Draw the Lewis structure for a bromide monoiodide (BrI) molecule.
The Correct Answer and Explanation is:
To draw the Lewis structure for the bromide monoiodide (BrI) molecule, follow these steps:
1. Count the Total Valence Electrons
- Bromine (Br) is in Group 17 of the periodic table, so it has 7 valence electrons.
- Iodine (I) is also in Group 17, so it has 7 valence electrons.
Total valence electrons:
7(Br)+7(I)=147 (Br) + 7 (I) = 147(Br)+7(I)=14 valence electrons.
2. Determine the Central Atom
Since Br and I are both halogens, they can be placed in either position. However, halogens tend to form single bonds and are not typically central atoms in molecular structures. Either bromine or iodine can be placed in the center of the structure, but let’s place Br in the center and I as the bonded atom. This arrangement helps us visualize the bonding clearly.
3. Draw the Skeleton Structure
Place Br and I next to each other, connected by a single bond (Br—I).
4. Distribute the Electrons
Each single bond (Br—I) uses up 2 electrons (one from Br and one from I). Since we have used 2 electrons, we now have:
14−2=1214 – 2 = 1214−2=12 electrons remaining.
Distribute these remaining electrons as lone pairs around the atoms, following these rules:
- Each halogen typically forms 3 lone pairs (6 electrons) to complete their octet (8 electrons).
- Place 3 lone pairs on both Br and I.
After distributing the lone pairs, both atoms now have 8 electrons in their valence shells, completing their octets.
5. Verify the Structure
Ensure all atoms follow the octet rule:
- Br has 2 electrons in the bond with I and 6 electrons in 3 lone pairs, totaling 8 electrons.
- I has 2 electrons in the bond with Br and 6 electrons in 3 lone pairs, also totaling 8 electrons.
Thus, the structure is complete.
Lewis Structure for BrI:
- A single bond (Br—I).
- Three lone pairs around both Br and I.
This structure shows that BrI is a simple diatomic molecule with a single bond, and both atoms satisfy the octet rule.
