Q.(a)
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🔒 Start your 14-day free trial to unlock the full solution →Part (a)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. …
Part (b)Concept understanding — Magnetic Moment Calculation
From a Paperclip to a Magnet: The Intuition
You already know that a magnet can pick up iron nails. But what is actually happening inside that nail when it gets near the magnet? And why does a plastic comb, rubbed on hair, pick up tiny bits of paper — but never iron filings?
The answer lies in magnetization — the process by which a material becomes magnetic.
Think of a piece of iron as a chaotic crowd of tiny compass needles. Each needle is an atomic magnetic moment (a tiny magnet, arising from the spin of electrons). In unmagnetized iron, these needles point in random directions. Their magnetic effects cancel out, so the iron as a whole shows no net magnetism.
Now bring a strong magnet close. Its magnetic field acts like a command: "Line up!" The tiny compass needles inside the iron start rotating, aligning themselves with the external field. The more they align, the stronger the iron's own magnetic field becomes. This alignment is magnetization.
Magnetization is not the same as inducing a current. It is a purely magnetic reorientation of atomic dipoles inside a material.
The Precise Definition
Magnetization (M) is the net magnetic dipole moment per unit volume of a material. It tells you how strongly a material is magnetized — how many tiny atomic magnets are aligned, and in which direction.
If a material has N atoms per unit volume, each with an average magnetic moment μavg, then:
M=Nμavg
The SI unit of M is amperes per metre (A/m). Why? Because a magnetic dipole moment has units of A·m², and dividing by volume (m³) gives A/m.
M=volumetotal magnetic dipole moment
How Magnetization Connects to the Magnetic Field
When a material gets magnetized, it produces its own magnetic field. The total magnetic field B inside the material is the sum of:
- The external applied field H (caused by free currents, like the current in a solenoid)
- The material's response — the magnetization M
The fundamental relation is:
B=μ0(H+M)
where μ0=4π×10−7T⋅m/A is the permeability of free space.
Do not confuse H (magnetic field intensity, or "magnetizing field") with B (magnetic flux density). H is what you apply; M is what the material does; B is the total field you measure.
The Three Kinds of Magnetic Materials
Not all materials respond the same way to an external field. The magnetization M is proportional to H for most materials (at least for small fields):
M=χmH
where χm is the magnetic susceptibility — a dimensionless number that tells you how easily a material magnetizes.
| Material Type | χm | Behaviour | Example |
|---|---|---|---|
| Diamagnetic | Small and negative (≈−10−5) | Weakly repelled by a magnet; M opposes H | Water, copper, bismuth |
| Paramagnetic | Small and positive (≈10−5 to 10−3) | Weakly attracted; M aligns with H | Aluminium, oxygen gas |
| Ferromagnetic | Large and positive (≫1) | Strongly attracted; M can be huge and persists even after H is removed | Iron, nickel, cobalt |
Why this formula?
Magnetic Moment Calculation: Why the Formula Holds
Let's build this from first principles — understanding the why before the formula.
1. What is Magnetic Moment?
A magnetic moment (μ) is a measure of the strength and orientation of a magnet or current loop. It tells us how strongly an object will interact with an external magnetic field.
The core idea: any moving charge creates a magnetic field. A loop of current is like a tiny bar magnet — its magnetic moment quantifies this.
2. The Fundamental Formula: Current Loop
The Setup
Consider a planar loop of wire carrying a steady current I, enclosing an area A.
Why μ=IA?
Step 1: Force on a moving charge
A charge q moving with velocity v in a magnetic field B experiences:
F=q(v×B)
Step 2: Torque on a current loop
For a rectangular loop of sides a and b (A=ab), placed in a uniform B:
- Current I means charge flows. On side of length a, the force magnitude is F=IaB (since I=tq and v=ta).
- These forces on opposite sides form a couple (equal, opposite, not collinear).
