rechargeable.

It is generally used in torches, transistors, radios, calculators, tape recorders, etc. It consists of a hollow zinc cylinder which is filled with a paste of NH₄Cl and a little ZnCl₂. This paste is made with the help of water. The zinc cylinder acts as anode while cathode is a graphite rod (Carbon). The carbon rod is surrounded by a black paste of MnO₂ and carbon powder. The zinc cylinder has an outer insulation of cardboard case.

Dry cells are sealed with wax or other material to protect the moisture from evaporation. When the electrodes are connected, the cell operates.

The electrode reactions are complex. Metallic Zn is oxidized to Zn²⁺ and the electrons liberated are left on the container. The reactions which take place at electrodes can be represented as :

This reaction prevents polarization due to formation of ammonia. It also prevents the substantiaL increase of concentration of Zn²⁺ ions which would decrease the cell potential. This potential of dry cell is approximately 1.5 V.

Defect: Due to acidic nature of NH₄Cl zinc container corrodes due to which holes develop through which the chemicals come out.

Nowadays, the cells are made leakage resistant. In it KOH is used in place of NH₄Cl by which zinc does not corrode.

Question 7.
What is Kohlrausch law ? Give its two applications.
Answer:
Kohlrausch in 1875 gave a generalisation known as Kohlrausch’s law, “At infinite dilution when the dissociation of the electrolyte is complete, each ion makes a definite contribution towards molar conductance of the electrolyte irrespective of the nature of the other ion with which it is associated.”
Or
“The value of molar conductance at infinite dilution is given by the sum of the contributions of ions (cation and anion).”

Where, λ^∞₊ and λ^∞_– are ionic contributions or ionic conductances of cation and anion while v₊ and v_– are the number of cations and anions in the formula unit of electrolyte.

Applications of Kohlrausch’s law :

(i) Calculation of molar conductance at infinite dilution for weak electrolytes :

Molar conductance or equivalent conductance of weak electrolytes cannot be obtained graphically by extrapolation method, since these are feebly ionized. Kohlrausch’s law enables indirect evaluation in such cases. For example, molar conductances of acetic acid can be obtained from the knowledge of molar conductances at infinite dilution of HCl, CH₃COONa and NaCl which are strong electrolytes.

From Kohlrausch’s law, it is clear that

(ii) Determination of degree of dissociation :

Question 8.
Draw a labelled diagram of Daniel cell and explain cell reaction.
Or,
Draw a labelled diagram of electrochemical cell and write cell reaction.
Answer:
Electrochemical cell: In the redox reactions, the transfer of electrons between oxidizing and reducing agents occurs through wire and thus chemical energy changes into electrical energy. The device on which chemical energy changes into electrical energy is called electro chemical cell. These are also known as galvanic or voltaic cells. Working of these cells can be understood with the example of Daniel cell.

Daniel cell : In this cell, Zn rod is dipped in ZnSO4 solution and Cu rod in copper sulphate solution. Both solutions are connected through KC1 salt bridge. When Zn and Cu electrodes are connected by wire and galvanometer, flow of electrons from Zn to Cu occurs. Zinc atoms change into Zn2+ and electrons reach at Cu electrode, where Cu2+ changes into Cu metal and this copper deposits on electrode.

Electrochemistry Numerical Questions

Question 1.
On passing 5 ampere electric current for 30 minute through a container filled with AgNO₃ 10.07 gram silver is deposited, then determine the chemical equivalent of silver. If electrochemical equivalent of hydrogen is 0-00001036 then calculate the equivalent mass of Silver.
Solution:
Given : W = 10-07 gm, i = 5 ampere, t = 30 × 60 second
According to Faraday’s first law W = Zit
∴ 10.7 = Z × 30 × 60

Question 2.
What are weak electrolytes ? Give one example. Find out molar conductivity of LiBr aqueous solution infinite dilution when joint conductance of Li⁺ ion and Br^– ion are 38.7 Scm² mol^-1 and 78.40 Scm² mol^-1 respectively.
Solution:
Weak electrolytes : These are the substances which dissociate only to a small extent.
Examples: CH₃COOH,NH₄OH

Question 3.
What are strong electrolytes ? Find out the molar conductivity of aqueous solution of BaCl₂ at infinite dilution when ionic conductance of Ba⁺² ion and Cl^– ion are 127.30 Scm² mol^-1 and 76.34 Scm² mol^-1 respectively.
Solution:
Strong electrolytes : These are substances which dissociate almost completely into ions under all dilutions.
Examples: NaCl, HCl,CH₃COONa

Question 4.
Calculate the molar conductance of Al₂(SO₄)₃ if


Solution:
By the ionisation of Al₂(SO₄)₃ solution

Question 5.
Molar conductivity of \(\frac { M }{ 30 } \) CH₃COOH is 9.625 mho and molar conductance of CH₃COOH at infinite dilution (Λ^∞_m) is 385 mho. Calculate the percentage of dissociation of \(\frac { M }{ 30 } \) CH₃COOH.
Solution:
Degree of dissociation

Question 6.
Calculate molar conductance of acetic acid at infinite dilution from following values:

Solution:
From Kohlrausch’s law,

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Chemistry Solutions Chapter 1 The Solid State

The Solid State NCERT Intext Exercises

Question 1.
Why are solids rigid ?
Answer:
Solids are rigid due to presence of strong inter- molecular forces between the constituent particles.

Question 2.
Why do solids have a definite volume ?
Answer:
In solids, the constituent particles are bonded together by strong attractive forces between them. By the increase or decrease of pressure, the intermolecular space remains unaffected. Thus, volume of solids is definite.

Question 3.
Classify the following as amorphous or crystalline solids : Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, poly vinyl chloride, fibre glass, copper.
Answer:
Amorphous solids: Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass.
Crystalline solids : Benzoic acid, potassium nitrate, copper.

Question 4.
Why is glass considered a super cooled liquid ?
Answer:
Like liquids, glass has tendency to flow but very slowly. Therefore, it is called super cooled liquid.

Question 5.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property ?
Answer:
Solid is amorphous because amorphous solids are isotropic in nature. No, it would not show the cleavage property.

Question 6.
Classify the following solids in different categories based on the nature of intermolecular forces operating in them : Potassium sulphate, tin, benzene, urea, ammonia, water, zinc sulphide, graphite, rubidium, argon, silicon carbide.
Answer:
Potassium sulphate : Ionic, Tin : Metallic, Benzene : Molecular (non-polar), Urea: Molecular (polar), Ammonia: Molecular (H-bonded), Water: Molecular (H-bonded), Zinc sulphide: Ionic, Graphite: Covalent or network, Rubidium: Metallic, Argon: Molecular (non-polar), Silicon carbide : Covalent or network.

Question 7.
Solid A is a very hard electrical insulator in solid as well as in molten state and melts at extremely high temperature. What type of solid is it ?
Answer:
Covalent solid.

Question 8.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In Solid state, ions are not free therefore ionic solids are bad conductor. How¬ever, in molten state, the ions become free to conduct electric current.

Question 9.
What type of solids are electrical conductors, malleable and ductile ?
Answer:
Metallic solids, they are conductor due to the presence of free electron in them.

Question 10.
Give the significance of a ‘lattice point’.
Answer:
Each lattice point represents one constituent particle of the solid. This constituent particle may be an atom, a molecule (group of atoms) or an ion.

Question 11.
Name the parameters that characterise a unit cell.
Answer:
A unit cell is characterized by :

1. The dimensions along the three edges. These are represented by a, b, and c.
2. The angle between the edges. These are represented by α,β, and γ. The angle α is between b and c,β is between a and c and γ is between a and b (For diagram refer book).

Question 12.
Distinguish between (i) Hexagonal and monoclinic unit cells (ii) Face centred and end centred unit cells.
Answer:

1. For Hexagonal unit cell, a = b ≠ c, α = β = 90°, γ = 120°.
   For monoclinic unit cell a ≠ b ≠ c, α = γ = 90°, β = 90°.
2. Face centred unit cell has points at the comers as well as the centre of each face. It has 4 atoms per unit cell.
   End centred unit cell has points at all the comers and at the centre of any two opposite faces. It has 2 atoms per unit cell.

