CBSE Class 12 Chemistry Coordination Compounds Important Questions Set 08

Very Short Answer Questions (1 Mark)

Question 1. When a coordination compound CrCl₃.6 H₂O is mixed with AgNO₃, two moles of AgCl are precipitated per mole of the compound. What is the structural formula of the coordination compound ? (CBSE 2019)
Answer:
Formation of two moles of AgCl shows that there are two ionizable chloride ions. Hence, the formula is [Cr(H₂O)₅Cl] Cl₂.H₂O.

Teacher's Note:
1. Only Cl⁻ outside the square bracket is precipitated by AgNO₃; with CN = 6, one Cl⁻ must be inside along with 5 H₂O.

Question 2. Arrange the following in increasing order of crystal field splitting energy [Cr(CN)₆]³⁻, [Cr(NH₃)₆]³⁺, [CrCl₆]³⁻ (CBSE 2019)
Answer:
From spectrochemical series, CFSE is in the order : Cl⁻ < NH₃ < CN⁻
i.e., [CrCl₆]³⁻ < [Cr(NH₃)₆]³⁺ < [Cr(CN)₆]³⁻.

Teacher's Note:
1. Same metal ion, same geometry: only the ligand changes, so the spectrochemical series alone decides the order.

Question 3. Write the hybridisation and magnetic character of [CoF₆]³⁻ (At. No. of Co = 27). (CBSE Foreign 2017)
Answer:
sp³d², highly paramagnetic (4 unpaired electrons).

Teacher's Note:
1. F⁻ is a weak field ligand, so the 3d electrons of Co³⁺ (3d⁶) do not pair up and outer 4d orbitals are used.

Question 4. Among octahedral and tetrahedral crystal fields, in which case the magnitude of crystal field splitting is larger ?
Answer:
It is larger in case of octahedral crystal field (Δ_t = 4/9 Δ_o or Δ_o = 9/4 Δ_t).

Teacher's Note:
1. Tetrahedral complexes have fewer ligands (4) and none of them points directly at the d-orbitals, so the splitting is smaller.

Question 5. A coordination compound has the formula, CoCl₃.4NH₃. It does not liberate ammonia but precipitates chloride ions as silver chloride. Give the IUPAC name of the complex and write its structural formula.
Answer:
Remembering that coordination number of Co is 6, the formula of the complex will be [Co(NH₃)₄Cl₂]Cl. The name will be tetraamminedichloridocobalt (III) chloride.

Teacher's Note:
1. No free NH₃ means all four NH₃ are ligands; the remaining two positions must be filled by Cl⁻, leaving only one ionisable Cl⁻.

Question 6. The molar conductivity of CoCl₃.4NH₃.2H₂O is found to be same as that of 3 : 1 electrolyte. What is the structural formula of the complex ?
Answer:
The complex should be of the type A₃B or AB₃. As coordination number of Co is 6, hence its structure formula will be [Co(NH₃)₄(H₂O)₂]Cl₃.

Teacher's Note:
1. A 3 : 1 electrolyte gives 4 ions per formula unit, so all three Cl⁻ must be outside the coordination sphere.

Question 7. What happens when potassium ferrocyanide solution is added to a ferric salt solution ?
Answer:
A prussian blue colour complex (ferri-ferrocyanide) is formed :
4Fe³⁺ + 3[Fe(CN)₆]⁴⁻ → Fe₄[Fe(CN)₆]₃ (Prussian blue)

Teacher's Note:
1. This is the confirmatory test for Fe³⁺ ions in qualitative analysis.

Question 8. What is the oxidation state of Ni in Ni(CO)₄ ?
Answer:
Zero

Teacher's Note:
1. CO is a neutral ligand and the complex is neutral, so Ni must be in the zero oxidation state.

Question 9. Name the central metal atom present in haemoglobin, chlorophyll and vitamin-B₁₂. (HP Board 2011, J & K Board 2011)
Answer:
Fe, Mg and Co respectively.

