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Calculate percentage ionic character of HCl given that observed dipole moment is 1.03D and bond length is 1.275 angstrom.

Calculate percentage ionic character of HCl given that observed dipole moment is 1.03D and bond length is 1.275 angstrom.

Grade:

2 Answers

Arun
25763 Points
3 years ago
Dear student
 
Formula used :
μ = e × d
For complete separation of unit charge (100% ionic character) the dipole moment calculated as:
μ = 1.602 × 10–19 coulomb × 1.275 × 10–10 m
= 2.04 C m
Actual dipole moment of HCl = 1.03 D = 0.34 x 10–29 C m (1 C m = 2.9979×1029 D.)
% ionic character of HCl = Actual dipole moment/calculated dipole moment × 100 =
0.34 x 10-29 C m / 2.04 × 10–29 C m ×100 = 17%
 
Regards
Arun (askIITians forum expert)
SR Roy
128 Points
3 years ago
Formula used :
μ = e × d
For complete separation of unit charge (100% ionic character) the dipole moment calculated as:
μ = 1.602 × 10–19 coulomb × 1.275 × 10–10 m
= 2.04 C m
Actual dipole moment of HCl = 1.03 D = 0.34 x 10–29 C m (1 C m = 2.9979×1029 D.)
% ionic character of HCl = Actual dipole moment/calculated dipole moment × 100 =
0.34 x 10-29 C m / 2.04 × 10–29 C m ×100 = 17%

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