Crystal Field Theory & Spectrochemical Series

Suggested timing. Almost every question here reduces to one chain: oxidation state → d-electron count → compare Δ_o with P → write t_2g/e_g → read off unpaired electrons or CFSE. Target 40–45 seconds per question, about 22 minutes for the set. Writing the configuration explicitly is faster than trying to hold it in your head, and it prevents most CFSE sign errors.

Log your attempt before opening the solutions. Record each wrong answer against one of three causes — this is more useful than the raw score.

Section	Q	Correct	Didn't know	Misread	Too slow
Splitting	01–08
Spectrochemical	09–15
High / low spin	16–23
CFSE & limits	24–30

Crystal Field Splitting

08 Q

[ 01 ]^A NEET styleeg set

In an octahedral crystal field, the d orbitals that constitute the e_g set are:

(a)d_xy and d_yz
(b)d_xy, d_yz and d_xz
(c)d_x²−y² and d_z²
(d)d_xz and d_z²

Answer — (c) d_x²−y² and d_z²

Given

Six ligands approaching the metal along the ±x, ±y and ±z axes.

Asked

The two orbitals forming the higher-energy e_g set.

Concept

Repulsion between metal d electrons and ligand electrons is greatest when the orbital lobes point directly at the incoming ligands. Those orbitals are raised in energy and form the e_g set.

Formula

Axial orbitals → e_g (raised); inter-axial orbitals → t_2g (lowered)

Baby steps

The ligands approach along the three Cartesian axes.
d_x²−y² has lobes along x and y; d_z² has its main lobes along z → head-on repulsion.
d_xy, d_yz and d_xz point between the axes → less repulsion → lowered.
So the e_g set is d_x²−y² and d_z².

ShortcutRead the subscript: names containing only axis symbols (x²−y², z²) are e_g; two-letter names (xy, yz, xz) are t_2g. In a tetrahedral field the labels swap round.

[ 02 ]^A NEET styleBarycentre

Relative to the barycentre, the energies of the e_g and t_2g orbitals in an octahedral field change by:

(a)+0.6Δ_o and −0.4Δ_o
(b)+0.4Δ_o and −0.6Δ_o
(c)+0.5Δ_o and −0.5Δ_o
(d)+Δ_o and −Δ_o

Answer — (a) +0.6Δ_o and −0.4Δ_o

Given

Octahedral splitting of the five degenerate d orbitals into t_2g and e_g.

Asked

Energy change of each set relative to the barycentre.

Concept

The barycentre (centre of gravity of energy) is conserved: the total energy gained by the two e_g orbitals must exactly equal the total energy lost by the three t_2g orbitals.

Formula

2 × (0.6Δ_o) = 3 × (0.4Δ_o) and x + y = Δ_o

Baby steps

Let e_g rise by x and t_2g fall by y, with x + y = Δ_o.
Barycentre conservation: 2x = 3y.
Solving: x = 0.6Δ_o, y = 0.4Δ_o.
So e_g is raised by 3/5 Δ_o and t_2g lowered by 2/5 Δ_o.

ShortcutTwo orbitals up by 3/5, three orbitals down by 2/5 — the numbers cross over. That crossing is the memory hook.

[ 03 ]^A NEET styleNumerical

For a certain octahedral complex Δ_o = 200 kJ mol⁻¹. By how much is each e_g orbital raised above the barycentre?

(a)80 kJ mol⁻¹
(b)100 kJ mol⁻¹
(c)120 kJ mol⁻¹
(d)200 kJ mol⁻¹

Answer — (c) 120 kJ mol⁻¹

Given

Δ_o = 200 kJ mol⁻¹ for an octahedral complex.

Asked

Energy rise of each e_g orbital.

Concept

Each e_g orbital sits 0.6Δ_o above the barycentre and each t_2g orbital 0.4Δ_o below it, so that the total energy is unchanged.

Formula

E(e_g) = +0.6Δ_o; E(t_2g) = −0.4Δ_o

Baby steps

Rise of each e_g orbital = 0.6 × Δ_o.
= 0.6 × 200 = 120 kJ mol⁻¹.
Check the balance: each t_2g falls by 0.4 × 200 = 80 kJ mol⁻¹.
Total up = 2(120) = 240; total down = 3(80) = 240. Barycentre preserved.

