Chemistry 9701/23 — May/June 2013

Background Concept

A dot-and-cross diagram (Lewis structure) represents the valence electrons in a molecule. The central atom is typically the least electronegative (excluding hydrogen) and is surrounded by terminal atoms. Bonds are shown as shared pairs (or multiple pairs for double/triple bonds) between atoms, while lone pairs (non-bonding electrons) are placed on the outer atoms. The octet rule generally applies: atoms tend to gain, lose, or share electrons to have eight electrons in their outer shell.

Understanding the Question

You are asked to draw the dot-and-cross diagram for carbon disulfide ($\text{CS}_2$), showing only outer (valence) electrons. Carbon has 4 outer electrons; sulfur has 6. The sequence is S–C–S.

Approach

1. Place carbon in the centre and sulfur atoms on either side.
2. Share electrons to form double bonds between C and each S to satisfy the octet rule for all atoms.
3. Add remaining lone pairs to the sulfur atoms.

Step-by-Step Reasoning

• Carbon ($\text{C}$) is in Group 4, so it has 4 valence electrons.
• Sulfur ($\text{S}$) is in Group 6, so each has 6 valence electrons.
• To satisfy the octet for carbon (needs 4 more electrons), it forms two double bonds, one with each sulfur atom. Each double bond consists of 4 shared electrons (2 from C, 2 from S).
• Each sulfur atom uses 4 electrons for bonding, leaving 4 non-bonding electrons (2 lone pairs) on each sulfur.
• Carbon has no lone pairs remaining.

Key Takeaways

Dot-and-cross diagrams visually demonstrate how atoms share electrons to achieve stable electron configurations. Double bonds involve four shared electrons.

Common Mistakes

• Drawing single bonds and leaving carbon with an incomplete octet.
• Placing lone pairs on the central carbon atom (carbon has no lone pairs in $\text{CS}_2$).
• Using the wrong number of electrons for sulfur (must be 6 valence electrons total per S atom).

Things to Be Careful About

Ensure the diagram clearly distinguishes between bonding electrons (shared between atoms) and lone pairs (localized on sulfur). The central carbon must have exactly 8 electrons around it (from the two double bonds), and each sulfur must have 8 (4 from bonding + 4 lone pairs).

Background Concept

Valence Shell Electron Pair Repulsion (VSEPR) theory states that electron pairs (bonding and lone pairs) around a central atom arrange themselves to minimize repulsion. The geometry depends on the number of bonding pairs and lone pairs on the central atom.

Understanding the Question

Determine the molecular shape and bond angle of $\text{CS}_2$ based on its Lewis structure.

Approach

Count the electron domains (bonding regions and lone pairs) on the central carbon atom. Two double bonds count as two bonding domains. Zero lone pairs on carbon.

Step-by-Step Reasoning

• Central atom: Carbon.
• Bonding domains: 2 (the two C=S double bonds).
• Lone pairs on central atom: 0.
• Total electron domains: 2.
• Electron geometry: linear.
• Molecular shape: linear.
• Bond angle: $180^\circ$.

Key Takeaways

Molecules with two electron domains and no lone pairs on the central atom are linear with a bond angle of $180^\circ$. Double bonds count as a single domain for shape prediction.

Common Mistakes

• Confusing electron domains with total number of bonds (e.g., thinking 4 bonds means tetrahedral).
• Forgetting that lone pairs on outer atoms do not affect the molecular shape, only lone pairs on the central atom do.

Things to Be Careful About

Ensure you only count domains on the central atom. The lone pairs on sulfur do not influence the overall shape of the molecule.

Background Concept

Combustion of a compound containing C, S, and O (or just C and S) in excess oxygen produces carbon dioxide ($\text{CO}_2$) and sulfur dioxide ($\text{SO}_2$). Balancing equations ensures the number of atoms of each element is conserved.

Understanding the Question

Write a balanced symbol equation for the complete combustion of $\text{CS}_2$ to give $\text{CO}_2$ and $\text{SO}_2$.

Approach

1. Write the unbalanced equation: $\text{CS}_2 + \text{O}_2 \rightarrow \text{CO}_2 + \text{SO}_2$.
2. Balance C and S first, then O.

Step-by-Step Reasoning

• Carbon: 1 on left, 1 on right (balanced).
• Sulfur: 2 on left, so need 2 $\text{SO}_2$ on right.
• Oxygen on right: $2 (\text{from } \text{CO}_2) + 2 \times 2 (\text{from } 2\text{SO}_2) = 6$ oxygen atoms.
• Oxygen on left: Need 3 $\text{O}_2$ molecules to provide 6 oxygen atoms.
• Balanced equation: $\text{CS}_2 + 3\text{O}_2 \rightarrow \text{CO}_2 + 2\text{SO}_2$.

Key Takeaways

Always balance atoms other than O and H first, then balance oxygen last in combustion reactions.

Common Mistakes

• Forgetting to multiply the subscript in $\text{SO}_2$ by the coefficient when counting oxygen atoms.
• Writing $\text{SO}_3$ instead of $\text{SO}_2$ (complete combustion of sulfur gives $\text{SO}_2$, not $\text{SO}_3$, unless specified).

Things to Be Careful About

State symbols are often required in Cambridge A-Level chemistry. $\text{CS}_2$ is a liquid (l), $\text{O}_2$ and products are gases (g). Check if the mark scheme requires them; if not, they are good practice.

