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Oxidation number of S in is :
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By Expert Faculty of Sri Chaitanya
a
6
b
3
c
2
d
-2
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Correct option is A
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)Similar Questions
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