Inorganic Chemistry · Part 3 of 9

Valence Bond & Crystal Field Theory

Coordination Chemistry, Part 3 · 8 sections · about 20,828 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

Parts 1 and 2 described coordination compounds completely without explaining a single thing about them. This part begins the explanation. It follows the two models in the order history produced them: valence bond theory, which gets the shapes right and the colours hopelessly wrong, and crystal field theory, which throws away covalency altogether and, by that single crude assumption, explains colour, magnetism, ionic radii, hydration energies and the shape of [Ni(CN)₄]²⁻ in one stroke. By the end you can take any complex, split its d orbitals, decide whether it is high or low spin, compute its CFSE, and predict whether it will distort. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 8 sections in Part 3

1Valence bond theory — hybridisation and the dative bond Free below
2What valence bond theory cannot do
3The crystal field model and the octahedral splitting
4Tetrahedral, tetragonal and square-planar fields
5Crystal field stabilisation energy
6High spin and low spin — the Δ versus P criterion
7The spectrochemical series — of ligands and of metals
8Jahn–Teller distortion and the thermodynamic evidence

Valence bond theory — hybridisation and the dative bond

Free extract

Section E.1 of Part 3, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

The seven-step recipe that turns a formula into a hybridisation scheme, a geometry and a magnetic moment. This is still the fastest way to answer a large class of exam questions.

Valence bond theory takes the Lewis picture literally. A ligand is a Lewis base: it has a lone pair. A metal ion is a Lewis acid: it has empty orbitals. A coordinate (dative) bond forms when the lone pair goes into an empty metal orbital and is then shared between the two atoms, exactly like an ordinary covalent bond except in its origin.

Coordinate (dative) bond: a two-electron, two-centre covalent bond in which both electrons came from the same atom — the ligand donor. Once formed, it is indistinguishable from any other single bond; the distinction is one of bookkeeping, not of physics.

The empty metal orbitals cannot be used raw. An unhybridised 4s orbital is spherical and the three 4p orbitals are mutually perpendicular; nothing about that set points at the six corners of an octahedron. So Pauling did what he had already done for carbon: he hybridised them. Mixing the right number of s, p and d orbitals produces an equivalent set of directed hybrids pointing exactly where the ligands are.

Hybridisation: a mathematical recombination of atomic orbitals on one atom into an equal number of new, equivalent, directed orbitals whose lobes point towards the bonded neighbours. It is a construction on paper, not a physical process the atom undergoes.

The hybridisation schemes you must know

CN	Hybridisation	Geometry	Typical example	Comment
2	sp	linear	[Ag(NH₃)₂]⁺, [CuCl₂]⁻	Almost confined to d¹⁰ ions of Cu(I), Ag(I), Au(I), Hg(II)
3	sp²	trigonal planar	[HgI₃]⁻	Rare
4	sp³	tetrahedral	[NiCl₄]²⁻, [CoCl₄]²⁻, [Zn(NH₃)₄]²⁺	Uses the outer ns and np only
4	dsp²	square planar	[Ni(CN)₄]²⁻, [PtCl₄]²⁻, [Cu(NH₃)₄]²⁺	Uses one inner (n−1)d orbital — specifically d_x²−y²
5	dsp³ or sp³d	trigonal bipyramidal	[Fe(CO)₅]	sp³d with d_z²
5	d⁴s or sp³d	square pyramidal	[Ni(CN)₅]³⁻	The two CN 5 geometries are close in energy
6	d²sp³	octahedral (inner orbital)	[Co(NH₃)₆]³⁺, [Fe(CN)₆]³⁻	Uses two (n−1)d orbitals — these must be emptied first
6	sp³d²	octahedral (outer orbital)	[CoF₆]³⁻, [FeF₆]³⁻, [Ni(H₂O)₆]²⁺	Uses two nd orbitals, which are higher in energy and more diffuse

Note that the two CN 6 rows give the same geometry. The octahedron does not distinguish them; only the magnetic moment does. That single observation is the whole content of the inner/outer distinction, and, as E.2 will argue, it is also VBT’s deepest weakness.

