Question

Identify three compounds that lie on the boundary between ionic and covalent bonding, and calculate the difference in electronegativity and %ionic character for each of those compounds.

Answer 1

THe difference of dipole moment leads to the polarisation which induces the covalency in the otherwise ionic bond. When there is no polarisation ,no overlapping zone exists and the bond is purely ionic.

The dipole moment is a very direct indicator of the percent ionic character of a given bond. For a given bond, the percent ionic character is given simply by

Ionic character= 100*

Observed is the dipole moment of the actual molecule and the (ionic) is the dipole moment if the bonds were 100% ionic.

	DIPOLE MOMENT	% OF IONIC CHARACTER
HF	3.98(F)-2.20(H)=1.78	39
HCl	3.16(Cl)-2.20(H)=0.96	18
AgCl	3.16(CI)-1.93(Ag)=1.23	25

% OF IONIC CHARACTER is actually calculated from the product of electronic charge* distance seperating the atoms.

Here I have used the Hanay and Smith formula in which we have, Percent ionic character

=(0.16*en+0.035*en^2)∗100

For the HF

[0.16*1.78+0.035*(1.78)^2]*100 = 39%

HCl

[0.16*0.96+0.035*(0.96)^2]*100 = 18%

AgCl

[0.16*1.23+0.035*(1.23)^2]*100 =25%

Thus we can see the the difference in electronegativity actually parallels to the percentage of IONIC character OR the higher the difference in electronegativity the higher the IONIC character.