ShortcutAlways verify with the balance check 2 × (0.6Δ) = 3 × (0.4Δ). If the two totals disagree, an arithmetic slip has crept in.

[ 04 ]^A NEET styleTetrahedral

For the same metal, the same ligands and the same metal–ligand distance, the tetrahedral and octahedral splittings are related by:

(a)Δ_t = (4/9)Δ_o
(b)Δ_t = (9/4)Δ_o
(c)Δ_t = (2/3)Δ_o
(d)Δ_t = 2Δ_o

Answer — (a) Δ_t = (4/9)Δ_o

Given

Identical metal, ligands and bond distance in two geometries.

Asked

Relation between Δ_t and Δ_o.

Concept

A tetrahedral field is much weaker because there are only four ligands instead of six, and none of them points directly at any d orbital. The splitting pattern is also inverted.

Formula

Δ_t = (4/9)Δ_o ≈ 0.44Δ_o

Baby steps

Factor 2/3 arises from having four ligands rather than six.
A further factor 2/3 arises because no ligand lies along a d orbital lobe.
Net factor = (2/3) × (2/3) = 4/9.
So Δ_t is roughly 44% of Δ_o — always considerably smaller.

ShortcutThe consequence matters more than the fraction: because Δ_t is small, tetrahedral complexes are essentially always high spin.

[ 05 ]^A NEET styleTetrahedral

In a tetrahedral crystal field, the splitting pattern compared with the octahedral case is:

(a)Identical, with t_2g below e_g
(b)There is no splitting at all
(c)Inverted, with the t₂ set below the e set
(d)Inverted, with the e set below the t₂ set

Answer — (d) Inverted, with the e set below the t₂ set

Given

Four ligands at the alternate corners of a cube around the metal.

Asked

Order of the split d orbital sets.

Concept

In a tetrahedron the ligands approach between the axes, so the inter-axial orbitals (d_xy, d_yz, d_xz) now suffer more repulsion and are raised. The pattern is therefore the inverse of the octahedral one.

Formula

Tetrahedral: e (lower, 2 orbitals) < t₂ (higher, 3 orbitals)

Baby steps

Tetrahedral ligands do not lie on the Cartesian axes.
They approach closer to the lobes of d_xy, d_yz and d_xz.
Those three are raised and form the t₂ set.
d_x²−y² and d_z² are lowered and form the e set → pattern inverted.

ShortcutOctahedral: 3 below, 2 above. Tetrahedral: 2 below, 3 above. The set with fewer orbitals is always the one nearer the ligands.

[ 06 ]^A NEET styleNotation

The subscript ‘g’ is not used with the energy levels of tetrahedral complexes because:

(a)Tetrahedral complexes are always high spin
(b)Tetrahedral complexes have no d orbitals
(c)The splitting is too small to label
(d)Tetrahedral complexes have no centre of symmetry

Answer — (d) Tetrahedral complexes have no centre of symmetry

Given

Labelling of energy levels in different geometries.

Asked

Reason for dropping the ‘g’ subscript.

Concept

The subscript g stands for gerade, a symmetry label that can only be applied when the structure possesses a centre of symmetry. Octahedral and square planar complexes have one; tetrahedral complexes do not.

Formula

Octahedral / square planar → t_2g, e_g. Tetrahedral → t₂, e.

Baby steps

A centre of symmetry means every point has an identical point directly opposite through the centre.
An octahedron satisfies this; a tetrahedron does not.
Without a centre of symmetry, the gerade/ungerade classification is meaningless.
Hence tetrahedral levels are written e and t₂, without the g.

ShortcutIf you see t₂ and e written without subscript g, the complex is tetrahedral — and therefore high spin.

[ 07 ]^A NEET styleDegeneracy

The degeneracy of the five d orbitals of a metal ion is retained when the ion is surrounded by:

(a)Six ligands in an octahedral arrangement
(b)Four ligands in a tetrahedral arrangement
(c)A spherically symmetrical field of negative charge
(d)Four ligands in a square planar arrangement

Answer — (c) A spherically symmetrical field of negative charge

Given

Different arrangements of negative charge around a metal ion.

Asked

The arrangement that preserves degeneracy.

Concept

Degeneracy is lifted only when the field is asymmetrical. A perfectly spherical field raises all five d orbitals equally, so they remain degenerate at a higher energy.

Formula

Asymmetrical field ⇒ splitting; spherical field ⇒ uniform rise only

Baby steps

In a spherical field every direction is equivalent, so every d orbital feels identical repulsion.
All five rise in energy by the same amount → still degenerate.
In octahedral, tetrahedral and square planar fields the ligands occupy specific directions.
That asymmetry lifts the degeneracy → only option C preserves it.

ShortcutThis is the reason the barycentre exists as a reference: it represents the energy the orbitals would have in the hypothetical spherical field.

[ 08 ]^A NEET styleModel

Crystal field theory treats the metal–ligand bond as:

(a)Purely covalent
(b)A three-centre bond
(c)Partly ionic and partly covalent
(d)Purely ionic, arising from electrostatic interactions

Answer — (d) Purely ionic, arising from electrostatic interactions

Given

The basic assumption of CFT.

Asked

Nature of the metal–ligand interaction in this model.

Concept

CFT is an electrostatic model. Ligands are treated as point negative charges (for anions) or point dipoles (for neutral molecules), and the whole interaction is taken as ionic.

Formula

—

Baby steps

Anionic ligands are modelled as point charges.
Neutral ligands such as NH₃ and H₂O are modelled as point dipoles, negative end facing the metal.
The d orbital splitting arises purely from electrostatic repulsion.
Hence the bond is treated as purely ionic.

ShortcutThis assumption is also CFT's biggest weakness: ignoring covalency is why it cannot explain the spectrochemical series correctly.

Spectrochemical Series

07 Q

[ 09 ]^B NEET styleOrder

The correct order of increasing crystal field strength is:

(a)CN⁻ < NH₃ < H₂O < F⁻ < I⁻
(b)I⁻ < F⁻ < H₂O < NH₃ < CN⁻
(c)F⁻ < I⁻ < NH₃ < H₂O < CN⁻
(d)H₂O < I⁻ < F⁻ < CN⁻ < NH₃

Answer — (b) I⁻ < F⁻ < H₂O < NH₃ < CN⁻

Given

Five common ligands.

Asked

Increasing order of field strength.

Concept

The spectrochemical series is an experimentally determined order based on the absorption spectra of complexes. It cannot be predicted from ligand charge.

Formula

I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < EDTA⁴⁻ < NH₃ < en < CN⁻ < CO

Baby steps

Halides occupy the weak end, weakening down the group, so I⁻ is weakest.
F⁻ is the strongest halide but still weaker than the neutral O-donor H₂O.
N-donors (NH₃, en) are stronger than O-donors.
C-donors (CN⁻, CO) are strongest → order is option B.

ShortcutLearn the donor-atom trend rather than the whole list: I < Br < Cl < F < O < N < C. It reproduces the series almost perfectly.

[ 10 ]^B NEET styleStrongest

The strongest field ligand among the following is:

(a)H₂O
(b)NH₃
(c)CN⁻
(d)CO

Answer — (d) CO

Given

Four ligands from the spectrochemical series.

Asked

The strongest field ligand.

Concept

CO sits at the very top of the spectrochemical series because, besides donating a sigma pair, it accepts electron density into its empty π* orbitals (pi back bonding), which greatly increases the splitting.

Formula

Field strength: H₂O < NH₃ < en < CN⁻ < CO

Baby steps

H₂O is a moderate O-donor.
NH₃ is a stronger N-donor.
CN⁻ is a strong C-donor with pi acceptor ability.
CO is the strongest, being a neutral but powerful pi acceptor.

ShortcutNote the paradox: CO is neutral yet the strongest, while I⁻ is anionic yet the weakest. That reversal is exactly what CFT fails to explain.

[ 11 ]^B NEET styleLargest split

For which of the following will Δ_o be the largest?

(a)[Cr(H₂O)₆]³⁺
(b)[Cr(NH₃)₆]³⁺
(c)[Cr(CN)₆]³⁻
(d)[CrCl₆]³⁻

Answer — (c) [Cr(CN)₆]³⁻

Given

Four Cr(III) octahedral complexes with different ligands.

Asked

The largest crystal field splitting.

Concept

With the metal and its oxidation state held constant, Δ_o depends solely on the ligand's position in the spectrochemical series.

Formula

Δ_o ∝ ligand field strength

Baby steps

All four contain Cr³⁺ in an octahedral field, so only the ligand differs.
Rank the four: Cl⁻ < H₂O < NH₃ < CN⁻.
CN⁻ is the strongest of the four.
Hence [Cr(CN)₆]³⁻ has the largest Δ_o.

