Important Questions for CBSE Class 11 Chemistry Chapter 6 - Equilibrium

Very Short Answer Questions  1 Mark

1. Define dynamic equilibrium. 

Ans: Dynamic equilibrium is that stage of the solution when the reactants are present in a closed vessel at any particular temperature reacts with each other to give rise the products and in this process the concentrations of the reactants goes on decreasing while of products it keep on increasing and after sometime a stage occurs where we don’t find any change in the concentration of reactants or products. This stage is called the dynamic equilibrium.

2. What is physical equilibrium? Give an example.

Ans: The equilibrium which occurs between two different physical states of same substances. Example of physical equilibrium can be shown as below:

\[P=\frac{n}{V}RT=CRT\] 

3. What is meant by the statement ‘Equilibrium is dynamic in nature’?

Ans: That stage of the reaction at which reaction will not stop rather than stopping it will be in a continuous manner then equilibrium is said to be dynamic in nature. In this case reaction appears to stop because the rate of forward reaction is equal to the rate of backward reaction.

4. How does dilution with water affect the pH of a buffer solution?

Ans: Rate of dilution shows no effect on the pH of buffer solution as pH depends upon the ratio of salt, acid or salt, base and dilution will not affect this ratio.

5. On what factor does the boiling point of the liquid depends?

Ans: The boiling point of liquid generally depends upon the atmospheric pressure of the given liquid.

6. State Henry’s law.

Ans: Henry’s law states that the mass of a gas dissolved at any temperature in a given mass of any solvent is directly proportional to the gas present in the solvent.

7. What happens to the boiling point of water at high altitude?

Ans: Boiling point generally depends upon the depth of the substance and at high altitude boiling point of water goes on decreasing.

8. On which factor does the concentration of solute in a saturated solution depends?

Ans: Concentration of solute in a saturated solution generally depends upon the temperature of that solution. If the temperature is high then the solute dissolves easily.

9. What conclusion is drawn from the following –

$Solid\rightleftharpoons Liquid$

${{H}_{2}}O(s)\rightleftharpoons {{H}_{2}}O(l)$

Ans: This states that melting point is fixed at any constant pressure as there is proper melting point for solid at given pressure like ice melts at 0 degree Celsius. 

10. State the law of chemical equilibrium.

Ans: At any temperature product of the concentrations of the reaction raised to the power of their respective stoichiometric coefficient in the balanced chemical equation which is further divided by the product of concentrations of the reactants which are also raised to their individual stoichiometric coefficients has a constant value. This is known as the equilibrium law or law of chemical equilibrium.

11. Write the equilibrium constant for the following equation:

$aA+bB\rightleftharpoons cC+dD$

Ans: The equilibrium constant of this reaction can be written as:

${{K}_{c}}=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$, Here [A], [B], [C] and [D] are said to be the equilibrium concentrations of the reactants and products.

12. Write the chemical equation for the following chemical constant.

${{K}_{c}}=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$

Ans: The chemical equation is given by the following reaction:

${{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)$

13. Write the expression for equilibrium constant ${{K}_{p}}$ for the reaction

$3Fe(s)+4{{H}_{2}}O(g)\rightleftharpoons F{{e}_{3}}{{O}_{4}}(s)+4{{H}_{2}}(g)$

Ans: The equilibrium constant for the given equation can be written as follows:

${{K}_{p}}=\frac{{{(P{{H}_{2}})}^{4}}}{{{(P{{H}_{2}}O)}^{4}}}=\frac{P{{H}_{2}}}{P{{H}_{2}}O}$

14. The equilibrium constant for the reaction ${{H}_{2}}O+CO\rightleftharpoons {{H}_{2}}+C{{O}_{2}}$. Is 0.44 at 1260 K. What will be the value of the equilibrium constant for the reaction: 

$2{{H}_{2}}(g)+2CO(g)\rightleftharpoons 2CO(g)$ at 1260 K.

Ans: The value of equilibrium get reversed and also becomes double which can be shown as:

${{K}_{c}}={{(\frac{1}{0.44})}^{2}}=5.16$

15. Define reaction quotient.

Ans: Reaction quotient generally represented by the letter Q is used to measure the relative amounts of products and reactants present in a reaction at any given time.

16. If ${{Q}_{c}}>{{K}_{c}}$, what would be the type of reaction?

Ans: This resembles reaction in the side of reactants which tells that reaction is of reverse type i.e., on the left hand side or we can say it as backward reaction.

