Concept: The hybridization state of a central atom can be determined by calculating its steric number, which is the sum of the number of sigma bonds and the number of lone pairs around the central atom. The steric number corresponds to a specific hybridization:

• Steric number 2: sp
• Steric number 3: sp²
• Steric number 4: sp³
• Steric number 5: sp³d
• Steric number 6: sp³d²

Why (B) is correct:

Let's determine the hybridization of the central iodine atom in I₃^-.

1. Step 1: Determine the central atom and total valence electrons.
   In I₃^-, one iodine atom acts as the central atom, and the other two are terminal atoms. The total number of valence electrons is $3\times 7$ (for three iodine atoms) $+1$ (for the negative charge) $=22$ valence electrons.

2. Step 2: Draw the Lewis structure.
   The central iodine atom forms single bonds with the two terminal iodine atoms. This uses $2\times 2=4$ electrons. The remaining $22-4=18$ electrons are distributed as lone pairs.

   Each terminal iodine atom gets 3 lone pairs (6 electrons each), totaling $2\times 6=12$ electrons. The remaining $18-12=6$ electrons are placed on the central iodine atom as 3 lone pairs.

   So, the central iodine atom has 2 sigma bonds and 3 lone pairs.

3. Step 3: Calculate the steric number.
   Steric number = (Number of sigma bonds) + (Number of lone pairs)
   Steric number = $2+3=5$

4. Step 4: Determine the hybridization.
   A steric number of 5 corresponds to sp³d hybridization.

Option Analysis:

• A) sp²: This corresponds to a steric number of 3 (e.g., BF₃). Incorrect.
• B) sp³d: This corresponds to a steric number of 5, which matches our calculation for I₃^-. Correct.
• C) sp³: This corresponds to a steric number of 4 (e.g., CH₄, NH₃, H₂O). Incorrect.
• D) sp³d²: This corresponds to a steric number of 6 (e.g., SF₆). Incorrect.

Correct Answer: (B)