Chemistry · Chemical Bonding

JEE Main 2025 — 2 April, Morning Shift — Question 8

A molecule with the formula $A X_{4} Y$ has all its elements from p-block. Element $A$ is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among $A$ , $X$ and $Y$ . Elements $X$ and $Y$ have first and second highest electronegativity values respectively among all the known elements. The shape of the molecule is:

Option A:
Trigonal bipyramidal
Option B:
Square pyramidal
Correct
Option C:
Octahedral
Option D:
Pentagonal planar

Answer: B

Step-by-step solution

A is xenon

$X$ is fluorine

Y is oxygen

$\mathrm{XeOF}_{4}$

Shape: Square pyramidal

Hybridisation: $s p^{3} d^{2}$

Answer key and solution verified before publishing.

Practise Chemical Bonding

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Chemistry
Chapter: Chemical Bonding
Topic: Hybridization and VSEPR

Among SO 2, NF 3, NH 3, XeF 2, CIF 3 and SF 4 , the hybridization of the molecule with non-zero dipole moment and highest number of…

Answer: B

Step-by-step solution

Practise Chemical Bonding

More Chemical Bonding questions from this paper