Q.Give the electronic configuration of the following complexes on the basis of Crystal Field Splitting theory.
[CoF6]3−, [Fe(CN)6]4− and [Cu(NH3)6]2+.
🔒You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.
🔒 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. …
Concept: Crystal Field Splitting – The arrangement of d-electrons in an octahedral complex depends on the strength of the ligand field (weak-field → high-spin; strong-field → low-spin) and the pairing energy.
Step 1: Identify metal ion and dn configuration
- [CoF6]3−: Co is +3, d6 (Co: 3d7, remove 3e⁻ → 3d6).
- [Fe(CN)6]4−: Fe is +2, d6 (Fe: 3d6, remove 2e⁻ → 3d6).
- [Cu(NH3)6]2+: Cu is +2, d9 (Cu: 3d10, remove 2e⁻ → 3d9).
Step 2: Apply ligand field strength
- F− is a weak-field ligand → small Δo, high-spin.
- CN− is a strong-field ligand → large Δo, low-spin.
- NH3 is intermediate but for d9, no spin-pairing choice (only one hole in t2g).
Step 3: Fill t2g and eg orbitals …
Crystal Field Splitting theory explains how ligands affect d-orbital energies, determining whether a complex is high-spin or low-spin. For [CoF6]3−: t2g4eg2 (high-spin); for [Fe(CN)6]4−: t2g6eg0 (low-spin); for [Cu(NH3)6]2+: t2g6eg3 (one unpaired electron).
The Core Idea: Why Ligands Matter
Crystal Field Splitting (CFS) is about how the arrangement of ligands around a central metal ion breaks the degeneracy of the d-orbitals. In an octahedral field, the five d-orbitals split into two sets: the lower-energy t2g set (dxy,dxz,dyz) and the higher-energy eg set (dz2,dx2−y2). The energy gap between them is called Δo (or 10 Dq).
The key question is: how do electrons fill these orbitals? That depends on two things:
- The strength of the ligand (how large Δo is).
- The pairing energy (P) — the energy cost to put two electrons in the same orbital.
If Δo is small (weak field ligands like F−), electrons prefer to occupy all five orbitals singly first (Hund's rule) — this gives a high-spin configuration. If Δo is large (strong field ligands like CN−), electrons pair up in the lower t2g set before occupying eg — this gives a low-spin configuration.
A common mistake is to forget that the oxidation state of the metal changes the d-electron count. Always determine the metal's oxidation state first, then count d-electrons.
Let's work through each complex step by step.
1. [CoF6]3−
Step 1: Find the oxidation state of Co.
The complex has a net charge of 3−. Each F− ligand carries a −1 charge. Let the oxidation state of Co be x.
x+6(−1)=−3⟹x−6=−3⟹x=+3
So, cobalt is in the +3 oxidation state: Co3+.
Step 2: Determine the d-electron count.
Cobalt (atomic number 27) has the ground state configuration [Ar]3d74s2. For Co3+, we remove three electrons (the two 4s electrons and one 3d electron), leaving 3d6. So, Co3+ has 6 d-electrons.
Step 3: Identify the ligand and its field strength.
F− (fluoride) is a weak field ligand. It lies near the end of the spectrochemical series. This means Δo is small, and Δo<P (pairing energy). Electrons will not pair up unless forced.
Step 4: Fill the orbitals.
With 6 electrons and a small Δo, we follow Hund's rule: fill all five d-orbitals singly first, then pair up.
- t2g set (3 orbitals): Put one electron in each → 3 electrons.
- eg set (2 orbitals): Put one electron in each → 2 electrons.
- Now we have placed 5 electrons. The 6th electron must go into the t2g set, pairing with an existing electron.
The configuration is: t2g4eg2.
For d6 in a weak field, the high-spin configuration always has 4 unpaired electrons. You can quickly check: t2g4 has 2 unpaired (since one orbital is doubly occupied), and eg2 has 2 unpaired — total 4 unpaired electrons.
Final configuration for [CoF6]3−: t2g4eg2 (high-spin).
2. [Fe(CN)6]4−
Step 1: Find the oxidation state of Fe.
Net charge = 4−. Each CN− ligand carries a −1 charge. Let Fe's oxidation state be x.
x+6(−1)=−4⟹x−6=−4⟹x=+2
So, iron is Fe2+.
Step 2: Determine the d-electron count.
Iron (atomic number 26) has ground state [Ar]3d64s2. For Fe2+, remove two 4s electrons → 3d6. So, Fe2+ has 6 d-electrons.
Step 3: Identify the ligand and its field strength.
CN− (cyanide) is a strong field ligand. It lies near the top of the spectrochemical series. This means Δo is large, and Δo>P. Electrons will pair up in the lower t2g set before occupying eg.
Step 4: Fill the orbitals.
With 6 electrons and a large Δo, we fill the t2g set completely first.
- t2g set: Put 2 electrons in each of the 3 orbitals → 6 electrons.
- eg set: Empty.
The configuration is: t2g6eg0.
This is a low-spin d6 configuration. All 6 electrons are paired, so the complex is diamagnetic (no unpaired electrons). This is a classic example of how a strong ligand can force pairing.
Final configuration for [Fe(CN)6]4−: t2g6eg0 (low-spin).
