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Chemical Bonding and Molecular Structure question

2023 · 31 Jan · Shift 1 · Q15
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Chemical Bonding and Molecular Structure question

2023 · 31 Jan · Shift 1 · Q15

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Match List I with List II

List I List II
A. XeF4\mathrm{XeF_4}XeF4​ I. See-saw
B. SF4\mathrm{SF_4}SF4​ II. Square-planar
C. NH4+\mathrm{NH_{4}^{+}}NH4+​ III. Bent T-shaped
D. BrF3\mathrm{BrF_3}BrF3​ IV. Tetrahedral

Choose the correct answer from the options given below :

  1. A
    A - II, B - I, C - III, D - IV
  2. B
    A - II, B - I, C - IV, D - III
  3. C
    A - IV, B - I, C - II, D - III
  4. D
    A - IV, B - III, C - II, D - I
View written solutionFree

Correct answer: B

  1. Determine the shape of each species using VSEPR theory

We match each compound/ion in List I with its molecular geometry in List II.

  1. Species A: XeF4\mathrm{XeF_4}XeF4​
  • Central atom: Xe
  • Xe has 8 valence electrons.
  • It forms 4 bonds with F atoms, leaving 2 lone pairs.
  • Total electron pairs around Xe = 6
  • Electron pair geometry = octahedral
  • With 2 lone pairs opposite each other, the molecular shape is square planar.

So, A→IIA \to \text{II}A→II

  1. Species B: SF4\mathrm{SF_4}SF4​
  • Central atom: S
  • S has 6 valence electrons.
  • It forms 4 bonds with F atoms and has 1 lone pair.
  • Total electron pairs around S = 5
  • Electron pair geometry = trigonal bipyramidal
  • With one equatorial lone pair, the molecular shape is see-saw.

So, B→IB \to \text{I}B→I

  1. Species C: NH4+\mathrm{NH_4^+}NH4+​
  • Central atom: N
  • In NH4+\mathrm{NH_4^+}NH4+​, nitrogen forms 4 bonds and has no lone pair.
  • Total electron pairs around N = 4
  • Molecular shape = tetrahedral.

So, C→IVC \to \text{IV}C→IV

  1. Species D: BrF3\mathrm{BrF_3}BrF3​
  • Central atom: Br
  • Br has 7 valence electrons.
  • It forms 3 bonds with F atoms and has 2 lone pairs.
  • Total electron pairs around Br = 5
  • Electron pair geometry = trigonal bipyramidal
  • Two equatorial lone pairs leave a T-shaped structure, often described here as bent T-shaped in the given list.

So, D→IIID \to \text{III}D→III

  1. Final matching

Thus the correct matching is: A−II,  B−I,  C−IV,  D−IIIA-\mathrm{II},\; B-\mathrm{I},\; C-\mathrm{IV},\; D-\mathrm{III}A−II,B−I,C−IV,D−III

This corresponds to Option B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.

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