- Torque τ=force×perpendicular distance=(IaB)×(bsinθ)
Step 3: Recognize the pattern
τ=I(ab)Bsinθ=IABsinθ
This looks exactly like:
τ=μBsinθ
Comparing, we identify:
μ=IA
Why this works: The torque on a current loop is proportional to the current and the area — this product naturally defines the magnetic moment.
3. For a Single Moving Charge (Orbital Magnetic Moment)
The Setup
An electron of charge −e moves in a circular orbit of radius r with speed v.
Why μ=2evr?
Step 1: Treat orbit as a current loop
- Time for one revolution: T=v2πr
- Current (charge per unit time): I=Te=2πrev
Step 2: Apply μ=IA
- Area of orbit: A=πr2
- So: μ=(2πrev)(πr2)=2evr
Step 3: Express in terms of angular momentum
- Orbital angular momentum: L=mvr
- Therefore: μ=2meL
Why this matters: The magnetic moment is directly proportional to angular momentum. The factor 2me is called the gyromagnetic ratio — it links mechanics to magnetism.
4. For a Solenoid (Many Turns)
The Setup
A solenoid of N turns, length l, carrying current I, cross-sectional area A.
Why μ=NIA? …
Part (a)
(i) d5, Δo<P (weak field / high spin): all five orbitals singly filled →
t2g3eg2(5 unpaired electrons)
(ii) [Fe(CN)6]3− vs [Fe(CN)6]4−: CN− is strong field → low spin.
- [Fe(CN)6]3−: Fe3+, d5 → t2g5 → 1 unpaired → weakly paramagnetic.
- [Fe(CN)6]4−: Fe2+, d6 → t2g6 → 0 unpaired → diamagnetic. …
Part (a): (i) high-spin t2g3eg2; (ii) [Fe(CN)6]3− = 1 unpaired (weakly paramagnetic), [Fe(CN)6]4− = 0 (diamagnetic); (iii) 3 ions. Part (b): (i) μ=4.90 BM; (ii) tetraaquadichloridochromium(III) chloride; (iii) [Fe(C2O4)3]3− (chelate effect).
Part (a)
(i) d5 configuration when Δo<P
Weak field (Δo<P) means it is cheaper to place electrons in eg than to pair, so all five d orbitals are singly occupied (Hund's rule) — the high-spin case:
t2g3eg2(5 unpaired)
(ii) [Fe(CN)6]3− vs [Fe(CN)6]4−
CN− is a strong-field ligand → large Δo → low spin.
- [Fe(CN)6]3−: Fe is +3, 3d5 → t2g5eg0 → one unpaired electron → weakly paramagnetic.
- [Fe(CN)6]4−: Fe is +2, 3d6 → t2g6eg0 → no unpaired electrons → diamagnetic.
(iii) Ions from [Co(NH3)6]Cl2
[Co(NH3)6]Cl2→[Co(NH3)6]2++2Cl− …
Showing the 12 most recent of 36 on this concept.
- CBSE 2026Set ANNUAL1 markQ.Write the formula for calculating 'spin only' magnetic moment.
›Reveal solutionSolution
The spin-only formula estimates a transition metal ion's magnetic moment purely from its number of unpaired electrons, ignoring orbital contribution.
…
- CBSE 2026Set ANNUAL1 markQ.Write any one example of low spin complex.
›Reveal solutionSolution
A low-spin complex forms when a strong-field ligand causes the d electrons to pair up in the lower-energy t2g set rather than spreading into eg, reducing the number of unpaired electrons.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Assertion [A]: [Ni(CN)4]2- is a square-planar and diamagnetic. Reason [R]: It has no unpaired electrons due to presence of strong field.(a) Both [A] and [R] are true and [R] is the correct explanation of [A].(b) Both [A] and [R] are true, but [R] is not the correct explanation of [A].(c) [A] is true, but [R] is false.(d) [A] is false, but [R] is true.
›Reveal solutionSolution
[Ni(CN)4]2− is indeed square planar and diamagnetic, and this is correctly explained by CN⁻ being a strong field ligand that forces electron pairing, leaving no unpaired electrons.
