Coordination Compounds in Pharmaceutical Chemistry

1

Naming Coordination Compounds

Name the given coordination compounds. Determine the ligands dentation, the coordination number of the complexing agent and hybridization type for them. Draw the structures: a. K₃[Co(C₂O₄)₃] b. [Pt(C₂O₄)₂(NH₃)₂] c. Na₂[Mn(OAc)₂(OH)₂]

Part A: K₃[Co(C₂O₄)₃]

Name:

Potassium tris(oxalato)cobaltate(III)

Ligand dentation:

Oxalate (C₂O₄²⁻) is a bidentate ligand (binds at two points)

Coordination number:

3 ligands × 2 binding sites = 6

Hybridization type:

For coordination number 6, hybridization is d²sp³

Part B: [Pt(C₂O₄)₂(NH₃)₂]

Name:

Diamminedioxalatoplatinum(IV)

Ligand dentation:

Oxalate (bidentate), Ammonia (monodentate)

Coordination number:

(2 × 2) + (2 × 1) = 6

Hybridization type:

d²sp³ for octahedral geometry

Part C: Na₂[Mn(OAc)₂(OH)₂]

Name:

Sodium diacetatodihydroxomanganate(II)

Ligand dentation:

Acetate (OAc⁻) is bidentate, Hydroxide (OH⁻) is monodentate

Coordination number:

(2 × 2) + (2 × 1) = 6

Hybridization type:

d²sp³ for octahedral geometry

Coordination compounds are widely used in pharmaceuticals as contrast agents (e.g., gadolinium complexes for MRI), anticancer drugs (e.g., cisplatin), and in diagnostics. Understanding their nomenclature and structure is essential for pharmaceutical applications.

a) Potassium tris(oxalato)cobaltate(III)
b) Diamminedioxalatoplatinum(IV)
c) Sodium diacetatodihydroxomanganate(II)

2

Formulas from Names

With use of the given names write the formulas of the coordination compounds. Determine type of the given compounds: a. diiododiaquacalcium b. sodium difluoroglycinatomercurate(II) c. hexaamminenickel(III) bromide

Part A: diiododiaquacalcium

Identify components:

Central metal: Calcium (Ca)

Ligands: 2 I⁻ (iodo), 2 H₂O (aqua)

Determine charge:

2 I⁻ = -2, 2 H₂O = 0, Ca = +2 → Complex is neutral

Write formula:

\([\text{CaI}_2(\text{H}_2\text{O})_2]\)

Type:

Neutral complex

Part B: sodium difluoroglycinatomercurate(II)

Identify components:

Central metal: Mercury (Hg²⁺)

Ligands: 2 F⁻ (fluoro), glycinate (bidentate ligand)

Counterion: Sodium (Na⁺)

Determine charge:

Hg²⁺ = +2, 2 F⁻ = -2, glycinate = -1 → Complex ion charge = -1

Write formula:

\(\text{Na}[\text{HgF}_2(\text{gly})]\)

Type:

Anionic complex with Na⁺ counterion

Part C: hexaamminenickel(III) bromide

Identify components:

Central metal: Nickel (Ni³⁺)

Ligands: 6 NH₃ (ammine)

Counterion: Bromide (Br⁻)

Determine charge:

Ni³⁺ = +3, 6 NH₃ = 0 → Complex ion charge = +3

Write formula:

\([\text{Ni}(\text{NH}_3)_6]\text{Br}_3\)

Type:

Cationic complex with Br⁻ counterions

Many coordination compounds have pharmaceutical applications. For example, mercury complexes were historically used as antiseptics, and nickel complexes are studied for potential anticancer properties. Understanding their formulas is crucial for pharmaceutical research and development.

a) [CaI₂(H₂O)₂] (neutral)
b) Na[HgF₂(gly)] (anionic)
c) [Ni(NH₃)₆]Br₃ (cationic)

3

Ionization Isomers

For the coordination compound which contains Co³⁺, 2Br⁻, NO₃⁻ and 4NH₃ write formula of all possible ionization isomers. What is the coordination number of the complexing agent in the resulted isomers?

