Draw the Lewis structure for ICl3. What is the hybridization on the I (iodine) atom?
The correct answer and explanation is:
Correct Answer:
- Lewis Structure for ICl₃:
Iodine (I) is the central atom, bonded to three chlorine (Cl) atoms, with two lone pairs on iodine. - Hybridization of iodine in ICl₃:
sp³d
Explanation (300 words):
Step 1: Count the total valence electrons
- Iodine (I) has 7 valence electrons.
- Each chlorine (Cl) also has 7 valence electrons.
- Total valence electrons:
7(I)+3×7(Cl)=287 (I) + 3 \times 7 (Cl) = 28 electrons
Step 2: Draw the skeleton structure
Place iodine in the center with three single bonds to chlorine atoms:
Cl
|
Cl–I–Cl
This uses up 3×2=63 \times 2 = 6 electrons for bonding.
Remaining electrons: 28−6=2228 – 6 = 22
Step 3: Complete octets of chlorine
Each Cl atom needs 6 more electrons to complete an octet:
- 3 Cl × 6 electrons = 18 electrons
Remaining electrons: 22−18=422 – 18 = 4 electrons
Step 4: Place remaining electrons on iodine
Place the remaining 4 electrons (2 lone pairs) on iodine.
Now iodine has:
- 3 bonding pairs (from Cl)
- 2 lone pairs
This gives 5 regions of electron density, meaning trigonal bipyramidal electron geometry.
Step 5: Determine hybridization
The number of electron regions (bonding + lone pairs) around iodine is 5 → hybridization is sp³d.
Molecular Shape:
Due to two lone pairs on iodine, the actual molecular shape of ICl₃ is T-shaped, not trigonal bipyramidal.
Summary:
- Lewis structure: Iodine in center, three Cl atoms bonded, two lone pairs on I.
- Electron regions: 5 (3 bonds + 2 lone pairs)
- Hybridization: sp³d
- Molecular geometry: T-shaped due to lone pairs causing electron repulsion.