17. What inference you get when ${{Q}_{c}}>{{K}_{c}}$ ?

Ans: This relation suggests that reaction is at equilibrium.

18. State Le chatelier’s principle.

Ans: Le chatelier’s principle is generally to predict the changes occurring in chemical equilibrium of any reaction.

19. Can a catalyst change the position of equilibrium in a reaction?

Ans: Catalyst cannot change the position of equilibrium it just affects the rate of reaction.

20. What is the effect of reducing the volume on the system described below?

$2C(s)+{{O}_{2}}(g)\rightleftharpoons 2CO(g)$

Ans: By reducing the volume the equilibrium will shift in the forward direction.

21. What happens when temperature increases for a reaction?

Ans: The value of equilibrium constant will decrease with the rise in temperature.

22. Can a catalyst change the position of equilibrium in a reaction?

Ans: Catalyst cannot change the position of equilibrium it just affects the rate of reaction.

23. If ${{Q}_{c}}>{{K}_{c}}$, when we continuously remove the product, what would be the direction of the reaction?

Ans: When we continuously remove the product and the value of ${{Q}_{c}}>{{K}_{c}}$ then reaction will continue in forward direction i.e. in the direction of products.

24. Define strong and weak electrolyte.

Ans: Electrolyte which dissociates completely into ions in an aqueous solution is called Strong electrolyte while which is not able to dissociate completely is called weak electrolyte.

25. Write the conjugate acids for the following Bronsted bases: $N{{H}_{2}},N{{H}_{3}},HCO{{O}^{-}}$.

Ans: Conjugate acid of $N{{H}_{2}},N{{H}_{3}},HCO{{O}^{-}}$ are $N{{H}_{3}}^{+},N{{H}_{4}}^{+},HCOOH$ respectively.

26. Which conjugate base is stronger $C{{N}^{-}}\ \text{or }{{\text{F}}^{-}}$?

Ans: $C{{N}^{-}}$ is a stronger base.

27. What is the difference between a conjugate acid and a conjugate base?

Ans: The main difference between conjugate acid and base is proton.

28. Select Lewis acid and Lewis base from the following:

\[C{{u}^{2+}},{{H}_{2}}O,B{{F}_{3}},O{{H}^{-}}\]

Ans: Lewis acid: $C{{u}^{2+}}$ and $B{{F}_{3}}$

Lewis base: ${{H}_{2}}O$ and $O{{H}^{-}}$.

29. The dimethyl ammonium ion, ${{(C{{H}_{3}})}_{2}}N{{H}_{2}}^{+}$, is a weak acid and ionizes to a slight degree in water what is its conjugate base?

Ans: The conjugate base is ${{(C{{H}_{3}})}_{2}}NH$.

30. If pH of a solution is 7, calculate its pOH value.

Ans: We know the relation between pH and pOH is given as

$pH+pOH=14$

pH = 7 given then pOH = 14 -7 = 7.

31. What happens to the pH if a few drops of acid are added to $C{{H}_{3}}COON{{H}_{4}}$ solution?

Ans: pH remains unchanged or constant.

32. What is the concentration of ${{H}_{3}}{{O}^{+}}$ and $O{{H}^{-}}$ ions in water at 298K?

Ans: $[{{H}_{3}}{{O}^{+}}]=[O{{H}^{-}}]=1\times {{10}^{-7}}mo{{l}^{-1}}$.

33. The ${{p}^{ka}}$ of acetic acid and ${{p}^{kb}}$ of ammonium hydroxide are 4.76 and 4.75

respectively. Calculate the pH of ammonium acetate solution.

Ans: The relation between pH, ${{p}^{ka}}$ and ${{p}^{kb}}$ is given as follows:

\[pH=7+\frac{1}{2}({{p}^{ka}}-{{p}^{kb}})\]

\[pH=7+\frac{1}{2}(4.76-4.75)\]

\[pH=7+\frac{1}{2}(0.01)=7+0.005=7.005\]

34. Calculate the pH of the solution of 0.002 M HBr

Ans: pH value of 0.002 M HBr is given by the following reaction:

\[HBr+{{H}_{2}}O\to {{H}_{3}}{{O}^{+}}+B{{r}^{-}}\]

Now at 0.002 M, the concentration of ${{H}_{3}}{{O}^{+}}$ can be written as: $2\times {{10}^{-3}}$

The formula of pH will be $pH=-(\log {{H}_{3}}{{O}^{+}})=-\log (2\times {{10}^{-3}})$

This will be written (3-log2) and the value of log2 is 0.3010 and the value will be 2.7.

