Chemistry 9701/22 — October/November 2024

Background Concept

Every atom has a specific arrangement of electrons that corresponds to its lowest possible total energy. This arrangement is called the ground state electronic configuration. When an atom absorbs energy (for example, from heat or light), one or more electrons can be promoted to higher energy levels. The atom is then said to be in an excited state. Excited states are unstable, and the electrons will eventually fall back to lower energy levels, releasing energy in the process (often as photons of light, which is the basis of emission spectra).

Understanding the Question

The question asks for a definition of the term 'ground state' in the context of an atom's electronic configuration. The command word is 'State', which requires a concise, direct definition without elaboration.

Approach

Recall the standard IUPAC-accepted definition of ground state for an atom: the state where all electrons occupy the lowest available energy orbitals, meaning no external energy has been added to promote them.

Step-by-Step Reasoning

1. Identify the key concept: the energy state of an atom's electrons.
2. Formulate the definition: The ground state is the lowest energy state of the atom.
3. Add qualifying context (optional but good for clarity): This is the state with no external energy added, or where electrons are not promoted to higher energy levels.
4. The mark scheme accepts 'lowest energy state' as the primary creditable point.

Key Takeaways

• Ground state: Lowest energy configuration of electrons.
• Excited state: One or more electrons have absorbed energy and moved to a higher energy orbital.

Common Mistakes

• Writing 'the state with the most electrons' (all atoms have a fixed number of electrons regardless of state).
• Writing 'the state with no energy' (atoms always possess internal energy; it is the lowest possible energy state, not zero energy).
• Confusing ground state with 'ground level' or 'ground state of the nucleus'.

Things to Be Careful About

• Keep the definition concise. 'Lowest energy state' is sufficient and exactly what the mark scheme rewards. Avoid over-elaborating unless asked to explain.

Background Concept

The electrons-in-boxes (or orbital filling) diagram represents the distribution of electrons in the orbitals of an atom's outer shells. Each box represents one orbital, and arrows (↑ and ↓) represent electrons with opposite spins (Pauli exclusion principle).

Two key rules govern how these boxes are filled:

1. Aufbau principle: Orbitals are filled from lowest to highest energy. For the 3d and 4s subshells, the 4s orbital is lower in energy than the 3d orbitals, so 4s fills first.
2. Hund's rule: Within a subshell of degenerate orbitals (like the five 3d orbitals), electrons occupy empty orbitals singly with parallel spins before pairing up. This minimises electron-electron repulsion.

Understanding the Question

The question provides the shorthand configuration of vanadium: $[\text{Ar}]3\text{d}^34\text{s}^2$. It asks to draw this using the electrons-in-boxes notation. The provided diagram (Fig 1.1) shows five boxes for the 3d subshell and one box for the 4s subshell, all initially empty.

Approach

1. Fill the 4s box first with 2 electrons (paired, opposite spins).
2. Distribute the 3 electrons in the 3d boxes according to Hund's rule: one electron in each of the first three boxes, all with upward spin.

Step-by-Step Reasoning

• The configuration is $[\text{Ar}]3\text{d}^34\text{s}^2$.
• The $4\text{s}^2$ part means the 4s orbital (the single box on the right) contains 2 electrons. By the Pauli exclusion principle, they must have opposite spins: one up (↑) and one down (↓).
• The $3\text{d}^3$ part means there are 3 electrons in the 3d subshell (the five adjacent boxes on the left).
• By Hund's rule, these 3 electrons will occupy separate orbitals with parallel spins to minimise repulsion. So, the first three 3d boxes each get one up arrow (↑), and the remaining two 3d boxes are empty.
• The final diagram has [Ar] on the left, then three boxes with ↑, two empty boxes, and finally the 4s box with ↑↓.

Key Takeaways

• Always fill 4s before 3d when drawing the configuration, even though 3d is written first in the shorthand notation.
• Apply Hund's rule to degenerate orbitals: fill singly before pairing.

Common Mistakes

• Pairing electrons in the 3d subshell too early (e.g., ↑↓ in the first box, ↑ in the second, empty others). This violates Hund's rule.
• Putting the paired electrons in the 3d subshell and leaving 4s with one electron. This violates the Aufbau principle (4s is lower energy than 3d for neutral atoms in this region).
• Forgetting that the 4s box is separate from the 3d boxes in the diagram.

Things to Be Careful About

• The mark scheme allows the 3 d electrons to go in any of the d orbitals (e.g., they could be in boxes 3, 4, 5 instead of 1, 2, 3), as long as they are unpaired and parallel. However, the standard convention is to fill from left to right. Ensure the 4s box clearly contains a pair.

Background Concept

The noble gas shorthand notation $[\text{Ar}]$ represents the full electronic configuration of argon, which accounts for the first 18 electrons. To count specific types of electrons (like p-electrons), one must expand the noble gas core or carefully track which subshells contain them.

The subshells are s, p, d, f. Each s holds 2, each p holds 6, each d holds 10, each f holds 14. Argon's configuration is $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6$.

Understanding the Question

The question asks for the total number of electrons in the p sub-shells of a vanadium atom. Vanadium's full configuration is $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3 4\text{s}^2$.

Approach

1. Identify all p-subshells in the full electronic configuration.
2. Sum the electrons in those subshells.

Step-by-Step Reasoning

• Expand $[\text{Ar}]$: $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6$.
• Add the vanadium valence electrons: $3\text{d}^3 4\text{s}^2$.
• Full configuration: $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^6 3\text{d}^3 4\text{s}^2$.
• Identify p-subshells: $2\text{p}^6$ and $3\text{p}^6$.
• Count electrons: $6 + 6 = 12$.
• The $3\text{d}$ and $4\text{s}$ subshells contain d and s electrons respectively, so they contribute 0 to the p-electron count.

Key Takeaways

• When asked to count electrons in a specific subshell type (s, p, d, f) for an element with a noble gas core, expand the core to identify all such subshells.
• Argon ($Z=18$) has 12 p-electrons ($2\text{p}^6 + 3\text{p}^6$). Any element in period 3 or later that doesn't add p-electrons (like transition metals in period 4) will still have 12 p-electrons.

Common Mistakes

• Forgetting to count the p-electrons in the noble gas core (answering 0 or just counting valence p-electrons if any).
• Miscounting the capacity of p-subshells (thinking p holds 2 or 10 instead of 6).
• Including d-electrons in the p-count.

Things to Be Careful About

• Ensure you are counting electrons, not orbitals. A p-subshell has 3 orbitals but holds up to 6 electrons. Here, both filled p-subshells hold 6 electrons each.

Background Concept

Isotope notation is written as $^{A}_{Z}\text{X}$, where:

• X is the chemical symbol.
• Z (subscript) is the proton number (atomic number), which defines the element.
• A (superscript) is the mass number (nucleon number), which is the total number of protons and neutrons in the nucleus.

