Question

For H 2 O, the H-O-H bond angle is 104.5 o and the measure dipole moment is 1.85 Debye. What is the magnitude of the O-H bond dipole moment for H 2 O? Estimate the bond angle in H 2 S given that the H-S bond dipole moment is 0.67 Debye and the resultant molecular dipole moment is 0.95 Debye.

Answer 1

Answer:

Dipole moment of H₂O: 2.26 Debye

Bond angle in H₂S: 89.7°

Explanation:

The total dipole moment(μr) for H₂O can be calculated by the sum of the dipole moments of each bond of O-H. Because the dipole moment is a vector, the sum of these vectors can be calculated by the cosine law.

μr² = (μO-H)² + (μO-H)² + 2*(μO-H)*(μO-H)*cos(104.5°)

μr² = 1.85² + 1.85² + 2*1.85*1.85*cos(104.5°)

μr² = 3.4225 + 3.4225 + 6.845*(-0.2504)

μr² = 5.13115

μr = √5.13115

μr = 2.26 Debye.

For H₂S:

0.95² = 0.67² + 0.67² + 2*0.67*0.67*cosθ

0.9025 = 0.4489 + 0.4489 + 0.8978cosθ

0.8978cosθ = 0.0047

cosθ = 0.00523

θ = arcos0.00523

θ = 89.7°