What is the Lewis Structure for BI3?

The Lewis structure for BI₃ (Boron Triiodide) can be drawn by following these steps:

1. Determine the total number of valence electrons: Boron (B) has 3 valence electrons, and each Iodine (I) atom has 7 valence electrons. Since there are three Iodine atoms, the total number of valence electrons is 3 (from Boron) + 3 × 7 (from Iodine) = 24 valence electrons.
2. Place the least electronegative atom in the center: Boron is less electronegative than Iodine, so it will be placed in the center of the Lewis structure.
3. Connect the outer atoms to the central atom with single bonds: Each Iodine atom will be connected to the Boron atom with a single bond. This uses up 3 × 2 = 6 electrons, leaving 24 – 6 = 18 electrons.
4. Distribute the remaining electrons: The remaining 18 electrons are distributed as lone pairs on the Iodine atoms. Each Iodine atom will have 3 lone pairs, which accounts for all 18 electrons.
5. Check the octet rule: Boron has only 6 electrons around it (3 from the single bonds), which is an exception to the octet rule. Iodine atoms, however, have 8 electrons each (2 from the single bond and 6 from the lone pairs), satisfying the octet rule.

Here is the Lewis structure for BI₃:

    I
    |
I -- B -- I
    |
    I

In this structure, Boron is the central atom with three single bonds to Iodine atoms, and each Iodine atom has three lone pairs of electrons.