Question

Oxidation state of cobalt in $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4\right({\mathrm{H}}_2\mathrm{O}\left)\mathrm{Cl}\right]{\mathrm{SO}}_4$ ?

• A. 0
• B. -4
• C. +2
• D. +3

Answer 1

The correct option is D

+3

Oxidation Number:

• The total number of electrons that an atom either gains or loses in order to form a chemical bond with another atom is known as the Oxidation Number.
• Generally, the Oxidation number is the charge present on the ions.

Explanation for Correct answer:

option(d) +3:

Let the oxidation number of $\mathrm{Co}$ be x.

Also. we know that the oxidation state of ${\mathrm{NH}}_3$, ${\mathrm{H}}_2\mathrm{O}$ , ${\mathrm{Cl}}^{-}$ and ${{\mathrm{SO}}_4}^{2-}$ are 0, 0, -1, and -2 respectively.

For $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4\right({\mathrm{H}}_2\mathrm{O}\left)\mathrm{Cl}\right]{\mathrm{SO}}_4$ ,

Oxidation state of $\mathrm{Co}$ + 4 x Oxidation state of ${\mathrm{NH}}_3$ + Oxidation state of ${\mathrm{H}}_2\mathrm{O}$ + Oxidation state of ${\mathrm{Cl}}^{-}$ + Oxidation state of ${{\mathrm{SO}}_4}^{2-}$= 0 ( As the charge on $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4\right({\mathrm{H}}_2\mathrm{O}\left)\mathrm{Cl}\right]{\mathrm{SO}}_4$ is 0)

x + 4(0) + 0 + (-1) + (-2) = 0

x = +3

Therefore, the oxidation state of $\mathrm{Co}$ in $\left[\mathrm{Co}{\left({\mathrm{NH}}_3\right)}_4\right({\mathrm{H}}_2\mathrm{O}\left)\mathrm{Cl}\right]{\mathrm{SO}}_4$ is +3.