Question 13.
Explain, how much portion of an atom located at (i) corner and (ii) body centre of a cubic unit cell is part of its neighbouring unit cell ?
Answer:

1. 1/8th part of an atom located at comer belongs to neighbouring unit cell.
2. The atom at the body centre of a cubic unit cell is not shared by any other unit cell. Hence, it belongs fully to the unit cell.

Question 14.
What is the two-dimensional co-ordination number of a molecule in square close packed layer ?
Answer:
4.

Question 15.
A compound forms hexagonal close packed structure. What is the total number of voids in 0-5 mol of it ? How many of these are tetrahedral voids ?
Answer:
An atom in hep structure has three voids, one octahederal and two tetrahederal.
Number of atoms in 0-5 mol = 0.5 × 6.022 × 10²³ = 3.011 × 10²³
Total number of voids = 3 × 3.011 × 10²³ = 9.033 × 10²³
Number of tetrahedral voids = 2 × 3.011 × 10²³ = 6.022 × 10²³.

Question 16.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupy 1/3rd of tetrahedral voids. What is the formula of the compound ?
Answer:
Since, N forms ccp arrangement, it will have 4 atoms in a unit cell.
Number of N atoms in unit cell = 4
For each atom, there are two tetrahedral voids so that there are 8 tetrahedral voids per unit cell.
No. of M atoms = \(\frac { 1 }{ 3 } \) × 8 = \(\frac { 8 }{ 3 } \)
Formula = M_8/3 N₄ or M₂N₃.

Question 17.
Which of the following lattices has the highest packing efficiency (i) simple cubic (ii) body centred cubic and (iii) hexagonal close packed lattice ?
Answer:
The packing efficiencies are :
Simple cubic = 52.4%
Body centred cubic = 68%
Hexagonal close packed = 74%
∴ Hexagonal close packed lattice has highest packing efficiency.

Question 18.
An element with molar mass 2.7 × 10^-2 kg mol^-1 forms a cubic unit cell with edge length 405 pm. If its density is 2.7 × 10³ kg m^-3, what is the nature of the cubic unit cell ?
Answer:
We know that Z = \(\frac{a^{3} \times \mathrm{N}_{\mathrm{A}} \times d}{\mathrm{M}}\)
Where a = 405pm = 405 x 10^-10
d = 2.7 × 10³ kg m^-3
M = 2.7 × 10^-2 kg m^-1
= 2.7g mol^-1
N_A = 6.203 x 10²³

Z =  4
∴ The element has fee (cep) unit cell.

Question 19.
What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what way ?
Answer:
Vacancy defect is created when a solid is heated. This is because on heating some atoms or ions leave the lattice site completely. As a result the density of substance decreases.

Question 20.
What type of stoichiometric defect is shown by : (i) ZnS, (ii) AgBr.
Answer:
(i) ZnS, shows Frenkel defect due to large difference in size of ions.
(ii) AgBr, shows both Frenkel defect and Schottky defect.

Question 21.
Explain, how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it ?
Answer:
When a cation of higher valency is added as an impurity in the ionic solid, some of the site of the original cations are occupied by the cations of higher valency. For example, Sr⁺² in NaCl. Each Sr⁺² replaces two Na⁺ ions. It occupies the site of one Na⁺ ion and the other site remains vacant. The cation vacancies thus produced are equal in number to that Sr⁺² ions.

Question 22.
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour ? Explain with the help of a suitable example.
Answer:
The anionic vacancies due to metal excess defect in ionic solids are occupied by free electrons to maintain the electrical neutrality. These impart colour by excitation of these electrons when they absorb energy from the visible light falling on the crystals. For example :
When NaCl is heated in presence of sodium vapours, Na⁺ ions are in excess, Cl^– ions leave their normal site and come to the surface. The vacant site of anion is occupied by electron forming F-centre. They absorb light from visible region and radiate complementary colour.

Question 23.
A group-14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ?
Answer:
n-type semiconductors are obtained by doping of a higher group impurity. Hence, to convert group 14 element to n-type semiconductor, it should be doped with a group 15 element.

Question 24.
What type of substances would make better permanent magnets, ferro-magnetic or ferrimagnetic ? Justify your answer.
Answer:
Ferromagnetic materials would make better permanent magnets than ferrimagnetic materials because in ferromagnetic solids, the magnetic moments of unpaired electrons spontaneously align themselves in same direction. However, in ferrimagnetic solids the magnetic moments of the domains are aligned in parallel and anti-parallel direction in unequal numbers.

The Solid State NCERT TextBook Exercises

Question 1.
Define the term ‘amorphous’. Give a few examples of amorphous solids.
Answer:
In non-crystalline solids constituent particles like atoms, molecules or ions do,not have a definite ordered structure. These do not have a definite geometry, thus they are also known as pseudo solid.
Example : Glass, rubber, plastic.

Question 2.
What makes a glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass ?
Answer:
Quartz is a crystalline solid whereas glass is a amorphous solid. Quartz can be converted into glass by melting and rapid cooling.

Question 3.
Classify each of the following solids as ionic, metallic, molecular, network (covalent) or amorphous.
(i) Tetra phosphorus decoxide (P₄O₁₀)
(ii) Ammonium phosphate (NH₄)₃PO₄
(iii) SiC
(iv) I₂
(v) P₄
(vi) Plastic
(vii) Graphite
(viii) Brass
(ix) Rb
(x) LiBr
(xi) Si.
Answers:
Ionic : (NH₄)3PO₄, LiBr
Metallic : Brass, Rb
Molecular : P₄O₁₀,I₂,P₄
Network : Graphite, SiC, Si
Amorphous : Plastics.

Question 4.
(i) What is meant by the term ‘co-ordination number’ ?
(ii) What is the co-ordination number of atoms :
(a) In a cubic close packed structure ?
(b) In a body centred cubic structure ?
Answer:
(i) The number of nearest neighbours of a particle in its close packing is called its coordination number.
(ii) (a) 12, (b) 8.

Question 5.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell ? Explain.
Answer:
Atomic mass (M)

Question 6.
‘Stability of a crystal is reflected in the magnitude of its melting points’, comment. Collect melting points of solid water, ethyl alcohol, diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules ?
Answer:
Stability of a crystal depends on the force of attraction so : Higher the melting point of crystal, stronger the intermolecular force of attraction, hence greater is the stability of a crystal.

Melting point of the H₂O, C₂H₅OH, diethyl ether and methane given as below :

On the basis of melting point, the strength of the intermolecular forces between these molecules follows the order :
Water > Diethyl ether > Ethyl alcohol > Methane.

Question 7.
How will you distinguish between the following pairs of terms :
(i) Hexagonal close packing and cubic close packing ?
(ii) Crystal lattice and unit cell ?
(iii) Tetrahedral void and octahedral void ?
Answer:
(i) Refer to NCERT Text-Book (Close packing in crystals).

(ii) Crystal lattice :
Regular arrangement of the consti-tuent particle (atoms, molecules, or ions) of a crystal in three dimentional space.

Unit cell :
It is the smallest repeating unit in three-dimensional space, which is repeated again and again to give the complete lattice.

(iii)Tetrahedral void :

1. It is the open space between four touching spheres of two layers of atoms.
2. The radius of tetrahedral void relative to radius of sphere is 0-225.

Octahedral void :

1. It is the open space between six touching spheres of two layers of atoms.
2. The radius of octahedral void relative to radius of sphere is 0-414.

Question 8.
How many lattice points are there in one unit cell of each of the following lattice:
(i) Face centred cubic
(ii) Face centred tetragonal
(iii) Body centred.
Answer:
(i) In face centred cubic arrangement, number of lattice points are :
= 8 (at comers) + 6 (at face centres)
Lattice points per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) = 4.

(ii) In face centred tetragonal, number of lattice points are :
= 8 (at comers) + 6 (at face centres)
Lattice points per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) = 4

(iii) In body centred cubic arrangement, number of lattice points are :
= 8 (at comers) + 1 (at body centres)
Lattice points per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 1 = 2

Question 9.
Explain:
(i) The basis of similarities and differences between metallic and ionic crystals.
(ii) Ionic solids are hard and brittle.
Answer:
(i) Metallic and Ionic crystal:
(a) Both metallic and ionic solids have high melting points.
(b) Ionic solids are hard and brittle but metallic solids are hard but not brittle, metals are malleable and ductile.
(c) Ionic solids are bad conductor but good conductor in a molten state and in solution. Metallic solids are good conductor in solid and liquid as well as in vapour state.
(d) Constituent units in ionic solids are cations and anions. In metallic solids, constituent units are Kernel (positively charged ions) surrounded by a sea of delocalized electrons.

