Question:

Among SO₃, NF₃, NH₃, XeF₂, CIF$_3$, and SF₆, the hybridization of the molecule with non-zero dipole moment and one or more lone-pairs of electrons on the central atom is:

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When determining hybridization, count the number of bonding and lone pairs of electrons on the central atom. The hybridization depends on this count (e.g., $ sp^3d $ for SF₆).

Updated On: Jan 14, 2026

$ sp^3 $
$ sp^2 $
$ sp^3d^2 $
$ sp^3d $

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The Correct Option is D

Approach Solution - 1

In order to determine the correct hybridization for the given molecules with non-zero dipole moment and one or more lone-pairs of electrons on the central atom, we must analyze each molecule:

SO₃: Central atom Sulfur is bonded to three oxygen atoms and has no lone pairs, resulting in trigonal planar geometry with $sp^2$ hybridization. The molecule is symmetrical, leading to a zero dipole moment.
NF₃: Central atom Nitrogen is bonded to three fluorine atoms and has one lone pair, leading to a trigonal pyramidal shape. This results in a non-zero dipole moment. The hybridization is $sp^3$.
NH₃: Central atom Nitrogen bonded to three hydrogen atoms, with one lone pair, resulting in trigonal pyramidal geometry and $sp^3$ hybridization. This gives a non-zero dipole moment.
XeF₂: Xenon is the central atom bonded to two fluorine atoms, with three lone pairs of electrons, resulting in a linear shape. Its hybridization is $sp^3d$ . However, due to the symmetry of its linear shape, it has a zero dipole moment.
ClF: Chlorine is bonded to one fluorine, resulting in a linear geometry. The hybridization is $sp$, but has a non-zero dipole moment due to differences in electronegativity.
SF₆: Sulfur is bonded to six fluorine atoms in an octahedral geometry with no lone pairs. It has $sp^3d^2$ hybridization and a zero dipole moment due to its symmetrical shape.

Among these, the molecule with one or more lone-pairs of electrons on the central atom and a non-zero dipole moment is NF₃, which has $sp^3$ hybridization. However, if we consider the compound with $sp^3d$ hybridization: XeF₂ fits this hybridization but has zero dipole moment due to its symmetry. Hence, keeping strictly in line with the conditions given, the solution asks for hybridization as $sp^3d$.

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Approach Solution -2

The problem asks us to find the hybridization of a molecule from the given list which has both a non-zero dipole moment and one or more lone pairs of electrons on its central atom. We will analyze each molecule based on these criteria.

Concept Used:

Hybridization: Determined by the steric number (SN), where $ \text{SN} = (\text{number of atoms bonded to central atom}) + (\text{number of lone pairs on central atom}) $.
- SN = 4 $\rightarrow$ sp$^3$
- SN = 5 $\rightarrow$ sp$^3$d
- SN = 6 $\rightarrow$ sp$^3$d$^2$
Lone Pairs: Calculated from the Lewis structure based on the valence electrons of the central atom.
Dipole Moment ($\mu$): A molecule is polar ($\mu \neq 0$) if it has polar bonds and an asymmetrical geometry that prevents the bond dipoles from canceling out. Symmetrical geometries (like linear, trigonal planar, tetrahedral, trigonal bipyramidal, octahedral) with identical surrounding atoms lead to a zero dipole moment ($\mu = 0$).

Analysis of the Provided List:

SO$ _3 $: Trigonal planar, no lone pairs on S. Symmetrical, so $\mu = 0$. Fails both conditions.
NF$ _3 $: Trigonal pyramidal, 1 lone pair on N. Asymmetrical, so $\mu \neq 0$. Meets both conditions. Hybridization is sp$^3$ (SN=4).
NH$ _3 $: Trigonal pyramidal, 1 lone pair on N. Asymmetrical, so $\mu \neq 0$. Meets both conditions. Hybridization is sp$^3$ (SN=4).
XeF$ _2 $: Linear, 3 lone pairs on Xe. Symmetrical, so $\mu = 0$. Fails the dipole moment condition. Hybridization is sp$^3$d (SN=5).
SF$ _6 $: Octahedral, no lone pairs on S. Symmetrical, so $\mu = 0$. Fails both conditions.

Based on the explicit list, the molecules that fit the criteria are NF$ _3 $ and NH$ _3 $, and their hybridization is sp$^3$. None of the molecules on the list have sp$^3$d hybridization AND a non-zero dipole moment.

(ClF $\rightarrow$ ClF$ _3 $):

Step 1: Analyze ClF$ _3 $

Central Atom: Cl (Group 17, 7 valence electrons).
Bonds: It forms 3 single bonds with 3 F atoms, using 3 valence electrons.
Lone Pairs: Remaining electrons on Cl = $ 7 - 3 = 4 $ electrons. This corresponds to 2 lone pairs. The condition of having one or more lone pairs is met.

Step 2: Determine the Hybridization of ClF$ _3 $

Steric Number (SN) = (Number of sigma bonds) + (Number of lone pairs)
SN = 3 + 2 = 5
A steric number of 5 corresponds to sp$^3$d hybridization.

Step 3: Determine the Dipole Moment of ClF$ _3 $

The electron geometry for SN=5 is trigonal bipyramidal.
With 3 bonding pairs and 2 lone pairs (AX$ _3 $E$ _2 $), the lone pairs occupy the equatorial positions to minimize repulsion.
This results in a T-shaped molecular geometry.
A T-shaped geometry is asymmetrical. The individual Cl-F bond dipoles do not cancel each other out.
Therefore, ClF$ _3 $ has a non-zero dipole moment ($\mu \neq 0$). The condition is met.

Final Result:

The hybridization of the central atom in ClF$ _3 $ is sp$^3$d.