
Assertion: The sulfur in \[S{O_2}\] is \[s{p^2}\] hybridized.
Reason: \[S{O_2}\] has linear electron pair geometry.
A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
C. Assertion is correct but reason is not correct.
D. Assertion is not correct but Reason is correct
E. Both the assertion and reason are not correct
Answer
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Hint: In Sulphur dioxide, the central atom Sulphur is bonded to each oxygen atom with a double bond and a lone pair is present above the central atom Sulphur. During hybridization a total of three orbitals takes part.
Complete step by step answer:
In sulfur dioxide \[S{O_2}\], the central atom sulfur is bonded to two oxygen atoms. The structure of \[S{O_2}\] is represented by \[O = S = O\]. Here one sigma and one pi bond is present between the Sulphur and oxygen atom and one lone pair is localized at the central atom sulfur. The atomic weight of sulfur is 16, the electronic configuration of sulfur is \[[Ne]3{s^2}3{p^4}\]. The atomic weight of oxygen is 8, the electronic configuration of oxygen is \[[He]2{s^2}2{p^4}\]. The ground state of Sulphur contains 6 valence electrons where 4 electrons are present in 3p orbital and two electrons are present in 3s orbital. To form the four bonds with oxygen, sulfur needs four unpaired electrons. To generate four unpaired electrons, the electrons move to the excited state, where one electron from 3px jumps to the 3d orbital. As a result one unpaired electron is present in the 3d orbital and three electrons are present in 3p orbital. During hybridization, two 3p orbital and one 3s orbital take part which leads to total 3 orbital. The resulting hybridization will be \[s{p^2}\]. The electron geometry is trigonal planar.
Thus, assertion is correct but reason is not correct.
Therefore, the correct option is C.
Note: The hybridization of oxygen is Sulphur dioxide is also \[s{p^2}\]. The molecular geometry of sulfur dioxide is V-shaped or bent-shaped.
Complete step by step answer:
In sulfur dioxide \[S{O_2}\], the central atom sulfur is bonded to two oxygen atoms. The structure of \[S{O_2}\] is represented by \[O = S = O\]. Here one sigma and one pi bond is present between the Sulphur and oxygen atom and one lone pair is localized at the central atom sulfur. The atomic weight of sulfur is 16, the electronic configuration of sulfur is \[[Ne]3{s^2}3{p^4}\]. The atomic weight of oxygen is 8, the electronic configuration of oxygen is \[[He]2{s^2}2{p^4}\]. The ground state of Sulphur contains 6 valence electrons where 4 electrons are present in 3p orbital and two electrons are present in 3s orbital. To form the four bonds with oxygen, sulfur needs four unpaired electrons. To generate four unpaired electrons, the electrons move to the excited state, where one electron from 3px jumps to the 3d orbital. As a result one unpaired electron is present in the 3d orbital and three electrons are present in 3p orbital. During hybridization, two 3p orbital and one 3s orbital take part which leads to total 3 orbital. The resulting hybridization will be \[s{p^2}\]. The electron geometry is trigonal planar.
Thus, assertion is correct but reason is not correct.
Therefore, the correct option is C.
Note: The hybridization of oxygen is Sulphur dioxide is also \[s{p^2}\]. The molecular geometry of sulfur dioxide is V-shaped or bent-shaped.
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