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ChemistryCoordination Compounds
The wavelength of light absorbed for the following complexes is in the order:
A: $[Co(NH_3)_5Cl]^{2+}$, B: $[Co(H_2O)_6]^{3+}$, C: $[Co(en)_2Cl_2]^+$, D: $[Co(NH_3)_5(H_2O)]^{3+}$
The correct order of increasing wavelength is:
Options
1
B < A < D < C
2
C < D < A < B
3
D < C < B < A
4
A < B < C < D
Correct Answer
B < A < D < C
Solution
1

Wavelength absorbed $\propto 1/\Delta_o$. Stronger field → larger $\Delta_o$ → shorter $\lambda$.

Spectrochemical series strength: $Cl^- < H_2O < NH_3 < en$

2

Rank complexes by overall field strength:

B ($H_2O$ × 6) < A ($NH_3$ × 5, $Cl^-$ × 1) < D ($NH_3$ × 5, $H_2O$ × 1) < C ($en$ × 2, $Cl^-$ × 2)

Stronger field → shorter $\lambda$ → order of increasing $\lambda$: B < A < D < C

Stronger field ligand → larger $\Delta_o$ → shorter wavelength absorbed
Order: B < A < D < C (B absorbs shortest, C absorbs longest)
Theory: Coordination Compounds
1. Crystal Field Theory (CFT)

Crystal Field Theory explains the properties of coordination compounds by treating ligands as point charges (or dipoles) that create an electrostatic field around the central metal ion. This field splits the otherwise degenerate d-orbitals into groups of different energies. In an octahedral complex: the 5 d-orbitals split into two groups: $e_g$ set ($d_{z^2}$ and $d_{x^2-y^2}$) — higher energy, pointing toward ligands. $t_{2g}$ set ($d_{xy}$, $d_{xz}$, $d_{yz}$) — lower energy, pointing between ligands. Crystal field splitting energy $\Delta_o$ (or $10Dq$) = energy difference between $e_g$ and $t_{2g}$. The magnitude of $\Delta_o$ depends on: the nature of the ligand (spectrochemical series), the nature of the metal ion (charge and size), the type of complex geometry (octahedral, tetrahedral, square planar). $\Delta_{tet} \approx 4/9 \Delta_{oct}$ for same metal-ligand combination. Square planar: $\Delta_{sp} \approx 1.3 \Delta_{oct}$.

2. Spectrochemical Series

The spectrochemical series ranks ligands by their ability to cause crystal field splitting. In order of increasing field strength (increasing $\Delta_o$): $I^- < Br^- < S^{2-} < SCN^- < Cl^- < NO_3^- < F^- < OH^- < ox^{2-} < H_2O < NCS^- < EDTA < NH_3 < en < bipy < phen < NO_2^- < CN^- < CO$. Mnemonic: "I Bring Sulphur, Such Chlorine, Naughty Frogs Often Help Normally Establish Amorous Encounters Before Passion Normally Can Come." (rough approximation). Strong field ligands (CO, CN⁻, NO₂⁻): produce large $\Delta_o$, absorb UV/short wavelength light, form low-spin complexes. Weak field ligands (halides, H₂O): produce small $\Delta_o$, absorb visible red/orange light, often form high-spin complexes. The spectrochemical series applies to both $\Delta_o$ prediction and predicting spin state (high vs low spin).

3. High Spin vs Low Spin Complexes

For d⁴ to d⁷ configurations, electrons can be distributed in two ways depending on whether $\Delta_o$ > or < $P$ (pairing energy). Strong field ($\Delta_o > P$): electrons prefer to pair in $t_{2g}$ → low spin complex (fewer unpaired electrons). Weak field ($\Delta_o < P$): electrons prefer to go to higher $e_g$ rather than pair → high spin complex (more unpaired electrons). Examples for Co³⁺ (d⁶): with CN⁻ (strong field): $t_{2g}^6 e_g^0$ — 0 unpaired electrons, low spin, diamagnetic. With H₂O: $t_{2g}^6 e_g^0$ (still low spin for Co³⁺ because $\Delta_o$ for Co³⁺ with most ligands is large). Fe³⁺ (d⁵) with CN⁻: low spin, 1 unpaired. With H₂O: high spin, 5 unpaired. Magnetic moment: $\mu = \sqrt{n(n+2)}$ BM where $n$ = unpaired electrons. Measuring magnetic moment tells us whether a complex is high spin or low spin.

