Coordination Compounds

The chapter on Coordination Compounds in the NCERT 12th standard chemistry textbook covers the following topics:

Werner’s theory of coordination compounds: This theory explains the structure and properties of coordination compounds.
Nomenclature of coordination compounds: This section explains how to name coordination compounds.
Isomerism in coordination compounds: This section describes the different types of isomerism that can occur in coordination compounds.
Bonding in coordination compounds: This section explains the different types of bonding that can occur in coordination compounds.
Valence bond theory: This theory explains the bonding in coordination compounds in terms of the overlap of atomic orbitals.
Crystal field theory: This theory explains the bonding in coordination compounds in terms of the interaction between the metal ion and the ligands.
Applications of coordination compounds: This section describes the various applications of coordination compounds.

Overall, the chapter provides a comprehensive overview of coordination compounds. It is an important chapter for students who are interested in chemistry.

Here are some of the key concepts covered in the chapter:

Central metal atom: The atom in a coordination compound that is surrounded by ligands.
Ligands: Ions or molecules that are bonded to the central metal atom.
Coordinate bond: A type of bond in which both electrons in the bond come from the same atom.
Coordination number: The number of ligands that are bonded to the central metal atom.
Isomerism: The phenomenon in which two or more compounds have the same chemical formula but different structures.
Valence bond theory: A theory that explains the bonding in coordination compounds in terms of the overlap of atomic orbitals.
Crystal field theory: A theory that explains the bonding in coordination compounds in terms of the interaction between the metal ion and the ligands.

Exercise

1. Explain the bonding in coordination compounds in terms of Werner’s postulates.

Ans :

Two types of valencies: Metals have primary (ionizable, ionic bonds) and secondary (non-ionizable, coordinate covalent bonds) valencies.

Fixed coordination number: Each metal has a fixed number of secondary valencies, determining the number of ligands.

Directional secondary valencies: These dictate the geometry of the complex.

2. FeSO4 solution mixed with (NH4)2SO4 solution in 1 : 1 molar ratio gives the test of Fe2+ ion but CuSO4 solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu2+ ion. Explain why?

Ans :

FeSO₄ + (NH₄)₂SO₄: These combine to form Mohr’s salt, FeSO₄.(NH₄)₂SO₄.6H₂O. This is a double salt. In solution, it completely dissociates into its constituent ions, including Fe²⁺ ions. Because Fe²⁺ ions are present in the solution, they can be detected by standard tests.
CuSO₄ + NH₃: Copper sulfate reacts with ammonia to form a complex ion, [Cu(NH₃)₄]²⁺. This is a complex. In this complex, the Cu²⁺ ion is tightly bound to the ammonia molecules (ligands). It’s no longer a “free” Cu²⁺ ion in the solution. Because the copper is tied up in the complex, it doesn’t react with the reagents used in tests for Cu²⁺, and thus, you won’t get a positive result.

In short: Mohr’s salt releases Fe²⁺ ions, while the copper-ammonia complex keeps the Cu²⁺ ions bound, preventing them from reacting in standard tests.

3. Explain with two examples each of the following: coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.

Ans :

1. Coordination Entity: A coordination entity is the central metal atom or ion bonded to a specific number of ligands. It’s enclosed in square brackets.

Example 1: [Co(NH₃)₆]Cl₃ (Hexaamminecobalt(III) chloride)

Example 2: [Ni(CN)₄]²⁻ (Tetracyanonickelate(II) ion)

2. Ligand: A ligand is an ion or molecule that binds to the central metal atom/ion in a coordination entity by donating electron pairs.

Example 1: NH₃ (ammonia) in [Co(NH₃)₆]Cl₃

Example 2: CN⁻ (cyanide) in [Ni(CN)₄]²⁻

Example 3: Cl⁻ (chloride) can also act as a ligand, although in the first example it is a counter-ion. In a compound like [PtCl₄]²⁻ it is a ligand.

Example 4: H₂O (water) can act as a ligand, such as in [Cr(H₂O)₆]³⁺

3. Coordination Number: The coordination number is the number of ligand atoms directly bonded to the central metal atom/ion.

