Chapter 9: Coordination Compounds
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What is a coordination compound? Give an example.
Coordination Compound: A compound in which a central metal atom or ion is bonded to a specific number of neutral molecules or anions (ligands) by coordinate covalent bonds.
Example: $K_4[Fe(CN)_6]$ (Potassium hexacyanoferrate(II)).
2. Differentiate between a double salt and a coordination complex.
| Double Salt | Coordination Complex |
|---|---|
| It completely dissociates into simple ions when dissolved in water. | It does not completely dissociate into simple ions in water. The complex ion remains intact. |
| It loses its identity in an aqueous solution. | It retains its identity in an aqueous solution. |
| Example: Mohr's salt ($FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O$) | Example: Potassium ferrocyanide ($K_4[Fe(CN)_6]$) |
3. Define Ligand. What type of bond is formed between a ligand and a central metal ion?
Ligand: An atom, ion, or molecule that can donate at least one pair of electrons to the central metal atom/ion to form a coordinate bond.
Bond Type: Coordinate covalent bond (dative bond). The ligand acts as a Lewis base (electron pair donor) and the metal ion acts as a Lewis acid (electron pair acceptor).
4. Classify ligands based on their denticity with one example each.
- Monodentate (Unidentate): Ligands that bind through only one donor atom. Example: $NH_3, Cl^-, H_2O$.
- Bidentate (Didentate): Ligands that bind through two donor atoms. Example: Ethylenediamine (en), Oxalate ion ($C_2O_4^{2-}$).
- Polydentate: Ligands that bind through multiple donor atoms (tri, tetra, hexa). Example: $EDTA^{4-}$ (hexadentate, binds through 2 Nitrogen and 4 Oxygen atoms).
5. What is an ambidentate ligand? Give two examples.
Ambidentate Ligand: A monodentate ligand that contains more than one coordinating atom, but uses only one of them at a time to bind to the central metal ion.
Examples:
- $NO_2^-$ ion: can bind through Nitrogen (nitrito-N) or through Oxygen (nitrito-O).
- $SCN^-$ ion: can bind through Sulfur (thiocyanato-S) or through Nitrogen (thiocyanato-N).
6. What are chelating ligands? Define the chelate effect.
Chelating Ligands: Polydentate or bidentate ligands that use two or more donor atoms simultaneously to bind to the same central metal ion, forming a ring structure. Example: Ethylenediamine (en).
Chelate Effect: Complexes containing chelate rings are exceptionally more stable than complexes containing similar monodentate ligands. This extra thermodynamic stability is called the chelate effect (driven by an increase in entropy).
7. Define Coordination Number (C.N.). Find the C.N. of Cobalt in $[Co(en)_3]^{3+}$.
Coordination Number: It is the total number of coordinate bonds formed by the ligands with the central metal atom or ion.
In $[Co(en)_3]^{3+}$, 'en' (ethylenediamine) is a bidentate ligand. It forms 2 coordinate bonds per molecule.
Since there are 3 'en' molecules: C.N. = $3 \times 2 = \mathbf{6}$.
8. Distinguish between Coordination Sphere and Ionization Sphere.
- Coordination Sphere: It consists of the central metal ion and the ligands directly attached to it. It is enclosed in square brackets $[ \ ]$. It behaves as a single unit and does not ionize in water.
- Ionization Sphere (Counter ions): The ionizable groups written outside the square brackets. They separate into ions when dissolved in water.
Example: In $K_4[Fe(CN)_6]$, $K^+$ are in the ionization sphere, and $[Fe(CN)_6]^{4-}$ is the coordination sphere.
9. Calculate the oxidation number of Iron (Fe) in $[Fe(CN)_6]^{3-}$.
Let the oxidation number of Fe be $x$.
The charge on the cyanide ligand ($CN^-$) is $-1$.
The total charge on the complex ion is $-3$.
$x + 6(-1) = -3$
$x - 6 = -3 \implies x = +3$.
The oxidation number of Iron is +3.
