The correct answer is $\begin{array}{llll} A & B & C & D \end{array}$
The correct answer is
$\begin{array}{llll}
A & B & C & D
\end{array}$
III IV II V
II I III IV
V II III IV
II I V III
Solution
(a) $\mathrm{B}_2 \mathrm{H}_6$ is an electron deficient compound in Lewis terms, since two electrons are less available per bond and has 3-centre-2-electron bond in it.
(b) $\mathrm{CH}_4$ is an electron precise hybrid since in this the number of electrons present is equal to the number of electrons required.
(c) $\mathrm{PH}_3$ is an electron rich hybride in which the number of electrons present are more than the number of electrons required for making octet.
(d) $\mathrm{CaH}_2$ is a saline hybride, since its structure is salt like.
