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# What is the hybridization of Carbon atoms present in haloarenes?(A) $SP$(B) $S{{P}^{2}}$(C) $S{{P}^{3}}$(D) $S{{P}^{3}}d$

Last updated date: 20th Jun 2024
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Hint: This question is based on VSEPR theory. In haloarenes each carbon atom forms one $\pi$ bond and as carbon is tetravalent so remaining three bonds will be sigma bonds and also carbon doesn’t have a lone pair, so its degree of hybridization will be three and the ring will be having planar structure.

This is the structure of haloarene where one can clearly observe each carbon atom and find that it is forming one pi bond and three sigma bonds. Electronic configuration of carbon atom
$1{{S}^{2}}2{{S}^{2}}2{{P}^{2}}$
One electron of 2S orbital shifts into vacant 2P orbital so carbon can be tetravalent. So new electronic configuration of carbon atom
$1{{S}^{2}}2{{S}^{1}}2{{P}^{3}}$
Now carbon has four half filled orbitals, out of which one P-orbital takes part in pi-bond formation and remaining three will form sigma bonds. Among the orbitals those are forming sigma bonds, one is S orbital and other two are P orbitals.
Number of sigma bonds= 3
Number of lone pairs=0
Degree of hybridization= Number of sigma bonds + Number of lone pairs
Degree of hybridization= 3+0
Degree of hybridization=3
As the degree of hybridization is three and one S and 2 P orbitals are taking part in sigma bond formation, that’s why hybridization of carbon atoms over here is SP.