One of the following conversion results in the change of hybridization and geometry :

For the conversion, \(\mathrm{CH}_{4} \rightarrow \mathrm{C}_{2} \mathrm{H}_{6}\),
There is no change in the hybridisation of the carbon atom. Each carbon loses an atom of hydrogen and the two carbon hydrogen bonds are replaced by a carbon-carbon bond.
Both \(\mathrm{CH}_{4}\) and \(\mathrm{C}_{2} \mathrm{H}_{6}\) have \(\mathrm{sp}^{3}\) hybridisation and tetrahedral shape.
For the conversion, \(\mathrm{NH}_{3} \rightarrow \mathrm{NH}_{4}^{+}\)
Nitrogen in \(\mathrm{NH}_{3}\) has 3 bonds and one lone pair. It has \(\mathrm{sp}^{3}\) hybridisation, thus has a tetrahedral geometry but due to lone pair of electrons, its shape is pyramidal. On conversion to \(\mathrm{NH}_{4}^{+}\), a proton attaches to the lone pair of electrons. The hybridisation remains the same, that is \(\mathrm{pp}^{3}\) but the shape of the molecule becomes tetrahedral.
For the conversion, \(\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{H}_{3} \mathrm{O}^{+}\),
Oxygen in \(\mathrm{H}_{2}\) Oforms two bonds with two hydrogen atoms and has two lone pairs of electrons.
Thus, it has a hybridisation of \({s p^{3}}^{3}\), a tetrahedral geometry but a bent shape. On conversion to
\(\mathrm{H}_{3} \mathrm{O}^{+}\), a proton attaches to the lone pair of electrons. The hybridisation remains the same,
that is \(\operatorname{sp}^{3}\) but the shape of the molecule now becomes pyramidal.
For the conversion, \(\mathrm{BF}_{3} \rightarrow \mathrm{BF}_{4}^{-}\)
Boron in \(\mathrm{BF}_{3}\) forms three bonds with three fluorine atoms and has no lone pair of electrons.
Thus, it has the hybridisation \(\mathbf{s p}^{2}\) and trigonal planar shape. On conversion to \(\mathrm{BF}_{4}^{-}\), a fluoride
ion puts its lone pair in the empty orbital of boron atom changing the hybridisation to \(\mathrm{sp}^{3}\) and shape to tetrahedral.