The number of neutrons is calculated as $A - Z$.

Understanding the Question

The question provides a table with two isotopes: $^{93}_{41}\text{Nb}$ and $^{181}_{73}\text{Ta}$. The relative isotopic masses are given. The candidate must fill in the number of protons and neutrons for each.

Approach

1. Read the proton number directly from the subscript (Z).
2. Calculate the neutron number by subtracting the proton number from the mass number (A - Z).

Step-by-Step Reasoning

• For $^{93}_{41}\text{Nb}$:
  • Proton number (Z) = 41 (from the subscript).
  • Mass number (A) = 93 (from the superscript).
  • Neutron number = $93 - 41 = 52$.
• For $^{181}_{73}\text{Ta}$:
  • Proton number (Z) = 73 (from the subscript).
  • Mass number (A) = 181 (from the superscript).
  • Neutron number = $181 - 73 = 108$.

Key Takeaways

• Isotope notation $^{A}_{Z}\text{X}$: Z is protons, A is protons + neutrons.
• Neutrons = A - Z.
• The relative isotopic mass is close to the mass number but not exactly an integer due to binding energy and the mass of neutrons vs protons.

Common Mistakes

• Confusing the mass number (superscript) with the neutron number.
• Subtracting incorrectly (e.g., $181 - 73 = 108$, but writing 118).
• Forgetting that the proton number is the subscript, not the relative isotopic mass.

Things to Be Careful About

• Ensure the values are placed in the correct columns. The table asks for 'number of protons' and 'number of neutrons', not mass number.
• The relative isotopic mass (92.91, 180.95) is not used in this part; it is provided for part (iii). Ignore it here.

Background Concept

Relative masses are used because actual masses of atoms are incredibly small (e.g., $\sim 10^{-26}$ kg). The unified atomic mass unit (u or Da) is defined as $\frac{1}{12}$ of the mass of a carbon-12 atom.

• Relative isotopic mass: The mass of a specific isotope of an element relative to $\frac{1}{12}$ of the mass of a carbon-12 atom.
• Relative atomic mass ($A_r$): The weighted average mass of an atom of the element (considering all naturally occurring isotopes and their abundances) relative to $\frac{1}{12}$ of the mass of a carbon-12 atom.

Understanding the Question

The question asks for the definition of 'relative isotopic mass'. The command word is 'Define', requiring a precise, textbook definition.

Approach

State the two key components of the definition:

1. What is being measured: the mass of an atom of a specific isotope.
2. What it is compared to: $\frac{1}{12}$ of the mass of a carbon-12 atom (or the unified atomic mass unit).

Step-by-Step Reasoning

• The mark scheme requires two marks (M1 and M2).
• M1: Must mention 'mass of an (atom of an) isotope'.
• M2: Must mention 'relative/compared to $\frac{1}{12}$ of the mass of a carbon-12 atom' OR 'on a scale where C-12 is exactly 12'.
• Combining these: 'The mass of an atom of an isotope relative to $\frac{1}{12}$ of the mass of a carbon-12 atom.'

Key Takeaways

• Distinguish between relative isotopic mass (single isotope) and relative atomic mass (average of isotopes).
• The reference standard is always $\frac{1}{12}$ of a carbon-12 atom.

Common Mistakes

• Defining relative atomic mass instead of relative isotopic mass (mentioning 'average' or 'naturally occurring' is wrong for isotopic mass).
• Saying 'relative to hydrogen' (this is an old, incorrect definition).
• Forgetting the fraction $\frac{1}{12}$ (just saying 'relative to carbon-12' is incomplete; it must be 1/12th of it).
• Saying 'mass number' instead of 'relative mass' (mass number is an integer count of nucleons; relative isotopic mass is a precise measured value like 92.91).

Things to Be Careful About

• The mark scheme accepts alternative valid routes, such as defining it in terms of 1 mole of substance (mass of 1 mol of isotope relative to 1/12 mass of 1 mol of C-12, which is 12 g). However, the single-atom definition is more direct and standard for this level.

Background Concept

The relative atomic mass ($A_r$) of an element is the weighted average mass of its atoms relative to $\frac{1}{12}$ of the mass of a carbon-12 atom. When an element exists as a mixture of isotopes (like pelopium, which is a mixture of Nb and Ta), the $A_r$ is calculated using the relative isotopic masses and their percentage abundances.

Formula:

or equivalently, using fractional abundance (percentage / 100):

Understanding the Question

Pelopium is a mixture of 90.9% $^{93}_{41}\text{Nb}$ (relative isotopic mass 92.91) and 9.1% $^{181}_{73}\text{Ta}$ (relative isotopic mass 180.95). The question asks to calculate the theoretical relative atomic mass of pelopium to two decimal places.

Approach

1. Convert percentage abundances to fractional abundances (divide by 100).
2. Multiply each isotope's relative isotopic mass by its fractional abundance.
3. Sum the results.
4. Round to the required significant figures (two decimal places).

Step-by-Step Reasoning

• Fractional abundance of Nb = $90.9\% / 100 = 0.909$.
• Fractional abundance of Ta = $9.1\% / 100 = 0.091$.
• Check: $0.909 + 0.091 = 1.000$ (correct, abundances sum to 100%).
• Contribution from Nb = $92.91 \times 0.909 = 84.45519$.
• Contribution from Ta = $180.95 \times 0.091 = 16.46645$.
• Total $A_r = 84.45519 + 16.46645 = 100.92164$.
• Round to two decimal places: $100.92$.

Key Takeaways

• Relative atomic mass is a weighted average, not a simple average.
• Always convert percentages to decimals (or divide the final sum by 100) before calculating.
• The result should be between the masses of the lightest and heaviest isotopes (100.92 is between 92.91 and 180.95, and closer to 92.91 because Nb is more abundant).

Common Mistakes

• Using percentage values directly without dividing by 100 (e.g., $92.91 \times 90.9 + 180.95 \times 9.1 = 10092$, which is wrong).
• Calculating a simple average: $(92.91 + 180.95) / 2 = 136.93$ (ignores abundance).
• Rounding too early in the calculation (keep full precision until the final step).
• Rounding to the wrong number of decimal places (question asks for two decimal places).

Things to Be Careful About

• The mark scheme awards 1 mark for the correct calculation setup ($92.91 \times 0.909 + 180.95 \times 0.091$) and 1 mark for the correct final answer (100.92).
• Error carried forward (ecf) is allowed: if a candidate used wrong abundances but applied the formula correctly, they can still get the second mark.
• Ensure the final answer has no units (relative atomic mass is dimensionless).

Background Concept

When elements react with oxygen, they undergo combustion. Group 1 metals like sodium react vigorously to form ionic oxides ($\text{Na}_2\text{O}$), while non-metals like sulfur in Period 3 react to form covalent molecular oxides ($\text{SO}_2$). Writing these equations requires knowing the correct stoichiometry and including the appropriate state symbols: (s) for solid sodium and sulfur, (g) for oxygen gas, and (s) for the solid oxide products.

