- A$s{p^2}$
- ✓$s{p^3}d$
- C$s{p^3}$
- D$s{p^3}{d^2}$
Hybridisation of a atom with $3$ lone pairs and $2$ bond pairs is $sp ^3 d ^2$.
So, the hybridisation of $Xe$ is $sp ^3 d$.
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$\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_3}(excess) + B{r_2}\xrightarrow{{hv}}}
\end{array}$ $\mathop {\begin{array}{*{20}{c}}
{\,\,\,\,\,C{H_3}} \\
| \\
{C{H_3} - C - C{H_3}} \\
| \\
{\,\,Br}
\end{array}}\limits_{(A)} $ + $\mathop {\begin{array}{*{20}{c}}
{C{H_3}\,\,\,\,\,\,\,\,\,\,} \\
{\,\,|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{C{H_3} - CH - C{H_2} - Br}
\end{array}}\limits_{(B)} $
the percentage yields of the products $(A)$ and $(B)$ are expected to be
