Chemistry 9701/23 — October/November 2024

Background Concept

The shorthand (or abbreviated) electronic configuration uses the symbol of the preceding noble gas in square brackets to represent all the electrons in the inner shells. For elements in Period 4, [Ar] represents the 18 electrons of argon, which is the noble gas at the end of Period 3. The remaining electrons are then written explicitly.

Understanding the Question

The question states that cobalt's shorthand configuration is [Ar]3d⁷4s² and asks you to identify what [Ar] means by writing out argon's full electronic configuration. This is a direct recall task.

Approach

Argon has atomic number 18, so it has 18 electrons. Fill orbitals in order of increasing energy: 1s, 2s, 2p, 3s, 3p.

Step-by-Step Reasoning

• 1s holds 2 electrons → 1s²
• 2s holds 2 electrons → 2s²
• 2p holds 6 electrons → 2p⁶
• 3s holds 2 electrons → 3s²
• 3p holds 6 electrons → 3p⁶
• Total: 2 + 2 + 6 + 2 + 6 = 18 electrons ✓

Key Takeaways

Noble gas shorthand is a convention that saves writing out inner-shell electrons. Always know the full configuration of the noble gases up to krypton (Z = 36) for A-Level work.

Common Mistakes

• Writing 3d before 3p (3d is higher in energy than 3p for argon, so it is empty).
• Forgetting to include all subshells up to 3p.

Things to Be Careful About

The configuration must be written in the correct filling order and must total 18 electrons. Do not include any 3d electrons.

Background Concept

Atomic orbitals have characteristic shapes determined by the angular momentum quantum number $l$. For s orbitals ($l = 0$), the shape is spherical — the electron density is distributed equally in all directions around the nucleus. This applies to all s orbitals (1s, 2s, 3s, etc.), though higher s orbitals have radial nodes.

Understanding the Question

The question asks you to draw the shape of the 1s orbital. Since 1s is an s-type orbital, it is spherical. A simple circle (representing the 2D projection of a sphere) is sufficient.

Approach

Recall that all s orbitals are spherical. Draw a circle to represent the boundary of the 1s orbital.

Step-by-Step Reasoning

The 1s orbital has $l = 0$, giving it a spherically symmetric shape. In a 2D drawing, this is represented by a circle. No axes, labels, or additional detail are required — just the circular outline.

Key Takeaways

• s orbitals: spherical
• p orbitals: dumbbell-shaped (two lobes)
• d orbitals: cloverleaf (mostly)

Common Mistakes

• Drawing a dumbbell shape (that is a p orbital).
• Adding unnecessary detail such as axes or labels when only the shape is required.

Things to Be Careful About

The question asks for the shape only — a simple circle is the expected answer. Do not confuse the 1s orbital with a Bohr model circle (which represents a shell, not an orbital).

Background Concept

Hund's rule states that electrons occupy degenerate orbitals (orbitals of the same energy, such as the five 3d orbitals) singly before pairing begins. This maximises the number of unpaired electrons in the ground state, minimising electron-electron repulsion.

Understanding the Question

Cobalt has the configuration [Ar]3d⁷4s². The 4s electrons are paired (2 in one orbital), so all unpaired electrons must come from the 3d⁷ arrangement. You need to determine how many of the 7 d-electrons are unpaired.

Approach

Draw the five 3d orbitals as boxes. Fill 7 electrons following Hund's rule (singly first, then pair), and count the unpaired electrons.

Step-by-Step Reasoning

Five 3d orbitals: _ _ _ _ _

Place 7 electrons:

1. ↑ _ _ _ _ (1st electron)
2. ↑ ↑ _ _ _ (2nd electron)
3. ↑ ↑ ↑ _ _ (3rd electron)
4. ↑ ↑ ↑ ↑ _ (4th electron)
5. ↑ ↑ ↑ ↑ ↑ (5th electron — all singly occupied)
6. ↑↓ ↑ ↑ ↑ ↑ (6th electron pairs in first orbital)
7. ↑↓ ↑↓ ↑ ↑ ↑ (7th electron pairs in second orbital)

Orbitals 3, 4, and 5 each have one unpaired electron → 3 unpaired electrons.

