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Question

(A) The resultant dipole moment $\mu$ of a triatomic molecule $XY_2$ is given by the formula: $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos 	heta

Vedclass · Accepted Answer

$	heta = 90^{\circ}$

Vedclass · Answer

(A) The resultant dipole moment $\mu$ of a triatomic molecule $XY_2$ is given by the formula: $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos 	heta}$,where $\mu_1$ and $\mu_2$ are the bond dipole moments and $	heta$ is the bond angle.Assuming $\mu_1 = \mu_2 = \mu_0$,the formula becomes $\mu = \sqrt{2\mu_0^2(1 + \cos 	heta)} = 2\mu_0 \cos(	heta/2)$.For $	heta = 180^{\circ}$,$\mu = 2\mu_0 \cos(90^{\circ}) = 0$.For $	heta = 120^{\circ}$,$\mu = 2\mu_0 \cos(60^{\circ}) = \mu_0$.For $	heta = 90^{\circ}$,$\mu = 2\mu_0 \cos(45^{\circ}) = \sqrt{2}\mu_0 \approx 1.414\mu_0$.Comparing the values,the dipole moment is maximum when the bond angle is smallest among the given options,which is $90^{\circ}$.

Which bond angle $\theta$ gives maximum dipole moment for a triatomic molecule $XY_2$?

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