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In which of the following molecules the central atom does not have sp3dsp^3d hybridization?

Hybridisation
NEET
1

BrF₅

2

XeF₂

3

ClF₃

4

SF₄

Solution:

To determine the hybridization of the central atom in each molecule, we can use the steric number (SN) method, where SN = (number of sigma bonds) + (number of lone pairs on the central atom).

  1. BrF₅ (Bromine Pentafluoride):

    • Central atom: Br (Bromine), Group 17, has 7 valence electrons.

    • Number of F atoms bonded: 5 (forming 5 sigma bonds).

    • Electrons used in bonding = 5.

    • Remaining valence electrons = 7 - 5 = 2.

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

    • Steric Number (SN) = 5 (sigma bonds) + 1 (lone pair) = 6.

    • Hybridization for SN=6 is sp3d2sp^3d^2.

  2. XeF₂ (Xenon Difluoride):

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

    • Number of F atoms bonded: 2 (forming 2 sigma bonds).

    • Electrons used in bonding = 2.

    • Remaining valence electrons = 8 - 2 = 6.

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

    • Steric Number (SN) = 2 (sigma bonds) + 3 (lone pairs) = 5.

    • Hybridization for SN=5 is sp3dsp^3d.

  3. ClF₃ (Chlorine Trifluoride):

    • Central atom: Cl (Chlorine), Group 17, has 7 valence electrons.

    • Number of F atoms bonded: 3 (forming 3 sigma bonds).

    • Electrons used in bonding = 3.

    • Remaining valence electrons = 7 - 3 = 4.

    • Number of lone pairs = 4 / 2 = 2.

    • Steric Number (SN) = 3 (sigma bonds) + 2 (lone pairs) = 5.

    • Hybridization for SN=5 is sp3dsp^3d.

  4. SF₄ (Sulfur Tetrafluoride):

    • Central atom: S (Sulfur), Group 16, has 6 valence electrons.

    • Number of F atoms bonded: 4 (forming 4 sigma bonds).

    • Electrons used in bonding = 4.

    • Remaining valence electrons = 6 - 4 = 2.

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

    • Steric Number (SN) = 4 (sigma bonds) + 1 (lone pair) = 5.

    • Hybridization for SN=5 is sp3dsp^3d.

The question asks which molecule's central atom does not have sp3dsp^3d hybridization. From the analysis, BrF₅ has sp3d2sp^3d^2 hybridization, while the others have sp3dsp^3d hybridization.

The final answer is 1\boxed{1}

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