Teacher's Note:
1. Remember: blood pigment - iron, plant pigment - magnesium, vitamin B₁₂ (cyanocobalamin) - cobalt.

Question 10. Give names of two complexes which are used in medicines. (Assam Board 2012)
Answer:
(i) EDTA, used in the treatment of lead poisoning.
(ii) cis-platin, [Pt(NH₃)₂Cl₂], used in the treatment of cancer.

Teacher's Note:
1. Only the cis isomer of [Pt(NH₃)₂Cl₂] is an anticancer drug; the trans isomer is inactive.

Question 11. Write the name of an organometallic compound which is used as a homogeneous catalyst.
Answer:
Wilkinson's catalyst.

Teacher's Note:
1. Wilkinson's catalyst is [(Ph₃P)₃RhCl]; it is used for the hydrogenation of alkenes.

Short Answer Questions (2 Marks)

Teacher's Note:
1. Both are inner orbital complexes; count unpaired electrons left in 3d after pairing to decide magnetic character.

Teacher's Note:
1. Whatever the hybridisation model, Cu²⁺ (d⁹) always has one unpaired electron, so all its complexes are paramagnetic (about 1.73 BM).

Question 14. Give an example of each of the following :
(i) Outer orbital octahedral complex (ii) Tetrahedral complex (iii) Inner orbital octahedral complex
Answer:
(i) [CoF₆]³⁻ (sp³d²)
(ii) [NiCl₄]²⁻ or [Ni(CO)₄] (sp³)
(iii) [Co(NH₃)₆]³⁺ or [Fe(CN)₆]⁴⁻ (d²sp³)

Teacher's Note:
1. Weak field ligands (F⁻, H₂O) usually give outer orbital complexes; strong field ligands (CN⁻, NH₃ with Co³⁺) give inner orbital complexes.

Question 15. Compare the magnetic behaviour of the complex entities [Fe(CN)₆]⁴⁻ and [FeF₆]³⁻ (Fe = 26).
Answer:
In [Fe(CN)₆]⁴⁻, Fe²⁺ (3d⁶) : CN⁻ is a strong field ligand, all six electrons pair up ; d²sp³. No unpaired electron, so it is diamagnetic.
In [FeF₆]³⁻, Fe³⁺ (3d⁵) : F⁻ is a weak field ligand, no pairing occurs ; sp³d². It has 5 unpaired electrons, so it is strongly paramagnetic (μ = \( \sqrt{5 \times 7} \) = 5.92 BM).

Teacher's Note:
1. Use μ = √n(n + 2) BM to convert unpaired electrons into magnetic moment: n = 5 gives 5.92 BM.

Teacher's Note:
1. Different hybridisation (dsp² vs sp³) gives different shapes, but magnetic behaviour depends only on unpaired electrons.

Teacher's Note:
1. Five electron pairs from five CO ligands need five hybrid orbitals: one 3d + 4s + three 4p = dsp³.

Teacher's Note:
1. Ni(CO)₄ and [Ni(CN)₄]²⁻ are both diamagnetic, but only Ni(CO)₄ is tetrahedral.

Teacher's Note:
1. Compare with [Co(NH₃)₆]³⁺: same Co³⁺, but NH₃ forces pairing and gives a diamagnetic inner orbital complex.

Teacher's Note:
1. The magnetic moment of [NiCl₄]²⁻ is about √8 = 2.83 BM (two unpaired electrons).

Teacher's Note:
1. Both metal ions are d⁶; the ligand strength alone decides inner versus outer orbital complex.

Teacher's Note:
1. No pairing is needed here, so the inner/outer decision for d³ does not depend on ligand strength.

Question 23. Name two properties of the central metal ion which enable it to form stable complex entities.
Answer:
(i) High charge on the central metal ion.
(ii) Small size of the central metal ion.
Thus, higher the charge density (charge/radius ratio) of the metal ion, greater is the stability of its complexes; e.g., complexes of Fe³⁺ are more stable than those of Fe²⁺. The availability of vacant d-orbitals of suitable energy also helps.

