Question

In the coordination compound $N{a_4}\left[ {Fe{{\left( {CN} \right)}_5}NOS} \right]$ oxidation state of Fe is :
A.$+ 1$
B.$+ 2$
C.$+ 3$
D.$+ 4$

Answer 1

To answer this question, you should recall the concept of oxidation numbers. The oxidation state of an atom is defined as the number of electrons lost and therefore, describes the extent of oxidation of the atom. For example, the oxidation state of carbon in ${\text{C}}{{\text{O}}_{\text{2}}}$ would be $+ 4$ since the hypothetical charge held by the carbon atom if both of the carbon-oxygen double bonds were completely ionic would be equal to $+ 4$.

Complete step by step answer:
The most frequent terms Oxidation state and oxidation number are terms frequently used interchangeably. They are defined and described as the number of electrons lost in an atom. We determine the oxidation number of iron in the given complex as $N{a_4}\left[ {Fe{{\left( {CN} \right)}_5}NOS} \right]$ dissociating in water to give $Fe{[{(CN)_5}NO]^{ - 4}}$ ion and $N{a^ + }$- ions. Charge on $NO$ = $+ 1$, ${H_2}O$ = Neutral ligands, Let the oxidation state of iron be ${\mathbf{X}}$
${\mathbf{X}} + {\mathbf{5}}\left( { - {\mathbf{1}}} \right) + {\mathbf{1}}{\text{ }} = - {\mathbf{2}}$
${\mathbf{X}}{\text{ }} - {\mathbf{5}} + {\mathbf{1}} = - 2$
Solving this:
${\mathbf{X}} = {\text{ }} + {\mathbf{2}}$
Hence, the oxidation number of iron in the given complex is $+ 2$.

So,the correct option is B.

Note:
In most of the compounds, the oxidation number of oxygens is $- 2$. There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $- 1$. Example, $N{a_2}{O_2}$
Superoxide- Every oxygen atom is allocated an oxidation number of $- \dfrac{{{\text{ }}1}}{2}$. Example, $K{O_2}$
Oxygen is bonded to fluorine- Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $+ 1$.