Q.What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Crystal Field Splitting
Crystal Field Splitting: From Intuition to Precision
Imagine you are a negatively charged electron sitting on a metal ion. All around you, the space is perfectly spherical — every direction feels the same. Your energy depends only on how far you are from the nucleus, not on which way you face.
Now imagine that six negative ions (or the negative ends of polar molecules) march in from the x, y, and z axes and stop close to you. Suddenly, the space around you is no longer uniform. If you try to move straight toward one of these approaching ions, you feel a strong repulsion — that path costs extra energy. If you move between the axes (say, along a diagonal), you feel less repulsion because you are farther from the incoming charges.
This is the core intuition: when ligands approach a metal ion, they break the spherical symmetry of the space around the metal. Different directions in space are no longer equivalent. Electrons in orbitals that point directly at the ligands get pushed up in energy; electrons in orbitals that point between the ligands stay lower.
The Precise Statement
Crystal Field Splitting is the splitting of degenerate d orbitals of a transition metal ion into two or more sets of different energies, caused by the electrostatic repulsion between the metal's d electrons and the negative charge (or dipole) of surrounding ligands.
For the most common geometry — octahedral — here is what happens:
- Six ligands sit at the corners of an octahedron, along the +x, −x, +y, −y, +z, −z axes.
- The dx2−y2 and dz2 orbitals point their lobes directly along these axes. These are the eg set. They feel maximum repulsion → higher energy.
- The dxy, dxz, and dyz orbitals point their lobes between the axes (into the octahedral faces). These are the t2g set. They feel less repulsion → lower energy.
The energy gap between these two sets is denoted by Δo (or 10Dq). The t2g set drops by 0.4Δo and the eg set rises by 0.6Δo, keeping the average energy unchanged (the "barycentre" rule).
The labels eg and t2g come from group theory — they describe how the orbitals transform under the symmetry operations of an octahedron. You do not need to memorise the derivation, but the notation is standard in every exam.
Why This Matters
Crystal field splitting explains three things you will see repeatedly:
- Colour — electrons can jump from t2g to eg by absorbing visible light. The gap Δo determines the colour you see.
- Magnetism — if Δo is large, electrons pair up in the lower t2g set (low spin). If Δo is small, electrons spread out (high spin). This changes the number of unpaired electrons. …
Why this formula?
Crystal Field Splitting: Why the Energy Splitting Occurs
Crystal Field Theory (CFT) explains how the d-orbitals of a transition metal ion split in energy when placed in an electrostatic field created by surrounding ligands (anions or polar molecules). The key result is that five degenerate d-orbitals split into two or more sets with different energies. Let's understand why this happens.
1. The Starting Point: Degenerate d-Orbitals
In a free transition metal ion (no ligands), all five d-orbitals have the same energy (degenerate). Their shapes are:
- dxy, dxz, dyz — lobes lie between the x, y, z axes (called t2g set in octahedral symmetry)
- dx2−y2, dz2 — lobes point directly along the x, y, z axes (called eg set)
Key idea: The spatial orientation of each orbital determines how it interacts with approaching ligands.
2. The Octahedral Case: Why eg Orbitals Are Higher in Energy
Imagine six ligands approaching along the +x, –x, +y, –y, +z, –z axes (octahedral geometry).
What happens to dx2−y2 and dz2?
- Their lobes point directly at the ligands.
- The negatively charged ligands repel the electron density in these orbitals.
- This repulsion raises the energy of these orbitals — they become less stable (higher energy).
What happens to dxy, dxz, dyz?
- Their lobes point between the axes (e.g., dxy lobes lie in the xy-plane but at 45° to x and y).
- They avoid the ligands — less repulsion.
- Their energy is lower than the eg set.
The Splitting Pattern
Δoct=E(eg)−E(t2g)
Where:
- E(eg) = energy of dx2−y2 and dz2 (higher)
- E(t2g) = energy of dxy, dxz, dyz (lower)
- Δoct is called the crystal field splitting energy (CFSE)
Why the name? The eg orbitals are "doubly degenerate" (2 orbitals), t2g are "triply degenerate" (3 orbitals). The letters come from group theory symmetry labels.