Key Takeaways

For any dⁿ configuration (n = 1–9), the number of unpaired electrons follows: n ≤ 5 gives n unpaired; n > 5 gives (10 − n) unpaired. For d⁷: 10 − 7 = 3.

Common Mistakes

• Forgetting that the 4s² electrons are paired and counting them as unpaired.
• Pairing electrons before all orbitals have one electron (violating Hund's rule).
• Confusing the number of electrons with the number of unpaired electrons.

Things to Be Careful About

The question asks about a cobalt atom (neutral), not an ion. The configuration given is for the neutral atom, so use 3d⁷4s² directly.

Background Concept

Relative isotopic mass is the mass of a specific isotope expressed on a scale where one atom of carbon-12 has a mass of exactly 12 units. It is dimensionless (a ratio). This differs from relative atomic mass ($A_r$), which is the weighted average of all isotopes of an element.

Understanding the Question

You must give a precise definition of relative isotopic mass worth 2 marks. The mark scheme requires two key ideas: (1) mass of an atom of an isotope, and (2) comparison to 1/12 of the mass of a carbon-12 atom.

Approach

State both components of the definition clearly.

Step-by-Step Reasoning

• Component 1: It refers to a specific isotope (not an average), so say 'mass of an atom of an isotope'.
• Component 2: The reference standard is 1/12 of the mass of one atom of carbon-12. This is the unified atomic mass unit (amu).

Key Takeaways

• Relative isotopic mass → one specific isotope.
• Relative atomic mass → weighted average over all isotopes.
• Both are dimensionless quantities referenced to carbon-12.

Common Mistakes

• Saying 'compared to hydrogen' or 'compared to oxygen' (outdated standards).
• Omitting 'of an isotope' and giving the definition of relative atomic mass instead.
• Saying 'mass of 1/12 of carbon-12' without specifying 'an atom of'.

Things to Be Careful About

The phrase 'compared to' or 'relative to' must be present — it is a ratio, not an absolute mass. The mark scheme also accepts 'on a scale on which a carbon-12 atom has a mass of exactly 12 units' or 'divided by 1/12 the mass of a carbon-12 atom'.

Background Concept

The relative atomic mass of an element is the weighted average of the relative isotopic masses, weighted by their relative abundances. When an alloy contains multiple elements, you must isolate the isotopes of the element in question and renormalise the abundances to sum to 100% for that element alone.

Understanding the Question

Table 1.2 gives abundances for all three isotopes in the alloy (Rh and two Ir isotopes). The question asks specifically for the $A_r$ of iridium in the alloy, so only the two iridium isotopes are used. Their abundances (15.18 and 34.82) sum to 50.00, which becomes the denominator.

Approach

1. Identify the iridium isotopes: $^{191}\text{Ir}$ (mass 190.96, abundance 15.18) and $^{193}\text{Ir}$ (mass 192.96, abundance 34.82).
2. Apply the weighted mean formula using only these two.
3. Divide by the sum of their abundances (50.00).

Step-by-Step Reasoning

Numerator: $(190.96 \times 15.18) + (192.96 \times 34.82) = 2898.77 + 6718.87 = 9617.64$

Denominator: $15.18 + 34.82 = 50.00$

$A_r = 9617.64 / 50.00 = 192.3528 ≈ 192.35$ (to 2 d.p.)

Key Takeaways

When calculating $A_r$ from alloy data, use only the isotopes of the element being asked about and divide by the sum of their abundances (not by 100).

Common Mistakes

• Including the rhodium isotope in the calculation (the question asks for iridium's $A_r$ only).
• Dividing by 100 instead of 50.00.
• Rounding to the wrong number of decimal places.

Things to Be Careful About

The answer must be given to two decimal places as specified. The mark scheme requires 192.35 (cao).

Background Concept

In a hydrated salt $\text{MX}_n \cdot x\text{H}_2\text{O}$, the water molecules are incorporated into the crystal lattice. The percentage by mass of water is $\frac{18x}{M_r(\text{anhydrous}) + 18x} \times 100\%$. Setting this equal to the given percentage allows $x$ to be solved algebraically.

