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The oxidation state of $\mathrm{Xe}$ in $\mathrm{XeO}_3$ and the bond angle in it respectively are:
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The correct answer is:
$+6,103^{\circ}$
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 $=103^{\circ}$
$\mathrm{XeO}_3, x+(-2 \times 3)=0$
$x=+6$
$\mathrm{XeO}_3$ has $s p^3$-hybridisation with bond angle $=103^{\circ}$
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