The trigonal bipyramidal geometry results from the hybridization:

Why is the bond angle in $CH_4$ not $90^{\circ}$,even though the outermost $p$-orbitals of carbon are at $90^{\circ}$ to each other?

The types of hybridisation on the five carbon atoms from left to right in $pent-1-en-4-yne$ are

Identify the correct set from the following table:

Molecule	Hybridisation of central atom	Shape
$(a)$ $PCl_5$	$dsp^3$	square pyramidal
$(b)$ $[Ni(CN)_4]^{2-}$	$sp^3$	tetrahedral
$(c)$ $SF_6$	$sp^3d^2$	octahedral
$(d)$ $IF_3$	$dsp^3$	pyramidal

$sp^3d^2$ hybrid orbitals result in which geometry?

Which among the following molecules have $sp^3d$ hybridisation with one lone pair of electrons on the central atom?
$(i)\ SF_4$
$(ii)\ [PCl_4]^+$
$(iii)\ XeO_2F_2$
$(iv)\ ClOF_3$