Give the number of lone pairs around the central atom and the molecular geometry of IBr5.
The correct answer and explanation is:
Correct Answer:
- Number of lone pairs around the central atom (Iodine): 1 lone pair
- Molecular geometry of IBr₅: Square pyramidal
Explanation:
Step 1: Determine the total number of valence electrons
Iodine (I) belongs to group 17 and has 7 valence electrons.
Each bromine (Br) atom also has 7 valence electrons.
IBr₅ has one iodine atom and five bromine atoms: Total valence electrons=7 (from I)+5×7 (from Br)=7+35=42 valence electrons\text{Total valence electrons} = 7\ (\text{from I}) + 5 \times 7\ (\text{from Br}) = 7 + 35 = 42\ \text{valence electrons}
Step 2: Choose the central atom
Iodine is less electronegative than bromine and can expand its octet (since it’s in period 5), so it will be the central atom.
Step 3: Form bonds
Five Br atoms form single bonds with iodine, using 2 electrons per bond, for a total of: 5×2=10 electrons5 \times 2 = 10\ \text{electrons}
Subtract these from the total: 42−10=32 electrons left42 – 10 = 32\ \text{electrons left}
Step 4: Distribute remaining electrons
Each Br needs 6 more electrons (to complete their octets), and with 5 Br atoms: 5×6=30 electrons used for Br lone pairs5 \times 6 = 30\ \text{electrons used for Br lone pairs}
Now: 32−30=2 electrons left32 – 30 = 2\ \text{electrons left}
These remaining 2 electrons (1 lone pair) go on the central iodine atom.
Step 5: Determine electron geometry and molecular geometry
With 5 bonding pairs and 1 lone pair on iodine (6 total regions of electron density), the electron geometry is octahedral.
However, the molecular geometry (which considers only atoms, not lone pairs) is square pyramidal — 4 atoms form a square base around iodine, 1 atom is above the central atom, and the lone pair is opposite to that.
Conclusion:
- Lone pairs on iodine: 1
- Molecular geometry: Square pyramidal (AX₅E, based on VSEPR theory)