The dipole moment of H2S is 0.95 D. If the bond angle is 97° , the S-H bond moment would be (cos 48.5 =0.662)

Question

The dipole moment of H2S is 0.95 D. If the bond angle is 97° , the S-H bond moment would be (cos 48.5 =0.662) (A) 0.95 D (B) 0.662 D (C) 48.5 D (D) 0.72 D

Tardigrade Admin · Accepted Answer

μobs=μS-HCos97° 2μS-Hcos48.5 0.95=2μS-H×0.662 μS-H=(0.95/2×0.662)=0.72D

Q. The dipole moment of $H_{2} S$ is 0.95 D. If the bond angle is $9 7^{\circ}$ , the S-H bond moment would be (cos 48.5 =0.662)

0.95 D

0.662 D

48.5 D

0.72 D

7.2 D

$μ_{o b s} = μ_{S - H} C os 9 7^{\circ} 2 μ_{S - H} cos 48.5$
$0.95 = 2 μ_{S - H} \times 0.662$
$μ_{S - H} = \frac{0.95}{2 \times 0.662} = 0.72 D$

Q. The dipole moment of H2​S is 0.95 D. If the bond angle is 97∘, the S-H bond moment would be (cos 48.5 =0.662)