# What is the oxidation number of Sn in the compound Na_2SnO_2?

May 29, 2016

$+ 2$

#### Explanation:

To find the oxidation number of $S n$ in $N {a}_{2} S n {O}_{2}$ we should know that:

The oxidation number for $N a$ is always $+ 1$.

The oxidation number of $O$ is always $- 2$ except in ${H}_{2} {O}_{2}$ it is $- 1$ and in $O {F}_{2}$ it is $+ 2$.

Let the oxidation number of $S n$ be $x$, therefore,

$2 \times \left(+ 1\right) + x + 2 \times \left(- 2\right) = 0$

$\implies 2 + x - 4 = 0$

$\implies x - 2 = 0$

$\implies x = + 2$

Thus, the oxidation number of $S n$ in $N {a}_{2} S n {O}_{2}$ is $+ 2$.