Question
Explain the hybridization in PCl₅ and SF₆. Why does phosphorus form 5 bonds while nitrogen cannot? Describe the geometry of each molecule, including bond angles and bond lengths.
(NCERT Class 11, Chemical Bonding)
Solution — Step by Step
PCl₅: P has 5 bond pairs, 0 lone pairs. Steric number = 5. This requires sp³d hybridization.
SF₆: S has 6 bond pairs, 0 lone pairs. Steric number = 6. This requires sp³d² hybridization.
Phosphorus (Z = 15) has the configuration . It has vacant 3d orbitals available. During hybridization:
One 3s + three 3p + one 3d orbital mix to form 5 sp³d hybrid orbitals.
These 5 orbitals arrange themselves in a trigonal bipyramidal geometry to minimise repulsion.
The trigonal bipyramidal shape has two distinct positions:
- 3 equatorial bonds: 120° apart in the central plane
- 2 axial bonds: 90° to the equatorial plane
Key detail: axial bonds in PCl₅ are longer (219 pm) than equatorial bonds (204 pm). Why? Axial bonds experience more repulsion from 3 equatorial pairs at 90°, while equatorial bonds face only 2 axial pairs at 90°. More repulsion → bond stretches slightly.
Sulphur (Z = 16) has . Two 3d orbitals participate:
One 3s + three 3p + two 3d orbitals mix to form 6 sp³d² hybrid orbitals.
These arrange in a perfect octahedral geometry. All 6 S-F bonds are equivalent with bond angles of exactly 90° between adjacent bonds. Bond length = 156 pm (all equal).
Nitrogen is in the second period — it has only 2s and 2p orbitals in its valence shell. There are no 2d orbitals. Without d orbitals, N cannot expand its octet beyond 4 bond pairs. This is why NF₃ exists (sp³, 3 bonds + 1 lone pair) but NF₅ does not.
The availability of low-energy d orbitals in period 3 and beyond is what allows P, S, and heavier elements to form hypervalent compounds.
Why This Works
VSEPR theory tells us that electron pairs repel each other and arrange to maximise the distance between them. For 5 pairs, the optimal arrangement is trigonal bipyramidal. For 6 pairs, it’s octahedral. Hybridization is the VBT explanation for the same geometry — mixing atomic orbitals to create hybrid orbitals pointing in the right directions.
The non-equivalence of axial vs equatorial positions in trigonal bipyramidal geometry is crucial. Lone pairs (if present) always prefer equatorial positions because they experience less 90° repulsion there. This explains the shapes of molecules like SF₄ (see-saw) and ClF₃ (T-shaped).
Alternative Method
You can predict the shape without thinking about hybridization at all — just use VSEPR:
- Count bond pairs + lone pairs around the central atom
- 5 electron pairs → trigonal bipyramidal arrangement
- 6 electron pairs → octahedral arrangement
For JEE, remember this shortcut: if the central atom has no lone pairs, the molecular geometry = electron geometry. PCl₅ (no lone pairs) = trigonal bipyramidal. SF₆ (no lone pairs) = octahedral. Lone pairs distort the shape — that’s when geometry and electron arrangement differ.
Common Mistake
Students often assume all 5 bonds in PCl₅ are identical. They are not. The 2 axial P-Cl bonds are longer and weaker than the 3 equatorial P-Cl bonds. This non-equivalence also explains why PCl₅ in the solid state actually exists as — a tetrahedral cation and an octahedral anion — rather than as discrete trigonal bipyramidal molecules.