# 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.

• There are no axial and equatorial positions here, to begin with. – Ivan Neretin Jan 26 '19 at 16:26

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