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Expanded Hybridisation: sp3d, sp3d2 and sp3d3 (PCl5, SF6, XeF6)

Chemistry · Chemical Bonding · NEET

When a central atom has more than 4 electron pairs, it uses empty d-orbitals too. sp3d means 5 electron pairs (like PCl5, trigonal bipyramidal), sp3d2 means 6 pairs (like SF6, octahedral), and sp3d3 means 7 pairs (like XeF6, distorted octahedral). Memory hook: just count the number after "d" plus 4 to get the pairs — sp3d = 4+1 = 5, sp3d2 = 4+2 = 6, sp3d3 = 4+3 = 7.
Expanded Hybridisation: count the electron pairssp3d = 5 pairsPCl5 (0 lone pair)Trigonal bipyramidalsp3d2 = 6 pairsSF6 (0 lone pair)Octahedralsp3d3 = 7 pairsXeF6 (1 lone pair)Distorted octahedrallone pair
Count total electron pairs to get the hybridisation: 5 pairs = sp3d (PCl5, trigonal bipyramidal), 6 pairs = sp3d2 (SF6, octahedral), 7 pairs = sp3d3 (XeF6, distorted octahedral). XeF6 has one lone pair, which is why it reaches 7 pairs.

Your doubts, answered

What exactly does sp3d, sp3d2 and sp3d3 mean?

The name just tells you which atomic orbitals mixed to form the new hybrid orbitals. sp3d = one s + three p + one d orbital mix to give 5 hybrid orbitals. sp3d2 = one s + three p + two d = 6 hybrid orbitals. sp3d3 = one s + three p + three d = 7 hybrid orbitals. Each hybrid orbital holds one electron pair (bond pair or lone pair). So the number of hybrid orbitals equals the number of electron pairs around the central atom.

How do I quickly count the electron pairs to get the hybridisation?

Count the total electron pairs around the central atom (bond pairs + lone pairs). 2 pairs = sp, 3 = sp2, 4 = sp3, 5 = sp3d, 6 = sp3d2, 7 = sp3d3. For expanded ones just remember: 5 -> sp3d, 6 -> sp3d2, 7 -> sp3d3. Example: PCl5 has P bonded to 5 Cl and no lone pair = 5 pairs = sp3d.

Why can P, S and Xe expand their octet but C and N cannot?

Atoms in Period 3 and below (P, S, Cl, Br, I, Xe) have empty d-orbitals in their valence shell that they can use to hold extra electron pairs. Carbon and nitrogen are in Period 2 and have no d-orbitals in their valence shell, so they can never go beyond 4 pairs (sp3). That is why PCl5 exists but NCl5 does not.

What is the difference between sp3d and sp3d2?

sp3d has 5 electron pairs and a trigonal bipyramidal electron geometry (like PCl5). sp3d2 has 6 electron pairs and an octahedral electron geometry (like SF6). The difference is one extra electron pair and one extra d-orbital used.

Why is XeF6 sp3d3 and not sp3d2?

Xenon in XeF6 forms 6 bonds to fluorine, but Xe also keeps 1 lone pair. So total electron pairs = 6 bond pairs + 1 lone pair = 7 pairs = sp3d3. The extra lone pair pushes the shape into a distorted octahedral, not a perfect octahedron. This is a favourite NEET trap: XeF6 has 7 pairs, not 6.

Do lone pairs count when deciding sp3d, sp3d2 or sp3d3?

Yes. Always count bond pairs AND lone pairs to decide the hybridisation. The hybridisation is fixed by total electron pairs. But the shape (what you see) is decided only by the atoms, because lone pairs are invisible. Example: BrF5 has 5 bonds + 1 lone pair = 6 pairs = sp3d2, and its shape is square pyramidal, not octahedral.

Is the shape the same as the hybridisation geometry?

Not always. Hybridisation gives the electron-pair geometry (counting lone pairs). The molecular shape only counts atoms. When there are no lone pairs, they match (PCl5 = trigonal bipyramidal, SF6 = octahedral). When lone pairs exist, the shape is different (XeF4 is sp3d2 but square planar; BrF5 is sp3d2 but square pyramidal).

⚠️ The NEET trap
✗ XeF6 has 6 bonds, so it must be sp3d2 and octahedral like SF6.
✓ XeF6 has 6 bond pairs PLUS 1 lone pair = 7 electron pairs = sp3d3, giving a distorted octahedral shape.
🧠 Never stop at bonds — always add the lone pairs. SF6 (0 lone pair) = 6 pairs = sp3d2, but XeF6 (1 lone pair) = 7 pairs = sp3d3. This exact pair fooled students in 2025.