35. Define Buffer solution.

Ans: The solutions which resist the change in pH on dilution or by addition of small amounts of acid or alkali are called Buffer solutions.

36. When is a solution called unsaturated?

Ans: The solution is said to be unsaturated when the ionic product of the solution is less than the solubility product.

37. Give an example of acidic buffer.

Ans: $C{{H}_{3}}COOH+C{{H}_{3}}COONa$

38. Calculate the solubility of AgCl (s) in pure water.

Ans: First suppose the solubility of AgCl in pure water is $S\ mol{{L}^{-1}}$.

$AgCl\rightleftharpoons A{{g}^{+}}+C{{l}^{-}}$

$A{{g}^{+}}=[S],C{{l}^{-}}=[S]$

Then, Solubility product will be ${{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]$

$2.8\times {{10}^{-10}}=s\times s$

Or $s=\sqrt{2.8\times {{10}^{-10}}}=1.673\times {{10}^{-5}}mo{{l}^{-1}}$.

39. Name a basic buffer having pH around 10.

Ans: Basic buffer which have pH around 10 is Borax sodium hydroxide shown by the following composition $N{{a}_{2}}{{B}_{4}}{{O}_{7}}+NaOH$.

Short Answer Questions   2 Marks

1. Mention the general characteristics of equilibria involving physical processes.

Ans: It can be explained by four types of equilibrium:

a. Solid to Liquid Equilibrium: There is only one temperature i.e. 1 atm at which two phases coexist with each other or if there is no exchange of heat around surroundings then mass of these two phases remain constant.

b. Liquid Vapors Equilibrium: In this case vapors pressure is constant at any given temperature.

c. Dissolution of Solids in Liquids: In this case the solubility remains constant at any given temperature.

d. Dissolution of Gases in Liquids: In this case the concentration of a gas in liquid is proportional to the pressure of the gas over the liquid.

2. Write the expression for the equilibrium constant for the reaction:

$4N{{H}_{3}}(g)+5{{O}_{2}}(g)\rightleftharpoons 4NO(g)+6{{H}_{2}}O(g)$

Ans: Equilibrium constant can be represented by:

${{K}_{c}}=\frac{{{[NO]}^{4}}{{[{{H}_{2}}O]}^{5}}}{{{[N{{H}_{3}}]}^{4}}{{[{{O}_{2}}]}^{5}}}$

3. When the total number of moles of product and reactants are equal, K has no unit. Give reason.

Ans: When the total number of moles of products is equal to the total number of moles of reactants the equilibrium constant k has no unit this can be shown with the help of an example as given below:

\[{{H}_{2}}(g)+{{I}_{2}}(g)\rightleftharpoons 2HI(g)\]

Equilibrium can be shown as:

$K=\frac{{{[HI]}^{2}}}{[{{H}_{2}}][{{I}_{2}}]}$

Then units of K will be given as

$K=\frac{mol/L\times mol/L}{mol/L\times mol/L}=No\ \text{units}$

4. What is the unit of equilibrium for the reaction ${{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)$.

Ans: Equilibrium of this reaction can be shown a follows:

\[K=\frac{{{[N{{H}_{3}}]}^{2}}}{[{{N}_{2}}]{{[{{H}_{2}}]}^{3}}}\]

Then units of K will be shown as:

$K=\frac{{{[mol/L]}^{2}}}{[mol/L]{{[mol/L]}^{3}}}=\frac{1}{{{(mol/L)}^{2}}}$

= ${{L}^{2}}mo{{l}^{-1}}$

5. Give the relation ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$.

Ans: Assume a reaction:

$aA+bB\rightleftharpoons cC+dD$

Now equilibrium constant for this reaction can be written as:

${{K}_{c}}=\frac{{{[C]}^{c}}{{[D]}^{d}}}{{{[A]}^{a}}{{[B]}^{b}}}$

Now assume that these gaseous compounds behave ideally and follows the equation

$PV=nRT$

Now we can write the equation as:

