To draw the Lewis dot structure of nitrogen trifluoride (NF3), we first need to determine the total number of valence electrons in the molecule. Nitrogen (N) has 5 valence electrons, and each fluorine (F) atom has 7 valence electrons, giving us:
- 1 Nitrogen: 5 electrons
- 3 Fluorines: 3 x 7 = 21 electrons
Therefore, the total number of valence electrons for NF3 is 5 + 21 = 26 electrons.
Next, we place the nitrogen atom in the center, as it is less electronegative than fluorine. Then, we attach the three fluorine atoms to nitrogen:
F
|
F -- N -- F
After connecting the nitrogen atom to each fluorine atom with single bonds, we allocate the remaining electrons to complete the octets of the fluorine atoms. Each F atom needs 8 electrons surrounding it, and they already have 2 (from the bond with N), so we place 6 more electrons around each F:
:F:
|
:F: -- N -- :F:
Now each fluorine has a complete octet, and nitrogen has one lone pair of electrons. There are no formal charges on any of the atoms because:
- N has 5 valence electrons and is bonded to 3 fluorine atoms (3 bonds, each counted as 1), and has 1 lone pair (2 electrons), resulting in:
- Formal Charge of N = 5 – (0 + 3) = +2
- Each F has 7 valence electrons, is bonded with one bond, and has 6 non-bonding electrons, thus:
- Formal Charge of F = 7 – (0 + 1) = 0
Since there are no non-zero formal charges for the fluorine atoms, we can represent the formal charge of nitrogen in the final structure as zero too:
:F:
|
: F: -- N -- :F:
Now addressing resonance forms, the structure of NF3 does not have any significant resonance structures due to the nature of the nitrogen and fluorine bonds. The lone pair on nitrogen cannot be used as a bonding pair, and there are no other ways to distribute the electrons to create equivalent structures while maintaining the octet rule for fluorine. Thus, the most stable and correct representation of NF3 is the one shown above:
:F:
|
: F: -- N -- :F: