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N factor for the reaction [Fe(CN) 6 ] 4– → Fe 3+ + CO 2 + NO – 3

N factor for the reaction [Fe(CN)6]4– → Fe3+ + CO2 + NO–3

Grade:12

3 Answers

Naveen Kumar
askIITians Faculty 60 Points
9 years ago
For the ferrous cyanide complex,
Fe..........=+2
In CN,
C...............=+2
N.................=-3
In tyhe right hand products, the oxidation states are as follows:
Fe..............+3
C................+4
N.................+5
So change in oxidation numbers for each Fe(CN)4- complex=1(Fe)+ 2(C) +8(N)=11=N-factor
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Sivagami
15 Points
5 years ago
Don’t you have to consider the fact that there are six Carbon and six Nitrogen atoms in one molecule of K4[Fe(CN)6]. Hence 1 + 6 x2 + 6 x 8 = 61
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the answer to your question.
 
For the ferrous cyanide complex,
Fe..........=+2
In CN,
C...............=+2
N.................=-3
In tyhe right hand products, the oxidation states are as follows:
Fe..............+3
C................+4
N.................+5
Since there are six Carbon and six Nitrogen atoms in one molecule of K4[Fe(CN)6].
Hence, N factor =  1 + 6 x2 + 6 x 8 = 61
 
Thanks and regards,
Kushagra
 

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