Which of the following is the best Lewis dot structure for the BeF2 molecule?

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

Answer: (A)

Verified step by step guidance

1. Step 1: Determine the total number of valence electrons available for bonding in BeF2. Beryllium (Be) has 2 valence electrons, and each fluorine (F) atom has 7 valence electrons. Since there are two fluorine atoms, total valence electrons = 2 (from Be) + 2 × 7 (from F) = 16 electrons.
2. Step 2: Arrange the atoms with Be as the central atom because it is less electronegative than fluorine. Connect Be to each F atom with a single bond initially. Each single bond represents 2 electrons.
3. Step 3: Distribute the remaining valence electrons to satisfy the octet rule for the fluorine atoms first. Each fluorine needs 8 electrons total (including bonding electrons). After forming single bonds, assign lone pairs to fluorine atoms to complete their octets.
4. Step 4: Check the central atom (Be) for its electron count. Beryllium is an exception to the octet rule and is stable with only 4 electrons around it (two single bonds). Therefore, Be does not have lone pairs and does not form double bonds in this molecule.
5. Step 5: Confirm that the Lewis structure has no formal charges or the smallest possible formal charges, and that all atoms (especially fluorine) have complete octets. The best Lewis structure will have Be in the center with two single bonds to F atoms, each F atom having three lone pairs, and Be having no lone pairs.