The recipe

Step 1. Find the oxidation state of the metal from the overall charge and the ligand charges.
Step 2. Write the dⁿ count of the ion: group number minus oxidation state, ignoring the 4s electrons (they are lost first).
Step 3. Count donor atoms to get the coordination number.
Step 4. Draw the valence orbital boxes: (n−1)d, ns, np, and nd if you need it.
Step 5. Put the dⁿ electrons in by Hund’s rule — singly first, parallel spins.
Step 6. Ask whether enough inner d orbitals are already empty. If yes, use them. If not, decide from the measured magnetic moment whether the electrons pair up to empty them (inner orbital) or the metal reaches up to the outer nd set instead (outer orbital).
Step 7. Fill every hybrid orbital with a ligand lone pair. Count the remaining unpaired electrons and quote μ_s.o. = √[n(n + 2)] BM.

μ_s.o. = √[n(n + 2)] BM (n = number of unpaired electrons)
n = 1 → 1.73 · 2 → 2.83 · 3 → 3.87 · 4 → 4.90 · 5 → 5.92

Note: Step 6 is where the honesty is. VBT does not predict whether a complex is inner or outer orbital; it accommodates whichever answer the magnetometer gives. Hold that thought — it is the first of the four charges laid against the theory in E.2.

Worked case 1 — the two cobalt(III) complexes

The classic pair. Both are Co(III), both octahedral, both d⁶. [Co(NH₃)₆]³⁺ is diamagnetic; [CoF₆]³⁻ is paramagnetic with four unpaired electrons (μ_s.o. = 4.90 BM; the measured moment runs a little higher, for the ground-term reason given in Part 3 E.6). Same metal, same oxidation state, same geometry, completely different magnetism — and VBT can draw both.

Inner-orbital (low-spin) Co(III). The six 3d electrons are forced into three of the five 3d orbitals, which empties two of them. Those two, plus 4s and the three 4p, hybridise to d²sp³ and receive the six ammonia lone pairs. No unpaired electrons remain, so the complex is diamagnetic — which is exactly what is measured. Note the price paid: three electron pairs have been forced into single orbitals, and that costs energy.

Outer-orbital (high-spin) Co(III). The 3d electrons keep their Hund arrangement — one pair and four singles — so no 3d orbital is free. The metal instead uses the 4d orbitals: 4s + 4p + two 4d hybridise to sp³d², still octahedral. Four unpaired electrons survive, giving μ_s.o. = √24 = 4.90 BM. The 4d orbitals are large and high in energy, so these bonds are longer and weaker — and, as Part 8 shows, such complexes are kinetically labile while inner-orbital ones are inert.

Inner-orbital complex (also low spin, spin-paired, or in Pauling’s word hyperligated): the two d orbitals used in hybridisation come from the (n−1)d shell, so the d electrons must pair up to vacate them.
Outer-orbital complex (also high spin, spin-free, hypoligated): the two d orbitals come from the higher nd shell, so the d electrons keep their maximum-multiplicity arrangement.

Worked case 2 — the two nickel(II) complexes, and why shape follows magnetism

Coordination number 4 is where VBT looks best, because here the hybridisation choice changes the geometry and not merely the electron count. Ni(II) is d⁸. [NiCl₄]²⁻ has μ ≈ 2.8 BM (two unpaired) and is tetrahedral; [Ni(CN)₄]²⁻ is diamagnetic and square planar.

Tetrahedral Ni(II). Chloride is a weak-field ligand and does not force pairing, so the d⁸ configuration keeps two unpaired electrons and no 3d orbital is vacated. Only 4s and 4p are available, and four of those hybridise to sp³ — which is tetrahedral. The paramagnetism and the tetrahedral shape are the same fact seen twice.

Square-planar Ni(II). Cyanide forces the eight d electrons into four orbitals, emptying one — and the one emptied is d_x²−y², whose lobes lie along the x and y axes. That orbital plus 4s, 4p_x and 4p_y give dsp², four hybrids at 90° in the xy plane. The 4p_z orbital is left empty and perpendicular to the plane. Diamagnetism and square-planar geometry, again the same fact seen twice. This is the argument Part 2 promised: the existence of two isomers of [Pt(NH₃)₂Cl₂] proved the square plane; VBT now says why a d⁸ ion should choose it.