Teacher's Note:
1. Transition metal ions form so many complexes because they combine small size, high charge and vacant d-orbitals.

Question 24. Describe the nature of bonding :
(i) in metal carbonyls and
(ii) in π complexes.
Answer:
(i) Metal carbonyls : The metal-carbon bond in metal carbonyls possesses both σ and π character. The M–C σ bond is formed by the donation of the lone pair of electrons on the carbonyl carbon into a vacant orbital of the metal. The M–C π bond is formed by the donation of a pair of electrons from a filled d orbital of the metal into the vacant antibonding π* orbital of carbon monoxide (back bonding). The metal to ligand bonding creates a synergic effect which strengthens the bond between CO and the metal. Because of back bonding, the C–O bond order decreases and the C–O bond becomes slightly longer than in free CO, e.g., [Ni(CO)₄] (tetrahedral), [Fe(CO)₅] (trigonal bipyramidal), [Cr(CO)₆] (octahedral).
(ii) π complexes : In π complexes, the metal is bonded to an unsaturated organic molecule (alkene, alkyne or aromatic ring) through its π electrons. The filled π orbital of the ligand donates electron density to a vacant orbital of the metal, and a filled d orbital of the metal donates electron density back into the vacant π* orbital of the ligand, e.g., Zeise's salt K[PtCl₃(η²-C₂H₄)], ferrocene [Fe(η⁵-C₅H₅)₂] and dibenzene chromium [Cr(η⁶-C₆H₆)₂].

Teacher's Note:
1. Synergic bonding: σ donation (CO → M) makes the metal electron-rich, which increases π back donation (M → CO), and each strengthens the other.

Question 25. Giving a suitable example, describe the importance of the formation of complex compounds in
(i) the estimation of hardness of water
(ii) the extraction of a particular metal from its natural source.
Answer:
(i) Hardness of water is estimated by simple titration with Na₂EDTA. The Ca²⁺ and Mg²⁺ ions form stable complexes with EDTA, [Ca(EDTA)]²⁻ and [Mg(EDTA)]²⁻. The selective estimation of these ions can be done due to difference in the stability constants of their complexes.
(ii) Gold (and silver) are extracted by the cyanide process. The metal is dissolved by forming a soluble cyano complex in the presence of air, and is then recovered by adding zinc :
4Au + 8CN⁻ + 2H₂O + O₂ → 4[Au(CN)₂]⁻ + 4OH⁻
2[Au(CN)₂]⁻ + Zn → [Zn(CN)₄]²⁻ + 2Au

Teacher's Note:
1. In the cyanide process zinc displaces gold because Zn is more electropositive and forms the stable [Zn(CN)₄]²⁻.

Question 26. The formation of complex compounds finds application in the extraction of some metals. Furnish one example to support this statement.
Answer:
Silver is extracted from its ore (argentite) by leaching with dilute NaCN solution in the presence of air, when it forms a soluble complex :
4Ag + 8CN⁻ + 2H₂O + O₂ → 4[Ag(CN)₂]⁻ + 4OH⁻
Silver is then precipitated by adding zinc : 2[Ag(CN)₂]⁻ + Zn → [Zn(CN)₄]²⁻ + 2Ag

Teacher's Note:
1. Gold is extracted by exactly the same complex-forming steps; either example earns the marks.

Question 27. (a) Give reason : [CoF₆]³⁻ is a high spin complex ion.
(b) Draw the two geometrical isomers of the complex compound, [Pt(NH₃)₂Cl₂]. (Assam Board 2013)

Answer:
(a) F⁻ is a weak field ligand, so Δ_o < P. The 3d⁶ electrons of Co³⁺ do not pair up (t_2g⁴e_g²), leaving 4 unpaired electrons. Hence it is a high spin (outer orbital, sp³d²) complex.
(b) [Pt(NH₃)₂Cl₂] is square planar and exists as cis and trans isomers :

Teacher's Note:
1. Ligands sit at the corners of the square; in the cis form like ligands share one side, in the trans form they lie on a diagonal.

