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ChemistryCoordination Compounds
Which of the following complexes are paramagnetic?
A: $[NiCl_4]^{2-}$
B: $[Ni(CO)_4]$
C: $[Ni(CN)_4]^{2-}$
D: $[Ni(H_2O)_6]^{2+}$
Options
1
A and B only
2
B and C only
3
A and D only
4
C and D only
Correct Answer
A and D only
Solution
1

Check oxidation state and d-electron count:

A: Ni²⁺ (d⁸), tetrahedral, weak field (Cl⁻) → 2 unpaired → paramagnetic

B: Ni⁰ (d¹⁰), all paired → diamagnetic

2

C: Ni²⁺ (d⁸), square planar, strong field (CN⁻), all paired → diamagnetic

D: Ni²⁺ (d⁸), octahedral, weak field (H₂O) → 2 unpaired → paramagnetic

Answer: A and D only

Paramagnetic = has unpaired electrons
A (Ni²⁺, d⁸, weak field tet.) and D (Ni²⁺, d⁸, weak field oct.) both have 2 unpaired e⁻
Theory: Coordination Compounds
1. Magnetic Properties — Paramagnetism and Diamagnetism

Paramagnetism: substances with unpaired electrons are attracted to a magnetic field. Magnetic moment $\mu = \sqrt{n(n+2)}$ BM where $n$ = number of unpaired electrons. n=1: $\mu=1.73$ BM. n=2: $\mu=2.83$ BM. n=3: $\mu=3.87$ BM. n=4: $\mu=4.90$ BM. n=5: $\mu=5.92$ BM (high spin Mn²⁺, Fe³⁺). Diamagnetism: all electrons paired → weak repulsion from magnetic field. The spin-only formula gives reasonable approximation for first-row transition metals; orbital contribution becomes significant for second and third row. Measuring magnetic moment by Gouy method tells us the number of unpaired electrons and hence the spin state (high spin vs low spin).

2. Crystal Field Theory and Magnetic Properties

Octahedral weak field (small Δo): d electrons fill orbitals with maximum unpaired spins (Hund rule). d1: 1 unpaired. d2: 2. d3: 3. d4: 4 (high spin). d5: 5 (high spin, Fe3+, Mn2+). d6: 4 (high spin, Fe2+). d7: 3 (Co2+). d8: 2 (Ni2+). d9: 1 (Cu2+). d10: 0 (Zn2+). Octahedral strong field (large Δo): electrons pair in t2g first. d4 → 2 unpaired (low spin). d5 → 1. d6 → 0 (Co3+ with CN-). d7 → 1. d8 → 2 (same as weak field — t2g6 eg2 in both cases!). d9 → 1. d10 → 0. So d8 is special: same number of unpaired (2) in both weak and strong field octahedral! The magnetic moment alone cannot distinguish high spin from low spin for d8.

3. Geometry and Hybridisation of Ni Complexes

Ni2+ (d8) forms complexes with different geometries: Tetrahedral (sp3): with weak field ligands (Cl-, Br-, I-, SCN-). [NiCl4]2-: tetrahedral, 2 unpaired, paramagnetic. μ = 2.83 BM. Square planar (dsp2): with strong field ligands (CN-, CO, en). [Ni(CN)4]2-: square planar, 0 unpaired, diamagnetic. The geometry preference: in tetrahedral, the crystal field stabilisation is smaller (Δt = 4/9 Δo) — insufficient to force pairing. In square planar: very large crystal field stabilisation for d8 configuration (can be > pairing energy) → all electrons pair in 4 lower orbitals → dsp2 hybridisation. Ni0 (d10): tetrahedral with CO → [Ni(CO)4]. d10 has all electrons paired regardless of field strength.

4. Electronic Spectra of Transition Metal Complexes

Transition metal complexes absorb visible light by d-d transitions (electron jumps from t2g to eg in octahedral complex). Selection rules: Laporte rule: transitions between orbitals of same parity (g→g or u→u) are forbidden. For octahedral: d-d transitions are g→g (forbidden) but occur weakly due to vibrational distortion. Spin selection rule: transitions between states of different spin multiplicity are forbidden. [Ti(H2O)6]3+: simplest case, d1, one d-d transition, one absorption band. [Cu(H2O)6]2+: d9, one "hole" in d9 acts like d1. Many-electron cases (d2-d8): multiple electronic states, complex spectra described by Tanabe-Sugano diagrams. Charge transfer (CT) spectra: much more intense than d-d (Laporte allowed). MnO4- intense purple = CT (not d-d).

