JEE MAIN - Chemistry (2025 - 2nd April Morning Shift - No. 8)
A molecule with the formula $\mathrm{AX}_4 \mathrm{Y}$ has all it's elements from p-block. Element A is rarest, monoatomic, non-radioactive from its group and has the lowest ionization enthalpy value among $\mathrm{A}, \mathrm{X}$ 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:
Pentagonal planar
Square pyramidal
Trigonal bipyramidal
Octahedral
Explanation
Given A is rarest, monoatomic, non-radioactive p-block element and form $\mathrm{AX}_4 \mathrm{Y}$ type of molecule.
$\therefore$ It is concluded that it is Xe
It is given the electronegativity of A is less than X & Y
It is given the electronegativity of $\mathrm{X} \& \mathrm{Y}$ is highest and second highest respectively among all element.
$\therefore \mathrm{X}$ & Y are F & O
$\therefore$ Compound is consider as $\mathrm{XeOF}_4$ with square pyramidal shape.
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