$P=\frac{n}{V}RT=CRT$

Then, ${{K}_{p}}=\frac{{{p}_{C}}^{c}{{p}_{D}}^{d}}{{{p}_{A}}^{a}{{p}_{B}}^{b}}=\frac{{{([C]RT)}^{c}}\times {{([D]RT)}^{d}}}{{{([A]RT)}^{a}}\times {{([B]RT)}^{b}}}$

= $\frac{{{[C]}^{c}}\times {{[D]}^{d}}}{{{[A]}^{a}}\times {{[B]}^{b}}}\times (RT)[(c+d)-(a+b)]$

Where, $[(c+d)-(a+b)]=\Delta n$

This represents that ${{K}_{p}}={{K}_{c}}{{(RT)}^{\Delta n}}$

6. The value of ${{K}_{c}}$ for the reaction $2A\rightleftharpoons B+C$ is $2\times {{10}^{-3}}$. At a given time, the composition of the reaction mixture is [A] = [B] =[C] = $3\times {{10}^{-4}}$ M. In which direction the reaction will proceed?

Ans: For this reaction ${{Q}_{c}}$ will be given as follows:

${{Q}_{c}}=\frac{[B][C]}{{{[A]}^{2}}}$

According to question [A] = [B] =[C] = $3\times {{10}^{-4}}$ M, Put the values in equation

${{Q}_{c}}=\frac{[3\times {{10}^{-4}}][3\times {{10}^{-4}}]}{{{[3\times {{10}^{-4}}]}^{2}}}=1$

This suggests that ${{Q}_{c}}$ is greater than ${{K}_{c}}$ so the reaction will proceed in reverse reaction.

7. Write the equilibrium constant expression for each of the following reactions. In each case, indicate which of the reaction is homogeneous or heterogeneous.

a. $2CO(g)+{{O}_{2}}(g)\rightleftharpoons 2C{{O}_{2}}(g)$

Ans: ${{K}_{c}}=\frac{{{[C{{O}_{2}}]}^{2}}}{{{[CO]}^{2}}[{{O}_{2}}]}$

b. ${{N}_{2}}{{O}_{5}}(g)\rightleftharpoons N{{O}_{2}}(g)+N{{O}_{3}}(g)$

Ans: ${{K}_{c}}=\frac{[N{{O}_{2}}][N{{O}_{3}}]}{[{{N}_{2}}{{O}_{5}}]}$

c. $Zn(s)+2HCl(g)\rightleftharpoons ZnC{{l}_{2}}(s)+{{H}_{2}}(g)$

Ans: ${{K}_{c}}=\frac{[{{H}_{2}}]}{{{[HCl]}^{2}}}$

d. $2{{H}_{2}}O(l)\rightleftharpoons 2{{H}_{2}}O(l)+{{O}_{2}}(g)$

Ans: ${{K}_{c}}=[{{O}_{2}}]$

a and b are said to be homogeneous in nature while c and d are heterogeneous in nature.

8. The dissociation of HI is independent of pressure, while dissociation of $PC{{l}_{5}}$ depends upon the pressure applied. Why?

Ans: The reaction of HI can be shown as:

$2HI\rightleftharpoons {{H}_{2}}+{{I}_{2}}$

Equilibrium constant for this reaction can be shown as:

${{K}_{c}}=\frac{{{x}^{2}}}{4(1-{{x}^{2}})}$, Here x is degree of dissociation.

The reaction of $PC{{l}_{5}}$ can be shown as:

X $PC{{l}_{5}}\rightleftharpoons PC{{l}_{3}}+C{{l}_{2}}$

${{K}_{c}}=\frac{{{x}^{2}}}{(1-x)}$, Here x is degree of dissociation.

Since ${{K}_{c}}$ for HI does not have any volume terms and thus dissociation of HI is independent of pressure. On the other hand ${{K}_{c}}$ for $PC{{l}_{5}}$ has volume in denominator and increase in pressure reduces volume. Hence to remain ${{K}_{c}}$ constant, x decrease.

9. On what factors does the value of the equilibrium constant of a reaction depend?

Ans: The equilibrium constant of a reaction depends upon the temperature, pressure and stoichiometry of the reaction.

10. Why the addition of inert gas does not change the equilibrium?

Ans: Because the addition of an inert gas at constant volume is not able to change the partial pressures or the molar concentrations of the substance which are involved in the reaction.

11. The equilibrium constant of a reaction increases with rise in temperature. Is the reaction exo – or endothermic?

Ans: Equilibrium constant of reaction increase with increase in temperature this represents the reaction is endothermic.

