Which of the following is the correct number of valence electrons for the neutral compound SiF4, and which Lewis structure best represents it?

• A. 28 valence electrons; Si at the center with double bonds to each F atom, each F with two lone pairs
• B. 24 valence electrons; Si at the center with single bonds to F atoms, each F with two lone pairs
• C. 36 valence electrons; Si at the center with single bonds to F atoms, each F with four lone pairs
• D. 32 valence electrons; Si at the center with four single bonds to F atoms, each F with three lone pairs

Answer: (D)

Verified step by step guidance

1. Determine the total number of valence electrons by adding the valence electrons of each atom in the molecule. Silicon (Si) is in group 14 and has 4 valence electrons, and each fluorine (F) atom is in group 17 with 7 valence electrons. Since there are 4 fluorine atoms, calculate the total as: \$4 + 4 \times 7$.
2. Sum the valence electrons to find the total number available for bonding and lone pairs in the molecule.
3. Draw the Lewis structure by placing the silicon atom at the center because it is less electronegative and can form multiple bonds, then connect each fluorine atom to silicon with a single bond initially.
4. Distribute the remaining valence electrons as lone pairs around the fluorine atoms to complete their octets. Each fluorine should have three lone pairs (6 electrons) after forming a single bond with silicon.
5. Check that the silicon atom has a complete octet by counting the electrons shared in bonds (each single bond counts as 2 electrons). Confirm that the total number of electrons used in bonds and lone pairs equals the total valence electrons calculated.