Question

A diatomic molecule, AB, has a dipole moment of 1.2 D. If its bond distance is 1.0 Å, calculate the percentage ionic character in the bond? (1D = 3.34 x 10⁻³⁴ cm electronic charge = 1.6 x 10⁻¹⁹ C)

Answer 1

To find the percentage ionic character of the diatomic molecule AB, calculate the actual dipole moment in C·m, find the dipole moment if the bond were 100% ionic, and then divide the actual by the hypothetical limit and multiply by 100%. The AB molecule has a 25% ionic character in its bond.

Step-by-step explanation:

To calculate the percentage ionic character of the diatomic molecule AB, we first need to convert the given dipole moment from Debye (D) to coulombs-meter (C·m). We know that 1 Debye = 3.34 x 10−30 C·m. Therefore, the dipole moment of AB is 1.2 D x 3.34 x 10−30 C·m/D = 4.008 x 10−30 C·m.

Next, we calculate the bond dipole moment if AB were 100% ionic, which is the charge of an electron (1.6 x 10−19 C) multiplied by the bond distance (1.0 Å = 1.0 x 10−10 m): Mlim = 1.6 x 10−19 C x 1.0 x 10−10 m = 1.6 x 10−29 C·m.

Finally, the percentage ionic character can be found by taking the ratio of the measured dipole moment to the hypothetical 100% ionic dipole moment and multiplying by 100: Percentage Ionic Character = (Measured Dipole Moment / Limiting Dipole Moment) x 100% = (4.008 x 10−30 C·m / 1.6 x 10−29 C·m) x 100% ≈ 25%.

Therefore, the AB molecule has a 25% ionic character in its bond.