10. Define Effective Atomic Number (EAN). Calculate EAN for Cobalt in $[Co(NH_3)_6]^{3+}$. (Atomic no. of Co = 27).
EAN Rule: Sidgwick proposed that a metal ion continues to accept electron pairs from ligands until the total number of electrons around the metal ion equals the atomic number of the next noble gas. This total number of electrons is the EAN.
$$EAN = Z - X + Y$$
Where $Z$ = Atomic number, $X$ = Oxidation number (electrons lost), $Y$ = Electrons gained from ligands ($2 \times \text{Coordination Number}$).
For $[Co(NH_3)_6]^{3+}$: $Z = 27$, $X = +3$ (since $NH_3$ is neutral), $Y = 6 \times 2 = 12$.
$EAN = 27 - 3 + 12 = \mathbf{36}$ (which is the atomic number of Krypton, a noble gas).
11. What are homoleptic and heteroleptic complexes? Give examples.
- Homoleptic Complex: A complex in which the metal is bound to only one kind of donor group (ligand). Example: $[Co(NH_3)_6]^{3+}$.
- Heteroleptic Complex: A complex in which the metal is bound to more than one kind of donor group. Example: $[Co(NH_3)_4Cl_2]^+$.
12. State briefly the IUPAC rules for naming a coordination complex.
- The cation is named first, followed by the anion.
- Within the complex ion, ligands are named in alphabetical order before the metal ion.
- Prefixes di-, tri-, tetra- are used for simple ligands. Bis-, tris-, tetrakis- are used for complex ligands (like en).
- Anionic ligands end in '-o' (chloro, cyano). Neutral ligands retain their name (except aqua for $H_2O$, ammine for $NH_3$, carbonyl for $CO$).
- The oxidation state of the metal is written in Roman numerals in parentheses.
- If the complex is an anion, the metal name ends in '-ate' (e.g., ferrate, cuprate).
13. Write the IUPAC name for $K_4[Fe(CN)_6]$.
The cation is Potassium. The complex is an anion $[Fe(CN)_6]^{4-}$.
Oxidation state of Fe: $x + 6(-1) = -4 \implies x = +2$.
Ligand is cyano (6 of them = hexacyano). Since it's an anionic complex, Iron becomes Ferrate.
IUPAC Name: Potassium hexacyanoferrate(II)
14. Write the IUPAC name for $[Co(NH_3)_5Cl]Cl_2$.
The complex is a cation $[Co(NH_3)_5Cl]^{2+}$. The counter anion is Chloride.
Ligands: 5 ammine (pentaammine) and 1 chloro (chlorido). Alphabetical order: ammine before chlorido.
Oxidation state of Co: $x + 5(0) + 1(-1) = +2 \implies x = +3$.
IUPAC Name: Pentaamminechloridocobalt(III) chloride
15. Write the IUPAC name for $[Pt(NH_3)_2Cl_2]$.
This is a neutral complex molecule.
Ligands: 2 ammine (diammine) and 2 chloro (dichlorido).
Oxidation state of Pt: $x + 2(0) + 2(-1) = 0 \implies x = +2$.
IUPAC Name: Diamminedichloridoplatinum(II)
16. Write the IUPAC name for $K_3[Al(C_2O_4)_3]$.
Cation is Potassium. The complex is an anion $[Al(C_2O_4)_3]^{3-}$.
Ligand: Oxalate ($C_2O_4^{2-}$), named oxalato. (3 of them = trioxalato).
Oxidation state of Al: $x + 3(-2) = -3 \implies x = +3$.
Since it's an anion, Aluminum becomes Aluminate.
IUPAC Name: Potassium trioxalatoaluminate(III)
17. State the main postulates of Werner's Theory of Coordination Compounds.
- Metals in coordination compounds exhibit two types of valency: Primary valency and Secondary valency.
- Primary Valency: It is ionizable and relates to the oxidation state of the metal. It is satisfied by negative ions.
- Secondary Valency: It is non-ionizable and relates to the coordination number. It is satisfied by neutral molecules or negative ions (ligands).
- The secondary valencies have directed spatial arrangements around the central metal atom, which determine the geometry of the complex.
