Q) plz.. clarify how oxidation number of Fe in Q9 is +2?????
Asked by araima2001
27th September 2017,
10:40 AM
Answered by Expert
Answer:
Na4[Fe(CN)5(NOS)]
Let us consider the oxidation state of Fe as x in the given compound:
The overall charge on the compound is zero.
Hence,
4(+1) + x + (5x-1) + (1x-1) = 0
4 + x - 5 - 1 = 0
x = +2
Thus the oxidation state of Fe in the given compound is +2.
Answered by Expert
27th September 2017,
3:09 PM
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