CSIR-NET Chemical Bonding Theories Compared — VBT, MOT, CFT and LFT

By Aniket Bhardwaj · 21 September 2026 · CSIR-NET Chemistry

A large family of CSIR-NET inorganic questions is not really asking you to calculate anything. It is asking which theory accounts for a given observation — and, just as often, which theory cannot. If you know that valence bond theory has nothing to say about colour, or that crystal field theory cannot explain why neutral CO sits at the strong-field end of the spectrochemical series, you can eliminate two options before doing any work. This article puts the four theories side by side, works the numbers each one actually produces, and states plainly where each one breaks.

The four theories in one table

Theory	Core picture	Explains well	Fails at
Valence bond theory (VBT)	Hybridised metal orbitals overlap with filled ligand orbitals; the bond is a localised electron pair	Geometry, hybridisation, magnetic moment (inner vs outer orbital complexes)	Colour, spectra, the magnitude of splitting, temperature dependence of magnetism
Molecular orbital theory (MOT)	Atomic orbitals combine into delocalised molecular orbitals over the whole species	Bond order, paramagnetism of O₂, bonding in odd-electron species, spectra	Requires more work; simple diagrams assume symmetry that real molecules may not have
Crystal field theory (CFT)	Ligands are point negative charges; the d orbitals split purely electrostatically	Splitting patterns, high/low spin, CFSE, colour, Jahn–Teller distortion	Spectrochemical order, nephelauxetic effect, π-bonding, charge-transfer bands
Ligand field theory (LFT)	CFT corrected by real orbital overlap — an MO treatment of the complex	Everything CFT does, plus the spectrochemical series, backbonding, covalency	Heaviest to apply by hand; needs symmetry labels and interaction diagrams

Valence bond theory — what it still gets right

VBT survives in NET papers because it answers one question quickly: is the complex inner-orbital (using (n−1)d orbitals) or outer-orbital (using nd orbitals)? That single choice fixes the hybridisation and the number of unpaired electrons.

Worked example 1 — two cobalt(III) complexes. Co is Z = 27, so Co³⁺ is [Ar]3d⁶: six d electrons.

[Co(NH₃)₆]³⁺: ammonia forces pairing, so the six d electrons occupy three 3d orbitals as three pairs. Two 3d orbitals are freed, giving d²sp³ hybridisation — an inner-orbital octahedral complex with zero unpaired electrons, diamagnetic.

[CoF₆]³⁻: fluoride does not force pairing, so the 3d level keeps four unpaired electrons (t_2g⁴e_g² in field language). The metal must use 4d orbitals: sp³d², an outer-orbital complex with four unpaired electrons.

Spin-only moment for the second: μ = √(n(n+2)) = √(4 × 6) = √24 = 4.90 BM. (Check the arithmetic: 4 × 6 = 24, and 4.9² = 24.01, so √24 = 4.899 ≈ 4.90.)

Notice what VBT never told us: both complexes are coloured, and their colours differ. VBT has no orbital-energy gap in it, so it cannot produce an absorption wavelength at all. That is the standard "which theory fails" answer.

Molecular orbital theory — bond order is the deliverable

Bond order = (number of bonding electrons − number of antibonding electrons) ÷ 2

For second-row diatomics the filling order depends on s–p mixing, and this is the single most common slip in NET answers:

B₂, C₂, N₂ (Z ≤ 7): s–p mixing is strong, so the π_2p pair lies below σ_{2p_z}.
O₂, F₂, Ne₂ (Z > 7): mixing is weak, so σ_{2p_z} lies below the π_2p pair.

Worked example 2 — O₂, N₂ and NO.

O₂ has 8 + 8 = 16 electrons. Filling σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹ uses all 16. Bonding electrons = 2 + 2 + 2 + 2 + 2 = 10; antibonding = 2 + 2 + 1 + 1 = 6. Bond order = (10 − 6)/2 = 2, with two unpaired electrons — O₂ is paramagnetic. No Lewis structure predicts this; MOT does it in one line.

N₂ has 14 electrons; the π orbitals fill before σ2p. Bonding = 10, antibonding = 4, bond order = (10 − 4)/2 = 3, diamagnetic.

NO has 7 + 8 = 15 electrons — one more than N₂, and it goes into a π* orbital. Bonding = 10, antibonding = 5, bond order = (10 − 5)/2 = 2.5. Removing that electron gives NO⁺ with bond order 3, which is why NO⁺ has a shorter, stronger bond than NO — a favourite two-species comparison.

Crystal field theory — where the arithmetic lives

CFT replaces the ligands by point charges. In an octahedral field the five d orbitals split into a lower t_2g set (three orbitals, −0.4Δ_o each) and an upper e_g set (two orbitals, +0.6Δ_o each). The centre of gravity is preserved: 3(−0.4) + 2(+0.6) = −1.2 + 1.2 = 0.

CFSE (octahedral) = (−0.4 × n_t2g + 0.6 × n_eg) Δ_o + (extra pairs formed) × P

Here P is the mean pairing energy. Low spin is favoured when Δ_o > P; high spin when Δ_o < P.

Worked example 3 — d⁶ iron(II), both spin states.

[Fe(H₂O)₆]²⁺, high spin: t_2g⁴e_g². CFSE = (−0.4 × 4) + (0.6 × 2) = −1.6 + 1.2 = −0.4 Δ_o, with 4 unpaired electrons, μ = √(4 × 6) = 4.90 BM.

