Question

the dipole moments of diatomic molecules ab and cd are 10.41D and 10.27Drespectively while their bond distances are 2.82Aand 2.67A calcute their ionic bond charecter

Answer 1

answer : their Ionic characters are ; 78.51 % and 80.17%

case 1 : Bond distance, d = 2.82A = 2.82 × 10^-10 m

charge on each molecule, q = 1.6 × 10^-19 C dipole moment, P = q × d

= 1.6 × 10^-19 C × 2.82 × 10^-10 m

= (1.6 × 2.82) × 10^-29 C.m

we know, 1 Debye = 3.335 × 10^-30 Cm

so, dipole moment , P = (1.6 × 2.82/3.335) × 10¹

= 13.529 D

ionic character = μ/P × 100

= 10.41/13.529 × 100

= 78.51 %

case 2 : bond distance, d = 2.67A = 2.67 × 10^-10 m

charge on each molecule, q = 1.6 × 10^-19 C

dipole moment, P = q × d

= 1.6 × 10^-19 C × 2.67 × 10^-10 m

= (1.6 × 2.67) × 10^-29 Cm

= (1.6 × 2.67/3.335) × 10¹ D

= 12.8095 D

ionic character = μ/P × 100

= 10.27 D/12.8095 D × 100

= 80.17 %