Question

Find the number of molecule(s) having trigonal planar geometry from the following:
${\text{OCC}}{{\text{l}}_{2{\text{ }}}}$, ${\text{CH}}_3^ +$, ${\text{CH}}_3^ -$, ${\text{Cl}}{{\text{F}}_3}$, ${\text{I}}_3^ -$

Answer 1

Hint: As a general rule, molecules having ${\text{s}}{{\text{p}}^2}$ hybridization have trigonal planar geometry.

Complete step by step answer:
First we will see the electronic configurations of the given elements-
Carbon –${\text{1}}{{\text{s}}^2}{\text{ 2}}{{\text{s}}^2}{\text{ 2p}}_x^1{\text{ 2p}}_y^1{\text{ 2p}}$
Oxygen –${\text{1}}{{\text{s}}^2}{\text{ 2}}{{\text{s}}^2}{\text{ 2p}}_x^2{\text{ 2p}}_y^1{\text{ 2p}}_z^1$
Chlorine –${\text{1}}{{\text{s}}^2}{\text{ 2}}{{\text{s}}^2}{\text{ 2}}{{\text{p}}^6}{\text{ 3}}{{\text{s}}^2}{\text{ 3}}{{\text{p}}^5}$
Fluorine –${\text{1}}{{\text{s}}^2}{\text{ 2}}{{\text{s}}^2}{\text{ 2p}}$
Iodine - $[{\text{Kr}}]{\text{ 4}}{{\text{d}}^{10}}{\text{ 5}}{{\text{s}}^2}{\text{ 5}}{{\text{d}}^5}$
We will now look at the structure and hybridization of each of the given molecules.
${\text{CH}}_3^ +$
Carbon is the central atom with 4 valence electrons. Since, one electron is removed we have 3 valence electrons left. These are one s and two p orbitals. Therefore one 's' and two 'p' orbitals of carbon combine to overlap with 's' orbitals of 3 individual H atoms resulting the hybridization of ${\text{s}}{{\text{p}}^2}$. Hence, the shape of ${\text{CH}}_3^ +$molecule will be a trigonal planar.
${\text{CH}}_3^ -$
Carbon is the central atom with 4 valence electrons. Here, along with 3 hydrogen atoms one extra electron is also present. So, the central carbon atom undergoes ${\text{s}}{{\text{p}}^3}$hybridization. So, the shape of the molecule is distorted tetrahedral.
${\text{Cl}}{{\text{F}}_3}$
Chlorine is the central atom containing 7 valence electrons. One 3s, three 3p and one of the 3d orbitals of Cl participate in the hybridization and five ${\text{s}}{{\text{p}}^3}{\text{d}}$ hybrid orbitals are formed. Due to the two lone pairs the molecule has a T-shaped geometry.
${\text{I}}_3^ -$
Iodine is the central atom. In case of triiodide ions, the lone pairs there are 3 such pairs while the number of atoms donating valence electrons is 2. Adding these numbers, we get 3 + 2 =5. So, the hybridization will be ${\text{s}}{{\text{p}}^3}{\text{d}}$ and the shape of the molecule is linear.
${\text{OCC}}{{\text{l}}_{2{\text{ }}}}$
In molecules, s and p orbitals hybridize to form ${\text{s}}{{\text{p}}^2}$ hybrid orbitals. Hence, the shape of the molecule is trigonal planar.

Note: The different types of hybridization influence the bond strength and hence the geometry of the molecules. In a trigonal planar molecule, three molecules are attached to a central atom and are arranged in such a manner that repulsion between the electrons is minimal.