(ii) Ionic crystals are hard because there are strong electrostatic forces of attraction among the oppositely charged ions. They are brittle because the ionic bond is non-directional.

Question 10.
Calculate the efficiency of packing in case of a metal crystal for:
(i) Simple cubic
(ii) Body centred cubic
Solution:
(i) Packing effiency of simple cube =

(ii) Packing efficiency in body centred cubic structure

(iii) Packing efficiency of face centred cubic structure

Question 11.
Silver crystallizes in fee lattice. If edge length of the cell is 4.077 × 10^-8 cm and density is 10-5 gem-3, calculate the atomic mass of silver.
Solution:

Z = 4 (fcc lattice), d = 10.5 gcm, N = 6.022 x 10, (a = 4.077 × 10^-8 cm)

Question 12.
A cubic solid is made of two elements P and Q. Atoms of Q are at the corners of the cube and P at the body centre. What is the formula of the compound ? What are the co-ordination numbers of P and Q ?
Answer:
As atom Q are present at the 8 comers of the cube, therefore, number of atoms of Q in the unit cell = 8 × \(\frac { 1 }{ 8 } \) = 1.
As atoms P are present at the body centre, therefore number of atoms P in the unit cell = 1.
∴ Formula of the compound = PQ
Co-ordination number of each P and Q = 8.

Question 13.
Niobium crystallizes in body centred cubic structure. If density is 8.55 gem-3, calculate atomic radius of niobium using its atomic mass 93u.
Solution:
Density = 8.55 g cm^-3
Let, length of the edge = a cm
Number of atoms per unit cell, Z = 2 (bcc)
Atomic mass, M = 93 g mol^-1
density, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)

= 1.431 x 10^-10m = 0.143 nm

Question 14.
If the radius of the octahedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
Solution:
Atoms covering the octahedral void from the top and below are not shown in the figure. Centre of octahedral void is C and its radius is equal to r. Atoms surrounding the voids are of radius R.

According to the figure,
Radius of atom surrounding the void BA = R
BC = Radius of void + Radius of outer atom
= R + r
∠ABC = 45°
in triangle ABC \(\frac {AB}{BC} \) = cos 45° = \(\frac{1}{\sqrt{2}}\) = 0.707
or \(\frac {AB}{BC} \) = 0.707
or 0.707 R + 0.707 r = R
or o.293 = 0.707r
\(\frac {r}{R} \) = \(\frac {0.293}{0.707} \) = 0.414

Question 15.
Copper crystallizes into a fee lattice with edge length 3.61 × 10^-8 cm. Show that the calculated density is in agreement with its measured value of 8.92 gcm^-3.
Solution:
We know that, d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)
For fcc Z = 4, Atomic mass of copper = 63.5, a = 3.16 x 10^-8

= 8.96 g cm^-3
This value is close to measured value.

Question 16.
Analysis shows that nickel oxide has the formula NiO_0.98 O_1.00. What fractions of nickel exist as Ni²⁺ and Ni³⁺ ions ?
Solution:
Let there are x ions of Ni²⁺ and (0.98 – x) ions of Ni³⁺.
For electrical neutrality of the compound :
Total positive charge contributed by Ni²⁺ and Ni³⁺ ions = Total negative charge contributed by
O^2- ions
(+2 × x) + {+3 × (0-98 – x)} = 2
2x + 2.94 – 3x = 2
x = 0.94.
Thus, fraction of Ni²⁺ = \(\frac { 0.94 }{ 0.98 } \) = 0.96 or 96%
Fraction of Ni³⁺ = (1 – 0.96) = 0.04 or 4%.

Question 17.
What is a semiconductor ? Describe the two main types of semiconductors and contrast their conduction mechanism.
Answer:
Substance whose conductance likes in between that of metals (Conductors) and insulators are called semiconductors. There are two main types of semiconductors :

(i) n-type semiconductors : Silicon and germanium belong to group 14 of the periodic table and have four valence electrons each. In their crystals each atom forms four covalent bonds with its neighbours. When doped with a group 15 element like P or As, which contains five valence electrons, they occupy some of the lattice sites in silicon or germanium crystal. Four out of five electrons are used in the formation of four covalent bonds with the four neighbouring silicon atoms. The fifth electron in extra and becomes delocalised.

These delocalised electrons increase the conductivity of doped silicon (or germanium). Here, the increase in conductivity is due to the negatively charged electron, hence silicon doped with electron with impurity is called n-type semiconductor.

(ii) p-type semiconductors: Silicon or germanium can also be doped with a group 13 element like B, A1 or Ga which contains only three valence electrons. The place where the fourth valence electron is missing is called electron-hole or electron vacancy. An electron from a neighbouring atom can come and fill the electron hole, but in doing so it would leave an electron hole at its original position.

If it happens it would appear as if the electron hole has moved in the direction opposite to that of the electron that filled it under the influence of electric field, electrons would move towards the positively charged plate through electronic holes, but it would appear as if electron holes are positively charged and are moving towards negatively charged plate. This type of semiconductors are called p-type semiconductors.

Question 18.
Non-stoichiometric cuprous oxide, Cu₂O can be prepared in laboratory. In this oxide, copper to oxygen ratio is slightly less than 2:1. Can you account for the fact that this substance is a p-type semiconductor ?
Answer:
The ratio less than 2: 1 in Cu₂O shows that some cuprous (Cu⁺) ions have been replaced by cupric (Cu²⁺) ions. To maintain electrical neutrality, every two Cu⁺ ions will be replaced by one Cu²⁺ ion thereby creating a hole. As conduction will be due to presence of these positive hole, hence it is a p type semiconductor.

Question 19.
Ferric oxide crystallizes in a hexagonal close packed array of oxide ions with two out of every three octahedral holes occupied by ferric ions. Derive the formula of the ferric oxide.
Solution:
Number of oxide (O^2-) ions = n
Number of octahedral voids = n
Number of Fe³⁺ ions = \(\frac { 2 }{ 3 } \) n
Fe³⁺ : O^2- = \(\frac { 2 }{ 3 } \)n : n
= 2 : 3
Formula, Fe₂O₃.

Question 20.
Classify each of the following as being either a p-type or an-type semiconductor :
(i) Ge doped with In
(ii) B doped with Si.
Answer:
(i) Ge belongs to group 14 and In belongs to group 13, therefore an electron-deficient hole is created and hence it is n-type semiconductor.
(ii) B belongs to group 13 and Si belongs to group 14, therefore there will be a free electron and hence it is n-type semiconductor.

Question 21.
Gold (atomic radius = 0.144 nm) crystallizes in a face-centred unit cell. What is the length of a side of the cell ?
Solution:
According to the question, r = 0.144 nm.
For fee structure:
Edge length (a) = 2\(\sqrt { 2 }\) × Radius of atom= 2 × 1.414 × 0.144
= 0.407 nm.

Question 22.
In terms of band theory, what is the difference :
(i) between a conductor and an insulator ?
(ii) between a conductor and a semiconductor ?
Answer:
(i) The energy gap between the valence band and the conduction band in an insulator is very large whereas in a conductor the energy gap is either very small or there is overlapping between valence band and conduction band.

(ii) In a conductor, the energy gap between the valence band and conduction band is very small or there is overlapping between valence band and conduction band. But in a semiconductor, there is always a small energy gap between them.

Question 23.
Explain the following terms with suitable examples :
(i) Schottky defect
(ii) Frenkel defect
(iii) Interstitials and
(iv) F-centres.
Answer:
(i) Schottky defect: This defect arises in the crystal when one cation and one anion is missing from their normal lattice site and as a result vacancies are created. Since, the number of missing positive ions is equal to the missing negative ions, the crystal as a whole is electrically neutral. Due to this defect, the density of the crystal decreases.

This defect generally occurs in strongly ionic compounds with high coordination number and where positive and negative ions are almost of similar sizes, e.g., NaCl and CsCl.

