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I found on internet that the shape of $\ce{XeF4}$ is square planar but what if one place $2$ of $\ce{F}$ on the equitorial position and rest two on axial position and the two lone pair on the leftout axial position then its shape will not be square planar but will change to something else

Can someone please explain this to me.

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closed as unclear what you're asking by Mithoron, Mathew Mahindaratne, DrMoishe Pippik, Todd Minehardt, Tyberius Jan 28 at 15:17

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  • $\begingroup$ There are no axial and equatorial positions here, to begin with. $\endgroup$ – Ivan Neretin Jan 26 at 16:26
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The shape of $\ce{XeF4}$ is indeed square planar as you mentioned it is so because:

  • The lone pairs are kept in anti position to minimise repulsion between them and also from the electrons in fluorine atoms. Keeping the lone pair anti results in 4 empty positions which are filled with fluorine, giving rise to a square planar shape and octahedral geometry.

Please fell free to comment any doubt, and give the tick if you are satisfied.

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