Question

Consider, \(PF_5,BrF_5,PCl_3,SF_6,[ICl_4]^–,ClF_3 and\ IF_5\).
Amongst the above molecule(s)/ion(s), the number of molecule(s)/ion(s) having \(sp^3d^2\) hybridisation is _____.

Answer 1

Correct Answer: 4

To determine the hybridization of the given molecules/ions, we first count the number of electron pairs—both bonding and lone pairs—around the central atom in each species. Hybridization state \(sp^3d^2\) corresponds to a central atom surrounded by 6 electron pairs (octahedral geometry).

1. \(PF_5\): Phosphorus has 5 bonds with fluorine, resulting in 5 bonding pairs and no lone pairs. The hybridization is \(sp^3d\), not \(sp^3d^2\).
2. \(BrF_5\): Bromine is the central atom with 5 bonds to fluorine and 1 lone pair, totaling 6 pairs. Hence, the hybridization is \(sp^3d^2\).
3. \(PCl_3\): Phosphorus forms 3 bonds with chlorine and has 1 lone pair, totaling 4 pairs. The hybridization is \(sp^3\), not \(sp^3d^2\).
4. \(SF_6\): Sulfur forms 6 bonds with fluorine without lone pairs, totaling 6 pairs. The hybridization is \(sp^3d^2\).
5. \([ICl_4]^–\): Iodine forms 4 bonds with chlorine and has 2 lone pairs, totaling 6 pairs. The hybridization is \(sp^3d^2\).
6. \(ClF_3\): Chlorine forms 3 bonds with fluorine and has 2 lone pairs, totaling 5 pairs. The hybridization is \(sp^3d\), not \(sp^3d^2\).
7. \(IF_5\): Iodine forms 5 bonds with fluorine and has 1 lone pair, totaling 6 pairs. The hybridization is \(sp^3d^2\).

The molecules/ions with \(sp^3d^2\) hybridization are \(BrF_5, SF_6, [ICl_4]^–,\) and \(IF_5\).
The total number is 4, which fits within the range of 4,4.