Question

Oxidation state of cobalt in $\left[Co\right(N{H}_3{)}_4\left(H_{2}O\right)Cl]{SO}_4$ ?

• A. $0$
• B. $+4$
• C. $-2$
• D. $+3$

Answer 1

Answer: (D)

$+3$

$\left[Co{\left(N{H}_3\right)}_4\left(H_{2}O\right)Cl\right]S{O}_4$

In the above compound,

Oxidation state of ammonia, $\left(N{H}_3\right)=0$

Oxidation state of water, $\left(H_{2}O\right)=0$

Oidation state of chlorine, $\left(Cl\right)=-1$

Oxidation state of sulphate ion, $\left(S{O}_4\right)=-2$

Let the oxidation state of cobalt, $\left(Co\right)$ be x.

As we know that sum of oxidation states of all atoms is equal to the overall charge on the compound.

$\therefore x+(4\times 0)+(1\times 0)+(-1)+(-2)=0$

$\Rightarrow x+0+0-1-2=0$

$\Rightarrow x-3=0$

$\Rightarrow x=+3$

Hence, the oxidation state of cobalt, $\left(Co\right)$ in the compound $\left[Co{\left(N{H}_3\right)}_4\left(H_{2}O\right)Cl\right]S{O}_4$ is +3.