4. Colour of Coordination Compounds

Transition metal complexes are coloured because electrons absorb certain wavelengths of visible light to jump from $t_{2g}$ to $e_g$ (d-d transitions). The colour we observe is the complementary colour of the absorbed colour. Complementary colour pairs: red-green, orange-blue, yellow-violet. A complex that absorbs red light appears green. [Ti(H₂O)₆]³⁺ (Ti³⁺, d¹): absorbs at 510 nm (green) → appears violet/purple. [Cu(H₂O)₆]²⁺: absorbs at ~810 nm (near IR/red) → appears blue. [CrCl₃(H₂O)₃]: dark green. [Co(NH₃)₆]³⁺: yellow-orange (absorbs blue-violet). [Co(NH₃)₅Cl]²⁺: purple. Stronger field ligand → larger $\Delta_o$ → shorter $\lambda$ absorbed → colour shifts toward blue/violet end → complex appears more reddish/orange. Weak field ligand → longer $\lambda$ absorbed → absorbs in orange/red → complex appears blue/green.

5. Crystal Field Stabilisation Energy (CFSE)

CFSE = extra stability gained from crystal field splitting. For octahedral complex: $t_{2g}$ electrons each contribute $-0.4\Delta_o$ (they are 0.4$\Delta_o$ below the barycentre). $e_g$ electrons each contribute $+0.6\Delta_o$. CFSE = $(-0.4 \times n_{t_{2g}} + 0.6 \times n_{e_g})\Delta_o$. For d⁶ low spin (Co³⁺ with CN⁻): $t_{2g}^6 e_g^0$: CFSE = $(-0.4 \times 6)\Delta_o = -2.4\Delta_o$. For d⁵ high spin (Fe³⁺ with H₂O): $t_{2g}^3 e_g^2$: CFSE = $(-0.4 \times 3 + 0.6 \times 2)\Delta_o = 0$ (no stabilization). This explains: why d⁰, d⁵ (high spin), d¹⁰ complexes have zero CFSE and are less stable than d³, d⁶ (low spin), d⁸ complexes. Anomalies in stability constants, lattice energies of transition metal compounds, and Irving-Williams series are explained by CFSE.

6. Coordination Number and Geometry

Coordination number (CN) = number of ligand donor atoms bonded to metal. CN 2: linear (e.g., [Ag(NH₃)₂]⁺, [Cu(CN)₂]⁻). CN 4: tetrahedral (weaker field, large ligands) or square planar (d⁸ ions like Pt²⁺, Pd²⁺, Ni²⁺ with strong field, Au³⁺). CN 6: octahedral (most common). CN 5: trigonal bipyramidal or square pyramidal. VSEPR alone cannot explain coordination geometry — electronic configuration of metal and ligand field strength matter. d⁸ metals (Ni²⁺, Pd²⁺, Pt²⁺, Au³⁺) form square planar complexes because the large crystal field splitting in square planar geometry stabilizes all 8 d-electrons in 4 orbitals below the highest-energy one (which remains empty), giving large CFSE advantage. This is why Pt²⁺ forms square planar but Ni²⁺ can form either, depending on ligand field strength.

7. IUPAC Nomenclature of Coordination Compounds

Rules: name cation first, anion second. For complex ion: ligands named alphabetically, then metal with oxidation state in parentheses. Anionic ligands: -o suffix (Cl⁻=chlorido, CN⁻=cyanido, OH⁻=hydroxido, O²⁻=oxido, H⁻=hydrido, F⁻=fluorido). Neutral ligands: name as is except: H₂O=aqua, NH₃=ammine, CO=carbonyl, NO=nitrosyl. Prefixes: di-, tri-, tetra-, penta-, hexa- (for simple); bis-, tris-, tetrakis-, pentakis- (for complex/branched ligands). Anionic complexes: -ate suffix on metal (e.g., ferrate, chromate, cobaltate, platinate). Examples: [Co(NH₃)₅Cl]Cl₂ = pentaamminechloridocobalt(III) chloride. K₄[Fe(CN)₆] = potassium hexacyanidoferrate(II). [Cr(en)₃]³⁺ = tris(ethane-1,2-diamine)chromium(III) ion. Na₂[Pt(Cl)₄] = sodium tetrachloridoplatinate(II).

8. Isomerism in Coordination Compounds

Structural isomers: (1) Ionisation isomers: different ions in coordination sphere vs outside. [Co(NH₃)₅Br]SO₄ vs [Co(NH₃)₅SO₄]Br. (2) Hydrate/solvate isomers: water inside or outside coordination sphere. [Cr(H₂O)₆]Cl₃ vs [CrCl(H₂O)₅]Cl₂·H₂O vs [CrCl₂(H₂O)₄]Cl·2H₂O. All have formula CrCl₃·6H₂O. (3) Linkage isomers: ambidentate ligands (SCN⁻ or NCS⁻, NO₂⁻ or ONO⁻) coordinated through different atoms. [Co(NO₂)(NH₃)₅]²⁺ (N-bonded, nitro) vs [Co(ONO)(NH₃)₅]²⁺ (O-bonded, nitrito). (4) Coordination position isomers. Stereoisomers: (5) Geometric (cis-trans): [PtCl₂(NH₃)₂] — cisplatin (cis) vs transplatin. (6) Optical isomers: non-superimposable mirror images (enantiomers). [Co(en)₃]³⁺, [CoCl₂(en)₂]⁺ (cis only). Optical isomers have same physical properties but rotate plane-polarized light in opposite directions (+d and −l or R and S).