Background Concept

Standard enthalpy changes are defined with specific reference states and conditions. The standard state is the most stable form of a substance at 298 K (25 °C) and 100 kPa (1 bar). Enthalpy of combustion specifically refers to burning in oxygen.

Understanding the Question

Define $\Delta H_c^\ominus$. This is a recall question testing precise terminology.

Approach

Identify the four key components of the definition:

1. Quantity: 1 mol of substance.
2. Process: Complete combustion (burning in excess oxygen).
3. Conditions: Standard conditions.

Step-by-Step Reasoning

• Must specify 1 mol of the substance (not per gram or arbitrary amount).
• Must specify complete combustion or burning in excess oxygen (ensures full oxidation to $\text{CO}_2$ and $\text{SO}_2$, not partial oxidation to CO).
• Must specify standard conditions (temperature and pressure).

Key Takeaways

Definitions in chemistry are precise. Missing '1 mol', 'excess oxygen', or 'standard conditions' will cost marks.

Common Mistakes

• Saying 'burning in oxygen' without specifying 'excess' or 'complete'.
• Forgetting to mention '1 mol'.
• Confusing with enthalpy of formation (which is from elements in standard states).

Things to Be Careful About

The symbol $\ominus$ denotes standard conditions. Ensure you mention all three criteria: 1 mol, complete combustion/excess oxygen, standard conditions.

Background Concept

Qualitative analysis clues (colours, states) help identify products in a reaction. 'Yellow solid' is a classic description for elemental sulfur (S). 'Colourless gases' could be $\text{CO}_2$, $\text{N}_2$, $\text{CO}$, or $\text{N}_2\text{O}$.

Understanding the Question

$\text{CS}_2$ reacts with NO in a 1:2 molar ratio. Products: yellow solid (S) and two colourless gases. Write a balanced equation.

Approach

1. Identify the yellow solid as sulfur (S).
2. The gases must contain C, N, and O from the reactants ($\text{CS}_2$ and NO).
3. Possible gas pairs: $\text{CO}_2$ and $\text{N}_2$, or $\text{CO}$ and $\text{N}_2\text{O}$.
4. Balance the equation for the most likely products ($\text{CO}_2$ and $\text{N}_2$).

Step-by-Step Reasoning

• Reactants: 1 $\text{CS}_2$ + 2 NO.
• Atoms available: 1 C, 2 S, 2 N, 2 O.
• Yellow solid: 2 S (elemental sulfur is yellow).
• Remaining atoms for gases: 1 C, 2 N, 2 O.
• Possible combination 1: $\text{CO}_2$ (1 C, 2 O) + $\text{N}_2$ (2 N). Both are colourless gases. This balances perfectly.
• Equation: $\text{CS}_2 + 2\text{NO} \rightarrow \text{CO}_2 + 2\text{S} + \text{N}_2$.
• Alternative combination: $\text{CO}$ (1 C, 1 O) + $\text{N}_2\text{O}$ (2 N, 1 O). Also balances: $\text{CS}_2 + 2\text{NO} \rightarrow \text{CO} + 2\text{S} + \text{N}_2\text{O}$.

Key Takeaways

Use qualitative observations (colour, state) to identify specific elements or compounds. Sulfur is characteristically yellow.

Common Mistakes

• Identifying the yellow solid as something else (e.g., $\text{S}_8$ is the standard state, but writing 'S' is acceptable and standard for balancing).
• Not balancing the equation correctly (e.g., forgetting the coefficient 2 for S or NO).

Things to Be Careful About

The mark scheme accepts either $\text{CO}_2 + \text{N}_2$ or $\text{CO} + \text{N}_2\text{O}$ as the gas products. Ensure the equation is fully balanced for both alternatives.

Background Concept

Oxidation number (oxidation state) is the charge an atom would have if all bonds were ionic. Rules: elemental form = 0; O is usually -2; sum of oxidation numbers in a neutral molecule = 0. Electronegativity determines bond polarity in covalent molecules.

Understanding the Question

Find the change in oxidation number of sulfur in the reaction: $\text{CS}_2 + 2\text{NO} \rightarrow \text{CO}_2 + 2\text{S} + \text{N}_2$.

Approach

1. Calculate O.N. of S in $\text{CS}_2$.
2. Calculate O.N. of S in the product (elemental S).
3. Find the difference.

Step-by-Step Reasoning

• In $\text{CS}_2$: Sulfur (EN = 2.58) is slightly more electronegative than Carbon (EN = 2.55). Therefore, S takes the electrons. Each S is assigned -2. (C is +4). Check: $+4 + 2(-2) = 0$. Correct.
• In the product, sulfur is elemental ($\text{S}$), so its oxidation number is $0$.
• Change: from $-2$ to $0$.
• (Note: Sulfur is oxidised, losing 2 electrons per atom).

Key Takeaways

Oxidation number in an element is 0. In compounds, use electronegativity to assign electrons to the more electronegative atom. S is more electronegative than C.

Common Mistakes

• Assuming C is more electronegative than S (they are very close, but S is slightly higher: 2.58 vs 2.55). If you assumed C was more electronegative, you'd get S as +2, which is wrong.
• Forgetting that elemental sulfur has an oxidation number of 0.

Things to Be Careful About

The question asks for the change 'from ... to ...'. Ensure the order is correct: reactant ($-2$) to product ($0$).