Chemistry 9701/41 — October/November 2016

Background Concept

In coordination chemistry, ligands are ions or molecules that donate a lone pair of electrons to a central metal ion to form a coordinate (dative) bond. When a complex reacts with a reagent that provides a different ligand, the original ligands can be replaced by the new ones. This process is called ligand exchange, ligand substitution, or ligand replacement.

Understanding the Question

The equilibrium shows $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ reacting with $\text{Cl}^-$ to form $[\text{CuCl}_4]^{2-}$ and water. The water ligands are being replaced by chloride ligands.

Approach

Identify the process: water molecules (ligands) are being replaced by chloride ions (new ligands). This is a classic ligand exchange reaction.

Step-by-Step Reasoning

• The reactant complex has water ligands.
• The product complex has chloride ligands.
• Water is displaced by chloride. This is a ligand exchange (or substitution) reaction.

Key Takeaways

• Replacing one set of ligands with another in a complex ion is called ligand exchange.

Common Mistakes

• Calling it a 'precipitation' or 'redox' reaction. No change in oxidation state occurs here (Cu remains +2).

Things to Be Careful About

• Any of the terms 'ligand exchange', 'substitution', 'replacement', or 'displacement' are acceptable.

Background Concept

The colour of transition metal complexes arises from d-d electron transitions. When ligands approach a central metal ion, they split the degenerate $3\text{d}$ orbitals into two energy levels (e.g., $t_{2g}$ and $e_g$ in an octahedral field). Electrons can absorb visible light to jump from the lower to the higher energy $\text{d}$ orbitals. The colour observed is the complementary colour to the absorbed light.

The magnitude of the splitting ($\Delta E$) depends on the nature of the ligands. Chloride ions ($\text{Cl}^-$) are weaker-field ligands than water molecules ($\text{H}_2\text{O}$), meaning they cause a smaller splitting of the $\text{d}$ orbitals. This changes the wavelength of light absorbed, and thus the colour of the complex.

Understanding the Question

State the colours of the hexaaquacopper(II) ion and the tetrachlorocuprate(II) ion.

Approach

Recall the standard colours of these common copper(II) complexes.

Step-by-Step Reasoning

• $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ is the standard aqueous copper(II) ion, which is blue.
• $[\text{CuCl}_4]^{2-}$ is formed when concentrated hydrochloric acid or chloride ions are added to a blue copper solution. It is yellow (often described as yellow/green or green/yellow in practice due to mixtures with the blue ion).

Key Takeaways

• Aqueous copper(II) is blue. Adding concentrated chloride turns it yellow/green.

Common Mistakes

• Confusing the colours of different transition metal complexes (e.g., saying cobalt is yellow).

Things to Be Careful About

• The mark scheme accepts 'yellow', 'yellow/green', or 'green/yellow' for $[\text{CuCl}_4]^{2-}$. Just 'green' alone might not score if the exact shade isn't specified, but 'yellow' or 'yellow/green' is standard.

Background Concept

The shape of a complex ion is determined by its coordination number (the number of ligand donor atoms bonded to the central metal ion).

• Coordination number 4: tetrahedral (e.g., $[\text{CuCl}_4]^{2-}$, $[\text{Zn}(\text{NH}_3)_4]^{2+}$)
• Coordination number 6: octahedral (e.g., $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$, $[\text{Co}(\text{H}_2\text{O})_6]^{2+}$)

Understanding the Question

State the shape of the $[\text{CuCl}_4]^{2-}$ ion.

Approach

Count the number of chloride ligands attached to copper. There are 4 $\text{Cl}^-$ ligands, so the coordination number is 4. Complexes with coordination number 4 are tetrahedral.

Step-by-Step Reasoning

• The formula is $[\text{CuCl}_4]^{2-}$.
• There are 4 chloride ligands.
• Coordination number = 4.
• Shape = tetrahedral.

Key Takeaways

• CN 4 = tetrahedral, CN 6 = octahedral.

Common Mistakes

• Saying 'square planar' for $[\text{CuCl}_4]^{2-}$. While some $\text{d}^8$ complexes like $[\text{PtCl}_4]^{2-}$ or $[\text{Ni}(\text{CN})_4]^{2-}$ are square planar, $[\text{CuCl}_4]^{2-}$ is tetrahedral.

Things to Be Careful About

• Ensure you spell 'tetrahedral' correctly.