(ii) Frenkel defect: This defect is due to vacancy at a cation site. The cation leaves its correct lattice site and moves to another position between the two. Frenkel defects are common in ionic compound which possess low co-ordination number in which there is large difference between the size of positive and negative ions. As there is no absence of ions from the lattice the density remains the same, e.g. ZnS, AgCl.

(iii) Interstitial defect: When some constituent particles (atoms or molecules) occupy an interstitial site the crystal is said to have interstitial defect. This defect increases the density of the substance. Non ionic solids are example of this defect.

(iv) F-centres : Alkalihalides like NaCl and KCl show this type of defect when crystals of NaCl are heated in a atmosphere of sodium vapour the sodium atoms are deposited normal on the surface of the crystal. The Cl^– ions diffuse to the surface of the crystal and combine with Na atoms to give NaCl. This happens by the loss of electrons by sodium atoms to form Na⁺ ions. The released electron diffuse into the crystal and occupy anionic sites.

As a result crystal has now excess of sodium. The anionic sites occupied by unpaired electrons are called F-centres. They impart yellow colour to the crystal of NaCl. The colour results by excitation of these electrons when they absorb energy from the visible light falling on the crystals. Another examples are : LiCl, KCl etc.

Question 24.
Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125 pm.
(i) What is the length of the side of the unit cell ?
(ii) How many unit cells are there in 1.00 cm³ of aluminium ?
Solution:
(i) For a cubic close packed structure, length of the side of unit cell is related to radius.

Question 25.
If NaCI is doped with 10-3 mol % of SrCl₂, what is the concentration of cation vacancies ?
Solution:
We know that, doping of SrCl₂ to NaCI brings in replacement of two Na⁺ ions by each Sr²⁺ ions, but each Sr²⁺ occupies only one lattice point. This produces one cation vacancy.
Thus, doping of 10^-3 mole of SrCl₂ in 100 moles of NaCI.
NaCI will produce cation vacancies = 10³ mol
∵ 100 mole of NaCI will have cation vacancies after doping = 10^-3 mol
∴ 1 mole of NaCI will have cation vacancies after doping = \(\frac{10^{-3}}{100}=10^{-5}\) mol
Total cationic vacancies after doping = 10^-5 × NA
= 10^-5 × 6.023 × 10²³
= 6.023 × 10¹⁸ vacancies.

Question 26.
Explain the following with suitable examples :
(i) Ferromagnetism
(ii) Paramagnetism
(iii) Ferrimagnetism
(iv) Anti-ferromagnetism
(v) 12-16 and 13-15 group compounds.
Answer:
On the basis of magnetic properties solids are classified into following categories :
(i) Ferromagnetic : A substance which shows unusually large paramagnetism and shows permanent magnetism even in absence of a magnetic field is called ferromagnetic.
Examples : Fe, Co, Ni, CrO₂, Fe₃O₄, alnico (alloy of Al, Ni, Co, Fe and Cu)
A ferromagnetic substance if once magnetised remains magnetised permanently. Fer-romagnetism arises due to spontaneous alignment of magnetic moments (due to unpaired electrons) in the same direction.

(ii) Paramagnetic : A substance which is attracted by a magnetic field is called para-magnetic. Paramagnetism arises due to the presence of permanent dipoles due to unpaired electrons in atoms, ions or molecules.

Example : Cu⁺², Fe³⁺, TiO, CuO, O₂ etc.
(iii) Ferrimagnetic : A substance which shows fairly good paramagnetic character is called ferrimagnetic.
In ferrimagnetic substances the alignment of magnetic moments in opposite directions are not equal. As a result the substance attains a net magnetic moments, e.g. Fe₃O₄ isaferrimag- netic substance which on heating upto 850 K becomes paramagnetic.

(iv) Antiferromagnetic : A substance which shows much reduced paramagnetism than expected is called antiferromagnetic.
In antiferromagnetic substances the alignment of magnetic moments are equal and in opposite directions. Hence, the net magnetic moment is zero.
Examples: V₂O₃,Cr₂O₃,MnO,Mn₂O₃,MnO₂,FeO,Fe₂O₃,CoO, Co₃O₄,NiO

(v) 12-16 and 13-15 group compounds: Various types of compounds are formed by the mixture of elements of group 13 and 15 and group 12 and 16 whose average valency is like Ge and Si is 4. Of these specific compounds of group 13-15 are InSb, AlP and GaAs. Gallium arsenious are accelerated sensitive semiconductors. Semiconductors brought a revolutionary change in the manufacture of devices. ZnS, CdS, CdSe and HgTe are examples of compounds of group 12-16. Bonds of these compounds are not totally covalent and their ionic properties depend on the electronegativity of both the elements present.

The Solid State Other Important Questions and Answers

The Solid State Objective Type Questions

Question 1.
Choose the correct answer:

Question 1.
Due to Frenkel defect, density of ionic solids :
(a) Decreases
(b) Increases
(c) Does not change
(d) It changes.

Question 2.
In CsCl each Cl is surrounded by how many Cs :
(a) 8
(b) 6
(c) 4
(d) 2.

Question 3.
Frenkel defect is not shown by :
(a) AgBr
(b)AgCl
(c) KBr
(d) ZnS.

Question 4.
In NaCl crystal number of oppositely charged ions situated at equal distance are:
(a) 8
(b) 6
(c) 4
(d) 2.

Question 5.
Best conductor of electricity is :
(a) Diamond
(b) Graphite
(c) Silicon
(d) Carbon (Non-crystalline).

Question 6.
Which type of point defect is found in NaCI crystal or KCl crystal:
(a) Frenkel defect
(b) Schottky defect
(c) Lattice defect
(d) Impurity defect.

Question 7.
How many space lattices (Bravais lattice) can be obtained from various crystal systems:
(a) 7
(b) 14
(c) 32
(d) 230.

Question 8.
Diamond is a:
(a) H-bond solid
(b) Ionic solid
(c) Covalent solid
(d) Glass.

Question 9.
The Co-ordination number of Ca²⁺ ions in fluoride structure is :
(a) 4
(b) 6
(c) 8
(d) 3.

Question 10.
8 : 8 Co-ordination number is found in which compound :
(a) MgO
(b) Al₂O₃
(C) CsCl
(d) All of these.

Question 11.
Co-ordination number of body centred cubic cell is :
(a) 8
(b) 12
(c) 6
(d)4.

Question 12.
Density of unit cell is :

Question 13.
The number of tetrahedral voids in unit cell of cubic close packing :
(a) 4
(b) 8
(c) 6
(d) 2.

Question 14.
Intra-ionic distance of CsCl will be :
(a) a
(b) \(\frac { a }{ 2 } \)
(c) \(\frac{\sqrt{3}}{2} a\)
(d) \(\frac{2 a}{\sqrt{3}}\)

Question 15.
Number of atoms in body centred cubic unit cell is :
(a) 1
(b) 2
(c) 3
(d) 4

Question 16.
Which of the following is Bragg equation :
(a) nλ = 2Φ sin θ
(b) nλ = 2d sin θ
(c) nλ = sin θ
(d) \(n \frac{\theta}{2}=\frac{d}{2} \sin \theta\)

Question 17.
Constituents of covalent crystal is :
(a) Atom
(b) Molecule
(c) Ion
(d) All of these.

Question 18.
Number of Na atom present in the unit cell of NaCl crystal is :
(a) 1
(b) 2
(c) 3
(d) 4.

Question 19.
What type of magnetic substance are Fe, Co, Ni:
(a) Paramagnetic
(b) Ferromagnetic
(c) Diamagnetic
(d) Antiferromagnetic.

Question 20.
The correct example of Frenkel defect is :
(a) NaCI
(b) CsCl
(c) KCl
(d) AgCI.

Question 21.
Dry ice (solid CO₂) is a/an :
(a) Ionic crystal
(b) Covalent crystal
(c) Molecular crystal
(d) Metallic crystal.

Question 22.
Co-ordination number of Cs in CsCl:
(a) Like Cl i.e., 6
(b) Like Cl i.e., 8
(c) Unlike Cl i.e., 8
(d) Unlike Cl i.e., 6.

Question 23.
Structure of NaCl crystal:
(a) Tetragonal
(b) Cubic
(c) Orthorhombic
(d) Monoclinic.