18. Distinguish between Primary Valency and Secondary Valency.
| Primary Valency | Secondary Valency |
|---|---|
| Corresponds to the Oxidation State. | Corresponds to the Coordination Number. |
| It is ionizable. | It is non-ionizable. |
| Satisfied only by anions. | Satisfied by neutral molecules, anions, or sometimes cations. |
| Non-directional (does not affect geometry). | Directional (determines the spatial geometry). |
19. What is Structural Isomerism? Name its four main types in coordination compounds.
Structural Isomerism: Isomers having the same molecular formula but different structural arrangements of atoms or groups of atoms around the central metal ion.
The four main types are:
- Ionization Isomerism
- Linkage Isomerism
- Coordination Isomerism
- Hydrate (Solvate) Isomerism
20. Explain Ionization Isomerism with a suitable example.
Ionization Isomerism: Occurs when the counter ion in a coordination compound can act as a ligand, and it exchanges places with a ligand inside the coordination sphere. As a result, the compounds yield different ions in solution.
Example:
- $[Co(NH_3)_5Br]SO_4$ (Yields $SO_4^{2-}$ ions in solution, precipitates with $BaCl_2$)
- $[Co(NH_3)_5SO_4]Br$ (Yields $Br^-$ ions in solution, precipitates with $AgNO_3$)
21. Explain Linkage Isomerism with a suitable example.
Linkage Isomerism: Arises when a coordination compound contains an ambidentate ligand (which has two different donor atoms), and the ligand binds to the metal via different donor atoms in different isomers.
Example: Involving the nitrite ion ($NO_2^-$).
- $[Co(NH_3)_5(NO_2)]Cl_2$ : Ligand is bonded through Nitrogen (yellow complex).
- $[Co(NH_3)_5(ONO)]Cl_2$ : Ligand is bonded through Oxygen (red complex).
22. Explain Coordination Isomerism with a suitable example.
Coordination Isomerism: Occurs in compounds where both the cation and the anion are complex ions. Isomerism arises due to the interchange of ligands between the cationic and anionic coordination spheres.
Example:
- $[Co(NH_3)_6][Cr(CN)_6]$ (Cobalt is bonded to ammonia, Chromium to cyanide)
- $[Cr(NH_3)_6][Co(CN)_6]$ (Chromium is bonded to ammonia, Cobalt to cyanide)
23. What is Hydrate (Solvate) Isomerism? Give an example.
Hydrate Isomerism: A special case of ionization isomerism where water molecules are involved. It occurs when water molecules can either act as ligands inside the coordination sphere or exist as water of hydration outside the sphere.
Example: $CrCl_3 \cdot 6H_2O$ exists in three isomeric forms:
- $[Cr(H_2O)_6]Cl_3$ (Violet, 0 water molecules outside)
- $[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$ (Light green, 1 water molecule outside)
- $[Cr(H_2O)_4Cl_2]Cl \cdot 2H_2O$ (Dark green, 2 water molecules outside)
24. What is Stereoisomerism? Name its two types.
Stereoisomerism: Isomers having the same chemical formula and the same chemical bonds but differing in the spatial arrangement of ligands around the central metal ion.
It is classified into two types:
- Geometrical Isomerism (Cis-Trans isomerism)
- Optical Isomerism (Enantiomerism)
25. Explain Geometrical Isomerism in Square Planar complexes with an example.
In square planar complexes of the type $MA_2B_2$ (where M is metal, A and B are monodentate ligands), two arrangements are possible:
- Cis isomer: The two identical ligands occupy adjacent positions (bond angle $90^\circ$).
- Trans isomer: The two identical ligands occupy opposite positions (bond angle $180^\circ$).
Example: $[Pt(NH_3)_2Cl_2]$ exists as Cis-platin (used as an anti-cancer drug) and Trans-platin.
26. Explain fac and mer isomerism in Octahedral complexes.
In octahedral complexes of the type $MA_3B_3$ (e.g., $[Co(NH_3)_3Cl_3]$), geometrical isomerism gives rise to Facial (fac) and Meridional (mer) isomers:
- Facial (fac) isomer: The three identical ligands occupy the corners of one triangular face of the octahedron.
- Meridional (mer) isomer: The three identical ligands occupy positions around the meridian (a plane bisecting the octahedron) of the octahedron.
