Question

The oxidation number of underlined element in $K[\underset{–}{Co}{\left(C_2{O}_4\right)}_2.{\left(N{H}_3\right)}_2]$ compound is :

• A. $+3$
• B. $+2$
• C. $+1$
• D. $0$

Answer 1

Answer: (A)

$+3$

Let $x$ be the oxidation number of Co in $K[\underset{–}{Co}{\left(C_2{O}_4\right)}_2.{\left(N{H}_3\right)}_2]$.
The overall charge on the complex is $0$. Therefore, sum of oxidation state sof all elements should be equal to zero.
Charge on $K$ is $+1$.
Oxidation state of $C_2{O}_4$ is $-2$.
Oxidation state of $N{H}_3$ is $0$.
Therefore, $+1+x+2\times (-2)+0=0$
or, $x=+3$

Hence, the correct option is A.