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Chemistry MCQs for NEET — Practice Questions with Answers

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Which of the following oxidation states are the most characteristics for lead and tin respectively?

+4, +2

+2, +4

+4, +4

+2, +2

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Explanation

(b) The tendency to form +2 ionic state increase on moving down the group due to the inert pair effect. Most characteristic oxidation state for lead and tin are +2, +4 respectively.

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Which one of the following orders correctly represents the increasing acid strengths of the given acids?

HOCl<HOClO<HOClO₂<HOClO₃

HOClO<HOCl<HOClO₃<HOClO₂

HOClO₂<HOClO₃<HOClO<HOCl

HOClO₃>HOClO₂<HOClO<HOCl

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Explanation

The acidic character of oxyacids increases with increase in oxidation number. The acidity of oxyacids of chlorine decreases in the following order.

$H \overset{+ 7}{Cl} O_{4} > H \overset{+ 5}{Cl} O_{3} > H \overset{+ 3}{Cl} O_{2} > H \overset{+ 1}{Cl} O$

So relative acidic character increases

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The valency of Cr in the complex [Cr(H₂O)₄Cl₂]⁺ [MP PMT 2000]

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Explanation

${[\overset{*}{Cr} {(H_{2} O)}_{4} {Cl}_{2}]}^{+}$

$x +0+2 (-1)=+ 1$ ; $x -2=+ 1$

x = +3 for Cr in complex.

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In the conversion ${Br}_{2} \to {BrO}_{3}^{-},$ the oxidation state of bromine changes from [EAMCET 1990; AMU 1999; RPMT 2002]

– 1 to – 1

0 to – 1

0 to + 5

0 to – 5

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Explanation

$\overset{0}{B r_{2}} \to \overset{+ 5}{B r O_{3}^{-}}$ , in this reaction oxidation state change from 0 to + 5.

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Oxidation number of cobalt in K[Co(CO)₄] is [KCET 1996]

+ 1

+ 3

– 1

– 3

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Explanation

$K[\overset{*}{C} o {(C O)}_{4}]$

$1 + x +0=0; x =-1$ .

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Oxidation number of nickel in Ni(CO)₄ [AIIMS 1984; MNR 1985; CPMT 1997; MP PET/PMT 1998;]

+ 4

– 4

+ 2

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Explanation

If any central metal atom combined with corbonyl group than central metal atom shows always zero oxidation state.

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Sn⁺⁺ loses two electrons in a reaction. What will be the oxidation number of tin after the reaction

+ 2

Zero

+ 4

– 2

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Explanation

$S n^{2 +} \to S n^{4 +} + 2 e^{-}$

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The oxidation number of S in H₂S₂O₈is [MP PET 2002]

+ 2

+ 4

+ 6

+ 7

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Explanation

The chemical structure of H₂S₂O₈ is as follows:-

$H - O - {S_{| |}^{| |}}_{O}^{O} - O - O - {S_{| |}^{| |}}_{O}^{O} - O - H$

So the oxidation number of S should be :

$\underset{(f o r H)}{2 \times (+ 1)} + \underset{(f o r S)}{2 \times X} + \underset{(f o r O)}{6 \times (- 2)} + \underset{(f o r O - O)}{2 \times (- 1)} = 0$ or X = +6.

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The oxidation state of nitrogen in N₃H is [NCERT 1977, 81]

$+ \frac{1}{3}$

+ 3

–1

$- \frac{1}{3}$

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Explanation

In hydrazoic acid (N₃H) nitrogen shows $- \frac{1}{3}$ oxidation state.

$\overset{*}{N_{3} H}$

$3 x + 1 = 0$ , 3x = –1, $x = - \frac{1}{3}$ .

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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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