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Redox Reactions MCQs for NEET — Chemistry Questions with Answers

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The oxidation state of Cr in [Cr(NH₃)₄Cl₂]⁺ is [AIEEE 2005]

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Explanation

${[Cr {{(NH}_{3})}_{4} {Cl}_{2}]}^{+}$

$x + 4 \times (0) - 2 = 1$

⇒ $x + 0 - 2 = 1$

⇒ x = 1 + 2 = + 3.

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A brown ring complex compound is formulated as $[F e {(H_{2} O)}_{5} N O] S O_{4} .$ The oxidation state of iron is

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Explanation

In this complex iron is a central metal atom showing +1 oxidation state.becoz NO is showing +1 charge here.

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Oxidation number of P in Mg₂P₂O₇ is [CPMT 1989; MP PMT 1995]

+ 3

+ 2

+ 5

– 3

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Explanation

$\overset{*}{M g_{2} P_{2} O_{7}}$

$4 + 2 x - 2 \times 7 = 0$ ; $2 x = 14 - 4 = 10$

2x = 10; $x = \frac{10}{2} = + 5$ .

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The oxidation number of phosphorus in Ba(H₂PO₂)₂ is [Kurukshetra CEE 1998; DCE 2004]

– 1

+ 1

+ 2

+ 3

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Explanation

$B a {(H_{2} \overset{*}{P} O_{2})}_{2}$ ; $B a H_{4} \overset{*}{P_{2}} O_{4}$

$2 + 4 + 2 x - 8 = 0$ ; 2x = 2

$x = \frac{2}{2} = + 1$ .

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What is the oxidation number of sulphur in $N a_{2} S_{4} O_{6}$ [AIIMS 1998; DCE 1999]

$\frac{2}{3}$

$\frac{3}{2}$

$\frac{3}{5}$

$\frac{5}{2}$

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Explanation

$N a_{2} {\overset{*}{S}}_{4} O_{6}$

$2 + 4 x - 12 = 0$

$4 x = 10$

$\Rightarrow x = \frac{10}{4}$

$\Rightarrow x = \frac{5}{2}$

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What is the oxidation number of Co in $[C o {(N H_{3})}_{4} C l N O_{2}]$ [BHU 1999]

+ 2

+ 3

+ 4

+ 5

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Explanation

$[\overset{*}{C o} {(N H_{3})}_{4} C l N O_{2}]$

$x + 4 (0) + 1 (- 1) + 1 (- 1) = 0$

$x + 0 - 1 - 1 = 0$

x – 2 = 0; x = +2.

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When KMnO₄ acts as an oxidising agent and ultimately forms ${[M n O_{4}]}^{- 2}, M n O_{2}, M n_{2} O_{3}, M n^{+ 2}$ then the number of electrons transferred in each case respectively is [AIEEE 2002]

4, 3, 1, 5

1, 5, 3, 7

1, 3, 4, 5

3, 5, 7, 1

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Explanation

Number of e^– transferred in each case is 1, 3, 4, 5.

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For the redox reaction $M n O_{4}^{-} + C_{2} O_{4}^{- 2} + H^{+} \to M n^{2 +} + C O_{2} + H_{2} O$ the correct coefficients of the reactants for the balanced reaction are [IIT 1988, 92; BHU 1995; CPMT 1997; RPMT 1999; DCE 2000; MP PET 2003]

	$M n O_{4}^{-}$	$C_{2} O_{4}^{2 -}$	$H^{+}$

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Explanation

$M n O_{4}^{-} + 8 H^{+} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O \times 2$

$C_{2} O_{4}^{2 -} \to 2 C O_{2} + 2 e^{-} \times 5$

$\bar{2 M n O_{4}^{-} + 5 C_{2} O_{4}^{2 -} + 16 H^{+} \to 2 M n^{2 +} + 10 C O_{2} + 8 H_{2} O}$

Thus the coefficient of $M n O_{4}^{-}$ , $C_{2} O_{4}^{2 -}$ and H⁺ in the above balanced equation respectively are 2, 5, 16.

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In the balanced chemical reaction,

$I O_{3}^{-} + a I^{-} + b H^{+} \to c H_{2} O + d I_{2}$

a, b, c and d respectively correspond to [AIIMS 2005]

5, 6, 3, 3

5, 3, 6, 3

3, 5, 3, 6

5, 6, 5, 5

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Explanation

IO₃^– + aI^–+ bH⁺ → cH₂O + dI₂

Step 1 : I^–1 → I₂ (oxidation)

IO₃^– → I₂ (reduction)

Step 2 : 2IO₃^– + 12H⁺ → I₂ + 6H₂O

Step 3 : 2IO₃^– + 12H⁺ + 10e → I₂ + 6H₂O

2I^– → I₂ + 2e

Step 4 : 2IO₃^– + 12H⁺ + 10e^– → I₂ + 6H₂O

[2I^– → I₂ + 2e]5

Step 5 : 2IO₃^– + 10I^– + 12H⁺ → 6I₂ + 6H₂O

IO₃^– + 5I^– + 6H⁺ → 3I₂ + 3H₂O

On comparing, a = 5, b = 6, c = 3, d = 3

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The number of moles of KMnO₄ reduced by one mole of KI in alkaline medium is: [CBSE PMT 2005]

One fifth

five

One

Two

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Explanation

In alkaline medium

$2 K M n O_{4} + K I + H_{2} O \to 2 M n O_{2} + 2 K O H + K I O_{3}$ .

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