The covalent radii of Li & F are 123 pm & 64 pm respectively. The electronegativity of Li & F are 1.0 & 4.0 respectively. The dipole moment of LiF was experimentally determined to be 6.4D. Assuming th

Question

The covalent radii of Li & F are 123 pm & 64 pm respectively. The electronegativity of Li & F are 1.0 & 4.0 respectively. The dipole moment of LiF was experimentally determined to be 6.4D. Assuming that Schomaker and Stevenson's formula gives accurate value of Li-F bond length, calculate the percentage ionic character in LiF molecule. Round off your answer to nearest integer.

TestHub · Accepted Answer

The bond length of Li–F is calculated as $d_{L i F} = r_{L i} + r_{F} - 9∣Δ EN ∣ = 123 + 64 - 9∣4 - 1∣ = 160 p m$ .

Therefore, the Li–F bond length is $160 p m$ . The observed dipole moment $μ = 6.4 D$ .

The percentage of ionic character is calculated as:

$% i o ni cc ha r a c t er = \frac{6.4 \times 3.33 \times 1 0 ^{- 30}}{( 1.6 \times 1 0 ^{- 19} ) ( 160 \times 1 0 ^{- 12} )} \times 100 \approx 83.25%$

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