Which of the following is the correct Lewis dot structure for boron trichloride (BCl_3)?

• A. B atom in the center with three double bonds to Cl atoms; each Cl atom has two lone pairs; B has no lone pairs.
• B. B atom in the center with three single bonds to Cl atoms; each Cl atom has three lone pairs; B has no lone pairs.
• C. B atom in the center with three single bonds to Cl atoms; each Cl atom has two lone pairs; B has one lone pair.
• D. B atom in the center with three single bonds to Cl atoms; each Cl atom has one lone pair; B has two lone pairs.

Answer: (B)

Verified step by step guidance

1. Step 1: Identify the total number of valence electrons available for bonding. Boron (B) has 3 valence electrons, and each chlorine (Cl) atom has 7 valence electrons. Since there are three Cl atoms, total valence electrons = 3 (from B) + 3 × 7 (from Cl) = 24 electrons.
2. Step 2: Determine the bonding framework. Boron is the central atom because it is less electronegative than chlorine. Connect the boron atom to each chlorine atom with a single bond initially. Each single bond represents 2 electrons.
3. Step 3: Distribute the remaining electrons as lone pairs on the chlorine atoms to complete their octets. Each Cl atom needs 8 electrons total (including bonding electrons). Since each Cl is bonded with one single bond (2 electrons), add 6 electrons (3 lone pairs) to each Cl atom.
4. Step 4: Check the boron atom's electron count. Boron is an exception to the octet rule and is stable with 6 electrons around it (3 single bonds × 2 electrons each). Therefore, boron has no lone pairs and does not form double bonds in BCl₃.
5. Step 5: Confirm the Lewis structure: Boron in the center with three single bonds to chlorine atoms; each chlorine atom has three lone pairs; boron has no lone pairs. This matches the correct Lewis dot structure for BCl₃.