JEE Chemistry 2025 Question with Solution

Given: A molecule with formula $AX_2Y_2$ is described using periodic trends. X has the highest electronegativity, Y has the second highest electronegativity, and A is the rarest monatomic non-radioactive element of its group with the lowest ionization energy among A, X, and Y.

Find: The shape of the molecule.

From the solution working, the most electronegative element is identified as fluorine, so $X = F$. The second most electronegative element is identified as oxygen, so $Y = O$. The rarest monatomic non-radioactive p-block element of its group with lower ionization energy than $F$ and $O$ is identified as xenon, so $A = Xe$.

Thus, the molecule considered in the solution is XeOF${}_4$. This indicates a discrepancy with the printed formula $AX_2Y_2$ in the question text; the solution clearly works with XeOF${}_4$ and concludes the geometry from that identified molecule.

Apply VSEPR theory to XeOF${}_4$:

There are 5 bonded atoms around xenon and 1 lone pair on xenon.

A steric number of $6$ gives an octahedral electron-pair geometry. With one lone pair, the molecular shape becomes square pyramidal.

Therefore, the correct option is A.

Given: Use periodic trends and VSEPR theory.

Find: The shape of the molecule identified from the clues.

1. The highest electronegativity belongs to fluorine, hence $X = F$.
2. The second highest electronegativity belongs to oxygen, hence $Y = O$.
3. The rarest monatomic non-radioactive element from the p-block noble gases is xenon. Also, xenon has lower first ionization energy than oxygen and fluorine, so $A = Xe$.

So the molecule used in the solution is XeOF${}_4$.

For xenon in XeOF${}_4$:

• Xenon has $8$ valence electrons.
• It forms four bonds with fluorine and one bond domain with oxygen.
• One lone pair remains on xenon.

Hence,

which corresponds to square pyramidal molecular shape.

Therefore, the molecule has square pyramidal geometry, so the correct option is A.