Question

The oxidation state of \[Fe\] in \[Fe{\left( {CO} \right)_{5\;}}\] is
A. $0$
B. $ + 2$
C. $ - 2$
D. $ + 6$

Answer 1

Hint: let's consider, the oxidation number of Fe is x. We know that CO is a neutral ligand and the neutral ligand does not have any charge. substitute the value of Fe and CO in the equation to calculate the oxidation state $Fe(CO)_{5}$

Complete step by step answer: In the given question we have to find the oxidation state of the iron in the complex.
To give the answer for the given question we have to conclude some steps and facts along :
-\[Fe{\left( {CO} \right)_{5\;}}\] is a neutral compound as it carries no charge. Therefore we can say that the total oxidation state of \[Fe{\left( {CO} \right)_{5\;}}\] is $0$ .
- Apart from that let's just consider the oxidation number of \[Fe\] theoretically to be $x$.
-\[CO\] is a neutral ligand as it carries no charge. That is the reason why the oxidation state of $CO$ ligand is $0$ in that state.
iv. Now we just have to put the above value in the equation and after then let's try to find out the oxidation state of \[Fe\] .
Equation would be :
\[{\text{x }} + {\text{ }}0\left( 5 \right){\text{ }} = {\text{ }}0\]
The value of $x$ will also become zero and thus the oxidation state of $Fe$ is equal to zero or will turn to it.
Hence, the correct option will be A, $0$.

Note: Fe has its oxidation state to be $0$ . This is because the ligand \[\left[ {CO} \right]\] also falls under the category of the neutral monodentate ligand. Since it is neutral, it has no overall charge i.e. $0$ , and also the compound itself is neutral. So,
We can conclude by saying that the oxidation state for both Fe is $0$.