Question

Observed dipole moment of LiF is 6.32 D. Calculate percentage ionic character of LiF if

bond length (Li–F) is 0.156 nm.​

Answer 1

The percent ionic character of LiF is 84.35%.

Explanation:

The percent ionic character is the ratio of observed dipole moment to the dipole moment.

Here we are given with observed dipole moment which is 6.32 D.

Know we will calculate the dipole moment.

we will convert the bond length into cm.

1.56×10⁻⁸cm

dipole moment = 4.8 ×10⁻¹⁰esu × 1.56×10⁻⁸cm = 7.48×10⁻¹⁸ esu.cm

As,

1 D = 1×10⁻¹⁸ esucm

thus,

7.48D

Percent ionic character = 6.32/ 7.48 × 100 = 84.35%

Answer 2

Answer:

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