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Electronic Configuration and Oxidation States

Oxidation sta...

Question

Oxidation state of S in $H_{2} S O_{3}$ is :

A

6

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B

4

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C

2

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D

3

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Solution

The correct option is B 4
$H_{2} S O_{3}$
Let, Oxidation state of S is $x$
So, $x + 3 \times (- 2) + 2 \times (+ 1) = 0$
$\Rightarrow x = + 4$
Hence, the correct option is (B).

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Similar questions

When $H_{2} S O_{3}$ is converted into $H_{2} S O_{4}$ the change in the oxidation state of sulphur is from:

Q. List-I has compounds and list-II has oxidation state of

S

in these compounds. The correct match is:

List-I	List-ll
A $) H_{2} S O_{4}$	1 $) + 4$
B $) H_{2} S_{2} O_{6}$	2 $) + 3$
C $) H_{2} S O_{3}$	3 $) + 2$
D $) H_{2} S_{2} O_{4}$	4 $) + 6$
	5 $) + 5$

Q. Assertion :The oxidation state of

S

H_{2} S_{2} O_{8}

6

. Reason: Maximum oxidation state of

S

6

because the maximum oxidation state of an element is its number of valence electron.

Q. Balance the following equation by oxidation number method.

F e_{2} (S O_{4})_{3} + H_{2} S O_{3} + H_{2} O \to F e S O_{4} + H_{2} S O_{4}

Q. Oxidation state of

S

{S O_{4}}^{2 -}

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