--- …
Crystal Field Splitting – Electronic Configuration Method
Method Name: Crystal Field Theory (CFT) – Filling of t2g and eg orbitals based on ligand field strength.
Steps to determine electronic configuration:
- Determine the oxidation state of the central metal ion (using charge balance).
- Write the d-electron count for the metal ion.
- Identify the geometry (here all are octahedral).
- Classify the ligand as weak field (high spin) or strong field (low spin) using the spectrochemical series.
- Fill the t2g and eg orbitals following Hund’s rule and the energy gap (Δo):
- Weak field (small Δo): electrons fill all orbitals singly before pairing (high spin).
- Strong field (large Δo): electrons pair in t2g first (low spin).
1. [CoF6]3−
- Oxidation state: Co: x+6(−1)=−3⇒x=+3 → Co(III)
- d-count: Co (atomic number 27) → Co3+ = [Ar]3d6
- Ligand: F− is a weak field ligand → small Δo → high spin
- Filling: t2g gets 4 electrons (one pair), eg gets 2 electrons (unpaired)
Electronic configuration:
t2g4eg2(high spin)
2. [Fe(CN)6]4−
- Oxidation state: Fe: x+6(−1)=−4⇒x=+2 → Fe(II)
- d-count: Fe (atomic number 26) → Fe2+ = [Ar]3d6
- Ligand: CN− is a strong field ligand → large Δo → low spin
- Filling: All 6 electrons pair up in t2g; eg is empty
Electronic configuration:
t2g6eg0(low spin)
3. [Cu(NH3)6]2+
- Oxidation state: Cu: x+6(0)=+2⇒x=+2 → Cu(II)
- d-count: Cu (atomic number 29) → Cu2+ = [Ar]3d9 …
Here are the most common mistakes students make when applying Crystal Field Splitting Theory (CFT) to determine electronic configurations, along with how to avoid each.
Mistake 1: Forgetting to Determine the Oxidation State of the Central Metal Ion
This is the most fundamental error. The electronic configuration of the free metal ion is the starting point, and you cannot get it right without the correct charge on the metal.
- The Mistake: Students directly use the configuration of the neutral atom (e.g., Fe for [Fe(CN)6]4−) instead of the ion (Fe2+).
- How to Avoid:
- Always calculate the oxidation state first. Let the charge on the complex = (charge on metal) + (charge on ligands × number of ligands).
- For [Fe(CN)6]4−: CN⁻ is a -1 ligand. So, x+6(−1)=−4⟹x=+2. The metal ion is Fe2+.
- Write the configuration of the free ion. Fe (atomic number 26): [Ar]3d64s2. For Fe2+, remove the 4s electrons first: [Ar]3d6.
- Key Exam Tip: The 4s orbital is always filled before 3d in the atom, but emptied before 3d when forming ions.
Mistake 2: Confusing Strong-Field and Weak-Field Ligands
The entire splitting pattern (high-spin vs. low-spin) depends on whether the ligand is a strong-field (low-spin) or weak-field (high-spin) ligand.
- The Mistake: Assuming all ligands behave the same way, or misidentifying ligands like CN⁻ (strong) and F⁻ (weak).
- How to Avoid:
- Memorize the Spectrochemical Series (partial list):
- Strong field (low spin): CN−>CO>en>NH3
- Weak field (high spin): H2O>F−>Cl−>Br−>I−
- Apply it:
- [CoF6]3−: F− is a weak field ligand → high spin complex.
- [Fe(CN)6]4−: CN− is a strong field ligand → low spin complex.
- [Cu(NH3)6]2+: NH3 is a strong field ligand → low spin complex.
- Memorize the Spectrochemical Series (partial list):
Mistake 3: Incorrectly Filling Electrons in the t2g and eg Orbitals
This is where the actual configuration is written. The rules of Hund's rule and Pauli's exclusion principle still apply, but the order of filling depends on the ligand field.
- The Mistake: For d4, d5, d6, or d7 configurations, students often fill all t2g orbitals first (pairing them) even for weak-field ligands, or they spread electrons out in eg before pairing in t2g for strong-field ligands.
- How to Avoid:
- For weak field (small Δo): Electrons fill all five orbitals singly first (Hund's rule) before pairing. This gives the maximum number of unpaired electrons.
- Example: Co3+ is d6. For [CoF6]3− (weak field): t2g4eg2 (4 unpaired electrons).
- For strong field (large Δo): Electrons fill the lower-energy t2g orbitals completely first (pairing them) before any electron goes into the higher-energy eg orbitals. This gives the minimum number of unpaired electrons.
- Example: Fe2+ is d6. For [Fe(CN)6]4− (strong field): t2g6eg0 (0 unpaired electrons).
- For weak field (small Δo): Electrons fill all five orbitals singly first (Hund's rule) before pairing. This gives the maximum number of unpaired electrons.
Mistake 4: Forgetting Jahn-Teller Distortions for d9 and d4 (high-spin) / d7 (low-spin)
For [Cu(NH3)6]2+, the Cu2+ ion is d9. A perfectly octahedral d9 complex is unstable and undergoes a Jahn-Teller distortion. …
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. …
🎓Unlock everything free for 14 days
- ✓Full step-by-step solutions
- ✓Concept-first explanations
- ✓Methods, shortcuts & mistakes
- ✓PYQ mapping + timed mock tests
Full access for 14 days. No credit card required.