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Coordination Compounds · Sections 5.5.2 & 5.5.5

Magnetism
& Colour

Thirty NEET-pattern questions on magnetic moments, the d4–d7 anomalies, d–d transitions and complementary colour. Each is worked through Given, Asked, Concept, Formula, Baby Steps and a Shortcut. Attempt with solutions closed, then open to check the reasoning — not just the answer.

30Questions
0701
0702
0803
0804
30Shortcuts

Suggested timing. Two short chains cover almost every question here. Magnetism: oxidation state → d count → ligand strength → n → μ. Colour: ligand strength → Δo → wavelength absorbed → complementary colour. Both should run in 30–40 seconds, about 18 minutes for the set. The commonest error is not arithmetic — it is stopping one step early and naming the absorbed colour instead of the observed one.

Log your attempt before opening the solutions. Record each wrong answer against one of three causes — this is more useful than the raw score.
SectionQCorrectDidn't knowMisreadToo slow
Magnetic moment01–07
Anomalies08–14
Origin of colour15–22
Ligand & colour23–30
01

Magnetic Moment & Unpaired Electrons

07 Q
[ 01 ]A NEET styleSusceptibility

Magnetic susceptibility measurements on coordination compounds are used to obtain information about:

  • (a)The wavelength of light absorbed
  • (b)The stability constant of the complex
  • (c)The number of unpaired electrons and hence the structure adopted
  • (d)The oxidation state of the ligand

Answer — (c) The number of unpaired electrons and hence the structure adopted

Given
Magnetic susceptibility experiments performed on a coordination compound.
Asked
What information such measurements provide.
Concept
Susceptibility measurements give the magnetic moment, which is determined solely by the number of unpaired electrons. That number then reveals whether pairing occurred, and therefore which structure the complex has adopted.
Formula
μ = √n(n+2) BM
Baby steps
  1. The experiment measures how strongly the sample is attracted into a magnetic field.
  2. This gives the magnetic moment μ of the complex.
  3. Inverting μ = √n(n+2) gives n, the number of unpaired electrons.
  4. Comparing n with the free-ion value shows whether pairing occurred, fixing the structure.
ShortcutMagnetism is the experimental route to structure. Colour is the experimental route to Δo. Keep the two probes separate.
[ 02 ]A NEET styled5 high spin

The magnetic moment of a high spin d5 octahedral complex is:

  • (a)1.73 BM
  • (b)3.87 BM
  • (c)4.90 BM
  • (d)5.92 BM

Answer — (d) 5.92 BM

Given
Octahedral d5 ion in a weak field.
Asked
Spin-only magnetic moment.
Concept
In a weak field the five d electrons occupy all five orbitals singly, obeying Hund's rule, giving the maximum possible number of unpaired electrons for the 3d series.
Formula
μ = √n(n+2) BM
Baby steps
  1. High spin d5 → t2g3eg2.
  2. All five electrons are unpaired → n = 5.
  3. μ = √5(5+2) = √35.
  4. μ = 5.92 BM, the largest value obtainable in the first transition series.
Shortcut5.92 BM in the options is a strong signal for Mn2+ or Fe3+ with a weak ligand — the two common high spin d5 ions.
[ 03 ]A NEET styleReverse

A complex of a first-series metal has a spin-only magnetic moment of 3.87 BM. The central metal ion could be:

  • (a)Ti3+
  • (b)Cr3+
  • (c)Fe3+
  • (d)Zn2+

Answer — (b) Cr3+

Given
μ = 3.87 BM; four candidate metal ions.
Asked
Which ion gives this moment.
Concept
Convert the moment into an unpaired electron count first, then find the ion whose d configuration produces that count.
Formula
μ = 3.87 BM ⇒ n(n+2) = 15 ⇒ n = 3
Baby steps
  1. 3.87 BM corresponds to n = 3 unpaired electrons.
  2. Ti3+ = 3d1 → n = 1 → 1.73 BM.
  3. Cr3+ = 3d3 → three electrons in three t2g orbitals → n = 3. Match.
  4. Fe3+ = 3d5 and Zn2+ = 3d10 → n = 5 and 0 respectively.
ShortcutConvert μ to n first, then hunt for the ion. Testing every option against the formula wastes time.
[ 04 ]A NEET styleOrigin

Paramagnetism in a coordination compound arises because of:

