Which of the following is the correct number of valence electrons for the ion SnF6^{2-}, and which Lewis structure best represents it?

• A. SnF6^{2-} has 54 valence electrons; the Lewis structure shows Sn at the center with six single bonds to F atoms, each F atom with four lone pairs.
• B. SnF6^{2-} has 48 valence electrons; the Lewis structure shows Sn at the center with six single bonds to F atoms, each F atom with three lone pairs.
• C. SnF6^{2-} has 36 valence electrons; the Lewis structure shows Sn at the center with six double bonds to F atoms, each F atom with two lone pairs.
• D. SnF6^{2-} has 42 valence electrons; the Lewis structure shows Sn at the center with six single bonds to F atoms, each F atom with two lone pairs.

Answer: (B)

Verified step by step guidance

1. Determine the total number of valence electrons by adding the valence electrons from each atom and adjusting for the charge on the ion. Tin (Sn) is in group 14, so it has 4 valence electrons. Each fluorine (F) atom is in group 17, so each has 7 valence electrons. Since there are 6 fluorine atoms, multiply 7 by 6. Then, add 2 extra electrons because the ion has a 2- charge.
2. Write the total valence electrons calculation as: \$4 + (6 \times 7) + 2$.
3. Draw the Lewis structure with Sn as the central atom bonded to six F atoms. Each bond represents 2 electrons, so 6 single bonds use \$6 \times 2 = 12$ electrons.
4. Subtract the bonding electrons from the total valence electrons to find the remaining electrons, which will be placed as lone pairs on the fluorine atoms. Each fluorine atom should complete its octet by having 3 lone pairs (6 electrons) after forming a single bond.
5. Verify that the total number of electrons used in bonds and lone pairs equals the total valence electrons calculated, and confirm that the central atom and fluorine atoms satisfy the octet rule or expanded octet if applicable.