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JEE Advanced 2021 Paper 1 · Chemistry · Chemical Bonding and Molecular Structure · easy

Consider the species SF4, ClF3, I3^-, XeF2, XeF4, BrF5, XeO3 and PCl5. For those species in which the central atom is sp^3d hybridised and all the…

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Consider the species SF4\mathrm{SF_4}, ClF3\mathrm{ClF_3}, I3−\mathrm{I_3^-}, XeF2\mathrm{XeF_2}, XeF4\mathrm{XeF_4}, BrF5\mathrm{BrF_5}, XeO3\mathrm{XeO_3} and PCl5\mathrm{PCl_5}. For those species in which the central atom is sp3dsp^3d hybridised and all the atoms lie in one plane (a linear species is taken to satisfy this), what is the total number of lone pairs on the central atoms?
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Answer: 8

Hybridisation and shape from VSEPR: SF4\mathrm{SF_4} is sp3dsp^3d see-saw (1 lone pair, non-planar). ClF3\mathrm{ClF_3} is sp3dsp^3d T-shaped (2 lone pairs, planar). I3−\mathrm{I_3^-} is sp3dsp^3d linear (3 lone pairs on the central I). XeF2\mathrm{XeF_2} is sp3dsp^3d linear (3 lone pairs). XeF4\mathrm{XeF_4} and BrF5\mathrm{BrF_5} are sp3d2sp^3d^2, so they are excluded; XeO3\mathrm{XeO_3} is sp3sp^3 pyramidal, excluded; PCl5\mathrm{PCl_5} is sp3dsp^3d trigonal bipyramidal with no lone pair, but it is non-planar and so excluded. The qualifying species are ClF3\mathrm{ClF_3}, I3−\mathrm{I_3^-} and XeF2\mathrm{XeF_2}, with lone pairs 2+3+3=82+3+3=8. The see-saw SF4\mathrm{SF_4} would add one lone pair if the planarity condition were ignored, giving the incorrect total of 99.
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