To understand the Lewis structures for NF₃ (nitrogen trifluoride) and PF₅ (phosphorus pentafluoride), we first need to consider the valence electrons available for bonding.
For NF₃, nitrogen (N) has 5 valence electrons and each fluorine (F) has 7 valence electrons. In total, NF₃ has:
- 1 Nitrogen: 5 electrons
- 3 Fluorines: 3 × 7 = 21 electrons
- Total: 5 + 21 = 26 electrons
The Lewis structure shows nitrogen at the center with three single bonds to each fluorine atom:
N: F | F
This arrangement gives nitrogen a complete octet, satisfying the octet rule, while each fluorine also achieves a stable configuration with 8 electrons.
Now, for PF₅, phosphorus (P) has 5 valence electrons and each fluorine contributes 7, making a total of:
- 1 Phosphorus: 5 electrons
- 5 Fluorines: 5 × 7 = 35 electrons
- Total: 5 + 35 = 40 electrons
In the Lewis structure of PF₅, phosphorus is at the center bonded to five fluorine atoms:
F
|
F - P - F
|
F
This arrangement allows phosphorus to utilize its d orbitals for bonding, accommodating more than 8 electrons around it, which contributes to its stability.
Next, let’s discuss the stability of these molecules. NF₃ is stable due to its complete octet, while PF₃ also follows a similar structure, benefiting from phosphorus’s ability to expand its octet. PF₅ is stable because phosphorus can use d orbitals to hold additional electrons from the five fluorine atoms.
On the contrary, NF₅ does not exist because nitrogen cannot expand its octet due to the absence of d orbitals. Its maximum coordination is three; therefore, attempting to bond with five fluorine atoms results in an unstable configuration that cannot be formed.
In summary, while NF₃, PF₃, and PF₅ are stable due to their respective valence electron arrangements and orbital utilizations, NF₅ fails to exist because nitrogen is unable to accommodate more than eight electrons.