Background Concept

The stability constant ($K_{\text{stab}}$) is the equilibrium constant for the formation of a complex ion from its central metal ion and ligands. It measures the stability of the complex; a larger $K_{\text{stab}}$ indicates a more stable complex.

For a general reaction:

$$\text{M} + n\text{L} \rightleftharpoons [\text{ML}_n]$$ $$K_{\text{stab}} = \frac{[[\text{ML}_n]]}{[\text{M}][\text{L}]^n}$$

Crucial rule: Pure liquids (like $\text{H}_2\text{O}$ in aqueous solution) and pure solids are not included in the equilibrium expression. Their concentrations are effectively constant and are incorporated into the value of $K_{\text{stab}}$.

Understanding the Question

Write the $K_{\text{stab}}$ expression for:

$$[\text{Cu}(\text{H}_2\text{O})_6]^{2+} + 4\text{Cl}^- \rightleftharpoons [\text{CuCl}_4]^{2-} + 6\text{H}_2\text{O}$$

Approach

1. Products over reactants.
2. Omit $\text{H}_2\text{O}$ (it is a pure liquid/solvent).
3. Raise concentrations to the power of their stoichiometric coefficients.

Step-by-Step Reasoning

• Numerator: $[[\text{CuCl}_4]^{2-}]$ (coefficient 1)
• Denominator: $[[\text{Cu}(\text{H}_2\text{O})_6]^{2+}]$ (coefficient 1) and $[\text{Cl}^-]^4$ (coefficient 4)
• Water is omitted.

$$K_{\text{stab}} = \frac{[\text{CuCl}_4^{2-}]}{[\text{Cu}(\text{H}_2\text{O})_6^{2+}][\text{Cl}^-]^4}$$

Key Takeaways

• Always omit water in aqueous equilibrium expressions.
• Square brackets denote concentration in mol dm$^{-3}$.

Common Mistakes

• Including $[\text{H}_2\text{O}]$ in the expression.
• Forgetting the power of 4 on $[\text{Cl}^-]$.
• Writing the expression for the reverse reaction (formation constant vs dissociation constant). $K_{\text{stab}}$ is for formation of the complex, so the complex is in the numerator.

Things to Be Careful About

• Ensure charges are correct inside the brackets: $\text{CuCl}_4^{2-}$, $\text{Cu}(\text{H}_2\text{O})_6^{2+}$, $\text{Cl}^-$.

Background Concept

Ligands are classified by the number of donor atoms they use to bond to the central metal ion (their denticity).

• Monodentate: One donor atom (e.g., $\text{H}_2\text{O}$, $\text{NH}_3$, $\text{Cl}^-$).
• Bidentate: Two donor atoms (e.g., ethane-1,2-diamine 'en', oxalate ion $\text{C}_2\text{O}_4^{2-}$).
• Multidentate: Three or more donor atoms (e.g., EDTA$^{4-}$, which is hexadentate).

A bidentate ligand must have two atoms, each with a lone pair of electrons, positioned such that they can both simultaneously bond to the same metal ion to form a ring structure (a chelate).

Understanding the Question

Define the term 'bidentate ligand'.

Approach

State the two key requirements: two lone pairs (or two donor atoms) and the formation of coordinate bonds.

Step-by-Step Reasoning

• Must mention two lone pairs (or two donor atoms).
• Must mention they form coordinate (dative covalent) bonds (to a metal ion/atom).

Key Takeaways

• 'Bi' means two, 'dentate' relates to teeth (donating electron pairs like biting).

Common Mistakes

• Saying 'two bonds' without specifying 'coordinate' or 'dative'.
• Saying 'two atoms' without mentioning the lone pairs.
• Forgetting to mention the metal ion/atom.

Things to Be Careful About

• Both points (two lone pairs AND coordinate bond) are required for full marks. 'Two lone pairs' alone is not enough.

Background Concept

The stability constant $K_{\text{stab}}$ is a measure of the extent to which the complex ion forms. A larger $K_{\text{stab}}$ value means the equilibrium lies further to the right (products side), indicating a more stable complex.

This question also introduces the chelate effect: when a bidentate ligand (like 'en') replaces monodentate ligands (like $\text{NH}_3$), the $K_{\text{stab}}$ is significantly larger. This is because the reaction produces more product molecules (increasing entropy, $\Delta S > 0$), making $\Delta G$ more negative ($\Delta G = -T\Delta S$ at constant enthalpy).

