Question

The correct sequence of the oxidation state of Fe, Ta, P, S and N elements in the given compounds are respectively:
${\text{N}}{{\text{a}}_{\text{2}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{5}}}{\text{NO}}} \right]$,${{\text{K}}_{\text{2}}}{\text{Ta}}{{\text{F}}_7}$,${\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$,${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_4}{{\text{O}}_6}$,${{\text{N}}_3}{\text{H}}$
A. $ + 3,\, + 5,\, + 5,\, + \,2.5,\,\, - \dfrac{1}{3}$
B. $ + 5,\, + 3,\, + 5,\, + \,3,\,\, + \dfrac{1}{3}$
C. $ + 3,\, + 3,\, + 5,\, + \,5,\,\, - \dfrac{1}{3}$
D. $ + 5,\, + 5,\, + 3,\, + \,2.5,\,\, + \dfrac{1}{3}$

Answer 1

Hint:The charge represents the oxidation state of an atom. It represents the number of electrons gained or lost by the atoms. In neutral molecules, the sum of the oxidation number is equal to zero.

Complete step by step solution:
An atom loses or gains electrons to form a bond with other atoms. So, the atoms get charged known as ions. The superscript of the ions represents the oxidation number of that atom.
The alkali metals have the oxidation number $ + 1$.
The alkaline earth metals have the oxidation number $ + 2$.
Halogens have oxidation number $ - 1$.
Oxygen has an oxidation number $ - 2$.
Hydrogen has an oxidation number $ + 1$.
The oxidation number of Fe in ${\text{N}}{{\text{a}}_{\text{2}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{5}}}{\text{NO}}} \right]$ is as follows:
The oxidation number of cyanide is $ - 1$ and nitronium ion is zero.
$\left( { + 1 \times 2} \right)\,\left( {x \times 1} \right) + \left( { - 1 \times 5} \right)\, + \,\left( {0 \times 1} \right)\, = \,0$
$\Rightarrow \,x = \, + 3$
The oxidation number of Fe is $ + 3$.
The oxidation number of Ta in ${{\text{K}}_{\text{2}}}{\text{Ta}}{{\text{F}}_7}$ is as follows:
$\left( { + 1 \times 2} \right) + \,\left( {x \times 1} \right)\, + \,\left( { - 1 \times 7} \right)\, = \,0$
$\Rightarrow \,x = \, + 5$
The oxidation number of Ta is$ + 5$.
The oxidation number of P in ${\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}$ is as follows:
$\left( { + 2 \times 2} \right) + \,\left( {x \times 2} \right)\, + \,\left( { - 2 \times 7} \right)\, = \,0$
$\Rightarrow \,x = \, + 5$
The oxidation number of P is $ + 5$.
The oxidation number of S in ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_4}{{\text{O}}_6}$ is as follows:
$\left( { + 1 \times 2} \right) + \,\left( {x \times 4} \right)\, + \,\left( { - 2 \times 6} \right)\, = \,0$
$\Rightarrow \,x = \, + 2.5$
The oxidation number of S is$ + 2.5$.
The oxidation number of N in ${{\text{N}}_3}{\text{H}}$ is as follows:
$\,\left( {x \times 3} \right)\, + \,\left( { + 1 \times 1} \right)\, = \,0$
$\Rightarrow \,x = - \dfrac{1}{3}$
The oxidation number of N is$ - \dfrac{1}{3}$.
So, correct sequence of the oxidation state of Fe, Ta, P, S and N elements in the given compounds are $ + 3,\, + 5,\, + 5,\, + \,2.5,\,\, - \dfrac{1}{3}$respectively.

Therefore, option (A) $ + 3,\, + 5,\, + 5,\, + \,2.5,\,\, - \dfrac{1}{3}$, is correct.

Note:
In charged molecules, the sum of the oxidation number is equal to the charge of the molecule. The atoms in elemental numbers have zero oxidation number. Oxygen in peroxides always has $ - 1$ oxidation number. In a molecule, the more electronegative atom has a negative oxidation number, and the less electronegative has a positive oxidation number. Hydrogen with non-metal shows $ - 1$ oxidation number.