Example 1: 6 in [Co(NH₃)₆]Cl₃ (Cobalt is bonded to six ammonia molecules)

Example 2: 4 in [Ni(CN)₄]²⁻ (Nickel is bonded to four cyanide ions)

4. Coordination Polyhedron: The coordination polyhedron is the three-dimensional geometric shape formed by the arrangement of ligands around the central metal atom/ion.

Example 1: Octahedral in [Co(NH₃)₆]Cl₃ (Six ligands arranged octahedrally around Cobalt)

Example 2: Square planar in [Ni(CN)₄]²⁻ (Four ligands arranged in a square plane around Nickel)

5. Homoleptic: A complex in which the metal is bound to only one type of ligand.

Example 1: [Co(NH₃)₆]Cl₃ (Cobalt is only bound to ammonia ligands)

Example 2: [Ni(CN)₄]²⁻ (Nickel is only bound to cyanide ligands)

6. Heteroleptic: A complex in which the metal is bound to more than one type of ligand.

Example 1: [Co(NH₃)₄Cl₂]Cl (Cobalt is bound to both ammonia and chloride ligands)

Example 2: [Pt(NH₃)₂Cl₂] (Platinum is bound to both ammonia and chloride ligands)

4. What is meant by unidentate didentate and ambidentate ligands? Give two examples for each.

Ans :

1. Unidentate Ligands:

Meaning: These ligands have only one “donor atom” through which they can bind to the central metal ion. Think of it like having only one hand to hold onto the metal.
Examples:
- Ammonia (NH₃): The nitrogen atom has a lone pair of electrons that it can donate to the metal.
- Chloride ion (Cl⁻): The chloride ion has a negative charge and can donate its lone pair of electrons to the metal.

2. Didentate Ligands:

Meaning: These ligands have two donor atoms, meaning they can form two coordinate bonds with the central metal ion. It’s like having two hands to hold onto the metal.
Examples:
- Ethylenediamine (en): This molecule has two nitrogen atoms, each with a lone pair of electrons, allowing it to bind to the metal through both nitrogens.
- Oxalate ion (C₂O₄²⁻): This ion has two oxygen atoms, each with a lone pair of electrons, that can coordinate with the metal.

3. Ambidentate Ligands:

Meaning: These ligands have two (or more) potential donor atoms, but they can only bind to the metal through one of those atoms at a time. It’s like having two hands, but you can only hold with one at any given moment.
Examples:
- Cyanide ion (CN⁻): This ion can bind to the metal through either the carbon atom or the nitrogen atom.
- Thiocyanate ion (SCN⁻): This ion can bind to the metal through either the sulfur atom or the nitrogen atom.

5. Specify the oxidation numbers of the metals in the following coordination entities:

(i) [Co(H2O)(CN)(en)2]2+

(ii) [CoBr2(en)2]+

(iii) [PtCl4]2-

(iv) K3[Fe(CN)6]

(v) [Cr(NH3)3CI3]

Ans :

(i) [Co(H₂O)(CN)(en)₂]²⁺

H₂O (water) is a neutral ligand (oxidation number 0).
CN⁻ (cyanide) has an oxidation number of -1.
en (ethylenediamine) is a neutral ligand (oxidation number 0).
The overall charge of the complex is +2.

Let ‘x’ be the oxidation number of Co.

x + 0 + (-1) + 2(0) = +2 x – 1 = +2 x = +3

Therefore, the oxidation number of Co is +3.

(ii) [CoBr₂(en)₂]⁺

Br⁻ (bromide) has an oxidation number of -1.
en (ethylenediamine) is a neutral ligand (oxidation number 0).
The overall charge of the complex is +1.

Let ‘x’ be the oxidation number of Co.

x + 2(-1) + 2(0) = +1 x – 2 = +1 x = +3

Therefore, the oxidation number of Co is +3.

(iii) [PtCl₄]²⁻

Cl⁻ (chloride) has an oxidation number of -1.
The overall charge of the complex is -2.

Let ‘x’ be the oxidation number of Pt.

x + 4(-1) = -2 x – 4 = -2 x = +2

Therefore, the oxidation number of Pt is +2.

