Calculate the percentage ionic character of $H C l$ molecule. The bond length is $1.275 A^{o}$ and observed dipole moment is $1.03 D$ .

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Solution

Given, Observed dipole moment = 1.03 D
Theoretical bond length = 4.8 $\times$ bond length
Percentage ionic character = $\frac{μ_{e x p e r i m e n t a l}}{μ_{t h e o r e t i c a l}} \times 100 = \frac{1.03}{1.275 \times 4.8} \times 100 ≃ 17 %$

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˚ A

The experimental value of dipole moment of HCl is 1.03 D. The length of H−Cl bond is 1.275 A. The percentage ionic character of HCl is:

H C l

1.275 ˚ A

1.03 D

= 4.8 \times 10^{- 10} e s u

L i F

6.32 D

L i - F

0.156

1.27 \circ A

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