In one question, I found that it is given $\ce{XeF2}$ and $\ce{CO2}$ have the same shape.

$\ce{CO2}$ is linear. In $\ce{XeF2}$, if the fluorine is placed in an axial position, then only shape is linear. But in most compounds, lone pairs are placed in an axial position.

Is the lone pair or the electronegative element more appropriate for placement in an axial position?


So in $\ce{XeF2}$ there will be three lone pairs on the xenon, and the xenon will have two covalent bonds to fluorines. The xenon is of course the central atom, on to that we apply VSEPR-theory. We count five electron dense areas (two bonds and three lone pairs) and so VSEPR-theory predicts a trigonal bipyramidal shape for $\ce{XeF2}$

Trigonal bipyramidal molecule

The lone pairs want to stay as far away from each other as possible (in VSEPR). Indeed would there have been just two lone pairs putting them axially would have introduced the maximal possible angle of 180°. Instead we have three lone pairs, so the equatorial positions introduce the best possible angle of 120°. Compare this to the 90° angle introduced if we had put two lone pairs axially and just one equatorial.

What is left now are the axial positions to which we allocate the fluorines, obtaining a linear molecule. In this case the VSEPR prediction has been confirmed experimentally.

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  • $\begingroup$ I dislike this explanation extremely. A much better explanation would not apply VSEPR but rather say that it is a four-electron three-centre bond with a p-orbital of xenon which means that it must follow a p-orbital’s shape namely linear. VSEPR is a simple concept meant for simple cases. $\endgroup$ – Jan Aug 28 '15 at 15:56

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