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

X e O F_{2}

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X e O F_{4}

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X e F_{2}

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X e O_{2} F_{2}

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Solution

The correct options are
A $X e O F_{2}$
C $X e F_{2}$
D $X e O_{2} F_{2}$
We had earlier seen that $X e F_{2}$ 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 $X e O F_{2}$ . 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 $X e O_{2} F_{2}$ :
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 $X e F_{2}$ structure. It is a good guess!

As for the geometry, $X e F_{2}$ , $X e O F_{2}$ and $X e O_{2} F_{2}$ all have the same trigonal bipyramidal geometry. $X e F_{2}$ has a linear shape. $X e O F_{2}$ is T - Shaped. $X e O_{2} F_{2}$ is see-saw shaped.
Earlier we had seen that the geometry of $X e F_{4}$ 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 $X e O F_{4}$ . The four F atoms line in a plane perpendicular to the Xe – O axial sigma bond. This geometry is still the same as $X e F_{4}$ but the shape will be square pyramidal!

So correct options are a c and d

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