Question

Oxidation number of Fe in $N{a_2}[Fe{(CN)_5}(NO)].2{H_2}O$____________.
A) -2
B) +1
C) +3
D) +5

Answer 1

Hint: Oxidation number of an atom is the charge that would exist on the atom if the bonding were completely ionic. It may be positive, negative or zero. We should have the knowledge of charge on ligands to calculate the oxidation number.

Complete answer:
In this coordination compound Fe has coordination number 6 because 6 ligands are attached.
Two types of ligands present are (CN) and (NO) which are called cyanido and nitroso.
Cyanido has -1 charge on it and Nitroso has 0 charge on it.
To calculate the oxidation no. first let us convert this into an ionic form.
${[Fe{(CN)_5}(NO)]^{2 - }}$
Now we will calculate the oxidation number. Let us suppose the oxidation number of Fe be x.
x - 5 = - 2
x = + 3
This is our required oxidation number.

So our correct option is C.

Additional information: Let us discuss spectrochemical series: This series is a list of ligands ordered on ligand strength and a list of metal ions based on oxidation number, group and its identity.
$I^−$ < $Br^−$ < $S^{2−}$ < $SCN^−$ < $Cl^−$ < $NO^{3−}$ < $N^{3−}$ < $F^−$ < $OH^−$ < $C_2O_4^{2−}$ < $H_2O$ < $NCS^−$ < $CH_3CN$ < $py$ < $NH$

Note: To solve this type of question we should have knowledge of charge on ligands. Also remember there are some neutral ligands which have 0 charge on it. Remember the water molecule attached to the compound does not affect the oxidation number of the metal.