In [Ni(CN)4]2−, nickel is in the +2 oxidation state: Ni2+ has configuration 3d8 (8 electrons: t2g6eg2 in a free-ion sense, or 3d8=↑↓↑↓↑↓↑ ↑).
…
- CBSE 2026Set ANNUAL1 markQ.Give one example of a complex having tetrahedral geometry and paramagnetic in nature.
›Reveal solutionSolution
[NiCl4]2− is the standard example of a tetrahedral, paramagnetic complex, arising from sp3 hybridisation of Ni2+ with the weak-field Cl− ligand.
Why [NiCl4]2− fits
Ni has configuration [Ar]3d84s2; in Ni2+, this becomes 3d8. Cl− is a weak-field ligand (low in the spectrochemical series), so it does not force pairing of the 3d electrons. With four ligands and no d-orbital freed by pairing, nickel uses one 4s and three 4p orbitals — sp3 hybridisation — giving a **tetrahedr …
- CBSE 2026Set ANNUAL1 markQ.Which one is an inner-orbital complex? [Co(NH3)6]3+ or [CoF6]3−
›Reveal solutionSolution
Because NH3 is a strong-field ligand, Co3+'s d-electrons pair up and the complex uses the inner (n−1)d orbitals for hybridisation — making [Co(NH3)6]3+ the inner-orbital complex, unlike [CoF6]3−.
Analysis
Co3+ has the configuration 3d6 in both complexes; the difference lies in the field strength of the ligand.
- In [Co(NH3)6]3+: NH3 is a strong-field ligand. It forces all 6 d-electrons to pair up within three 3d orbitals (t2g6), freeing the other two 3d orbitals for hybridisation. Cobalt then hybridises as d2sp3, using inner (n−1)d, i.e. 3d, orbitals — this is an inner-orbital (low-spin) complex, diamagnetic. …
- CBSE 2026Set ANNUAL1 markQ.According to VBT, which one has the highest paramagnetic character? [Cr(H2O)6]3+ or [Fe(H2O)6]2+
›Reveal solutionSolution
Counting unpaired d-electrons for each ion under VBT shows Fe2+ (d6, high-spin, 4 unpaired) is more paramagnetic than Cr3+ (d3, always 3 unpaired).
[Cr(H2O)6]3+
Cr (Z=24) is [Ar]3d54s1; Cr3+ removes 3 electrons to give 3d3. With only 3 electrons for the three t2g orbitals, Hund's rule places one electron in each — t2g3 — giving 3 unpaired electrons, regardless of whether the ligand is weak- or strong-field (there's no way to pair up 3 electrons across 3 orbitals to reduce this further).
[Fe(H2O)6]2+
…
- CBSE 2026Set ANNUAL1 markQ.Write True or False: Value of magnetic moment of a divalent ion in aqueous solution having atomic number 25, will be 5.92 B.M.
›Reveal solutionSolution
Mn2+ has 5 unpaired electrons, giving a spin-only moment of 5.92 B.M., so the statement is true.
Atomic number 25 = manganese, [Ar] 3d5 4s2. The divalent ion Mn2+ = [Ar] 3d5, which has 5 unpaired electrons.
…
- CBSE 2026Set ANNUAL1 markMCQQ.Which one of the following metal ions is likely to have a magnetic moment of 1.73 BM?(a) Fe²⁺(b) Mn²⁺(c) Cr²⁺(d) Cu²⁺
›Reveal solutionSolution
Using μ = √(n(n+2)) BM, 1.73 BM means n = 1 unpaired electron; Cu²⁺ (d⁹) is the only ion with one unpaired electron — option (D).
The spin-only magnetic moment is μ=n(n+2) BM, where n is the number of unpaired electrons. A value of 1.73 BM gives 1(1+2)=3=1.73, so n=1 unpaired electron.
