Discussion

A few species are given in Column I. Column II contains the hybrid orbitals used by the central atom of the species for bonding.

The CORRECT match for the species to their central atom hybridization is

(Given: Atomic numbers of B: 5; C: 6; O: 8; F: 9; P: 15; Cl: 17; I: 53)

Column I Species	Column II Hybrid orbitals used by the central atom for bonding
i. I₃⁻	a. sp
ii. PCl₃	b. sp²
iii. BF₃	c. sp³
iv. CO₂	d. sp³d

Answer: A

The correct match is i-d, ii-c, iii-b, iv-a.

Count the electron domains (bond pairs plus lone pairs) on the central atom; that number fixes the hybridization.

I₃⁻: the central I has 2 bonds and 3 lone pairs, 5 domains, so sp³d (linear ion, lone pairs equatorial).
PCl₃: 3 bonds and 1 lone pair, 4 domains, so sp³ (pyramidal).
BF₃: 3 bonds and no lone pair, so sp² (trigonal planar).
CO₂: two double bonds, 2 domains, so sp (linear).

The other options give I₃⁻ sp (counting only its two bonds), or swap PCl₃ with BF₃, or give CO₂ a trigonal hybrid; each fails one of these counts.

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