Understanding the Question

The question asks for balanced chemical equations for the reactions of sodium with oxygen and sulfur with oxygen. It is a straightforward recall and balancing task.

Approach

Identify the reactants and products. Sodium burns in oxygen to form sodium oxide ($\text{Na}_2\text{O}$). Sulfur burns in oxygen to form sulfur dioxide ($\text{SO}_2$). Balance the atoms on both sides and add state symbols.

Step-by-Step Reasoning

1. Sodium and oxygen: Sodium is a Group 1 metal with a +1 oxidation state, and oxygen is Group 16 with a -2 oxidation state, giving the formula $\text{Na}_2\text{O}$. Balancing the equation: $4\text{Na} + \text{O}_2 \rightarrow 2\text{Na}_2\text{O}$. Add states: $4\text{Na(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{Na}_2\text{O(s)}$.
2. Sulfur and oxygen: Sulfur burns in oxygen to predominantly form sulfur dioxide, $\text{SO}_2$. The equation is already balanced: $\text{S} + \text{O}_2 \rightarrow \text{SO}_2$. Add states: $\text{S(s)} + \text{O}_2\text{(g)} \rightarrow \text{SO}_2\text{(g)}$.

Key Takeaways

Always remember to balance equations and include state symbols (s, l, g, aq) as they are often required for full marks in CIE Chemistry. Group 1 metals form normal oxides ($\text{M}_2\text{O}$), while sulfur forms $\text{SO}_2$ (not $\text{SO}_3$) upon direct combustion.

Common Mistakes

• Writing $\text{NaO}$ instead of $\text{Na}_2\text{O}$ (ignoring the +1/-2 charge balance).
• Forgetting state symbols, especially (g) for $\text{O}_2$.
• Writing $\text{SO}_3$ for the sulfur reaction; direct combustion of sulfur yields $\text{SO}_2$.

Things to Be Careful About

Ensure the equations are fully balanced. For sodium, 4 atoms are needed on the left to match 4 on the right in $2\text{Na}_2\text{O}$.

Background Concept

Aluminium oxide ($\text{Al}_2\text{O}_3$) is an ionic compound formed between a metal (Al, Group 13) and a non-metal (O, Group 16). Aluminium has 3 outer electrons and loses them to form $\text{Al}^{3+}$. Oxygen has 6 outer electrons and gains 2 to complete its octet, forming $\text{O}^{2-}$. In a dot-and-cross diagram for ionic compounds, we show the outer electrons of the anion, using different symbols (dots and crosses) to distinguish which atom contributed them. Cations are shown with a charge and no outer electrons (or empty brackets), while anions are shown with brackets, a charge, and 8 outer electrons.

Understanding the Question

Draw a dot-and-cross diagram for $\text{Al}_2\text{O}_3$ showing only outer electrons. This means drawing one aluminium ion and one oxide ion (representing the 1:3 ratio conceptually for the diagram, or just the species present).

Approach

Aluminium loses 3 electrons to become $\text{Al}^{3+}$ (no outer electrons shown). Oxygen gains 2 electrons to become $\text{O}^{2-}$ (8 outer electrons: 6 original + 2 from Al). Draw $\text{Al}^{3+}$ and an oxide ion with 8 outer electrons (e.g., 6 dots, 2 crosses), surrounded by brackets with a $2-$ charge.

Step-by-Step Reasoning

1. Aluminium ion: $\text{Al}$ atom has 3 outer electrons. It loses all 3 to achieve a noble gas configuration. Draw $\text{Al}^{3+}$ with no dots or crosses around it.
2. Oxide ion: $\text{O}$ atom has 6 outer electrons. It gains 2 electrons (from Al) to have 8 outer electrons. Draw an oxygen symbol 'O' surrounded by 8 electrons (e.g., 6 blue dots and 2 red crosses). Place square brackets around the ion and write $2-$ as the charge outside the top right bracket.

Key Takeaways

For ionic dot-and-cross diagrams, the cation has no outer electrons shown (it lost them), and the anion has a full octet (8 electrons). Always include charges and brackets for the anion.

Common Mistakes

• Drawing covalent bonds (lines or shared pairs) instead of ionic transfer.
• Forgetting the brackets and charge on the oxide ion.
• Showing 7 electrons on the oxide ion instead of 8.

Things to Be Careful About

The question says 'draw outer electrons only'. Do not draw inner shell electrons for oxygen (the 1s2 pair). The aluminium ion has no outer electrons, so nothing is drawn around it.

Background Concept

Oxidation state (or oxidation number) in compounds reflects the number of electrons an atom loses, gains, or shares. For main group elements in Period 3, the maximum oxidation state corresponds to the number of electrons in the outer shell (valence electrons). Sodium has 1, magnesium has 2, aluminium has 3, silicon has 4, phosphorus has 5, sulfur has 6, and chlorine has 7.

Understanding the Question

State and explain how the maximum oxidation state of Period 3 elements varies in their oxides as you move from left to right across the period.

Approach

First, state the trend: it increases. Then, explain why: the number of outer (valence) electrons increases, and these are the electrons available for bonding (lost in ionic oxides or shared in covalent oxides).

Step-by-Step Reasoning

1. State the trend: Moving from Na to Cl, the maximum oxidation state increases (e.g., +1 in $\text{Na}_2\text{O}$, +2 in $\text{MgO}$, +3 in $\text{Al}_2\text{O}_3$, +4 in $\text{SiO}_2$, +5 in $\text{P}_4\text{O}_{10}$, +6 in $\text{SO}_3$, +7 in $\text{Cl}_2\text{O}_7$).
2. Explain the variation: The maximum oxidation state is determined by the number of valence electrons an atom can use. As you move across Period 3, the number of outer shell electrons increases from 1 to 7. Therefore, more electrons can be involved in bonding (lost to form cations or shared to form covalent bonds), leading to a higher maximum oxidation state.

Key Takeaways

The maximum oxidation state of a main group element equals its group number (or number of valence electrons). Across a period, this number increases, so the maximum oxidation state increases.

Common Mistakes

• Saying 'oxidation state increases because atomic number increases' (too vague; must mention valence electrons).
• Forgetting to state the trend clearly before explaining it.

Things to Be Careful About

Use precise terminology: 'valence electrons' or 'outer shell electrons', not just 'electrons'. Specify that these are the electrons available for bonding.

Background Concept

Period 3 oxides show a transition from basic to acidic as you move across the period. Magnesium oxide ($\text{MgO}$) is a basic oxide that reacts with water to form the weak base magnesium hydroxide, $\text{Mg(OH)}_2$. Phosphorus(V) oxide ($\text{P}_4\text{O}_{10}$) is an acidic oxide that reacts vigorously with water to form phosphoric acid, $\text{H}_3\text{PO}_4$.

