For the IBr2- Lewis structure, what is the number of pairs of nonbonding electrons on the central iodine atom
The Correct Answer and Explanation is:
Correct Answer: 3 pairs of nonbonding electrons
Explanation:
To determine the number of nonbonding electron pairs (lone pairs) on the central iodine atom in the IBr₂⁻ ion, we start by drawing its Lewis structure using the following steps:
- Count total valence electrons:
- Iodine (I) has 7 valence electrons.
- Each bromine (Br) atom also has 7 valence electrons.
- The negative charge (⁻1) adds 1 more electron.
- Total = 7 (I) + 7 (Br) × 2 + 1 (charge) = 7 + 14 + 1 = 22 valence electrons
- Choose the central atom:
- Iodine is less electronegative than bromine and can expand its octet, so it is the central atom.
- Form single bonds:
- Connect iodine to two bromine atoms with single bonds.
- Each bond uses 2 electrons, so 2 bonds use 4 electrons.
- Remaining electrons = 22 – 4 = 18 electrons
- Distribute remaining electrons:
- Complete the octet on the bromine atoms first. Each Br needs 6 more electrons to complete its octet.
- Two Br atoms × 6 electrons = 12 electrons.
- Remaining electrons = 18 – 12 = 6 electrons
- Place remaining electrons on iodine:
- The remaining 6 electrons are placed on iodine as lone pairs.
- 6 electrons = 3 pairs
So, the central iodine atom has 3 lone pairs (nonbonding pairs).
These lone pairs influence the shape of the molecule. With 2 bonding pairs and 3 lone pairs, the electron geometry is trigonal bipyramidal, but the molecular shape is linear, since the 3 lone pairs occupy the equatorial positions and cancel out their repulsions, leaving the two bromine atoms 180 degrees apart.
Therefore, iodine in IBr₂⁻ has 3 pairs of nonbonding electrons.