27. What are optical isomers (enantiomers)? What is the essential condition for a complex to be optically active?
Optical Isomers: Isomers that are non-superimposable mirror images of each other. They rotate the plane of plane-polarized light in opposite directions. The one rotating light to the right is dextro ($d$), and to the left is laevo ($l$).
Essential Condition: The molecule must lack any plane of symmetry or center of symmetry. It must be chiral (asymmetric).
28. Does $[Co(en)_3]^{3+}$ show optical isomerism? Why?
Yes, $[Co(en)_3]^{3+}$ shows optical isomerism. It forms $d$ and $l$ enantiomers.
Because it contains three bidentate ligands (ethylenediamine) arranged octahedrally around the Cobalt ion, the molecule adopts a propeller-like chiral shape. It lacks a plane of symmetry and its mirror image is non-superimposable.
29. State the main postulates of Valence Bond Theory (VBT) for coordination compounds.
- The central metal atom/ion provides a number of empty orbitals equal to its coordination number.
- These empty atomic orbitals (s, p, d) undergo hybridization to form a new set of equivalent hybridized orbitals (e.g., $sp^3, dsp^2, d^2sp^3, sp^3d^2$) which dictate the geometry.
- Ligands have at least one lone pair of electrons. Coordinate bonds are formed by the overlap of empty hybrid orbitals of the metal with filled orbitals of the ligands.
- If the complex contains unpaired electrons, it is paramagnetic; if all are paired, it is diamagnetic.
30. Using VBT, explain the geometry and magnetic property of $[NiCl_4]^{2-}$.
Ni is in +2 oxidation state. Configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$Cl^-$ is a weak field ligand, so it does not force pairing of the unpaired 3d electrons.
To accommodate 4 $Cl^-$ ligands, $Ni^{2+}$ uses one 4s and three 4p empty orbitals. Therefore, it undergoes $sp^3$ hybridization.
Geometry: Tetrahedral.
Since there are 2 unpaired electrons remaining in the 3d subshell, the complex is Paramagnetic.
31. Using VBT, explain the geometry and magnetic property of $[Ni(CN)_4]^{2-}$.
Ni is in +2 oxidation state. Configuration of $Ni^{2+}$ is $[Ar] 3d^8$.
$CN^-$ is a strong field ligand. It forces the two unpaired 3d electrons to pair up against Hund's rule, leaving one 3d orbital empty.
To accommodate 4 $CN^-$ ligands, $Ni^{2+}$ uses the empty 3d, one 4s, and two 4p orbitals. Therefore, it undergoes $dsp^2$ hybridization.
Geometry: Square Planar.
Since all electrons are paired, the complex is Diamagnetic.
32. What are inner orbital and outer orbital complexes?
- Inner Orbital Complex: An octahedral complex that uses the inner $(n-1)d$ orbitals for hybridization ($d^2sp^3$). This usually happens with strong field ligands (like $CN^-, NH_3$) which force electron pairing.
- Outer Orbital Complex: An octahedral complex that uses the outer $nd$ orbitals for hybridization ($sp^3d^2$). This usually happens with weak field ligands (like $F^-, Cl^-, H_2O$) where electron pairing does not occur.
33. Using VBT, explain the hybridization and magnetic property of $[Co(NH_3)_6]^{3+}$.
Co is in +3 state. $Co^{3+}$ is $[Ar] 3d^6$.
$NH_3$ acts as a strong field ligand for $Co^{3+}$, forcing the 6 electrons in the 3d subshell to pair up completely, leaving two 3d orbitals empty.
Hybridization involves two 3d, one 4s, and three 4p orbitals: $d^2sp^3$ (Inner orbital complex).
Geometry is Octahedral. Because all electrons are paired, it is Diamagnetic.
34. Using VBT, explain the hybridization and magnetic property of $[CoF_6]^{3-}$.
Co is in +3 state. $Co^{3+}$ is $[Ar] 3d^6$.
$F^-$ is a weak field ligand and cannot force electron pairing. The 3d electrons remain unpaired (4 unpaired electrons).
