MHTCET 11 Apr 2026 Shift 2 Paper

Concept:Apply VSEPR theory to determine the shape of each xenon compound from its bond pairs and lone pairs around xenon.Explanation:For $\mathrm{XeF}_{4}$:Xenon has 8 valence electrons, forms 4 $\mathrm{Xe}-\mathrm{F}$ bonds, and has 2 lone pairs.Its type is $AX_4E_2$.The electron-pair geometry is octahedral, with lone pairs opposite each other, giving a square planar shape.So, $\mathrm{XeF}_{4} \rightarrow$ square planar.For $\mathrm{XeF}_{6}$:It has 6 bond pairs and 1 lone pair, hence type $AX_6E$.The lone pair distorts the octahedral geometry, giving a distorted octahedral shape.So, $\mathrm{XeF}_{6} \rightarrow$ distorted octahedral.For $\mathrm{XeOF}_{4}$:Xenon is bonded to 1 oxygen and 4 fluorine atoms, with 1 lone pair.Therefore, the shape is square pyramidal.So, $\mathrm{XeOF}_{4} \rightarrow$ square pyramidal.For $\mathrm{XeO}_{3}$:Xenon is bonded to 3 oxygen atoms and has 1 lone pair, giving type $AX_3E$.This results in a pyramidal shape.So, $\mathrm{XeO}_{3} \rightarrow$ pyramidal.Thus, the correct matching is:$(a) \rightarrow (ii)$, $(b) \rightarrow (iii)$, $(c) \rightarrow (iv)$, $(d) \rightarrow (i)$.Answer:Option D.