Chapter 5: Coordination Compounds

Dear Class 12 Student! Transition metals have a unique superpower: the ability to form complex structures by accepting electron pairs from other molecules. These are Coordination Compounds. They are the reason our blood is red (Hemoglobin) and plants are green (Chlorophyll). This chapter is highly scoring. Focus intensely on IUPAC Naming, Isomerism, and Crystal Field Theory (CFT) to guarantee top marks in Boards and JEE. Let's complexify!

1. Introduction and Werner's Theory

When two or more stable compounds are mixed in stoichiometric amounts and crystallized, they form addition compounds. These are of two types:

Double Salts vs. Coordination Compounds 1. Double Salts: These compounds dissociate completely into simple ions when dissolved in water. They lose their identity in aqueous solution.
Examples: Mohr’s salt [$\text{FeSO}_4 \cdot (\text{NH}_4)_2\text{SO}_4 \cdot 6\text{H}_2\text{O}$], Potash alum [$\text{K}_2\text{SO}_4 \cdot \text{Al}_2(\text{SO}_4)_3 \cdot 24\text{H}_2\text{O}$], Carnallite [$\text{KCl} \cdot \text{MgCl}_2 \cdot 6\text{H}_2\text{O}$].

2. Coordination Compounds (Complexes): These do not dissociate completely into simple ions. The complex ion remains perfectly intact in the solution. They retain their identity in both solid and solution states.
Example: Potassium ferrocyanide $\text{K}_4[\text{Fe}(\text{CN})_6]$ dissociates into $4\text{K}^+$ and $[\text{Fe}(\text{CN})_6]^{4-}$. It does not give $\text{Fe}^{2+}$ or $\text{CN}^-$ ions.

Werner’s Theory of Coordination Compounds (1893)

Alfred Werner, the father of coordination chemistry, proposed that transition metals possess two types of valencies (linkages):

Primary Valency: It corresponds to the Oxidation State of the central metal ion. It is ionizable and is always satisfied by negative ions (anions). It is non-directional.
Secondary Valency: It corresponds to the Coordination Number (CN). It is non-ionizable and is satisfied by neutral molecules or negative ions. It is highly directional in space, determining the definite geometry (shape) of the complex.

Practice Problem 1 Question: When 1 mole of $\text{CoCl}_3 \cdot 6\text{NH}_3$ is treated with excess silver nitrate ($\text{AgNO}_3$) solution, 3 moles of $\text{AgCl}$ precipitate. However, when 1 mole of $\text{CoCl}_3 \cdot 5\text{NH}_3$ is treated, only 2 moles of $\text{AgCl}$ precipitate. Formulate the Werner complexes.

Solution:
Precipitated $\text{AgCl}$ comes only from ionizable $\text{Cl}^-$ ions (outside the square brackets).
1. For $\text{CoCl}_3 \cdot 6\text{NH}_3$: 3 moles of $\text{AgCl}$ means all 3 $\text{Cl}^-$ are outside the coordination sphere. The complex is $[\text{Co}(\text{NH}_3)_6]\text{Cl}_3$.
2. For $\text{CoCl}_3 \cdot 5\text{NH}_3$: 2 moles of $\text{AgCl}$ means 2 $\text{Cl}^-$ are outside, and 1 $\text{Cl}^-$ is inside satisfying the secondary valency. The complex is $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$.

2. Basic Terminology

Coordination Entity: A central metal atom or ion bonded to a fixed number of ions or molecules enclosed in square brackets. e.g., $[\text{CoCl}_3(\text{NH}_3)_3]$.
Central Metal Atom/Ion: The atom/ion to which a fixed number of ligands are bound in a definite geometrical arrangement. It acts as a Lewis Acid (electron pair acceptor).
Ligands: The ions or neutral molecules bound to the central atom. They act as Lewis Bases (electron pair donors).

Types of Ligands based on Denticity

Denticity is the number of coordinating (ligating) atoms present in a single ligand.

Unidentate (Monodentate): Ligands binding through only one donor atom. e.g., $\text{H}_2\text{O}$ (O is donor), $\text{NH}_3$ (N is donor), $\text{Cl}^-$, $\text{CN}^-$.
Didentate (Bidentate): Ligands binding through two donor atoms. e.g., ethane-1,2-diamine ('en') [$\text{H}_2\text{N-CH}_2\text{-CH}_2\text{-NH}_2$], oxalate ('ox') [$\text{C}_2\text{O}_4^{2-}$].
Polydentate: Ligands with several donor atoms. A famous hexadentate ligand is $\text{EDTA}^{4-}$ (Ethylenediaminetetraacetate), which can bind through two nitrogen and four oxygen atoms.