ShortcutSame metal + same oxidation state = the question is purely about the spectrochemical series. Rank the ligands and stop.

[ 12 ]^B NEET styleWavelength

The correct order of the wavelength of light absorbed by [Co(NH₃)₆]³⁺, [Co(CN)₆]³⁻ and [Co(H₂O)₆]³⁺ is:

(a)[Co(CN)₆]³⁻ > [Co(NH₃)₆]³⁺ > [Co(H₂O)₆]³⁺
(b)[Co(H₂O)₆]³⁺ > [Co(NH₃)₆]³⁺ > [Co(CN)₆]³⁻
(c)[Co(NH₃)₆]³⁺ > [Co(H₂O)₆]³⁺ > [Co(CN)₆]³⁻
(d)All three absorb at the same wavelength

Answer — (b) [Co(H₂O)₆]³⁺ > [Co(NH₃)₆]³⁺ > [Co(CN)₆]³⁻

Given

Three Co(III) octahedral complexes with ligands of different strength.

Asked

Decreasing order of absorbed wavelength.

Concept

A stronger ligand gives a larger Δ_o, so more energy is needed for the d–d transition, and higher energy means shorter wavelength. Wavelength order is the exact reverse of the field-strength order.

Formula

Δ_o = hc/λ ⇒ λ ∝ 1/Δ_o

Baby steps

Field strength: H₂O < NH₃ < CN⁻.
So Δ_o: H₂O complex < NH₃ complex < CN⁻ complex.
Wavelength varies inversely with Δ_o.
λ: H₂O > NH₃ > CN⁻ complex.

ShortcutWrite the spectrochemical order, then reverse it for wavelength. NCERT data confirms it: [Co(NH₃)₆]³⁺ absorbs at 475 nm but [Co(CN)₆]³⁻ at only 310 nm.

[ 13 ]^B NEET styleMetal trend

For the same ligand, the magnitude of Δ_o increases in the order:

(a)5d < 4d < 3d
(b)3d < 4d < 5d
(c)4d < 5d < 3d
(d)Δ_o is independent of the metal

Answer — (b) 3d < 4d < 5d

Given

Metals from different transition series with the same ligand.

Asked

Trend of Δ_o down a group.

Concept

Larger, more diffuse 4d and 5d orbitals extend further towards the ligands, giving stronger metal–ligand interaction and therefore much larger splitting.

Formula

Δ_o: 3d < 4d < 5d

Baby steps

Descending a group, d orbitals become larger and less shielded.
Overlap with ligand orbitals improves, so repulsion and splitting increase.
Δ_o rises by roughly 30–50% at each step.
Order = 3d < 4d < 5d.

ShortcutThe practical consequence: 4d and 5d complexes are almost always low spin, whatever the ligand, because Δ_o comfortably exceeds P.

[ 14 ]^B NEET styleCharge

Δ_o for [Fe(H₂O)₆]³⁺ compared with [Fe(H₂O)₆]²⁺ is:

(a)Larger, because the higher charge pulls the ligands closer
(b)Smaller, because Fe³⁺ has fewer d electrons
(c)The same, since the ligand is identical
(d)Zero

Answer — (a) Larger, because the higher charge pulls the ligands closer

Given

Two hexaaqua iron complexes differing only in oxidation state.

Asked

Comparison of Δ_o.

Concept

Crystal field splitting depends on both the ligand and the charge on the metal ion. A higher charge draws the ligands in closer, increasing the electrostatic interaction and hence the splitting.

Formula

Δ_o increases with the oxidation state of the metal

Baby steps

Both complexes contain the same ligand, H₂O, and the same metal.
Fe³⁺ is smaller and more highly charged than Fe²⁺.
The shorter metal–ligand distance means stronger repulsion with the d electrons.
So Δ_o is larger for the +3 ion.

ShortcutTwo factors control Δ_o: the ligand's place in the spectrochemical series, and the metal's charge and series. Check both before comparing.

[ 15 ]^B NEET styleNature

The spectrochemical series is best described as:

(a)A theoretical series derived from the charge on the ligands
(b)A series based on the sizes of the ligands
(c)An experimentally determined series based on the absorption of light by complexes
(d)A series based on the stability constants of complexes

Answer — (c) An experimentally determined series based on the absorption of light by complexes

Given

Nature of the spectrochemical series.