3. The Energy Conservation Rule
The total energy of all five d-orbitals must remain constant (no energy is created or destroyed). So:
- The center of gravity (average energy) of the split set equals the original degenerate energy.
- For octahedral splitting:
- 2 eg orbitals go up by +0.6Δoct each
- 3 t2g orbitals go down by −0.4Δoct each
Check:
2×(+0.6Δ)+3×(−0.4Δ)=1.2Δ−1.2Δ=0
This conservation of energy is a fundamental constraint — the splitting is not arbitrary.
4. The Tetrahedral Case: Why It's Opposite and Smaller
In a tetrahedral complex, four ligands approach from alternate corners of a cube. The axes are different:
- The dxy, dxz, dyz orbitals now point closer to the ligands (more repulsion).
- The dx2−y2 and dz2 orbitals point away from ligands (less repulsion).
Result:
- e set ( dx2−y2, dz2 ) — lower energy
- t2 set ( dxy, dxz, dyz ) — higher energy
The splitting is inverted compared to octahedral.
Magnitude:
Δtet≈94Δoct
Why smaller?
- Only 4 ligands (vs. 6) → less total repulsion.
- Ligands are not directly along axes → weaker interaction. …
The key idea is that the observed colour of a transition metal complex is the complementary colour of the light it absorbs. This follows from the absorption of specific wavelengths in the visible spectrum due to crystal field splitting.
- When white light falls on a complex, electrons in the d-orbitals absorb a specific wavelength (λabs) to undergo a d-d transition. The energy gap Δ determines this wavelength: Δ=λabshc.
- The transmitted (or reflected) light is white light minus the absorbed wavelength. Our eyes perceive this remaining mixture as the complementary colour of the absorbed light. …
The observed colour of a complex is the complementary colour of the light it absorbs — if a complex absorbs light of a certain wavelength, we see the colour opposite to that wavelength on the colour wheel.
The key to understanding this lies in how our eyes perceive colour. A complex appears coloured because it absorbs some wavelengths of visible light and transmits (or reflects) the rest. The colour we see is not the absorbed colour — it is the sum of all the remaining wavelengths that reach our eyes.
Why this happens: The concept of complementary colours
When white light (which contains all visible wavelengths) falls on a complex, certain wavelengths are absorbed by the d-electrons jumping between split energy levels (crystal field splitting). The light that is not absorbed is what reaches your eye. Your brain interprets this mixture of leftover wavelengths as a single colour — the complementary colour of the absorbed light.
For example, if a complex absorbs strongly in the red region (around 650–700 nm), the transmitted light will be rich in blue-green wavelengths. So the complex appears blue-green, not red.
Observed colour=Complementary colour of absorbed wavelength
Step-by-step reasoning
-
White light contains all colours. When it hits a complex, some wavelengths are absorbed by the d-orbital electrons (due to crystal field splitting). The absorbed wavelength corresponds to the energy gap Δ: Δ=λabsorbedhc.
-
The transmitted light is white light minus the absorbed wavelengths. This is a subtractive process — the absorbed colour is removed from the spectrum.
-
Our eye sees the mixture of remaining wavelengths. The brain does not see individual wavelengths; it averages the remaining spectrum into a single perceived colour.
-
This perceived colour is the complement of the absorbed colour. On a standard colour wheel, complementary colours are opposite each other:
- Red ↔ Cyan (blue-green)
- Orange ↔ Blue
- Yellow ↔ Violet
- Green ↔ Magenta
-
The relationship is inverse, not direct. A common mistake is to think the complex appears the same colour as the light it absorbs. That is wrong — the complex appears the opposite colour.
Never say "the complex is red because it absorbs red light." That is incorrect. If it absorbs red light, it appears cyan (the complement of red). The absorbed colour and observed colour are always complementary.
A concrete example
Consider the complex [Ti(H2O)6]3+. It absorbs light at about 500 nm (green region). The complementary colour of green is violet-red (magenta). Indeed, this complex appears violet-red in solution.
| Absorbed wavelength (nm) | Absorbed colour | Observed (complementary) colour |
|--------------------------|-----------------|----------------------------------| …
Concept: Electronic Transitions and the Colour Wheel
The colour we observe in a transition metal complex is complementary to the colour of light it absorbs. This is governed by the colour wheel — a circular diagram where opposite colours are complementary.
