Question
Question: The oxidation state of \(Fe\) in \(\mathop K\nolimits_4 \left[ {\mathop {Fe(CN)}\nolimits_6 } \right...
The oxidation state of Fe in K4[Fe(CN)6] is:
Solution
The first nitrosyl complex was discovered by J Priestley in 1790, K4[Fe(CN)6] . Here, we first write the oxidation state of each molecule individually leaving only Fe.
Complete step by step answer:
The oxidation state is the number of electrons that are lost, gained, or shared by a certain atom with other atoms in the same molecule
Here, we will focus on all the fixed components and then work out the oxidation state of Fe.
The net charge on the molecule is 0. Therefore, all the oxidation states will add up to 0 after they are multiplied by the number of molecules.
(4×oxidation state of K)+(oxidation state of Fe)+(6×oxidation state of CN)=0
Let the oxidation state of Fe be x
We know that almost all elements from group 1 have an oxidation state of +1. This rule never changes. Thus, we will assume that the oxidation state of K is +1. And the oxidation state of CN is −1.
Now plugging in the numbers in the equation given above, we get,
[4×(+1)]+[x]+[6×(−1)]=0
⇒+4+x−6=0
⇒x−2=0
∴x=+2
So, the oxidation state of Fe is +2..
Thus, the oxidation state of Fe is +2 when the oxidation states of K and CN are +1 and −1respectively.
Note:
Always check what the net charge on the given ion or molecule is, this will change the number present on the R.H.S. of the equation.
For example: If we consider SO42− then to calculate the oxidation state of S, we need to consider the equation.
(oxidation state of S)+(4×oxidation state of O)=−2
Let the oxidation state of S be x
Then, [x]+[4×(−2)]=−2
⇒x+(−8)=−2
⇒x=+6
Here the oxidation state of S will be +6. If we do not pay attention to the net charge, the answer will turn out to be +8, which is not possible.