Hybridization must use outer 4d orbitals to get 6 empty orbitals: one 4s, three 4p, and two 4d orbitals. Hybridization is $sp^3d^2$ (Outer orbital complex).
Geometry is Octahedral. Due to the presence of 4 unpaired electrons, it is highly Paramagnetic.
35. What are the limitations of Valence Bond Theory (VBT)?
- It does not explain why some complexes are colored.
- It does not give a quantitative interpretation of magnetic data.
- It does not explain why certain ligands form inner orbital complexes (strong field) while others form outer orbital complexes (weak field).
- It does not provide thermodynamic or kinetic stabilities of coordination compounds.
36. State the basic assumptions of Crystal Field Theory (CFT).
- The interaction between the metal ion and the ligands is purely electrostatic (ionic). No covalent bonding is considered.
- Ligands are treated as point charges (if anions like $Cl^-$) or point dipoles (if neutral like $H_2O, NH_3$).
- In the free metal ion, the five d-orbitals are degenerate. When ligands approach, the electrons in the d-orbitals experience repulsion from the ligand electrons, causing the degeneracy to break (Crystal Field Splitting).
37. Explain crystal field splitting in Octahedral complexes.
In an octahedral complex, 6 ligands approach the central metal ion along the x, y, and z axes.
The d-orbitals lying directly on the axes ($d_{x^2-y^2}$ and $d_{z^2}$) experience maximum repulsion and their energy increases. They form a higher energy set called the $e_g$ set.
The d-orbitals lying between the axes ($d_{xy}, d_{yz}, d_{xz}$) experience less repulsion and their energy decreases relative to the average energy. They form a lower energy set called the $t_{2g}$ set.
The energy difference between $e_g$ and $t_{2g}$ sets is the Crystal Field Splitting Energy ($\Delta_o$).
38. What is the value of energy for $t_{2g}$ and $e_g$ orbitals relative to the barycenter in an octahedral field?
Relative to the barycenter (average energy level):
- The energy of the three $t_{2g}$ orbitals is lowered by $-0.4 \Delta_o$ (or $-2/5 \Delta_o$).
- The energy of the two $e_g$ orbitals is raised by $+0.6 \Delta_o$ (or $+3/5 \Delta_o$).
39. What is the Spectrochemical Series?
Spectrochemical Series: It is an experimentally determined arrangement of common ligands in increasing order of their crystal field splitting power ($\Delta$).
Partial series: $I^- < Br^- < Cl^- < F^- < OH^- < H_2O < NH_3 < en < CN^- < CO$
Halogens are weak field ligands (produce small $\Delta$), while Nitrogen and Carbon donor ligands are strong field ligands (produce large $\Delta$).
40. Distinguish between High spin and Low spin complexes.
In an octahedral field, filling the 4th electron ($d^4$):
- High Spin Complex (Weak Field): If $\Delta_o < P$ (pairing energy), the 4th electron prefers to enter the higher $e_g$ orbital rather than pair up in $t_{2g}$. Configuration: $t_{2g}^3 e_g^1$. It has more unpaired electrons.
- Low Spin Complex (Strong Field): If $\Delta_o > P$, the 4th electron prefers to pair up in the lower $t_{2g}$ orbital rather than jump to $e_g$. Configuration: $t_{2g}^4 e_g^0$. It has fewer unpaired electrons.
41. What is Crystal Field Stabilization Energy (CFSE)?
CFSE: It is the net energy decrease (stabilization) resulting from the splitting of the d-orbitals and the placement of electrons in the lower energy orbitals compared to the unsplit average energy level.
For an octahedral complex: $\text{CFSE} = [-0.4 \times (\text{electrons in } t_{2g}) + 0.6 \times (\text{electrons in } e_g)] \Delta_o + (\text{pairing energy if applicable})$.
42. Calculate the CFSE for a $d^5$ configuration in a strong octahedral field.
In a strong field ($\Delta_o > P$), it forms a low spin complex.
All 5 electrons will pair up in the lower $t_{2g}$ orbitals. Configuration: $t_{2g}^5 e_g^0$.
