Chemistry 9701/23 — May/June 2023

Background Concept

Metals, including transition metals like copper, consist of a regular arrangement of positive ions (cations) in a lattice structure. The outer electrons are delocalised and form a 'sea' that moves freely throughout the lattice. This arrangement is known as a giant metallic lattice.

Understanding the Question

The question asks for the specific type of lattice structure present in solid copper. Given that copper is a metal, we can directly infer its bonding and structural type.

Approach

Recall the standard lattice types: giant ionic, giant covalent, simple molecular, and giant metallic. Since copper is a metal, it must be a giant metallic lattice.

Step-by-Step Reasoning

Copper is a transition metal. Metals do not form simple molecular structures or giant covalent networks (except for metalloids like silicon). They form giant metallic lattices where positive metal ions are held together by delocalised electrons. Therefore, the lattice structure is giant metallic.

Key Takeaways

Metals form giant metallic lattices. This is a fundamental structural classification that explains their physical properties like malleability and electrical conductivity.

Common Mistakes

Writing 'metallic bonding' instead of 'giant metallic lattice'. The question asks for the lattice structure, not the type of bonding.

Things to Be Careful About

Ensure the answer matches the exact terminology expected: 'giant metallic' or 'giant metallic lattice'.

Background Concept

Metallic bonding is the electrostatic attraction between a lattice of positive metal ions and a sea of delocalised electrons. To represent this in a 2D diagram, we show a regular arrangement of positive ions surrounded by smaller, freely moving electrons.

Understanding the Question

You are asked to draw a labelled diagram showing the bonding in copper. This means depicting the structural components (positive ions and delocalised electrons) and labelling them correctly.

Approach

Draw a regular grid or lattice of circles to represent the positive copper ions. Surround them with smaller circles or dots to represent the delocalised electrons. Ensure both components are clearly labelled.

Step-by-Step Reasoning

1. Draw at least four large circles arranged in a regular pattern (e.g., a 2x2 or 3x3 grid) to represent the positive copper ions. You can put a '+' inside them or label them as 'Cu^n+', 'positive ion', or 'cation'.
2. Draw smaller circles or dots around and between the large circles to represent the delocalised electrons. You can put a '-' inside them or label them as 'e^-' or 'delocalised electrons'.
3. The key is to show that the electrons are separate from and surrounding the positive ions, illustrating the 'sea of electrons' model.

Key Takeaways

A metallic bonding diagram must show both the positive ions (with appropriate labels) and the delocalised electrons (with appropriate labels) in a regular arrangement.

Common Mistakes

Forgetting to label the components. Drawing electrons inside the positive ions (like a dot-and-cross diagram for covalent bonding). Not showing a regular lattice arrangement.

Things to Be Careful About

The mark scheme allows circles with just a '+' to be unlabeled, but if you use empty circles or write 'Cu', you MUST label them as positive ions/cations. Similarly, electrons must be shown as 'e^-', '-', or labelled 'delocalised electrons'.

Background Concept

The unified atomic mass unit (u or Da) is a standard unit of mass that quantifies mass on an atomic or molecular scale. It is defined relative to the carbon-12 isotope, which is assigned an exact mass of 12 u.

Understanding the Question

The question asks for the formal definition of the unified atomic mass unit.

Approach

Recall the standard IUPAC definition of the unified atomic mass unit.

Step-by-Step Reasoning

The unified atomic mass unit is defined as exactly one-twelfth of the mass of an unbound neutral atom of carbon-12 in its nuclear and electronic ground state. In exam terms, stating 'one twelfth of the mass of a carbon-12 atom' is sufficient and correct.

Key Takeaways

The unified atomic mass unit is based on the carbon-12 isotope. 1 u = 1/12 × mass of a $^{12}\text{C}$ atom.

Common Mistakes

Saying 'one twelfth of a carbon atom' without specifying the isotope carbon-12. Saying 'atomic mass of carbon-12 divided by 12' is acceptable but the formal definition references 'the mass of a carbon-12 atom'.

Things to Be Careful About

Always specify 'carbon-12' or '$^{12}\text{C}$'. Do not just say 'carbon', as natural carbon is a mixture of isotopes.

Background Concept

Relative atomic mass ($A_r$) is a dimensionless quantity that represents the average mass of atoms of an element, taking into account the relative abundances of its naturally occurring isotopes, relative to 1/12 of the mass of a carbon-12 atom (the unified atomic mass unit).

Understanding the Question

The question asks for the definition of relative atomic mass, specifically in terms of the unified atomic mass unit.

Approach

State that it is the average mass of the isotopes of an element, and specify that this is compared to (or divided by) the unified atomic mass unit.

Step-by-Step Reasoning

Relative atomic mass is calculated by weighting the masses of each isotope by their relative abundances. The definition requires stating that it is the 'average mass of the isotopes of an element' and that this is 'compared to the unified atomic mass unit' (or 'relative to 1/12 of the mass of a carbon-12 atom').

Key Takeaways

$A_r$ is a weighted average of isotopic masses, expressed relative to the unified atomic mass unit.

Common Mistakes

Forgetting to mention 'average' or 'weighted average'. Forgetting to mention 'compared to the unified atomic mass unit' or 'relative to 1/12 of carbon-12'.

Things to Be Careful About

Ensure you use the exact phrasing 'average mass of the isotopes' and 'compared to the unified atomic mass unit' as this is what the mark scheme rewards.

Background Concept

The relative atomic mass ($A_r$) of an element is the weighted average of the relative isotopic masses of all its naturally occurring isotopes. The formula is:
$A_r = \sum \left( \frac{\% \text{ abundance}}{100} \times \text{relative isotopic mass} \right)$

Understanding the Question

You are given the relative isotopic masses and percentage abundances of two copper isotopes ($^{63}\text{Cu}$ and $^{65}\text{Cu}$) and asked to calculate the relative atomic mass of copper in this sample.

Approach

Multiply each isotopic mass by its percentage abundance (divided by 100) and sum the results.

Step-by-Step Reasoning

1. For $^{63}\text{Cu}$: mass = 62.930, abundance = 69.15%. Contribution = $62.930 \times 0.6915 = 43.518$.
2. For $^{65}\text{Cu}$: mass = 64.928, abundance = 30.85%. Contribution = $64.928 \times 0.3085 = 20.030$.
3. Sum the contributions: $43.518 + 20.030 = 63.548$.
4. Rounding to an appropriate number of decimal places (usually 2 for $A_r$), we get 63.55.

Key Takeaways

To calculate $A_r$, always use the formula: sum of (isotopic mass × fractional abundance). Ensure percentages are divided by 100.

Common Mistakes

Forgetting to divide the percentage abundance by 100. Using the wrong isotopic mass for the wrong abundance. Arithmetic errors.

Things to Be Careful About

Check your calculator input. The mark scheme accepts 63.55 or 63.546. Ensure you show your working clearly as requested.

Background Concept

Copper(II) ions ($\text{Cu}^{2+}$) are oxidising agents and can oxidise iodide ions ($\text{I}^-$) to iodine ($\text{I}_2$), while being reduced to copper(I) ($\text{Cu}^+$), which precipitates as copper(I) iodide ($\text{CuI}$).

Understanding the Question

You are given the reactants ($\text{Cu}^{2+}$ and $\text{I}^-$) and one product ($\text{CuI}$). You need to deduce the second product and balance the equation. The problem states there are only two products.

Approach

1. Identify the redox changes: $\text{Cu}^{2+}$ is reduced to $\text{Cu}^+$ (in $\text{CuI}$). $\text{I}^-$ must be oxidised. The only logical oxidation product of $\text{I}^-$ is $\text{I}_2$.
2. Write the half-equations and combine them, or balance by inspection.

Step-by-Step Reasoning

1. Reduction half-equation: $\text{Cu}^{2+} + \text{I}^- + \text{e}^- \rightarrow \text{CuI}$ (or simply $\text{Cu}^{2+} + \text{e}^- \rightarrow \text{Cu}^+$).
2. Oxidation half-equation: $2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^-$.
3. To balance electrons, multiply the reduction half by 2: $2\text{Cu}^{2+} + 2\text{e}^- \rightarrow 2\text{Cu}^+$.
4. Combine: $2\text{Cu}^{2+} + 2\text{I}^- \rightarrow 2\text{Cu}^+ + \text{I}_2$.
5. The $\text{Cu}^+$ ions combine with the remaining $\text{I}^-$ ions to form $\text{CuI}$ precipitate. We need 2 more $\text{I}^-$ for the $2\text{CuI}$.
6. Total $\text{I}^-$ on the left = $2 + 2 = 4$.
7. Balanced equation: $2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} + \text{I}_2$.