Method: Complementary Colour Rule
Steps
-
Identify the absorbed wavelength (λabs) from the complex's absorption spectrum or given data.
-
Determine the colour of absorbed light using the visible spectrum:
- 380−450 nm → violet
- 450−495 nm → blue
- 495−570 nm → green
- 570−590 nm → yellow
- 590−620 nm → orange
- 620−750 nm → red
-
Find the complementary colour from the colour wheel:
- Violet ↔ Yellow-green
- Blue ↔ Yellow
- Green ↔ Red
- Yellow ↔ Violet-blue
- Orange ↔ Blue-green
- Red ↔ Green
-
State the observed colour — this is the complementary colour from step 3.
Example …
Here are the most common mistakes students make when linking the observed colour of a coordination complex to the wavelength of light absorbed, along with clear ways to avoid each.
Mistake 1: Confusing “absorbed” with “observed” colour
The error:
Students often say “the complex is blue because it absorbs blue light.” This is wrong. A complex appears blue because it absorbs orange/red light (the complementary colour) and transmits/reflects blue light.
Why it happens:
The eye sees the light that is not absorbed. The absorbed wavelength is complementary to the observed colour.
How to avoid:
- Memorise the colour wheel for d-block complexes.
- Use the rule:
Observed colour = complementary colour of absorbed light.
- Example: If a complex absorbs at λ≈580 nm (yellow), it appears violet (complement of yellow).
Mistake 2: Forgetting the complementary colour relationship
The error:
Students try to guess the observed colour without knowing the complementary pair (e.g., green–red, blue–orange, yellow–violet).
Why it happens:
The relationship is not intuitive — it requires memorisation of the visible spectrum’s complementary pairs.
How to avoid:
- Draw or visualise the colour wheel:
- Red ↔ Green
- Orange ↔ Blue
- Yellow ↔ Violet
- Practice with examples:
- Absorbs blue → appears orange
- Absorbs green → appears red
- Absorbs violet → appears yellow
Mistake 3: Ignoring the wavelength range of visible light
The error:
Students treat “absorbed wavelength” as any number without checking if it falls in the visible range (400–700 nm). If a complex absorbs in the UV or IR, it appears white or colourless (no visible absorption).
Why it happens:
They assume all d–d transitions produce colour, but if Δ is very large or very small, absorption is outside visible range.
How to avoid:
- Remember:
- λabsorbed<400 nm (UV) → complex appears colourless (transmits all visible light).
- λabsorbed>700 nm (IR) → complex appears colourless (absorbs no visible light).
- Only when λabsorbed is between 400–700 nm does the complex show colour.
Mistake 4: Mixing up the effect of ligand field strength
The error:
Students think “stronger ligand → larger Δ → shorter λ absorbed” but then incorrectly predict the observed colour.
Why it happens:
They forget that shorter λ (higher energy) corresponds to a different complementary colour.
How to avoid:
- Use the sequence:
Strong field → large Δ → absorbs shorter λ (blue/violet) → appears orange/yellow
Weak field → small Δ → absorbs longer λ (red/orange) → appears green/blue
- Practice with examples:
- [Co(NH3)6]3+ (strong field) absorbs violet → appears yellow
- [CoF6]3− (weak field) absorbs red → appears green
--- …
Showing the 12 most recent of 25 on this concept.
- AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQQ.In which of the following, ratio of t2g and eg electrons is 3:2? (A) [FeF6]3− (B) [CoF6]3− (C) [Fe(CN)6]3− (D) [Co(NH3)6]3+
›Reveal solutionSolution
This tests crystal field electron configurations for octahedral complexes; the answer is [FeF6]3−, the only complex with t2g:eg=3:2.
Concept and Intuition
The distribution of d-electrons between t2g and eg orbitals in an octahedral complex depends on both the metal's d-electron count and whether the ligand is weak-field (high spin, obeying Hund's rule across all 5 orbitals before pairing) or strong-field (low spin, filling t2g completely before touching eg).
Step-by-Step Solution
- [FeF6]3−: Fe3+ is d5. F− is a weak-field ligand ⇒ high spin: electrons fill all 5 orbitals singly first — t2g3eg2. Ratio = 3:2. Matches.