Question 24.
Each Na⁺ ion in NaCl crystal is surrounded by :
(a) Three Cl^– ions
(b) Eight Cl^– ions
(c) Four Cl^– ions
(d) Six Cl^– ions.

Question 25.
For increasing of electro-conductivity in a solid crystal, mixing of impurities is known as:
(a) Schottky defect
(b) Frenkel defect
(c) Doping
(d) Electronic defect

Question 26.
Which type of lattice is found in KCl crystal:
(a) Face centred cubic
(b) Body centred cubic
(c) Simple cubic
(d) Simple tetragonal.

Question 27.
Number of atoms in a body centred cubic unit cell of a monoatomic substance is:
(a) 1
(b) 2
(c) 3
(d) 4.

Question 28.
Radius ratio limit for tetrahedral symmetry is :
(a) 0.155
(b) 0.414
(c) 0.732
(d) 0.225.

Question 29.
The defect produced due to a cation and an anion vacancy in a crystal lattice is known as:
(a) Schottky defect
(b) Frenkel defect
(c) Crystal defect
(d) Ionic defect.

Question 30.
If co-ordination number of Cs⁺ is 8 in CsCl then co-ordination number of Cl^– ion is :
(a) 8
(b) 4
(c) 6
(d) 12.

Answers:
1. (c), 2. (a), 3. (c), 4. (b), 5. (b), 6. (b), 7. (b), 8. (c), 9. (c), 10. (c), 11. (a), 12. (a), 13. (b), 14. (c), 15. (b), 16. (b), 17. (a), 18. (d),’ 19. (b), 20. (d), 21. (c), 22. (b), 23. (b), 24. (d), 25. (c), 26. (a), 27. (b), 28. (d), 29 (a), 30. (a).

Question 2.
Fill in the blanks :

1. The defect produced due to removal of a cation and an anion from a crystal lattice is called ………………..
2. If in a crystal lattice a cation leaves its lattice site and occupies a space in the interstitial site then the defect is called ……………….
3. The cause of electric conduction of NaCl in its molten state are its ………………..
4. Due to ……………….. defect the density of crystal decreases.
5. Total ……………….. types of crystal system are there.
6. ……………….. proposed the concept of atom for the first time.
7. The ratio of the cation and anion present in a crystal is known as ………………..
8. The process of adding small amount of impurities in an element or compound is called ………………..
9. Total 14 types of unit cells are there which are known as ………………..
10.  In NaCl crystal structure, co-ordination number of both Na⁺ and Cl^– ion is ………………..
11. ……………….. defect is found in ZnS and AgCl crystal.
12. Due to Schottky defect, density of crystal ………………..
13. In metallic solids, conductivity is due to the presence of …………..
14. Point defects are found in ………….. crystals.Substances which are attracted in magnetic field are called …………..
15. For a unit cell, if r = \(\frac{a}{\sqrt{8}}\) then it will be ………….. type of unit cell.
16. Conductivity of semiconductor ………….. on increasing temperature.

Answers:

1. Schottky defect
2. Frenkel defect
3. Free ions
4. Schottky
5. Seven
6. Kannad
7. Radius ratio
8. Doping
9. Bravais lattice
10. Six
11. Frenkel
12. Decreases
13. Free electron
14. Ionic
15. Paramagnetic substance
16. fcc
17. Increases.

Question 3.
Match the following

Answers:
1. (b)
2. (d)
3. (c)
4. (a)

Answers:
1. (c)
2. (d)
3. (a)
4. (b).

Answer:
1. (d)
2. (c)
3. (b)
4. (a).

Question 4.
Answer in one word / sentence :

1. Give two examples of metallic crystal.
2. Give two examples of covalent crystal.
3. Give two examples of ionic crystal.
4. What is the co-ordination number of F^– in CaF₂ ?
5. What type of crystal is SiC ?
6. What is the value of co-ordination number of hexagonal close packing structure ?
7. Write the formula of radius ratio.
8. What is the type of structure of NaCl crystal ?
9. Give an example of body centred cubic cell.
10. Give an example of a compound which has both Schottky and Frenkel type of defect.
11. Give two examples of amorphous or non-crystalline solid.
12. Write Bragg equation.
13. What is effect on the density of a substance or crystal due to Schottky defect ?
14. State the co-ordination number of CsCl and NaCl.
15. Write the formulae of two superconductors substance.
16. Give an example of Frenkel defect.
17. Give an example of superconductor.
18. Radius ratio of tetrahedral void is.

Answers:

1. Copper, Nickel
2. Diamond, Graphite
3. NaCl, NaNO₃
4. 4
5. Covalent
6. 12
7. 
8. Cubic
9. CsCl
10. AgBr
11. Glass,plastic,
12. nλ = 2d sin θ
13. Due to schottky defect, density of substance decreases
14. Co-ordination number of CsCl = 8 : 8, Co-ordination number of NaCl = 6:6
15. (i) λBa3- Cu₂O₇, (ii) Bi₂Ca₂Sr₂Cu₃O₁₀
16. AgCl
17. Ba_0.7K_0.3BIO₃
18. 0.225.

The Solid State Very Short Answer Type Questions

Question 1.
What are crystalline solids ? Crystalline solids are of how many types ?
Answer:
Solids in which the constituent particles like atoms, molecules or ions are in a definite order, with a definite geometry are known as crystalline solids.

Crystalline solids are of four types :

1. Ionic crystal
2. Covalent crystal
3. Molecular crystal
4. Metallic crystal.

Question 2.
Write the formula of density of unit cell.
Answer:

Question 3.
What is crystal lattice ?
Answer:
Geometry of a crystal in which unit cells are arranged in a definite order and forms a crystal like the shape of unit cell is known as crystal lattice.

Question 4.
What is a unit cell ?
Answer:
The smallest unit formed by the arrangement of constituent particles atoms, ions or molecules of a crystal in an ordered form is known as unit cell of the crystal.

Question 5.
Give two-two examples of each of the following :

1. Diamagnetic substance
2. Paramagnetic substance
3. Ferromagnetic substance
4. Antiferromagnetic substance
5. Ferrimagnetism.

Answer:

1. Diamagnetic substance – TiO₂, NaCl
2. Paramagnetic substance – Cu⁺², Fe⁺³
3. Ferromagnetic substance – Fe, Co
4. Antiferromagnetic substance – MnO₂, MnO
5. Ferrimagnetism – Fe₃O₄, Ferrite.

Question 6.
Write the structure and co-ordination number of the following:
(1) CsCl
(2) NaCl
(3) Zn.
Answer:
(1) CsCl – Structure : Cubic, Co-ordination number : 8.
(2) NaCl – Structure: Octahedral, Co-ordination number : 6.
(3) Zn – Structure : Tetrahedral, Co-ordination number : 4.

Question 7.
What makes a glass different from a solid such as quartz ? Under what conditions could quartz be converted into glass ?
Answer:
Glass is an amorphous solid in which constituent particles are orderly arranged in a short range order. Quartz is a crystalline form of silica in which SiO₄ units are orderly arranged in a long range order.
Quartz can be converted into glass by melting the glass and then cooling it rapidly.

Question 8.
Which are the seven fundamental crystal system on the basis of crystal geometry ?
Answer:
Seven types of crystal system are following:

1. Cubic
2. Tetragonal
3. Orthorhombic
4. Monoclinic
5. Hexagonal
6. Rhombohedral
7. Triclinic.

Question 9.
Write the names of different type of cubic system,
Answer:
Cubic system is of three types :

1. Simple cubic (see)
2. Body centred cubic (bcc)
3. Face centred cubic (fcc).

Question 10.
What is co-ordination number of Na⁺ and Cl^– in the structure of NaCl ?
Answer:
In NaCl structure each Na⁺ is surrounded by 6 Cl^– and each Cl^– is surrounded by 6 Na⁺ ions.
Thus, Co-ordination number of Na⁺ = 6
Co-ordination number of Cl^– = 6.

Question 11.
What is co-ordination number ? What is the effect of temperature and pressure on co-ordination number ?
Answer:
The number of neighbouring ions around the constituent particles of a crystal lattice is known as its co-ordination number. At high pressure co-ordination number increases and at low pressure it decreases.

Question 12.
What information is obtained by X-ray diffraction study of crystals ?
Answer:
By X-ray diffraction study of crystals, spacing between crystal planes of constituents particles of crystals is known.

Question 13.
What type of solids are electrical conductors : metallic or ductile ?
Answer:
Metallic solids.

Question 14.
Give the significance of a ‘lattice point’.
Answer:
Each lattice point represents one constituent particle of the solid. The constituent particle may be an atom, a molecule (group of atom) or an ion.

Question 15.
Refractive index of a solid is observed to have the same value along all directions. Comment on the nature of this solid. Would it show cleavage property ?
Answer:
As the solid has same value of refractive index along all directions, this means that it is isotropic and hence amorphous. Being an amorphous solid, it would not show a clean cleavage when cut with a knife. Instead it would break into pieces with irregular surfaces.

The Solid State Short Answer Type Questions

Question 1.
What is Schottky defect ?
Answer:
This type of defect is found in crystals in which both the cation and anion leave their normal lattice site and make their place vacant. In this defect, density decreases but r electrical neutrality is maintained. Like NaCl, CsCl etc.

Question 2.
What is radius ratio of ions ?
Answer:
Ratio of radii of cation and anion in any crystal is called radius ratio.

For example, in NaCl, ionic radii of Na⁺ and Cl- ions are 95 pm and 181 pm respectively.
Radius ratio in NaCl = \(\frac { 95 }{ 181 } \) = 0.52
Due to presence of other forces, in the crystal the observed value of radius ratio is less as in NaCl crystal it is 0414.

Question 3.
Describe briefly the structure of CsCl.
Answer:
It is AB type ionic crystal with body centred cubic structure. In this Cs⁺ ion in centre of cube and Cl^– ions at comers of cube (or vice versa). Co-ordination number of caesium chloride is 8 : 8 and radius ratio of Cs⁺ and Cl^– ions is 0.732.

In the unit cell of CsCl one Cs⁺ ion and one Cl^– ion are present.
Cs⁺ = 8 (at Centre) × 1 = 1
Cl^– = 8 (at corners) × \(\frac { 1 }{ 8 } \) = 1

Question 4.
Calculate the density of unit cell.
Answer:

If number of particles (atom, molecule, ions) of unit cell is Z and mass of each particle is m, then
Mass of unit cell = m × Z …….(1)
If molar mass of the substance is M, then mass of each particle :

Question 5.
Why is glass considered as super cooled liquid ?
Answer:
Glass is an amorphous solid. Like liquids it has tendency to flow, though very slowly. The proof of this fact is that the glass panes in the windows or doors of old buildings are invariably found to be slightly thicker at the bottom that at the top.

Question 6.
Ionic solids conduct electricity in molten state but not in solid state. Explain.
Answer:
In the molten state, ionic solids dissociate to give free ions and hence can conduct electricity. However in the solid state, as the ions are not free but remain held together by strong electrostatic forces of attraction, they cannot conduct electricity in the solid state.

Question 7.
What type of defect can arise when a solid is heated ? Which physical property is affected by it and in what way ?
Answer:
When a solid is heated, vacancy defect is produced in the crystal. This is because on heating some atoms or ions leave the lattice site completely, some lattice sites become vacant. As a result of this defect the density of the substance decreases because some atoms ions leave the crystal completely.

Question 8.
A compound consists of A and B exhibits cubic structure. A atoms are arranged at corners of cube while B are at the centre of faces. What will be the formula of the compound ?
Answer:
Atoms situated at corners are 8 A atoms which are shared by 8 cubes. So in the unit cell:
Number of A atom = 8 × \(\frac { 1 }{ 8 } \) = 1
B atoms are present in centre of 6 faces and each face is shared by two cubes
So number of B atoms = 6 × \(\frac { 1 }{ 2 } \) = 3
Formula of compound will be AB₃.

Question 9.
Prove that in face centred cubic (fee) structure, there are four atoms in an unit cell.
Answer:
Face Centred Cubic Structure : In this structure one atom is also situated in each face along with the corners of cube. The atom in each face is shared by two faces.
Thus, number of atom in each unit cell = 8 × \(\frac { 1 }{ 8 } \) (at 8 comers) + 6 × \(\frac { 1 }{ 2 } \) (at 6 faces)
= 1 + 3 = 4.

Question 10.
Write Bragg equation.
Answer:
Bragg equation is as follows : 2d sin θ = nλ.
Where, d = Distance between two consecutive planes in a crystal,
θ = Incident angle of X-rays
n = Simple whole number
λ. = Wavelength of X-rays.
By this distance d between the planes of the crystal is determined.

Question 11.
Name the parameters that characterise a unit cell.
Answer:
A unit cell is characterised by : (i) Its dimensions along the three edges a, b and c. These edges may or may not be mutually perpendicular
(ii) Angles between the edges, a, (between b and c), β (between a and c) and γ (between a and b). Thus, a unit cell is characterised by six parameters a, b, c, α, β and γ.

Question 12.
Window glass of old buildings appear milky. Why ?
Answer:
In day time, glass becomes hot and cools down at night. This way, the process of annealing takes place. Due to annealing in many years, glass develops crystalline property and window glass appear milky.

Question 13.
Common salt sometimes appear yellow instead of being colourless. Why ?
Answer:
In common salt, due to metal excess defect the anion Cl^– disappears from its lattice site but leaves an electron thereby which the crystal remains electrically neutral. A hole is formed at the vacant space of anion. This hole is known as F-centre. Due to this reason NaCl appear yellow.

Question 14.
With the increase in temperature, electrical conductivity of semiconductors increases. Why ?
Answer:
The energy gap between valence band and conduction band is less. Thus, with the increase in temperature, some electrons from the valence band jump to the conduction band due to which electrical conductivity increases with the increase in temperature.

Question 15.
Which of the following lattices has the highest packing efficiency :
(i) Simple cubic
(ii) Body centred cubic
(iii) Hexagonal close packed lattice ?
Answer:
Packing efficiency for simple cubic = 52-4%
Body centred cubic = 68%
Hexagonal close packed = 74%
Hence, Hexagonal close packed (hcp) has highest packing efficiency.

Question 16.
What type of stoichiometric defect is shown by (i) ZnS (ii) AgBr ?
Answer:
(i) ZnS shows Frenkel defect because its ions have a large difference in size.
(ii) AgBr shows both Frenkel and Schottky defects.

Question 17.
Explain how much portion of an atom located at: (i) Corner and (ii) Body centre of a cubic unit cell is part of its neighbouring unit cell ?
Answer:
(i) An atom at the corner is stand by eight adjacent unit cells. Hence, portion of the atom at the corner not belongs to one unit cell = \(\frac { 1 }{ 8 } \)
(ii) The atom at the body centre of a cubic unit cell is not stand by other unit cell. Hence, it belongs fully to the unit cell.

Question 18.
How can you determine the atomic mass of an unknown metal if you know its density and the dimension of its unit cell ? Explain.
Answer:
Let d – Density of the unit cell and volume of the unit cell = a3 (in case of cubic crystal)
Mass of an atom present in the unit cell

Question 19.
Stability of a crystal is reflected in the magnitude of its melting point. Comment. Collect melting point of solid water, ethyl alcohol,diethyl ether and methane from a data book. What can you say about the intermolecular forces between these molecules ?
Answer:
Higher the melting point, greater are the forces holding the constituent particles together and hence greater is the stability.
Melting points of the substance are given below :

Question 20.
A group 14 element is to be converted into n-type semiconductor by doping it with a suitable impurity. To which group should this impurity belong ?
Answer:
n-type semiconductor means conduction due to the presence of excess of negative charged electrons. Hence, to convert group 14 element into n-type semiconductor, it should be doped with group 15 element.

Long Answer Type Questions

Question 1.
Give difference between crystalline solid and amorphous solid.
Answer:
Differences between Crystalline solid and Amorphous solid:

Crystalline solid:

1. Structure of crystalline solid are of definite geometrical shape.
2. In the internal structure, particle are arranged systematically.
3. Crystalline solids are solids in real sense.
4. Melting points of these compounds are sharp and definite.
5. These exhibit anisotropy.
6. Cooling curve are not continuous.
7. Crystalline solids possess less energy.

