Explain the hybridisation involved in PCl5 molecule.

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> [!TLDR]
> PCl5 forms via sp3d hybridisation of phosphorus, producing five equivalent hybrid orbitals arranged trigonal-bipyramidally (3 equatorial + 2 axial).
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> **Ground state of P (Z = 15):** $[Ne]\,3s^2\,3p^3$ — with the outer shell configuration $3s^2\,3p_x^1\,3p_y^1\,3p_z^1$.
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> **Excitation:** To bond with five chlorine atoms, one of the 3s electrons is promoted to the empty 3d orbital (this is energetically possible because phosphorus is in period 3, where 3d orbitals are accessible), giving the excited configuration:
> $$3s^1\,3p_x^1\,3p_y^1\,3p_z^1\,3d^1$$
> now with **five unpaired electrons**.
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> **Hybridisation:** These one 3s, three 3p, and one 3d orbital mix to form **five equivalent sp3d hybrid orbitals**, each singly occupied.
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> **Geometry:** These five sp3d hybrid orbitals overlap with the 3p orbitals of five chlorine atoms to form five P–Cl σ-bonds. The resulting geometry is **trigonal bipyramidal**:
> - 3 P–Cl bonds lie in the equatorial plane (bond angle 120° between them). …

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	$109.5^\circ$	$\text{CH}_4$

Q.Explain the hybridisation involved in PCl5 molecule.

Concept understanding — Orbital Hybridization Theory

Orbital Hybridization Theory – From Intuition to Precision

The Problem That Started It All

The Core Intuition

The Precise Statement

The Three Common Hybridizations

How It Works: The Methane Example

Q.Explain the hybridisation involved in PCl5 molecule.

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