Understanding the Question

Complete a table with the missing oxide for the basic product $\text{Mg(OH)}_2$, and the missing product and pH for the acidic oxide $\text{P}_4\text{O}_{10}$.

Approach

Identify the oxide that forms $\text{Mg(OH)}_2$ with water (it must be $\text{MgO}$). Determine the pH of a magnesium hydroxide suspension (weak base, pH 8-12). Identify the product of $\text{P}_4\text{O}_{10}$ with water ($\text{H}_3\text{PO}_4$) and its pH (weak acid, pH 1-4).

Step-by-Step Reasoning

1. Row 1: The product is $\text{Mg(OH)}_2$. The oxide that reacts with water to form this is magnesium oxide, $\text{MgO}$. $\text{Mg(OH)}_2$ is sparingly soluble and acts as a weak base, so the pH of the solution will be alkaline, typically between 8 and 12.
2. Row 2: The oxide is $\text{P}_4\text{O}_{10}$. This is a strongly acidic oxide. It reacts with water to form phosphoric acid: $\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$. The product is $\text{H}_3\text{PO}_4$. Phosphoric acid is a weak acid, but at typical concentrations formed, the pH will be acidic, between 1 and 4.

Key Takeaways

Basic oxides (left side of Period 3) form alkaline solutions with water. Acidic oxides (right side) form acidic solutions. The pH ranges reflect the strength of the resulting base or acid.

Common Mistakes

• Writing $\text{MgO}_2$ or other incorrect formulas for the oxide.
• Giving a pH of exactly 14 for $\text{Mg(OH)}_2$ (it is sparingly soluble, so it's a weak base, not a strong base like NaOH).
• Writing $\text{HPO}_4$ or incorrect acid formulas for the phosphorus product.

Things to Be Careful About

The mark scheme accepts a range for pH (e.g., 8-12 and 1-4). Do not give a single exact value unless sure. Ensure the chemical formula for phosphoric acid is correct: $\text{H}_3\text{PO}_4$.

Background Concept

Nitriles ($\text{R-C}\equiv\text{N}$) can be hydrolysed in the presence of water and an acid catalyst (or base). Acidic hydrolysis of a nitrile produces a carboxylic acid and an ammonium ion. The reaction requires 2 molecules of water. The $\text{H}^+$ acts as a catalyst and is consumed to form $\text{NH}_4^+$.

Understanding the Question

Write the balanced equation for the reaction of propanenitrile ($\text{CH}_3\text{CN}$) with water and hydrogen ions. The template is given: $\text{CH}_3\text{CN} + \dots\text{H}_2\text{O} + \dots\text{H}^+ \rightarrow \dots$

Approach

Identify the products: carboxylic acid ($\text{CH}_3\text{COOH}$) and ammonium ion ($\text{NH}_4^+$). Balance the oxygen and hydrogen atoms. 2 $\text{H}_2\text{O}$ are needed to provide the oxygen for the carboxylic acid and the hydrogens for the ammonium ion. 1 $\text{H}^+$ is needed to balance the charge and provide the final hydrogen for $\text{NH}_4^+$.

Step-by-Step Reasoning

1. Reactants: $\text{CH}_3\text{CN}$, $\text{H}_2\text{O}$, $\text{H}^+$.
2. Products: The $\text{C}\equiv\text{N}$ group is converted to a $\text{COOH}$ group (carboxylic acid) and the nitrogen becomes $\text{NH}_4^+$ (ammonium ion). So products are $\text{CH}_3\text{COOH}$ and $\text{NH}_4^+$.
3. Balancing: Left side has 1 C (from CN) + 1 O (from water) -> need 2 O for COOH, so 2 $\text{H}_2\text{O}$. Left side H: 3 (from CH3) + 4 (from 2H2O) + 1 (from H+) = 8 H. Right side H: 3 (from CH3) + 1 (from COOH) + 4 (from NH4+) = 8 H. Balanced.
4. Final equation: $\text{CH}_3\text{CN} + 2\text{H}_2\text{O} + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} + \text{NH}_4^+$.

Key Takeaways

Acidic hydrolysis of nitriles yields carboxylic acids and ammonium ions. Basic hydrolysis yields carboxylate ions and ammonia gas. Always balance carefully; 2 waters are needed.

Common Mistakes

• Writing $\text{NH}_3$ instead of $\text{NH}_4^+$ (in acidic conditions, ammonia is protonated to ammonium).
• Forgetting the $\text{H}^+$ on the left or getting the coefficient of water wrong (must be 2).
• Writing $\text{HCOOH}$ instead of $\text{CH}_3\text{COOH}$ (forgetting the methyl group).

Things to Be Careful About

The question provides the template with dots for coefficients. Ensure you fill them correctly: 2 for $\text{H}_2\text{O}$, 1 (implicit) for $\text{H}^+$.

Background Concept

The reaction shown is the hydration of propene ($\text{C}_3\text{H}_6$) with steam ($\text{H}_2\text{O}$) over a phosphoric acid ($\text{H}_3\text{PO}_4$) catalyst. This is an electrophilic addition reaction. The double bond in propene can be attacked by the electrophile ($\text{H}^+$) at either carbon, leading to two possible carbocation intermediates and thus two different alcohol products: propan-1-ol (primary) and propan-2-ol (secondary). The major product is propan-2-ol due to the greater stability of the secondary carbocation (Markovnikov's rule), but the question asks for the structures of both alcohols formed.

Understanding the Question

Draw the structures of the two alcohols ($\text{C}_3\text{H}_8\text{O}$) formed from the reaction of $\text{C}_3\text{H}_6$ with $\text{H}_2\text{O}$.

Approach

Propene is $\text{CH}_2=\text{CH}-\text{CH}_3$. Adding $\text{H}_2\text{O}$ across the double bond gives either $\text{CH}_3-\text{CH}_2-\text{CH}_2\text{OH}$ (propan-1-ol) or $\text{CH}_3-\text{CH}(\text{OH})-\text{CH}_3$ (propan-2-ol). Draw both skeletal or displayed structures.

Step-by-Step Reasoning

1. Identify the reactant: $\text{C}_3\text{H}_6$ is propene, $\text{CH}_3\text{CH}=\text{CH}_2$.
2. Addition of water: Water adds across the C=C bond. The OH group can attach to carbon 1 or carbon 2.
3. Isomer 1 (anti-Markovnikov/primary): OH on C1: $\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ (propan-1-ol).
4. Isomer 2 (Markovnikov/secondary): OH on C2: $\text{CH}_3\text{CH}(\text{OH})\text{CH}_3$ (propan-2-ol).
5. Draw structures: Draw skeletal structures showing the 3-carbon chain with the OH group on the end carbon for propan-1-ol, and on the middle carbon for propan-2-ol.

