Question

Four atoms are arbitrarily labeled D, E, F, and G. Their electronegativities are as follows: D =3.8, E = 3.3, F = 2.8, and G = 1.3. If the atoms of these elements form the molecules DE, DG, EG, and DF, how would you arrange these molecules in order of increasing covalent bond character?

Answer 1

Answer: DG < EG< DF< DE

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.  

Lower is the electronegativity diffrence, more will be the covalent character. Higher is the electronegativity diffrence, more will be the ionic character.

1. DE:  Electronegativity difference = electronegativity of D - electronegativity of E = 3.8-3.3= 0.5

2. DG: Electronegativity difference = electronegativity of D - electronegativity of G= 3.8-1.3=  2.5

3. EG: Electronegativity difference = electronegativity of E - electronegativity of G= 3.3-1.3= 2.0

4. DF: Electronegativity difference = electronegativity of D - electronegativity of F= 3.8-2.8= 1.0

Thus the order of increasing covalent character is DG < EG< DF< DE