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Question: Hybridization in \({\text{S}}{{\text{O}}_2}\) molecule is: A. \({\text{sp}}\) B. \({\text{s}}{{\...

Hybridization in SO2{\text{S}}{{\text{O}}_2} molecule is:
A. sp{\text{sp}}
B. sp2{\text{s}}{{\text{p}}^2}
C. sp3{\text{s}}{{\text{p}}^3}
D. sp3d{\text{s}}{{\text{p}}^3}{\text{d}}

Explanation

Solution

Covalent compounds are formed when the electrons are shared to complete the octet. Sulfur and oxygen are non-metals. They have very high electronegativity. Thus they form a covalent bond. While an ionic bond is formed between metal and non-metal.

Complete step by step answer: There are four types of chemical bonding-ionic, covalent, hydrogen and metallic bonding. Covalent bonding is generally formed between two nonmetals. It is a type of chemical bond in which there is mutual sharing of electrons between two atoms. It is further classified into single, double and triple covalent bonds with respect to mutual sharing of one, two and three bonds respectively.
Consider SO2{\text{S}}{{\text{O}}_2} molecule in which sulfur has six electrons in a valence shell. So to form an octet, it is easy to gain two electrons than lose six electrons. Therefore its valency is two. Similarly, the valency of oxygen is two. The electronic configuration of sulfur is 1s22s22p63s23p41{{\text{s}}^2}2{{\text{s}}^2}2{{\text{p}}^6}3{s^2}3{{\text{p}}^4}. First two shells are completely filled. There are two electrons in 3s3{\text{s}} orbital four electrons in 3p3{\text{p}} orbital and it needs four unpaired electrons. One electron in 3px3{{\text{p}}_{\text{x}}} jumps to 3d3{\text{d}} orbital. This leads to formation of one unpaired electron in 3d3{\text{d}} orbital and three unpaired electrons in 3p3{\text{p}} orbital. Two 3p3{\text{p}} orbitals and one 3s3{\text{s}} orbital gets hybridized, thereby forming three sp2{\text{s}}{{\text{p}}^2} hybrid orbitals. Two hybrid orbitals form σ\sigma bonds with oxygen. While 3d3{\text{d}} and 3p3{\text{p}} orbitals are involved in the formation of π\pi bonds. Moreover, the hybridization of oxygen is also sp2{\text{s}}{{\text{p}}^2} . Therefore it forms sp2{\text{s}}{{\text{p}}^2} hybridization.
Hence the correct option is B.

Note: The molecular geometry of SO2{\text{S}}{{\text{O}}_2} is V-shaped or bent and the bond angle is 119{119^ \circ }. Two double bonds are there and a lone pair of electrons is there in one hybrid orbital, thereby forming bent shape. The structure of SO2{\text{S}}{{\text{O}}_2} molecule is given below: