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.