The four dsp² hybrids point at 0°, 90°, 180° and 270° in one plane, which is where the four cyanide carbon donors sit. Schematic — the Ni–C–N units are linear and coplanar with the metal.

⚠ Common mistakes & exam traps

Count donor atoms, never ligands. [Ni(en)₂]²⁺ is four-coordinate even though only two ligands are named; [Co(EDTA)]⁻ is six-coordinate with one.
The 4s electrons go first. Fe is [Ar]3d⁶4s², but Fe^2⁺ is 3d⁶, not 3d⁴4s². Every dⁿ count in this book is the ion’s, and for the ion the (n−1)d level lies below ns.
d²sp³ and sp³d² are both octahedral. The label does not encode the shape; it encodes which d shell was used. A question asking for the geometry of [FeF₆]³⁻ wants “octahedral”, not “sp³d²”.
Do not write sp³d² when the d orbitals are inner. The ordering of the letters is the convention that carries the meaning: d written before sp means the inner (n−1)d shell, d written after means the outer nd shell. Writing “d²sp³” for [CoF₆]³⁻ is a factual error, not a notational nicety.
Spin-only moments are a starting point, not the answer. The formula ignores any orbital contribution. It works well for first-row ions with A or E ground terms, and poorly for Co(II) and for the heavier metals. Part 6 handles this properly.
“Strong field” is not a VBT concept. VBT has no field and no Δ. If you find yourself saying “CN⁻ is a strong-field ligand so it uses inner orbitals” you have silently switched to crystal field theory — which is fine as chemistry and fatal in a question that asks you to argue within VBT.

[Fe(CN)₆]³⁻ and [FeF₆]³⁻ have measured moments of about 2.3 BM and 5.9 BM. Give the hybridisation, geometry and unpaired-electron count of each on VBT. Medium

Oxidation state and d count. CN⁻ and F⁻ are each −1; six of them with an overall −3 charge puts iron at +3. Fe(III) is d⁵.

[FeF₆]³⁻. μ_s.o. for five unpaired electrons is √35 = 5.92 BM, which matches. So the five d electrons stay in five separate 3d orbitals and none can be vacated. Hybridisation must use the outer shell: sp³d², octahedral, outer orbital, five unpaired.

[Fe(CN)₆]³⁻. 2.3 BM is close to the one-unpaired value of 1.73 BM (the excess is an orbital contribution — Part 6). One unpaired electron means the five d electrons occupy three orbitals as two pairs and one single, vacating two 3d orbitals. Hybridisation is d²sp³, octahedral, inner orbital.

The honest footnote. Nothing in this argument predicted the difference; the magnetic moments were given and the drawings were fitted to them. Ask yourself what VBT would have said if you had only been handed the two formulae. The answer is: nothing.

Easy

Predict the geometry, hybridisation and spin-only magnetic moment of [Cr(NH₃)₆]³⁺. Why is there no inner/outer ambiguity here?

Show solution

Ammonia is neutral and the charge is +3, so chromium is Cr(III) = d³. Six donors, so octahedral. Three d electrons occupy three of the five 3d orbitals singly by Hund’s rule, which leaves two 3d orbitals already empty without any pairing at all. Those two, plus 4s and 4p, give d²sp³ — inner orbital, octahedral. Three unpaired electrons remain: μ_s.o. = √15 = 3.87 BM. No ambiguity because the inner-orbital scheme costs nothing: d¹, d² and d³ ions always have at least two empty (n−1)d orbitals, so they are always inner-orbital and there is never a high/low spin choice. d⁸, d⁹ and d¹⁰ have no high/low-spin choice either, but for the mirror-image reason: every (n−1)d orbital is already occupied and none can be vacated without promoting an electron out of the d shell, so an octahedral complex of these ions is always outer-orbital, sp³d² — as the table’s [Ni(H₂O)₆]²⁺ entry shows.

Med

[Mn(CN)₆]³⁻ has μ ≈ 3.2 BM. Work out its VBT description, and compare it with [Mn(H₂O)₆]³⁺, μ ≈ 4.9 BM.

