Which bond angle theta gives maximum dipole m | vedclass

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Question

(A) The resultant dipole moment $\mu$ of two bond dipoles $X-Y$ inclined at an angle $	heta$ is given by the formula: $\mu = \sqrt{\mu_1^2 + \mu_2^2

Vedclass · Accepted Answer

$	heta = 90^{\circ}$

Vedclass · Answer

(A) The resultant dipole moment $\mu$ of two bond dipoles $X-Y$ inclined at an angle $	heta$ is given by the formula: $\mu = \sqrt{\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2 \cos 	heta}$.Since $\mu_1 = \mu_2 = \mu_0$,the expression simplifies to $\mu = \sqrt{2\mu_0^2(1 + \cos 	heta)}$.As the angle $	heta$ increases from $0^{\circ}$ to $180^{\circ}$,the value of $\cos 	heta$ decreases from $1$ to $-1$.Therefore,the resultant dipole moment $\mu$ is maximum when $\cos 	heta$ is maximum,which occurs at $	heta = 0^{\circ}$.Among the given options,$	heta = 90^{\circ}$ provides the largest value for $\cos 	heta$ (which is $0$),compared to $120^{\circ}$ $(-0.5)$ and $180^{\circ}$ $(-1)$.Thus,the dipole moment is maximum at $	heta = 90^{\circ}$.

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

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