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CHXI08:REDOX REACTIONS

315463 In ${MnO}_{4}^{2 -}$ and ${MnO}_{4}^{-}$ which one undergoes disproportionation reaction in acidic medium?

1

{MnO}_{4}^{2 -}

2

{MnO}_{4}^{-}

3 Both

{MnO}_{4}^{2 -}

and

{MnO}_{4}^{-}

4 Neither

{MnO}_{4}^{2 -}

nor

{MnO}_{4}^{-}

Explanation:

In ${MnO}_{4}^{-}, Mn$ is the highest oxidation state is +7 . Therefore, it does not undergo disproportionation. ${MnO}_{4}^{2 -}$ undergoes disproportionation as follows. $3 {MnO}_{4}^{2 -} + 4 H^{+} \to 2 {MnO}_{4}^{-} + {MnO}_{2} + 2 H_{2} O .$

CHXI08:REDOX REACTIONS

315464 Which of the following is not an example of disproportionation reaction?

1

{Cl}_{2} + 2 NaOH \to NaCl + NaOCl + H_{2} O

2

P_{4} + 3 NaOH + 3 H_{2} O \to {PH}_{3} + 3 {NaH}_{2} {PO}_{2}

3

2 NaOH + 2 F_{2} \to 2 NaF + {OF}_{2} + H_{2} O

4

2 H_{2} O_{2} \to 2 H_{2} O + O_{2}

CHXI08:REDOX REACTIONS

315465 Match Column I with Column II
supporting img

1

A - S,B - P,C - Q,D - R

2

A - P,B - Q,C - R,D - S

3

A - R,B - Q,C - P,D - S

4

A - Q,B - R,C - S,D - P

Explanation:

$A - S, B - P, C - Q, D - R$
$N_{2}$ and $O_{2}$ are combining to form NO .So, it is combination reaction
$Pb {({NO}_{3})}_{2}$ is decomposed to $PbO, {NO}_{2}$ and $O_{2}$ ' Na ' displaced hydrogen in its reaction with water.
${NO}_{2}$ with +4 oxidation state of nitrogen changed into ${NO}_{2}^{-} (+ 3)$ and ${NO}_{3}^{-} (+ 5)$ , so it is disproportionation. So, the correct option is (1).

JEE Main - 2024

CHXI08:REDOX REACTIONS

315466 When chlorine gas passes through a concentrated solution of alkali.
I. ${Cl}_{2}$ is oxidised to ${ClO}^{-}$
II. ${Cl}_{2}$ is reduced to ${Cl}^{-}$
III. Products formed are $5 {Cl}^{-}, {ClO}_{3}^{-}$ and $3 H_{2} O$ .
Which of the above statement is/are correct?

1 Only II

2 I and II

3 I and III

4 II and III

Explanation:

Statement II and III are correct, while the statement I is incorrect. Its correct form is as follows
When chlorine gas passes through a concentrated solution of alkali, the following chemical reaction occurs.
${Cl}_{2} + {OH}^{-} (aq) \to {Cl}^{- 1} (aq) + {ClO}_{3}^{-}$
Thus, ${Cl}_{2}$ is reduced.
After balancing this equation, we have
$\underset{3 mol}{3 {Cl}_{2}} + \underset{6 mol}{6 {OH}^{-}} (aq) \to 5 {Cl}^{-} (aq) + {ClO}_{3}^{-} (aq) + 3 H_{2} O$

CHXI08:REDOX REACTIONS

315463 In ${MnO}_{4}^{2 -}$ and ${MnO}_{4}^{-}$ which one undergoes disproportionation reaction in acidic medium?

1

{MnO}_{4}^{2 -}

2

{MnO}_{4}^{-}

3 Both

{MnO}_{4}^{2 -}

and

{MnO}_{4}^{-}

4 Neither

{MnO}_{4}^{2 -}

nor

{MnO}_{4}^{-}

Explanation:

In ${MnO}_{4}^{-}, Mn$ is the highest oxidation state is +7 . Therefore, it does not undergo disproportionation. ${MnO}_{4}^{2 -}$ undergoes disproportionation as follows. $3 {MnO}_{4}^{2 -} + 4 H^{+} \to 2 {MnO}_{4}^{-} + {MnO}_{2} + 2 H_{2} O .$

