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parveen35
@parveen35
September 2021
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calculate oxidation no. of N in (NH4)2SO4
Answers & Comments
sd202167KoushikDas
Let the oxidation number of N be x.
So, 2 [x + 4 × 1] + 6 + 4 × (-2)
= 2x + 8 + 6 - 8
= 2x + 6
Therefore, Oxidation number is
2x + 6 = 0
or 2x = -6
or x = -3
Hence, the O.N. of N in (NH4)²SO4 is -3.
53 votes
Thanks 99
parveen35
y oxidation no. of S is taken as 6?
sd202167KoushikDas
Yes.
sd202167KoushikDas
You can verify that the O.N of S is 6 or not by just taking O.N of S as x and then put the O.N values of other elements.
parveen35
I think oxidation no. of S is -2?
sd202167KoushikDas
VERIFY yourself
sd202167KoushikDas
SORRY if I can not help you
akshatseanjamip2hjto
Hope this helps with your question.
24 votes
Thanks 39
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Answers & Comments
So, 2 [x + 4 × 1] + 6 + 4 × (-2)
= 2x + 8 + 6 - 8
= 2x + 6
Therefore, Oxidation number is
2x + 6 = 0
or 2x = -6
or x = -3
Hence, the O.N. of N in (NH4)²SO4 is -3.