Show solution

Both are Mn(III), d⁴ (Mn is group 7; 7 − 3 = 4). [Mn(CN)₆]³⁻: 3.2 BM sits close to the two-unpaired value of 2.83 BM, so the four d electrons occupy three orbitals as one pair plus two singles, vacating two 3d orbitals — d²sp³, inner orbital, octahedral, low spin. [Mn(H₂O)₆]³⁺: 4.9 BM is the four-unpaired value, so all four d electrons stay in separate orbitals, no 3d orbital is free, and the scheme is sp³d², outer orbital, octahedral, high spin. Same metal, same oxidation state, same shape, same d count — and the only thing that decided the difference was the ligand. VBT can record that; it cannot explain it, because it has no way to say that CN⁻ is different from H₂O in any relevant respect. (Note also that high-spin d⁴ Mn(III) is the standard Jahn–Teller ion, which VBT does not anticipate either — E.8.)

Med

[Zn(NH₃)₄]²⁺ is tetrahedral while [Pt(NH₃)₄]²⁺ is square planar. Both metals are in the +2 state. Explain within VBT, and say what makes the platinum case different.

Show solution

Zn(II) is d¹⁰. Every 3d orbital is full, so none can be vacated for hybridisation at any price — emptying one would mean promoting an electron right out of the 3d shell. The only orbitals available are 4s and 4p, giving sp³ and a tetrahedron. This is why essentially all four-coordinate Zn(II), Cd(II) and Hg(II) complexes are tetrahedral, and why they are all colourless and diamagnetic. Pt(II) is d⁸. Pairing the eight electrons into four 5d orbitals frees one (d_x²−y²), giving dsp² and a square plane. What makes platinum different from nickel is that the 5d orbitals are far more spatially extended and the splitting they experience is much larger, so the pairing is worth it for every ligand: [PtCl₄]²⁻ is square planar although [NiCl₄]²⁻ is tetrahedral. Strictly this is a crystal field argument (E.4) — VBT can only say that Pt(II) ‘prefers’ dsp², which is a restatement of the observation.

Advanced / reference layer

Why exactly those orbitals and not others — the symmetry argument that VBT never made explicitly, and the two patches Pauling had to add to keep the theory alive.

Textbooks state the hybridisation schemes as though the choice of orbitals were arbitrary or empirical. It is neither. Which orbitals can combine into a set of ligand-directed hybrids is fixed entirely by symmetry, and working it out once explains all three schemes at a stroke — and, more usefully, hands you the bridge to crystal field theory.

The reducible representation of the σ framework

Take the six ligand lone-pair orbitals of an octahedral ML₆ as a basis and work out how they transform under the operations of O_h. The result is the reducible representation

Γ_σ(O_h) = a_1g + e_g + t_1u

Now read off which metal orbitals carry those same symmetry labels. The s orbital is a_1g. The three p orbitals together are t_1u. The five d orbitals split into two symmetry sets: d_z² and d_x²−y² are e_g, while d_xy, d_yz and d_xz are t_2g. Matching labels gives, with no freedom of choice:

a_1g → s · e_g → d_z², d_x²−y² · t_1u → p_x, p_y, p_z ⇒ d²sp³
The t_2g set has no σ partner at all — no combination of ligand lone pairs transforms as t_2g.

That last line is the single most important sentence in this section, and it is worth reading twice. The three t_2g orbitals are excluded from σ bonding by symmetry. Not by accident, not by energy — by geometry. They are left over. Which is precisely why, when you meet crystal field theory in E.3 and find t_2g lying below e_g, you should recognise it as the same fact wearing different clothes: the e_g pair points at the ligands and the t_2g trio does not.