(iv) K₃[Fe(CN)₆]

CN⁻ (cyanide) has an oxidation number of -1.
K⁺ (potassium) has an oxidation number of +1. The complex ion itself has a charge of -3 to balance the 3 K+ ions.

Let ‘x’ be the oxidation number of Fe.

x + 6(-1) = -3 x – 6 = -3 x = +3

Therefore, the oxidation number of Fe is +3.

(v) [Cr(NH₃)₃Cl₃]

NH₃ (ammonia) is a neutral ligand (oxidation number 0).
Cl⁻ (chloride) has an oxidation number of -1.
The overall charge of the complex is 0.

Let ‘x’ be the oxidation number of Cr.

x + 3(0) + 3(-1) = 0 x – 3 = 0 x = +3

Therefore, the oxidation number of Cr is +3.

6. Using IUPAC norms, write the formulae for the following : (C.B.S.E. Foreign 2015)

(a) tetrahydroxozincate(II)

(b) hexaammineplatinum (TV)

(c) potassiumtetrachloridopalladate(II)

(d) tetrabromidocuprate (II)

(e) hexaaminecobalt(III) sulphate

(f) potassiumtetracyanonicklate (II)

(g) potassiumtrioxalatochromate(III)

(h) pentaamminenitrito-O-cobalt(III)

(i) diamminedichloridoplatinum(II)

(j) pentaamminenitrito-N-cobalt (III).

Ans :

(a) tetrahydroxozincate(II)

[Zn(OH)₄]²⁻

(b) hexaammineplatinum(IV)

[Pt(NH₃)₆]⁴⁺

(c) potassiumtetrachloridopalladate(II)

K₂[PdCl₄]

(d) tetrabromidocuprate(II)

[CuBr₄]²⁻

(e) hexaamminecobalt(III) sulfate

[Co(NH₃)₆]₂(SO₄)₃

(f) potassiumtetracyanonickelate(II)

K₂[Ni(CN)₄]

(g) potassiumtrioxalatochromate(III)

K₃[Cr(C₂O₄)₃]

(h) pentaamminenitrito-O-cobalt(III)

[Co(NH₃)₅(ONO)]²⁺

(i) diamminedichloridoplatinum(II)

[Pt(NH₃)₂(Cl)₂]

(j) pentaamminenitrito-N-cobalt(III)

[Co(NH₃)₅(NO₂)]²⁺

7. Using IUPAC norms write the systematic names of the following:

(i) [Co(NH3)6]CI3,

(ii)[Pt(NH3)2CI (NH2CH3)] Cl

(iii) [Ti(H20)6]3+

(iv) [Co(NH3)4Cl(N02)]CI

(v)|Mn(H20)6]2+

(vi)[NiCl4]2-

(vii)[Ni(NH3)6]CI2

(viii)[Co(en)3]3+

(ix) [Ni(CO)4]

Ans :

(i) [Co(NH₃)₆]Cl₃:

Hexaamminecobalt(III) chloride

(ii) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl:

Diamminechlorido(methylamine)platinum(II) chloride

(iii) [Ti(H₂O)₆]³⁺:

Hexaaquatitanium(III) ion

(iv) [Co(NH₃)₄Cl(NO₂)]Cl:

Tetraamminechloridonitrocobalt(III) chloride

(v) [Mn(H₂O)₆]²⁺:

Hexaaquamanganese(II) ion

(vi) [NiCl₄]²⁻:

Tetrachloridonickelate(II) ion

(vii) [Ni(NH₃)₆]Cl₂:

Hexaamminenickel(II) chloride

(viii) [Co(en)₃]³⁺: (en = ethylenediamine)

Tris(ethylenediamine)cobalt(III) ion

(ix) [Ni(CO)₄]:

Tetracarbonylnickel(0)

8. List various types of isomerism possible for coordination compounds, giving an example of each.

Ans :

9. How many geometrical isomers are possible in . the following coordination entities?

(i) [Cr(C2O4)3]3- (ii) [CoCl3(NH3)3]

Ans :

[Cr(C₂O₄)₃]³⁻: 0 geometrical isomers

[CoCl₃(NH₃)₃]: 2 geometrical isomers (fac and mer)