- [CoF6]3−: Co3+ is d6. F− weak field ⇒ high spin: t2g4eg2. Ratio = 4:2 = 2:1. Does not match.
- [Fe(CN)6]3−: Fe3+ is d5. CN− is a strong-field ligand ⇒ low spin: t2g5eg0. Ratio = 5:0. Does not match. …
- AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQQ.Which of the following has maximum number of electrons in t2g orbitals? (A) [Fe(CN)6]3− (B) [FeCl6]3− (C) [CoCl6]3− (D) [Co(NH3)6]3+
›Reveal solutionSolution
This tests crystal field theory: assign the metal oxidation state, its dn configuration, and whether the ligand is strong- or weak-field, then fill t2g/eg. [Co(NH3)6]3+, a low-spin d6 complex, has the maximum of 6 t2g electrons.
Concept and Intuition
In an octahedral field, d-orbitals split into the lower-energy t2g set (3 orbitals) and higher-energy eg set (2 orbitals). Whether electrons pair up in t2g first (low spin) or spread out with parallel spins across both sets first (high spin, Hund's rule) depends on whether the ligand is strong-field (large Δo, causes pairing) or weak-field (small Δo, favours unpaired spread). CN− and NH3 are strong-field ligands; Cl− is a weak-field ligand.
Step-by-Step Solution
- [Fe(CN)6]3−: Fe3+ is 3d5. CN− is strong field ⇒ low spin: all 5 electrons pack into t2g first ⇒t2g5eg0. t2g count = 5.
- [FeCl6]3−: Fe3+ is 3d5. Cl− is weak field ⇒ high spin: electrons fill all 5 d-orbitals singly first ⇒t2g3eg2. t2g count = 3. …
- AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQQ.Match the following. List – I (Complex ion) : (A) [MnCl6]3− (B) [Mn(CN)6]3− (C) [Fe(CN)6]4− (D) [CoF6]3− List – II (d-Electronic configuration) : (I) t2g6eg0 (II) t2g4eg2 (III) t2g3eg1 (IV) t2g4eg0 The correct answer is: (A) A-III, B-IV, C-I, D-II (B) A-III, B-IV, C-II, D-I (C) A-II, B-III, C-IV, D-I (D) A-II, B-I, C-IV, D-III
›Reveal solutionSolution
Determining the metal's oxidation state and d-electron count, then applying strong-field (low spin) vs weak-field (high spin) splitting for each ligand, matches A-III, B-IV, C-I, D-II.
Concept and Intuition
For an octahedral complex, the d-electron filling pattern depends on (i) how many d-electrons the metal ion has, and (ii) whether the ligand is strong-field (causes pairing → low spin, maximises t2g filling before using eg) or weak-field (no forced pairing → high spin, electrons spread out following Hund's rule across both t2g and eg first). CN⁻ is a strong field ligand; Cl⁻ and F⁻ are weak field ligands.
Step-by-Step Solution
- A: [MnCl6]3− — Cl is −1 each (6×−1=−6); overall charge −3 ⟹ Mn is +3. Mn (Z=25): [Ar]3d54s2; Mn3+ removes the two 4s electrons and one 3d electron ⟹ d4. Cl⁻ is weak field ⟹ high spin: t2g3eg1 — matches (III).
- B: [Mn(CN)6]3− — same Mn3+, d4, but CN⁻ is strong field ⟹ low spin: electrons pair up in t2g before occupying eg: t2g4eg0 — matches (IV).
- C: [Fe(CN)6]4− — CN is −1 each (−6); overall charge −4 ⟹ Fe is +2. Fe (Z=26): [Ar]3d64s2; Fe2+ removes 4s2 ⟹ d6. CN⁻ strong field ⟹ low spin: t2g6eg0 — matches (I). …
- AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQQ.In which of the following sets, given species are not only diamagnetic in nature but also inner orbital complexes? I. [Ni(CN)4]2−,[Fe(CN)6]4− II. [Mn(CN)6]3−,[Fe(CN)6]3− III. [Co(en)3]3+,[Ni(CN)4]2− The correct answer is (only = only) (A) I, II only (B) I, III only (C) II, III only (D) I, II, III
›Reveal solutionSolution
"Diamagnetic + inner orbital" needs a strong-field ligand forcing all d-electrons to pair using (n−1)d orbitals. Sets I and III both satisfy this; set II fails because its complexes are paramagnetic.