5. Organometallic Chemistry

Organometallic compounds: direct metal-carbon bond. Examples: [Ni(CO)4] (nickel tetracarbonyl), [Fe(CO)5] (iron pentacarbonyl), [Cr(CO)6] (chromium hexacarbonyl), ferrocene [Fe(C5H5)2]. 18-electron rule: stable organometallics obey the 18-electron rule (all valence orbitals filled). [Ni(CO)4]: Ni0 has 10 d electrons, 4 CO each donate 2 electrons → 10 + 8 = 18. Stable. Ferrocene: Fe2+ (6 e) + 2 Cp- (5 e each) = 6+10 = 16? No: Fe2+ d6 = 6 electrons + each C5H5- donates 6 π electrons = 6+6+6=18. Sandwich compound: metal between two cyclopentadienyl rings. Important industrial organometallics: [RhCl(PPh3)3] (Wilkinson catalyst, hydrogenation), [TiCl4+Al(C2H5)3] (Ziegler-Natta, alkene polymerisation), [PdCl2] (Wacker, CH2=CH2 → CH3CHO).

6. Variable Valency and Colour of Transition Metals

Transition metals show variable valency due to small energy difference between (n-1)d and ns orbitals. Both lose electrons in ionisation. Fe: +2, +3 (can be +4, +6 in special cases like FeO4²⁻). Cu: +1, +2. Mn: +2, +3, +4, +6, +7. Cr: +2, +3, +6. The colour of transition metal complexes arises from d-d transitions. Ti3+ (d1): purple. Cr3+ (d3): green/blue. Mn2+ (d5 high spin): very pale pink (transition forbidden, very weak). Fe3+ (d5 high spin): pale yellow-brown. Co2+ (d7): pink/red. Ni2+ (d8): green/blue. Cu2+ (d9): blue. Zn2+ (d10): colourless (no d-d transitions possible). The colour observed is complementary to absorbed colour.

7. Magnetic Moment Calculation

Magnetic moment $\mu_{spin-only} = \sqrt{n(n+2)}$ Bohr Magnetons (BM) where n = number of unpaired electrons. For [NiCl4]2- (n=2): μ = √(2×4) = √8 = 2.83 BM. For [Ni(H2O)6]2+ (n=2): μ = 2.83 BM. For diamagnetic [Ni(CN)4]2- (n=0): μ = 0 BM. For [Ni(CO)4] (d10, n=0): μ = 0 BM. Experimental magnetic moment of [NiCl4]2- ≈ 3.2-3.4 BM (slightly higher than spin-only due to orbital contribution). This confirms the spin-only formula gives approximate values for first-row transition metals. Higher BM than spin-only: orbital contribution significant (e.g., Co2+ complex: spin-only gives 3.87 BM but experimental ≈ 4.1-5.2 BM for tetrahedral complexes due to large T1 ground state orbital contribution).

8. Hard and Soft Acid Base Theory (HSAB)

Pearson HSAB theory: Lewis acids (metal ions) and Lewis bases (ligands) classified as hard or soft. Hard acids: high charge, small size, low polarisability. Mg2+, Fe3+, Cr3+, Co3+, Al3+, H+. Hard bases: high electronegativity, small, low polarisability. F-, OH-, H2O, NH3, Cl-. Soft acids: low charge, large, highly polarisable. Ni0, Pt2+, Pd2+, Cu+, Au+, Hg2+. Soft bases: large, polarisable, low electronegativity. CN-, CO, PR3, I-, RS-. Rule: hard acid + hard base = stable (ionic interaction). Soft + soft = stable (covalent/back-bonding). Mixed = less stable. Applications: Hg2+ (soft acid) prefers CN-, I-, RS- (soft bases) — explains Hg toxicity (binds to thiol groups -SH in enzymes). Fe3+ (hard) binds F- strongly. Ni0 (soft) forms stable [Ni(CO)4]. HSAB explains selectivity in biological systems, extraction chemistry, catalysis.