The same exercise gives the other two schemes immediately:

Geometry	Point group	Γ_σ	Metal orbitals matched	Hybrid
Octahedral ML₆	O_h	a_1g + e_g + t_1u	s; d_z², d_x²−y²; p_x,y,z	d²sp³
Tetrahedral ML₄	T_d	a₁ + t₂	s; and t₂ is carried by both (p_x,y,z) and (d_xy, d_yz, d_xz)	sp³, or sd³, or any mixture
Square planar ML₄	D_4h	a_1g + b_1g + e_u	s; d_x²−y²; p_x, p_y	dsp²

Two things fall out. First, the square-planar row identifies the d orbital used as d_x²−y² specifically — the one whose lobes lie along the M–L directions — which is exactly the orbital that CFT will place highest and leave empty in a d⁸ ion. Second, the tetrahedral row shows why sp³ and sd³ are not distinguishable by symmetry in T_d: p and d(t₂) share a label, so real tetrahedral hybrids are always a mixture. Textbooks that insist tetrahedral complexes are “purely sp³” are stating a convention, not a result.

Why an inner-orbital complex should be more strongly bound

VBT gives a genuine, if qualitative, argument for the inner/outer choice, and it is worth stating because it survives into modern language. Two energies compete:

The cost: forcing d electrons to pair in the (n−1)d shell. Pairing two electrons in one orbital raises the energy by the coulombic repulsion between them and destroys some of the exchange stabilisation of the parallel-spin arrangement. In crystal field language this sum is the pairing energy P (E.6); VBT never named it.
The gain: a bond built from a compact, low-lying 3d orbital overlaps the ligand lone pair far better than one built from a diffuse 4d orbital that lies some hundreds of kJ mol⁻¹ higher. Six such bonds, each somewhat stronger, can repay the pairing cost.

This makes a real, checkable prediction, and the prediction holds: outer-orbital complexes should have longer, weaker M–L bonds and should therefore exchange their ligands faster. [CoF₆]³⁻ and [Co(H₂O)₆]³⁺ are labile; [Co(NH₃)₆]³⁺ and [Co(CN)₆]³⁻ are among the most kinetically inert species in the whole of inorganic chemistry. Part 8 shows that the modern explanation of that inertness is a crystal field one — the loss of crystal field stabilisation on reaching the transition state — but VBT got to the correlation first.

Patch one: the electroneutrality principle

A purely dative picture of [Co(NH₃)₆]³⁺ puts six electron pairs onto a Co^3⁺ ion. Taken literally that gives cobalt a formal charge of −3, which is absurd for the most electropositive atom in the molecule. Pauling patched this with the electroneutrality principle.

Electroneutrality principle (Pauling): the electron distribution in a stable molecule is such that the charge on any atom stays close to zero — conventionally within about ±1 unit. Bonds adjust their ionic character until this is satisfied.

Applied here, it says the Co–N bonds must have substantial ionic character: the donated pairs stay polarised towards nitrogen, so the real charge on cobalt is small and positive. Where a ligand can also accept electron density — CO, CN⁻, PR₃ — Pauling invoked back-donation from filled metal d orbitals into empty ligand orbitals, which relieves the build-up of negative charge. That is the π-acceptor idea, and it is correct; it is simply bolted onto VBT from outside rather than emerging from it. Part 4 derives it.

Patch two: the d-orbital contraction argument

A second, subtler problem. Free-ion 3d orbitals are compact and rather deeply buried, and a naive estimate makes their overlap with ligand lone pairs too small to build strong bonds. Pauling’s answer was that the effective size of the d orbitals is not fixed: a positive charge on the metal contracts them, and the approach of anionic ligands expands them again. The size that matters is the size in the complex. The same argument reappears, quantified and renamed, as the nephelauxetic effect in Part 5 — the experimental observation that interelectronic repulsion parameters (the Racah B) are always smaller in a complex than in the free ion, because the d electrons are spread over a larger volume.

Origins of the valence bond treatment of complexes.

J. E. Huheey, E. A. Keiter and R. L. Keiter, Inorganic Chemistry: Principles of Structure and Reactivity — the chapter on coordination chemistry gives the inner/outer orbital treatment and the hypoligated/hyperligated terminology used above.
C. E. Housecroft and A. G. Sharpe, Inorganic Chemistry — concise statement of VBT for complexes and of its failures, and the symmetry reduction of the σ framework.

Read the rest of Part 3

The remaining 7 sections of this part — What valence bond theory cannot do, The crystal field model and the octahedral splitting, Tetrahedral, tetragonal and square-planar fields, Crystal field stabilisation energy — and all nine parts of Coordination Chemistry are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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