Now count unpaired electrons for each ion:
- Fe2+: 3d6 → 4 unpaired (μ≈4.9 BM). …
- CBSE 2026Set ANNUAL1 markQ.The oxidation number of all the alkali metals in their compounds is ________.
›Reveal solutionSolution
[!TLDR]
+1
Method
Alkali metals (Group 1) have one valence electron and invariably show a +1 oxida …
- CBSE 2025Set 56/5/11 markMCQQ.In which of the following groups are both ions coloured in aqueous solution ? I. Cu+ II. Ti4+ III. Co2+ IV. Fe2+ [Atomic number : Cu = 29, Ti = 22, Co = 27, Fe = 26] (A) I and II (B) II and III (C) III and IV (D) I and IV
›Reveal solutionSolution
The colour of a transition metal ion in aqueous solution depends on the presence of unpaired d-electrons, which allow d-d transitions. Both Co2+ and Fe2+ have unpaired d-electrons and are coloured, while Cu+ and Ti4+ have fully filled or empty d-subshells and are colourless. The correct pair is III and IV, i.e., option (C).
The question asks which two ions among the given four are coloured in aqueous solution. Colour in transition metal ions arises from the absorption of visible light due to electronic transitions between split d-orbitals — the famous d-d transition. But this only happens if the d-subshell is partially filled (i.e., has at least one unpaired electron and at least one vacant orbital). If the d-subshell is completely empty (d0) or completely filled (d10), no d-d transition is possible, and the ion is colourless (or white) in solution.
Let’s examine each ion one by one.
-
Cu+ (Copper(I))
Atomic number of Cu = 29. Neutral Cu has configuration [Ar]3d104s1.
Cu+ loses the 4s electron, so its configuration becomes [Ar]3d10.
The d-subshell is completely filled. No d-d transitions possible.
Result: Colourless in aqueous solution.
-
Ti4+ (Titanium(IV))
Atomic number of Ti = 22. Neutral Ti has [Ar]3d24s2.
Ti4+ loses all four valence electrons (two from 4s and two from 3d), so its configuration becomes [Ar]3d0.
The d-subshell is completely empty. No d-d transitions possible.
Result: Colourless in aqueous solution.
-
Co2+ (Cobalt(II))
Atomic number of Co = 27. Neutral Co has [Ar]3d74s2.
Co2+ loses the two 4s electrons, giving [Ar]3d7.
The d-subshell is partially filled (7 electrons in 5 orbitals — there are unpaired electrons). In aqueous solution, Co2+ forms the pink [Co(H2O)6]2+ complex.
Result: Coloured (pink) in aqueous solution. …
-
- CBSE 2025Set JZ1 markMCQQ.Magnetic moment of a bivalent ion in aqueous solution will be, if its atomic number is 25(a) 1.73 BM(b) 2.83 BM(c) 4.96 BM(d) 5.92 BM
›Reveal solutionSolution
Mn2+ (3d5) has 5 unpaired electrons, so μ=5(5+2)=5.92 BM — option (d).
Concept. The magnetic moment of a transition-metal ion depends only on the number of unpaired d-electrons (n), through the spin-only formula μ=n(n+2) BM.
Step 1 — identify the ion. Atomic number 25 → manganese (Mn), configuration [Ar]3d54s2. A bivalent ion Mn2+ loses the two 4s electrons: Mn2+=[Ar]3d5.
…
- CBSE 2025Set D1 markMCQQ.The structure of complex ion [Ni(CN)4]2- is(a) Linear(b) Tetrahedral(c) Square planar(d) Octahedral
›Reveal solutionSolution
Ni2+ (d8) with strong-field CN- gives dsp2 hybridisation -> square planar [Ni(CN)4]2-.
Step 1 - oxidation state: In [Ni(CN)4]2-, four CN- (each -1) give -4; overall charge -2, so Ni is +2.
Step 2 - configuration: Ni2+ is 3d8.
Step 3 - ligand strength: CN- is a strong-field ligand. It pairs up the d electrons, freeing one 3d orbital. …
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