Asked

How the series is arrived at.

Concept

The series is built from measured absorption spectra: the wavelength absorbed gives Δ_o directly, and the ligands are then ranked by the Δ_o they produce. It is empirical, not predicted.

Formula

Δ_o = hc/λ_absorbed

Baby steps

Record the absorption spectrum of a series of complexes of the same metal.
Convert each absorption maximum into an energy, which is Δ_o.
Arrange the ligands in order of the Δ_o values obtained.
The result is the experimentally determined spectrochemical series.

ShortcutBecause it is empirical, it can contradict theory — and it does. CFT predicts anionic ligands should split most, yet they sit at the weak end.

High Spin vs Low Spin

08 Q

[ 16 ]^C NEET styleDelta vs P

A complex will be low spin when:

(a)Δ_o < P
(b)Δ_o > P
(c)Δ_o = P
(d)Δ_o = 0

Answer — (b) Δ_o > P

Given

Δ_o is the crystal field splitting and P the pairing energy.

Asked

Condition for a low spin configuration.

Concept

An electron entering the d orbitals chooses the cheaper option: pay the pairing energy P and stay in t_2g, or pay Δ_o and go up to e_g.

Formula

Δ_o > P → low spin; Δ_o < P → high spin

Baby steps

If the gap Δ_o is large, promotion to e_g is expensive.
The electron then prefers to pair up in the lower t_2g set.
Pairing reduces the number of unpaired electrons → low spin.
Strong field ligands produce exactly this situation.

ShortcutTreat it as a price comparison: the electron always takes the cheaper route. Large gap → pair up; small gap → climb.

[ 17 ]^C NEET styled4

The electronic configuration of a d⁴ ion in a weak octahedral field is:

(a)t_2g⁴e_g⁰
(b)t_2g³e_g¹
(c)t_2g²e_g²
(d)t_2g⁰e_g⁴

Answer — (b) t_2g³e_g¹

Given

d⁴ metal ion in a weak octahedral field, so Δ_o < P.

Asked

t_2g/e_g configuration.

Concept

In a weak field the splitting is smaller than the pairing energy, so the fourth electron avoids pairing by occupying the higher e_g level instead.

Formula

Δ_o < P ⇒ fill all five orbitals singly first

Baby steps

The first three electrons occupy the three t_2g orbitals singly (Hund's rule).
The fourth electron must choose: pair in t_2g (cost P) or climb to e_g (cost Δ_o).
Since Δ_o < P, climbing is cheaper.
Configuration = t_2g³e_g¹, 4 unpaired electrons, μ = 4.90 BM.

ShortcutHigh spin: fill all five orbitals singly, then pair. Low spin: fill t_2g completely, then move to e_g.

[ 18 ]^C NEET styleConfiguration

The t_2g and e_g configuration of [Fe(CN)₆]³⁻ is: (Z of Fe = 26)

(a)t_2g³e_g²
(b)t_2g⁵e_g⁰
(c)t_2g⁴e_g¹
(d)t_2g⁶e_g⁰

Answer — (b) t_2g⁵e_g⁰

Given

[Fe(CN)₆]³⁻, Z_Fe = 26, CN⁻ strong field.

Asked

Electronic configuration in the split d orbitals.

Concept

A strong field ligand gives Δ_o > P, so electrons fill the three t_2g orbitals completely, pairing as necessary, before any occupies e_g.

Formula

Strong field: fill t_2g (max 6) first

Baby steps

Oxidation state: x − 6 = −3 ⇒ Fe³⁺ = 3d⁵.
CN⁻ is strong field → low spin.
All five electrons crowd into t_2g: two pairs plus one single electron.
Configuration = t_2g⁵e_g⁰, 1 unpaired, μ = 1.73 BM.

ShortcutFor low spin ions, first fill t_2g up to six electrons, then put the remainder in e_g. Never place an electron in e_g while t_2g has a vacancy.

[ 19 ]^C NEET styleUnpaired

Using crystal field theory, the number of unpaired electrons in [Co(NH₃)₆]³⁺ is: (Z of Co = 27)

(a)4
(b)2
(c)1
(d)0

Answer — (d) 0

Given

[Co(NH₃)₆]³⁺, Z_Co = 27, NH₃ strong field for M³⁺.

