MCQ
Oxidation number of S in S2O32− is:
- A-2
- B+2
- C+6
- D0
✓
Answer
- +2
Explanation:
Let Oxidation number of S in S2O32− 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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