What is the oxidation number of $S n$ $Sn$ in the compound $N a_{2} S n O_{2}$ $Na_2SnO_2$ ?

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Answer 1 · 2016-05-29T12:14:46

$+ 2$ $+2$

To find the oxidation number of $S n$ $Sn$ in $N a_{2} S n O_{2}$ $Na_2SnO_2$ we should know that:

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

The oxidation number of $O$ $O$ is always $- 2$ $-2$ except in $H_{2} O_{2}$ $H_2O_2$ it is $- 1$ $-1$ and in $O F_{2}$ $OF_2$ it is $+ 2$ $+2$ .

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

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

$\Rightarrow 2 + x - 4 = 0$ $=>2+x-4=0$

$\Rightarrow x - 2 = 0$ $=>x-2=0$

$\Rightarrow x = + 2$ $=>x=+2$

Thus, the oxidation number of $S n$ $Sn$ in $N a_{2} S n O_{2}$ $Na_2SnO_2$ is $+ 2$ $+2$ .

What is the oxidation number of SnSnSn in the compound Na_2SnO_2Na2SnO2Na_2SnO_2?