Question

Determine the electron geometry (eg) and molecular geometry (mg) of $CH_3^+$.

a) eg = bent, mg = bent

b) eg = tetrahedral, mg = trigonal planar

c) eg = trigonal pyramidal, mg = trigonal pyramidal

d) eg = tetrahedral, mg = tetrahedral

e) eg = trigonal planar, mg = trigonal planar

Answer 1

Answer : The correct option is, (e) eg = trigonal planar, mg = trigonal planar

Explanation :

Formula used  :

$\text{Number of electron pair}=\frac{1}{2}[V+N-C+A]$

where,

V = number of valence electrons present in central atom

N = number of monovalent atoms bonded to central atom

C = charge of cation

A = charge of anion

The given molecule is, $CH_3^+$

$\text{Number of electron pair}=\frac{1}{2}\times [4+3-1]=3$

That means,

Bond pair = 3

Lone pair = 0

The number of electron pair are 3 that means the hybridization will be $sp^2$ and the electronic geometry of the molecule will be trigonal planar.

Hence, the electron geometry (eg) and molecular geometry (mg) of $CH_3^+$  is, trigonal planar and trigonal planar respectively.