Concept and Intuition
An inner orbital complex uses an (n−1)d orbital in its hybridisation (dsp2 for 4-coordinate square planar, or d2sp3 for 6-coordinate octahedral), which happens only with strong-field ligands (like CN−, en) that force pairing of the metal's d-electrons (low spin). Diamagnetism additionally requires that ALL electrons end up paired — so we must check the d-electron count of the actual oxidation state, not just "is the ligand strong field".
Step-by-Step Solution
- [Ni(CN)4]2−: Ni2+ is 3d8. With strong-field CN− in a 4-coordinate geometry, Ni2+ adopts square planar geometry via dsp2 hybridisation (using one 3d orbital) — this is the textbook example of an inner-orbital, diamagnetic complex (all 8 d-electrons pair up in the four lower-lying d-orbitals, leaving the dx2−y2 empty for the hybrid set).
- [Fe(CN)6]4−: Fe2+ is 3d6. Strong field CN− gives low-spin d2sp3 (inner orbital), t2g6eg0 — all 6 electrons paired → diamagnetic. Set I: both diamagnetic + inner orbital. ✓
- [Mn(CN)6]3−: Mn3+ is 3d4. Even low-spin (strong field, d2sp3, inner orbital), t2g4 across 3 orbitals means 3 are singly occupied first, and the 4th electron must pair into one of them: net 2 unpaired electrons → paramagnetic. This alone disqualifies set II (even without checking the second species). …
- AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQQ.Which of the following orders correctly represent the strength of ligands in the spectrochemical series? I. I−<Br−<S2−<SCN− II. H2O<NCS−<NH3<en III. Cl−<F−<N3−<OH− Correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III
›Reveal solutionSolution
Orders I and II match the standard spectrochemical series; order III has F− and N3− swapped, so it is incorrect.
Concept and Intuition
The spectrochemical series ranks ligands by the crystal field splitting energy (Δ) they produce, from weak field to strong field. It is a memorised empirical order (from spectroscopic data) rather than something derivable from first principles, so each proposed order must be checked against the standard sequence.
Step-by-Step Solution
- Standard spectrochemical series (weak → strong): I−<Br−<S2−<SCN−<Cl−<N3−<F−<OH−<C2O42−<H2O<NCS−<CH3CN<py≈NH3<en<…<CN−≈CO.
- I: I−<Br−<S2−<SCN− — matches the series exactly in this stretch. Correct.
- II: H2O<NCS−<NH3<en — matches the series exactly (water is weaker than NCS⁻, which is weaker than ammonia, which is weaker than ethylenediamine). Correct. …
- AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQQ.Match the following List-I (aquated ion) List-II (colour) A) Ni2+ I) violet B) Fe3+ II) blue C) Mn3+ III) yellow D) V4+ IV) red V) green Correct answer is (A) A-V, B-III, C-IV, D-II (B) A-IV, B-V, C-I, D-III (C) A-I, B-III, C-IV, D-V (D) A-V, B-III, C-I, D-II
›Reveal solutionSolution
Ni2+-green, Fe3+-yellow, Mn3+-violet, V4+-blue → option (D).
Concept and Intuition
Colours of hydrated transition-metal ions arise from d-d electronic transitions and are tabulated in NCERT. Matching each ion to its standard aqueous colour resolves the question.
Step-by-Step Solution
- Ni2+ (d8): green → V.
- Fe3+ (d5): yellow → III.
- Mn3+ (d4): violet → I.
- V4+, present as the vanadyl VO2+ ion (d1): blue → II. …
- AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQQ.In which of the following, complex ions are not in correct order with respect to their magnitude of crystal field splitting ? (A) [Fe(H2O)6]3+>[FeF6]3− (B) [Fe(en)3]3+>[Fe(NCS)6]3− (C) [Fe(CN)6]4−>[Fe(H2O)6]2+ (D) [Fe(H2O)6]2+>[Fe(NH3)6]2+
›Reveal solutionSolution
Tests the spectrochemical series ordering of common ligands and its effect on crystal field splitting energy, Δ₀.
