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Q.

Oxidation number of S in SO42− is :

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a

6

b

3

c

2

d

-2

answer is A.

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Detailed Solution

Oxidation number (ON.) of S in  SO42− is found as :O.N. of S +4 (O.N. of 0) = -2O.N. of S + 4(-2) = -2 O.N.ofS-8 = -2 O.N of S = -2+8 =+6.So, the correct answer is (1)
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