CHXI08:REDOX REACTIONS

315464 Which of the following is not an example of disproportionation reaction?

1

{Cl}_{2} + 2 NaOH \to NaCl + NaOCl + H_{2} O

2

P_{4} + 3 NaOH + 3 H_{2} O \to {PH}_{3} + 3 {NaH}_{2} {PO}_{2}

3

2 NaOH + 2 F_{2} \to 2 NaF + {OF}_{2} + H_{2} O

4

2 H_{2} O_{2} \to 2 H_{2} O + O_{2}

CHXI08:REDOX REACTIONS

315465 Match Column I with Column II
supporting img

1

A - S,B - P,C - Q,D - R

2

A - P,B - Q,C - R,D - S

3

A - R,B - Q,C - P,D - S

4

A - Q,B - R,C - S,D - P

Explanation:

$A - S, B - P, C - Q, D - R$
$N_{2}$ and $O_{2}$ are combining to form NO .So, it is combination reaction
$Pb {({NO}_{3})}_{2}$ is decomposed to $PbO, {NO}_{2}$ and $O_{2}$ ' Na ' displaced hydrogen in its reaction with water.
${NO}_{2}$ with +4 oxidation state of nitrogen changed into ${NO}_{2}^{-} (+ 3)$ and ${NO}_{3}^{-} (+ 5)$ , so it is disproportionation. So, the correct option is (1).

JEE Main - 2024

CHXI08:REDOX REACTIONS

315466 When chlorine gas passes through a concentrated solution of alkali.
I. ${Cl}_{2}$ is oxidised to ${ClO}^{-}$
II. ${Cl}_{2}$ is reduced to ${Cl}^{-}$
III. Products formed are $5 {Cl}^{-}, {ClO}_{3}^{-}$ and $3 H_{2} O$ .
Which of the above statement is/are correct?

1 Only II

2 I and II

3 I and III

4 II and III

Explanation:

Statement II and III are correct, while the statement I is incorrect. Its correct form is as follows
When chlorine gas passes through a concentrated solution of alkali, the following chemical reaction occurs.
${Cl}_{2} + {OH}^{-} (aq) \to {Cl}^{- 1} (aq) + {ClO}_{3}^{-}$
Thus, ${Cl}_{2}$ is reduced.
After balancing this equation, we have
$\underset{3 mol}{3 {Cl}_{2}} + \underset{6 mol}{6 {OH}^{-}} (aq) \to 5 {Cl}^{-} (aq) + {ClO}_{3}^{-} (aq) + 3 H_{2} O$

CHXI08:REDOX REACTIONS

315463 In ${MnO}_{4}^{2 -}$ and ${MnO}_{4}^{-}$ which one undergoes disproportionation reaction in acidic medium?

1

{MnO}_{4}^{2 -}

2

{MnO}_{4}^{-}

3 Both

{MnO}_{4}^{2 -}

and

{MnO}_{4}^{-}

4 Neither

{MnO}_{4}^{2 -}

nor

{MnO}_{4}^{-}

Explanation:

In ${MnO}_{4}^{-}, Mn$ is the highest oxidation state is +7 . Therefore, it does not undergo disproportionation. ${MnO}_{4}^{2 -}$ undergoes disproportionation as follows. $3 {MnO}_{4}^{2 -} + 4 H^{+} \to 2 {MnO}_{4}^{-} + {MnO}_{2} + 2 H_{2} O .$

CHXI08:REDOX REACTIONS

315464 Which of the following is not an example of disproportionation reaction?

1

{Cl}_{2} + 2 NaOH \to NaCl + NaOCl + H_{2} O

2

P_{4} + 3 NaOH + 3 H_{2} O \to {PH}_{3} + 3 {NaH}_{2} {PO}_{2}

3

2 NaOH + 2 F_{2} \to 2 NaF + {OF}_{2} + H_{2} O

4

2 H_{2} O_{2} \to 2 H_{2} O + O_{2}

CHXI08:REDOX REACTIONS

315465 Match Column I with Column II
supporting img

1

A - S,B - P,C - Q,D - R

2

A - P,B - Q,C - R,D - S

3

A - R,B - Q,C - P,D - S

4

A - Q,B - R,C - S,D - P

Explanation:

$A - S, B - P, C - Q, D - R$
$N_{2}$ and $O_{2}$ are combining to form NO .So, it is combination reaction
$Pb {({NO}_{3})}_{2}$ is decomposed to $PbO, {NO}_{2}$ and $O_{2}$ ' Na ' displaced hydrogen in its reaction with water.
${NO}_{2}$ with +4 oxidation state of nitrogen changed into ${NO}_{2}^{-} (+ 3)$ and ${NO}_{3}^{-} (+ 5)$ , so it is disproportionation. So, the correct option is (1).