Chelate Effect & Ambidentate Ligands Chelating Ligands: When a di- or polydentate ligand uses its two or more donor atoms to bind to a single metal ion, it forms a ring structure. Such complexes are exceptionally stable. This extra thermodynamic stability is called the Chelate Effect.

Ambidentate Ligands (Important): Unidentate ligands which contain more than one coordinating atom and can ligate through two different atoms.
Examples:
- Nitrite ion: Can bind through Nitrogen ($-\text{NO}_2$, nitrito-N) or through Oxygen ($-\text{ONO}$, nitrito-O).
- Thiocyanate ion: Can bind through Sulfur ($-\text{SCN}$, thiocyanato) or through Nitrogen ($-\text{NCS}$, isothiocyanato).

Coordination Number (CN): The total number of ligand donor atoms directly bonded to the central metal ion. (e.g., in $[\text{Pt}(\text{en})_2]^{2+}$, CN is 4, not 2, because 'en' is didentate).
Oxidation Number: The charge the central metal would carry if all the ligands were removed along with the electron pairs that were shared with the central atom.

Practice Problem 2 Question: Calculate the oxidation number and the coordination number of the central metal in $\text{K}_3[\text{Fe}(\text{C}_2\text{O}_4)_3]$.

Solution:
1. Coordination Number: The ligand is oxalate ($\text{C}_2\text{O}_4^{2-}$), which is a didentate ligand. Since there are 3 oxalate ligands, the total number of donor atoms $= 3 \times 2 = \mathbf{6}$. Thus, CN = 6.
2. Oxidation Number: Let the oxidation number of $\text{Fe}$ be $x$.
Potassium ($\text{K}$) is $+1$. Oxalate ($\text{C}_2\text{O}_4^{2-}$) is $-2$.
Total charge of the complex is zero: $3(+1) + x + 3(-2) = 0$
$3 + x - 6 = 0 \implies x - 3 = 0 \implies x = \mathbf{+3}$.

3. IUPAC Nomenclature (Guaranteed Board Question)

Rules for Naming Coordination Compounds 1. Order of naming ions: The cation is named first, followed by the anion, just like simple salts (e.g., Sodium chloride).
2. Naming the coordination sphere: The ligands are named in strictly alphabetical order before the name of the central metal atom/ion.
3. Names of ligands:
- Anionic ligands end in '-o' (e.g., $\text{Cl}^-$ is chlorido/chloro, $\text{CN}^-$ is cyanido/cyano, $\text{C}_2\text{O}_4^{2-}$ is oxalato).
- Neutral ligands keep their normal names with a few crucial exceptions: $\text{H}_2\text{O}$ is aqua, $\text{NH}_3$ is ammine (note the double 'm'), $\text{CO}$ is carbonyl, $\text{NO}$ is nitrosyl.
4. Numerical prefixes:
- Use di, tri, tetra, penta, hexa to indicate the number of simple ligands.
- If the ligand name already includes a numerical prefix (like ethylenediamine), use bis, tris, tetrakis, and put the ligand name in parentheses.
5. Naming the central metal:
- If the complex is a cation or neutral, the metal name remains unchanged (e.g., Cobalt, Platinum).
- If the complex is an anion (has a negative charge), the metal name must end in '-ate' (e.g., Cobaltate, Platinate, Ferrate, Cuprate, Argentate).
6. Oxidation state: Written immediately after the metal name in Roman numerals enclosed in parentheses (e.g., (II), (III), (0)).

Practice Problem 3 (Formula to Name) Question: Write the IUPAC names for the following complexes:
(i) $[\text{Co}(\text{NH}_3)_5\text{Cl}]\text{Cl}_2$
(ii) $\text{K}_3[\text{Fe}(\text{CN})_6]$
(iii) $[\text{Pt}(\text{NH}_3)_2\text{Cl}(\text{NO}_2)]$

Solution:
(i) Cation is the complex. Ligands: ammine (A), chlorido (C). 5 ammines = pentaammine. Oxidation state of Co: $x + 5(0) + (-1) = +2$ (from outside Cl) $\implies x = +3$.
Name: Pentaamminechloridocobalt(III) chloride.

