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Chemistryd-Block Elements
The calculated 'spin-only' magnetic moment of Ti²⁺ (3d²) is :
Options
1
2·84 BM
2
5·92 BM
3
4·90 BM
4
3·87 BM
Correct Answer
Option 1 : 2·84 BM
Step-by-Step Solution
1

Electronic configuration of Ti (Z=22):

Ti: [Ar] 3d² 4s²

Ti²⁺: Remove 2 electrons from 4s first → [Ar] 3d²

2

3d² — apply Hund's rule:

5 d-orbitals: ↑ ↑ _ _ _

Both electrons go to separate orbitals with parallel spins

n = 2 unpaired electrons

3

Spin-only magnetic moment formula:

μ = √n(n+2) BM

μ = √2(2+2) = √(2×4) = √8 = 2√2 ≈ 2·84 BM

Ti²⁺: [Ar] 3d² → n = 2 unpaired electrons

μ = √n(n+2) = √(2×4) = √8 ≈ 2·83 BM ≈ 2·84 BM

Theory: Magnetic Properties of Transition Metals
1. Spin-Only Magnetic Moment Formula

The magnetic moment of a transition metal ion depends on the number of unpaired electrons (n). The spin-only formula: μ = √n(n+2) BM (Bohr Magnetons). This formula assumes only spin contribution (no orbital contribution). For first-row transition metals, orbital contribution is often quenched, making the spin-only formula approximately valid. Each unpaired electron has a spin magnetic moment of approximately 1·73 BM (= √3 BM = √1(1+2) BM). Diamagnetic species (n=0): μ = 0, repelled by magnetic field. Paramagnetic species (n≥1): μ > 0, attracted to magnetic field.

2. Magnetic Moments — Complete Table
Unpaired e⁻ (n)μ = √n(n+2)μ (BM)Example ions
0√00Zn²⁺, Cu⁺, Sc³⁺, Ti⁴⁺
1√31·73Ti³⁺, Cu²⁺, V⁴⁺
2√82·84Ti²⁺, V³⁺, Ni²⁺
3√153·87V²⁺, Cr³⁺, Co³⁺(hs)
4√244·90Cr²⁺, Mn³⁺, Fe²⁺(hs)
5√355·92Mn²⁺, Fe³⁺(hs)
3. Electronic Configurations of First-Row Transition Metal Ions

📌 Ti (Z=22): [Ar]3d²4s² → Ti²⁺: [Ar]3d² (n=2, μ=2·84 BM)

📌 V (Z=23): [Ar]3d³4s² → V²⁺: [Ar]3d³ (n=3, μ=3·87 BM)

📌 Cr (Z=24): [Ar]3d⁵4s¹ → Cr³⁺: [Ar]3d³ (n=3, μ=3·87 BM)

📌 Mn (Z=25): [Ar]3d⁵4s² → Mn²⁺: [Ar]3d⁵ (n=5, μ=5·92 BM) ← maximum!

📌 Fe (Z=26): [Ar]3d⁶4s² → Fe²⁺: [Ar]3d⁶ (n=4, μ=4·90 BM); Fe³⁺: [Ar]3d⁵ (n=5, μ=5·92 BM)

📌 Co (Z=27): [Ar]3d⁷4s² → Co²⁺: [Ar]3d⁷ (n=3, μ=3·87 BM)

📌 Ni (Z=28): [Ar]3d⁸4s² → Ni²⁺: [Ar]3d⁸ (n=2, μ=2·84 BM) — same as Ti²⁺!

📌 Cu (Z=29): [Ar]3d¹⁰4s¹ → Cu²⁺: [Ar]3d⁹ (n=1, μ=1·73 BM)

📌 Zn (Z=30): [Ar]3d¹⁰4s² → Zn²⁺: [Ar]3d¹⁰ (n=0, μ=0, diamagnetic)

4. Why are 4s Electrons Removed Before 3d in Ion Formation?

In neutral atoms, 4s fills before 3d (Aufbau principle — 4s is lower energy than 3d for neutral atoms). But in ion formation, 4s electrons are removed first because once the atom starts losing electrons, the 3d orbitals become lower in energy than 4s (due to increased effective nuclear charge). So Ti (3d²4s²) loses 2 electrons from 4s to give Ti²⁺ (3d²), NOT from 3d. This is a highly tested NEET concept — always remove ns electrons before (n-1)d electrons when forming positive ions.