12. Using Le – chatelier principle, predict the effect of

(a) Decreasing the temperature

Ans: ${{N}_{2}}(g)+3{{H}_{2}}(g)\rightleftharpoons 2N{{H}_{3}}(g)+\Delta$

In an exothermic reaction increase in temperature shifts the equilibrium to the left hand side and decrease in temperature shifts the reaction to right.

(b) Increasing the temperature

in each of the following equilibrium systems:

Ans: ${{N}_{2}}(g)+{{O}_{2}}(g)+\Delta \rightleftharpoons 2NO(g)$

For an endothermic reaction increase in temperature shifts the equilibrium to the right and decrease in temperature shifts the reaction towards right.

13. (i) In the reaction equilibrium $A+B\rightleftharpoons C+D$, What will happen to the concentrations of A, B and D if concentration of C is increased.

Ans: Equilibrium constant of the reaction can be shown as:

${{K}_{c}}=\frac{[C][D]}{[A][B]}$

If the concentration of product is increased then the concentration of other components changes which decreases the concentration of C and vice – versa.

And if the concentration of C is increased then concentration of D will decrease and in case of A and B concentration increases which shifts the equilibrium towards left.

(ii) What will happen if concentration of A is increased?

Ans: If the concentration of A is increased the concentration of B will decrease and the concentration of C and D will also increase and in this case ${{K}_{c}}$ is the same and vice – versa which shifts the equilibrium to the right.

14. Give two examples of actions which can act as Lewis acids.

Ans: The chemicals which are not going to act as Lewis acid are $A{{g}^{+}},{{H}^{+}}$.

15. Justify the statement that water behaves like an acid and also like a base on the basis of protonic concept.

Ans: Ionization of water is shown as:

${{H}_{2}}O+{{H}_{2}}O\rightleftharpoons {{H}_{3}}{{O}^{+}}+O{{H}^{-}}$

Water behave as a base with strong acid and accept the proton which is lost by the acid

${{H}_{2}}O+HCl\rightleftharpoons {{H}_{3}}{{O}^{+}}+C{{l}^{-}}$

Water behave as an acid with strong base 

${{H}_{2}}O+N{{H}_{3}}\rightleftharpoons N{{H}_{4}}^{+}+O{{H}^{-}}$

16. The degree of dissociation of ${{N}_{2}}{{O}_{4}}$, ${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)$, at temperature T and total pressure is $\alpha $. Find the expression for the equilibrium constant of this reaction at this temperature and pressure?

Ans: ${{N}_{2}}{{O}_{4}}(g)\rightleftharpoons 2N{{O}_{2}}(g)$

If p is the total pressure then

\[{{P}_{{{N}_{2}}{{O}_{4}}}}=\frac{(1-\alpha )}{(1+\alpha )}P\]

\[{{P}_{N{{O}_{2}}}}=\frac{2\alpha }{(1+\alpha )}P\]

Then ${{K}_{P}}=\frac{{{p}^{2}}_{N{{O}_{2}}}}{{{P}_{{{N}_{2}}{{O}_{4}}}}}=\frac{{{[2\alpha p/(1+\alpha )]}^{2}}}{[(1-\alpha )p/(1+\alpha )]}=\frac{4{{\alpha }^{2}}p}{1-{{\alpha }^{2}}}$

17. A solution gives the following colors with different indicators. Methyl orange – yellow, methylred – yellow, and bromothymol blue - Orange. What is the pH of the solution?

Ans: Methyl orange if given yellow color then it is said to be basic in nature and have pH 3.2-4.5.

Methyl red which shows yellow color is also of basic nature showing pH of 4.4-6.5.

Bromothymol blue gives orange yellowish color in acidic medium and has a pH range from 6.0-7.8.

18. Calculate the pH of the solution 0.002 M HBr.

Ans: pH value of 0.002 M HBr is given by the following reaction:

$HBr+{{H}_{2}}O\to {{H}_{3}}{{O}^{+}}+B{{r}^{-}}$

Now at 0.002 M, the concentration of ${{H}_{3}}{{O}^{+}}$ can be written as: $2\times {{10}^{-3}}$

The formula of pH will be $pH=-(\log {{H}_{3}}{{O}^{+}})=-\log (2\times {{10}^{-3}})$

This will be written (3-log2) and the value of log2 is 0.3010 and the value will be 2.7.

