Question

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)

• A. i–d, ii–c, iii–b, iv–a
• B. i–a, ii–d, iii–b, iv–c
• C. i–d, ii–c, iii–a, iv–b
• D. i–d, ii–b, iii–c, iv–b

Hint:

To determine the hybridization of the central atom, count the number of bonding pairs and lone pairs of electrons around the atom. The hybridization is typically sp, sp², or sp³ depending on the number of regions of electron density.

Answer 1

The Correct Option is A

To match the species in Column I with the correct hybrid orbitals from Column II, we need to analyze the bonding in each molecule:
- i. \( I_3^- \): The central iodine atom has three bonding pairs and one lone pair, which corresponds to sp hybridization. Hence, the correct match is \( i \)–d.
- ii. \( PCl_3 \): Phosphorus has three bonding pairs and one lone pair, so it uses sp² hybridization. Hence, the correct match is \( ii \)–c.
- iii. \( BF_3 \): Boron forms three bonds with fluorine and has no lone pairs, so it uses sp² hybridization. Hence, the correct match is \( iii \)–b.
- iv. \( CO_2 \): Carbon in \( CO_2 \) has two bonding pairs and no lone pairs, so it uses sp hybridization. Hence, the correct match is \( iv \)–a.
Thus, the correct answer is option (A).