(ii) Cation is Potassium. Anion is the complex $[\text{Fe}(\text{CN})_6]^{3-}$. Since complex is an anion, metal is Ferrate. Oxidation state: $3(+1) + x + 6(-1) = 0 \implies x = +3$.
Name: Potassium hexacyanoferrate(III).

(iii) Neutral complex. Ligands: ammine (A), chlorido (C), nitrito-N (N). Oxidation state of Pt: $x + 2(0) + (-1) + (-1) = 0 \implies x = +2$.
Name: Diamminechloridonitrito-N-platinum(II).

Practice Problem 4 (Name to Formula) Question: Write the formulas for the following coordination compounds:
(i) Tetracarbonylnickel(0)
(ii) Potassium tetracyanidocuprate(II)

Solution:
(i) Central metal is Ni. Oxidation state is 0. Ligands are 4 carbonyls ($\text{CO}$).
Formula: $[\text{Ni}(\text{CO})_4]$.

(ii) Cation is Potassium ($\text{K}^+$). Complex anion has Cu(II) and 4 cyanides ($\text{CN}^-$).
Charge on complex sphere $= +2 + 4(-1) = -2$. To balance this, we need two $\text{K}^+$ ions.
Formula: $\text{K}_2[\text{Cu}(\text{CN})_4]$.

4. Isomerism in Coordination Compounds (Crucial for JEE & Boards)

Isomers are two or more compounds that have the same chemical formula but a different arrangement of atoms. They are broadly divided into Structural and Stereoisomerism.

A. Structural Isomerism

Ionization Isomerism: Arises when the counter ion in a complex salt is itself a potential ligand and can displace a ligand which can then become the counter ion. They give different ions in solution.
Example: $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ (Red-violet, gives $\text{SO}_4^{2-}$ precipitate with $\text{BaCl}_2$) vs. $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$ (Red, gives $\text{AgBr}$ precipitate with $\text{AgNO}_3$).
Linkage Isomerism: Arises exclusively in complexes containing ambidentate ligands. The ligand can bind through two different atoms.
Example: $[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]\text{Cl}_2$ (bonded through N, yellow) vs. $[\text{Co}(\text{NH}_3)_5(\text{ONO})]\text{Cl}_2$ (bonded through O, red).
Coordination Isomerism: Arises when both cation and anion are complex ions. There is an interchange of ligands between the cationic and anionic entities.
Example: $[\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6]$ vs. $[\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6]$.
Solvate/Hydrate Isomerism: Similar to ionization isomerism, but differs by whether a solvent molecule (like water) is directly bonded to the metal ion or merely present as free crystal lattice water.
Example: $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ (Violet) vs. $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}$ (Grey-green).

B. Stereoisomerism

1. Geometrical Isomerism (cis-trans): Arises due to different possible geometric arrangements of the ligands.

Coordination Number 4 (Square Planar): Form $Ma_2b_2$ and $Ma_2bc$. If similar ligands are adjacent ($90^\circ$), it is cis. If they are opposite ($180^\circ$), it is trans.
Note: Tetrahedral complexes NEVER show geometrical isomerism because all positions are adjacent to each other.
Coordination Number 6 (Octahedral): Form $Ma_4b_2$ (cis/trans). For the form $Ma_3b_3$, we get two special isomers: Facial (fac) where three similar ligands occupy the corners of one triangular face, and Meridional (mer) where they occupy positions around the meridian (equator) of the octahedron.

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Geometrical isomers of an octahedral complex $[Co(NH_3)_4Cl_2]^+$. Left side: 'cis' isomer. Show Cobalt in center of an octahedron. Two Cl ligands are placed on adjacent corners (e.g., top and one equatorial spot, 90 degrees apart), with four NH3 filling the rest. Right side: 'trans' isomer. Two Cl ligands are placed on exactly opposite polar ends (top and bottom, 180 degrees apart), with four NH3 making a square plane around the equator. Clean educational vector graphic."

2. Optical Isomerism (d and l forms): Arises when the complex and its mirror image are non-superimposable (Chiral). They rotate plane-polarized light in opposite directions.

