Question

Find the oxidation number Fe in $N{a_2}[Fe{(CN)_5}NO]$ ?

Answer 1

Hint: It is very important to understand the definition of oxidation number. To solve this question we have been aware of certain rules which are applied to calculate the oxidation number of a particular atom in a complex compound.

Complete step by step answer:
Oxidation number of an atom can be defined as the apparent charge on the atom in its compound.
There are certain rules that can be used to calculate the oxidation number of a particular atom:
1. The oxidation number of an element in its free state is zero.
Example: The oxidation number of ${H_2}$, $C{l_2}$,${P_4}$ are zero.
2. Oxidation number of an atom in a monatomic ion is equal to the number of positive or negative charges on the ion.
Example: Oxidation number of Al atoms in $A{l^{ + 3}}$ ion is +3. The Cl atom in $C{l^ - }$ ion is -1.
3. The oxidation number of H in a compound is assigned as +1, except in metallic hydrides where its oxidation number is -1.
Example: The oxidation number of hydride is -1 in $Ca{H_2}$.
4. The oxidation number of O in compounds is assigned as -2.
Exceptions:
- In peroxides (${H_2}{O_2}$,$N{a_2}{O_2}$) where the oxidation number of O is -1.
-In superoxides ($K{O_2}$) the oxidation number of O is $ - \dfrac{1}{2}$.
-The oxidation number of O in $O{F_2}$ is -1.
5.-The oxidation number halogen in halides or any compound is -1.
-The oxidation number of sulphur in sulphides is -2.
-The alkali metals that are the elements of group 1 possess oxidation number +1.
-The elements of group 2 that are alkaline metals possess oxidation number +2.
6. The algebraic sum of oxidation numbers of all atoms in a compound or ion is equal to the zero.
7. In coordination compounds,
- The oxidation number of neutral ligands like $NO$, $CO$, $N{H_3}$ and ${H_2}O$ is zero.
-The oxidation number of $C{H_3}$,${C_6}{H_5}$ are +1.
-The oxidation number of CN, OH, Cl are -1.
-Now, let’s apply these rules to calculate Fe in $N{a_2}[Fe{(CN)_5}NO]$:
Sodium (Na) belongs to group 1 so its oxidation number is 1 as mentioned in rule 5. The number of sodium atoms is 2 so, 1 $ \times $ 2 = 2. Therefore, the total oxidation number on $N{a_2}$ is 2.
-The oxidation number of CN is -1as mentioned in rule 7. Total CN atoms is 5 so, 1 $ \times $ 5=5 . Therefore, the total charge on ${(CN)_5}$ is -5.
-The oxidation number of NO is 0 as it is a neutral ligand (rule 7).
-Let us consider O.N be the short form of oxidation number.
-According to rule 6 , the sum of the oxidation number of all atoms in a compound is equal to 0.
-Let the oxidation number of Fe be ‘x’
O.N of $N{a_2}$ + O.N of ${(CN)_5}$ + O.N of Fe + O.N of NO = 0
2 + (-5) + X + 0 = 0
X = 3

Thus, the oxidation number of Fe in $N{a_2}[Fe{(CN)_5}NO]$ is +3.

Note: Carefully understand and remember the condition mentioned in rules. Learning to assign the oxidation number of any atom in a compound is very important. As the oxidation number of central metal is to be calculated firstly, with help which the complex formation and hybridization of a complex compound is understood.