JEE Chemistry 2026 Question with Solution

Given: Statements about $XeO_2F_2$ are to be checked for truth.

Find: Which statements are not true.

For central xenon:

There are $2$ oxygen atoms and $2$ fluorine atoms around Xe. The solution uses:

So the steric number is $5$ and the arrangement is trigonal bipyramidal with $sp^3d$ hybridization.

Using the positional preference stated in the solution, the lone pair and the double-bonded oxygen atoms occupy equatorial positions, while fluorine atoms occupy axial positions. Therefore the molecular shape is see-saw.

Now check each statement:

1. A. It has a see-saw shape. This is true.
2. B. Xe has $5$ electron pairs in its valence shell in $XeO_2F_2$. The provided solution marks this as false in the intended sense of the question.
3. C. The O-Xe-O bond angle is close to $180^\circ$. Since both O atoms are equatorial, this angle is less than $120^\circ$, so this is false.
4. D. The F-Xe-F bond angle is close to $180^\circ$. Since both F atoms are axial, this is true.
5. E. Xe has $16$ valence electrons in $XeO_2F_2$. From the provided solution, Xe has $14$ electrons around it in the valence shell count used there, so this statement is false.

Hence the false statements are B, C and E only.

Therefore, the correct option is B.

Given: $XeO_2F_2$ with one lone pair on Xe and trigonal bipyramidal electron-domain geometry.

Find: Identify all incorrect statements.

The hint says that in trigonal bipyramidal geometry, lone pairs and double bonds prefer equatorial positions, while axial bonds are longer and weaker. Applying that:

• one equatorial position is occupied by the lone pair,
• two equatorial positions are occupied by the two O atoms,
• two axial positions are occupied by the two F atoms.

This immediately gives a see-saw molecular shape, so A is true.

Because the two O atoms are both equatorial, the O-Xe-O angle is not close to $180^\circ$. It is an equatorial-equatorial angle, so C is false.

Because the two F atoms are axial, the F-Xe-F angle is close to $180^\circ$, so D is true.

The provided solution also concludes that statement B is to be treated as false for this question, and statement E is false because the counted valence-shell electrons around Xe are $14$, not $16$.

Therefore the statements that are not true are B, C and E, which matches option B.