5. Special Electronic Configurations — Exceptions to Aufbau

Cr (Z=24): Expected [Ar]3d⁴4s² but actual [Ar]3d⁵4s¹. Reason: half-filled d⁵ configuration (all 5 d orbitals singly occupied) has extra stability due to symmetry and exchange energy. Cu (Z=29): Expected [Ar]3d⁹4s² but actual [Ar]3d¹⁰4s¹. Reason: completely filled d¹⁰ configuration has extra stability. These two exceptions are extremely important for NEET — must be memorised. Their ions: Cr³⁺ = [Ar]3d³; Cu²⁺ = [Ar]3d⁹; Cu⁺ = [Ar]3d¹⁰ (diamagnetic).

6. Colour of Transition Metal Compounds

Transition metal compounds are coloured because of d-d electronic transitions. When ligands coordinate to the metal, the d orbitals split into two groups (in octahedral field: t₂g and eg). The energy difference between these groups (Δ = crystal field splitting energy) falls in the visible light range (~400–700 nm). The compound absorbs certain wavelengths (complementary colour is observed). Species with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) configurations are colourless — no d-d transitions possible. Ti³⁺ (d¹) is purple, Cu²⁺ (d⁹) is blue, Fe³⁺ (d⁵ high spin) is yellow.

7. High Spin vs Low Spin Complexes

In octahedral complexes, d orbitals split by crystal field: lower t₂g (3 orbitals) and upper eg (2 orbitals). Electrons can pair up in t₂g (low spin — strong field ligands: CN⁻, CO, NO⁺, en, NH₃) or fill all orbitals before pairing (high spin — weak field ligands: F⁻, Cl⁻, Br⁻, I⁻, OH⁻, H₂O). High spin → more unpaired electrons → higher magnetic moment. Low spin → more paired electrons → lower magnetic moment. Example: Fe²⁺ (d⁶): high spin has 4 unpaired (μ = 4·90 BM); low spin has 0 unpaired (μ = 0, diamagnetic).

8. Applications of Magnetic Measurements

Magnetic moment measurements help determine: (1) Number of unpaired electrons → electronic configuration of the ion. (2) Oxidation state of the metal. (3) Whether a complex is high spin or low spin. (4) Geometry of complex (square planar complexes of d⁸ metal ions like Ni²⁺, Pd²⁺, Pt²⁺ are diamagnetic — all electrons paired; tetrahedral Ni²⁺ complexes are paramagnetic with 2 unpaired electrons). Practical: MRI (magnetic resonance imaging) uses magnetic properties of hydrogen nuclei (protons) in body tissues.

