Question

In carbonium ion, the carbon bearing the positive charge is:
(A) $sp$ hybridised
(B) $s{p^2}$ hybridised
(C) $s{p^3}$ hybridise
(D) unhybridized

Answer 1

Hint: Carbonium ions are the cations which contain the positive charge on the carbon atom with 6 electrons in the valence shell. So, fill these 6 electrons in the orbitals according to Aufbau principle and then you will get the hybridisation of the carbon in carbonium ion.

Complete step by step solution:
Carbonium ion is an organic cation having a pentavalent carbon atom. The simplest carbonium ion is methanium ion (CH5+), which has a carbon atom with +1 charge and five hydrogen atoms. Moreover, there are five C-H bonds (usually carbon forms 4 bonds with its 4 electrons in the valence shell) and When carbon forms 5 bonds, it has to lose its 1 electron and thus will have a +1 charge on it.
Every bond pair shares 2 electrons with the bonding atom and lone pairs consist of two unpaired electrons. Carbon in carbonium ion has 3 electrons in the valence shell and thus it can form only 3 sigma bonds with the bonding atoms. Therefore, it needs three hybridised orbitals.
$sp$ hybridisation: It involves the mixing of one s and one p-orbital, resulting in the formation of two equivalent $sp$ hybrid orbitals.
$s{p^2}$ hybridisation: in this hybridisation, there is the involvement of one s and two p-orbitals in order to form three hybridised orbitals.
$s{p^3}$ hybridisation: in this, there is the mixing of one s orbital and three p orbitals resulting in the four $s{p^3}$ hybridised orbitals of equivalent energies.
And we know, carbonium ion needs 3 hybridised orbitals, so it would be $s{p^2}$hybridised.

Thus the correct option is B.

Note: Hybridisation is always of the central atom and the type of hybridisation indicates the geometry of the molecules. Carbonium ion is $s{p^2}$ hybridised. One s and two p orbitals are oriented in a trigonal planar arrangement. Therefore, the geometry of the carbonium ion is trigonal planar with a bond angle of ${120^{\text{o}}}$.