  • (a)The presence of unpaired electrons
  • (b)The presence of paired electrons
  • (c)The high oxidation state of the metal
  • (d)The large size of the ligands

Answer — (a) The presence of unpaired electrons

Given
Origin of paramagnetic behaviour.
Asked
The cause of paramagnetism.
Concept
An unpaired electron has a net spin and behaves as a tiny magnet. Paired electrons have opposite spins whose magnetic effects cancel exactly, contributing nothing.
Formula
μ = √n(n+2), where n counts only unpaired electrons
Baby steps
  1. Each electron possesses spin angular momentum and an associated magnetic moment.
  2. In a pair, the two opposite spins cancel one another completely.
  3. Only unpaired electrons leave a residual moment.
  4. More unpaired electrons → larger μ → more strongly paramagnetic.
ShortcutThis is exactly why the formula is called the spin-only magnetic moment — it counts spins and ignores orbital contribution.
[ 05 ]A NEET styleDiamagnetic

A diamagnetic coordination compound is one in which:

  • (a)All electrons are paired and it is weakly repelled by a magnetic field
  • (b)All electrons are unpaired and it is attracted by a magnetic field
  • (c)There is exactly one unpaired electron
  • (d)The metal is in the zero oxidation state

Answer — (a) All electrons are paired and it is weakly repelled by a magnetic field

Given
Definition of diamagnetism.
Asked
Characteristic of a diamagnetic complex.
Concept
With every electron paired, the spin moments cancel completely, so n = 0 and μ = 0. Such substances are not attracted into a magnetic field; they are weakly repelled by it.
Formula
n = 0 ⇒ μ = 0
Baby steps
  1. Count the unpaired electrons in the complex.
  2. If none remain, all spin moments cancel.
  3. μ = √0(0+2) = 0 BM.
  4. The substance is diamagnetic and weakly repelled by the field.
ShortcutA metal in the zero oxidation state is not automatically diamagnetic — but the common carbonyls Ni(CO)4, Fe(CO)5 and Cr(CO)6 happen to be.
[ 06 ]A NEET styled1

The spin-only magnetic moment of [Ti(H2O)6]3+ is: (Z of Ti = 22)

  • (a)0 BM
  • (b)1.73 BM
  • (c)2.83 BM
  • (d)3.87 BM

Answer — (b) 1.73 BM

Given
[Ti(H2O)6]3+, ZTi = 22.
Asked
Spin-only magnetic moment.
Concept
For d1, d2 and d3 ions the unpaired electron count is the same in weak and strong fields, since the t2g orbitals fill singly either way. Ligand strength need not be considered.
Formula
μ = √n(n+2) BM
Baby steps
  1. Oxidation state of Ti = +3 ⇒ Ti3+ = [Ar]3d1.
  2. The single electron occupies one t2g orbital → n = 1.
  3. μ = √1(1+2) = √3.
  4. μ = 1.73 BM, paramagnetic.
ShortcutThis same ion is the standard example for colour too — d1 gives the simplest possible d–d transition, t2g1 → eg1.
[ 07 ]A NEET styleFree ion

The number of unpaired electrons in the free Fe2+ ion is: (Z of Fe = 26)

  • (a)6
  • (b)5
  • (c)4
  • (d)2

Answer — (c) 4

Given
Free (uncomplexed) Fe2+ ion, ZFe = 26.
Asked
Number of unpaired electrons.
Concept
In a free ion there is no ligand field, so the five d orbitals are degenerate and Hund's rule alone governs the filling: every orbital is singly occupied before pairing begins.
Formula
Hund's rule: fill all five orbitals singly, then pair
Baby steps
  1. Fe = [Ar]3d64s2; losing two electrons from 4s gives Fe2+ = [Ar]3d6.
  2. Place the first five electrons singly, one in each d orbital.
  3. The sixth must pair with one of them.
  4. Unpaired electrons = 4, so μ = 4.90 BM.
ShortcutElectrons are removed from 4s before 3d when forming transition metal cations. Getting this backwards corrupts the entire d count.
02

Magnetic Behaviour & Anomalies

07 Q
[ 08 ]B NEET styled1-d3

For metal ions with up to three d electrons, such as Ti3+, V3+ and Cr3+, the magnetic behaviour of the free ion and of its coordination entities is:

  • (a)Always different
  • (b)Zero in both cases
  • (c)Different only for strong field ligands
  • (d)Similar, because two vacant d orbitals are available without any pairing

Answer — (d) Similar, because two vacant d orbitals are available without any pairing

Given
Metal ions with d1, d2 and d3 configurations.
Asked
Comparison of free ion and complex magnetic behaviour.
Concept
With three or fewer d electrons, the electrons occupy three separate orbitals under Hund's rule. Two d orbitals are therefore already vacant, so no pairing is ever forced and the unpaired count is unchanged on complexation.
Formula
d1–d3: n is the same in weak and strong fields
Baby steps
  1. Ti3+ (d1), V3+ (d2) and Cr3+ (d3) fill t2g singly.
  2. Two vacant d orbitals remain regardless of ligand strength.
  3. No electron is ever forced to pair, whatever ligand approaches.
  4. So the free ion and its complexes have the same magnetic moment.
ShortcutComplications begin only at d4. From d4 to d7, the ligand decides the spin state; outside that range it does not.
[ 09 ]B NEET styleAnomaly

[Mn(CN)6]3− has a magnetic moment corresponding to two unpaired electrons whereas [MnCl6]3− corresponds to four. This is because:

  • (a)CN− forms an inner orbital complex with pairing, while Cl− forms an outer orbital complex without pairing
  • (b)Manganese has different oxidation states in the two complexes
  • (c)The chloride complex is tetrahedral
  • (d)Cl− is a stronger ligand than CN−

Answer — (a) CN− forms an inner orbital complex with pairing, while Cl− forms an outer orbital complex without pairing

Given
Two Mn(III) octahedral complexes, both d4, with different magnetic moments.
Asked
Explanation of the difference.
Concept
Both contain Mn3+ (d4). The strong field CN− forces pairing and inner (3d) orbital participation; the weak field Cl− does not, so outer (4d) orbitals are used.
Formula
μ = √n(n+2)
Baby steps
  1. Oxidation state in both: x − 6 = −3 ⇒ Mn3+ = 3d4.
  2. CN− strong field → pairing → n = 2 → μ = 2.83 BM, inner orbital, d2sp3.
  3. Cl− weak field → no pairing → n = 4 → μ = 4.90 BM, outer orbital, sp3d2.
  4. The difference is entirely due to ligand field strength.
ShortcutSame metal, same charge, different μ ⇒ the answer is always inner versus outer orbital. Oxidation state is never the explanation in these pairs.
[ 10 ]B NEET styleAnomaly

[Fe(CN)6]3− has a magnetic moment of one unpaired electron while [FeF6]3− has five. The numbers of unpaired electrons and the hybridisations are:

  • (a)1 with sp3d2 and 5 with d2sp3
  • (b)1 with d2sp3 and 5 with sp3d2
  • (c)Both with d2sp3
  • (d)Both with sp3d2

Answer — (b) 1 with d2sp3 and 5 with sp3d2

Given
Two Fe(III) octahedral complexes, both d5.
Asked
Unpaired electron count and hybridisation of each.
Concept
A strong field ligand pairs the d electrons, freeing inner 3d orbitals for d2sp3. A weak field ligand leaves the electrons unpaired, so outer 4d orbitals must be used, giving sp3d2.
Formula
μ = √n(n+2); inner = d2sp3, outer = sp3d2
Baby steps
  1. Both contain Fe3+ = 3d5.
  2. [Fe(CN)6]3−: strong field → pairing → 1 unpaired, two 3d orbitals freed → d2sp3, μ = 1.73 BM.
  3. [FeF6]3−: weak field → no pairing → 5 unpaired, no 3d free → sp3d2, μ = 5.92 BM.
  4. So the cyanide complex is weakly and the fluoride complex strongly paramagnetic.
ShortcutMatch the words to the numbers: weakly paramagnetic means few unpaired electrons (strong ligand); strongly paramagnetic means many (weak ligand).
[ 11 ]B NEET styleAnomaly

[CoF6]3− is paramagnetic with four unpaired electrons while [Co(C2O4)3]3− is diamagnetic. Both are octahedral Co(III) complexes because:

  • (a)Oxalate is a stronger field ligand than fluoride and causes complete pairing
  • (b)Cobalt is in different oxidation states
  • (c)The oxalate complex is square planar
  • (d)Fluoride causes greater splitting than oxalate

Answer — (a) Oxalate is a stronger field ligand than fluoride and causes complete pairing

Given
Two octahedral Co(III) complexes, both d6, with opposite magnetic behaviour.
Asked
Explanation for the difference.
Concept
For a d6 ion, a strong field produces t2g6eg0 with zero unpaired electrons, whereas a weak field leaves t2g4eg2 with four.
Formula
μ = √n(n+2)
Baby steps
  1. Both contain Co3+ = 3d6.
  2. F− is a weak field ligand → high spin → 4 unpaired → μ = 4.90 BM.
  3. Oxalate lies higher in the spectrochemical series → complete pairing → 0 unpaired.
  4. Hence one is paramagnetic and the other diamagnetic.
ShortcutLow spin d6 is the only octahedral configuration that is fully diamagnetic. It is also the one with maximum CFSE, −2.4Δo.
[ 12 ]B NEET styleIdentification

Which of the following complexes is paramagnetic?

  • (a)[Ni(CO)4]
  • (b)[Ni(CN)4]2−
  • (c)[Zn(NH3)4]2+
  • (d)[NiCl4]2−

Answer — (d) [NiCl4]2−

Given
Four four-coordinate complexes.
Asked
The paramagnetic one.
Concept
Determine the metal's d configuration and whether the ligand forces pairing. Only an unpaired electron produces paramagnetism.
Formula
μ = √n(n+2)
Baby steps
  1. [Ni(CO)4]: Ni(0) becomes 3d10 under the strong CO field → n = 0, diamagnetic.
  2. [Ni(CN)4]2−: d8 square planar, all paired → n = 0, diamagnetic.
  3. [Zn(NH3)4]2+: Zn2+ = 3d10 → n = 0, diamagnetic.
  4. [NiCl4]2−: d8 with weak Cl− → tetrahedral, 2 unpaired → μ = 2.83 BM, paramagnetic.
ShortcutThree of the four are d10 or square planar d8, both of which are automatically diamagnetic. Recognising those two patterns solves this instantly.
[ 13 ]B NEET styleOrdering

The correct order of increasing spin-only magnetic moment is:

  • (a)[Fe(CN)6]3− < [Ni(H2O)6]2+ < [Cr(H2O)6]3+ < [Mn(H2O)6]2+
  • (b)[Mn(H2O)6]2+ < [Cr(H2O)6]3+ < [Ni(H2O)6]2+ < [Fe(CN)6]3−
  • (c)[Cr(H2O)6]3+ < [Fe(CN)6]3− < [Ni(H2O)6]2+ < [Mn(H2O)6]2+
  • (d)All four have equal magnetic moments

Answer — (a) [Fe(CN)6]3− < [Ni(H2O)6]2+ < [Cr(H2O)6]3+ < [Mn(H2O)6]2+

Given
Four octahedral complexes with the stated ligands.
Asked
Increasing order of magnetic moment.
Concept
Work out the unpaired electron count for each after applying the ligand field strength, then order them. The moment rises monotonically with n.
Formula
μ = √n(n+2)
Baby steps
  1. [Fe(CN)6]3−: d5 strong field → n = 1 → 1.73 BM.
  2. [Ni(H2O)6]2+: d8 → n = 2 → 2.83 BM.
  3. [Cr(H2O)6]3+: d3 → n = 3 → 3.87 BM.
  4. [Mn(H2O)6]2+: d5 weak field → n = 5 → 5.92 BM. Order = option A.
ShortcutConvert every complex to a bare number n first, then sort. Comparing the complexes directly, without reducing them to n, is where errors creep in.
[ 14 ]B NEET styleStrong ligand

[Cr(CN)6]3− is paramagnetic with three unpaired electrons, even though CN− is a strong field ligand. The reason is:

  • (a)Cyanide behaves as a weak ligand towards chromium
  • (b)The complex is tetrahedral
  • (c)Cr3+ is d3, and three electrons occupy three separate t2g orbitals with no pairing possible
  • (d)Chromium is in the +2 oxidation state

Answer — (c) Cr3+ is d3, and three electrons occupy three separate t2g orbitals with no pairing possible

Given
[Cr(CN)6]3− with a strong field ligand, yet three unpaired electrons.
Asked
Reason the strong ligand does not reduce the moment.
Concept
Pairing can only occur if two electrons are competing for the same region. With just three electrons and three t2g orbitals available, each takes its own orbital and no ligand can force them together.
Formula
d3 ⇒ t2g3 whatever the field strength
Baby steps
  1. Oxidation state: x − 6 = −3 ⇒ Cr3+ = 3d3.
  2. The three lower t2g orbitals can hold three electrons singly.
  3. Hund's rule keeps them unpaired; no pairing energy needs to be paid.
  4. n = 3 regardless of ligand → μ = 3.87 BM.
ShortcutA strong field ligand only reduces μ for d4 to d7 ions. Outside that range the answer is the same either way.
03

Origin of Colour

08 Q
[ 15 ]C NEET styled-d transition

Crystal field theory attributes the colour of coordination compounds to:

  • (a)Charge transfer from ligand to metal
  • (b)d–d transition of an electron
  • (c)Vibration of the metal–ligand bond
  • (d)Ionisation of the counter ion

Answer — (b) d–d transition of an electron

Given
Origin of colour in transition metal complexes.
Asked
The transition responsible.
Concept
When ligands split the d orbitals, an electron in the lower set can absorb visible light and jump to the higher set. This d–d transition removes part of the visible spectrum, and the transmitted light is coloured.
Formula
Δo = hν = hc/λ
Baby steps
  1. The ligand field splits the five d orbitals into t2g and eg.
  2. The energy gap Δo happens to correspond to visible light.
  3. An electron absorbs a photon and is promoted from t2g to eg.
  4. The remaining transmitted wavelengths give the observed colour.
ShortcutA d–d transition needs a partly filled d subshell. If the ion is d0 or d10, no such transition exists and the compound is colourless.
[ 16 ]C NEET styleComplementary

If a coordination compound absorbs green light, it will appear:

  • (a)Green
  • (b)Blue
  • (c)Red
  • (d)Colourless

Answer — (c) Red

Given
A complex whose absorption falls in the green region.
Asked
The colour observed.
Concept
The colour seen is the complementary colour of the light absorbed — it is generated by the wavelengths that were left over and transmitted.
Formula
Observed colour = complement of absorbed colour
Baby steps
  1. White light passes through the sample.
  2. The green component is absorbed by the d–d transition.
  3. The remaining wavelengths reach the eye.
  4. The complement of green is red, so the complex appears red.
ShortcutCommon pairs from NCERT Table 5.3: absorbs yellow → violet; blue-green → red; blue → yellow-orange; red → blue.
[ 17 ]C NEET styleTi(III)

[Ti(H2O)6]3+ absorbs light of wavelength 498 nm in the blue-green region. The colour of the complex is:

  • (a)Blue-green
  • (b)Violet
  • (c)Yellow
  • (d)Colourless

Answer — (b) Violet

Given
[Ti(H2O)6]3+ with an absorption maximum at 498 nm (blue-green).
Asked
Observed colour.
Concept
Ti3+ is 3d1. The single electron absorbs blue-green light and is promoted from t2g to eg. The transmitted light gives the complementary colour.
Formula
t2g1eg0 → t2g0eg1
Baby steps
  1. Ti3+ = 3d1, ground state t2g1eg0.
  2. Absorption of 498 nm light promotes the electron to eg.
  3. Blue-green is thereby removed from the transmitted light.
  4. The complementary colour is violet, which is what the solution appears.
ShortcutThis is the textbook illustration of colour in CFT, because d1 allows only one possible transition — no ambiguity about which electron moves.
[ 18 ]C NEET styleColourless

Which of the following complexes is colourless?

  • (a)[Cu(H2O)4]2+
  • (b)[Ni(H2O)6]2+
  • (c)[Sc(H2O)6]3+
  • (d)[Ti(H2O)6]3+

Answer — (c) [Sc(H2O)6]3+

Given
Four aqua complexes of first-series metals.
Asked
The colourless one.
Concept
A d–d transition requires an electron in the lower set and a vacancy in the upper set. A d0 ion has no electron to promote; a d10 ion has no vacancy to promote into. Both are colourless.
Formula
Coloured ⇒ d1 to d9
Baby steps
  1. Cu2+ = d9 → coloured (blue).
  2. Ni2+ = d8 → coloured (green).
  3. Sc3+ = d0 → no d electron to excite → colourless.
  4. Ti3+ = d1 → coloured (violet).
ShortcutColourless ions to know on sight: Sc3+, Ti4+ (d0) and Zn2+, Cu+, Ag+, Cd2+ (d10).
[ 19 ]C NEET styleHydration

Anhydrous CuSO4 is white but CuSO4·5H2O is blue. The reason is:

  • (a)Copper is reduced to Cu+ in the anhydrous salt
  • (b)Water molecules are themselves blue
  • (c)In the absence of ligands there is no crystal field splitting, so no d–d transition occurs
  • (d)The sulphate ion absorbs visible light in the hydrate

Answer — (c) In the absence of ligands there is no crystal field splitting, so no d–d transition occurs

Given
Anhydrous and hydrated copper(II) sulphate.
Asked
Reason for the colour difference.
Concept
Splitting of the d orbitals requires a ligand field. Without coordinated ligands the five d orbitals remain degenerate, no energy gap exists, and no visible light is absorbed.
Formula
No ligand ⇒ Δ = 0 ⇒ no d–d transition
Baby steps
  1. Cu2+ is d9 in both compounds, so the electron count is not the cause.
  2. Anhydrous CuSO4 has no aqua ligands → degenerate d orbitals → white.
  3. In the pentahydrate, [Cu(H2O)4]2+ forms and the d orbitals split.
  4. It absorbs red light near 600 nm → appears blue.
ShortcutBeware the trap: the metal ion is d9 in both. The difference is the presence of ligands, not a change in oxidation state.
[ 20 ]C NEET styleDehydration

On heating, [Ti(H2O)6]Cl3 loses water and becomes:

  • (a)More intensely violet
  • (b)Colourless
  • (c)Yellow
  • (d)Green

Answer — (b) Colourless

Given
[Ti(H2O)6]Cl3, violet, heated to remove water.
Asked
Colour after dehydration.
Concept
Removing the aqua ligands removes the ligand field. With no field there is no splitting, no energy gap, and therefore no d–d transition — even though the d1 electron is still present.
Formula
Δo = 0 ⇒ no absorption in the visible region
Baby steps
  1. The violet colour comes from the t2g → eg transition in [Ti(H2O)6]3+.
  2. Heating drives off the coordinated water molecules.
  3. With no ligands present, the d orbitals become degenerate again.
  4. No visible light is absorbed → the solid turns colourless.
ShortcutSame principle as anhydrous CuSO4. Both examples make the point that colour needs ligands, not merely d electrons.
[ 21 ]C NEET styleTransition

The electronic transition responsible for the colour of [Ti(H2O)6]3+ is:

  • (a)t2g1eg0 → t2g0eg1
  • (b)t2g0eg1 → t2g1eg0
  • (c)3d → 4s
  • (d)4s → 4p

Answer — (a) t2g1eg0 → t2g0eg1

Given
[Ti(H2O)6]3+, a 3d1 octahedral complex.
Asked
The transition producing the colour.
Concept
Absorption promotes an electron from the lower t2g level to the empty eg level. The reverse process is emission, not absorption, and s or p transitions require far more energy than visible light.
Formula
Energy absorbed = Δo
Baby steps
  1. Ground state of a d1 ion: the electron sits in t2g.
  2. The next available level is the empty eg set.
  3. A visible photon of energy Δo is absorbed and the electron is promoted.
  4. Transition = t2g1eg0 → t2g0eg1.
ShortcutAbsorption always runs low to high. Any option showing the electron falling to a lower level describes emission and can be eliminated at once.
[ 22 ]C NEET styleEnergy

The energy of the light absorbed by an octahedral complex during a d–d transition is equal to:

  • (a)The crystal field splitting energy Δo
  • (b)The pairing energy P
  • (c)The ionisation energy of the metal
  • (d)The lattice energy of the complex

Answer — (a) The crystal field splitting energy Δo

Given
A d–d transition in an octahedral complex.
Asked
What the absorbed energy corresponds to.
Concept
The electron jumps from t2g to eg, and the gap between those two levels is by definition Δo. Measuring the absorbed wavelength therefore measures Δo directly.
Formula
Δo = hc/λabsorbed
Baby steps
  1. The t2g and eg sets are separated by Δo.
  2. Promoting an electron across that gap requires exactly Δo of energy.
  3. The photon absorbed must carry that energy: hc/λ = Δo.
  4. So the absorbed energy equals Δo.
ShortcutThis relation is what makes the spectrochemical series possible — it is built entirely from measured λ values converted into Δo.
04

Ligand Effect on Colour

08 Q
[ 23 ]D NEET styleWavelength order

The correct order of the wavelength of light absorbed by [Ni(H2O)6]2+, [Ni(NH3)6]2+ and [Ni(NO2)6]4− is:

  • (a)[Ni(NO2)6]4− > [Ni(NH3)6]2+ > [Ni(H2O)6]2+
  • (b)[Ni(NH3)6]2+ > [Ni(H2O)6]2+ > [Ni(NO2)6]4−
  • (c)[Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4−
  • (d)All three absorb identical wavelengths

Answer — (c) [Ni(H2O)6]2+ > [Ni(NH3)6]2+ > [Ni(NO2)6]4−

Given
Three Ni(II) octahedral complexes with ligands of increasing field strength.
Asked
Decreasing order of absorbed wavelength.
Concept
Stronger ligand → larger Δo → more energy needed → shorter wavelength absorbed. The wavelength order is therefore the exact reverse of the spectrochemical order.
Formula
Δo = hc/λ ⇒ λ ∝ 1/Δo
Baby steps
  1. Field strength: H2O < NH3 < NO2−.
  2. Therefore Δo: aqua < ammine < nitro complex.
  3. Wavelength varies inversely with Δo.
  4. λ: H2O > NH3 > NO2− complex.
ShortcutWrite the spectrochemical order, then flip it. The weakest ligand always absorbs the longest wavelength.
[ 24 ]D NEET styleen series

When ethane-1,2-diamine is added progressively to a green solution of [Ni(H2O)6]2+ in en : Ni ratios of 1:1, 2:1 and 3:1, the colours observed are:

  • (a)Pale blue, blue/purple, violet
  • (b)Yellow, orange, red
  • (c)Violet, blue/purple, pale blue
  • (d)Colourless throughout

Answer — (a) Pale blue, blue/purple, violet

Given
[Ni(H2O)6]2+ (green) with en added in increasing molar ratios.
Asked
Sequence of colours.
Concept
Each en replaces two water molecules, since it is didentate. As the stronger-field en accumulates, Δo rises, shifting absorption to shorter wavelength and moving the observed colour across the spectrum.
Formula
Δo increases as en replaces H2O
Baby steps
  1. 1:1 → [Ni(H2O)4(en)]2+, pale blue.
  2. 2:1 → [Ni(H2O)2(en)2]2+, blue/purple.
  3. 3:1 → [Ni(en)3]2+, violet.
  4. The colour deepens steadily as Δo grows.
ShortcutNote the stoichiometry as well as the colours: one en displaces two water molecules, because en is didentate. That detail is often the real question.
[ 25 ]D NEET styleUV absorption

[Co(CN)6]3− absorbs at 310 nm, which lies in the ultraviolet region. The complex therefore appears:

  • (a)Deep blue
  • (b)Bright red
  • (c)Pale yellow
  • (d)Black

Answer — (c) Pale yellow

Given
[Co(CN)6]3− with absorption at 310 nm (ultraviolet).
Asked
Observed colour.
Concept
CN− produces a very large Δo, pushing the absorption out of the visible range entirely. With almost no visible light removed, the complex is only faintly coloured.
Formula
Large Δo ⇒ short λ ⇒ absorption moves into the UV
Baby steps
  1. CN− is near the top of the spectrochemical series → very large Δo.
  2. The transition energy is high, so λ is short: 310 nm.
  3. 310 nm is ultraviolet, so scarcely any visible light is absorbed.
  4. The complex appears only faintly tinted — pale yellow.
ShortcutVery strong ligands push absorption into the UV, giving pale complexes. Very weak ligands absorb far red, also giving weak colour. Mid-range ligands give the most intense colours.
[ 26 ]D NEET styleColour table

[Co(NH3)6]3+ absorbs blue light at 475 nm. Its observed colour is:

  • (a)Blue
  • (b)Violet
  • (c)Green
  • (d)Yellow-orange

Answer — (d) Yellow-orange

Given
[Co(NH3)6]3+ with absorption maximum at 475 nm in the blue region.
Asked
Observed colour.
Concept
The observed colour is complementary to the absorbed colour. Blue light removed from white light leaves the yellow-orange region dominant.
Formula
Observed = complement of absorbed
Baby steps
  1. Absorbed light: blue, at 475 nm.
  2. Blue is removed from the transmitted beam.
  3. The complement of blue is yellow-orange.
  4. Hence the complex appears yellow-orange, as listed in NCERT Table 5.3.
ShortcutDo not answer with the absorbed colour itself — that is the commonest error on this question type. Always take the complement.
[ 27 ]D NEET styleField strength

A complex absorbs light of a shorter wavelength than another complex of the same metal ion. This indicates that its ligand:

  • (a)Produces a smaller Δo
  • (b)Carries a higher negative charge
  • (c)Has a larger size
  • (d)Produces a larger Δo

Answer — (d) Produces a larger Δo

Given
Two complexes of the same metal ion, one absorbing at shorter wavelength.
Asked
What the shorter wavelength implies about the ligand.
Concept
Shorter wavelength means higher photon energy, and that energy equals Δo. So a shorter absorption wavelength signals a stronger ligand field.
Formula
Δo = hc/λ
Baby steps
  1. Photon energy E = hc/λ, so E rises as λ falls.
  2. The absorbed energy equals the splitting Δo.
  3. A shorter λ therefore corresponds to a larger Δo.
  4. That means the ligand lies higher in the spectrochemical series.
ShortcutLigand charge is a decoy here. CFT predicts anionic ligands should split most, but experimentally I− is weakest while neutral CO is strongest.
[ 28 ]D NEET styleGemstones

The red colour of ruby and the green colour of emerald both arise from:

  • (a)Fe2+ ions in a silicate lattice
  • (b)Ti3+ ions in an alumina lattice
  • (c)Charge transfer between aluminium and oxygen
  • (d)d–d transitions of Cr3+ ions in octahedral sites

Answer — (d) d–d transitions of Cr3+ ions in octahedral sites

Given
Ruby is Al2O3 containing about 0.5–1% Cr3+; emerald is beryl containing Cr3+.
Asked
Origin of the colour in both gemstones.
Concept
In both minerals Cr3+ (d3) occupies octahedral sites and behaves as an octahedral chromium(III) complex built into the lattice. The colour comes from d–d transitions at those centres.
Formula
Δo = hc/λ
Baby steps
  1. Ruby: Cr3+ replaces some Al3+ in the alumina lattice → octahedral Cr(III) centres.
  2. Emerald: Cr3+ occupies octahedral sites in beryl, Be3Al2Si6O18.
  3. The different lattices give slightly different Δo values.
  4. In emerald the absorption bands shift to longer wavelength, so green is transmitted rather than red.
ShortcutThe striking point is that the same ion gives two different colours. Only the surrounding lattice, and hence Δo, has changed.
[ 29 ]D NEET styleColour table

[Cu(H2O)4]2+ absorbs at 600 nm in the red region. The colour of the solution is:

  • (a)Red
  • (b)Blue
  • (c)Yellow
  • (d)Green

Answer — (b) Blue

Given
[Cu(H2O)4]2+ with absorption maximum at 600 nm (red).
Asked
Observed colour of the solution.
Concept
Copper(II) is d9, so a d–d transition is possible. Red light is absorbed and the complementary colour is transmitted.
Formula
Observed = complement of absorbed
Baby steps
  1. Cu2+ = 3d9 → a partly filled d subshell, so the ion is coloured.
  2. Light of 600 nm, in the red region, is absorbed.
  3. Red is therefore removed from the transmitted light.
  4. The complement of red is blue — the familiar colour of aqueous copper sulphate.
ShortcutLongest wavelength in NCERT Table 5.3 belongs to this complex (600 nm), which is consistent with H2O being a relatively weak ligand.
[ 30 ]D NEET styleComparison

Among the following, the complex that absorbs light of the longest wavelength is:

  • (a)[Co(CN)6]3−
  • (b)[Co(NH3)6]3+
  • (c)[Co(en)3]3+
  • (d)[CoF6]3−

Answer — (d) [CoF6]3−

Given
Four Co(III) octahedral complexes with different ligands.
Asked
The one absorbing the longest wavelength.
Concept
The longest wavelength corresponds to the smallest energy gap, so the answer is the complex with the weakest field ligand.
Formula
λ ∝ 1/Δo
Baby steps
  1. Rank the ligands: F− < NH3 < en < CN−.
  2. The smallest Δo belongs to the fluoride complex.
  3. Smallest gap → lowest photon energy → longest wavelength.
  4. Answer = [CoF6]3−.
ShortcutTwo questions hide in one here: ‘largest Δo’ and ‘longest λ’ have opposite answers. Read which one is being asked before ranking.