Question

Which of the following are isolobal?
A.$B{H_4}^ - $ and $N{H_4}^ + $
B.$C{O_2}$ and $S{O_2}$
C.${N_2}$ and $NO$
D.All of these

Answer 1

Hint:According to Roald Hoffmann, those molecular fragments are called isolobal in which the number, symmetry properties, approximate energy and shape of the frontier orbitals and the number of electrons in them are similar, but not identical.

Complete step by step answer:
Let us first consider the bonding and the structure of the above molecules.
In $B{H_4}^ - $, the hybridization of the central atom boron is $s{p^3}$ . It has a tetrahedral structure with the four hydrogen atoms pointed in the four corners of the tetrahedron. In case of $N{H_4}^ + $, the hybridization of the central is also $s{p^3}$. Hence, here also the four hydrogen atoms should occupy the four corners of a tetrahedron. So, these are isolobal.
Coming to $C{O_2}$ and $S{O_2}$, the hybridization of the central atom in $C{O_2}$ is $sp$ hence, the shape is linear. But in $S{O_2}$, the hybridization of sulphur is $s{p^2}$. So the structure should be triangular, but as per the VSEPR theory, the structure is angular. Hence, these two are not isolobal.
Coming to ${N_2}$ and $NO$, the hybridization of nitrogen in nitrogen molecules is $sp$, so it is a linear structure. In the case of $NO$, the hybridization of nitrogen is $2.5$ and that of oxygen is three. Hence, the structure is linear. But there is a lone electron on nitrogen which makes it different from ${N_2}$.

Hence, the isolobal pair is $B{H_4}^ - $ and $N{H_4}^ + $. So, the correct answer is A.

Note:
The isolobal analogy or the isolobal principle is a strategy used in organometallic chemistry to relate the structure of organic and inorganic molecular fragments to predict bonding properties of organometallic compounds.