- ✓$d_{x^2-y^2}$
- B$d_{z^2}$
- C$d_{xy}$
- D$d_{xz}$
When $d_{z^{2}}$ orbital is involved in the hybridization, the geometry is trigonal bipyramidal.
When $d_{x^{2}-y^{2}}$ orbital is involved in the hybridization, the geometry is square pyramidal.
Hence the correct option is A.
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(Nearest integer)
$\left(\mathrm{h}=6.63 \times 10^{-34}\, \mathrm{Js}, \mathrm{c}=3.00 \times 10^{8} \,\mathrm{~ms}^{-1}\right)$
$\begin{array}{*{20}{c}}
{BrC{H_2} - CH - CO - C{H_2} - C{H_2}C{H_3}} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{CON{H_2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$
is :
$(I)$ $S_2O_4^{2-}$ $(II)$ $S_2O_5^{2-}$ $(III)$ $S_2O_6^{2-}$