Asked

Number of unpaired electrons.

Concept

For a d⁶ ion in a strong field, all six electrons occupy the three t_2g orbitals as three pairs, leaving none unpaired.

Formula

μ = √n(n+2); n = 0 ⇒ diamagnetic

Baby steps

Co³⁺ = [Ar]3d⁶.
NH₃ with a +3 metal gives Δ_o > P → low spin.
Six electrons fill t_2g completely → t_2g⁶e_g⁰.
Unpaired electrons = 0; the complex is diamagnetic.

ShortcutLow spin d⁶ is the only octahedral configuration that is completely diamagnetic. It also has the maximum CFSE, −2.4Δ_o.

[ 20 ]^C NEET styleTetrahedral

Low spin configurations are rarely observed in tetrahedral complexes because:

(a)Δ_t is small and almost always less than the pairing energy
(b)Tetrahedral complexes have no d electrons
(c)Δ_t is larger than Δ_o
(d)The pairing energy is zero in tetrahedral fields

Answer — (a) Δ_t is small and almost always less than the pairing energy

Given

Tetrahedral coordination entities.

Asked

Reason for the rarity of low spin tetrahedral complexes.

Concept

Since Δ_t = (4/9)Δ_o, the splitting energy is small. It rarely exceeds the pairing energy, so electrons prefer to spread out singly across all five orbitals.

Formula

Δ_t < P ⇒ high spin

Baby steps

Only four ligands are present, and none points directly at a d orbital.
Δ_t is therefore only about 44% of the corresponding Δ_o.
The pairing energy P almost always exceeds this small splitting.
Electrons occupy the upper t₂ set singly → high spin.

ShortcutTreat every tetrahedral complex as high spin unless a question states otherwise. It saves a step and is right virtually every time at this level.

[ 21 ]^C NEET styleComparison

[Mn(H₂O)₆]²⁺ has five unpaired electrons while [Mn(CN)₆]⁴⁻ has only one. The crystal field explanation is:

(a)H₂O gives Δ_o < P so no pairing occurs, while CN⁻ gives Δ_o > P so pairing does occur
(b)Manganese has different oxidation states in the two complexes
(c)The cyanide complex is tetrahedral
(d)H₂O produces a larger Δ_o than CN⁻

Answer — (a) H₂O gives Δ_o < P so no pairing occurs, while CN⁻ gives Δ_o > P so pairing does occur

Given

Two Mn(II) octahedral complexes, both d⁵, with different magnetic behaviour.

Asked

The CFT explanation.

Concept

The metal and oxidation state are identical, so only the ligand field strength can differ. Weak field keeps all five electrons unpaired; strong field pairs them down to one.

Formula

Δ_o vs P decides the spin state

Baby steps

Mn²⁺ = 3d⁵ in both complexes.
H₂O is weak field → Δ_o < P → t_2g³e_g² → 5 unpaired.
CN⁻ is strong field → Δ_o > P → t_2g⁵e_g⁰ → 1 unpaired.
The difference is entirely due to ligand field strength.

ShortcutSame metal, same charge, different μ ⇒ the answer is always ‘different Δ_o relative to P’. Never oxidation state.

[ 22 ]^C NEET styleWhich dn

The high spin and low spin distinction for octahedral complexes arises only for the configurations:

(a)d¹ to d³
(b)d⁴ to d⁷
(c)d⁸ to d¹⁰
(d)d¹ to d¹⁰

Answer — (b) d⁴ to d⁷

Given

Octahedral dⁿ configurations.

Asked

The range showing a high spin / low spin choice.

Concept

A choice only exists when an electron could either pair in t_2g or occupy e_g. For d¹–d³ the t_2g orbitals are still singly occupied; for d⁸–d¹⁰ the filling order is forced.

Formula

Choice exists for d⁴, d⁵, d⁶, d⁷

Baby steps

d¹–d³: electrons enter three separate t_2g orbitals — no decision to make.
d⁴: the fourth electron faces the pair-or-climb choice → two possible configurations.
This continues through d⁵, d⁶ and d⁷.
d⁸–d¹⁰: t_2g is already full, so filling is forced → only one configuration.

ShortcutAlso worth knowing: calculations show d⁴ to d⁷ entities are more stable in a strong field than in a weak one.

[ 23 ]^C NEET styleIdentification

Which of the following is a high spin complex?

