The bonding of the HF molecule is approximately 43.6% ionic.If its equilibrium separation is 0.917 nm,then its electric dipole moment must be approximately A)3.20 $\times$ 10^-30 C · m B)6.40 $\times$ 10^-30 C · m C)8.76 $\times$ 10^-30 C · m D)11.4 $\times$ 10^-30 C · m E)13.2 $\times$ 10^-30 C · m

Question

Quizplus · Accepted Answer

The electric dipole moment (μ) can be calculated using the equation:

μ = Q × r

where Q is the magnitude of the charge separation and r is the distance between the charges.

In a covalent bond, the charge separation can be calculated as the product of the ionic character and the bond length. The ionic character of the HF molecule is 43.6%, so the charge separation is:

Q = 0.436 × 1.6 × 10^-19 C = 6.976 × 10^-20 C

The bond length is given as 0.917 nm or 0.917 × 10^-9 m.

So, μ = (6.976 × 10^-20 C) × (0.917 × 10^-9 m) = 6.40 × 10^-30 C·m, which is option B.

The Bonding of the HF Molecule Is Approximately 43 ×\times× 10-30 C · M B)6

The Bonding of the HF Molecule Is Approximately 43 $\times$ 10^-30 C · M
B)6