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Q)

In a non polar $AX_2$, the bond moment of $AX$ is 0.56D then the bond angle of $AX_2$ is

$\begin{array}{1 1}(a)\;180^{\large\circ}&(b)\; 90^{\large\circ}\\(c)\;45^{\large\circ}&(d)\;30^{\large\circ}\end{array}$

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A)
Net dipole moment $\mu$=2(bond moment)$\times \cos \large\frac{\theta}{2}$
But given $AX_2$ is a non polar molecule.Hence $\mu=0$
$0=2(0.56)D\times \cos\large\frac{\theta}{2}$
$\cos\large\frac{\theta}{2}$$=0$
$\theta=180^{\large\circ}$
Hence (a) is the correct answer.
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