Question

Identify the correct order of bond angle in the following species $CH_{3}^{+},\text{ }C{{H}_{4}},\text{ }CH_{3}^{-}$
(A) $CH_{3}^{-}>C{{H}_{4}}>CH_{3}^{+}$
(B) $C{{H}_{4}}>CH_{3}^{-}>CH_{3}^{+}$
(C) $CH_{3}^{+}>C{{H}_{4}}>CH_{3}^{-}$
(D) $CH_{3}^{+}>CH_{3}^{-}>CH$

Answer 1

We know that the electronegativity difference between the bonded atoms decides the existence of electrons on the respective atoms. The closer the electrons exist around the central atom the more is the repulsion between them. Here we have given hydrocarbons so that with the help of hybridization we have to find their bond order.

Complete step by step solution:
If the molecule does not have any lone pair then the structure of the molecule is symmetric. Due to the symmetric structure, the bond angle remains the same. If the molecule has lone pairs, the bond pairs will experience repulsion hence the bonds will come closer.
It has a total of five electrons in the outermost shell with two electrons in and three electrons in the shell. Out of the five electrons three are bonded to three halide atoms and two electrons remain non-bonded. The central atom phosphorus is hybridized.
In $CH_{3}^{+},$ three bond pairs and zero lone pairs are present which result in $s{{p}^{2}}$ hybridization of $C$ atom with bond angle of ${{120}^{0}}.$ In methane, $s{{p}^{3}}$ hybridization results in tetrahedral geometry with a bond angle of ${{109}^{0}}.$ In $CH_{3}^{-},$ the lone pair of electrons repels the bond pair of electrons. Hence, the bond angle is smaller than the tetrahedral angle.
So, the Correct option is (C).

Note:
Remember that the hybridisation concept has to be made through for assigning the bond angles which is based on geometry and make sure that you know which atoms possess lone pairs and which do not because this plays an important role in deciding the geometry of molecules.