Phosphorus is 3s² 3p³. Explain how it forms five bonds in PCl₅ and name the hybridisation.

Step 1. Five half-filled orbitals. One $3s$ electron is promoted to an empty $3d$ orbital: $3s^1 3p^3 3d^1$.

Step 2. Hybridisation. One $s$, three $p$ and one $d$ orbital mix into five $sp^3d$ hybrids, pointing to the corners of a trigonal bipyramid.

Step 3. Two kinds of position. Three equatorial bonds lie in a plane. Two axial bonds point above and below it.

Step 4. Bond length. Each axial bond has three bonds at $90°$ to it. Each equatorial bond has only two. The axial bond pairs feel more repulsion and sit slightly farther out.

Step 5. Reactivity. The longer axial bonds are weaker, so $\text{PCl}_5$ loses chlorine easily and is very reactive.