Real NEET questions

2021

Match List-I with List-II. List-I: (a) PCl5, (b) SF6, (c) BrF5, (d) BF3. List-II: (i) Square pyramidal, (ii) Trigonal planar, (iii) Octahedral, (iv) Trigonal bipyramidal.

A · (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
B · (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)
C · (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii) ✓
D · (a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)
Solution: Count electron pairs. PCl5: 5 bond pairs, no lone pair = sp3d = trigonal bipyramidal (iv). SF6: 6 bond pairs = sp3d2 = octahedral (iii). BrF5: 5 bonds + 1 lone pair = 6 pairs = sp3d2, shape is square pyramidal (i). BF3: 3 bonds, no lone pair = sp2 = trigonal planar (ii). So (a)-(iv), (b)-(iii), (c)-(i), (d)-(ii), option C.
2025

Match List-I with List-II. List-I: A. XeO3; B. XeF2; C. XeOF4; D. XeF6. List-II: I. sp3d; linear; II. sp3; pyramidal; III. sp3d3; distorted octahedral; IV. sp3d2; square pyramidal.

A · A-IV, B-II, C-III, D-I
B · A-IV, B-II, C-I, D-III
C · A-II, B-I, C-IV, D-III ✓
D · A-II, B-I, C-III, D-IV
Solution: XeO3: 3 bonds + 1 lone pair = 4 pairs = sp3, pyramidal (II). XeF2: 2 bonds + 3 lone pairs = 5 pairs = sp3d, linear (I). XeOF4: 5 bonds + 1 lone pair = 6 pairs = sp3d2, square pyramidal (IV). XeF6: 6 bonds + 1 lone pair = 7 pairs = sp3d3, distorted octahedral (III). So A-II, B-I, C-IV, D-III, option C. Note XeF6 is the key sp3d3 example.
2016

The correct geometry and hybridization for XeF4 are:

A · octahedral, sp3d2 ✓
B · trigonal bipyramidal, sp3d
C · planar triangle, sp3d3
D · square planar, sp3d2
Solution: Xe in XeF4 has 4 bond pairs + 2 lone pairs = 6 electron pairs = sp3d2. The electron-pair geometry is octahedral, and the molecular shape (atoms only) is square planar. The official 2016 key marked the electron-pair geometry, so the answer is octahedral, sp3d2 (option A). Both A and D describe XeF4 correctly; the key chose the electron-pair geometry.

Solved Chemical Bonding NEET PYQs

Try the real previous-year questions from this chapter — each with the answer and a full solution.

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Frequently asked

How many electron pairs are in sp3d, sp3d2 and sp3d3?

sp3d has 5 electron pairs, sp3d2 has 6, and sp3d3 has 7. Just add 4 to the number written after d: 4+1=5, 4+2=6, 4+3=7.

What is the shape of PCl5, SF6 and XeF6?

PCl5 is trigonal bipyramidal (sp3d), SF6 is octahedral (sp3d2), and XeF6 is distorted octahedral (sp3d3, because of one lone pair on Xe).

Which orbitals mix in expanded hybridisation?

sp3d uses 1 s, 3 p and 1 d orbital. sp3d2 uses 1 s, 3 p and 2 d orbitals. sp3d3 uses 1 s, 3 p and 3 d orbitals. The number of hybrid orbitals formed equals the number of pure orbitals mixed.

Why does NEET love these expanded hybridisation questions?

NEET repeats matching-type questions on PCl5, SF6, XeF4, XeF6, BrF5 and interhalogens almost every year. Knowing that hybridisation = total electron pairs, and that lone pairs change the shape, lets you solve all of them fast.

Can carbon show sp3d hybridisation?

No. Carbon is in Period 2 and has no d-orbitals in its valence shell, so it can never exceed 4 electron pairs (sp3). Only atoms from Period 3 onward, like P, S, Cl and Xe, can use d-orbitals for sp3d, sp3d2 or sp3d3.

Related concepts:
How to Find Hybridisation of Any Central Atom (Quick Trick)Hybridisation: sp, sp2 and sp3 ExplainedVSEPR Theory: How to Predict Molecular ShapeShapes of Common Molecules (Linear, Trigonal, Tetrahedral, Octahedral)Structure of Xenon and Interhalogen Compounds
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Chemical Bonding PYQs