JEE MAIN - Chemistry (2017 - 8th April Morning Slot - No. 15)
sp3d2 hybridization is not displayed by :
BrF5
SF6
[CrF6]3$$-$$
PF5
Explanation
Hybridization (X) = $${1 \over 2}$$ [ VE + MA – c + a ]
where, VE = No. of valence electrons of central atom
MA = No. of monovalent atoms/groups surrounding the central atom,
c = Charge on the cation,
a = Charge on the anion
(A) [BrF5] : X = $${1 \over 2}$$[ 7 + 5 - 0 + 0] = 6 = sp3d2 hybridized
(B) [SF6] : X = $${1 \over 2}$$[ 6 + 6 - 0 + 0] = 6 = sp3d2 hybridized
(C) [CrF6]3$$-$$ : Cr+3 = [Ar]3d3
(D) [PF5] : X = $${1 \over 2}$$[ 5 + 5 - 0 + 0] = 5 = sp3d hybridized
Note : [CrF6]3$$-$$ shows d2sp3 hybridization not sp3d2. So option (C) should also be the correct answer.
where, VE = No. of valence electrons of central atom
MA = No. of monovalent atoms/groups surrounding the central atom,
c = Charge on the cation,
a = Charge on the anion
(A) [BrF5] : X = $${1 \over 2}$$[ 7 + 5 - 0 + 0] = 6 = sp3d2 hybridized
(B) [SF6] : X = $${1 \over 2}$$[ 6 + 6 - 0 + 0] = 6 = sp3d2 hybridized
(C) [CrF6]3$$-$$ : Cr+3 = [Ar]3d3
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(D) [PF5] : X = $${1 \over 2}$$[ 5 + 5 - 0 + 0] = 5 = sp3d hybridized
Note : [CrF6]3$$-$$ shows d2sp3 hybridization not sp3d2. So option (C) should also be the correct answer.
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