The given compound SeFClBrH is optically active .
Because this compound does not have di-symmetry . And also forms non superimposable mirror image . the compound is optically active .
It has chiral center .
As the atomic no of selenium is 34 .
We know,
Argon electronic configuration is [Ar] 4s2 3d10 4p4
Excited state electronic configuration is [ Ar ] 4s2 3d10 4p3 5s1 .
thus it allows 4 distinct groups ( F, Cl, Br , H) to bond with it .
As, Any tetrahedral atom with four different groups attached can be a chiral center. so the given compound satisfies all the conditions.
thus the compound is optically active.
Optically active means , the central atom should be sp3 hybridised and it is bonded to 4 distinct groups or atoms. also it forms non-superimpaossable mirror image.
The central atom is Sp3 hybridized (4sigma bond ) .
4 distinct group is attached to the central atom.
The compound forms non -superimposable mirror image .
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Disclaimer : the question is incomplete . here is the complete question .
Question : Any tetrahedral atom with four different groups attached can be a chiral center . which of these species is optically active ? (a) CHClBrF ,(b) NBrCl2H+ , (c) PFClBrI+ , (d) SeFClBrH .
