Answer
$C_2H_3Cl$ has 18e- in total, distribute to 4 single bonds (3 C-H, 1 C-Cl), 1 double bond C=C, and 3 lone pairs on Cl
Work Step by Step
C: 4 valence electrons x 2 atoms = 8e-
H has 1 valence electrons x 3 atoms = 3e-
Cl has 7 valence e- x 1 atoms = 7 e-
Total of: 8+2+7=18 e-
• 4 single bonds = 4x2e- =8e-
• 1 double bonds C=C: 4e-
• 3 lone pairs for Cl, total of 3 x 2e- = 6e-
