In which of the following, \(\mathrm{CO}_{3}^{2-}, \mathrm{SO}_{2}, \mathrm{CCl}_{4}, \mathrm{CO}\) \(\mathrm{NO}_{2}^{-},\) would you expect to find \(s p^{2}\) hybridization of the central atom? Explain.

Short Answer

Expert verified
Only \(\mathrm{CCl}_{4}\) and \(\mathrm{NO}_{2}^{-}\) display \(sp^{2}\) hybridization of the central atom.

Step by step solution

01

Identify the Central Atom

The first step is to identify the central atom in each molecule. In \(\mathrm{CO}_{3}^{2-}\) the central atom is C, in \(\mathrm{SO}_{2}\) the central atom is S, in \(\mathrm{CCl}_{4}\) and \(\mathrm{CO}\) the central atom is C, and in \(\mathrm{NO}_{2}^{-}\) the central atom is N.
02

Determine the Hybridization

Next, determine the hybridization of each identified central atom. This can be done by using the formula for hybridization: \( H = 1/2 [V + X – C + A]\), where V = number of valence electrons of the central atom, X = number of monovalent atoms surrounding the central atom, C = charge of the cation, and A = charge of the anion.
03

Calculation for each molecule

Assess each molecule separately using the formula to find out which have \(sp^{2}\) hybridization. \(H = 1/2 [4 + 6 – 0 + 2] = 6\), which corresponds to \(sp^{3}\) for \(\mathrm{CO}_{3}^{2-}\). \(H = 1/2 [6 + 4 – 0 + 0] = 5\), which corresponds to \(sp^{3}\) for \(\mathrm{SO}_{2}\). \(H = 1/2 [4 + 4 – 0 + 0] = 4\), which corresponds to \(sp^{2}\) for \(\mathrm{CCl}_{4}\). \(H = 1/2 [4 + 2 – 0 + 0] = 3\), which corresponds to \(sp\) for \(\mathrm{CO}\). \(H = 1/2 [5 + 1 – 0 + 1] = 4\), which corresponds to \(sp^{2}\) for \(\mathrm{NO}_{2}^{-}\).
04

Result and Explanation

So, \(\mathrm{CCl}_{4}\) and \(\mathrm{NO}_{2}^{-}\) display \(sp^{2}\) hybridization of the central atom. This type of hybridization occurs when the central atom forms three sigma bonds, which results in a trigonal planar molecular geometry.

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