Short Answer Questions (3 Marks)

Teacher's Note:
1. Ni²⁺ (d⁸) with 4 ligands: strong field (CN⁻) gives dsp²/square planar/diamagnetic; weak field (Cl⁻) gives sp³/tetrahedral/paramagnetic.

Question 29. Consider the coordination compound, [Co(NH₃)₅SO₄] Br
(i) Write IUPAC name of the above coordination compound
(ii) What is the primary valence and secondary valence of the central metal, cobalt, in the above coordination compound?
(iii) Which type of structural isomerism is exhibited by the above coordination compound? (Kerala Board 2012)
Answer:
(i) Pentaamminesulphatocobalt (III) bromide
(ii) Primary valence = +3, Secondary valence = 6
(iii) Ionization isomerism

Teacher's Note:
1. Primary valence = oxidation state (x − 2 = +1 gives x = +3); secondary valence = coordination number.

Question 30. Write the name, state of hybridisation, shape and magnetic behaviour of the following complexes :
[CoCl₄]²⁻, [Ni(CN)₄]²⁻, [Cr(H₂O)₂(C₂O₄)₂]⁻ (At. nos. : Co = 27, Ni = 28, Cr = 24) (CBSE 2010)
Answer:
(i) [CoCl₄]²⁻ : Tetrachloridocobaltate (II) ion; Co²⁺ (3d⁷), Cl⁻ weak field; sp³, tetrahedral, paramagnetic (3 unpaired electrons).
(ii) [Ni(CN)₄]²⁻ : Tetracyanonickelate (II) ion; dsp², square planar, diamagnetic.
(iii) [Cr(H₂O)₂(C₂O₄)₂]⁻ : Diaquadioxalatochromate (III) ion; Cr³⁺ (3d³) has two vacant 3d orbitals; d²sp³, octahedral, paramagnetic (3 unpaired electrons).

Teacher's Note:
1. Cr³⁺ (d³) is always inner orbital and has 3 unpaired electrons whatever the ligand, since two 3d orbitals are already empty.

Question 31. How does crystal field theory explain
(i) High spin and low spin states of complexes (ii) Magnetism of complexes (iii) Colour of the complexes ?
Answer:
(i) In an octahedral field the d-orbitals split into lower t_2g and higher e_g sets. For d⁴ to d⁷ ions, the fourth electron may pair in t_2g or enter e_g. If Δ_o < P (pairing energy), as with weak field ligands, electrons enter e_g giving maximum unpaired electrons, i.e., a high spin complex. If Δ_o > P, as with strong field ligands, electrons pair in t_2g, giving a low spin complex.
(ii) The magnetic behaviour depends on the number of unpaired electrons left after filling the split orbitals. Complexes with unpaired electrons are paramagnetic (μ = \( \sqrt{n(n+2)} \) BM) and those with all electrons paired are diamagnetic. e.g., [CoF₆]³⁻ (t_2g⁴e_g²) is paramagnetic while [Co(NH₃)₆]³⁺ (t_2g⁶) is diamagnetic.
(iii) When light falls on the complex, an electron from t_2g absorbs energy equal to Δ_o and jumps to e_g (d-d transition). The absorbed radiation lies in the visible region and the complex shows the complementary colour, e.g., [Ti(H₂O)₆]³⁺ absorbs green-yellow light and appears violet. In the absence of ligands no splitting occurs and the substance is colourless.

Teacher's Note:
1. The colour seen is the complementary colour of the light absorbed, not the absorbed colour itself.

