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? (Use the appropriate <, =, or > symbol to separate substances in the list.)

Answer 1

Answer:

Thus the order of covalent character will be

DG < EG < DF < DE

Explanation:

The electronegativity decides the nature of bond formed between two atoms.

More the difference in electronegativity more the ionic character in the bond formed.

Let us check the electronegativity difference between the elements forming molecules

a) DE :

3.8-3.3 = 0.5

b) DG :

3.8-1.3 = 2.5

c) EG

3.3-1.3 = 2

d) DF

3.8-2.8 = 1.3

So the maximum ionic character will be in DG and minimum in DE

Thus the order of covalent character will be

DG < EG < DF < DE