# Confusion regarding the Shape of XeF4 [closed]

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.

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• There are no axial and equatorial positions here, to begin with. – Ivan Neretin Jan 26 at 16:26

The shape of $$\ce{XeF4}$$ is indeed square planar as you mentioned it is so because: