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Valence Bond Theory: Hybridisation and Geometry of Complexes

Chemistry · Coordination Compounds · NEET

Valence Bond Theory (VBT) says the metal makes empty orbitals mix together (hybridise) and the ligand lone pairs fill them. The type of hybridisation decides the shape: sp3 = tetrahedral, dsp2 = square planar, and both d2sp3 and sp3d2 = octahedral. Memory hook: count the ligands first (4 or 6), then check if the ligand is strong (pairs electrons) to pick the exact hybridisation.
Hybridisation decides the geometry (VBT)sp3Tetrahedrale.g. [Ni(CO)4]dsp2Square planare.g. [Ni(CN)4]2-d2sp3 / sp3d2Octahedrale.g. [Mn(CN)6]3-d2sp3 = strong ligand (inner, low spin) | sp3d2 = weak ligand (outer, high spin)
Each hybridisation type maps to one shape: sp3 to tetrahedral, dsp2 to square planar, and d2sp3/sp3d2 to octahedral. For coordination number 6, the ligand strength decides inner-orbital (d2sp3, low spin) versus outer-orbital (sp3d2, high spin) — both stay octahedral.

Your doubts, answered

How do I find the hybridisation of a complex step by step?

Do it in a fixed order every time. Step 1: find the oxidation state of the metal from the charge. Step 2: write the d-electron count (d-configuration) of that metal ion. Step 3: check the ligand — strong field (CN-, CO, NH3) pairs up the d electrons and frees inner d orbitals; weak field (Cl-, F-, H2O) does not pair. Step 4: count the empty orbitals the metal needs (4 for CN=4, 6 for CN=6). Step 5: name the hybridisation from which orbitals are used. This exact order is what NEET tests.

What is the difference between d2sp3 and sp3d2?

Both use 2 d + 1 s + 3 p orbitals and both give an octahedral shape, so the geometry is the SAME. The difference is WHICH d orbitals. In d2sp3 the two d orbitals are inner (3d) — these are used only when a strong ligand pairs the electrons and empties 3d. This is called an inner-orbital (low-spin) complex. In sp3d2 the two d orbitals are outer (4d) — used with weak ligands that do not pair electrons. This is an outer-orbital (high-spin) complex. Same octahedral shape, different magnetism.

Which hybridisation gives which shape? (quick table)

Remember only four. Coordination number 4: sp3 = tetrahedral; dsp2 = square planar. Coordination number 6: d2sp3 = octahedral (inner, low spin); sp3d2 = octahedral (outer, high spin). Coordination number 5 (like Fe(CO)5): sp3d = trigonal bipyramidal. For NEET, first count ligands to get 4/5/6, then decide the fine type.

Why is [Ni(CO)4] tetrahedral but [Ni(CN)4]2- is square planar?

In [Ni(CO)4] nickel is Ni(0), 3d10 — all d orbitals are full, so no d orbital is free. Only 4s and 4p are used, giving sp3 = tetrahedral. In [Ni(CN)4]2- nickel is Ni(2+), 3d8, and strong CN- pairs the electrons to free one inner 3d orbital. Now one 3d + 4s + 2 4p mix as dsp2 = square planar. Same metal, different oxidation state and ligand, so different shape.

How do I know if a ligand is strong field or weak field?

Use the short order of the spectrochemical series: weak I- < Br- < Cl- < F- < H2O < NH3 < en < CN- < CO strong. For NEET, remember CN-, CO, NH3 (and en, NO2-) are strong and force pairing (so d2sp3, low spin). Cl-, F-, H2O, Br- are weak and do not pair (so sp3d2, high spin). This one rule decides inner vs outer orbital.

What are the main limitations of VBT?

VBT gives shape and magnetism well, but it fails to explain: (1) why complexes are coloured, (2) the exact order of ligand strength (spectrochemical series), and (3) it does not give a quantitative measure of stability. For colour and detailed splitting you need Crystal Field Theory (CFT). NEET may ask 'which theory explains colour' — answer is CFT, not VBT.

⚠️ The NEET trap
✗ Marking [Mn(CN)6]3- as sp3d2 hybridised because the coordination number is 6.
✓ It is d2sp3 (inner-orbital, octahedral). CN- is a strong-field ligand, so the 3d4 electrons of Mn(III) pair up and free TWO inner 3d orbitals; these mix with 4s and 4p as d2sp3.
🧠 Coordination number 6 alone never tells you d2sp3 vs sp3d2 — always check the ligand. Strong ligand (CN-, NH3) = d2sp3; weak ligand (F-, H2O) = sp3d2. Both are octahedral.

Real NEET questions

2017

Pick out the correct statement with respect to [Mn(CN)6]3-:

A · It is sp3d2 hybridised and octahedral
B · It is sp3d2 hybridised and tetrahedral
C · It is d2sp3 hybridised and octahedral ✓
D · It is dsp2 hybridised and square planar
Solution: Each CN- is -1 and there are 6, giving -6 ligand charge; for overall -3 the metal is Mn(III). Mn is [Ar]3d5 4s2, so Mn(III) is 3d4. CN- is a strong-field ligand, so it pairs the 3d4 electrons and empties two inner 3d orbitals. These two 3d orbitals + 4s + three 4p give d2sp3 hybridisation, which is octahedral. So option C. (Option A would need a weak ligand; option D is for coordination number 4.)
2018

The geometry and magnetic behaviour of the complex [Ni(CO)4] are:

A · Square planar geometry and paramagnetic
B · Tetrahedral geometry and diamagnetic ✓
C · Square planar geometry and diamagnetic
D · Tetrahedral geometry and paramagnetic
Solution: CO is a neutral ligand, so nickel is Ni(0). Ni is [Ar]3d8 4s2; in Ni(0) the electrons rearrange to 3d10 4s0 under strong-field CO. With a full 3d10 set, all d electrons are paired, so the complex is diamagnetic. The empty 4s and three 4p orbitals hybridise as sp3, giving tetrahedral geometry. So option B.
2026

Match List-I (Complex/ion) with List-II (Shape/geometry): (a) [PtCl2(NH3)2] (b) [Co(NH3)6]Cl3 (c) [NiCl4]2- (d) [Fe(CO)5]; (i) Octahedral (ii) Trigonal bipyramidal (iii) Square planar (iv) Tetrahedral

A · (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)
B · (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii) ✓
C · (a)-(iv), (b)-(i), (c)-(iii), (d)-(ii)
D · (a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
Solution: Use hybridisation to get each shape. (a) [PtCl2(NH3)2]: Pt(II) 5d8, strong-field, dsp2 = square planar (iii). (b) [Co(NH3)6]3+: Co(III) 3d6, NH3 strong-field, d2sp3 = octahedral (i). (c) [NiCl4]2-: Ni(II) 3d8, Cl- weak-field, sp3 = tetrahedral (iv). (d) [Fe(CO)5]: Fe(0), sp3d = trigonal bipyramidal (ii). So (a)-iii, (b)-i, (c)-iv, (d)-ii = option B.

Solved Coordination Compounds NEET PYQs

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Frequently asked

What does Valence Bond Theory say in simple words?

It says the metal ion provides empty orbitals that mix into equal hybrid orbitals, and each ligand donates a lone pair into one of these hybrid orbitals to form a coordinate bond. The kind of hybridisation fixes the shape of the complex.

Does VBT explain why complexes are coloured?

No. VBT explains shape and magnetic behaviour but cannot explain colour or the exact ligand strength order. Colour is explained by Crystal Field Theory (CFT) through d-d transitions.

Is d2sp3 the same shape as sp3d2?

Yes, both are octahedral. The only difference is that d2sp3 uses inner (3d) orbitals with strong ligands (low spin), while sp3d2 uses outer (4d) orbitals with weak ligands (high spin).

How do I quickly get the geometry for NEET?

Count ligands first. 4 ligands means tetrahedral (sp3) or square planar (dsp2); 5 means trigonal bipyramidal (sp3d); 6 means octahedral (d2sp3 or sp3d2). Then use the ligand strength to pick the exact type.

Why is [Ni(CO)4] diamagnetic?

Because Ni is in the 0 oxidation state and becomes 3d10, so every d electron is paired. No unpaired electrons means it is diamagnetic, and sp3 hybridisation makes it tetrahedral.

Related concepts:
Inner vs Outer Orbital Complexes and Magnetic BehaviourHow to Calculate Spin-Only Magnetic MomentHigh Spin vs Low Spin: Strong and Weak Field LigandsCrystal Field Theory: d-Orbital SplittingHow to Find Coordination Number and Oxidation State
Practise Chemistry for NEET:
Chemistry MCQs (free, chapter-wise)Coordination Compounds PYQs