Key Takeaways

Hydration of unsymmetrical alkenes produces a mixture of alcohol isomers. Both must be drawn if asked for 'the two alcohols'. Propan-2-ol is the major product, but propan-1-ol is also formed.

Common Mistakes

• Drawing only one alcohol (usually the major product, propan-2-ol).
• Drawing incorrect carbon chains (e.g., 4 carbons).
• Forgetting the OH group or drawing an ether.

Things to Be Careful About

The question asks for 'structures', so skeletal or displayed formulas are acceptable. Ensure the molecular formula is $\text{C}_3\text{H}_8\text{O}$ (saturated alcohol).

Background Concept

Acidity is the ability to donate a proton ($\text{H}^+$). For alcohols ($\text{R-OH}$) and water ($\text{H-OH}$), the acidity depends on the ease of breaking the O-H bond and the stability of the resulting anion (alkoxide $\text{RO}^-$ vs hydroxide $\text{OH}^-$). Alkyl groups are electron-donating groups. They push electron density towards the oxygen atom via the sigma bonds (positive inductive effect).

Understanding the Question

Explain why alcohols are less acidic than water. This is a conceptual explanation requiring the use of the inductive effect.

Approach

Compare the alkyl group in alcohols to the hydrogen in water. The alkyl group donates electron density to oxygen. This affects the O-H bond.

Step-by-Step Reasoning

1. Identify the difference: Alcohols have an alkyl group (R) attached to the oxygen, while water has a hydrogen atom.
2. Inductive effect: Alkyl groups are electron-donating (positive inductive effect). They push electron density towards the electronegative oxygen atom.
3. Effect on O-H bond: This increased electron density on oxygen strengthens the O-H bond (makes it less polar, or reduces the partial positive charge on the hydrogen atom).
4. Result on acidity: A stronger O-H bond means it is harder to break, and the hydrogen is less likely to be donated as $\text{H}^+$. Therefore, alcohols are weaker acids than water.

Key Takeaways

Electron-donating groups (like alkyls) decrease acidity by strengthening the bond to the acidic proton. Electron-withdrawing groups (like halogens) increase acidity by weakening the bond and stabilizing the conjugate base.

Common Mistakes

• Saying 'alkyl groups are electron-withdrawing' (wrong direction of inductive effect).
• Saying 'alkyl groups make the O-H bond weaker' (opposite effect).
• Not mentioning the inductive effect or electron donation explicitly.

Things to Be Careful About

Use precise terms: 'electron-donating', 'positive inductive effect', 'strengthens the O-H bond', 'less likely to donate H+'. Do not just say 'it makes it less acidic' without explaining why.

Background Concept

Boiling points of molecular substances depend on the strength of intermolecular forces. For Group 16 hydrides ($\text{H}_2\text{S}$, $\text{H}_2\text{Se}$, $\text{H}_2\text{Te}$), the primary intermolecular forces are instantaneous dipole–induced dipole forces (also called London dispersion forces or van der Waals forces). The strength of these forces increases with the number of electrons in the molecule (or the size of the electron cloud), as larger electron clouds are more polarisable.

Understanding the Question

Explain the trend in boiling points from $\text{H}_2\text{S}$ to $\text{H}_2\text{Te}$ as shown in the bar chart (boiling point increases).

Approach

State that the number of electrons increases down the group. Explain that more electrons lead to stronger London dispersion forces. Conclude that more energy is needed to overcome these forces.

Step-by-Step Reasoning

1. Trend observation: Boiling point increases from $\text{H}_2\text{S}$ (213 K) to $\text{H}_2\text{Te}$ (271 K).
2. Electron count: Moving down Group 16 from S to Te, the atoms get larger and have more electrons. $\text{H}_2\text{S}$ (18 e-), $\text{H}_2\text{Se}$ (34 e-), $\text{H}_2\text{Te}$ (54 e-).
3. Polarisability: Larger electron clouds are more easily distorted (more polarisable), leading to stronger temporary (instantaneous) dipoles.
4. Intermolecular forces: Stronger instantaneous dipole–induced dipole (London) forces exist between the molecules.
5. Energy required: More thermal energy (higher temperature) is required to overcome these stronger intermolecular forces, hence the higher boiling point.

Key Takeaways

For molecules without hydrogen bonding, boiling point increases down a group due to increasing London dispersion forces from greater electron count and polarisability.

Common Mistakes

• Saying 'molecular mass increases' without linking it to electrons/polarisability (mark scheme prefers electrons/dispersion forces).
• Calling them 'van der Waals forces' without specifying 'instantaneous dipole-induced dipole' or 'London dispersion' (though 'van der Waals' is often accepted, be precise).
• Forgetting to mention that more energy is required to overcome the forces.

Things to Be Careful About

Use the term 'instantaneous dipole–induced dipole' or 'London dispersion forces'. 'Van der Waals' is a broader term that can include dipole-dipole, so be specific.

Background Concept

Water ($\text{H}_2\text{O}$) has an anomalously high boiling point compared to the other Group 16 hydrides ($\text{H}_2\text{S}$, $\text{H}_2\text{Se}$, $\text{H}_2\text{Te}$). This is because oxygen is highly electronegative and small, allowing it to form strong hydrogen bonds with the hydrogen atoms of neighbouring water molecules. Hydrogen bonding is a particularly strong type of dipole-dipole interaction.

Understanding the Question

Explain why the boiling point of $\text{H}_2\text{O}$ (373 K) is much higher than that of $\text{H}_2\text{S}$ (213 K), despite $\text{H}_2\text{S}$ having a higher molecular mass.

Approach

Identify that water has hydrogen bonding, which the others lack. State that hydrogen bonds are stronger than the London dispersion forces in the other hydrides.

Step-by-Step Reasoning

1. Anomaly observation: $\text{H}_2\text{O}$ has a much higher boiling point than expected from the trend of $\text{H}_2\text{S}$ to $\text{H}_2\text{Te}$.
2. Hydrogen bonding: Oxygen is highly electronegative (3.5) and has lone pairs. This allows water molecules to form hydrogen bonds (O-H...O) with each other. Sulfur is less electronegative (2.5) and larger, so $\text{H}_2\text{S}$ does not form significant hydrogen bonds.
3. Strength comparison: Hydrogen bonds are significantly stronger (typically 10-40 kJ/mol) than the instantaneous dipole–induced dipole (London dispersion) forces that are the main intermolecular forces in $\text{H}_2\text{S}$, $\text{H}_2\text{Se}$, and $\text{H}_2\text{Te}$ (typically 1-10 kJ/mol).
4. Energy required: Much more thermal energy is required to break the extensive hydrogen bonding network in water, resulting in a much higher boiling point.