Concept and Intuition
Crystal field splitting energy Δ₀ for an octahedral complex depends chiefly on the field strength of the ligand (for a fixed metal ion/oxidation state). The spectrochemical series ranks ligands by this field strength, largely independent of the metal: I−<Br−<S2−<SCN−<Cl−<N3−,F−<OH−<C2O42−<H2O<NCS−<py,NH3<en<NO2−<CN−<CO.
Step-by-Step Solution
- (A): [Fe(H2O)6]3+ vs [FeF6]3− — H2O lies above F− in the series, so H2O gives greater splitting: order is correct.
- (B): [Fe(en)3]3+ vs [Fe(NCS)6]3− — en (a strong bidentate N-donor) is well above NCS⁻: order is correct.
- (C): [Fe(CN)6]4− vs [Fe(H2O)6]2+ — CN⁻ is one of the strongest-field ligands, far above H₂O: order is correct. …
- AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQQ.In which one of the following complexes the metal ion has t2g3eg2 configuration ? (A) [Mn(H2O)6]2+ (B) [Fe(H2O)6]2+ (C) [Co(NH3)6]3+ (D) [Ni(H2O)6]2+
›Reveal solutionSolution
This tests d-electron counting plus crystal field splitting: t2g3eg2 is the high-spin d5 pattern, and only Mn2+ among the choices is d5.
Concept and Intuition
In an octahedral field the five d orbitals split into the lower t2g (3 orbitals) and upper eg (2 orbitals) sets. The configuration t2g3eg2 has 3+2=5 electrons distributed one-per-orbital across all five d orbitals — this is only possible for a d5 ion in the high-spin (weak field) case, where electrons singly occupy every orbital before any pairing (Hund's rule) because the ligand field splitting Δo is too small to force pairing.
Step-by-Step Solution
- Count total electrons in the configuration: t2g3eg2=5 electrons ⇒ the metal ion is d5.
- Find each ion's dn count:
- Mn2+: Mn (Z=25) is [Ar]3d54s2; losing the two 4s electrons gives Mn2+=3d5.
- Fe2+: Fe (Z=26) is [Ar]3d64s2; Fe2+=3d6.
- Co3+: Co (Z=27) is [Ar]3d74s2; removing 3 electrons gives Co3+=3d6.
- Ni2+: Ni (Z=28) is [Ar]3d84s2; Ni2+=3d8.
- Only Mn2+ is d5. …
- AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQQ.Arrange the following complexes in increasing order of number of unpaired electrons present in central metal ion I. [Fe(CN)6]3− II. [Co(C2O4)3]3− III. [Mn(CN6)]3− (A) II < III < I (B) III < I < II (C) II < I < III (D) I < II < III
›Reveal solutionSolution
Tests counting unpaired d-electrons in low-spin octahedral complexes; the increasing order of
unpaired electrons is II (Co, 0) < I (Fe, 1) < III (Mn, 2).
Concept and Intuition
In an octahedral crystal field, d orbitals split into a lower t2g set (3 orbitals) and a
higher eg set (2 orbitals). Strong-field ligands (like CN−) and chelating ligands
with high effective field strength for a high-oxidation-state metal (like oxalate on Co3+)
favour the low-spin arrangement, in which electrons fill and pair up in t2g before
occupying eg at all.
Step-by-Step Solution
- [Fe(CN)6]3−: Fe is +3, so Fe3+ is d5. CN− is a strong-field ligand → low spin: t2g5eg0. Filling t2g's 3 orbitals with 5 electrons gives 2 fully paired orbitals + 1 singly occupied orbital → 1 unpaired electron.
- [Co(C2O4)3]3−: Co is +3, so Co3+ is d6. Co3+ complexes are strongly disposed to low spin even with moderately strong chelating ligands like oxalate → t2g6eg0, all three t2g orbitals fully paired → 0 unpaired electrons.
