Question

Oxidation state of Ag in $N{a_3}[Ag{({S_2}{O_3})_2}]$ is:
A. +2
B. -2
C. 0
D. +1

Answer 1

Hint: To solve this question, we need to understand the concept of oxidation state. As we know that the oxidation state indicates the degree of oxidation for an atom in a chemical compound. We know that the hypothetical charge that an atom would have if all bonds to atoms of different elements were completely ionic. We can represent oxidation states as integers, which can be positive, negative, or zero.

Complete step by step answer:
We can define oxidation number as the degree of oxidation of an atom in a chemical compound.
As we know that in coordination compounds, oxidation number is always calculated for the central metal.
Here in $N{a_3}[Ag{({S_2}{O_3})_2}]$, we know that $Ag$ is the central metal.
To find the charge on Ag we will find the oxidation number of other elements.
O. N. of $Na$= +1
O. N. of ${S_2}{O_3}$= -2
Let’s consider the O.N. of Ag as ‘x’.
We know that the summation of oxidation numbers is 0 as there is no net charge. And the charge on the complex ion $[Ag({S_2}{O_3})]$ is -3 because the total charge on a cation is +3.
Now we calculate the oxidation state of this compound as,
$x + \left( {2 \times \left( { - 2} \right)} \right) = - 3$
$\left( {x - 4} \right) = - 3$
On simplification we get,
$ \Rightarrow x = + 1$

So, the correct answer is Option D.

Note: When we talk about oxidation state and oxidation number we know that both quantities are the same for atoms in a molecule and can be used interchangeably. Mostly, it doesn't matter if the term oxidation state or oxidation number is used. As oxidation state refers to the degree of oxidation of an atom in a molecule.
We have to remember that in the oxidation process the electrons are lost from an atom. A compound that gains electrons during oxidation is known as oxidizing agent.
We have to remember that the reduction is the gain of electrons by an atom and a compound that loses electrons during reduction is called a reducing agent.