JEE Main 2017 — Chemical Bonding And Molecular Structure Question with Solution

Hybridization (X) = $$\frac{1}{2}$$ [ 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 = $$\frac{1}{2}$$[ 7 + 5 - 0 + 0] = 6 = sp³d² hybridized

(B) [SF₆] : X = $$\frac{1}{2}$$[ 6 + 6 - 0 + 0] = 6 = sp³d² hybridized

(C) [CrF₆]^3$$-$$ : Cr⁺³ = [Ar]3d³

(D) [PF₅] : X = $$\frac{1}{2}$$[ 5 + 5 - 0 + 0] = 5 = sp³d hybridized

Note : [CrF₆]^3$$-$$ shows d²sp³ hybridization not sp³d². So option (C) should also be the correct answer.