Understanding the Question

Given $K_{\text{stab}}$ values:

• $[\text{Cu}(\text{NH}_3)_2(\text{H}_2\text{O})_4]^{2+}$: $7.94 \times 10^7$
• $[\text{Cu}(\text{en})(\text{H}_2\text{O})_4]^{2+}$: $3.98 \times 10^{10}$

What do these tell us about the positions of the two equilibria?

Approach

Compare the magnitudes of the $K_{\text{stab}}$ values. The larger value corresponds to the equilibrium that lies further to the right.

Step-by-Step Reasoning

• $K_{\text{stab}}$ for equilibrium 2 ($10^{10}$) is much larger than for equilibrium 1 ($10^7$).
• Therefore, equilibrium 2 lies more to the right (or the forward reaction is more favoured).
• The complex with 'en' is more stable.

Key Takeaways

• Larger $K_{\text{stab}}$ = equilibrium lies further to the right = more stable complex.

Common Mistakes

• Saying 'equilibrium 1 is faster' (kinetics vs thermodynamics; $K$ values are about position, not speed).
• Confusing which equilibrium has which complex.

Things to Be Careful About

• Simply stating 'equilibrium 2 is more stable' is vague. Say 'lies more to the right' or 'favours the forward reaction more'.

Background Concept

Optical isomerism occurs in molecules/complexes that are non-superimposable on their mirror images (chiral).

In coordination chemistry, octahedral complexes of the type $[\text{M}(\text{bidentate})_3]$ (where M is the metal and bidentate is a chelating ligand like ethane-1,2-diamine) exhibit optical isomerism. The three bidentate ligands form a propeller-like shape. One enantiomer has a right-handed twist, and the other has a left-handed twist. They cannot be superimposed.

Other complexes that show optical isomerism include those with bidentate and monodentate ligands like $[\text{M}(\text{bidentate})_2(\text{monodentate})_2]$ (cis isomer only).

Understanding the Question

Name the type of stereoisomerism in $[\text{Ni}(\text{en})_3]^{2+}$.

Approach

Recognise the pattern $[\text{M}(\text{bidentate})_3]$ and recall it gives optical isomers.

Step-by-Step Reasoning

• The complex is octahedral with three bidentate ligands.
• This arrangement is chiral.
• Therefore, it displays optical isomerism.

Key Takeaways

• $[\text{M}(\text{bidentate})_3]$ -> optical isomers.

Common Mistakes

n- Saying 'geometric' or 'cis-trans'. Geometric isomerism requires at least two different types of monodentate ligands (e.g., $[\text{Ma}_4\text{b}_2]$). With three identical bidentate ligands, only optical isomerism is possible.

Things to Be Careful About

• Just 'optical' is sufficient.

Background Concept

Drawing optical isomers of $[\text{M}(\text{bidentate})_3]$ requires showing an octahedral geometry with 3D perspective. The bidentate ligands ('en') form five-membered rings with the metal.

To draw them:

1. Draw an octahedron. Use solid lines for bonds in the plane, a solid wedge (▲) for a bond coming out towards you, and a dashed wedge (---) for a bond going away.
2. For the first isomer (say, right-handed), arrange the three 'en' ligands in a clockwise or counter-clockwise propeller fashion.
3. For the second isomer, draw its exact mirror image. Ensure you can demonstrate they are non-superimposable (e.g., by rotating one, you cannot make all three rings align with the other).

Understanding the Question

Draw 3D diagrams of the two stereoisomers (enantiomers) of $[\text{Ni}(\text{en})_3]^{2+}$.

Approach

Draw two octahedral structures. Show the Ni atom in the centre. Draw three 'en' ligands (represented as curved lines or explicit $\text{NH}_2\text{CH}_2\text{CH}_2\text{NH}_2$ chains) coordinating to Ni. Use 3D bonds (wedges/dashes) to show the tetrahedral-like arrangement of the donor atoms around the octahedral centre. Draw the mirror image for the second isomer.

Step-by-Step Reasoning

• Structure 1: Octahedral Ni. Three 'en' ligands. Use 3D bonds to show the 3D arrangement. For example, one 'en' in the plane, one coming out, one going back. Arrange them to form a right-handed helix.
• Structure 2: Mirror image of Structure 1. Left-handed helix.
• Both must clearly show the octahedral shape and the 3D nature (wedges/dashes).

