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. 23The 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 23).
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
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.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
The colour of a coordination compound is due to:
Which of the following complexes is expected to be colourless?
The correct order of field strength in the spectrochemical series is:
[Fe(CN)₆]⁴⁻ contains Fe²⁺ (d⁶). Given that CN⁻ is a strong-field ligand, the number of unpaired electrons and magnetic moment are:
[CoF₆]³⁻ contains Co³⁺ (d⁶). F⁻ is a weak-field ligand. The magnetic moment of this complex is approximately:
Two octahedral complexes of Cr³⁺ (d³) are: [Cr(NH₃)₆]³⁺ and [Cr(H₂O)₆]³⁺. Which statement about their colours is correct?
[Mn(CN)₆]⁴⁻ and [Mn(H₂O)₆]²⁺ both contain Mn²⁺ (d⁵). Calculate the difference in their magnetic moments (μ_high − μ_low):
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:
Pattern: P.CHE.U12.CRYSTAL_FIELD_HIGH_LOW_SPIN (observed NEET 2021, 2024)
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
Required
Determine (a) whether the complex is high-spin or low-spin, (b) the number of unpaired electrons, and (c) the magnetic moment.
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.
Formula
μ = √(n(n+2)) BM
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
Calculation
μ = 0 BM. The complex is diamagnetic.
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.
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.
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.) ---
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. 23Tetrahedral splitting is smaller than octahedral due to fewer/farther ligands.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Delta_o | octahedral splitting | J or eV |
| Delta_t | tetrahedral splitting | J or eV |
Same spin-only formula but n depends on high-spin/low-spin from CFT.
| Symbol | Quantity | SI Unit |
|---|---|---|
| n | unpaired electrons | - |
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.
Coordination compound with given ligand asking for magnetic moment, color, or spin state.
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.
Root cause: concept gap
Strong-field ligand (CN⁻, CO, NH₃ when applicable): Δ > P → low-spin. Weak-field (F⁻, Cl⁻, H₂O): Δ < P → high-spin. Use spectrochemical series.
11 questions from NEET 2021, 2022, 2023, 2024, 2025. Answers verified against NTA official keys.
Homoleptic complex from the following complexes is
Identify the incorrect statement from the following.
The IUPAC name of the complex- [Ag(H O) ][Ag(CN) ] is: 2 2 2
Ethylene diaminetetraacetate (EDTA) ion is :
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
forgets weak vs strong ligand
Default low-spin always
wrong order ligands
Uses non-alphabetical order
counts counter ions as ligands
Treats all attached species as ligands
Test yourself on this topic with real past-paper questions:
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