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Question: In \(SO_3^{2 - }\) : A.\(d\pi - p\pi \) bond between \(S\) and \(O\) is delocalized. B.Bonds bet...

In SO32SO_3^{2 - } :
A.dπpπd\pi - p\pi bond between SS and OO is delocalized.
B.Bonds between SS and OO are equivalent.
C.There is sp3s{p^3} hybridized sulphur atom
D.All of the facts given above are true.

Explanation

Solution

SO32SO_3^{2 - } known as sulphite is a sulphur oxoanion. They are naturally occurring substances present in some foods. The oxidation state of sulphur in sulphite is +4 + 4 . Sulphite also shows resonance effect.

Complete step by step answer:
The Lewis structure of SO32SO_3^{2 - } is as follows:

There is one double bond, two single bonds and one lone pair on the central atom that is sulphur.
The molecular structure of sulfite anion is trigonal pyramidal, with bond angle of 109.5{109.5^ \circ } .
A.dπpπd\pi - p\pi bond between SS and OO is delocalized.
Reason: The d-orbital of sulphur overlaps with the p-orbital of oxygen in order to form dπpπd\pi - p\pi bond.
Therefore the above statement is true.
B.the bonds between sulphur and oxygen are same:
Reason: The bonds between sulphur and oxygen are the same because of the resonance effect.
Resonance effect is the shifting of double bonds.
The resonance effect of sulphite is given as follows in the diagram:
C.There is sp3s{p^3} hybridized sulphur atom.
In order to check the hybridization of atom we will use the following formula:
Total number of attached atoms ++ number of lone pairs .
For spsp hybridization: Total number of attached atoms ++ number of lone pair =2 = 2
For sp2s{p^2} hybridization: Total number of attached atoms ++ number of lone pair =3 = 3
For sp3s{p^3} hybridization: Total number of attached atoms ++ number of lone pair =4 = 4
Now we will calculate the hybridization sulphur in SO32SO_3^{2 - } :
Total number of attached atoms present in sulphur is 33 .
And the number of lone pairs is 11.
\Rightarrow Total number of attached atoms to sulphur ++ number of lone pair on sulphur
Substituting the values we get,
=3+1= 3 + 1
=4= 4
Therefore the hybridization of sulphur in sulphite is sp3s{p^3} .
Hence the above statement is true.
Therefore all the statements about sulphite are true.
So the correct answer is option D i.e all of the above.
So the appropriate option would be option C.

Note: Due to the presence of lone pairs of electrons on sulphur, they try to push the bonds thus causing the structure to be trigonal pyramidal. Sulfite ion is the conjugate base of bisulphite.