Key Takeaways

• Optical isomers are mirror images that cannot be superimposed.
• 3D drawing is essential; flat 2D drawings often fail to earn the mark.

Common Mistakes

• Drawing flat 2D structures without 3D bonds.
• Drawing geometric isomers (which don't exist for this formula).
• Not showing the continuation of the 'en' ligand properly (must be a 5-membered ring: Ni-N-C-C-N).

Things to Be Careful About

• The mark scheme awards 1 mark for a correct 3D octahedral shape, 1 for one correct structure with 3D, and 1 for the second correct structure with 3D.
• Ensure the 'en' ligands are drawn as chelate rings (N-C-C-N attached to the central metal).

Background Concept

A Brønsted-Lowry base is a proton ($\text{H}^+$) acceptor. Amines are basic because the nitrogen atom has a lone pair of electrons that can form a coordinate bond with a proton.

The basicity of amines depends on the availability of this lone pair. Alkyl groups are electron-donating (by induction), which increases the electron density on nitrogen, making the lone pair more available to accept a proton. Thus, aliphatic amines are generally stronger bases than ammonia.

Understanding the Question

Explain how the amino groups in ethane-1,2-diamine allow it to act as a Brønsted-Lowry base.

Approach

Connect the definition of a Brønsted-Lowry base (proton acceptor) with the structural feature of amines (lone pair on nitrogen).

Step-by-Step Reasoning

• Ethane-1,2-diamine has two amino groups ($-\text{NH}_2$).
• The nitrogen atom in each amino group has a lone pair of electrons.
• This lone pair can accept a proton ($\text{H}^+$) to form a coordinate bond.
• Therefore, it acts as a Brønsted-Lowry base.

Key Takeaways

• Amines are bases because of the lone pair on nitrogen.
• Brønsted-Lowry base = proton acceptor.

Common Mistakes

• Saying 'donates a proton' (that's an acid).
• Saying 'lone pair forms a covalent bond with H' (must specify coordinate/dative or accepting a proton).

Things to Be Careful About

• Both 'lone pair' and 'accepts a proton' are required for the 2 marks.

Background Concept

Amines react with acids to form ammonium salts. Since ethane-1,2-diamine has two amino groups, it is a dibasic molecule. In the presence of excess acid, both amino groups will be protonated.

The reaction is:

$$\text{R-NH}_2 + \text{H}^+ \rightarrow \text{R-NH}_3^+$$

Understanding the Question

Write an equation for the reaction of ethane-1,2-diamine with an excess of hydrochloric acid.

Approach

• Reactant: $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ + 2 $\text{HCl}$ (because there are 2 $-\text{NH}_2$ groups and acid is in excess).
• Product: Both nitrogens become $-\text{NH}_3^+$. The chloride ions are counter-ions.
• Write the balanced equation.

Step-by-Step Reasoning

• Molecular equation: $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 + 2\text{HCl} \rightarrow \text{ClH}_3\text{NCH}_2\text{CH}_2\text{NH}_3\text{Cl}$
• Ionic equation: $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2 + 2\text{H}^+ \rightarrow \text{H}_3\text{N}^+\text{CH}_2\text{CH}_2\text{N}^+\text{H}_3$
• Both are acceptable.

Key Takeaways

• Dibasic amines react with 2 moles of acid.
• Excess acid ensures complete protonation.

Common Mistakes

• Only protonating one amino group (forgetting 'excess' or 'two amino groups').
• Writing $\text{H}_3\text{N}^+\text{CH}_2\text{CH}_2\text{NH}_3^{2+}$ (incorrect charge; each N is +1, total is +2, balanced by 2 Cl-).

Things to Be Careful About

• Ensure the equation is balanced. 2 moles of HCl are needed.

Background Concept

Polyamides are formed by the condensation polymerisation of a diamine and a dicarboxylic acid (or diacyl chloride). The reaction forms amide linkages ($-\text{CO}-\text{NH}-$) and releases a small molecule (usually water or HCl).

Ethane-1,2-diamine: $\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2$
Ethanedioic acid: $\text{HOOC}-\text{COOH}$

When they react, the $-\text{OH}$ from the carboxylic acid and an $-\text{H}$ from the amine combine to form water, leaving an amide bond: $-\text{CO}-\text{NH}-$.

Understanding the Question

Draw the structure of polymer Z formed from ethane-1,2-diamine and ethanedioic acid, showing two repeat units.