$\text{CFSE} = (-0.4 \times 5) \Delta_o + (0.6 \times 0) \Delta_o$
$\text{CFSE} = \mathbf{-2.0 \Delta_o}$ (ignoring pairing energy terms for simplicity).
43. Explain crystal field splitting in tetrahedral complexes.
In a tetrahedral complex, 4 ligands approach between the axes.
Therefore, the d-orbitals lying between the axes ($d_{xy}, d_{yz}, d_{xz}$) experience more repulsion, and their energy is raised (forming the $t_2$ set). The orbitals on the axes ($d_{x^2-y^2}, d_{z^2}$) experience less repulsion, and their energy is lowered (forming the $e$ set).
This is exactly opposite to the splitting pattern of octahedral complexes.
44. What is the relation between $\Delta_o$ and $\Delta_t$? Why are tetrahedral complexes rarely low spin?
Relation: $$\Delta_t = \frac{4}{9} \Delta_o$$
The crystal field splitting in a tetrahedral field ($\Delta_t$) is always much smaller than in an octahedral field ($\Delta_o$). Because $\Delta_t$ is almost always smaller than the pairing energy ($P$), electrons prefer to jump to the higher orbitals rather than pair up. Therefore, tetrahedral complexes are predominantly high spin.
45. How does CFT explain the color of coordination compounds?
According to CFT, color arises due to d-d transitions. When white light falls on a complex, an electron from a lower energy d-orbital (e.g., $t_{2g}$) absorbs a specific wavelength of light to jump to a higher energy d-orbital (e.g., $e_g$). The energy gap $\Delta_o$ corresponds to the energy of the visible light absorbed.
The color observed is the complementary color of the light absorbed.
46. Anhydrous $CuSO_4$ is white, but $CuSO_4 \cdot 5H_2O$ is blue. Explain using CFT.
In anhydrous $CuSO_4$, there are no ligands (water) to cause the crystal field splitting of the d-orbitals of $Cu^{2+}$. Without splitting, d-d transitions cannot occur, so it appears white.
In hydrated $CuSO_4 \cdot 5H_2O$, water molecules act as ligands, causing d-orbital splitting. The $Cu^{2+}$ ($d^9$) ion absorbs red-orange light for d-d transition, and transmits the complementary color, which is blue.
47. What is meant by the stability of a coordination compound in solution? On what factors does it depend?
Stability refers to the degree of association between the metal ion and ligands in an equilibrium state. A highly stable complex does not dissociate easily.
It depends on:
- Charge on metal ion: Higher charge = greater stability.
- Size of metal ion: Smaller size = greater stability (higher charge density).
- Nature of ligand: Strong field ligands and chelating ligands form highly stable complexes.
48. What is the overall stability constant ($\beta_n$)? How is it related to stepwise stability constants ($K$)?
The formation of a complex $ML_n$ occurs in stepwise additions of ligands, each with a stepwise stability constant ($K_1, K_2, \dots K_n$).
The overall stability constant ($\beta_n$) is the equilibrium constant for the overall formation reaction $M + nL \rightleftharpoons ML_n$.
Relation: The overall stability constant is the product of all stepwise stability constants.
$$\beta_n = K_1 \times K_2 \times K_3 \times \dots \times K_n$$
49. Mention an application of coordination compounds in medicine.
- Cisplatin ($cis-[Pt(NH_3)_2Cl_2]$) is widely used as a highly effective drug in the treatment of cancer (chemotherapy). It binds to DNA and inhibits cell division.
- EDTA is used in the treatment of heavy metal poisoning (like Lead poisoning) because it forms highly stable, water-soluble chelates with the metal, which are then excreted from the body.
50. State two biological importance of coordination compounds.
Many essential biological molecules are coordination complexes:
- Chlorophyll: It is the green pigment in plants responsible for photosynthesis. It is a coordination complex of Magnesium ($Mg^{2+}$) with a porphyrin ring.
- Hemoglobin: It is the red pigment in blood responsible for oxygen transport. It is a coordination complex of Iron ($Fe^{2+}$) with a heme group.
- Vitamin B12: Essential for preventing pernicious anemia, it is a complex of Cobalt ($Co^{3+}$).
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