7.3 Lewis symbols and structures  (Page 3/15)

Writing lewis structures

Solution

1. Calculate the number of valence electrons.
   HCN: (1 $\times$ 1) + (4 $\times$ 1) + (5 $\times$ 1) = 10
   H ₃ CCH ₃ : (1 $\times$ 3) + (2 $\times$ 4) + (1 $\times$ 3) = 14
   HCCH: (1 $\times$ 1) + (2 $\times$ 4) + (1 $\times$ 1) = 10
   NH ₃ : (5 $\times$ 1) + (3 $\times$ 1) = 8
2. Draw a skeleton and connect the atoms with single bonds. Remember that H is never a central atom:
   
3. Where needed, distribute electrons to the terminal atoms:
   
   HCN: six electrons placed on N
   H ₃ CCH ₃ : no electrons remain
   HCCH: no terminal atoms capable of accepting electrons
   NH ₃ : no terminal atoms capable of accepting electrons
4. Where needed, place remaining electrons on the central atom:
   
   HCN: no electrons remain
   H ₃ CCH ₃ : no electrons remain
   HCCH: four electrons placed on carbon
   NH ₃ : two electrons placed on nitrogen
5. Where needed, rearrange electrons to form multiple bonds in order to obtain an octet on each atom:
   HCN: form two more C–N bonds
   H ₃ CCH ₃ : all atoms have the correct number of electrons
   HCCH: form a triple bond between the two carbon atoms
   NH ₃ : all atoms have the correct number of electrons
   

Check your learning

Answer: