Coordination Compounds
Complete Revision Guide for CBSE Board Exams
1. Introduction & Werner's Theory
Coordination compounds are highly complex molecules where a central metal atom or ion is surrounded by a number of oppositely charged ions or neutral molecules. These are uniquely different from simple double salts (like Mohr's salt or Alum), which lose their identity in an aqueous solution. Coordination compounds retain their identity even in dissolved states.
Werner's Theory of Coordination Compounds
Alfred Werner proposed this theory to explain the structure of coordination compounds. Key postulates:
- In coordination compounds, metals show two types of linkages (valencies): Primary valency and Secondary valency.
- Primary valency is normally ionizable and is satisfied by negative ions. It corresponds to the Oxidation State of the central metal atom.
- Secondary valency is non-ionizable. These are satisfied by neutral molecules or negative ions. It corresponds to the Coordination Number of the metal.
- The ions/groups bound by the secondary linkages have characteristic spatial arrangements corresponding to different coordination numbers (determining the geometry like octahedral, tetrahedral, or square planar).
1.2 Crucial Terminology: Ligands
The ions or molecules bound to the central atom/ion in the coordination entity are called Ligands. They act as Lewis bases (electron pair donors).
- Unidentate: A ligand binding through a single donor atom. (e.g., $Cl^-$, $H_2O$, $NH_3$).
- Didentate: A ligand binding through two donor atoms. (e.g., ethane-1,2-diamine ($en$), oxalate ion ($C_2O_4^{2-}$)).
- Polydentate: Binding through several donor atoms. E.g., EDTA$^{4-}$ (ethylenediaminetetraacetate) is an important hexadentate ligand binding through two nitrogen and four oxygen atoms.
CBSE Favorites: Ambidentate & Chelate Ligands
Ambidentate Ligand: A unidentate ligand that contains more than one coordinating atom and can coordinate through either of them.
Examples: $NO_2^-$ (can link through N or O) and $SCN^-$ (can link through S or N).
Chelate Ligand: When a di- or polydentate ligand uses its two or more donor atoms to bind a single metal ion, it forms a ring structure. This is called a chelate ring, and the ligand is a chelating ligand. Chelating ligands form much more stable complexes than similar unidentate ligands (Chelate Effect).
2. IUPAC Nomenclature & Key Formulas
A. Rules for IUPAC Naming
- Order of Naming: The cation is named first in both positively and negatively charged coordination entities.
- Naming the Complex Sphere: Ligands are named in an alphabetical order before the name of the central metal atom/ion.
- Ligand Names: Anionic ligands end in '-o' (e.g., chlorido, cyanido, oxalato). Neutral ligands have specific names (e.g., $H_2O$ is aqua, $NH_3$ is ammine, $CO$ is carbonyl).
- Numerical Prefixes: Use di, tri, tetra for simple ligands. If the ligand name already includes a numerical prefix (like ethylenediamine), use bis, tris, tetrakis, and put the ligand name in parentheses.
- Metal Name: If the complex is a cation or neutral, the metal name is the same as the element. If the complex is an anion, the name of the metal ends with the suffix -ate (e.g., Cobaltate, Ferrate, Cuprate, Argentate).
- Oxidation State: The oxidation state of the metal is indicated by a Roman numeral in parentheses right after the metal name.
Example: $[Co(NH_3)_5Cl]Cl_2$ is named Pentaamminechloridocobalt(III) chloride.
B. Mathematical Relations in CFT
- Spin-Only Magnetic Moment ($\mu$):
$$\mu = \sqrt{n(n+2)} \text{ B.M.}$$(Where $n$ = number of unpaired electrons).
- Relation between Octahedral ($\Delta_o$) and Tetrahedral ($\Delta_t$) splitting:
$$\Delta_t = \frac{4}{9} \Delta_o$$(Tetrahedral splitting is always much smaller than octahedral splitting, hence tetrahedral complexes are rarely low-spin.)
3. Theories of Coordination Compounds
3.1 Isomerism in Coordination Compounds
Compounds that have the same chemical formula but different structural arrangements are called isomers. They are divided into two main categories: Structural and Stereoisomerism.
I. Structural Isomerism
- Linkage Isomerism: Occurs when an ambidentate ligand is present. For example, $[Co(NH_3)_5(NO_2)]Cl_2$ (yellow, bonded through N) and $[Co(NH_3)_5(ONO)]Cl_2$ (red, bonded through O).
- Coordination Isomerism: Arises from the interchange of ligands between cationic and anionic entities of different metal ions present in a complex. E.g., $[Co(NH_3)_6][Cr(CN)_6]$ and $[Cr(NH_3)_6][Co(CN)_6]$.
- Ionisation Isomerism: Occurs when the counter ion in a complex salt is itself a potential ligand and can displace a ligand from the coordination sphere. E.g., $[Co(NH_3)_5Br]SO_4$ (red-violet, gives $SO_4^{2-}$ precipitate with $BaCl_2$) and $[Co(NH_3)_5SO_4]Br$ (red, gives $Br^-$ precipitate with $AgNO_3$).
- Solvate (Hydrate) Isomerism: A special case of ionisation isomerism where water is involved as a solvent. E.g., $[Cr(H_2O)_6]Cl_3$ (violet) vs $[Cr(H_2O)_5Cl]Cl_2\cdot H_2O$ (grey-green).
II. Stereoisomerism
- Geometrical Isomerism: Common in heteroleptic complexes. In a square planar $MA_2B_2$ complex, if similar ligands are adjacent, it's cis; if opposite, it's trans. In octahedral $MA_3B_3$ complexes, if three similar ligands occupy the corners of one triangular face of the octahedron, it is facial (fac). If they are around the meridian, it is meridional (mer).
- Optical Isomerism: Optical isomers are non-superimposable mirror images called enantiomers (dextro and laevo). They are common in octahedral complexes involving didentate ligands. E.g., the $[Co(en)_3]^{3+}$ ion shows optical isomerism. Note: Trans isomers of $[M(en)_2X_2]$ type are optically inactive due to a plane of symmetry, but Cis isomers are optically active.
3.2 Valence Bond Theory (VBT)
VBT explains the structure and magnetic properties of coordination compounds based on the hybridization of the metal's empty atomic orbitals (s, p, d).
| Coordination Number | Hybridization | Geometry |
|---|---|---|
| 4 | $sp^3$ | Tetrahedral |
| 4 | $dsp^2$ | Square Planar |
| 6 | $sp^3d^2$ (Outer orbital) | Octahedral |
| 6 | $d^2sp^3$ (Inner orbital) | Octahedral |
Applying VBT (Crucial for Exams):
1. Determine the oxidation state and electronic configuration of the central metal ion.
2. Identify the nature of the ligand. Strong field ligands (like $CO$, $CN^-$, $NH_3$, $en$) force the pairing of unpaired d-electrons against Hund's rule, creating empty inner $(n-1)d$ orbitals for hybridization. These form Inner Orbital / Low Spin complexes.
3. Weak field ligands (like $F^-$, $Cl^-$, $H_2O$) generally cannot force pairing. The metal uses outer $nd$ orbitals for hybridization. These form Outer Orbital / High Spin complexes.
3.3 Crystal Field Theory (CFT)
CFT considers the metal-ligand bond to be purely ionic arising from electrostatic interactions. Ligands are treated as point charges. In a free metal ion, all five d-orbitals have the same energy (degenerate). When ligands approach, this degeneracy is broken due to repulsion between ligand electrons and metal d-electrons.
Crystal Field Splitting in Octahedral Complexes
In an octahedral field, six ligands approach along the x, y, and z axes. The d-orbitals lying along these axes ($d_{x^2-y^2}$ and $d_{z^2}$) experience maximum repulsion and their energy increases. They form the $e_g$ set. The orbitals lying between the axes ($d_{xy}$, $d_{yz}$, $d_{zx}$) experience less repulsion, and their energy drops. They form the $t_{2g}$ set.
The energy difference between $t_{2g}$ and $e_g$ is the Crystal Field Splitting Energy ($\Delta_o$).
- If $\Delta_o < P$ (Pairing energy): The ligand is weak field. The 4th electron will jump to the $e_g$ orbital. Configuration: $t_{2g}^3 e_g^1$. (High spin complex).
- If $\Delta_o > P$: The ligand is strong field. The 4th electron will pair up in the $t_{2g}$ orbital rather than jumping the large energy gap. Configuration: $t_{2g}^4 e_g^0$. (Low spin complex).
Spectrochemical Series: A series of ligands arranged in order of increasing crystal field splitting capability:
$I^- < Br^- < SCN^- < Cl^- < S^{2-} < F^- < OH^- < C_2O_4^{2-} < H_2O < NCS^- < edta^{4-} < NH_3 < en < CN^- < CO$
3.4 Color in Coordination Compounds
CFT beautifully explains color. In a complex, when light falls on it, a d-electron can jump from the lower energy $t_{2g}$ level to the higher energy $e_g$ level. This requires absorption of a specific wavelength of light (energy = $\Delta_o$). This is called the d-d transition. The complex transmits the complementary color. E.g., $[Ti(H_2O)_6]^{3+}$ absorbs yellow-green light and appears violet.
3.5 Bonding in Metal Carbonyls (Synergic Effect)
Homoleptic metal carbonyls (e.g., $Ni(CO)_4$, $Fe(CO)_5$) contain only carbonyl ligands. The Metal-Carbon bond in metal carbonyls possesses both $\sigma$ and $\pi$ character.
1. The M-C $\sigma$ bond is formed by the donation of a lone pair from carbon into a vacant metal orbital.
2. The M-C $\pi$ bond is formed by the donation of electrons from a filled metal d-orbital into the vacant antibonding $\pi^*$ orbital of Carbon monoxide.
This back-bonding creates a Synergic Effect, which greatly strengthens the bond between CO and the metal.
4. Most Important CBSE Board Questions
Strictly aligned with the latest CBSE sample papers and marking scheme.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. The IUPAC name for $[Pt(NH_3)_2Cl(NO_2)]$ is: 1 Mark
- Diamminechloridonitrito-N-platinum(II)
- Diamminechloridonitrito-N-platinate(II)
- Diamminechloridonitrito-O-platinum(II)
- Chloridodiamminenitrito-N-platinum(II)
Explanation: Alphabetical order: ammine comes before chlorido, which comes before nitrito. Oxidation state of Pt is $x + 0 + (-1) + (-1) = 0 \implies x = +2$. Since the complex is neutral, we use 'platinum', not 'platinate'. The ligand is $NO_2^-$, so it binds through N (nitrito-N).
Q2. Which of the following species is NOT expected to be a ligand? 1 Mark
- $NO$
- $NH_4^+$
- $NH_2CH_2CH_2NH_2$
- $CO$
Explanation: A ligand must be a Lewis base, meaning it must have at least one unshared (lone) pair of electrons to donate to the central metal. The Nitrogen atom in the ammonium ion ($NH_4^+$) has already shared all its valence electrons to form four bonds with hydrogen. It has no lone pair left to donate.
Assertion-Reasoning
Q3. Assertion (A): $[Co(en)_3]^{3+}$ is more stable than $[Co(NH_3)_6]^{3+}$.
Reason (R): Ethane-1,2-diamine (en) is a chelating ligand that forms rings with the metal ion. 1 Mark
Explanation: This describes the "Chelate Effect." When a multidentate ligand forms cyclic structures (chelate rings) around the central metal ion, the resulting complex is thermodynamically much more stable than an analogous complex formed by unidentate ligands (like $NH_3$).
Section B: 2 Mark Questions (Very Short Answer)
Q4. Write the formulas for the following coordination compounds:
(i) Potassium tetracyanidonickelate(II)
(ii) Tetraammineaquachloridocobalt(III) chloride 2 Marks
Logic: Nickel is in +2 state. 4 Cyanides provide -4 charge. Overall coordination sphere has -2 charge. Thus, it needs two $K^+$ ions outside.
(ii) $[Co(NH_3)_4(H_2O)Cl]Cl_2$
Logic: Cobalt is +3. $NH_3$ and $H_2O$ are neutral. Inside Cl is -1. Total charge of sphere = $+3 + 0 + 0 + (-1) = +2$. Therefore, two $Cl^-$ ions are needed outside to balance.
Q5. Explain the term 'linkage isomerism' with the help of an example. 2 Marks
Example: In the complex $[Co(NH_3)_5(NO_2)]Cl_2$, the nitrite ligand is bound through the Nitrogen atom (forming a yellow complex). In its linkage isomer, $[Co(NH_3)_5(ONO)]Cl_2$, the ligand is bound through the Oxygen atom (forming a red complex).
Section C: 3 Mark Questions (Short Answer)
Q6. Using Valence Bond Theory, predict the hybridization, geometry, and magnetic behaviour of the $[CoF_6]^{3-}$ complex ion. (Given atomic number of Co = 27) 3 Marks
$Co$ atomic number is 27. Electronic config: $[Ar] 3d^7 4s^2$.
In $[CoF_6]^{3-}$, let oxidation state of Co be $x$. $x + 6(-1) = -3 \implies x = +3$.
$Co^{3+}$ configuration is $[Ar] 3d^6$.
2. Nature of Ligand:
$F^-$ is a weak field ligand. It does not possess enough energy to force the pairing of the 3d electrons against Hund's rule.
3. Hybridization & Geometry:
Since pairing doesn't occur, the inner 3d orbitals are occupied. To accommodate six $F^-$ ligands, $Co^{3+}$ must use its outer 4d orbitals. The hybridization is $s + p_x + p_y + p_z + d_{z^2} + d_{x^2-y^2} \rightarrow \mathbf{sp^3d^2}$ (outer orbital complex).
The geometry for coordination number 6 is Octahedral.
4. Magnetic Behaviour:
The $3d^6$ configuration (without pairing) has 4 unpaired electrons ($|\uparrow\downarrow|\uparrow|\uparrow|\uparrow|\uparrow|$).
Because there are unpaired electrons, the complex is highly Paramagnetic.
Q7. Based on Crystal Field Theory (CFT), what will be the electronic configuration of $d^4$ ions if:
(i) $\Delta_o < P$
(ii) $\Delta_o > P$
Also state which of these configurations will result in a high spin complex. 3 Marks
(i) If $\Delta_o < P$ (Weak field ligand):
The energy required to jump to the $e_g$ level is less than the energy required to pair up in the $t_{2g}$ level. The 4th electron enters the $e_g$ orbital.
Configuration: $\mathbf{t_{2g}^3 e_g^1}$.
(ii) If $\Delta_o > P$ (Strong field ligand):
The energy gap is too large to jump. It is energetically more favorable for the 4th electron to pair up in the $t_{2g}$ orbital.
Configuration: $\mathbf{t_{2g}^4 e_g^0}$.
High Spin Complex: The configuration $\mathbf{t_{2g}^3 e_g^1}$ has 4 unpaired electrons (maximum spin), so it results in a High Spin complex. (The other gives a low spin complex with only 2 unpaired electrons).
Section D: 4 Mark Case-Based Questions
Q8. Read the passage given below and answer the following questions:
(i) Name the specific electronic transition responsible for color in transition metal complexes. (1 Mark)
(ii) Why is anhydrous $CuSO_4$ white, while $CuSO_4 \cdot 5H_2O$ is blue? (1 Mark)
(iii) Explain why zinc complexes like $[Zn(H_2O)_6]^{2+}$ are colorless, regardless of the ligand present. (2 Marks)
(ii) In anhydrous $CuSO_4$, there are no water molecules to act as ligands. Without ligands, there is no crystal field splitting of the d-orbitals (they remain degenerate). Hence, d-d transitions cannot occur, making it white. In $CuSO_4 \cdot 5H_2O$, water acts as a ligand, causing crystal field splitting, which allows d-d transitions resulting in the blue color.
(iii) Zinc has an atomic number of 30. The $Zn^{2+}$ ion has a perfectly completely filled d-orbital configuration ($3d^{10}$). Because all the d-orbitals ($t_{2g}$ and $e_g$ alike) are completely full, there is no empty space for an electron to jump into. Without the possibility of a d-d transition, it cannot absorb visible light, rendering all its complexes colorless.
Section E: 5 Mark Long Answer Questions
Q9. (a) Discuss the nature of bonding in metal carbonyls with the help of a suitable diagram/explanation. What is this special bonding effect called? (3 Marks)
(b) Write the hybridization and magnetic character of the complex $[Fe(CN)_6]^{4-}$. [Atomic number of Fe = 26] (2 Marks)
The metal-carbon bond in metal carbonyls (e.g., $Ni(CO)_4$) possesses both s and p character. The bonding occurs in two simultaneous steps:
1. Formation of M-C $\sigma$ bond: This bond is formed by the donation of a lone pair of electrons from the carbonyl carbon into a vacant d-orbital of the metal.
2. Formation of M-C $\pi$ bond (Back-bonding): This bond is formed by the donation of a pair of electrons from a filled d-orbital of the metal into the vacant anti-bonding $\pi^*$ orbital of the carbon monoxide molecule.
This two-way electron transfer (ligand to metal $\sigma$, metal to ligand $\pi$) creates a self-strengthening interaction called the Synergic Effect. This effect makes the bond between CO and the metal exceptionally strong.
(b) Hybridization and Magnetic Character of $[Fe(CN)_6]^{4-}$:
1. Metal State: Iron (Fe) is Z=26 ($[Ar] 3d^6 4s^2$). Let O.S. be $x$. $x + 6(-1) = -4 \implies x = +2$. $Fe^{2+}$ is $3d^6$.
2. Ligand Nature: $CN^-$ is a strong field ligand. It causes forced pairing of all 6 electrons in the 3d orbital against Hund's rule. This results in 3 pairs of electrons and leaves two inner 3d orbitals completely empty.
3. Hybridization: Since two inner 3d orbitals are available, it undergoes $\mathbf{d^2sp^3}$ hybridization forming an inner-orbital, octahedral complex.
4. Magnetic Character: Because all electrons are paired up due to the strong ligand, there are zero unpaired electrons ($n=0$). Therefore, the complex is entirely Diamagnetic.
Mastering 'Coordination Compounds'
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Nomenclature Pitfall: The most common mistake in naming is forgetting the '-ate' suffix when the complex sphere is negatively charged (e.g., $K_4[Fe(CN)_6]$ is Potassium hexacyanidoferrate(II), not iron). Also, write the ligand names strictly in alphabetical order.
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VBT vs CFT: If a question asks to predict hybridization, geometry, and magnetic moment, use VBT. If it asks to explain color, write electronic configurations ($t_{2g}, e_g$), or asks about pairing energy, you must use CFT.
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Learn the Extremes of the Spectrochemical Series: You don't need to memorize the whole series, but you MUST know the very weak ones ($I^-, Br^-, Cl^-, F^-, H_2O$) and the very strong ones ($NH_3, en, CN^-, CO$). This decides whether pairing happens or not.
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Isomerism Guarantee: Structural isomerism (especially Linkage and Ionisation) is almost guaranteed to appear as a 1 or 2-mark question. Be ready to write definitions and provide standard examples.
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