MCQ
Oxidation number of S in S2​O32− is:
  • A
    -2
  • B
    +2
  • C
    +6
  • D
    0

Answer

  1. +2

Explanation:

Let Oxidation number of S in S2​O32−​ be x.

Thus,

$2\text{x} + (-2 × 3) = -2 $

$2\text{x} -6 = -2$

$2\text{x} = -2 + 6$

$2\text{x} = 4$

$\text{x}=\frac{4}{2}$

$\text{x}=2$

So the oxidation state of sulfur is +2.

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