Key Takeaways

When balancing redox equations involving precipitation, ensure all atoms and charges are balanced. Copper(II) oxidises iodide to iodine and is reduced to copper(I) iodide.

Common Mistakes

Writing $\text{Cu}^{2+} + 2\text{I}^- \rightarrow \text{CuI} + \text{I}$ (incorrect iodine species). Forgetting to balance the iodide ions needed for the $\text{CuI}$ precipitate.

Things to Be Careful About

The equation must be fully balanced with correct stoichiometric coefficients. The second product is molecular iodine, $\text{I}_2$, not $\text{I}^-$ or $\text{I}$.

Background Concept

An oxidising agent is a species that accepts electrons from another species, thereby causing the other species to be oxidised. The oxidising agent itself is reduced.

Understanding the Question

You need to identify the oxidising agent in the reaction $2\text{Cu}^{2+} + 4\text{I}^- \rightarrow 2\text{CuI} + \text{I}_2$ and explain your choice using electron transfer terminology.

Approach

1. Identify which species is reduced (gains electrons).
2. State that this species is the oxidising agent.
3. Explain that it gained electrons from the other reactant.

Step-by-Step Reasoning

1. In the reaction, $\text{Cu}^{2+}$ changes oxidation state from +2 to +1 (in $\text{CuI}$). This is a reduction (gain of electrons).
2. $\text{I}^-$ changes oxidation state from -1 to 0 (in $\text{I}_2$). This is an oxidation (loss of electrons).
3. Since $\text{Cu}^{2+}$ gains electrons, it is the oxidising agent.
4. Explanation: $\text{Cu}^{2+}$ has gained / taken electron(s) from iodide ($\text{I}^-$) ions.

Key Takeaways

The oxidising agent is the species that is reduced. Always explain in terms of electron transfer: 'gained electrons' or 'took electrons'.

Common Mistakes

Identifying $\text{I}^-$ as the oxidising agent. Explaining using oxidation states instead of electron transfer (e.g., 'Cu decreased in oxidation state' is not sufficient; must say 'gained electrons').

Things to Be Careful About

The mark scheme accepts $\text{Cu}^{2+}$ or $\text{CuSO}_4$ as the oxidising agent. The explanation MUST mention electron transfer ('gained' or 'taken' electrons).

Background Concept

Copper (atomic number 29) has an anomalous electron configuration: $[\text{Ar}] 3\text{d}^{10} 4\text{s}^1$ (or $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1$). This is because a fully filled 3d subshell is particularly stable.

When forming ions, transition metals lose their 4s electrons before their 3d electrons.

Understanding the Question

You need to write the full electronic configuration of the $\text{Cu}^{2+}$ ion.

Approach

1. Write the configuration for neutral Cu.
2. Remove 2 electrons to form $\text{Cu}^{2+}$, starting from the 4s subshell.

Step-by-Step Reasoning

1. Neutral Cu (29 electrons): $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^{10} 4\text{s}^1$.
2. To form $\text{Cu}^{2+}$, remove 2 electrons. The first electron removed is from 4s (leaving 0 electrons in 4s). The second electron is removed from 3d (leaving 9 electrons in 3d).
3. Configuration of $\text{Cu}^{2+}$ (27 electrons): $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^9$.

Key Takeaways

Transition metals lose 4s electrons before 3d electrons when forming positive ions. Copper's anomalous configuration must be remembered.

Common Mistakes

Writing $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^8 4\text{s}^2$ (removing from 3d first, which is wrong). Forgetting the anomalous configuration of neutral Cu and using $3\text{d}^9 4\text{s}^2$ instead.

Things to Be Careful About

The question asks for the FULL electronic configuration, not the abbreviated $[\text{Ar}] ...$ form. Ensure all subshells up to 3d are written out with correct superscripts.