Question

Determine if a bond between each pair of atoms would be pure covalent, polar covalent, or ionic.

a. c and n
b. n and s
c. k and f
d. n and n

Answer 1

Types of Bonds can be predicted by calculating the difference in electronegativity.
If, Electronegativity difference is,

Less than 0.4 then it is Non Polar Covalent

Between 0.4 and 1.7 then it is Polar Covalent

Greater than 1.7 then it is Ionic

For C and N,
E.N of Nitrogen = 3.04
E.N of Carbon = 2.55
________
E.N Difference 0.49 (Weakly Polar Covalent)

For N and S,
E.N of Nitrogen = 3.04
E.N of Sulfur = 2.58
________
E.N Difference 0.46 (Weakly Polar Covalent)

For K and F,
E.N of Fluorine = 3.98
E.N of Potassium = 0.28
________
E.N Difference 3.70 (Ionic)

For N and N,
E.N of Nitrogen = 3.04
E.N of Nitrogen = 3.04
________
E.N Difference 0 (Non Polar Covalent)