Question

Oxidation state of silver in ${\text{N}}{{\text{a}}_{\text{2}}}\left[ {{\text{Ag}}{{\left( {{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}} \right)}_2}} \right]$ is:
A. $0$
B. $+ 1$
C. $+ 2$
D. $- 1$

Answer 1

Oxidation number is the number of electrons gained or lost by the atoms. The charge of an atom represents the oxidation number of that atom. Some atoms show fixed oxidation states so, by adding or subtracting the oxidation number of known and unknown we can calculate the oxidation number of the desired atom.

Complete answer
An atom loss or gain 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.

In neutral molecules, the sum of the oxidation number is equal to zero. 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. The alkali metals have the oxidation number$+ 1$. The transition metals show a variable oxidation number.

The oxidation number of silver in ${\text{N}}{{\text{a}}_{\text{2}}}\left[ {{\text{Ag}}{{\left( {{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}} \right)}_2}} \right]$ is as follows:
Sodium is an alkali metal so, the oxidation state of sodium is$+ 1$and the oxidation number of ${{\text{S}}_{\text{2}}}{\text{O}}_3^{2 - }$ligand is $- 2$.
$\Rightarrow \,\left( {x \times 1} \right)\, + \,\left( { - 2 \times 2} \right)\, = \, - 2$
$\Rightarrow \,x - 4\, = \, - 2$
$\Rightarrow \,x\, = \, - 2 + 4$
$\therefore \,x = \, + 2$

The oxidation number of silver is$+ 2$. So, the oxidation state of silver in ${\text{N}}{{\text{a}}_{\text{2}}}\left[ {{\text{Ag}}{{\left( {{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}} \right)}_2}} \right]$is$+ 2$.

Therefore, the correct answer is option (C) $+ 2$

Note: In a molecule, the more electronegative atom has a negative oxidation number, and less electronegative has a positive oxidation number. The atoms present out of the bracket are known as counter ions. Counter ions are used to balance the charge of complex only. Counter ions do not contribute to the oxidation state of the central metal atom. If we remove the two sodium from the given complex then the complex will get $- 2$ , so the sum of oxidation state of metal and ligand is placed equal to oxidation state of counter ion sodium.