Sketch the Lewis structure for the polyatomic ion PF4^-1 and select the choice from below that is correct. The Lewis structure has how many double bonds and how many lone pairs on the central atom?
The Correct Answer and Explanation is:
The correct Lewis structure for the polyatomic ion PF₄⁻ has:
- Zero double bonds
- Two lone pairs on the central atom (phosphorus)
Explanation
To sketch the Lewis structure of PF₄⁻, we start by counting the total number of valence electrons:
- Phosphorus (P) is in Group 15 and has 5 valence electrons.
- Fluorine (F) is in Group 17 and each fluorine atom has 7 valence electrons.
- There are 4 fluorine atoms: 4 × 7 = 28 valence electrons.
- The ion has a negative charge of -1, which adds 1 extra electron.
Total valence electrons = 5 (from P) + 28 (from F) + 1 (extra for charge) = 34 electrons.
Next, we draw phosphorus (P) as the central atom since it is less electronegative than fluorine. Each fluorine is connected to phosphorus with a single bond. This uses 4 bonds × 2 electrons = 8 electrons.
Now subtract the bonding electrons from the total:
34 – 8 = 26 electrons remaining.
These remaining electrons are placed around the fluorine atoms to complete their octets. Each fluorine needs 6 more electrons to complete its octet (they each already have one bond or 2 electrons from the P–F bond). For 4 fluorine atoms: 4 × 6 = 24 electrons.
26 – 24 = 2 electrons remaining.
The leftover 2 electrons go on the central atom, phosphorus, as a lone pair. Phosphorus now has 4 bonding pairs and 1 lone pair.
Phosphorus can expand its octet since it is in Period 3 and has d-orbitals. The structure is valid with phosphorus having 10 electrons.
Since all fluorine atoms form only single bonds (which satisfies their need for 8 electrons), there are no double bonds in the structure. The phosphorus atom ends up with 2 lone pairs total (one added pair and one from its initial electron count).
So, the correct answer:
Zero double bonds and two lone pairs on the central atom.