19. The concentration of ${{H}^{+}}$ in a soft drink is $3.8\times {{10}^{-3}}$ M. What is its pH?

Ans: Concentration of $[{{H}^{+}}]=3.8\times {{10}^{-3}}$

We know that $pH=-\log [{{H}^{+}}]$, put the value of ${{H}^{+}}$

$pH=-\log [3.8\times {{10}^{-3}}]=-\log 3.8-\log {{10}^{-3}}$

$=-0.5798+3=2.42$.

20. Define solubility product.

Ans: Solubility product of a salt is defined as that at any given temperature is equal to the product of the concentration of its ions which is raised to the power equal to the number of ions produced by dissociation of one mole of the substance.

21. ${{K}_{sp}}$ for $HgS{{O}_{4}}$ is $6.4\times {{10}^{-5}}$. What is the solubility of the salt?

Ans: We know that solubility of salt is given by the formula

$S={{({{K}_{sp}})}^{1/2}}$, Put the value of ${{K}_{sp}}$ in the equation

$S={{(6.4\times {{10}^{-5}})}^{1/2}}={{(64\times {{10}^{-6}})}^{1/2}}$

$=8\times {{10}^{-3}}$.

22. Calculate the pH of a buffer solution containing 0.1 mole of acetic acid and 0.15 mole of sodium acetate. Ionisation constant for acetic acid is $1.75\times {{10}^{-5}}$.

Ans: $pH={{p}_{Ka}}+\log \frac{Salt}{Acid}$

Put the values given in the equation

$pH=-\log 1.75\times {{10}^{-5}}+\log \frac{0.15}{0.10}$

Or $pH=-\log 1.75\times {{10}^{-5}}+\log 1.5=4.9$

Long Answer Questions    3 Marks

1. Name the three group into which chemical equilibrium can be classified.

Ans: Chemical equilibrium generally classified into three different groups which can be explained as follows:

(i)The reaction which proceeds nearly to completion and has only negligible concentrations of the reactants are left.

(ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium stage.

(iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium.

2. Give the generalizations concerning the composition of equilibrium mixtures.

Ans: Equilibrium mixture which is represented by ${{K}_{c}}$ tells us about the type of reaction i.e.

1. If the value of ${{K}_{c}}>{{10}^{3}}$ then reaction will be dominated towards reactants and if the value is very large then it is about to complete.

2. If the value of ${{K}_{c}}<{{10}^{3}}$ then reaction will be dominated towards the product and if the value is very small then reaction rarely proceeds.

3. If ${{K}_{c}}$ is in between the range of ${{10}^{-3}}-{{10}^{3}}$ then concentration of both reactants as well as products are appreciated.

3. Predict if the solutions of the following salts are neutral, acidic or basic:

$NaCl,KBr,NaCN,NaOH,{{H}_{2}}S{{O}_{4}},NaN{{O}_{2}},N{{H}_{4}}N{{O}_{3}},KF$.

Ans: Neutral - $NaCl,KBr$

Acidic - ${{H}_{2}}S{{O}_{4}},N{{H}_{4}}N{{O}_{3}},KF$

Basic: $NaCN,NaOH,NaN{{O}_{2}}$

Long Answer Questions  5 Marks

1. Find the oxidation state of sulphur in the following compounds:

${{H}_{2}}S,{{H}_{2}}S{{O}_{4}},{{S}_{2}}{{O}_{4}}^{2-},{{S}_{2}}{{O}_{8}}^{2-}and\ HS{{O}_{3}}^{-}$

Ans: In ${{H}_{2}}S$, oxidation state of hydrogen is one then for sulphur oxidation state will be $2+x=0;x=-2$

In ${{H}_{2}}S{{O}_{4}}$, oxidation state of hydrogen is one and oxygen is -2 then sulphur will be $2+x-8=0;x=6$

In ${{S}_{2}}{{O}_{4}}^{2-}$, oxidation state of oxygen is -2 then oxidation state of sulphur will be $2x-8=-2;x=3$

In ${{S}_{2}}{{O}_{8}}^{2-}$, oxidation state of oxygen is -2 then oxidation state of sulphur will be $2x-16=-2;x=7$

In $HS{{O}_{3}}^{-}$, the oxidation state of hydrogen is one and oxygen is -2 then sulphur will be $1+x-6=-1;x=4$.

Important Related Links for CBSE Class 11 Chemistry