Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer as compared to equatorial bonds

NCERT Exemplar · Q67

Q. Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer as compared to equatorial bonds in PCl5 whereas in SF6 both axial bonds and equatorial bonds have the same bond length. Explain.

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Concept understanding — Orbital Hybridization Theory

Orbital Hybridization Theory – From Intuition to Precision

The Problem That Started It All

Imagine you are looking at a methane molecule, $\text{CH}_4$ . Carbon has four valence electrons: two in the $2s$ orbital and two in the $2p$ orbitals. If carbon used its pure atomic orbitals to bond, you would expect two bonds from the $2s$ (identical, but one direction) and two from the $2p$ (at $90^\circ$ to each other). That would give you three different bond types and bond angles of $90^\circ$ and something else.

But experiment says methane is perfectly tetrahedral: all four bonds are identical in length, strength, and energy, and the bond angle is $109.5^\circ$ , not $90^\circ$ . Something is fundamentally wrong with the "pure orbital" picture.

This is the puzzle that hybridization theory solves.

The Core Intuition

Think of atomic orbitals as shapes that an electron can occupy. The $s$ orbital is a sphere. The $p$ orbitals are dumbbells along the $x$ , $y$ , and $z$ axes. When an atom forms bonds, it wants to mix these shapes together to create new, hybrid shapes that point in directions that maximise bond strength and minimise repulsion.

It is like mixing primary colours to get new colours. You don't have to use red, blue, and yellow separately — you can blend them to get green, orange, or purple. Similarly, an atom can blend its $s$ and $p$ orbitals to get new hybrid orbitals that are better suited for bonding.

The key insight: hybridization is a mathematical mixing of atomic orbitals on the same atom to produce an equal number of new, equivalent hybrid orbitals. The number of hybrid orbitals formed always equals the number of atomic orbitals mixed.

The Precise Statement

s

The theory rests on three pillars:

Conservation of orbitals: Mixing $n$ atomic orbitals gives exactly $n$ hybrid orbitals. No orbitals are created or destroyed.
Energy averaging: The hybrid orbitals have energies that are intermediate between the original $s$ and $p$ energies.
Directionality: Hybrid orbitals point in specific directions to minimise electron pair repulsion, which directly determines molecular shape.

The Three Common Hybridizations

Hybridization	Orbitals Mixed	Number of Hybrids	Geometry	Bond Angle	Example
$sp$	one $s$ + one $p$	2	Linear	$180^\circ$	$\text{BeCl}_2$
$sp^2$	one $s$ + two $p$	3	Trigonal planar	$120^\circ$	$\text{BF}_3$
$sp^3$	one $s$ + three $p$	4	Tetrahedral	$109.5^\circ$	$\text{CH}_4$

Note

The superscript in $sp^2$ or $sp^3$ tells you how many $p$ orbitals were mixed. $sp^3$ means one $s$ and three $p$ orbitals were blended. It does not mean there are three $s$ orbitals — there is only one $s$ orbital per shell.

How It Works: The Methane Example

Carbon in its ground state has the configuration $1s^2 2s^2 2p_x^1 2p_y^1$ . Only two unpaired electrons — it should form only two bonds. But we know carbon forms four bonds.

Step 1: Promotion. One electron from the $2s$ orbital is promoted (excited) to the empty $2p_z$ orbital. This costs a small amount of energy, but it is more than compensated by the energy released when four strong bonds form instead of two.

Step 2: Hybridization. The one $2s$ orbital and three $2p$ orbitals mix to form four equivalent $sp^3$ hybrid orbitals. Each hybrid has 25% $s$ character and 75% $p$ character.

Step 3: Bonding. Each $sp^3$ hybrid overlaps with the $1s$ orbital of a hydrogen atom, forming four identical $\sigma$ bonds. The hybrids point to the corners of a tetrahedron, giving the $109.5^\circ$ angle. …

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Phosphorus in $\ce{PCl5}$ undergoes $sp^3d$ hybridisation (trigonal bipyramidal), while sulfur in $\ce{SF6}$ uses $sp^3d^2$ hybridisation (octahedral). Axial bonds in $\ce{PCl5}$ are longer because they experience greater repulsion from three equatorial electron pairs at 90°, whereas all bonds in $\ce{SF6}$ are equivalent by symmetry.

The heart of this question lies in understanding how atomic orbitals mix to accommodate more than four electron pairs, and why geometry alone doesn't guarantee equal bond lengths.

When an atom forms more bonds than its ground-state valence electrons would suggest, it promotes electrons into higher-energy d orbitals and then hybridises. The resulting hybrid orbitals point in specific directions that minimise electron-pair repulsion. But here's the key insight: not all positions in a geometry are equivalent. The local environment around each bond—how many neighbouring electron pairs sit at 90° versus 120°—determines the repulsion that bond experiences, and repulsion lengthens bonds.

Hybridisation in $\ce{PCl5}$

Electronic configuration and promotion

Phosphorus has the ground state $[\ce{Ne}] 3s^2 3p^3$ . To form five bonds, one $3s$ electron is promoted to an empty $3d$ orbital, giving five unpaired electrons: $3s^1 3p^3 3d^1$ .
Hybridisation scheme

These five orbitals—one $3s$ , three $3p$ , and one $3d$ —mix to form five $sp^3d$ hybrid orbitals.
Geometry

The five $sp^3d$ hybrids arrange themselves in a trigonal bipyramidal geometry: three orbitals lie in an equatorial plane (120° apart), and two point along the axis perpendicular to that plane (the axial positions).
Bond-length difference

Now consider the local environment:
- Each equatorial $\ce{P-Cl}$ bond has two neighbouring electron pairs at 120° (the other two equatorial bonds) and two at 90° (the two axial bonds).
- Each axial $\ce{P-Cl}$ bond has three neighbouring electron pairs at 90° (all three equatorial bonds) and one at 180° (the other axial bond).
Repulsion is strongest at 90°. The axial bonds suffer repulsion from three equatorial pairs at 90°, whereas equatorial bonds experience only two such 90° interactions. This extra repulsion pushes the axial chlorine atoms farther from phosphorus.

d_{\text{axial}} > d_{\text{equatorial}} \quad \text{in } \ce{PCl5}

Hybridisation in $\ce{SF6}$

Electronic configuration and promotion

Sulfur starts as $[\ce{Ne}] 3s^2 3p^4$ . To form six bonds, one $3s$ and one $3p$ electron are promoted into two empty $3d$ orbitals, yielding six unpaired electrons: $3s^1 3p^3 3d^2$ .
Hybridisation scheme

One $3s$ , three $3p$ , and two $3d$ orbitals hybridise to form six $sp^3d^2$ hybrid orbitals.
Geometry

These six hybrids point toward the vertices of a regular octahedron: four bonds lie in a square plane (equatorial), and two point along the axis perpendicular to that plane (axial).
Bond-length equality

In an octahedron, every position is equivalent by symmetry: …

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Q.Describe hybridisation in the case of PCl5 and SF6. The axial bonds are longer as compared to equatorial bonds in PCl5 whereas in SF6 both axial bonds and equatorial bonds have the same bond length. Explain.

Hybridisation in PClX5\ce{PCl5}PClX5​

Hybridisation in SFX6\ce{SF6}SFX6​

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Hybridisation in $\ce{PCl5}$

Hybridisation in $\ce{SF6}$

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