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JEE Advanced 2019 Paper 1 · Chemistry · Chemical Bonding and Molecular Structure · medium

Consider the species XeF4, XeO3, ICl4^- and SF4.

One or more options correct

Consider the species XeF4\mathrm{XeF_4}, XeO3\mathrm{XeO_3}, ICl4−\mathrm{ICl_4^-} and SF4\mathrm{SF_4}. The correct statement(s) is(are)
  1. A. XeF4\mathrm{XeF_4} is planar and non-polar, and the central atom carries two lone pairs.
  2. B. XeO3\mathrm{XeO_3} is trigonal planar with no lone pair on Xe\mathrm{Xe}.
  3. C. ICl4−\mathrm{ICl_4^-} is square planar, with two lone pairs on iodine in axial positions of the octahedral electron geometry.
  4. D. SF4\mathrm{SF_4} is non-polar because the four S−F\mathrm{S{-}F} bond dipoles cancel.
Show answer and solution

Answer: A, C

Use VSEPR. XeF4\mathrm{XeF_4}: 8+4=12 electrons, six pairs, octahedral arrangement with 2 lone pairs trans to each other, hence square planar and non-polar; (A) is correct. XeO3\mathrm{XeO_3}: three bond regions plus one lone pair on Xe gives a trigonal pyramid, so (B) is wrong. ICl4−\mathrm{ICl_4^-}: 7+4+1=12 electrons, six pairs, again octahedral with two trans lone pairs, i.e. square planar with the lone pairs occupying the two axial positions; (C) is correct. SF4\mathrm{SF_4}: five pairs, one lone pair in an equatorial site, see-saw shape; the axial and equatorial bond dipoles do not cancel, so the molecule is polar and (D) is wrong. Answer: A and C.
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Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake