Key Fact
Spectrochemical series
Ligands ordered by increasing Δ: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. Predicts colour and magnetic properties of complexes.
-- NCERT Class 12 Chemistry, Ch. 5, p. 132Colour Magnetic Coordination
8 MCQs1 revision card9-step worked example
Source: NCERT Coordination CompoundsPYQ coverage: NEET 2021, 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Colour Magnetic Coordination, explained for NEET
The trap: Students default to one spin state when asked about colour or magnetic behaviour of a complex — they forget that the ligand's field strength determines whether a d-orbital configuration is high-spin or low-spin, which directly decides the number of unpaired electrons (magnetic moment) and the magnitude of Δ₀ (colour).
Colour in coordination compounds arises because the d–d transition absorbs a specific wavelength from visible light; the complementary colour is what we observe. The energy gap Δ₀ between t₂g and e_g sets determines which wavelength is absorbed. A larger Δ₀ (strong-field ligand) shifts absorption toward violet/blue → complex appears yellow/orange. A smaller Δ₀ (weak-field ligand) shifts absorption toward red → complex appears green/blue (NCERT Class 12 Chemistry Chapter 5, page 133).
Magnetic properties depend on unpaired electrons. The spin-only magnetic moment is μ = √(n(n+2)) BM. The critical question: what is n? That depends on whether the complex is high-spin or low-spin — controlled entirely by the ligand field strength relative to pairing energy P.
The spectrochemical series decides everything: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO
- Strong-field (CN⁻, CO, NH₃/en for certain metals): Δ₀ > P → electrons pair in t₂g → low-spin → fewer unpaired electrons → lower μ.
- Weak-field (F⁻, Cl⁻, H₂O): Δ₀ < P → electrons spread across both sets → high-spin → more unpaired electrons → higher μ.
Watch-out: A d⁶ metal with CN⁻ has n = 0 (diamagnetic, low-spin), but the same d⁶ metal with H₂O has n = 4 (paramagnetic, high-spin). One ligand swap changes μ from 0 to 4.90 BM. NEET exploits this gap regularly.
Can you answer these Colour Magnetic Coordination MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The colour of a coordination compound is due to:
Show answer and why every option is right or wrong
Answer: C. Colour arises from d–d transitions where an electron in a lower-energy d-orbital set absorbs visible light and jumps to the higher-energy set; the complementary wavelength is observed (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because σ-bonding forms the coordinate bond but does not selectively absorb visible wavelengths responsible for colour.
Why B is wrong: B is wrong because ionisation of counter ions produces free ions in solution but does not cause selective absorption of visible light by the metal centre.
Why D is wrong: D is wrong because charge-transfer bands exist in some complexes but the standard NCERT explanation for colour in coordination compounds is the d–d transition mechanism.
MCQ 2Easy RecallPractice
Which of the following complexes is expected to be colourless?
Show answer and why every option is right or wrong
Answer: D. Zn²⁺ has a d¹⁰ configuration — all d-orbitals are completely filled. No vacant d-orbital exists for a d–d transition, so no visible light is absorbed and the complex is colourless (NCERT Class 12 Chemistry Chapter 5, page 133).
Why A is wrong: A is wrong because Ti³⁺ is d¹ — one electron can undergo a d–d transition, absorbing visible light (appears purple/red).
Why B is wrong: B is wrong because Cu²⁺ is d⁹ — one hole allows a d–d transition, absorbing red light (appears deep blue).
Why C is wrong: C is wrong because Fe²⁺ is d⁶ — multiple d–d transitions are possible, giving a green/pale colour in aqueous solution.
MCQ 3Easy RecallPractice
The correct order of field strength in the spectrochemical series is:
Show answer and why every option is right or wrong
Answer: B. The spectrochemical series places ligands in increasing field strength: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. Option B correctly orders the four given ligands (NCERT Class 12 Chemistry Chapter 5, page 132).
Why A is wrong: A is wrong because it reverses the series — CN⁻ is among the strongest field ligands, not the weakest (trap: confusing charge with field strength).
Why C is wrong: C is wrong because it places Cl⁻ above H₂O, violating the established series order where H₂O is stronger than Cl⁻.
Why D is wrong: D is wrong because it places weak-field ligands (Cl⁻, H₂O) above strong-field ligands (NH₃, CN⁻), which is the reverse of the correct order.
MCQ 4Direct ApplicationPractice
[Fe(CN)₆]⁴⁻ contains Fe²⁺ (d⁶). Given that CN⁻ is a strong-field ligand, the number of unpaired electrons and magnetic moment are:
Show answer and why every option is right or wrong
Answer: B. CN⁻ is a strong-field ligand (Δ₀ > P), so Fe²⁺ (d⁶) adopts low-spin configuration: all 6 electrons pair in t₂g (t₂g⁶ e_g⁰). n = 0, μ = √(0(0+2)) = 0 BM — the complex is diamagnetic.
Why A is wrong: A is wrong because n = 4 corresponds to the high-spin d⁶ configuration (t₂g⁴ e_g²), which occurs with weak-field ligands like H₂O, not strong-field CN⁻ (trap: defaulting to high-spin without checking ligand field strength).
Why C is wrong: C is wrong because n = 2 does not correspond to any valid octahedral d⁶ configuration — neither high-spin (n=4) nor low-spin (n=0) gives 2 unpaired electrons.
Why D is wrong: D is wrong because n = 1 is not a valid electron count for octahedral d⁶ in either spin state.
MCQ 5Direct ApplicationPractice
[CoF₆]³⁻ contains Co³⁺ (d⁶). F⁻ is a weak-field ligand. The magnetic moment of this complex is approximately:
Show answer and why every option is right or wrong
Answer: A. F⁻ is weak-field (Δ₀ < P), so Co³⁺ (d⁶) is high-spin: t₂g⁴ e_g². This gives n = 4 unpaired electrons. μ = √(4(4+2)) = √24 ≈ 4.90 BM.
Why B is wrong: B is wrong because 1.73 BM corresponds to n = 1, which is not achievable in any octahedral d⁶ configuration.
Why C is wrong: C is wrong because 0 BM (diamagnetic) corresponds to low-spin d⁶ (all paired in t₂g), which requires a strong-field ligand — F⁻ is weak-field (trap: assuming all d⁶ complexes are diamagnetic).
Why D is wrong: D is wrong because 3.87 BM corresponds to n = 3, which would require a d³ or specific d⁷ configuration, not d⁶ high-spin.
MCQ 6Direct ApplicationPractice
Two octahedral complexes of Cr³⁺ (d³) are: [Cr(NH₃)₆]³⁺ and [Cr(H₂O)₆]³⁺. Which statement about their colours is correct?
Show answer and why every option is right or wrong
Answer: A. NH₃ is higher than H₂O in the spectrochemical series, producing a larger Δ₀. Larger Δ₀ means higher energy (shorter wavelength) absorption. Hence [Cr(NH₃)₆]³⁺ absorbs shorter-wavelength light and they display different colours.
Why B is wrong: B is wrong because even though both have d³, the ligand field strength differs (NH₃ > H₂O), so Δ₀ differs and absorption wavelength differs.
Why C is wrong: C is wrong because it reverses the spectrochemical series — NH₃ is a stronger-field ligand than H₂O, not weaker (trap: confusing ligand field order).
Why D is wrong: D is wrong because d³ (t₂g³) still permits d–d transitions to e_g; half-filled t₂g does not prevent colour (unlike d⁰ or d¹⁰).
MCQ 7CalculationPractice
[Mn(CN)₆]⁴⁻ and [Mn(H₂O)₆]²⁺ both contain Mn²⁺ (d⁵). Calculate the difference in their magnetic moments (μ_high − μ_low):
Show answer and why every option is right or wrong
Answer: C. H₂O is weak-field → high-spin d⁵: all 5 electrons unpaired, μ = √(5×7) = √35 ≈ 5.92 BM. CN⁻ is strong-field → low-spin d⁵: t₂g⁵ e_g⁰, n = 1, μ = √(1×3) = √3 ≈ 1.73 BM. Difference = 5.92 − 1.73 = 4.19 BM.
Why A is wrong: A is wrong because it uses n = 3 for low-spin d⁵, which is incorrect — low-spin d⁵ has configuration t₂g⁵ with 1 unpaired electron, not 3.
Why B is wrong: B is wrong because it uses n = 4 for high-spin (incorrect for d⁵) and n = 0 for low-spin (incorrect — d⁵ low-spin still has 1 unpaired electron in t₂g).
Why D is wrong: D is wrong because neither 3.87 BM (n=3) nor 1.73 BM (n=1) corresponds to the high-spin d⁵ case — high-spin d⁵ has all 5 unpaired giving 5.92 BM.
MCQ 8CalculationPractice
An octahedral complex of Fe³⁺ (d⁵) with a strong-field ligand shows μ = 1.73 BM. If the ligand is replaced by a weak-field ligand, the new magnetic moment and the change in number of unpaired electrons are:
Show answer and why every option is right or wrong
Answer: B. Strong-field Fe³⁺ (d⁵): low-spin t₂g⁵ e_g⁰, n = 1, μ = 1.73 BM. Weak-field Fe³⁺ (d⁵): high-spin t₂g³ e_g², n = 5, μ = √(5×7) = 5.92 BM. Change = 5 − 1 = +4 unpaired electrons.
Why A is wrong: A is wrong because n = 4 (μ = 4.90 BM) is not a valid high-spin d⁵ configuration — d⁵ high-spin has all 5 electrons unpaired, not 4.
Why C is wrong: C is wrong because n = 3 (μ = 3.87 BM) corresponds to d³, not high-spin d⁵. The maximum spread of d⁵ fills all five orbitals singly.
Why D is wrong: D is wrong because n = 2 (μ = 2.83 BM) does not correspond to any standard octahedral d⁵ configuration.
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Colour Magnetic Coordination: quick recall before you leave
How do you solve a Colour Magnetic Coordination question? A worked example
Pattern: P.CHE.U12.CRYSTAL_FIELD_HIGH_LOW_SPIN (observed NEET 2021, 2024)
- 1
Given
• Complex: [Co(NH₃)₆]³⁺• Metal ion: Co³⁺• d-electron count: d⁶ (Co is atomic number 27; Co³⁺ = 27 − 3 = 24 electrons; [Ar]3d⁶)• Ligand: NH₃
- 2
Required
Determine (a) whether the complex is high-spin or low-spin, (b) the number of unpaired electrons, and (c) the magnetic moment.
- 3
Concept
The spectrochemical series classifies NH₃ as a strong-field ligand. For octahedral complexes, if Δ₀ > P (pairing energy), electrons preferentially pair in t₂g before occupying e_g → low-spin.
- 4
Formula
μ = √(n(n+2)) BM
- 5
Substitution
NH₃ is strong-field → Δ₀ > P → low-spin d⁶ configuration:• t₂g⁶ e_g⁰ → all 6 electrons paired in t₂g• n = 0
μ = √(0(0+2)) = √0 = 0 BM - 6
Calculation
μ = 0 BM. The complex is diamagnetic.
- 7
Final answer
[Co(NH₃)₆]³⁺ is a low-spin, diamagnetic complex with μ = 0 BM.
Note on exact values: The integers 6 (electron count) and 0 (unpaired count) are exact counting numbers and do not limit significant figures in the magnetic moment calculation. - 8
Common trap
Defaulting to high-spin: if you forgot NH₃ is strong-field and assumed high-spin d⁶ (t₂g⁴ e_g²), you would get n = 4, μ = 4.90 BM — the exact wrong answer NTA places as a distractor.
- 9
Similar NEET-style question
"Calculate the magnetic moment of [Fe(H₂O)₆]²⁺. Given: Fe²⁺ is d⁶, H₂O is a weak-field ligand."
(Answer: high-spin, n = 4, μ = √(4×6) = √24 ≈ 4.90 BM — paramagnetic.)
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What to remember before solving Colour Magnetic Coordination questions
Which Colour Magnetic Coordination formulas do you need for NEET?
Magnetic moment of coordination complex
Same spin-only formula but n depends on high-spin/low-spin from CFT.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
Valid when
- High vs low spin determined by Δ_o vs P
- Octahedral (or tetrahedral with Δ_t)
Where do students lose marks on Colour Magnetic Coordination?
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Student defaults to one spin state. Strong-field ligand (CN⁻, CO, NH₃ for some) → low-spin (Δ > P, electrons pair). Weak-field (F⁻, H₂O, Cl⁻) → high-spin.
When it triggers
Coordination compound with given ligand asking for magnetic moment, color, or spin state.
How to avoid
Memorise spectrochemical series: I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < CN⁻ < CO. NH₃, CN⁻, CO usually strong-field. F⁻, H₂O, Cl⁻ usually weak-field.
More in Coordination Compounds: 1 exam trap or mistake · 1 formula · 2 question patterns from its other lessons.
Colour Magnetic Coordination questions from past NEET papers
2 questions from NEET 2021, 2026. Answers verified against NTA official keys.
NEET 2026
1[Mn(CN)₆]³⁻
2[Fe(CN)₆]³⁻
3[Co(NH₃)₆]³⁺
4[Ti(H₂O)₆]³⁺
NTA Answer: Option 1(final)
NEET 2021
1(a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
2(a)-(iv), (b)-(ii), (c)-(i), (d)-(iii)
3(a)-(ii), (b)-(iv), (c)-(iii), (d)-(i)
4(a)-(i), (b)-(iii), (c)-(iv), (d)-(ii)
NTA Answer: Option 1(final)
How does NEET ask about Colour Magnetic Coordination?
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
Determine high-spin or low-spin from ligand field strength. Strong-field (CN, CO): low-spin. Weak-field (F, H2O): high-spin.
InterpretationMedium
Common distractors
forgets weak vs strong ligand
Default low-spin always
Sources
NCERT refs: Class 12 Chemistry Chapter 5, p.133
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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