Question

Which of the following conversions involves change in both shape and hybridisation?
A. ${\text{C}}{{\text{H}}_{\text{4}}} \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$
B. ${\text{N}}{{\text{H}}_{\text{3}}} \to {\text{NH}}_{\text{4}}^{\text{ + }}$
C. ${{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$
D. ${\text{B}}{{\text{F}}_{\text{3}}} \to {\text{BF}}_{\text{4}}^{\text{ - }}$

Answer 1

Hint: In order to solve this question, you must recall the VSEPR (Valence shell electron pair repulsion) theory. It suggests that all valence shell electron pairs surrounding the central atom arrange themselves in such a manner so as to be as far away from each other as possible.

Complete step by step solution:
For the conversion, ${\text{C}}{{\text{H}}_{\text{4}}} \to {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$,
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 ${\text{C}}{{\text{H}}_{\text{4}}}$ and ${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}}$ have ${\text{s}}{{\text{p}}^3}$ hybridisation and tetrahedral shape.
For the conversion, ${\text{N}}{{\text{H}}_{\text{3}}} \to {\text{NH}}_{\text{4}}^{\text{ + }}$
Nitrogen in ${\text{N}}{{\text{H}}_{\text{3}}}$ has 3 bonds and one lone pair. It has ${\text{s}}{{\text{p}}^3}$ hybridisation, thus has a tetrahedral geometry but due to lone pair of electrons, its shape is pyramidal. On conversion to ${\text{NH}}_{\text{4}}^{\text{ + }}$ , a proton attaches to the lone pair of electrons. The hybridisation remains the same, that is ${\text{s}}{{\text{p}}^3}$ but the shape of the molecule becomes tetrahedral.
For the conversion, ${{\text{H}}_{\text{2}}}{\text{O}} \to {{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$,
Oxygen in ${{\text{H}}_2}{\text{O}}$forms two bonds with two hydrogen atoms and has two lone pairs of electrons. Thus, it has a hybridisation of ${\text{s}}{{\text{p}}^3}$, a tetrahedral geometry but a bent shape. On conversion to ${{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}$, a proton attaches to the lone pair of electrons. The hybridisation remains the same, that is ${\text{s}}{{\text{p}}^3}$ but the shape of the molecule now becomes pyramidal.
For the conversion, ${\text{B}}{{\text{F}}_{\text{3}}} \to {\text{BF}}_{\text{4}}^{\text{ - }}$
Boron in ${\text{B}}{{\text{F}}_{\text{3}}}$ forms three bonds with three fluorine atoms and has no lone pair of electrons. Thus, it has the hybridisation ${\text{s}}{{\text{p}}^2}$ and trigonal planar shape. On conversion to ${\text{BF}}_{\text{4}}^{\text{ - }}$, a fluoride ion puts its lone pair in the empty orbital of boron atom changing the hybridisation to ${\text{s}}{{\text{p}}^3}$ and shape to tetrahedral.

Thus, the correct option is D.

Note: During bond formation, the atomic orbitals of an atom are mixed in such a manner as to produce equivalent orbitals. This mixing of orbitals is known as hybridisation. The arrangement of these hybrid orbitals according to the VSEPR theory gives us the shape of the molecule.