A molecule in which \(s p^{2}\) hybrid orbitals are used by the central atom in forming covalent bonds is (a) \(\mathrm{PCl}_{5}\) (b) \(\mathrm{N}_{2} ;\) (c) \(\mathrm{SO}_{2} ;\) (d) \(\mathrm{He}_{2}\)

Short Answer

Expert verified
The molecule in which \(sp^{2}\) hybrid orbitals are used by the central atom in forming covalent bonds is \(\mathrm{SO}_{2}\).

Step by step solution

01

Structure of PCl5

The central atom in \(\mathrm{PCl}_{5}\) is Phosphorus (P). It has 5 valence electrons, and it forms 5 sigma bonds with 5 Chlorine (Cl) atoms. There are no lone pairs on Phosphorus. So, the hybridization is \(sp^{3}d\), not \(sp^{2}\).
02

Structure of N2

The molecule \(\mathrm{N}_{2}\) is composed of two Nitrogen (N) atoms. Each Nitrogen atom forms one sigma bond with the other Nitrogen atom and two pi bonds. So, no \(sp^{2}\) hybridization here. The hybridization is \(sp\).
03

Structure of SO2

The central atom in \(\mathrm{SO}_{2}\) is Sulphur (S). It forms two sigma bonds with two Oxygen (O) atoms and has one lone pair. Hence, it's hybridization is \(sp^{2}\). Therefore, \(\mathrm{SO}_{2}\) is the molecule in which \(sp^{2}\) hybrid orbitals are used by the central atom in forming covalent bonds.
04

Structure of He2

The molecule \(\mathrm{He}_{2}\) does not exist because Helium (He) is a noble gas and it does not form covalent bonds with other atoms. So, it doesn't undergo hybridization.

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