Explanation

Let us first determine the electron pair geometry of XeO₂F₂.

Xenon has 8 valence electrons.

In XeO₂F₂:

• Xe forms 2 Xe–O double bonds (each counts as one region)
• Xe forms 2 Xe–F single bonds
• Xe has 1 lone pair

Thus, total electron pair regions around Xe:

$$ 2 (\text{Xe–O}) + 2 (\text{Xe–F}) + 1 (\text{lone pair}) = 5 $$

So, the electron pair geometry is trigonal bipyramidal and the molecular shape is see-saw.

Statement A:
XeO₂F₂ has a see-saw shape.
This is true.

Statement B:
Xe has 5 electron pairs in its valence shell.

Electron pairs = 5 regions, but total electron pairs (bonding + lone pair electrons) are more than 5. Hence this statement is not correctly stated.
Statement B is false.

Statement C:
The two Xe=O bonds occupy equatorial positions and are separated by approximately 120°, not 180°.
So, this statement is false.

Statement D:
The two Xe–F bonds occupy axial positions, hence the F–Xe–F bond angle is close to 180°.
This statement is true.

Statement E:
Xe has 8 valence electrons, not 16. The expanded octet involves sharing but does not increase valence electron count.
Hence, Statement E is false.

Therefore, the statements which are NOT TRUE are B, C and E only.

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