Step-by-Step Solution:

Understand ionization isomers:

Ionization isomers differ in which ions are inside the coordination sphere versus outside as counterions.

Determine possible isomers:

With Co³⁺, 2Br⁻, NO₃⁻, and 4NH₃, the coordination number is 6 (4 NH₃ + 2 other ligands)

Possible isomers:

[Co(NH₃)₄Br₂]NO₃
[Co(NH₃)₄Br(NO₃)]Br
[Co(NH₃)₄(NO₃)₂]Br

Verify charge balance:

1. [Co(NH₃)₄Br₂]⁺NO₃⁻: Co³⁺ + 4(0) + 2(-1) = +1; NO₃⁻ = -1 → Balanced

2. [Co(NH₃)₄Br(NO₃)]⁺Br⁻: Co³⁺ + 4(0) + (-1) + (-1) = +1; Br⁻ = -1 → Balanced

3. [Co(NH₃)₄(NO₃)₂]⁺Br⁻: Co³⁺ + 4(0) + 2(-1) = +1; Br⁻ = -1 → Balanced

Coordination number:

In all isomers, the coordination number is 6 (4 NH₃ + 2 other ligands)

Ionization isomers can have different pharmaceutical properties despite having the same chemical formula. For example, one isomer might be effective as an anticancer drug while another is not, due to differences in how they interact with biological targets.

Isomers: [Co(NH₃)₄Br₂]NO₃, [Co(NH₃)₄Br(NO₃)]Br, [Co(NH₃)₄(NO₃)₂]Br
Coordination number: 6 for all isomers

4

Geometric Isomers of Diglycinatozinc

For the coordination compound diglycinatozinc draw the structures of cis- and trans- geometric isomers.

Step-by-Step Solution:

Identify the compound:

Diglycinatozinc: Zn(Gly)₂ where Gly = glycinate (H₂N-CH₂-COO⁻)

Glycinate is a bidentate ligand (binds through N and O)

Determine geometry:

Zinc typically forms tetrahedral complexes with coordination number 4

However, with bidentate ligands, it can form square planar complexes

For diglycinatozinc, it's typically square planar

Cis isomer:

Both glycinate ligands are adjacent to each other (90° angle between them)

\(\text{Structure: Zn at center, with two glycinate ligands arranged with their N and O atoms in adjacent positions}\)

Trans isomer:

Both glycinate ligands are opposite each other (180° angle between them)

\(\text{Structure: Zn at center, with two glycinate ligands arranged with their N and O atoms in opposite positions}\)

Pharmaceutical relevance:

Zinc complexes are used in pharmaceuticals for wound healing and as dietary supplements

Geometric isomers can have different biological activities due to their spatial arrangement

The cis and trans isomers of coordination compounds can have dramatically different pharmaceutical properties. For example, cisplatin (cis isomer) is an effective anticancer drug, while transplatin (trans isomer) is not. This highlights the importance of stereochemistry in drug design.

Cis isomer: Glycinate ligands adjacent
Trans isomer: Glycinate ligands opposite
Both have square planar geometry

5

Naming Coordination Compounds

Name the given coordination compounds. Propose the reagents for the following coordination compounds preparation. Write the molecular and ionic equations of the reactions. a. K₃[Fe(SCN)₆] b. [Cu(NH₃)₄]SO₄ c. K[Al(OH)₄] d. Fe₄[Fe(CN)₆]₃ e. Cu₂[Fe(SCN)₆]

Part A: K₃[Fe(SCN)₆]

Name:

Potassium hexathiocyanatoferrate(III)

Preparation reagents:

FeCl₃ + 6KSCN → K₃[Fe(SCN)₆] + 3KCl

Reaction equations:

