MHTCET 23 Apr 2025 Shift 1 Paper

Concept:The number of lone pairs on the central atom $\mathrm{Xe}$ is found by subtracting the electrons used in bonding from its 8 valence electrons and dividing the remainder by 2.Explanation:The valence shell of $\mathrm{Xe}$ contains 8 electrons.Only the electrons contributed by $\mathrm{Xe}$ toward bond formation are subtracted.The remaining electrons, divided by 2, give the lone pairs on $\mathrm{Xe}$.In $\mathrm{XeF}_{2}$, two $\mathrm{Xe-F}$ single bonds use 2 electrons. So, lone pairs $= \frac{8-2}{2} = 3$.In $\mathrm{XeF}_{4}$, four $\mathrm{Xe-F}$ single bonds use 4 electrons. So, lone pairs $= \frac{8-4}{2} = 2$.In $\mathrm{XeOF}_{2}$, two $\mathrm{Xe-F}$ bonds and one $\mathrm{Xe=O}$ double bond use 4 electrons. So, lone pairs $= \frac{8-4}{2} = 2$.In $\mathrm{XeOF}_{4}$, four $\mathrm{Xe-F}$ bonds and one $\mathrm{Xe=O}$ double bond use 6 electrons. So, lone pairs $= \frac{8-6}{2} = 1$.Thus, $\mathrm{XeOF}_{4}$ has exactly one lone pair in the valence shell of $\mathrm{Xe}$.Answer:$\mathrm{XeOF}_{4}$ (Option D)