Answer
$N_2F_4$ has total of 28e-, 4 F-N bonds, 1N-N bond, 3 lone pairs on each F, 1 lone pair on each N
Work Step by Step
N: 5 valence electrons x 2 atoms = 10 electrons
F has 7 valence electrons x 4 atoms = 28 electrons
Total of: 10+28=38 e-
• 5 bonds = 5x2e- =10 e-
• 3 lone pairs for each F, total of 3 x 2e- x 4F = 24 e-
• 1 lone pair for N, total of 2e- x 2N =4 e-
We draw the Lewis structure: see graph.
