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Q.

The oxidation number of Fe in  [Fe(CN)6]4Cr  in [Cr(NH3)3(NO2)3] and Ni in Ni(CO)4 are respectivey

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a

0,+3,+2

b

+3,+3,0

c

+3,+0,+3

d

+2,+3,0

answer is D.

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Detailed Solution

[Fe(CN)6]4

x+6(1)=4

x6=4;  x=64=2

oxidation number of Fe = +2

[Cr(NH3)3(NO2)3]

x+3(0)+3(1)=0

x3=0;  x=+3 oxidation number of Ni = 0

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