A) 51 B) 45 C) 47 D) 46 E) 48 Determine the number of valence electrons in SnF
and then draw the corresponding Lewis structure.
The Correct Answer and Explanation is:
To determine the number of valence electrons in SnF₆²⁻ (the hexafluorostannate(IV) ion), we begin by adding up the valence electrons from each atom and considering the extra electrons due to the negative charge.
Step 1: Count valence electrons
- Sn (tin) is in Group 14 of the periodic table, so it contributes 4 valence electrons.
- Each F (fluorine) atom is in Group 17 and contributes 7 valence electrons. Since there are 6 fluorine atoms:
6 × 7 = 42 valence electrons - The 2- charge means we add 2 more electrons.
Now we sum them up:
- 4 (Sn)
- 42 (F)
- 2 (charge)
= 48 valence electrons
So, the correct answer is:
E) 48
Step 2: Draw the Lewis Structure
- Place Sn at the center since it is less electronegative and can form more than one bond.
- Surround it with six fluorine atoms.
- Create a single bond between Sn and each F atom. This uses 6 bonds × 2 electrons = 12 electrons.
- Subtracting the bonding electrons from the total gives:
48 – 12 = 36 electrons left. - Distribute the remaining 36 electrons as lone pairs on the fluorine atoms. Each F atom needs 6 electrons to complete its octet (in addition to the 2 shared in the bond), so:
6 F atoms × 6 electrons = 36 electrons
All electrons are now accounted for, and each atom has a complete octet.
The Sn atom is surrounded by 6 bonding pairs, making it hypervalent, which is acceptable for elements in period 5 or higher like tin.
Summary
- Total valence electrons: 48
- Central atom: Sn
- Each F forms a single bond with Sn
- All F atoms have complete octets
- The structure obeys the octet rule for fluorine and expanded octet for tin
Correct answer: E) 48.