Amorphous solid

1. No any definite geometrical shape in the structure.
2. No any definite arrangement of particles in the internal structure. Amorphous solids are super cooled liquids.
3. Melting point of these compounds are not definite and sharp.
4. These exhibit isotropy.
5. Cooling curves are continuous.
6. Amorphous solids possess higher energy.

Question 2.
Distinguish between :
(i) Hexagonal and monoclinic unit cell
(ii) Face centred and end centred unit cell.
Answer:
(i) For hexagonal unit cell a = b ≠ c, α = β = 90°, γ = 120°
For monoclinic unit cell a ≠ b ≠ c, α = γ= 90°, β = 90°
(ii) A face centred unit cell has one constituent particle present at the centre of each face in addition to the particles present at the comers.
An end centred unit cell has one constituent particle each at the centre of any two opposite faces in addition to the particles present at the corners.

Question 3.
Give difference between Schottky and Frenkel defect.
Answer:
Differences between Schottky and Frenkel defect:

Schottky Defect:

1. Cation and anion completely leave their lattice sites.
2. Density of the crystal decreases by this defect.
3. This defect generally occur in ionic compounds with high co-ordination number where size cation and anion are nearly same.
4. No effect on dielectric constant.

Frenkel Defect:

1. Cation leave its normal lattice site and occupies the intestitial site.
2. No effect on the density of the crystal.
3. It is found in ionic compounds with low co-ordination number and where size of cation and anion differ largely.
4. Magnitude of dielectric constant increase.

Question 4.
What do you understand by imperfections in crystals ? What are the rea¬sons for imperfections ?
Answer:
It is generally supposed that arrangement of constituent particles in crystal struc¬ture is completely regular, but actually it is very hard to get a such complete ideal crystal. Crystal structures have many imperfections or defects. Reasons for these imperfections are following:
(i) Temperature: Crystal which have no imperfections or defects are known as ideal crystal. However such crystals exist only at absolute zero temperature because the energy of crystals at 0 K is minimum. At any temperature above 0 K, there are crystals which have some departure from complete order arrangement.

(ii) Presence of impurities : Sometimes presence of impurities causes disorder in regular arrangement of crystals which is responsible for imperfections and defects.

Question 5.
State the importance of radius ratio in crystal structure.
Answer:
Importance of radius ratio in crystal structure: Cations have tendency to get surrounded by maximum number of anions, hence larger be the size of cation greater will be its co-ordination number. We can understand this by taking examples of NaCl and CsCl. In NaCl, small sized Na⁺ has co-ordination number 6, while in CsCl large sized Cs⁺ ion has co-ordination number 8. Hence, radius ratio is closely related to co-ordination number.

Radius ratio, Co-ordination number and Structural arrangement

Question 6.
Ionic solids, which have anionic vacancies due to metal excess defect develop colour. Explain with the help of suitable example.
Answer:
Due to anion vacancies basic halides like NaCl, KCl etc., show this type of metal excess defect. Taking the example of NaCl, when its crystals are heated in presence of sodium vapour some chloride ion leave their lattice sites to combine with sodium to form NaCl. For this reaction to occur Na atoms lose electrons to form Na⁺ ions. The electron has released diffuse into the crystal to occupy the anion vacancies created by Cl^– ions. The crystal now has excess of sodium. The sites occupied by unpaired electrons are called F- centres. They impart yellow colour to the crystal because they absorb energy from the visible light and get excited.

Question 7.
What type of substances would make better permanent magnets : ferromagnetic or ferrimagnetic ? Justify your answer.
Answer:
Ferromagnetic substances make better permanent magnets. This is because the metal ions of a ferromagnetic substances are grouped into small regions called ‘domains’. Each domain acts as a tiny magnet. These domains are randomly oriented. When the substance is placed in a magnetic field all the domains get oriented in the direction of the magnetic field and a strong magnetic field is produced. This ordering of domains persists even when the external magnetic field is removed. Hence, the ferromagnetic substance becomes a permanent magnet.

Question 8.
How many lattice points are there in one unit cell of each of the following lattice:
(i) Face centred cubic
(ii) Face centred tetragonal
(iii) Body centred.
Solution:
(i) Lattice points in face-centred cubic lattice = 4.
(ii) Face centred tetragonal = 8 (at corners) + 6 (at the face centre) = 14.
However, particles per unit cell
= 8 × \(\frac { 1 }{ 8 } \) + 6 × \(\frac { 1 }{ 2 } \) = 1 + 3 = 4.
(iii) Lattice points in body centred cube
= 8 (at comers) + 1 (at the body centre)
= 9
However, particles per unit cell = 8 × \(\frac { 1 }{ 8 } \) + 1 = 2.

The Solid State Numerical Questions

Question 1.
A compound forms hexagonal close packed structure. What is the total number of voids in 0.5 mol of it ? How many of these are tetrahedral voids ?
Solution:
No. of atoms in the close packing = 0.5 mol
= 0.5 × 6.022 × 10²³
= 3.011 × 10²³
No. of octahedral voids = No. of atoms in the packing .
= 3.011 × 10²³
No. of tetrahedral voids = 2 × No. of atoms in the packing
= 2 × 3.011 × 10²³
= 6.022 × 10²³
Total No. of voids = 3.011 × 10²³ + 6.022 × 10²³
= 9.033 × 10²³.

Question 2.
A compound is formed by two elements M and N. The elements N forms ccp and atoms of M occupy l/3rd of the tetrahedral voids. What is the formula of the compound ?
Solution:
Suppose the atoms N in the ccp = n
∴ No. of tetrahedral voids = 2n
As 1/3rd of the tetrahedral voids are occupied by atoms M, therefore,
No of atoms M = \(\frac { 2n }{ 3 } \)
∴ Ratio of M : N = \(\frac { 2n }{ 3 } \) : n
Hence, the formula is M₂N₃

Question 3.
Structure of a solid AB is like NaCl. Radii of cation A is 100 pm, find out the radii of anion.
Solution:
For NaCl structure value of radius ratio (\(\frac{r^{+}}{r^{-}}\)) is in between 0.414 – 0.732.
Radius of cation is 100 pm.

Question 4.
The radii of A⁺ and B⁺ ions are 0.95 Å and 1.81 Å respectively. Find out the co-ordination number of A⁺.
Or,
The ionic radii of Na⁺ and Cl^– are 95 pm and 181 pm respectively.
What will be the co-ordination number of Na⁺?
Solution:
Radii of Na⁺ = 95 pm
Radii of Cl^– =181 pm
Radius ratio = \(\frac{r^{+}}{r^{-}}\) = \(\frac { 95 }{ 181 } \) = 0.524
Radius ratio is between 0.414 and 0.732. Thus, co-ordination number of Na⁺ or A⁺ will be 6.

Question 5.
Core length of a face centred cubic crystal is 400 pm calculate the density of element Atomic mass of element is 60.
Solution:
Density \((d)=\frac{Z \times M}{N_{0} \times a^{3}}\)
Where, Z = Number of atoms = 4, in face centred cubic structure.
M = Atomic mass = 60, N₀ = Avogadro number (6.023 × 10²³)
a = Core length (400 p.m.)

Question 6.
Silver crystallises in fee lattice. If edge length of the cell is 4.07 × 10^-8 cm and density is 10.5 g cm^-3, calculate the atomic mass of silver.
Solution:

Where d = Density of the material
a = Length of the edge of the cell
N_A = Avogadro number
Z = No. of atoms

∴ Atomic mass of silver = 107.08 g mol^-1.