10. Draw the structures of optical isomers of

(i) [Cr(C2O4)3]3-

(ii)[PtCI2(en)2]2+

(iii)[Cr(NH3)2CI2(en)]+

Ans :

11. Draw all the isomers (geometrical and optical) of

(i)[CoCl2(en)2]+

(ii)[Co(NH3) Cl (en)2]2+

(iii) [Co(NH3)2Cl2(en)]+

Ans :

12. Write all the geometrical isomers of [Pt(NH3)(Br)(Cl) (Py)] and how many of these will exhibit optical isomerism?

Ans :

These isomers are formed by keeping the position of one ligand, such as NH₃, fixed while rotating the positions of the other ligands. This type of isomerism does not exhibit optical isomerism. Optical isomerism is rarely observed in square planar or tetrahedral complexes and occurs only when they contain an asymmetrical chelating ligand.

13. Aqueous copper sulphate solution (blue in colour) gives: (i) a green precipitate with aqueous potassium fluoride and (ii)a bright green solution with aqueous potassium chloride. Explain these experimental results.

Ans :

(i) Reaction with Potassium Fluoride (KF):

Potassium fluoride (KF) reacts with aqueous copper sulfate to form a green precipitate of copper(II) fluoride (CuF₂):

[Cu(H₂O)₄]²⁺ (aq) + 2F⁻ (aq) → CuF₂ (s) + 4H₂O (l)

Explanation: Fluoride (F⁻) ions are strong field ligands. They replace the water ligands in the copper complex. CuF₂ is a sparingly soluble compound and precipitates out of the solution as a green solid.

(ii) Reaction with Potassium Chloride (KCl):

Potassium chloride (KCl) reacts with aqueous copper sulfate to form a bright green solution containing the tetrachloridocuprate(II) ion:

[Cu(H₂O)₄]²⁺ (aq) + 4Cl⁻ (aq) → [CuCl₄]²⁻ (aq) + 4H₂O (l)

Explanation: Chloride (Cl⁻) ions are also ligands, although they are weaker field ligands than fluoride. They replace the water ligands to form the tetrachloridocuprate(II) complex ion, [CuCl₄]²⁻. This complex ion is bright green in color, giving the solution its characteristic appearance. The reaction shows a shift in the equilibrium of the copper complex, leading to a change in color from blue to green.

14. What is the coordination entity formed when excess of aqueons KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H2S (g) is passed through this solution?

Ans :

The reaction produces the coordination entity [Cu(CN)₄]⁻. Because CN⁻ is a strong-field ligand, this complex ion is very stable and does not dissociate appreciably to produce Cu⁺ ions. Consequently, the concentration of Cu⁺ is too low to exceed the solubility product of CuS, and no precipitate forms when H₂S is introduced.

15. Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:

(i) [Fe(CN)6]4-

(ii) [FeF6]3-

(iii) [Co(C2O4)3]3-

(iv) [CoF6]3-

Ans :

16. Draw figure to show the splitting of d-orbitals in an octahedral crystal field.

Ans :

17. What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.

Ans :

Spectrochemical Series:

The spectrochemical series is an arrangement of ligands in order of their increasing ability to split the d-orbitals of a central metal ion in a coordination complex.In other words, it’s a list of ligands ranked according to the magnitude of the crystal field splitting (Δ₀) they produce.

Feature	Weak Field Ligands	Strong Field Ligands
Crystal Field Splitting (Δ₀)	Small	Large
d-orbital splitting	Small energy difference between t₂g and eg orbitals	Large energy difference between t₂g and eg orbitals
Spin State	Tend to form high-spin complexes (more unpaired e⁻)	Tend to form low-spin complexes (fewer unpaired e⁻)
Pairing of e⁻	Less likely to cause pairing	More likely to cause pairing
Examples	I⁻, Br⁻, Cl⁻, F⁻, OH⁻, H₂O	CN⁻, CO, NO₂⁻, NH₃, en (ethylenediamine)

18. What is crystal field splitting energy? How does the magnitude of Δ0 decide the actual configuration of d-orbitals in a coordination entity?

