The oxidation state of $\mathrm{Xe}$ in $\mathrm{XeO}_3$ and the bond angle in it respectively are:
The oxidation state of $\mathrm{Xe}$ in $\mathrm{XeO}_3$ and the bond angle in it respectively are:
$+6,109^{\circ}$
$+8,103^{\circ}$
$+6,103^{\circ}$
$+8,120^{\circ}$
Solution
The oxidation state of $\mathrm{Xe}$ in $\mathrm{XeO}_3$ can be calculate as
$\mathrm{XeO}_3, x+(-2 \times 3)=0$
$x=+6$
$\mathrm{XeO}_3$ has $s p^3$-hybridisation with bond angle $=103^{\circ}$