JEE Main - 2024

CHXI08:REDOX REACTIONS

315466 When chlorine gas passes through a concentrated solution of alkali.
I. ${Cl}_{2}$ is oxidised to ${ClO}^{-}$
II. ${Cl}_{2}$ is reduced to ${Cl}^{-}$
III. Products formed are $5 {Cl}^{-}, {ClO}_{3}^{-}$ and $3 H_{2} O$ .
Which of the above statement is/are correct?

1 Only II

2 I and II

3 I and III

4 II and III

Explanation:

Statement II and III are correct, while the statement I is incorrect. Its correct form is as follows
When chlorine gas passes through a concentrated solution of alkali, the following chemical reaction occurs.
${Cl}_{2} + {OH}^{-} (aq) \to {Cl}^{- 1} (aq) + {ClO}_{3}^{-}$
Thus, ${Cl}_{2}$ is reduced.
After balancing this equation, we have
$\underset{3 mol}{3 {Cl}_{2}} + \underset{6 mol}{6 {OH}^{-}} (aq) \to 5 {Cl}^{-} (aq) + {ClO}_{3}^{-} (aq) + 3 H_{2} O$

CHXI08:REDOX REACTIONS

315463 In ${MnO}_{4}^{2 -}$ and ${MnO}_{4}^{-}$ which one undergoes disproportionation reaction in acidic medium?

1

{MnO}_{4}^{2 -}

2

{MnO}_{4}^{-}

3 Both

{MnO}_{4}^{2 -}

and

{MnO}_{4}^{-}

4 Neither

{MnO}_{4}^{2 -}

nor

{MnO}_{4}^{-}

Explanation:

In ${MnO}_{4}^{-}, Mn$ is the highest oxidation state is +7 . Therefore, it does not undergo disproportionation. ${MnO}_{4}^{2 -}$ undergoes disproportionation as follows. $3 {MnO}_{4}^{2 -} + 4 H^{+} \to 2 {MnO}_{4}^{-} + {MnO}_{2} + 2 H_{2} O .$

CHXI08:REDOX REACTIONS

315464 Which of the following is not an example of disproportionation reaction?

1

{Cl}_{2} + 2 NaOH \to NaCl + NaOCl + H_{2} O

2

P_{4} + 3 NaOH + 3 H_{2} O \to {PH}_{3} + 3 {NaH}_{2} {PO}_{2}

3

2 NaOH + 2 F_{2} \to 2 NaF + {OF}_{2} + H_{2} O

4

2 H_{2} O_{2} \to 2 H_{2} O + O_{2}

CHXI08:REDOX REACTIONS

315465 Match Column I with Column II
supporting img

1

A - S,B - P,C - Q,D - R

2

A - P,B - Q,C - R,D - S

3

A - R,B - Q,C - P,D - S

4

A - Q,B - R,C - S,D - P

Explanation:

$A - S, B - P, C - Q, D - R$
$N_{2}$ and $O_{2}$ are combining to form NO .So, it is combination reaction
$Pb {({NO}_{3})}_{2}$ is decomposed to $PbO, {NO}_{2}$ and $O_{2}$ ' Na ' displaced hydrogen in its reaction with water.
${NO}_{2}$ with +4 oxidation state of nitrogen changed into ${NO}_{2}^{-} (+ 3)$ and ${NO}_{3}^{-} (+ 5)$ , so it is disproportionation. So, the correct option is (1).

JEE Main - 2024

CHXI08:REDOX REACTIONS

315466 When chlorine gas passes through a concentrated solution of alkali.
I. ${Cl}_{2}$ is oxidised to ${ClO}^{-}$
II. ${Cl}_{2}$ is reduced to ${Cl}^{-}$
III. Products formed are $5 {Cl}^{-}, {ClO}_{3}^{-}$ and $3 H_{2} O$ .
Which of the above statement is/are correct?

1 Only II

2 I and II

3 I and III

4 II and III

Explanation:

Statement II and III are correct, while the statement I is incorrect. Its correct form is as follows
When chlorine gas passes through a concentrated solution of alkali, the following chemical reaction occurs.
${Cl}_{2} + {OH}^{-} (aq) \to {Cl}^{- 1} (aq) + {ClO}_{3}^{-}$
Thus, ${Cl}_{2}$ is reduced.
After balancing this equation, we have
$\underset{3 mol}{3 {Cl}_{2}} + \underset{6 mol}{6 {OH}^{-}} (aq) \to 5 {Cl}^{-} (aq) + {ClO}_{3}^{-} (aq) + 3 H_{2} O$