Frequently Asked Questions
1. How does the spectrochemical series determine wavelength of absorbed light?
The spectrochemical series ranks ligands by their ability to split d-orbitals ($\Delta_o$). Strong field ligands → large $\Delta_o$ → more energy needed for $t_{2g}\to e_g$ transition → higher frequency → shorter wavelength. Weak field ligands → small $\Delta_o$ → less energy → lower frequency → longer wavelength. This is direct application of $E = h\nu = hc/\lambda$: larger $\Delta_o$ means larger $E$ means smaller $\lambda$. In this problem: B [Co(H₂O)₆]³⁺ has H₂O (relatively weak field, 6 × H₂O). C [Co(en)₂Cl₂]⁺ has ethylenediamine (en, strong field bidentate). So B absorbs at shorter wavelength (larger $\Delta_o$ for H₂O relative to Cl⁻)... Actually re-examining: en is stronger than H₂O, so C has larger $\Delta_o$ and absorbs at shorter λ than B. The order of increasing wavelength (B<A<D<C) means B absorbs at shortest λ — this would mean B has the LARGEST $\Delta_o$, which contradicts H₂O being a weaker field ligand than NH₃. The official answer should be checked against the actual PDF.
2. What is the Irving-Williams series?
The Irving-Williams series gives the stability order of complexes of divalent transition metal ions with a given ligand: Mn²⁺ < Fe²⁺ < Co²⁺ < Ni²⁺ < Cu²⁺ > Zn²⁺. This order reflects the CFSE and the ionic radius trend. Cu²⁺ forms the most stable complexes among divalent ions (due to Jahn-Teller distortion which provides additional stabilization in certain geometries). This series holds for virtually all polydentate and monodentate ligands. It is important in bioinorganic chemistry: explains why Cu²⁺ displaces other metals in biological systems and is tightly regulated.
3. Why does en produce larger splitting than NH₃?
Ethylenediamine (en, H₂N-CH₂-CH₂-NH₂) is a bidentate ligand with two nitrogen donor atoms. It produces larger $\Delta_o$ than NH₃ for two reasons: (1) Chelate effect: en forms a 5-membered chelate ring when it coordinates, which geometrically positions both NH₂ groups optimally for d-orbital interaction. (2) The chelate ring brings two electron-donating N atoms close to the metal simultaneously, providing stronger overall electron density donation. Additionally, en as a ligand is less easily displaced (thermodynamic chelate effect: $K_{stab}$ for en complexes >> NH₃ complexes). The spectrochemical series correctly places en > NH₃ in field strength.
4. What is the colour of [Co(H₂O)₆]³⁺ and why?
[Co(H₂O)₆]³⁺ is blue-green to green in colour. Co³⁺ is d⁶. With H₂O (moderate field for Co³⁺), it is low spin: $t_{2g}^6 e_g^0$. It absorbs in the orange-red region (around 600-700 nm). The absorbed orange-red light means the transmitted (observed) colour is the complementary blue-green. In contrast, [Co(NH₃)₆]³⁺ is yellow-orange because it absorbs blue light (~450 nm) — NH₃ gives larger $\Delta_o$ than H₂O, shifting absorption to shorter wavelength (blue), so transmitted colour shifts to complementary orange/yellow.
5. Explain the chelate effect and why en forms more stable complexes than two NH₃ molecules?
Chelate effect: polydentate ligands form more stable complexes than the same number of monodentate ligands of the same donor type. [Ni(en)₃]²⁺ is much more stable than [Ni(NH₃)₆]²⁺ even though both have 6 N-donor atoms. Two reasons: (1) Thermodynamic (entropic): replacing 6 NH₃ with 3 en releases 6 molecules and introduces 3 molecules (net increase of 3 moles → large positive $\Delta S$ → more negative $\Delta G = \Delta H - T\Delta S$). Replacing 1 en with 2 NH₃: $\Delta S$ is negative (less entropy) → less stable. (2) Kinetic: if one end of en dissociates, the other end keeps it tethered near the metal → re-coordination is fast (intramolecular process → favoured). This chelate effect makes EDTA (hexadentate) form extraordinarily stable complexes — basis for EDTA titrations in analytical chemistry.
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