Understanding the Question

You are told that $\text{RhCl}_3 \cdot x\text{H}_2\text{O}$ contains 20.5% water by mass. You must find the integer value of $x$. You need the $M_r$ of anhydrous $\text{RhCl}_3$ (using $A_r(\text{Rh}) = 102.9$ from the table, and $A_r(\text{Cl}) = 35.5$).

Approach

1. Calculate $M_r$ of $\text{RhCl}_3$.
2. Set up the equation: mass of water / total mass = 0.205.
3. Solve for $x$.

Step-by-Step Reasoning

$M_r(\text{RhCl}_3) = 102.9 + 3 \times 35.5 = 102.9 + 106.5 = 209.4$

Mass of water in formula = $18x$
Total $M_r$ of hydrate = $209.4 + 18x$

$\frac{18x}{209.4 + 18x} = 0.205$

Cross-multiply: $18x = 0.205(209.4 + 18x) = 42.927 + 3.69x$

$18x - 3.69x = 42.927$

$14.31x = 42.927$

$x = 42.927 / 14.31 = 3.0$

Key Takeaways

The water of crystallisation method always involves setting up a proportion between the mass contribution of water and the total formula mass, then solving for $x$.

Common Mistakes

• Using 102.91 (the relative isotopic mass from the table) instead of 102.9 (the $A_r$) — both give $x = 3$ here, but the mark scheme uses 102.9.
• Forgetting to include the water mass in the denominator (total mass of hydrate, not just anhydrous mass).
• Getting the algebra wrong when cross-multiplying.

Things to Be Careful About

The answer must be an integer (as stated in the question). The mark scheme accepts alternative routes (e.g., using the ratio of moles of anhydrous to water) but all must give $x = 3$.

Background Concept

Isomers are compounds with the same molecular formula but different arrangements of atoms. Structural isomers differ in the connectivity (which atoms are bonded to which). Stereoisomers have the same connectivity but differ in the spatial arrangement — this includes geometric (cis/trans or E/Z) isomers and optical isomers (enantiomers).

Understanding the Question

You must give a precise definition of stereoisomers worth 1 mark. The key distinction from structural isomers is that the connectivity is the same but the 3D arrangement differs.

Approach

State both required elements: same structural formula AND different spatial/3D arrangement.

Step-by-Step Reasoning

The definition must include:

1. Same structural formula (same molecular formula and same bonded sequence).
2. Different arrangement in 3D space (different spatial orientation of atoms/groups).

Key Takeaways

Stereoisomerism encompasses both geometric (cis/trans) and optical (enantiomer) isomerism. The defining feature is identical connectivity with different 3D arrangement.

Common Mistakes

• Saying 'different structural formula' (that describes structural isomers, not stereoisomers).
• Omitting 'same structural formula' and only saying 'different arrangement in space'.
• Confusing stereoisomers with conformational isomers (which interconvert by rotation about single bonds).

Things to Be Careful About

The mark scheme requires both 'same structural formula' and 'different 3D/spatial arrangement' — omitting either half loses the mark.

Background Concept

An addition reaction is one in which two (or more) reactant molecules combine to form a single product. No atoms are lost — all atoms in the reactants appear in the product. This contrasts with elimination (one reactant gives two products) and substitution (one group replaces another, giving two products).

Understanding the Question

The conversion is: $2\text{C}_2\text{H}_4 \rightarrow \text{C}_4\text{H}_8$ (but-2-ene). You must explain why this qualifies as an addition reaction.

Approach

Identify the key feature: two reactant molecules → one product molecule. This is the hallmark of addition.

Step-by-Step Reasoning

• Reactants: two molecules of ethene (C₂H₄).
• Product: one molecule of but-2-ene (C₄H₈).
• All atoms from both reactant molecules are incorporated into the single product.
• No small molecule (like H₂O or HCl) is eliminated.
• Therefore it is an addition reaction: two molecules add together to form one.

Key Takeaways

The defining criterion for an addition reaction is that two (or more) reactants combine to give a single product. The C=C double bond 'opens up' to form new single bonds.

