Question

The dipole moment AgCl is 6.08D in gaseous state. The Ag - Cl distance is 228.1pm . If the percent ionic character in AgCl is x%, what is x?​

Answer 1

Given: Dipole moment of AgCl is 6.08 D. The Ag-Cl distance is 228.1 pm

To find: Percent ionic character in AgCl

Explanation: Experimental dipole moment= 6.08D

Charge of electron= 1.6* 10^-19

Distance= 228.1 pm= 228.1* 10^-10 cm

Theoretical dipole moment= Charge of electron* distance= (1.6*10^-19)*(228.1*10^-10)

= 364.96 * 10^ -29 cm

= 3649.6 * 10^ -30 cm

Now, to convert cm into Debye(D), we use conversion formula: 1 D= 3.335* 10^-30 cm

$3649.6 \times {10}^{ - 30} = \frac{3649.6 \times {10}^{ - 30} cm}{3.335 \times {10}^{ - 30}cm }$

= 1094.3 D

$percent \: ionic \: character = \frac{experimemtal \: dipole \: moment}{theoretical \: dipole \: moment} \times 100$

= (6.08*100)/ 1094.3

= 0.55%

The percentage ionic character is 0.55% in this molecule.