(a)[Fe(CN)₆]⁴⁻
(b)[Co(NH₃)₆]³⁺
(c)[Fe(H₂O)₆]²⁺
(d)[Mn(CN)₆]³⁻

Answer — (c) [Fe(H₂O)₆]²⁺

Given

Four octahedral complexes with different ligands.

Asked

The high spin one.

Concept

High spin arises with weak field ligands, where Δ_o < P and Hund's rule prevails over pairing.

Formula

Weak field ⇒ high spin

Baby steps

CN⁻ is strong field → options A and D are low spin.
NH₃ with Co(III) is strong enough to pair → option B is low spin.
H₂O is a weak field ligand.
[Fe(H₂O)₆]²⁺ is d⁶ high spin, t_2g⁴e_g² → 4 unpaired, μ = 4.90 BM.

ShortcutScan the ligands first: halides, H₂O and OH⁻ signal high spin; CN⁻, CO and NO₂⁻ signal low spin.

CFSE & Limitations of CFT

07 Q

[ 24 ]^D NEET styleFormula

The crystal field stabilisation energy of an octahedral complex is calculated as:

(a)CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o
(b)CFSE = [0.4 n(t_2g) − 0.6 n(e_g)]Δ_o
(c)CFSE = [−0.6 n(t_2g) + 0.4 n(e_g)]Δ_o
(d)CFSE = n(t_2g) × n(e_g) × Δ_o

Answer — (a) CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o

Given

An octahedral complex with n(t_2g) and n(e_g) electrons in the two sets.

Asked

The expression for CFSE.

Concept

Each t_2g electron is stabilised by 0.4Δ_o (negative contribution) and each e_g electron is destabilised by 0.6Δ_o (positive contribution).

Formula

CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o

Baby steps

t_2g orbitals lie 0.4Δ_o below the barycentre → each electron there lowers the energy.
e_g orbitals lie 0.6Δ_o above → each electron there raises it.
Multiply each electron count by its energy shift and add.
A more negative CFSE means a more stable complex.

ShortcutMatch the coefficient to the orbital count: 3 t_2g orbitals go with 0.4, 2 e_g orbitals go with 0.6. The pairing-energy term is usually ignored at NEET level.

[ 25 ]^D NEET styleCalculation

The CFSE for a low spin d⁶ octahedral complex is:

(a)−0.4Δ_o
(b)−1.2Δ_o
(c)−1.6Δ_o
(d)−2.4Δ_o

Answer — (d) −2.4Δ_o

Given

Octahedral d⁶ ion in a strong field.

Asked

CFSE, ignoring the pairing energy term.

Concept

In a strong field, all six electrons occupy the stabilised t_2g set and none reaches e_g, so the stabilisation is maximal.

Formula

CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o

Baby steps

Low spin d⁶ → t_2g⁶e_g⁰.
n(t_2g) = 6, n(e_g) = 0.
CFSE = −0.4(6) + 0.6(0) = −2.4.
CFSE = −2.4Δ_o — the maximum possible for any octahedral configuration.

ShortcutThis maximal stabilisation is why low spin d⁶ complexes such as [Co(NH₃)₆]³⁺ and [Fe(CN)₆]⁴⁻ are so stable and kinetically inert.

[ 26 ]^D NEET styleCalculation

The CFSE of a high spin d⁵ octahedral complex is:

(a)0
(b)−0.4Δ_o
(c)−1.2Δ_o
(d)−2.0Δ_o

Answer — (a) 0

Given

High spin d⁵ octahedral ion, for example [Mn(H₂O)₆]²⁺.

Asked

CFSE.

Concept

With one electron in each of the five orbitals, the stabilisation from the three t_2g electrons is exactly cancelled by the destabilisation from the two e_g electrons.

Formula

CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o

Baby steps

High spin d⁵ → t_2g³e_g².
CFSE = −0.4(3) + 0.6(2).
= −1.2 + 1.2.
CFSE = 0 — there is no net crystal field stabilisation at all.

ShortcutUse this as an arithmetic checkpoint. High spin d⁵ and d¹⁰ both give CFSE exactly zero; if your working gives anything else, recheck the configuration.