\(\text{FeCl}_3 + 6\text{KSCN} \rightarrow \text{K}_3[\text{Fe(SCN)}_6] + 3\text{KCl}\)

\(\text{Fe}^{3+} + 6\text{SCN}^- \rightarrow [\text{Fe(SCN)}_6]^{3-}\)

Part B: [Cu(NH₃)₄]SO₄

Name:

Tetraamminecopper(II) sulfate

Preparation reagents:

CuSO₄ + 4NH₃ → [Cu(NH₃)₄]SO₄

Reaction equations:

\(\text{CuSO}_4 + 4\text{NH}_3 \rightarrow [\text{Cu(NH}_3)_4]\text{SO}_4\)

\(\text{Cu}^{2+} + 4\text{NH}_3 \rightarrow [\text{Cu(NH}_3)_4]^{2+}\)

Part C: K[Al(OH)₄]

Name:

Potassium tetrahydroxoaluminate

Preparation reagents:

AlCl₃ + 4KOH → K[Al(OH)₄] + 3KCl

Reaction equations:

\(\text{AlCl}_3 + 4\text{KOH} \rightarrow \text{K}[\text{Al(OH)}_4] + 3\text{KCl}\)

\(\text{Al}^{3+} + 4\text{OH}^- \rightarrow [\text{Al(OH)}_4]^-\)

Part D: Fe₄[Fe(CN)₆]₃

Name:

Iron(III) hexacyanoferrate(II) or Prussian blue

Preparation reagents:

4FeCl₃ + 3K₄[Fe(CN)₆] → Fe₄[Fe(CN)₆]₃ + 12KCl

Reaction equations:

\(4\text{FeCl}_3 + 3\text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 12\text{KCl}\)

\(4\text{Fe}^{3+} + 3[\text{Fe(CN)}_6]^{4-} \rightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3\)

Part E: Cu₂[Fe(SCN)₆]

Name:

Copper(II) hexathiocyanatoferrate(III)

Preparation reagents:

2CuCl₂ + Fe(SCN)₃ → Cu₂[Fe(SCN)₆]

Reaction equations:

\(2\text{CuCl}_2 + \text{Fe(SCN)}_3 \rightarrow \text{Cu}_2[\text{Fe(SCN)}_6]\)

\(2\text{Cu}^{2+} + [\text{Fe(SCN)}_6]^{3-} \rightarrow \text{Cu}_2[\text{Fe(SCN)}_6]\)

Many of these compounds have pharmaceutical applications. For example, Prussian blue (Fe₄[Fe(CN)₆]₃) is used to treat thallium and cesium poisoning. Understanding their preparation is crucial for pharmaceutical manufacturing.

a) Potassium hexathiocyanatoferrate(III)
b) Tetraamminecopper(II) sulfate
c) Potassium tetrahydroxoaluminate
d) Iron(III) hexacyanoferrate(II)
e) Copper(II) hexathiocyanatoferrate(III)

6

Color and Wavelength Relationship

For the following coordination compounds fill the table with data of wavelengths of the absorbed light and solution colors:

Step-by-Step Solution:

Understand color relationships:

A compound appears as the complementary color to the color it absorbs

Use the provided color chart to determine complementary colors

Fill the table based on absorption and complementary colors:

№	Formula	Wavelength, nm	Color
1	[Co(en)₂Cl₂]⁺	535	Red
2	[Co(NH₃)₆]³⁺		yellow
3	[Cr(H₂O)₆]³⁺	713	Green
4	[Co(H₂O)₆]²⁺		red
5	[Fe(H₂O)₆]²⁺	993	Green
6	[Cr(H₂O)₆]³⁺		violet

Explain color determination:

1. 535 nm absorption → green-yellow → complementary color = red

2. Yellow color → absorbs blue-violet (400-435 nm)

3. 713 nm absorption → red → complementary color = green

4. Red color → absorbs green (480-490 nm)

5. 993 nm absorption → infrared (beyond visible) → appears green

6. Violet color → absorbs yellow (560-580 nm)

Pharmaceutical relevance:

The color of coordination compounds is important in pharmaceuticals for identification and quality control. Many metal-based drugs have characteristic colors that help in their identification and purity assessment. The color also relates to the electronic structure, which affects the compound's reactivity and biological activity.