Question 32. (a) Draw a figure to show splitting of degenerate d- orbitals in an octahedral crystal field. How does the magnitude of Δ_o decide the actual configuration of d- orbitals in a complex entity ? (Hr. Board 2011)
(b) How is the magnitude of Δ_o affected by (i) nature of the ligand (ii) oxidation state of the metal ion ?
Answer:
(a) Splitting of d-orbitals in an octahedral field :
The first three electrons occupy t_2g singly. For the fourth electron : if Δ_o < P, it enters e_g giving t_2g³e_g¹ (high spin, weak field ligands); if Δ_o > P, it pairs in t_2g giving t_2g⁴e_g⁰ (low spin, strong field ligands).
(b) (i) Strong field ligands (higher in the spectrochemical series, e.g., CN⁻, CO) produce larger Δ_o than weak field ligands (e.g., I⁻, Cl⁻, F⁻).
(ii) Higher the oxidation state of the metal ion, larger is Δ_o, e.g., Δ_o for [Co(H₂O)₆]³⁺ > [Co(H₂O)₆]²⁺.

Teacher's Note:
1. In the diagram, t_2g lies 0.4 Δ_o below and e_g 0.6 Δ_o above the barycentre - keep these numbers for CFSE calculations.

Question 33. (a) What is the basis of formation of the spectrochemical series ?
(b) Draw the structures of geometrical isomers of the following coordination compounds : [Co(NH₃)₃Cl₃] and [CoCl₂(en)₂]⁺ (en = ethylene diamine and atomic number of cobalt is 27) (CBSE 2008)

Answer:
(a) The spectrochemical series is the arrangement of ligands in the order of increasing crystal field splitting (Δ_o) they produce. It is based on the experimentally determined values of Δ_o from the absorption spectra of complexes :
I⁻ < Br⁻ < SCN⁻ < Cl⁻ < S²⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ < edta⁴⁻ < NH₃ < en < CN⁻ < CO
(b) [Co(NH₃)₃Cl₃] shows fac and mer isomers, and [CoCl₂(en)₂]⁺ shows trans and cis isomers :

Teacher's Note:
1. The spectrochemical series is experimental; it cannot be predicted from ligand charge alone (neutral CO is the strongest, anionic I⁻ the weakest).

Question 34. (a) Give the electronic configuration of d-orbitals of Ti in [Ti(H₂O)₆]³⁺ ion and explain why this complex is coloured ? (At. No. of Ti = 22)
(b) Write the IUPAC name of [Cr(NH₃)₃(H₂O)₃]Cl₃.
Answer:
(a) Ti³⁺ : 3d¹. In the octahedral field the configuration is t_2g¹ e_g⁰. The single electron absorbs light of wavelength about 498 nm (green-yellow region) and is promoted from t_2g to e_g (d-d transition). Hence the complex shows the complementary colour, i.e., it appears violet.
(b) Triamminetriaquachromium (III) chloride

Teacher's Note:
1. A d-d transition needs at least one electron and a vacant higher orbital: d⁰ and d¹⁰ ions give colourless complexes.

Teacher's Note:
1. Memory aid for the two ends: halides are weak, CN⁻ and CO are strong; H₂O and NH₃ sit in the middle.

Question 36. For the complex [Fe(en)₂Cl₂] Cl, (en = ethylene diamine), identify :
(i) the oxidation number of iron
(ii) the hybrid orbitals and the shape of the complex
(iii) the magnetic behaviour of the complex
(iv) the number of geometrical isomers
(v) whether there is an optical isomer also, and
(vi) name of the complex (At. No. of Fe = 26). (CBSE 2009, 2011)
Answer:
(i) +3
(ii) d²sp³, octahedral
(iii) paramagnetic (one unpaired electron)
(iv) 2 (cis and trans)
(v) yes, for cis
(vi) dichloridobis(ethane-1,2-diamine)iron (III) chloride

Teacher's Note:
1. Fe³⁺ is d⁵; even after pairing in d²sp³ one electron stays unpaired, so the complex can never be diamagnetic.

