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?

Question

Dashawn

Chemistry

24 October, 03:43

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?

+2

Answers (1)

Know the Answer?

Conor Riddle · 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