Question

The oxidation state of $Xe$ in $Xe{O}_3$ and bond angle in it are:

• A. $+6,{109}^o$
• B. $+8,{103}^o$
• C. $+6,{103}^o$
• D. $+8,{120}^o$

Answer 1

Answer: (A)

$+6,{109}^o$

In $Xe{O}_3$, oxidation state of oxygen is $-2$.

So, $x+3(-2)=0$

$\Rightarrow x-6=0$

$\Rightarrow x=6$

where, $x$ is oxidation state of $Xe$.

$O$

$O=\stackrel{||}{Xe}=O$

Based on structure, it has a tetrahedral shape with bond angle nearly ${109}^o$ .