Highly common in octahedral complexes containing didentate ligands.
Examples: $[\text{Co}(\text{en})_3]^{3+}$ and the cis-isomer of $[\text{Co}(\text{en})_2\text{Cl}_2]^+$.
Crucial Note: The trans-isomer of $Ma_2b_2$ or $M(en)_2b_2$ is optically INACTIVE because it possesses a plane of symmetry.

Practice Problem 5 Question: Indicate the types of isomerism exhibited by the following complexes:
(i) $[\text{Co}(\text{en})_3]\text{Cl}_3$
(ii) $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$

Solution:
(i) $[\text{Co}(\text{en})_3]\text{Cl}_3$: It contains 3 didentate ligands in an octahedral geometry. It does not have a plane of symmetry. Hence, it exhibits Optical Isomerism (d and l forms). It does not show geometrical isomerism.
(ii) $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$: It is a square planar complex of the type $Ma_2b_2$. It exhibits Geometrical Isomerism (cis and trans forms). The cis-form is the famous cancer drug Cisplatin.

5. Valence Bond Theory (VBT)

Developed by Linus Pauling. Basic postulates:
1. The central metal ion provides a number of empty orbitals equal to its coordination number to accept electron pairs from ligands.
2. These empty orbitals hybridize to form a new set of equivalent orbitals of definite geometry.
3. The ligand orbitals overlap with the hybridized metal orbitals to form coordinate covalent bonds.

Coordination Number and Hybridization CN = 4:
- $sp^3$ hybridization $\implies$ Tetrahedral geometry (e.g., $[\text{NiCl}_4]^{2-}$).
- $dsp^2$ hybridization $\implies$ Square Planar geometry (e.g., $[\text{Ni}(\text{CN})_4]^{2-}$).

CN = 6:
- $sp^3d^2$ hybridization (uses outer 4d orbitals) $\implies$ Octahedral. Known as Outer Orbital or High Spin complex (e.g., $[\text{CoF}_6]^{3-}$).
- $d^2sp^3$ hybridization (uses inner 3d orbitals) $\implies$ Octahedral. Known as Inner Orbital or Low Spin complex (e.g., $[\text{Co}(\text{NH}_3)_6]^{3+}$).

Magnetic Properties: If the final hybridized complex contains any unpaired electrons, it is Paramagnetic. If all electrons are completely paired, it is Diamagnetic. Strong ligands (like $\text{CN}^-$, $\text{CO}$, $\text{NH}_3$) tend to force the pairing of electrons against Hund's rule, creating low spin, diamagnetic complexes. Weak ligands (like Halides, $\text{H}_2\text{O}$) do not cause pairing, resulting in high spin, paramagnetic complexes.

Practice Problem 6 Question: Using Valence Bond Theory, explain the geometry and magnetic behavior of $[\text{CoF}_6]^{3-}$ and $[\text{Co}(\text{NH}_3)_6]^{3+}$. (Atomic number of Co = 27).

Solution:
In both complexes, Cobalt is in the $+3$ oxidation state. $Co^{3+}$ has a $3d^6$ configuration.

1. $[\text{CoF}_6]^{3-}$: Fluoride ($\text{F}^-$) is a weak field ligand. It cannot force the pairing of the $3d^6$ electrons.
Configuration: $\uparrow\downarrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow$.
Since the inner 3d orbitals are occupied, it must use outer orbitals: one 4s, three 4p, and two 4d orbitals to hybridize as $sp^3d^2$.
Geometry: Octahedral (Outer orbital complex).
Magnetism: It has 4 unpaired electrons, so it is highly Paramagnetic (High spin).

2. $[\text{Co}(\text{NH}_3)_6]^{3+}$: Ammonia ($\text{NH}_3$) acts as a strong field ligand for $Co^{3+}$. It forces the 6 electrons to pair up completely in the inner 3d orbitals.
Configuration: $\uparrow\downarrow \quad \uparrow\downarrow \quad \uparrow\downarrow \quad (\text{empty}) \quad (\text{empty})$.
It uses the two empty 3d orbitals, one 4s, and three 4p orbitals to hybridize as $d^2sp^3$.
Geometry: Octahedral (Inner orbital complex).
Magnetism: All electrons are paired, so it is purely Diamagnetic (Low spin).

6. Crystal Field Theory (CFT) - High Board Priority

VBT had flaws: it couldn't explain the color of complexes or properly distinguish between strong and weak ligands. CFT treats the metal-ligand bond as purely electrostatic (ionic). The ligands are treated as point charges. When ligands approach the central metal ion, the electric field causes the 5 degenerate d-orbitals to split into different energy levels.

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Crystal Field Splitting diagram for Octahedral Complexes. Left side: 5 degenerate d-orbitals at ground state. Middle: 5 degenerate d-orbitals at a higher average energy spherical field. Right side: The 5 orbitals split into two sets. The lower set has 3 orbitals labeled 't2g' (dxy, dyz, dzx). The upper set has 2 orbitals labeled 'eg' (dx2-y2, dz2). The gap between them is labeled Delta_o (Crystal Field Splitting Energy). Center of gravity line (Barycenter) is drawn between them. White background."

Splitting in Octahedral Complexes ($\Delta_o$)

In an octahedral field, six ligands approach along the x, y, and z axes. The d-orbitals lying directly on the axes ($d_{x^2-y^2}, d_{z^2}$) experience more repulsion from the ligands and their energy rises. 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 gap between them is the Crystal Field Splitting Energy ($\Delta_o$).

Spectrochemical Series & Strong/Weak Ligands Ligands are arranged in increasing order of their field strength (ability to split d-orbitals):
$\text{I}^- < \text{Br}^- < \text{Cl}^- < \text{F}^- < \text{OH}^- < \text{H}_2\text{O} < \text{NH}_3 < \text{en} < \text{CN}^- < \text{CO}$

1. Weak Field Ligands (e.g., Halides, $\text{H}_2\text{O}$):
Produce a small splitting ($\Delta_o < P$, where P is the pairing energy). For a $d^4$ ion, the 4th electron will jump up to the $e_g$ level rather than pair up in $t_{2g}$. Configuration: $t_{2g}^3 e_g^1$. Forms High Spin complexes.

2. Strong Field Ligands (e.g., $\text{CN}^-, \text{CO}, \text{NH}_3$):
Produce a large splitting ($\Delta_o > P$). The gap is too huge to cross. For a $d^4$ ion, the 4th electron prefers to pair up in the lower $t_{2g}$ level. Configuration: $t_{2g}^4 e_g^0$. Forms Low Spin complexes.

Splitting in Tetrahedral Complexes ($\Delta_t$)

In a tetrahedral field, four ligands approach between the axes. Therefore, the splitting order is exactly reversed: the $e$ set is lower in energy, and the $t_2$ set is higher.
The splitting energy is much smaller: $\Delta_t = \frac{4}{9} \Delta_o$.
Because $\Delta_t$ is always smaller than the pairing energy $P$, orbital pairing is rare. Tetrahedral complexes are almost always high spin.

Practice Problem 7 Question: Based on Crystal Field Theory, write the electronic configuration for a $d^5$ ion if (i) $\Delta_o < P$, and (ii) $\Delta_o > P$.

Solution:
(i) $\Delta_o < P$ (Weak Field Ligand): The splitting gap is small. Electrons will fill the lower $t_{2g}$ and then jump to the higher $e_g$ before pairing up. The 5 electrons fill singly across all 5 orbitals.
Configuration: $t_{2g}^3 e_g^2$.

(ii) $\Delta_o > P$ (Strong Field Ligand): The splitting gap is very large. Electrons prefer to pair up in the lower $t_{2g}$ level rather than cross the gap. All 5 electrons will pack into the $t_{2g}$ level.
Configuration: $t_{2g}^5 e_g^0$.

Color in Coordination Compounds

According to CFT, color is attributed to d-d transitions. When white light falls on the complex, an electron from the lower energy d-orbital (e.g., $t_{2g}$) absorbs a photon of a specific wavelength (energy exactly equal to $\Delta_o$) and gets excited to the higher energy d-orbital (e.g., $e_g$). The remaining unabsorbed light is transmitted, giving the complex its complementary color.

Practice Problem 8 Question: Explain why anhydrous $\text{CuSO}_4$ is white, but hydrated $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$ is blue in color.

