Question

Which of the following has trigonal pyramidal geometry?
A.$\text{B}{\text{H}}_3$
B.${\text{H}}_2\text{O}$
C.$\text{C}{\text{H}}_4$
D.$$\text{N}{H}_3$$
E.$\text{AlC}{\text{l}}_3$

Answer 1

Hint:The geometry of a molecule or an ionic species depends on the electronic confirmation of the central atom of the molecule or the ion obeys the VSEPR theory and the geometry of the molecules is decided through hybridization of the central atom.
Formula used:
 Hybridization, H = ${\displaystyle \frac{\text{Z + M - C + A}}{\text{2}}}$ where, $\text{Z}$ is the no. of electrons in the valence shell, $\text{M}$ is the number of monovalent groups, $\text{C}$ is the cationic charge, and $\text{A}$ is the anionic charge.

Complete step by step answer:
The hybridization of the central atom for the above molecules is as follows:
In boron hydride or $\text{B}{\text{H}}_3$, the central atom boron has 3 electrons in the valence shell, there are three monovalent groups, therefore the hybridization is:
$\Rightarrow$${\displaystyle \frac{\text{3 + 3 - 0 + 0}}{\text{2}}}={\displaystyle \frac{6}{2}}=3$, $$\text{s}{\text{p}}^2$$, and the structure is trigonal.
In water or ${\text{H}}_2\text{O}$, the central atom oxygen has 6 electrons in the valence shell, there are two monovalent groups, therefore the hybridization is:
$\Rightarrow$${\displaystyle \frac{\text{6 + 2 - 0 + 0}}{\text{2}}}={\displaystyle \frac{8}{2}}=4$, $$\text{s}{\text{p}}^3$$, and the structure should be trigonal pyramidal. But from the VSEPR, as the lone pair-lone pair repulsion is maximum, due to the two lone pairs on oxygen, the shape becomes angular.
In methane or $\text{C}{\text{H}}_4$ , the central atom carbon has 4 electrons in the valence shell, there are four monovalent groups, therefore the hybridization is:
$\Rightarrow$${\displaystyle \frac{\text{4 + 4 - 0 + 0}}{\text{2}}}={\displaystyle \frac{8}{2}}=4$, $$\text{s}{\text{p}}^3$$, and the structure is tetrahedral.
In ammonia or $$\text{N}{H}_3$$, the central atom nitrogen has 5 electrons in the valence shell, there are three monovalent groups, therefore the hybridization is:
$\Rightarrow$${\displaystyle \frac{\text{5 + 3 - 0 + 0}}{\text{2}}}={\displaystyle \frac{8}{2}}=4$, $$\text{s}{\text{p}}^3$$, and the structure is trigonal pyramidal, the lone pair of nitrogen occupies the axial position.
In aluminium trichloride or$\text{AlC}{\text{l}}_3$, the central atom aluminum has 3 electrons in the valence shell, there are three monovalent groups, therefore the hybridization is:
$\Rightarrow$${\displaystyle \frac{\text{3 + 3 - 0 + 0}}{\text{2}}}={\displaystyle \frac{6}{2}}=3$, $$\text{s}{\text{p}}^2$$, and the structure is triangular.

Hence, the correct option is option D.

Note:
The VSEPR theory states the shape of a molecule will be such that there is minimum repulsion between the bond pairs and the lone pairs on the central atom.The hybridization along with the VSEPR theory both decide the shape of the molecules.