Ans :

Crystal Field Splitting Energy (Δ₀)

Crystal field splitting energy (Δ₀) is the difference in energy between the t₂g (lower energy) and eg (higher energy) sets of d orbitals in a coordination complex. This splitting occurs due to the electrostatic interactions between the ligands (or their negative charges) and the d orbitals of the central metal ion. The ligands create an electric field that raises the energy of some d orbitals more than others.

Small Δ₀ (weak field ligands): Favors high-spin complexes with more unpaired electrons. Electrons fill orbitals individually before pairing up.

Large Δ₀ (strong field ligands): Favors low-spin complexes with fewer unpaired electrons. Electrons pair up in the t₂g orbitals before occupying eg orbitals.

19. [Cr(NH3)6]3+ is paramagnetic while [Ni(CN)4]2- is diamagnetic. Explain why?

Ans :

20. A solution of [Ni(H20)6]2+ is green but a solution of [Ni(CN)4]2-is colourless. Explain.

Ans : The green color of [Ni(H₂O)₆]²⁺ is due to d-d transitions in the visible region, facilitated by the weak field water ligands and the octahedral geometry. The colorless nature of [Ni(CN)₄]²⁻ is due to the strong field cyanide ligands causing such a large crystal field splitting in the square planar complex that electronic transitions occur in the ultraviolet region, outside of the visible range.

21. [Fe(CN)6]4- and [Fe(H2O)6]2+ are of different colours in dilute solutions. Why?

Ans : The strong-field CN⁻ ligands in [Fe(CN)₆]⁴⁻ cause a large crystal field splitting, resulting in low-spin, and the complex absorbs light outside the visible region (so it appears nearly colorless). The weak-field H₂O ligands in [Fe(H₂O)₆]²⁺ cause a small crystal field splitting, resulting in high-spin, and the complex absorbs visible light, giving it a color.

22. Discuss the nature of bonding in metal carbonyls.

Ans :

1. Sigma (σ) Bond:

CO acts as a Lewis base, donating its lone pair of electrons on the carbon atom to the vacant d-orbital of the metal atom. This forms a sigma (σ) bond between the metal and carbon.

2. Pi (π) Back-bonding:

The metal atom, in turn, donates electrons from its filled d-orbitals (or a combination of d and s orbitals) back to the empty π* (pi-star, antibonding) molecular orbital of CO. This is called pi (π) back-bonding or back donation.

23. Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:

(i)K3[CO(C2O4)3I (ii) cis-[Cr(en)2Cl2]Cl (iii) (NH4)2[CoF4] (iv) [Mn(H20)6]SO4

Ans :

(i) K₃[Co(C₂O₄)₃]

Oxidation State: Let x be the oxidation state of Co. 3(+1) + x + 3(-2) = 0; x = +3
d-orbital Occupation: Co³⁺ is [Ar] 3d⁶.
Coordination Number: 6 (three bidentate oxalate ligands)

(ii) cis-[Cr(en)₂Cl₂]Cl

Oxidation State: Let x be the oxidation state of Cr. x + 2(0) + 2(-1) + (-1) = 0; x = +3
d-orbital Occupation: Cr³⁺ is [Ar] 3d³.
Coordination Number: 6 (two en ligands and two Cl ligands)

(iii) (NH₄)₂[CoF₄]

Oxidation State: Let x be the oxidation state of Co. 2(+1) + x + 4(-1) = 0; x = +2
d-orbital Occupation: Co²⁺ is [Ar] 3d⁷.
Coordination Number: 4 (four F ligands)

(iv) [Mn(H₂O)₆]SO₄

Oxidation State: Let x be the oxidation state of Mn. x + 6(0) + (-2) = 0; x = +2
d-orbital Occupation: Mn²⁺ is [Ar] 3d⁵.
Coordination Number: 6 (six H₂O ligands)

24. Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:

(i) K[Cr(H20)2(C204)2|-3H20 (ii) [Co(NH3)5CIlCl2

(iii) CrCI3(Py)3 (iv)Cs[FeCl4] (v)K4|Mn(CN)6|

Ans :

