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? place the symbols de, dg, eg, and df in the proper sequence, with the most covalent bond on the right.

(Less Covalent Bond)    DG    EG    DF    DE    (Most Covalent Bond)

Explanation:

1. List the electronegativities of the atoms:

D = 3.8

E = 3.3

F = 2.8

G = 1.3

2. Find the difference of electronegativities of the compounds formed:

DE = 3.8-3.3 = 0.5

DG = 3.8 - 1.3 = 2.5

EG = 3.3 - 1.3 = 2.0

DF = 3.8 - 2.8 = 1.0

3. Covalent bonds are formed with the smallest differrence of electronegativity between its atoms, so the order will be:

(Less Covalent Bond)    DG    EG    DF    DE    (Most Covalent Bond)

The strength covalent bond in acceding order of  DG < EG < DF < DE

The electronegativities if the given atoms,

D = 3.8

E = 3.3

F = 2.8

G = 1.3

As we know the covalent bond is stronger when the difference in electronegativities between atom is higher.

The difference in electronegativities of molecules,

DE = 3.8-3.3 = 0.5

DG = 3.8 - 1.3 = 2.5

EG = 3.3 - 1.3 = 2.0

DF = 3.8 - 2.8 = 1.0

Hence, we can conclude that the covalent bond of  DG < EG < DF < DE.

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