Question 7.
Atomic mass of a face centred cubic (fee) is 60 g mol^-1 and its edge of its face is 400 pm. Determine the density of the element
Solution:
Volume of unit cell (a³) = (Length of edge)³
= (400 × 10^-12 m)³
= 64 × 10^-30 m³
= 64 × 10^-30 (10² cm)³
= 64 × 10^-24 cm³
Number of atoms in fee unit cell = 4

Question 8.
Structure of CuCl like ZnS is cubic. If density of CuCl is 3-4 g cm^-3, then determine the length of edge of the unit cell.
Solution:
ZnS has fee structure, thus same structure will be of CuCl. If length of edge of unit cell is a, number of atoms in fee is Z, molecular mass M and Avogadro number is N_A, then,

MP Board Class 12th Chemistry Solutions

MP Board Class 12th Maths Solutions Chapter 11 प्रायिकता Ex 11.3

प्रश्न 1.
निम्नलिखित प्रश्नों में से प्रत्येक में समतल के अभिलम्ब का दिक् कोसाइन और मूल बिन्दु से दूरी ज्ञात कीजिए-
(a) z = 2
(b) x + y + z = 1
(c) 2x + 3y – 2 = 5
(d) 5y + 8 = 0
हल:
माना दिये गये समतल का कार्तीय समीकरण ax + by + cz =d तथा अभिलम्ब के दिक् अनुपात a, b, c हैं।
तब मूल बिन्दु से लम्बवत् दूरी = \(\left|\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|\)
(a) समतल का समी० z = 2
∴ दिक् कोसाइन 0, 0, 1 हैं।
(b) समतल x + 1 + z = 1
दिक अनुपात 1, 1, 1 हैं।
तब दिक् कोसाइन

प्रश्न 2.
उस समतल का सदिश समी० ज्ञात कीजिए जो मूल बिन्दु से 7 मात्रक दूरी पर है और सदिश \(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\) पर अभिलम्ब है।
हल:

प्रश्न 3.
निम्नलिखित समतलों का कार्तीय समी० ज्ञात कीजिए–

हल:

प्रश्न 4.
निम्न स्थितियों में, मूल बिन्दु से खींचे गए लम्ब के पाद के निर्देशांक ज्ञात कीजिए–
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
हल:
(a) समतल का समीकरण
2x + 3y + 4z – 12 = 0 ……(i)

(b) समतल का समीकरण


(c) समतल का समीकरण

(d) समतल का समीकरण

प्रश्न 5.
निम्नलिखित प्रतिबंधों के अन्तर्गत समतलों का सदिश व कार्तीय समीकरण ज्ञात कीजिए जो-
(a) बिन्दु (1,0,-2) से जाता हो और \(\hat{i}+\hat{j}-\hat{k}\) समतल पर अभिलंब हो।
(b) बिन्दु (1, 4, 6) से जाता हो और \(\hat{\boldsymbol{i}}-2 \hat{\boldsymbol{j}}+\hat{\boldsymbol{k}}\) समतल पर अभिलम्ब सदिश है।
हल:
(a) बिन्दु (1, 0, – 2) का स्थिति सदिश

यही अभीष्ट कार्तीय समीकरण है।

यही अभीष्ट कार्तीय समीकरण है।

प्रश्न 6.
उन समतलों का समीकरण ज्ञात कीजिए जो निम्नलिखित तीन बिन्दुओं से गुजरता है।
(a) (1, 1, – 1), (6, 4, – 5), (- 4, – 2, 3)
(b) (1, 1, 0), (1, 2, 1), (- 2, 2, – 1)
हल:
(a) माना
(x₁, y₁, z₁) ≡ (1, 1, – 1)
(x₂, y₂, z₂) ≡ (6, 4, – 5)
(x₃, y₃, z₃) ≡ ( – 4, – 2, 3)
इन तीन बिन्दुओं से जाने वाले समतल का समी०

समीकरण (1), (2), (3) में से a, b, c को लुप्त करने से, समतल का समीकरण
\(\left|\begin{array}{ccc}{x-1} & {y-1} & {z} \\ {0} & {1} & {1} \\ {3} & {-1} & {1}\end{array}\right|\) = 0
⇒ 2(x – 1) + 3 (y – 1) – 3z = 0
⇒ 2x – 2 + 3y – 3 – 3z = 0
⇒ 2x + 3y – 3z = 5

प्रश्न 7.
समतल 2x + y – x = 5 द्वारा काटे गये अन्तः खण्डों को ज्ञात कीजिए।
हल:
दिया गया समतल 2x + y – z = 5 समी० को अन्तः खण्ड रूप में लिखने पर
\(\frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{2}{(-5)}\) = 1
जो कि \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}\) = 1 के रूप की हैं।
अत: अक्षों से कटे अन्तः खण्ड, \(\frac{5}{2}\), 5, – 5 हैं।

प्रश्न 8.
उस समतल का समी० ज्ञात कीजिए जिसका y- अक्ष पर
अंतः खण्ड 3 और जो तल ZOX के समांतर है।
हल:
अक्ष ZOX का समी० y = 0 है।
∴ इस समतल के समांतर समी० y = a
∵ समतल का -अक्ष पर अंत: खण्ड 3 है इसलिए समतल y- अक्ष पर (0, 3, 0) पर मिलता हैं।
∴ 3 = a
अतः समतल का समी० y = 3

प्रश्न 9.
उस समतल का समीकरण ज्ञात कीजिए जो समतलों 3x – y + 22 – 4 = 0 और x + y + z – 2 = 0 के प्रतिच्छेदन तथा बिन्दु (2, 2, 1) से होकर जाता है।
हल:
समतल तथा उसके प्रतिच्छेदन बिन्दु से जाने वाले तल का समीकरण
(3x – y + 2z – 4) + λ (x + y + z – 2) = 0 …(i)
परन्तु तल बिन्दु (2.2.1) से होकर जाता है तब
(6 – 2 + 2 – 4) + (2 + 2 + 1 – 2) = 0
2 + 3λ = 0
λ = – \(\frac{2}{3}\)
λ का यह मान समीकरण (i) में रखने पर
(3x – y + 2z – 4) – \(\frac{2}{3}\) (x + y + z – 2) = 0
(9x – 3y + 6z – 12) – (2x + 2y + 2z – 4) = 0
7x – 5y + 4z – 8 = 0

प्रश्न 10.
उस समतल का सदिश समीकरण ज्ञात कीजिए जो समतलो \(\vec{r} \cdot(2 \hat{i}+2 \hat{j}-\hat{3} \hat{k})=7\), \(\vec{r} \cdot(2 \hat{i}+5 \hat{j}+3 \hat{k})=9\) के प्रतिच्छेदन रेखा और (2, 1, 3) से होकर जाता है।
हल:

प्रश्न 11.
तलों x + y+ z = 1 और 2x + 3y + 4z = 5 के प्रतिच्छेदन रेखा से होकर जाने वाले तथा तल x – y + z = 0 पर लम्बवत् तल का समीकरण ज्ञात कीजिए।
हल:
समतल x + y + z = 1 और 2x + 3y + 4z = 5 के प्रतिच्छेदन रेखा से होकर जाने वाली समतल का समीकरण
(x + y + z – 1) + λ (2x + 3y + 4z – 5) = 0
या (1+ 2λ) x + (1+3λ) y+ (1+4λ) z – 1 – 5λ = 0
यह तल x – y + z = 0 पर लम्बवत् है
∴ (1 + 2λ) . 1 + (1 + 3λ)(- 1) + (1 + 4λ) . 1 = 0 (∵ a₁ a₂ + b₁ b₂ + c₁ c₂) = 0
या 1 = 2λ – 1 – 3λ + 1 + 4λ = 0
⇒ 1 + 3λ = 0
⇒ λ = – \(\frac{1}{3}\)
λ का मान समी० (i) में रखने पर,

x – z + 2 = 0

प्रश्न 12.
समतलों जिनके सदिश समीकरण \(\vec{r} \cdot(2 \hat{i}+2 \hat{j}-3 \hat{k})=5\) और \(\vec{r}(3 \hat{i}-3 \hat{j}+5 \hat{k})=3\) हैं के बीच का कोण ज्ञात करो।
हल:
माना

प्रश्न 13.
निम्नलिखित प्रश्नों में ज्ञात कीजिए कि क्या दिए गए समतलों के युग्म समान्तर हैं अथवा लम्बवत् हैं और उस स्थिति में, जब ये न तो समान्तर हैं और न ही लम्बवत्, उनके बीच का कोण ज्ञात कीजिए।
(a) 7x + 5y + 6z + 30 = 0 और 3x -y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 और x – 2y + 5 = 0
(c) 2x – 2y + 4x + 5 = 0 और 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 और 2x – y + 3x + 3 =0
(e) 4x + 8y + 7 – 8 = 0 और y + 2 – 4 = 0
हल:


(b) दिए गए समतल

प्रश्न 14.
निम्न प्रश्नों में प्रत्येक दिये गये बिन्दु से दिये गये संगत समतलों की दूरी ज्ञात कीजिए।

हल:

MP Board Class 12th Maths Solutions