Question

The oxidation number of $Cr$ in $K_{2}C{r}_2{O}_7$ is :

• A. $+3$
• B. $-3$
• C. $+6$
• D. $-6$

Answer 1

The correct option is C $+6$

The sum of the oxidation numbers in a polyatomic ion is equal to the charge on the ion. The oxidation number of simple ions is equal to the charge on the ion. The metals in Group IA $\left(K\right)$ atom has an oxidation number of $+1$. Oxygen usually has an oxidation number of $-2$.

Oxidation no of $Cr$ in $K_{2}C{r}_2{O}_7$
$=2\times 1+2\times x+7\times (-2)=0$
$x=\frac{14-2}{2}=+6$