1. Red
2. 400-435 nm
3. Green
4. 480-490 nm
5. Green
6. 560-580 nm

7

Concentration in [Ni(en)₃]Cl₃ Solution

For the 0.05M solution of [Ni(en)₃]Cl₃ determine the concentration of Cl⁻, Ni³⁺ and ethylenediammine in the solution, if K_instab([Ni(en)₃]³⁺) = 5.6×10⁻¹⁹.

Step-by-Step Solution:

Write the dissociation equation:

\([\text{Ni(en)}_3]\text{Cl}_3 \rightarrow [\text{Ni(en)}_3]^{3+} + 3\text{Cl}^-\)

Determine initial concentrations:

0.05 M [Ni(en)₃]Cl₃ → 0.05 M [Ni(en)₃]³⁺ and 0.15 M Cl⁻

Write the instability constant expression:

\(K_{\text{instab}} = \frac{[\text{Ni}^{3+}][\text{en}]^3}{[[\text{Ni(en)}_3]^{3+}]} = 5.6 \times 10^{-19}\)

Assume complete dissociation of complex:

Let x = [Ni³⁺] = concentration of dissociated complex

Then [en] = 3x, and [[Ni(en)₃]³⁺] = 0.05 - x ≈ 0.05 (since K is very small)

Solve for x:

\(5.6 \times 10^{-19} = \frac{x(3x)^3}{0.05} = \frac{27x^4}{0.05}\)

\(x^4 = \frac{5.6 \times 10^{-19} \times 0.05}{27} = 1.04 \times 10^{-21}\)

\(x = (1.04 \times 10^{-21})^{1/4} = 1.01 \times 10^{-5}\)

Calculate final concentrations:

[Ni³⁺] = x = 1.01 × 10⁻⁵ M

[en] = 3x = 3.03 × 10⁻⁵ M

[Cl⁻] = 0.15 M (unchanged, as it's a counterion)

The extremely small dissociation constant shows that [Ni(en)₃]³⁺ is a very stable complex. In pharmaceutical applications, such stable complexes are often used to deliver metal ions in a controlled manner, preventing unwanted reactions with biological molecules.

[Cl⁻] = 0.15 M
[Ni³⁺] = 1.01 × 10⁻⁵ M
[en] = 3.03 × 10⁻⁵ M

8

Reaction Feasibility in Aqueous Solution

Estimate, is it possible to occur the following reactions in the aqueous solution? a. K[Ag(CN)₂] + 2KCl → K[AgCl₂] + 2KCN b. Na₂[HgBr₄] + 2NaI → Na₂[HgI₄] + 2NaBr

Part A: K[Ag(CN)₂] + 2KCl → K[AgCl₂] + 2KCN

Compare stability constants:

From the provided table:

\(K_{\text{stab}}[\text{Ag(CN)}_2^-] = 8.0 \times 10^8\)

\(K_{\text{stab}}[\text{AgCl}_2^-] = 1.8 \times 10^5\)

Calculate equilibrium constant:

\(K_{\text{eq}} = \frac{K_{\text{stab}}[\text{AgCl}_2^-]}{K_{\text{stab}}[\text{Ag(CN)}_2^-]} = \frac{1.8 \times 10^5}{8.0 \times 10^8} = 2.25 \times 10^{-4}\)

Interpret the result:

Since K_eq << 1, the reaction does not proceed to the right

The more stable complex [Ag(CN)₂]⁻ will not convert to the less stable [AgCl₂]⁻

Part B: Na₂[HgBr₄] + 2NaI → Na₂[HgI₄] + 2NaBr

Compare stability constants:

