Question

A polar covalent bond will form between which two atoms?

a. beryllium and fluorine (group 1 and group 7)
b. hydrogen and chlorine (group 7)
c. sodium and oxygen (group 1 and group 6)
d. fluorine and fluorine (group 7)

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 Be and F,

E.N of Fluorine = 3.98

E.N of Beryllium = 1.57

________

E.N Difference 2.41 (Ionic Bond)

For H and Cl,

E.N of Chorine = 3.16

E.N of Hydrogen = 2.20

________

E.N Difference 0.96 (Polar Covalent Bond)

For Na and O,

E.N of Oxygen = 3.44

E.N of Sodium = 0.93

________

E.N Difference 2.51 (Ionic Bond)

For F and F,

E.N of Fluorine = 3.98

E.N of Fluorine = 3.98

________

E.N Difference 0.00 (Non-Polar Covalent Bond)

Result:

A polar covalent bond is formed between Hydrogen and Chlorine atoms.