- [Mn(CN)6]3−: Mn is +3, so Mn3+ is d4. CN− forces low spin: …
- AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQQ.[Ni(H2O)6]2+ on reaction with ethane-1,2-diamine forms violet colour complex. Its formula and magnetic property respectively are (A) [Ni(H2O)4en]2+, Paramagnetic nature (B) [Ni(H2O)2en2]2+, Diamagnetic nature (C) [Ni(en)3]2+, Paramagnetic nature (D) [Ni(H2O)2en2]2+, Paramagnetic nature
›Reveal solutionSolution
[Ni(H2O)6]2+ reacts with excess ethylenediamine to fully substitute all six water ligands (chelate effect), forming the octahedral, violet [Ni(en)3]2+ complex, which is paramagnetic since octahedral d8 Ni(II) always retains 2 unpaired electrons.
Concept and Intuition
Ethane-1,2-diamine (en) is a bidentate ligand, and chelation with a multidentate ligand is entropically strongly favoured (the chelate effect) over monodentate water ligands. With enough en added, all six water molecules on [Ni(H2O)6]2+ are displaced by three en molecules (each en occupying two coordination sites), giving the well-known tris(ethylenediamine)nickel(II) complex. For Ni2+ (d8) in an octahedral field, the electron configuration is t2g6eg2 — the two eg electrons must occupy the two degenerate eg orbitals singly (Hund's rule), so an octahedral d8 complex is always paramagnetic (2 unpaired electrons), regardless of the ligand's field strength; only a square-planar geometry (as with CN− or in Pt2+/Pd2+ complexes) can force pairing to diamagnetism for d8.
Step-by-Step Solution
- Reaction: [Ni(H2O)6]2++3en→[Ni(en)3]2++6H2O (complete displacement, driven by the chelate effect).
- Geometry: octahedral (6-coordinate, three bidentate en ligands). …
- AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQQ.The paramagnetic complex ion which has no unpaired electrons in t2g orbitals is (A) [Fe(CN)6]4− (B) [Fe(CN)6]3− (C) [Zn(NH3)6]2+ (D) [Ni(NH3)6]2+
›Reveal solutionSolution
This tests crystal-field electron filling (t2g vs eg) across four octahedral complexes; the one paramagnetic ion with an empty t2g subshell of unpaired spins is [Ni(NH3)6]2+.
Concept and Intuition
In an octahedral field the five d-orbitals split into the lower t2g set (3 orbitals) and upper eg set (2 orbitals). Whether electrons pair up in t2g before occupying eg depends on whether the ligand is strong field (low spin) or weak/medium field (high spin), and on the metal's dn count.
Step-by-Step Solution
- [Fe(CN)6]4−: Fe2+=d6. CN− is a strong-field ligand ⇒ low spin: configuration t2g6eg0. All electrons paired — diamagnetic, so it is not even paramagnetic; eliminated.
- [Fe(CN)6]3−: Fe3+=d5, low spin with CN−: t2g5eg0. This has 1 unpaired electron, but it sits IN t2g — fails the "no unpaired electrons in t2g" condition.
- [Zn(NH3)6]2+: Zn2+=d10, fully filled t2g6eg4 regardless of field strength — diamagnetic, no unpaired electrons anywhere; eliminated. …
- AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQQ.How many of the following ligands are stronger than H2O? S2−,Br−,C2O42−,CN−,en,NH3,CO,OH− (A) 5 (B) 3 (C) 4 (D) 6
›Reveal solutionSolution
Checking each of the 8 ligands against the spectrochemical series relative to H₂O shows exactly 4 (CN⁻, en, NH₃, CO) are stronger field ligands than water.
Concept and Intuition
The spectrochemical series ranks ligands by their crystal-field splitting strength, from weak field (e.g. halides) to strong field (e.g. CO, CN⁻). Water sits roughly in the middle, so any given ligand must be individually checked against its position relative to H₂O.
Step-by-Step Solution
- Standard spectrochemical series (weak → strong): I−<Br−<S2−<SCN−<Cl−<N3−<F−<OH−<C2O42−<H2O<NCS−<NH3<en<NO2−<CN−≈CO.
- Classify each of the 8 given ligands relative to H2O:
- S2− — weaker than H2O.
- Br− — weaker than H2O.
- C2O42− — weaker than H2O (just below it in the series).
- CN− — stronger than H2O.
- en — stronger than H2O.
- NH3 — stronger than H2O.
- CO — stronger than H2O.
- OH− — weaker than H2O. …
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