From the provided table:

\(K_{\text{stab}}[\text{HgBr}_4^{2-}] = 2.0 \times 10^{22}\)

\(K_{\text{stab}}[\text{HgI}_4^{2-}] = 5.0 \times 10^{31}\)

Calculate equilibrium constant:

\(K_{\text{eq}} = \frac{K_{\text{stab}}[\text{HgI}_4^{2-}]}{K_{\text{stab}}[\text{HgBr}_4^{2-}]} = \frac{5.0 \times 10^{31}}{2.0 \times 10^{22}} = 2.5 \times 10^9\)

Interpret the result:

Since K_eq >> 1, the reaction proceeds to the right

The more stable complex [HgI₄]²⁻ forms from the less stable [HgBr₄]²⁻

Understanding reaction feasibility is crucial in pharmaceutical chemistry. For example, mercury compounds are used in some pharmaceuticals, and knowing which complexes are stable helps predict their behavior in the body and potential interactions with other compounds.

a) Reaction not possible (K_eq = 2.25 × 10⁻⁴)
b) Reaction possible (K_eq = 2.5 × 10⁹)

9

Reaction Schemes

Write down the reactions corresponded to the following schemes. a. HgBr₂ → K₂[HgBr₄] → K₂[HgI₄] b. Fe(NO₃)₃ → K₃[Fe(CN)₆] → Cu₃[Fe(CN)₆]₂

Part A: Mercury Reaction Scheme

First reaction: HgBr₂ → K₂[HgBr₄]

\(\text{HgBr}_2 + 2\text{KBr} \rightarrow \text{K}_2[\text{HgBr}_4]\)

This forms the tetrabromomercurate(II) complex

Second reaction: K₂[HgBr₄] → K₂[HgI₄]

\(\text{K}_2[\text{HgBr}_4] + 4\text{KI} \rightarrow \text{K}_2[\text{HgI}_4] + 4\text{KBr}\)

Iodide replaces bromide due to higher stability of [HgI₄]²⁻ complex

Part B: Iron Reaction Scheme

First reaction: Fe(NO₃)₃ → K₃[Fe(CN)₆]

\(\text{Fe(NO}_3)_3 + 3\text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{K}_3[\text{Fe(CN)}_6] + 3\text{KNO}_3 + \text{other products}\)

Actually, this is more complex - typically Fe³⁺ reacts with [Fe(CN)₆]⁴⁻ to form Prussian blue

But to form soluble K₃[Fe(CN)₆], we need: 4Fe(NO₃)₃ + 3K₄[Fe(CN)₆] → Fe₄[Fe(CN)₆]₃ + 12KNO₃

Second reaction: K₃[Fe(CN)₆] → Cu₃[Fe(CN)₆]₂

\(2\text{K}_3[\text{Fe(CN)}_6] + 3\text{CuSO}_4 \rightarrow \text{Cu}_3[\text{Fe(CN)}_6]_2 + 3\text{K}_2\text{SO}_4\)

This forms copper hexacyanoferrate(III)

Pharmaceutical relevance:

These reaction schemes demonstrate how coordination compounds can be transformed from one form to another, which is important in pharmaceutical manufacturing. For example, the iron-cyanide compounds are used in treating heavy metal poisoning, and understanding their chemistry helps in developing effective treatments.

a) HgBr₂ + 2KBr → K₂[HgBr₄]
K₂[HgBr₄] + 4KI → K₂[HgI₄] + 4KBr
b) 4Fe(NO₃)₃ + 3K₄[Fe(CN)₆] → Fe₄[Fe(CN)₆]₃ + 12KNO₃
2K₃[Fe(CN)₆] + 3CuSO₄ → Cu₃[Fe(CN)₆]₂ + 3K₂SO₄