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}=(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}=(1)/(2)* [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.