Key Takeaways

Hydrogen bonding occurs when H is bonded to N, O, or F. It causes anomalously high boiling points, melting points, and solubilities. Water is a classic example due to its ability to form up to 4 hydrogen bonds per molecule.

Common Mistakes

• Saying 'water has stronger van der Waals forces' (vague and incorrect; it's specifically hydrogen bonding).
• Not mentioning that hydrogen bonds are stronger than the forces in the other molecules.
• Forgetting to state that only water (or H2O) has hydrogen bonding in this series.

Things to Be Careful About

Be explicit: 'hydrogen bonding' is the key phrase. Do not just say 'dipole-dipole'; hydrogen bonding is a specific, stronger subset. Mention that it is 'much stronger' to explain the large difference in boiling points.

Background Concept

Nitrogen oxides ($\text{NO}_x$) are gases containing nitrogen and oxygen, most commonly $\text{NO}$ and $\text{NO}_2$. They are significant atmospheric pollutants involved in the formation of acid rain, photochemical smog, and the depletion of the ozone layer. Their sources are broadly categorised into natural processes and anthropogenic (man-made) activities.

Understanding the Question

The question asks for one example each of a natural and a man-made source of $\text{NO}_x$ in the atmosphere. This is a direct recall question testing knowledge of the environmental chemistry of nitrogen.

Approach

Recall the primary mechanisms by which nitrogen and oxygen combine in the atmosphere. Naturally, the extreme heat of lightning strikes provides the activation energy for $\text{N}_2$ and $\text{O}_2$ to react. Man-made sources primarily involve high-temperature combustion processes where air is used as the oxidant.

Step-by-Step Reasoning

1. Natural Source: During a lightning strike, the temperature in the air channel can exceed $30,000\text{ K}$. This immense energy breaks the strong triple bond in $\text{N}_2$ and the double bond in $\text{O}_2$, allowing nitrogen atoms to combine with oxygen atoms to form $\text{NO}$, which subsequently oxidises to $\text{NO}_2$ in the atmosphere.
2. Man-made Source: In internal combustion engines (petrol and diesel), air is drawn in and compressed. The high temperatures and pressures inside the cylinder cause atmospheric nitrogen and oxygen to react, forming $\text{NO}$ and $\text{NO}_2$. These are expelled in the exhaust gases. Other man-made sources include power stations and industrial furnaces.

Key Takeaways

• Lightning is the major natural source of atmospheric $\text{NO}_x$.
• Internal combustion engines are the major man-made source of atmospheric $\text{NO}_x$.
• Both sources rely on high temperatures to overcome the activation energy for the reaction between $\text{N}_2$ and $\text{O}_2$.

Common Mistakes

• Confusing sources of $\text{NO}_x$ with sources of $\text{SO}_2$ (volcanic eruptions, burning fossil fuels containing sulfur).
• Stating "factories" or "power stations" without specifying the combustion process, though these are generally accepted, "internal combustion engines" is the most precise and common answer expected.
• Stating "volcanic eruptions" as a source of $\text{NO}_x$; while they emit gases, they are primarily associated with $\text{SO}_2$.

Things to Be Careful About

• Ensure you provide one of each type (natural and man-made) as requested.
• Use the term "internal combustion engines" rather than just "cars" or "vehicles" for a more scientific answer.

Background Concept

Acid rain is caused by the dissolution of acidic oxides, primarily sulfur dioxide ($\text{SO}_2$) and nitrogen oxides ($\text{NO}_x$), in atmospheric water. Nitrogen dioxide ($\text{NO}_2$) is a particularly important contributor. When $\text{NO}_2$ reacts with water, it undergoes disproportionation (oxidation states of N change from +4 to +3 and +5) to form nitrous acid ($\text{HNO}_2$) and nitric acid ($\text{HNO}_3$). In the presence of oxygen, the $\text{HNO}_2$ can be further oxidised to $\text{HNO}_3$, or the overall reaction can be written to produce only nitric acid.

Understanding the Question

The question asks for an equation describing the role of $\text{NO}_2$ in the direct formation of acid rain. This means writing the reaction between $\text{NO}_2$ and water that produces acids.

Approach

Identify the reactants ($\text{NO}_2$ and $\text{H}_2\text{O}$) and the products ($\text{HNO}_2$ and $\text{HNO}_3$, or just $\text{HNO}_3$ if $\text{O}_2$ is included). Balance the equation ensuring conservation of mass and charge.

Step-by-Step Reasoning

1. Disproportionation in water: $\text{NO}_2$ reacts with water. Nitrogen is in the +4 oxidation state. It disproportionates to +3 (in $\text{HNO}_2$) and +5 (in $\text{HNO}_3$). $$2\text{NO}_2 + \text{H}_2\text{O} \rightarrow \text{HNO}_2 + \text{HNO}_3$$
2. Oxidation to nitric acid: Alternatively, in the presence of atmospheric oxygen, $\text{NO}_2$ can react to form only nitric acid. $$4\text{NO}_2 + \text{O}_2 + 2\text{H}_2\text{O} \rightarrow 4\text{HNO}_3$$ Both equations are accepted by the mark scheme. The first is the direct hydrolysis; the second represents the overall process in oxygen-rich air.

Key Takeaways

• $\text{NO}_2$ is an acidic oxide.
• Its reaction with water produces acids, lowering the pH of rainwater.
• $\text{NO}_2$ can undergo disproportionation.

Common Mistakes

• Writing the reaction of $\text{NO}$ with water (which does not occur directly to form acid).
• Forgetting to balance the equation.
• Writing $\text{HNO}_2$ as $\text{H}_2\text{NO}_2$ or incorrect formulas for the acids.
• Including state symbols incorrectly (though often ignored unless specified, aqueous/gas phases are implied).

Things to Be Careful About

• Ensure the equation is balanced.
• Do not confuse this with the formation of $\text{NO}_2$ from $\text{NO}$ ($2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2$), which is a step before acid rain formation.
• The mark scheme accepts either the disproportionation equation or the oxidation equation.

Background Concept

Photochemical smog is a mixture of pollutants formed when sunlight acts on nitrogen oxides and volatile organic compounds (VOCs) in the atmosphere. Peroxyacetyl nitrate (PAN) is a key component and a powerful lachrymator (tear-inducing). Its formation involves a complex series of radical reactions initiated by sunlight, where $\text{NO}_2$ plays a crucial role in oxidising hydrocarbons.

Understanding the Question

The question asks for a description of how PAN forms from $\text{NO}_2$. It is looking for the other reactant(s) required for this process.

Approach

Recall the definition and formation conditions of photochemical smog. Identify the two main precursor groups: nitrogen oxides and hydrocarbons.

Step-by-Step Reasoning

1. PAN is a secondary pollutant.
2. It forms in the atmosphere through photochemical reactions.
3. The essential reactants are nitrogen dioxide ($\text{NO}_2$) and unburned hydrocarbons (VOCs) emitted from vehicle exhausts and industrial solvents.
4. Sunlight provides the energy for the reaction.

