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?

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

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.