Question

select the correct lewis structure for nitrogen trifluoride, nf3
a)
: f - n - f:
: f :
b)
f = n = f
: f :
c)
: f - n = f
: f :
d)
f = n = f
f
e)

Explanation:

Step1: Calculate the total number of valence electrons

Nitrogen has 5 valence electrons and each fluorine has 7 valence electrons. For \(NF_3\), the total number of valence electrons is \(5 + 3\times7=5 + 21 = 26\)

Step2: Analyze each option based on valence - electron count and octet rule

• Option A:

The structure has \(3\) single \(N - F\) bonds. Each single bond has \(2\) electrons. Nitrogen has \(1\) lone pair (\(2\) electrons) and each fluorine has \(3\) lone pairs (\(3\times2 = 6\) electrons per \(F\)).
The total number of electrons: \(3\times2+2 + 3\times6=6 + 2+18 = 26\). Nitrogen has \(8\) electrons (\(3\) bonding pairs and \(1\) lone pair) and each fluorine has \(8\) electrons (\(1\) bonding pair and \(3\) lone pairs)

• Option B:

Double bonds are present. If there are double bonds (\(N = F\)), fluorine (which is in group 17) does not form double bonds as it needs only one electron to complete its octet. Also, the electron - count will be wrong. For example, if we assume double bonds, the electron count will be less than 26 (because double bonds “use” more electrons in bonding but we can calculate: let's say \(2\) double bonds (\(N = F\)) and \(1\) single \(N - F\) bond. The bonding electrons: \(2\times4+ 2=10\). Lone - pair electrons: assume wrong distribution, but even without full calculation, we know \(F\) doesn't form double bonds

• Option C:

A double bond (\(N = F\)) is present. Fluorine does not form double bonds. Also, if we calculate the electron count: bonding electrons (\(2\) single \(N - F\) bonds and \(1\) double \(N - F\) bond: \(2\times2+4 = 8\) bonding electrons). Lone - pair electrons: wrong distribution (e.g., if \(N\) has \(1\) lone pair (\(2\) electrons) and \(F\)s have wrong number of lone pairs), total electron count will not be 26

• Option D:

Double bonds (\(N = F\)) are present. Fluorine does not form double bonds. Also, electron - count (bonding electrons: \(2\) double bonds and \(1\) single bond: \(2\times4+2=10\) bonding electrons) will be wrong for getting 26 valence electrons

Answer:

A.