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Match the List-I with List-II.

List-I

A. XeO₃

B. XeF₂

C. XeOF₄

D. XeF₆

List-II

I. sp³d; linear

II. sp³; pyramidal

III. sp³d³; distorted octahedral

IV. sp³d²; square pyramidal

Choose the correct answer from the options given below.

Hybridisation
NEET 2025 PYQ
1

A-II, B-I, C-IV, D-III

2

A-II, B-I, C-III, D-IV

3

A-IV, B-II, C-III, D-I

4

A-IV, B-II, C-I, D-III

Solution:

To determine the hybridization and molecular geometry for each compound, we use the VSEPR theory and calculate the steric number (number of sigma bonds + number of lone pairs) around the central atom (Xenon).

A. XeO₃:

  • Central atom: Xe (Group 18, 8 valence electrons).

  • Oxygen forms double bonds. Each double bond consists of one sigma bond and one pi bond. For steric number calculation, we count only sigma bonds.

  • Number of sigma bonds = 3 (for 3 oxygen atoms).

  • Electrons used in sigma bonding = 3 × 2 = 6 (each oxygen forms a double bond, using 2 electrons from Xe).

  • Remaining valence electrons on Xe = 8 - 6 = 2 electrons.

  • Number of lone pairs = 2 / 2 = 1 lone pair.

  • Steric number = (sigma bonds) + (lone pairs) = 3 + 1 = 4.

  • Hybridization: sp³.

  • Molecular Geometry: Trigonal pyramidal (due to 3 bond pairs and 1 lone pair).

  • Match: A-II.

B. XeF₂:

  • Central atom: Xe (8 valence electrons).

  • Number of sigma bonds = 2 (for 2 fluorine atoms, each forming a single bond).

  • Electrons used in bonding = 2 × 1 = 2.

  • Remaining valence electrons on Xe = 8 - 2 = 6 electrons.

  • Number of lone pairs = 6 / 2 = 3 lone pairs.

  • Steric number = 2 + 3 = 5.

  • Hybridization: sp³d.

  • Molecular Geometry: Linear (due to 2 bond pairs and 3 lone pairs in a trigonal bipyramidal electron geometry, where lone pairs occupy equatorial positions to minimize repulsion).

  • Match: B-I.

C. XeOF₄:

  • Central atom: Xe (8 valence electrons).

  • Number of sigma bonds = 1 (for oxygen double bond) + 4 (for 4 fluorine single bonds) = 5.

  • Electrons used in bonding = 2 (for O) + 4 (for F) = 6.

  • Remaining valence electrons on Xe = 8 - 6 = 2 electrons.

  • Number of lone pairs = 2 / 2 = 1 lone pair.

  • Steric number = 5 + 1 = 6.

  • Hybridization: sp³d².

  • Molecular Geometry: Square pyramidal (due to 5 bond pairs and 1 lone pair in an octahedral electron geometry).

  • Match: C-IV.

D. XeF₆:

  • Central atom: Xe (8 valence electrons).

  • Number of sigma bonds = 6 (for 6 fluorine single bonds).

  • Electrons used in bonding = 6 × 1 = 6.

  • Remaining valence electrons on Xe = 8 - 6 = 2 electrons.

  • Number of lone pairs = 2 / 2 = 1 lone pair.

  • Steric number = 6 + 1 = 7.

  • Hybridization: sp³d³.

  • Molecular Geometry: Distorted octahedral (due to 6 bond pairs and 1 lone pair in a pentagonal bipyramidal electron geometry, the lone pair causes distortion from a perfect octahedron).

  • Match: D-III.

Combining the matches:

A-II, B-I, C-IV, D-III.

This corresponds to option (1).

The final answer is 1\boxed{1}

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