Solution:
In hydrated $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}$, water molecules act as ligands surrounding the $Cu^{2+}$ ion, causing crystal field splitting of the d-orbitals. The unpaired electron in the $3d^9$ configuration absorbs red light to undergo a d-d transition, transmitting the complementary blue color.
In anhydrous $\text{CuSO}_4$, there are no water ligands present. Without ligands, there is no crystal field splitting. Since all d-orbitals remain degenerate (same energy), d-d transitions are impossible. Hence, no light is absorbed, and it appears white.

7. Bonding in Metal Carbonyls

Complexes containing only carbon monoxide ($\text{CO}$) ligands are called homoleptic carbonyls (e.g., $\text{Ni}(\text{CO})_4$ is tetrahedral, $\text{Fe}(\text{CO})_5$ is trigonal bipyramidal, $\text{Cr}(\text{CO})_6$ is octahedral). The metal is generally in the zero oxidation state.

[AI Image Placeholder: Landscape (16:9) - Background: #FFFFFF]
AI Prompt: "Synergic Bonding in Metal Carbonyls. Show a Metal atom (M) and a Carbon atom (C) of Carbon Monoxide. Draw a forward arrow labeled 'sigma bond' showing electron donation from a filled Carbon orbital to an empty Metal d-orbital. Draw a backward curved arrow labeled 'pi-back bond' showing electron density flowing from a filled Metal d-orbital back into an empty antibonding pi* orbital of Carbon. Label the overall cycle 'Synergic Effect'. White background."

Synergic Bonding Effect (Crucial Concept) The Metal-Carbon bond in carbonyls possesses both $\sigma$ and $\pi$ character, creating an immensely strong, self-reinforcing bond called Synergic Bonding.

1. $\sigma$-bond Formation: The Carbon atom donates its lone pair of electrons into a vacant orbital of the metal to form a standard metal-carbon $\sigma$ bond.
2. $\pi$-back bond Formation: The metal now has excess negative charge. To stabilize itself, the metal donates electrons from its filled, electron-rich d-orbitals back into the empty anti-bonding $\pi^*$ orbitals of the Carbon monoxide ligand. This forms a $M \rightarrow C$ $\pi$ bond.

This two-way transfer of electrons reinforces each other, making the bond exceptionally strong.

Practice Problem 9 Question: How does synergic bonding affect the bond length of the C-O bond in metal carbonyls?

Solution:
In synergic bonding, the metal donates electron density into the anti-bonding $\pi^*$ orbitals of the CO ligand. According to molecular orbital theory, adding electrons to anti-bonding orbitals decreases the overall bond order of the molecule. Since the bond order of C-O decreases, the C-O bond weakens and the C-O bond length increases compared to free carbon monoxide gas.

8. Stability and Applications of Coordination Compounds

The stability of a complex in solution refers to the degree of association between the metal and ligands at equilibrium. It is expressed by the Stability Constant ($\beta_n$). A higher value of $\beta$ implies a more thermodynamically stable complex.

Importance and Applications

Metallurgy: Extraction of noble metals. Gold and Silver are extracted by the MacArthur-Forrest cyanide process, where they form water-soluble complexes $[Ag(CN)_2]^-$ and $[Au(CN)_2]^-$. Nickel is purified via the Mond process by forming volatile $[Ni(CO)_4]$.
Biological Systems: Life depends entirely on coordination compounds.
- Chlorophyll: The green pigment in plants responsible for photosynthesis is a complex of Magnesium.
- Hemoglobin: The red blood pigment that carries oxygen is a complex of Iron.
- Vitamin B12: Essential to prevent pernicious anemia, it is a complex of Cobalt (cyanocobalamin).
Analytical Chemistry: EDTA (ethylenediaminetetraacetic acid) is used in complexometric titrations to estimate the hardness of water (caused by $Ca^{2+}$ and $Mg^{2+}$ ions).

Practice Problem 10 Question: Name the coordination compound used in the treatment of cancer, and state its geometry. How is EDTA used in medicine?

Solution:
1. The coordination compound used highly effectively in the chemotherapy treatment of cancer (specifically testicular and ovarian tumors) is Cisplatin. Its IUPAC name is cis-diamminedichloridoplatinum(II), formula $\text{cis-}[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$. Its geometry is Square Planar.
2. In medicine, EDTA (in the form of calcium EDTA complex) is used as an antidote for Lead Poisoning. Because EDTA forms an extremely stable chelate complex with lead, it wraps around the toxic lead ions in the bloodstream, allowing them to be safely excreted through urine.