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Ainat [17]
2 years ago
15

Determine the electron geometry (eg) and molecular geometry (mg) of

1" title="CH_3^+" alt="CH_3^+" align="absmiddle" class="latex-formula">.
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
Chemistry
1 answer:
Yuri [45]2 years ago
7 0

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.

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