Frequently Asked Questions
1. How to quickly determine paramagnetism of a Ni complex?
Step 1: Find oxidation state of Ni. In [NiCl4]2-: 4(-1)+Ni=-2, Ni=+2 (d8). In [Ni(CO)4]: CO neutral, Ni=0 (d10). Step 2: For d10 (Ni0): all paired, always diamagnetic regardless of geometry or ligand. Step 3: For d8 (Ni2+): determine geometry. Tetrahedral (weak field Cl-): 2 unpaired, paramagnetic. Square planar (strong field CN-, CO): 0 unpaired, diamagnetic. Octahedral (moderate field H2O): 2 unpaired, paramagnetic. Rule for NEET: if Ni2+ (d8) + weak field ligand → paramagnetic. If Ni2+ (d8) + strong field → square planar → diamagnetic.
2. Why does Ni2+ prefer square planar over octahedral with strong field ligands?
For d8 configuration: square planar geometry gives very large Crystal Field Stabilisation Energy (CFSE) because in square planar, the dx2-y2 orbital is at very high energy and the remaining 8 electrons fill the 4 lower d-orbitals perfectly (completely filled with 0 unpaired electrons). This CFSE advantage can exceed the energy cost of steric strain. Octahedral d8: CFSE = (-0.4×6+0.6×2)Δo = -2.4+1.2 = -1.2Δo. Square planar d8: CFSE = -2.464Δo (much larger). For d8 + strong field: square planar is more stable. For d8 + weak field: tetrahedral is preferred (less steric strain, smaller CFSE difference). Pt2+, Pd2+, Rh+ are always square planar with any ligand because their Δo values are very large (5d series).
3. What is the significance of [Ni(CO)4] in industry?
[Ni(CO)4] (nickel tetracarbonyl) is key to the Mond process for purifying nickel: (1) Impure Ni + CO (50-60°C) → [Ni(CO)4] (volatile, bp 43°C). (2) Distil to separate from impurities. (3) Decompose at 230°C → pure Ni + 4CO (CO recycled). High-purity Ni obtained this way is 99.99+%. [Ni(CO)4] is extremely toxic (highly reactive with haemoglobin). The Mond process revolutionised Ni refining in the 1890s. The reaction is possible because Ni0 (d10 soft acid) + CO (soft base via carbon) = stable complex. Iron also forms Fe(CO)5 but at higher CO pressure; this complicates Ni purification from ore that also contains Fe.
4. Compare the four Ni complexes in this question systematically?
Complex | Ni OS | d-electrons | Geometry | Field | Unpaired | Magnetic. [NiCl4]2- | +2 | d8 | Tetrahedral | Weak (Cl-) | 2 | Paramagnetic. [Ni(CO)4] | 0 | d10 | Tetrahedral | Strong (CO) | 0 | Diamagnetic. [Ni(CN)4]2- | +2 | d8 | Square planar | Strong (CN-) | 0 | Diamagnetic. [Ni(H2O)6]2+ | +2 | d8 | Octahedral | Weak (H2O) | 2 | Paramagnetic. Key insight: same Ni2+ (d8) can be paramagnetic or diamagnetic depending on ligand field and geometry — demonstrating the power of Crystal Field Theory to explain magnetic behaviour.
5. How is the Gouy method used to measure magnetic susceptibility?
The Gouy method: a sample is weighed in and out of a magnetic field. Paramagnetic samples are attracted into the field (appear heavier); diamagnetic samples are repelled (appear lighter). The difference in weight gives the magnetic susceptibility χ. From χ, calculate molar susceptibility χm = χ × M/ρ. Effective magnetic moment μeff = 2.84√(χm × T) BM. Compare experimental μeff with spin-only formula √(n(n+2)) BM to find n (number of unpaired electrons). This tells us: spin state (high vs low), oxidation state, and geometry of the metal complex. Modern instruments use SQUID (Superconducting Quantum Interference Device) magnetometers for much greater sensitivity.
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