Question 37. (a) State the hybridisation and magnetic behaviour of [Cr(CO)₆].
(b) What are the various factors affecting crystal field splitting energy.
(c) Which of the two is more stable and why ? K₄[Fe(CN)₆] or K₃[Fe(CN)₆]
Answer:
(a) d²sp³, diamagnetic (Cr(0), 3d⁶ after pairing by strong field CO; all electrons paired).
(b) (i) Nature of the ligand : strong field ligands produce larger splitting. (ii) Charge (oxidation state) on the metal ion : higher the charge, larger the splitting. (iii) Geometry of the complex : Δ_t = 4/9 Δ_o. (iv) Position of the metal in the series : Δ increases in the order 3d < 4d < 5d.
(c) K₄[Fe(CN)₆] is more stable. In [Fe(CN)₆]⁴⁻, the effective atomic number of Fe²⁺ = 24 + 12 = 36, which is the atomic number of the noble gas krypton, whereas in [Fe(CN)₆]³⁻ it is 23 + 12 = 35.

Teacher's Note:
1. EAN = electrons of the metal ion + 2 × (number of ligand donor atoms); EAN equal to a noble gas number is linked with extra stability.

Question 38. Explain the following terms giving a suitable example in each case :
(i) Ambidentate ligand (ii) Denticity of a ligand (iii) Crystal field splitting in an octahedral field. (CBSE 2011)
Answer:
(i) Ambidentate ligand : A unidentate ligand which can coordinate through two different donor atoms, one at a time, e.g., NO₂⁻ (M–NO₂ or M–ONO), SCN⁻ (M–SCN or M–NCS).
(ii) Denticity of a ligand : The number of donor atoms through which a ligand binds the central metal ion, e.g., denticity of NH₃ = 1, en = 2, EDTA⁴⁻ = 6.
(iii) Crystal field splitting in an octahedral field : When six ligands approach the metal ion along the axes, the d_x²−y² and d_z² orbitals (e_g), pointing towards the ligands, are raised in energy, and the d_xy, d_yz, d_xz orbitals (t_2g), pointing between the axes, are lowered. The energy difference is Δ_o, e.g., in [Ti(H₂O)₆]³⁺ the electron goes into t_2g.

Teacher's Note:
1. The e_g orbitals rise because their lobes point straight at the ligands and feel the most repulsion.

Long Answer Questions (5 Marks)

Question 39. What is meant by stability of a coordination compound in solution ? State the factors which govern stability of complexes. In what ways the concept of stability of complexes has been helpful in analytical chemistry ?
Answer:
Stability : The stability of a complex in solution refers to the degree of association between the metal ion and the ligands in the state of equilibrium. It is expressed by the stability (formation) constant. For M + 4L ⇌ ML₄, \( \beta_4 = \frac{[ML_4]}{[M][L]^4} \). Larger the value of β, greater is the stability of the complex. The reciprocal of the stability constant is the instability (dissociation) constant.
Factors governing stability :
(i) Charge on the central metal ion : Greater the charge, greater the stability, e.g., Fe³⁺ complexes are more stable than Fe²⁺ complexes.
(ii) Size of the metal ion : Smaller the size (higher charge density), greater the stability.
(iii) Basic nature of the ligand : More basic ligands (better electron donors) form more stable complexes.
(iv) Chelation : Chelating ligands form more stable complexes than unidentate ligands (chelate effect), five- and six-membered rings being most stable.
Uses in analytical chemistry :
(i) Hardness of water is estimated by titration with EDTA, which forms stable complexes with Ca²⁺ and Mg²⁺; the difference in their stability constants allows selective estimation.
(ii) Separation of ions in qualitative analysis, e.g., AgCl dissolves in NH₃ forming the stable [Ag(NH₃)₂]⁺, separating it from Hg₂Cl₂ and PbCl₂; Cu²⁺ and Cd²⁺ are separated with KCN because [Cu(CN)₄]³⁻ is so stable that it is not precipitated by H₂S while [Cd(CN)₄]²⁻ gives CdS.
(iii) Detection of ions by formation of coloured stable complexes, e.g., Ni²⁺ with dimethylglyoxime (red), Fe³⁺ with [Fe(CN)₆]⁴⁻ (Prussian blue).

