If compound MX_4 is non-polar and has no lone | vedclass

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(D) For a molecule $MX_4$ to be non-polar with no lone pair on the central atom,it must have a highly symmetric geometry such that the dipole moments

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(D) For a molecule $MX_4$ to be non-polar with no lone pair on the central atom,it must have a highly symmetric geometry such that the dipole moments cancel out.$1$. If the geometry is tetrahedral ($sp^3$ hybridization),the bond angle is $109^o28'$.$2$. If the geometry is square planar ($dsp^2$ hybridization),the bond angle is $90^o$.$3$. If the geometry is linear ($sp^3d^2$ or similar symmetric arrangements),the bond angle can be $180^o$.Therefore,all the given bond angles are possible depending on the specific geometry of the $MX_4$ molecule.

If compound $MX_4$ is non-polar and has no lone pair at the central atom,then the bond angle $\angle XMX$ in the compound may be:

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