Question

Here, $Al{{H}_{3}}$ is:
(A)- pyramidal
(B)- trigonal planar
(C)- linear
(D)- tetrahedral

Answer 1

Hint: The geometry of the molecule is derived from the orientation of the hybrid orbitals, in order to minimize the repulsion between the adjacent orbitals, leading to ease in overlapping with the orbitals of the atoms to be bonded. Thus, forming a stable covalent bond.

Complete answer:
The shape of aluminium trihydride can be obtained by determining the hybridisation of the molecule, using the VSEPR theory. In the aluminium atom, with configuration $\left[ Ne \right]3{{s}^{2}}3{{p}^{1}}$ and having three valence electrons.
In order to bond with three hydrogen atoms, the aluminium atom in its excited state, one electron from the 3s-orbitals jumps to the 3p-orbital. Thus, having three unpaired electrons. It undergoes hybridisation of the one 3s and the two 3p-orbitals to form three $s{{p}^{2}}$ hybrid orbitals with one electron each and have similar energy. The $3{{p}_{z}}$ orbital is empty and is perpendicular to the $s{{p}^{2}}$hybrid orbitals.
These hybrid orbitals align themselves in the corners of the triangle (at ${{120}^{\circ }}$ angle) to minimise the repulsion, thus, forming the trigonal geometry.
                     

Then, the three $s{{p}^{2}}$orbitals overlap with the 1s orbital of the three hydrogen atoms, in the trigonal planar structure.

Therefore, the aluminium hydride has option (B)- trigonal planar geometry.

Note:
The octet of the aluminium atom is incomplete, having only six electrons. This is due to the empty $3{{p}_{z}}$ orbital. Hence, the stable hydride is an electron-deficient molecule and a good Lewis acid.
The hydride and its derivatives are often used in the reduction of carbonyl compounds like aldehyde, ketones, esters etc, to alcohols.