where,
H= Number of orbitals involved in hybridization
V=Valence electrons of central atom
M- Number of monovalent atoms linked to central atom
C= Charge of cation A = Charge of anion Now consider,
$\mathrm{XeF}_{3}^{+}$
$\mathrm{H}=\frac{1}{2}[8+3-1]=5 \Rightarrow \mathrm{sp}^{3} \mathrm{d}$ hybridized state with 2 lone pairs.
Hence, it is T shaped.
Now consider,
$\mathrm{XeF}_{5}^{+}$
$\mathrm{H}=\frac{1}{2}[8+5-1]=6 \Rightarrow \mathrm{sp}^{3} \mathrm{d}^{2}$ hybridized state with 1 lone pair.
It is square pyramidal in shape. Hence, option B is the right answer.
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$(A)$ $\sigma$ orbital has a total of two nodal planes.
$(B)$ $\sigma^*$ orbital has one node in the $x z$-plane containing the molecular axis.
$(C)$ $\pi$ orbital has one node in the plane which is perpendicular to the molecular axis and goes through the center of the molecule.
$(D)$ $\pi^*$ orbital has one node in the $x y$-plane containing the molecular axis.
