Referring to the most stable Lewis dot structure of I 1/3
, what is the hybridization of the central atom? a. sp
b. sp
c. sp d. sp
d
The Correct Answer and Explanation is:
Correct Answer: b. sp³
Explanation:
To determine the hybridization of the central atom in the triiodine cation (I₃⁺), we must first draw its most stable Lewis dot structure and then apply the Valence Shell Electron Pair Repulsion (VSEPR) theory.
Step 1: Calculate the total number of valence electrons.
Iodine (I) is in Group 17 of the periodic table, so each iodine atom has 7 valence electrons.
- Total valence electrons from 3 iodine atoms = 3 × 7 = 21
- The ion has a +1 charge, which means one electron has been removed.
- Total valence electrons for I₃⁺ = 21 – 1 = 20 electrons.
Step 2: Draw the Lewis structure.
- Arrange the atoms with one iodine atom in the center, bonded to the other two: I — I — I. This skeletal structure uses 2 bonds, which account for 4 electrons.
- Distribute the remaining electrons (20 – 4 = 16) to the terminal atoms first to satisfy their octets. Each terminal iodine atom needs 6 more electrons (3 lone pairs). This uses 2 × 6 = 12 electrons.
- Place the remaining electrons (16 – 12 = 4) on the central iodine atom. These 4 electrons form 2 lone pairs.
The resulting Lewis structure is:
[ :Ï—Ï—Ï: ]⁺
Where the central iodine has two lone pairs and the terminal iodines each have three lone pairs. The formal charge on the central iodine is +1, while the terminal iodines have a formal charge of 0, correctly summing to the overall +1 charge of the ion.
Step 3: Determine the hybridization.
The hybridization of the central atom is determined by its steric number, which is the total number of electron domains (bonding pairs + lone pairs) around it.
- Number of atoms bonded to the central atom (bonding domains) = 2
- Number of lone pairs on the central atom = 2
- Steric Number = 2 (bonding pairs) + 2 (lone pairs) = 4
A steric number of 4 indicates that the central atom uses four hybrid orbitals to accommodate its electron domains. These orbitals are formed by mixing one s orbital and three p orbitals, resulting in sp³ hybridization. The electron geometry around the central atom is tetrahedral, though the molecular geometry is bent or V-shaped due to the presence of the two lone pairs.
