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What is the n factor of K4[Fe(CN)6] in the given reaction
2 months ago

Answers : (3)

Arun
22540 Points
							
For the ferrous cyanide complex,Fe..........=+2In CN,C...............=+2N.................=-3In tyhe right hand products, the oxidation states are as follows:Fe..............+3C................+4N.................+5So change in oxidation numbers for each Fe(CN)4- complex=1(Fe)+ 2(C) +8(N)=11=N-factor
2 months ago
Vikas TU
8729 Points
							
n factor is the number of electron transfer for single atom or the oxidation state of the central atom
In the complex
K4[Fe(CN)6] 
4(1) + x + 6(−1) =0
x−2 = 0
x = +2 
so n factor =2
2 months ago
Naina Fathima
15 Points
							
In K4[Fe(CN)6]  n actor is given by:
total positive charge carried by K+ which is equal to 4
or
it is given by total negative charge carriedby 
anionic part [Fe(CN)6] - which is equal to 4. Therefore the answer is 4
2 months ago
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