Question

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

Answer 1

Hint: The angle between the two bonds originating from the same atom in a covalent species is known as the bond angle. In case of the molecules made up of three or more atoms, the bond angle is the average angle between the bonded orbitals (that is between the two covalent bonds).

Complete step by step solution:
-There are many factors which affect the bond angle in different ways and are given below-
(i) Repulsion present between the atoms or groups attached to the central atoms can increase or decrease the bond angle. Larger the repulsion, the greater will be the bond angle. More the attractions, lesser the bond angle.
(ii) When the hybridization is considered, the bond angle increases with the increase in the s-character of the s-hybrid bond. For example,
(a) A molecule with a central atom in $s{{p}^{3}}$has 25% s-character and thus has a lesser bond angle $109{}^\circ 28'$.
(b) A molecule with a central atom in $s{{p}^{2}}$have 33.33% s-character, and thus have a bond angle a bit greater than $s{{p}^{3}}$, that is $120{}^\circ $.
(c) A molecule with a central atom in sp has the maximum s-character as 50% and thus have a maximum bond angle as $180{}^\circ $
(iii) The bond angle decreases with the decrease in the electronegativity of the central atom. For example, the bond angle decreases as ${{H}_{2}}O>{{H}_{2}}S>{{H}_{2}}Se>{{H}_{2}}Te$ because the electronegativity of the central atom decreases. Oxygen is the most electronegative has the highest bond angle as $104.5{}^\circ $ followed by sulphur and selenium-containing compounds which have a bond angle as $92.2{}^\circ \text{ and }91.2{}^\circ $ respectively. The tellurium being the least electronegative has the least bond angle $89.5{}^\circ $.
In case, if the central metal atom remains the same, the bond angle increases with the decrease in the electronegativity of atoms surrounding the central atom. For example in $PC{{l}_{3}},PB{{r}_{3}},P{{I}_{3}}$ the bond angle of these compounds are $100{}^\circ ,101.5{}^\circ \text{ and 102}{}^\circ $ respectively.
-Following the knowledge, let us now consider the compounds given in the question.
-In methyl carbocation, the three bond pairs and the absence of lone pairs result in the hybridization of C atom as $s{{p}^{2}}$ and the bond angle as $120{}^\circ $.
-In the methane molecule, the C atom is $s{{p}^{3}}$ and bond angle is $109{}^\circ 28'$ .
-In methane carbanion, the presence of lone pair creates repulsion with the bond pair electrons and hence the bond angle is smaller than the tetrahedral angle.

So, the correct answer is option C.

Note: As of now, we read everything about the bond angle, its definition and influencing factors, but what is the significance of the bond angle? Determining the bond angle of a molecule helps in determining the structure of the molecule. The bond angle helps us to differentiate between the various geometries of compounds such as linear, trigonal planar, tetrahedral, trigonal bi-pyramidal or any other geometry.