(i) K[Cr(H₂O)₂(C₂O₄)₂]·3H₂O

IUPAC Name: Potassium diaquabis(oxalato)chromate(III) trihydrate
Oxidation State: Cr = +3
Electronic Configuration: Cr³⁺ = [Ar] 3d³
Coordination Number: 6
Stereochemistry: The complex can exhibit geometrical isomerism (cis and trans with respect to the water ligands). The cis isomer is chiral and exists as a pair of optical isomers (enantiomers). The trans isomer is achiral.
Magnetic Moment: Since Cr³⁺ has 3 unpaired electrons (high-spin or low-spin do not change the number of unpaired electrons in this case), the magnetic moment is expected to be close to that calculated using the spin-only formula: √n(n+2) BM = √3(3+2) BM ≈ 3.87 BM.

(ii) [Co(NH₃)₅Cl]Cl₂

IUPAC Name: Pentaamminechlorocobalt(III) chloride
Oxidation State: Co = +3
Electronic Configuration: Co³⁺ = [Ar] 3d⁶ (low-spin, t₂g⁶)
Coordination Number: 6
Stereochemistry: No geometrical isomers are possible.
Magnetic Moment: Since it is low-spin d⁶, there are no unpaired electrons. The complex is diamagnetic (μ = 0 BM).

(iii) CrCl₃(Py)₃ (Py = pyridine)

IUPAC Name: Trichloridotri(pyridine)chromium(III)
Oxidation State: Cr = +3
Electronic Configuration: Cr³⁺ = [Ar] 3d³
Coordination Number: 6
Stereochemistry: The complex can exhibit facial (fac) and meridional (mer) isomerism.
Magnetic Moment: Since Cr³⁺ has 3 unpaired electrons, the magnetic moment is expected to be close to 3.87 BM.

(iv) Cs[FeCl₄]

IUPAC Name: Cesium tetrachloridoferrate(III)
Oxidation State: Fe = +3
Electronic Configuration: Fe³⁺ = [Ar] 3d⁵
Coordination Number: 4
Stereochemistry: The complex is tetrahedral. Tetrahedral complexes do not show geometrical isomerism.
Magnetic Moment: Since Fe³⁺ is d⁵, it has 5 unpaired electrons (high-spin), the magnetic moment is expected to be close to 5.92 BM.

(v) K₄[Mn(CN)₆]

IUPAC Name: Potassium hexacyanomanganate(II)
Oxidation State: Mn = +2
Electronic Configuration: Mn²⁺ = [Ar] 3d⁵
Coordination Number: 6
Stereochemistry: No geometrical isomers are observed.
Magnetic Moment: CN⁻ is a strong field ligand, but Mn²⁺ is a d⁵ system. All five d electrons will remain unpaired. The magnetic moment is expected to be close to 5.92 BM.

25. Explain the violet colour of the complex [Ti(H2O)6] 3+ on the basis of crystal field theory.

Ans : The violet color of [Ti(H₂O)₆]³⁺ is due to the electronic transition of the single d electron from a t₂g orbital to an eg orbital. This transition absorbs light in the green-yellow region, and we observe the complementary color, which is violet. The crystal field splitting caused by the six water ligands determines the energy difference between the t₂g and eg orbitals, and thus the wavelength of light absorbed.

26. What is meant by the chelate effect? Give an example.

Ans : The chelate effect refers to the enhanced stability of coordination compounds containing chelating ligands (also called polydentate ligands) compared to complexes with similar monodentate ligands. Essentially, if a ligand can grab onto the central metal ion in multiple places (like a claw), it forms a more stable complex than if several separate ligands each only bond in one place.

Example:

Consider the formation of two cobalt(II) complexes:

[Co(NH₃)₆]²⁺ + 2en → [Co(en)₂(NH₃)₂]²⁺ + 4NH₃
[Co(NH₃)₆]²⁺ + 3en → [Co(en)₃]²⁺ + 6NH₃

27. Discuss briefly giving an example in each case the role of coordination compounds in :

(a) biological systems,

(b) analytical chemistry,

(c) medicinal chemistry, and

(d) extraction/metallurgy of metals.

Ans :

(a) Biological Systems:

Coordination compounds play crucial roles in many biological systems.

