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Question

In which of the following compounds, the Fluorine atoms are placed axially in a trigonal bipyramidal geometry?

A
XeOF2
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B
XeOF4
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C
XeF2
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D
XeO2F2
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Solution

The correct option is D XeO2F2
We had earlier seen that XeF2 has a linear shape. The 3 lone pairs of electrons of central Xe atom lie on the plane perpendicular to the axial Xe – F bonds. The molecule has a trigonal bipyramidal geometry

In this molecule, if one oxygen atom takes the place of a lone pair, we get the geometry of XeOF2. Do note that Xenon has a double bond with the oxygen atom. Let us look at the ground state and excited state:

Let us look at the geometry:

Now try to deduce the shape and geometry of XeO2F2:
A logical leap is to assume that the structure would be similar to replacing two of the three lone pairs with two oxygen atoms in the XeF2 structure. It is a good guess!

As for the geometry, XeF2, XeOF2 and XeO2F2 all have the same trigonal bipyramidal geometry. XeF2 has a linear shape. XeOF2 is T - Shaped. XeO2F2 is see-saw shaped.
Earlier we had seen that the geometry of XeF4 has a square bipyramidal geometry and square planar shape:


In the above geometry, if we replace a lone pair with an Oxygen atom such that there is a double bond between Xe and O, then we get the geometry of XeOF4. The four F atoms line in a plane perpendicular to the Xe – O axial sigma bond. This geometry is still the same as XeF4 but the shape will be square pyramidal!

So correct options are a c and d

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