Question

What is the oxidation number of $Cr$ in ${(N{H_3})_3}Cr{O_4}$or ${(N{H_3})_3}Cr{({O_2})_2}$?

Answer 1

Oxidation Number of an element or compound will depend on the number of electrons that are gained or lost by an atom that is present in a compound. Oxidation number is also known as the Oxidation state of a compound. We can determine the Oxidation number of the element by following certain rules.

Complete step by step solution:
Let us consider of the compound ${(N{H_3})_3}Cr{O_4}$ or ${(N{H_3})_3}Cr{({O_2})_2}$
The structure of ${(N{H_3})_3}Cr{O_4}$ or ${(N{H_3})_3}Cr{({O_2})_2}$ is a pentagonal bipyramidal. The structure is given below.

Let us now calculate the oxidation number of $Cr$ in ${(N{H_3})_3}Cr{O_4}$ or ${(N{H_3})_3}Cr{({O_2})_2}$.
Consider the oxidation number of Chromium as $x$
The total oxidation number of the compound is 0.
The oxidation number of the $N{H_3}$ group is 0.
Therefore, $3 \times 0 = 0$
The oxidation number of Oxygen is -1. Because the oxygen present in ${(N{H_3})_3}Cr{O_4}$ is in the peroxide group.
Therefore,
$3 \times 0 + x + 4 \times - 1 = 0$
$x - 4 = 0$
$x = 4$

Therefore, the oxidation number of Cr in ${(N{H_3})_3}Cr{O_4}$ or ${(N{H_3})_3}Cr{({O_2})_2}$ is +4.

Additional information:
We have to follow certain rules to assign the oxidation number to certain elements.
- Atoms that are in the free elemental state i.e. $({H_2},{O_2})$ will be 0.
- Monatomic ions $N{a^ + } = 1$
- Halogen like $C{l^ - } = - 1$
- Oxygen =-2
- Hydrogen =1
- Metal hydrides -1.

Note: We must always remember that the oxidation state of the oxygen atom in peroxide will always be -1, as the oxygen atom is linked to only the chromium atom through a single bond. The Oxygen which does not belong to the peroxide molecule will be having the oxidation number of -2.