Teacher's Note:
1. β values are very large, so they are usually quoted as log β; compare log β values directly to rank stabilities.

Question 40. Select a complex formation reaction and write an expression for the stability constant of the complex species. What information is conveyed regarding the strengths of ligands from the stability constant values of their complexes with a metal ion ? Illustrate your answer with examples of monodentate ligands.
Answer:
Consider the formation of tetraamminecopper(II) ion in aqueous solution :
Cu²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄]²⁺
Overall stability constant, \( \beta_4 = \frac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}][NH_3]^4} \)
The reaction takes place in steps, Cu²⁺ + NH₃ ⇌ [Cu(NH₃)]²⁺ (K₁), ..., [Cu(NH₃)₃]²⁺ + NH₃ ⇌ [Cu(NH₃)₄]²⁺ (K₄), and β₄ = K₁ × K₂ × K₃ × K₄.
Information conveyed : For the same metal ion, a ligand that forms a complex with a larger stability constant is the stronger ligand. In aqueous solution, the metal ion is already present as an aqua complex, so a ligand L replaces H₂O only if it is a stronger ligand than water.
Examples : Cu²⁺ in water exists as [Cu(H₂O)₄]²⁺ (pale blue). On adding NH₃, the deep blue [Cu(NH₃)₄]²⁺ (log β₄ = 13.1) is formed, showing NH₃ is a stronger ligand than H₂O. With CN⁻, the complex [Cu(CN)₄]²⁻ has log β₄ ≈ 27.3, showing CN⁻ is a much stronger ligand than NH₃. Similarly, [Fe(H₂O)₆]³⁺ reacts with SCN⁻ to give blood red [Fe(H₂O)₅(SCN)]²⁺, and on adding F⁻ the colour disappears due to formation of the more stable [FeF₆]³⁻; thus the strength order for Fe³⁺ is F⁻ > SCN⁻ > H₂O.

Teacher's Note:
1. A colour change on adding a new ligand (pale blue to deep blue with NH₃) is visible evidence that the new ligand has a larger stability constant.

Question 41. Briefly describe the importance of coordination compounds in
(i) Qualitative Analysis (ii) Extraction of metals (iii) Biological systems.
Answer:
(i) Qualitative analysis : Many metal ions are detected and separated through characteristic complexes, e.g., Ni²⁺ gives a red precipitate with dimethylglyoxime, [Ni(dmg)₂]; Cu²⁺ gives deep blue [Cu(NH₃)₄]²⁺ with excess NH₃; Fe³⁺ gives Prussian blue Fe₄[Fe(CN)₆]₃ with potassium ferrocyanide and blood red colour with SCN⁻; AgCl dissolves in NH₃ as [Ag(NH₃)₂]⁺.
(ii) Extraction of metals : Silver and gold are extracted by the cyanide process. The metal is dissolved in dilute NaCN in the presence of air as the soluble complex [Ag(CN)₂]⁻ or [Au(CN)₂]⁻, and then recovered by displacement with zinc :
4Au + 8CN⁻ + 2H₂O + O₂ → 4[Au(CN)₂]⁻ + 4OH⁻ ; 2[Au(CN)₂]⁻ + Zn → [Zn(CN)₄]²⁻ + 2Au
Similarly, impure nickel is purified by converting it to volatile [Ni(CO)₄], which on heating decomposes to give pure Ni.
(iii) Biological systems : Many biologically important compounds are coordination compounds : chlorophyll (pigment for photosynthesis) is a complex of Mg; haemoglobin (oxygen carrier in blood) is a complex of Fe; vitamin B₁₂ (cyanocobalamin, anti-pernicious anaemia factor) is a complex of Co. Enzymes like carboxypeptidase A and carbonic anhydrase contain coordinated metal ions.

Teacher's Note:
1. Link each biological complex to its metal: chlorophyll - Mg, haemoglobin - Fe, vitamin B₁₂ - Co.