Key Takeaways

• PAN formation requires $\text{NO}_2$, hydrocarbons (VOCs), and sunlight.
• It is a component of photochemical smog.

Common Mistakes

• Stating that PAN forms from $\text{NO}$ instead of $\text{NO}_2$.
• Forgetting to mention hydrocarbons or VOCs.
• Confusing PAN with smog caused by sulfur dioxide (London-type smog).

Things to Be Careful About

• The mark scheme specifically looks for the reaction with "unburned hydrocarbons" or "VOCs". Simply saying "pollutants" is too vague.

Background Concept

Nitric acid ($\text{HNO}_3$) is a strong acid. Calcium oxide ($\text{CaO}$) is a basic oxide. The reaction between an acid and a base (or basic oxide) is a neutralisation reaction, producing a salt and water. The salt formed is a nitrate. Since calcium is in Group 2, it forms a $2+$ ion, so the formula of calcium nitrate is $\text{Ca(NO}_3)_2$.

Understanding the Question

Write the balanced chemical equation for the reaction between nitric acid and calcium oxide.

Approach

1. Identify reactants: $\text{HNO}_3$ and $\text{CaO}$.
2. Identify products: Salt ($\text{Ca(NO}_3)_2$) and water ($\text{H}_2\text{O}$).
3. Write the skeleton equation.
4. Balance the atoms.

Step-by-Step Reasoning

1. Skeleton: $\text{HNO}_3 + \text{CaO} \rightarrow \text{Ca(NO}_3)_2 + \text{H}_2\text{O}$
2. Balance nitrate ions: There are 2 $\text{NO}_3$ groups on the right, so we need 2 $\text{HNO}_3$ on the left.
   $2\text{HNO}_3 + \text{CaO} \rightarrow \text{Ca(NO}_3)_2 + \text{H}_2\text{O}$
3. Balance hydrogen: 2 H on left, 2 H on right (in water). Balanced.
4. Balance oxygen (excluding nitrate): 1 O in $\text{CaO}$, 1 O in $\text{H}_2\text{O}$. Balanced.
5. Balance calcium: 1 Ca on each side. Balanced.

Key Takeaways

• Acid + Metal Oxide -> Salt + Water.
• Correct formula of calcium nitrate is $\text{Ca(NO}_3)_2$.

Common Mistakes

• Writing the formula of calcium nitrate as $\text{CaNO}_3$ (ignoring the 2+ charge on Ca).
• Failing to balance the equation.
• Writing $\text{Ca(OH)}_2$ as a reactant instead of $\text{CaO}$ (the question specifies calcium oxide).

Things to Be Careful About

• Ensure brackets are used correctly in the formula $\text{Ca(NO}_3)_2$.
• State symbols are not explicitly required by the mark scheme text provided, but are good practice.

Background Concept

Most metal nitrates (except those of alkali metals) decompose on strong heating to form the metal oxide, nitrogen dioxide, and oxygen. Nitrogen dioxide ($\text{NO}_2$) is a brown gas. Calcium nitrate decomposes according to the equation:

Understanding the Question

Describe what is seen when solid calcium nitrate is heated strongly. This asks for a qualitative observation.

Approach

Identify the gaseous products of the decomposition and their physical appearance.

Step-by-Step Reasoning

1. The decomposition produces $\text{NO}_2$ and $\text{O}_2$.
2. Oxygen is a colourless gas.
3. Nitrogen dioxide is a brown gas.
4. Therefore, the visible observation is the evolution of brown fumes.

Key Takeaways

• Thermal decomposition of Group 2 nitrates produces $\text{NO}_2$.
• $\text{NO}_2$ is brown.
• Observation questions require describing what is seen, not naming the products.

Common Mistakes

• Stating "nitrogen dioxide is produced" (this is an inference/conclusion, not an observation).
• Stating "gas is evolved" (too vague; must specify colour).
• Confusing with the decomposition of carbonates (which releases $\text{CO}_2$, a colourless gas).

Things to Be Careful About

• Use the term "fumes" or "gas" with the colour "brown".
• Do not mention the solid residue unless asked, as the change from white solid to white solid is not a distinctive observation.

Background Concept

The oxidation state (or oxidation number) is a measure of the degree of oxidation of an atom in a substance. For a polyatomic ion, the sum of the oxidation states of all atoms must equal the overall charge of the ion. Oxygen generally has an oxidation state of -2.

Understanding the Question

Deduce the oxidation state of nitrogen in the nitrate ion, $\text{NO}_3^-$.

Approach

Set up an algebraic equation based on the known oxidation state of oxygen and the overall charge of the ion.

Step-by-Step Reasoning

1. Let $x$ be the oxidation state of nitrogen.
2. The oxidation state of oxygen is -2.
3. There are 3 oxygen atoms.
4. The overall charge of the nitrate ion is -1.
5. Equation: $x + 3(-2) = -1$
6. Solve: $x - 6 = -1 \Rightarrow x = +5$

Key Takeaways

• Sum of oxidation states in an ion equals the ion's charge.
• Oxygen is usually -2.
• Nitrogen in nitrates is in the +5 state.

Common Mistakes

• Forgetting the overall charge of the ion and setting the sum to 0.
• Incorrectly assigning the oxidation state of oxygen.

Things to Be Careful About

• Include the sign (+) in the answer.

Background Concept

Oxidation is defined as an increase in oxidation number (loss of electrons). The species that is oxidised acts as the reducing agent. We need to track the oxidation numbers of the elements in the reactants and products to identify which one increases.

Understanding the Question

Identify the species that is oxidised in the given equation:

Approach

Compare the oxidation states of the key elements (N, Al, O, H) on both sides of the equation.

Step-by-Step Reasoning

1. Nitrogen: In $\text{NO}_3^-$, N is +5. In $\text{NH}_3$, H is +1, so N is -3. The oxidation number decreases (+5 to -3), so nitrogen is reduced.
2. Aluminium: In elemental Al, the oxidation number is 0. In $[\text{Al}(\text{OH})_4]^-$, OH is -1, so Al is +3. The oxidation number increases (0 to +3), so aluminium is oxidised.
3. Oxygen/Hydrogen: Remain -2 and +1 respectively throughout.

Key Takeaways

• Oxidation = Increase in oxidation number.
• Elemental metals have an oxidation number of 0.
• Al is oxidised to $\text{Al}^{3+}$ (in the complex).

Common Mistakes

• Identifying the species reduced ($\text{NO}_3^-$) instead of the one oxidised.
• Saying "Nitrogen" instead of "Nitrate ion" or "Aluminium" instead of "Al".

Things to Be Careful About

• The question asks for the species, so "Al" or "Aluminium" is correct.