Frequently Asked Questions
1. Why is Ti²⁺ configuration [Ar]3d² and not [Ar]3d⁰4s²?
When Ti (Z=22, configuration [Ar]3d²4s²) loses 2 electrons to form Ti²⁺, the 4s electrons are removed first. This is because in ions, the 3d orbitals are lower in energy than 4s (the increased nuclear charge stabilises 3d more than 4s). So Ti²⁺ = [Ar]3d², not [Ar]4s². If we incorrectly remove 3d electrons: Ti²⁺ would be [Ar]4s² (0 unpaired, μ=0, diamagnetic) — but experimentally Ti²⁺ is paramagnetic (μ≈2·84 BM), confirming the correct configuration is [Ar]3d².
2. Which ion has the maximum magnetic moment?
Mn²⁺ and Fe³⁺ both have d⁵ configuration (high spin) with 5 unpaired electrons: μ = √5(5+2) = √35 ≈ 5·92 BM. This is the maximum for first-row transition metals. Mn²⁺: [Ar]3d⁵. Fe³⁺: [Ar]3d⁵. Both have all 5 d orbitals singly occupied (Hund's rule: ↑ ↑ ↑ ↑ ↑). No pairing until all orbitals have one electron. 5·92 BM is the highest spin-only value for first-row transition metals.
3. Ni²⁺ and Ti²⁺ have the same magnetic moment — why?
Ni²⁺: [Ar]3d⁸. 3d⁸ has 8 electrons in 5 d orbitals: ↑↓ ↑↓ ↑↓ ↑ ↑ → 2 unpaired electrons. Ti²⁺: [Ar]3d² → 2 unpaired electrons (↑ ↑ _ _ _). Both have n=2 → μ = √8 ≈ 2·84 BM. So YES, both have identical spin-only magnetic moments despite very different electronic configurations. The magnetic moment depends only on the NUMBER of unpaired electrons, not on how the electrons are distributed.
4. What is the meaning of BM (Bohr Magneton)?
Bohr Magneton (BM or μB) is the unit of magnetic moment for atomic systems. 1 BM = eℏ/2mₑ = 9·274×10⁻²⁴ J/T. It arises from the spin of a single electron. The spin-only magnetic moment is expressed in BM because it's a multiple of the magnetic moment of one unpaired electron. A substance with μ = 2·84 BM has a magnetic moment equal to 2·84 times the Bohr magneton. Experimentally measured μ values close to calculated spin-only values confirm spin-only approximation is valid.
5. Why is Zn²⁺ diamagnetic?
Zn²⁺: [Ar]3d¹⁰ — all 10 d electrons are paired (↑↓ ↑↓ ↑↓ ↑↓ ↑↓). Zero unpaired electrons → μ = √0(0+2) = 0. Diamagnetic — all electrons paired, no net magnetic moment, weakly repelled by magnetic field. Zinc compounds are therefore generally white/colourless (no d-d transitions possible either). Zn²⁺ is an exception among transition metal ions in being diamagnetic. Similarly, Cu⁺ (d¹⁰) is diamagnetic.
6. What is the difference between paramagnetism and ferromagnetism?
Paramagnetism: individual atoms/ions with unpaired electrons are attracted to magnetic field, but alignment is random (thermal motion). Weak, temporary magnetism. All transition metal compounds with unpaired electrons are paramagnetic. Ferromagnetism: in certain metals (Fe, Co, Ni), unpaired electron spins in adjacent atoms align parallel spontaneously → strong, permanent magnetism (domain structure). Fe above Curie temperature (770°C) becomes paramagnetic. Antiferromagnetism: alternate atoms have opposite spin alignment → cancels out → very weak magnetism.
7. How does crystal field theory explain colour of transition metal complexes?
In an octahedral crystal field, the 5 degenerate d orbitals split: lower t₂g (d_xy, d_xz, d_yz) and higher eg (d_x²-y², d_z²). Crystal field splitting energy Δ_oct depends on the ligand (spectrochemical series). An electron absorbs a photon of energy = Δ_oct and jumps from t₂g to eg. The absorbed wavelength corresponds to a specific colour. The complementary colour is what we observe: [Ti(H₂O)₆]³⁺ absorbs green (510 nm) → appears purple-violet. Different Δ_oct values → different absorbed wavelengths → different colours.
8. What is the spectrochemical series?
The spectrochemical series arranges ligands by their ability to split d orbitals (Δ_oct): I⁻ < Br⁻ < Cl⁻ < F⁻ < OH⁻ < H₂O < NH₃ < en < NO₂⁻ < CN⁻ < CO. Weak field ligands (left): small Δ, high spin, more unpaired e⁻, higher μ. Strong field ligands (right): large Δ, low spin, fewer unpaired e⁻, lower μ. CO and CN⁻ are the strongest field ligands (π back-bonding). F⁻, Cl⁻ are weak field (mostly σ donors). This series is crucial for predicting spin state, magnetic moment, and colour of transition metal complexes.
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