[Fe(CN)₆]⁴⁻, low spin: t_2g⁶e_g⁰. CFSE = (−0.4 × 6) + (0.6 × 0) = −2.4 Δ_o, diamagnetic. Relative to the free ion the low-spin case has formed two extra pairs, so the honest comparison is −2.4Δ_o + 2P against −0.4Δ_o; low spin wins only when 2Δ_o > 2P, i.e. Δ_o > P.

Tetrahedral fields invert the pattern — the e set (two orbitals) lies below the t₂ set (three orbitals) — and the splitting is smaller, Δ_t ≈ (4/9)Δ_o. That single factor is why tetrahedral complexes of the first transition series are almost always high spin.

Worked example 4 — CFSE of a tetrahedral d³ ion. Filling is e²t₂¹; the e orbitals are at −0.6Δ_t and t₂ at +0.4Δ_t.
CFSE = (−0.6 × 2) + (0.4 × 1) = −1.2 + 0.4 = −0.8 Δ_t.
Expressed in octahedral units this is −0.8 × (4/9)Δ_o = −0.356Δ_o, much less stabilisation than the octahedral d³ value of (−0.4 × 3) = −1.2Δ_o. This is exactly why d³ ions such as Cr³⁺ so strongly prefer octahedral geometry.

Where CFT breaks, and what LFT adds

CFT treats ligands as bare negative charges. If that were true, the spectrochemical series should follow charge and size. It does not. The observed order runs roughly

I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO

Neutral CO outranks the anions, and iodide — the largest, softest anion — sits at the weak end. A point-charge model cannot produce this. Ligand field theory can, because it treats the metal–ligand interaction as real orbital overlap:

π-donor ligands (halides, OH⁻) have filled p orbitals that interact with the metal t_2g set and push it up, shrinking Δ_o — weak field.
Pure σ-donors (NH₃, en) leave t_2g essentially non-bonding — intermediate field.
π-acceptor ligands (CN⁻, CO, NO) have empty π* orbitals that accept t_2g electron density and pull it down, enlarging Δ_o — strong field. This is backbonding, and it is invisible to CFT.

LFT also accounts for the nephelauxetic effect: the Racah parameter B measured in a complex is smaller than in the free ion, because the d electrons are spread onto the ligands. A purely electrostatic model has no mechanism for that at all, so the nephelauxetic ratio β = B_complex/B_{free ion} being less than 1 is direct evidence of covalency. Charge-transfer bands — intense LMCT and MLCT absorptions such as the deep colour of MnO₄⁻, where the metal is d⁰ and can have no d–d transition at all — are the other standard proof.

Choosing the right theory in the exam

The question asks about…	Reach for
Shape, hybridisation, inner/outer orbital	VBT
Bond order, bond length or strength trend in a diatomic	MOT
Paramagnetism of O₂, stability of O₂⁺/O₂⁻	MOT
High spin vs low spin, μ_spin-only, CFSE, Jahn–Teller	CFT
Colour of a dⁿ complex, Δ_o from λ_max	CFT (Tanabe–Sugano for detail)
Spectrochemical order, backbonding, ν(CO) shifts, nephelauxetic effect	LFT / MO treatment
Intense colour of a d⁰ or d¹⁰ species	Charge transfer — LFT, never d–d

Mistakes that cost marks

Using the wrong MO order. Putting σ_2p below π_2p for N₂ gives the right bond order by luck but the wrong HOMO, and every photoelectron-spectrum question then fails. Remember the switch happens after nitrogen.
Counting d electrons from the neutral atom. The d count belongs to the ion. Co³⁺ is d⁶, not d⁷; Fe³⁺ is d⁵, not d⁶. Strip the 4s electrons first, then the required number of 3d.
Comparing CFSE without pairing energy. −2.4Δ_o looks better than −0.4Δ_o until you add the 2P cost of the extra pairs. Only the full comparison decides the spin state.
Explaining colour with VBT. VBT contains no energy gap; any answer that uses it to justify an absorption band is wrong on principle.
Assuming Δ_t = (4/9)Δ_o is exact. It is a point-charge estimate for identical metal, ligands and bond length. Treat it as a working approximation and say so.
Calling a strong-field ligand a strong base. CN⁻ and CO sit at the top of the spectrochemical series for π-acceptor reasons, not Brønsted basicity — CO is a very poor base.

Where this appears in the paper

Exam	Typical use of these theories
CSIR-NET Chemical Sciences	Part B and Part C inorganic: spin state, μ, CFSE ordering, spectrochemical reasoning, "which theory explains" items
GATE Chemistry	Bond order of diatomics and ions, magnetic moments, MO energy ordering
IIT-JAM / CUET-PG	Hybridisation and geometry, high vs low spin, basic MO diagrams
MSc coursework	Full LFT treatment, Tanabe–Sugano diagrams, nephelauxetic series

Part A, Part B and Part C are the three sections of the CSIR-NET paper; for the current marking scheme, number of questions and negative-marking rules, always read the official notification for your session rather than any coaching summary, including this one.

The arithmetic here is short but unforgiving. CFSE sums, spin-only moments and bond orders are all a handful of multiplications, and the marks are lost to slips rather than to theory. Work each one on paper and then recompute it independently — the suite's scientific calculator is enough for that. There is no dedicated bonding-theory tool in the suite, so this button honestly opens the suite itself rather than pretending a matching calculator exists.

Open the ABC Chemistry Calculator Suite →

Preparing for CSIR-NET, GATE, IIT-JAM or CUET-PG chemistry? ABC Chemistry runs dedicated competitive-exam batches at its coaching centre and fully online for students across India — details at abcchemistry.in.