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Chemical Bonding and Molecular Structure question

2022 · 25 Jul · Shift 2 · Q1
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Chemical Bonding and Molecular Structure question

2022 · 25 Jul · Shift 2 · Q1

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1

Match List I with List II:

List I
(molecule)		 List II
(hybridization ; shape)
(A) XeO 3_33​ (I) sp 3^33 d ; linear
(B) XeF 2_22​ (II) sp 3^33 ; pyramidal
(C) XeOF 4_44​ (III) sp 3^33 d 3^33 ; distorted octahedral
(D) XeF 6_66​ (IV) sp 3^33 d 2^22 ; square pyramidal

Choose the correct answer from the options given below:

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

Correct answer: A

  1. Determine hybridization and shape of each xenon compound using VSEPR.

(A) XeO3\mathrm{XeO_3}XeO3​

  • Central atom: Xe
  • Around Xe: 3 bonded atoms (3 O atoms)
  • In VSEPR, each multiple bond counts as one electron domain.
  • Xe also has 1 lone pair.
  • Total steric number =3+1=4= 3 + 1 = 4=3+1=4

So, hybridization is sp3sp^3sp3. With one lone pair and three bond pairs, shape is pyramidal.

Thus,

XeO3→sp3; pyramidal\mathrm{XeO_3} \rightarrow sp^3;\ \text{pyramidal}XeO3​→sp3; pyramidal

This matches (II).

So, A →\to→ II.

(B) XeF2\mathrm{XeF_2}XeF2​

  • Central atom: Xe
  • Bonded atoms: 2 F
  • Lone pairs on Xe: 3
  • Total steric number =2+3=5= 2 + 3 = 5=2+3=5

So, hybridization is sp3dsp^3dsp3d. Electron pair geometry is trigonal bipyramidal. Three lone pairs occupy equatorial positions, leaving two axial Xe–F bonds. Hence shape is linear.

Thus,

XeF2→sp3d; linear\mathrm{XeF_2} \rightarrow sp^3d;\ \text{linear}XeF2​→sp3d; linear

This matches (I).

So, B →\to→ I.

(C) XeOF4\mathrm{XeOF_4}XeOF4​

  • Central atom: Xe
  • Bonded atoms: 1 O and 4 F ⇒5\Rightarrow 5⇒5 bonded domains
  • Lone pairs on Xe: 1
  • Total steric number =5+1=6= 5 + 1 = 6=5+1=6

So, hybridization is sp3d2sp^3d^2sp3d2. Electron pair geometry is octahedral. With one lone pair and five bonded atoms, molecular shape is square pyramidal.

Thus,

XeOF4→sp3d2; square pyramidal\mathrm{XeOF_4} \rightarrow sp^3d^2;\ \text{square pyramidal}XeOF4​→sp3d2; square pyramidal

This matches (IV).

So, C →\to→ IV.

(D) XeF6\mathrm{XeF_6}XeF6​

  • Central atom: Xe
  • Bonded atoms: 6 F
  • Lone pairs on Xe: 1
  • Total steric number =6+1=7= 6 + 1 = 7=6+1=7

So, hybridization is sp3d3sp^3d^3sp3d3. For 7 electron domains, the arrangement is based on pentagonal bipyramidal geometry, but due to one lone pair the observed molecular shape is distorted octahedral (or monocapped octahedron / distorted structure).

Thus,

XeF6→sp3d3; distorted octahedral\mathrm{XeF_6} \rightarrow sp^3d^3;\ \text{distorted octahedral}XeF6​→sp3d3; distorted octahedral

This matches (III).

So, D →\to→ III.

  1. Final matching:
A→II,B→I,C→IV,D→IIIA\to II,\quad B\to I,\quad C\to IV,\quad D\to IIIA→II,B→I,C→IV,D→III

  1. Compare with options This corresponds to Option A.

  2. Comparison with stored answer Stored correct answer: A

Our derived answer also is A, so they agree.

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