Question

A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 Å. Find the percentage ionic character in the molecule.

Answer 1

Answer:

A polar molecule AB has dipole moment 3.2 D (Debye) while the bond length is 1.6 Å. Find the percentage ionic character

Answer 2

A polar molecule AB has dipole moment 3.2 D while the bond length is 1.6 A°.

we have to find the percentage ionic character in the molecule.

solution : AB is polar molecule. let A is electropositive element and B is electronegative element.

so charge on A⁺ ion = 1.6 × 10^-19 C

charge on B¯ ion = - 1.6 × 10^-19 C

so dipole moment of AB = magnitude of charge on each ion × bond length of AB

= 1.6 × 10^-19 C × 1.6 × 10^-10 m

= 2.56 × 10^-29 Cm

but we know, 1 Debye = 3.335 × 10^-30 Cm

so, dipole moment of AB = (25.6/3.335) D

= 7.676 D

percentage ionic character = practical value of dipole moment /theoretical value of dipole moment × 100

= 3.2/7.676 × 100

= 41.68 %

Therefore 41.68. % ionic character in the molecule.