Explain hybridisation of phosphorous in the formation of PCl5.

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> [!TLDR]
> One 3s electron of phosphorus is promoted to a 3d orbital, giving five unpaired electrons that hybridize as $sp^3d$, producing a trigonal bipyramidal $	ext{PCl}_5$ molecule.
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> **Ground state electronic configuration of P (Z=15):** $[	ext{Ne}]\,3s^2\,3p_x^1\,3p_y^1\,3p_z^1$ — only 3 unpaired electrons, which would allow only 3 bonds.
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> **Excitation:** to form 5 bonds (as needed for $	ext{PCl}_5$), one electron from the filled $3s$ orbital is promoted to an empty $3d$ orbital. The excited-state configuration becomes:
> $$3s^1\,3p_x^1\,3p_y^1\,3p_z^1\,3d^1$$
> giving 5 unpaired electrons, each available to pair with an electron from a chlorine atom.
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> **Hybridisation:** the one $3s$, three $3p$, and one $3d$ orbital mix together to form five equivalent $sp^3d$ hybrid orbitals. These five hybrid orbitals overlap with the $3p$ orbitals of five chlorine atoms to form five P–Cl sigma bonds.
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> **Geometry:** the five $sp^3d$ hybrid orbitals arrange themselves in a trigonal bipyramidal shape to minimize electron-pair repulsion:
> - 3 orbitals lie in a plane (equatorial), at $120°$ to each other. …

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 hybridisation of phosphorous in the formation of PCl5.

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 hybridisation of phosphorous in the formation of PCl5.

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