Q30 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Chemistry

The hybridisation of $\mathrm{Xe}$ in $\mathrm{XeO}_3$ is

  • Correct answer
    A
    $s p^3$
  • B
    $s p^3 d$
  • C
    $s p^3 d^3$
  • D
    $d s p^2$

Solution

Figure
$\begin{aligned} & H=\frac{1}{2}(V+M-C+A)=\frac{1}{2}(8+0-0+0)=4\left(s p^3\right) \\ & {[\because V=\text { number of valence electrons of } \mathrm{Xe}=8} \\ & M=\text { number of monovalent atoms }=0 \\ & C=\text { number of cationic charge }=0 \\ & A=\text { number of anionic charge }=0]\end{aligned}$

Hence, the correct answer is option: 'A'.

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