[ 27 ]^D NEET styleCalculation

The CFSE of [Cr(H₂O)₆]³⁺ is: (Z of Cr = 24)

(a)−0.4Δ_o
(b)−0.8Δ_o
(c)−1.2Δ_o
(d)−1.6Δ_o

Answer — (c) −1.2Δ_o

Given

[Cr(H₂O)₆]³⁺, Z_Cr = 24.

Asked

Crystal field stabilisation energy.

Concept

For d¹ to d³ ions the configuration is the same whether the field is weak or strong, so the ligand strength does not need to be considered.

Formula

CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o

Baby steps

Cr³⁺ = [Ar]3d³.
Three electrons occupy the three t_2g orbitals singly → t_2g³e_g⁰.
CFSE = −0.4(3) + 0.6(0).
= −1.2Δ_o.

Shortcutd³ has the largest CFSE of any high spin configuration, which is why Cr(III) complexes are unusually stable and slow to react.

[ 28 ]^D NEET styleMaximum

Among the following, the complex with the maximum crystal field stabilisation energy is:

(a)[Mn(H₂O)₆]²⁺
(b)[Fe(H₂O)₆]²⁺
(c)[Co(NH₃)₆]³⁺
(d)[Ni(H₂O)₆]²⁺

Answer — (c) [Co(NH₃)₆]³⁺

Given

Four octahedral complexes with differing d counts and ligand strengths.

Asked

The largest CFSE in magnitude.

Concept

Compute the configuration for each, then apply the CFSE expression. Strong field low spin configurations concentrate electrons in the stabilised t_2g set and give the largest values.

Formula

CFSE = [−0.4 n(t_2g) + 0.6 n(e_g)]Δ_o

Baby steps

[Mn(H₂O)₆]²⁺: high spin d⁵ → t_2g³e_g² → CFSE = 0.
[Fe(H₂O)₆]²⁺: high spin d⁶ → t_2g⁴e_g² → −0.4Δ_o.
[Co(NH₃)₆]³⁺: low spin d⁶ → t_2g⁶e_g⁰ → −2.4Δ_o.
[Ni(H₂O)₆]²⁺: d⁸ → t_2g⁶e_g² → −1.2Δ_o.

ShortcutScan for a low spin d⁶ option first. If one is present, it wins on CFSE almost every time.

[ 29 ]^D NEET styleLimitation

A major limitation of crystal field theory is that:

(a)It predicts anionic ligands to cause the greatest splitting, whereas experimentally they lie at the weak end of the series
(b)It cannot explain the colour of coordination compounds
(c)It cannot explain the magnetic properties of coordination compounds
(d)It cannot account for the geometry of octahedral complexes

Answer — (a) It predicts anionic ligands to cause the greatest splitting, whereas experimentally they lie at the weak end of the series

Given

Shortcomings of CFT.

Asked

A genuine limitation.

Concept

Because CFT treats ligands as point charges, it follows that the most negative ligands should split the d orbitals most. Experiment contradicts this: I⁻, Br⁻ and Cl⁻ are the weakest field ligands while neutral CO is the strongest.

Formula

—

Baby steps

CFT does explain colour through d–d transitions → option B is false.
It does explain magnetism through high and low spin configurations → option C is false.
It handles geometry perfectly well → option D is false.
The anionic ligand prediction genuinely fails → option A.

ShortcutDo not confuse the two theories' failures. VBT cannot explain colour; CFT can. CFT's failures are the anionic ligand ordering and the neglect of covalency.

[ 30 ]^D NEET styleLimitation

The weaknesses of crystal field theory are addressed by:

(a)Valence bond theory
(b)The effective atomic number rule
(c)Werner's theory
(d)Ligand field theory and molecular orbital theory

Answer — (d) Ligand field theory and molecular orbital theory

Given

Theories of bonding in coordination compounds.

Asked

Which theories overcome CFT's limitations.

Concept

CFT ignores the covalent character of the metal–ligand bond. Ligand field theory (LFT) and molecular orbital theory (MOT) include orbital overlap and so explain both the spectrochemical series and covalency.

Formula

—

Baby steps

CFT assumes a purely ionic, electrostatic interaction.
This ignores overlap between metal and ligand orbitals.
LFT and MOT build the bonding from combined molecular orbitals instead.
They therefore explain pi bonding, back bonding and the true ligand ordering.

ShortcutNCERT lists four theories in order of sophistication: VBT → CFT → LFT → MOT. Only the first two are examinable, but knowing the order answers this question type.