Which bond angle theta would result in the ma | vedclass

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Question

(A) The resultant dipole moment $\mu$ of two bond dipoles $\mu_1$ and $\mu_2$ inclined at an angle $	heta$ is given by the formula $\mu = \sqrt{\mu_1

Vedclass · Accepted Answer

$	heta = 90^{\circ}$

Vedclass · Answer

(A) The resultant dipole moment $\mu$ of two bond dipoles $\mu_1$ and $\mu_2$ inclined at an angle $	heta$ is given by the formula $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos 	heta}$.For a molecule $XY_2$,the bond dipoles are equal,so $\mu_1 = \mu_2 = \mu_0$. The formula becomes $\mu = \sqrt{2\mu_0^2(1 + \cos 	heta)}$.Using the trigonometric identity $1 + \cos 	heta = 2 \cos^2(	heta/2)$,we get $\mu = \sqrt{4\mu_0^2 \cos^2(	heta/2)} = 2\mu_0 \cos(	heta/2)$.As the angle $	heta$ increases from $0^{\circ}$ to $180^{\circ}$,the value of $\cos(	heta/2)$ decreases.Therefore,the dipole moment is maximum when the angle $	heta$ is smallest among the given options,which is $90^{\circ}$.

Which bond angle $\theta$ would result in the maximum dipole moment for the triatomic molecule $XY_2$ shown below?

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