Background Concept

According to the Brønsted-Lowry theory, a base is defined as a proton ($\text{H}^+$) acceptor. Ammonia ($\text{NH}_3$) has a lone pair of electrons on the nitrogen atom, which allows it to form a coordinate bond with a proton, producing the ammonium ion ($\text{NH}_4^+$).

Understanding the Question

Describe how $\text{NH}_3$ is able to act as a base.

Approach

State the Brønsted-Lowry definition of a base and apply it to ammonia.

Step-by-Step Reasoning

1. Identify the definition: Base = Proton acceptor.
2. Apply to $\text{NH}_3$: It accepts an $\text{H}^+$ ion.
3. (Optional context): It uses its lone pair to bond with the proton.

Key Takeaways

• Brønsted-Lowry bases accept protons.
• Ammonia is a classic example of a weak base.

Common Mistakes

• Defining a base as an electron pair donor (Lewis definition) without mentioning protons, unless the question allows it. However, "H+ acceptor" is the specific answer required here.
• Saying "donates OH-" (Arrhenius definition, which is incorrect for ammonia gas).

Things to Be Careful About

• Use the term "proton" or "H+".

Background Concept

The shape of a complex ion can be predicted using VSEPR (Valence Shell Electron Pair Repulsion) theory. In $[\text{Al}(\text{OH})_4]^-$, the central aluminium ion ($\text{Al}^{3+}$) is bonded to four hydroxide ligands. There are 4 bonding pairs of electrons and 0 lone pairs on the central atom.

Understanding the Question

Suggest the shape of the $[\text{Al}(\text{OH})_4]^-$ ion.

Approach

Count the number of electron domains around the central atom.

Step-by-Step Reasoning

1. Central atom: Al.
2. Ligands: 4 OH groups.
3. Total electron pairs around Al: 4 bonding pairs.
4. Arrangement that minimises repulsion for 4 pairs: Tetrahedral.
5. Bond angle: $109.5^\circ$.

Key Takeaways

• 4 bonding pairs, 0 lone pairs = Tetrahedral.
• Many $\text{Al}^{3+}$ complexes are tetrahedral or octahedral depending on coordination number.

Common Mistakes

• Confusing with trigonal planar (3 pairs) or octahedral (6 pairs).
• Miscounting the number of ligands.

Things to Be Careful About

• Ensure the spelling is correct.

Background Concept

The $n$-th ionisation energy is the energy required to remove one electron from each ion in one mole of gaseous ions with a charge of $(n-1)+$. Therefore, the third ionisation energy involves removing an electron from a gaseous $2+$ ion to form a gaseous $3+$ ion.

Understanding the Question

Construct an equation to represent the third ionisation energy of phosphorus (P).

Approach

1. Start with the species resulting from the second ionisation: $\text{P}^{2+}(\text{g})$.
2. Remove one electron.
3. End with the resulting species: $\text{P}^{3+}(\text{g})$.
4. Include state symbols (g).

Step-by-Step Reasoning

1. Reactant: $\text{P}^{2+}(\text{g})$
2. Product: $\text{P}^{3+}(\text{g}) + \text{e}^-$
3. Equation: $\text{P}^{2+}(\text{g}) \rightarrow \text{P}^{3+}(\text{g}) + \text{e}^-$

Key Takeaways

• IE1: $\text{X} \rightarrow \text{X}^+ + \text{e}^-$
• IE2: $\text{X}^+ \rightarrow \text{X}^{2+} + \text{e}^-$
• IE3: $\text{X}^{2+} \rightarrow \text{X}^{3+} + \text{e}^-$
• State symbols are crucial.

Common Mistakes

• Starting with neutral P.
• Omitting state symbols.
• Writing the electron as $\text{e}$ instead of $\text{e}^-$.

Things to Be Careful About

• Ensure the charge on the reactant is 2+ and the product is 3+.

Background Concept

Successive ionisation energies always increase because it becomes harder to remove an electron from an increasingly positive ion. However, there are "jumps" in magnitude when an electron is removed from a new, inner principal shell closer to the nucleus. Phosphorus (atomic number 15) has the electron configuration $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3$. The 3rd shell contains 5 electrons. Therefore, the first 5 ionisations remove electrons from the 3rd shell. The 6th ionisation removes an electron from the 2nd shell, resulting in a very large increase in energy.

Understanding the Question

Complete the graph for the 3rd to 6th ionisation energies of P.

Approach

1. Determine the electron configuration of P.
2. Identify the shell boundaries.
3. Plot the points: 3rd, 4th, 5th should be relatively low and increasing gradually (same shell). 6th should be much higher (new shell).

Step-by-Step Reasoning

1. Configuration: $1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^3$. Outer shell (n=3) has 5 electrons.
2. IE1-IE5: Removing electrons from the 3rd shell. Energy increases gradually.
3. IE6: Removing an electron from the 2nd shell (n=2). This is much closer to the nucleus and experiences greater effective nuclear charge. Huge jump in energy.
4. Graphing:
   • 3rd: Higher than 2nd.
   • 4th: Higher than 3rd.
   • 5th: Higher than 4th.
   • 6th: Significantly higher than 5th (the jump).
   • 7th-10th: Given on the graph, they are high and increasing gradually (removing from n=2).
   • The plotted points for 3, 4, 5 should follow the trend of the 1st and 2nd points (low, gradual increase). The 6th point should be high, comparable to or slightly lower than the 7th point shown on the graph.

Key Takeaways

• Large jumps in IE indicate a change in principal quantum number (shell).
• Number of electrons in outer shell = number of IEs before the first major jump.
• P has 5 valence electrons -> jump after IE5.

Common Mistakes

• Plotting the jump after IE3 (confusing with Group 3 elements).
• Not making the 6th IE high enough.
• Plotting the 3rd, 4th, 5th IEs too high.

Things to Be Careful About

• The relative height of the 6th point is the key discriminator for the second mark. It must show the large gap between the 5th and 6th ionisations.

Background Concept

A triple bond consists of one sigma ($\sigma$) bond and two pi ($\pi$) bonds. The sigma bond is formed by the end-to-end (head-on) overlap of atomic orbitals along the internuclear axis. The pi bonds are formed by the lateral (side-on) overlap of unhybridised p orbitals above and below the internuclear axis.

Understanding the Question

Describe the formation of the $\text{P}\equiv\text{P}$ bond in terms of orbital overlap.

Approach

Break down the triple bond into its components (1 sigma, 2 pi) and describe the overlap type for each.

Step-by-Step Reasoning

1. Sigma bond: Formed by the head-on overlap of orbitals (typically p-orbitals or hybrid orbitals) along the axis connecting the two nuclei.
2. Pi bonds: Formed by the side-on overlap of parallel p-orbitals. There are two such pairs of orbitals (e.g., $p_y$ and $p_z$), resulting in two pi bonds.