Example: Hemoglobin, the protein responsible for oxygen transport in the blood, is a coordination complex. It contains an iron(II) ion coordinated to a porphyrin ring. The iron ion’s ability to bind oxygen reversibly is essential for respiration. Similarly, chlorophyll, the pigment responsible for photosynthesis in plants, is a coordination complex of magnesium. Vitamin B₁₂ (cobalamin), essential for cell growth and DNA synthesis, is another example of a coordination compound containing cobalt.

(b) Analytical Chemistry:

Coordination compounds are widely used in qualitative and quantitative analysis.

Example: EDTA (ethylenediaminetetraacetate) is a powerful chelating agent used extensively in complexometric titrations to determine the concentration of metal ions in solution.
For instance, EDTA can be used to determine the hardness of water by titrating the calcium and magnesium ions present. Dimethylglyoxime (DMG) is a specific reagent for the detection and quantitative determination of nickel.

(c) Medicinal Chemistry:

Coordination compounds have applications in medicine, particularly in the treatment of certain diseases.

Example: Cisplatin, a platinum-containing complex, is a widely used anticancer drug. It works by binding to DNA in cancer cells, interfering with their replication and thus slowing down tumor growth. Other metal complexes are being investigated for their potential use as antiviral, antibacterial, and anti-inflammatory agents.

(d) Extraction/Metallurgy of Metals:

Coordination compounds are sometimes involved in the extraction and purification of metals from their ores.

Example: The Mond process for refining nickel involves the formation of a volatile nickel carbonyl complex, Ni(CO)₄. Impurities are left behind when the complex is vaporized and then decomposed at a higher temperature to deposit pure nickel. Similarly, the complexing ability of cyanide ions (CN⁻) is used in the extraction of gold and silver from their ores. The metals are leached with a cyanide solution to form soluble cyano complexes, which are then reduced to the pure metals.

28. How many ions are produced from the complex Co(NH3)6Cl2 in solution?

(i) 6

(ii) 4

(iii)3

(iv)2

Ans : The complex Co(NH3)6Cl2, when dissolved in water, dissociates into two ions:

The complex ion [Co(NH3)6]2+
Two chloride ions (Cl-)

Therefore, the correct answer is (iv) 2.

29. Amongst the following ions? Which one has the highest magnetic moment value:

(i) [Cr(H2O)6]3+

(ii) [Fe(H20)6]2+ (iii) [Zn(H20)6]2+

Ans :

[Cr(H₂O)₆]³⁺:

Chromium (Cr) is in group 6, so Cr³⁺ has a d³ configuration.
H₂O is a weak-field ligand, so it will be high spin.
There are 3 unpaired electrons.

[Fe(H₂O)₆]²⁺:

Iron (Fe) is in group 8, so Fe²⁺ has a d⁶ configuration.
H₂O is a weak-field ligand, so it will be high spin.
There are 4 unpaired electrons.

[Zn(H₂O)₆]²⁺:

Zinc (Zn) is in group 12, so Zn²⁺ has a d¹⁰ configuration.
All electrons are paired. There are 0 unpaired electrons.

[Fe(H₂O)₆]²⁺ has the highest number of unpaired electrons (4), so it will have the highest magnetic moment value among the given options.

30. Amongst the following, the most stable complex is:

(i) [Fe(H2O)6] (ii) [Fe(NH3)6]3+

(iii) [Fe(C2O4)3]3- (iv) [FeCl6]3-

Ans : The most stable complex among the options is (iii) [Fe(C₂O₄)₃]³⁻.

31. What will be the correct order for the wavelengths of absorption in the visible region for the following:[Ni(NO2)6]4-, [Ni(NH3)6]2+, [Ni(H20)6]2+?

Ans :

Order of Δ₀:

NO₂⁻ > NH₃ > H₂O

Order of Wavelength (λ):

Since wavelength is inversely proportional to Δ₀, the order of absorption wavelengths will be the opposite:

H₂O > NH₃ > NO₂⁻

Therefore, the correct order for the wavelengths of absorption in the visible region will be:

[Ni(H₂O)₆]²⁺ > [Ni(NH₃)₆]²⁺ > [Ni(NO₂)₆]⁴⁻