Approach

1. Identify the monomers: diamine and dicarboxylic acid.
2. Determine the linkage: amide ($-\text{CONH}-$).
3. Draw the chain: $-\text{CO}-\text{CO}-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-$ repeating.
4. Show two complete repeat units with continuation bonds at both ends.

Step-by-Step Reasoning

• Monomer 1: $\text{HOOC}-\text{COOH}$ (ethanedioic acid)
• Monomer 2: $\text{H}_2\text{N}-\text{CH}_2-\text{CH}_2-\text{NH}_2$ (ethane-1,2-diamine)
• Linkage: $-\text{CO}-\text{NH}-$
• Repeat unit: $-\text{CO}-\text{CO}-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-$
• Two repeat units: $-\text{CO}-\text{CO}-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-\text{CO}-\text{CO}-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-$
• Add continuation bonds (lines) at both ends.

Key Takeaways

• Diamine + dicarboxylic acid -> polyamide (nylon-type).
• Amide linkage is $-\text{CO}-\text{NH}-$.
• Always show continuation bonds for polymers.

Common Mistakes

• Drawing an ester linkage ($-\text{COO}-$) instead of amide.
• Forgetting the continuation bonds.
• Drawing only one repeat unit.
• Incorrectly connecting the atoms (e.g., N to C of carbonyl is correct, but ensure the chain is $-\text{C}(=\text{O})-\text{C}(=\text{O})-\text{NH}-\text{CH}_2-\text{CH}_2-\text{NH}-$).

Things to Be Careful About

• The question asks for two repeat units. Count them carefully.
• Displayed formula or skeletal is acceptable, but atoms must be correct.

Background Concept

There are two main types of polymerisation:

1. Addition polymerisation: Monomers add together with no loss of atoms. Typically involves alkenes (C=C double bonds opening up). Example: ethene -> polyethene.
2. Condensation polymerisation: Monomers join together with the loss of a small molecule (usually water, HCl, or methanol). Typically involves bifunctional monomers like diols + dicarboxylic acids (polyesters) or diamines + dicarboxylic acids (polyamides).

CIE also accepts the term addition-elimination for condensation polymerisation, as it describes the mechanism (addition of monomers with elimination of a small molecule).

Understanding the Question

Name the type of reaction occurring during the polymerisation of ethane-1,2-diamine and ethanedioic acid.

Approach

Water is lost when the amide bond forms. This is condensation polymerisation.

Step-by-Step Reasoning

• Two different bifunctional monomers react.
• A small molecule (water) is eliminated at each linkage.
• Therefore, it is condensation polymerisation.

Key Takeaways

• Diamine + dicarboxylic acid -> condensation polymerisation (forms polyamide).
• Alkene -> addition polymerisation.

Common Mistakes

• Saying 'addition' (no small molecule is lost in addition).
• Saying 'substitution' (not a polymerisation term in this context).

Things to Be Careful About

• 'Condensation' or 'addition-elimination' are both acceptable.

Background Concept

Biodegradable polymers can be broken down by microorganisms (bacteria, fungi) into natural substances like water, carbon dioxide, and biomass. They often contain hydrolyzable linkages (like ester or amide bonds) that enzymes can attack. Examples: poly(lactic acid) (PLA), poly(glycolic acid) (PGA), and polyamides/polyesters made from natural monomers.

Non-biodegradable polymers are typically addition polymers with strong, unreactive carbon-carbon backbones (C-C bonds) that microorganisms cannot break down. Examples: polyethene (PE), polypropene (PP), PVC, polystyrene (PS).

Understanding the Question

Name a polymer that is non-biodegradable.

Approach

Recall a common addition polymer.

Step-by-Step Reasoning

• Polyethene (polyethylene) is a classic non-biodegradable polymer.
• PVC (polyvinyl chloride) is another.
• Polystyrene is another.

Key Takeaways

• Addition polymers (C-C backbone) are generally non-biodegradable.
• Condensation polymers with hydrolyzable bonds are often biodegradable (though not always, e.g., some polyamides are durable).

Common Mistakes

• Naming a biodegradable polymer (like PLA or PHA) when asked for non-biodegradable.
• Saying 'plastic' (too vague, must name a specific polymer).

Things to Be Careful About

• 'Polyethene', 'PVC', 'polystyrene', 'Bakelite', or 'Kevlar' are all acceptable per the mark scheme.