Common Mistakes

• Saying 'a double bond is broken' — this describes what happens mechanistically but is not the definition of addition (elimination also involves bond changes).
• Saying 'atoms are added' without specifying that a single product is formed.

Things to Be Careful About

The mark scheme's point is simply 'a single product is made' — the key is that two reactant molecules give one product. Keep the answer concise.

Background Concept

But-2-ene (CH₃CH=CHCH₃) exhibits geometric (cis/trans) isomerism because each carbon of the double bond carries two different groups (H and CH₃). The restricted rotation about the C=C bond locks the groups into fixed positions. In the cis isomer, the two methyl groups are on the same side of the double bond; in the trans isomer, they are on opposite sides.

Understanding the Question

You must draw both stereoisomers of but-2-ene. Each isomer is worth 1 mark, for a total of 2. The drawings must clearly show the correct spatial arrangement around the double bond.

Approach

Draw the C=C double bond horizontally (or at an angle) with the four substituents positioned correctly: two CH₃ groups and two H atoms. Show cis (same side) and trans (opposite sides).

Step-by-Step Reasoning

cis-but-2-ene: Both CH₃ groups on the same side of the C=C bond. The molecule is bent (kinked) at the double bond.

trans-but-2-ene: The two CH₃ groups on opposite sides of the C=C bond. The molecule is roughly linear/extended.

Either displayed or skeletal formulae are acceptable, provided the geometry is clearly shown.

Key Takeaways

Geometric isomerism requires: (1) a C=C double bond (or ring), and (2) two different groups on each carbon of the double bond. The cis/trans labels refer to the relative positions of the identical (or chosen priority) groups.

Common Mistakes

• Drawing the same isomer twice (e.g., both with methyls on the same side).
• Drawing but-1-ene instead of but-2-ene (but-1-ene does not show cis/trans isomerism because one carbon has two identical H atoms).
• Not showing the double bond clearly or not indicating the spatial arrangement.

Things to Be Careful About

The double bond must be drawn explicitly (not as a single line). The relative positions of the CH₃ groups must be unambiguous — use a clear angle at the double bond to distinguish cis from trans.

Background Concept

The halogens are diatomic molecules with increasing numbers of electrons down the group, so van der Waals forces between molecules strengthen and the states change from gas (Cl2) to liquid (Br2) to solid (I2). The colour intensifies down the group as the molecules absorb light of progressively lower energy.

Understanding the Question

'State' type recall: give the colour of each halogen from chlorine to iodine at room temperature.

Approach

Recall the standard colours and their physical states.

Step-by-Step Reasoning

Chlorine is a pale yellow-green gas. Bromine is a red-brown/orange volatile liquid giving brown vapour. Iodine is a dark grey/black lustrous solid (violet vapour on sublimation).

Key Takeaways

Colour and volatility both increase down Group 17; learn the colours with the states.

Common Mistakes

Writing 'iodine is purple' — that describes its vapour or its solution in an organic solvent; the solid is silver-grey/black. Vague answers like 'dark' score nothing.

Things to Be Careful About

The mark scheme accepts orange/brown/red for bromine and silver-grey/black for iodine; give a colour plus state for clarity.

Background Concept

A halogen acts as an oxidising agent by gaining electrons: X2 + 2e- -> 2X-. Down the group, atomic radius and shielding increase, so the attraction of the nucleus for an incoming electron weakens and the tendency to gain electrons falls.

Understanding the Question

Describe the relative reactivity of Cl2, Br2 and I2 as oxidising agents.

Approach

State the trend in one clear sentence.

Step-by-Step Reasoning

Oxidising power is highest for the smallest halogen: Cl2 > Br2 > I2. Hence chlorine displaces bromine and iodine from their salts, bromine displaces iodine, and iodine displaces none.

Key Takeaways

Oxidising power decreases down Group 17 — the reverse of Group 1/2 metal reactivity trends.

Common Mistakes

Saying reactivity increases down the group (confusing with metals). Only stating 'decreases' without direction (must say from chlorine to iodine).

Things to Be Careful About

Use 'oxidising strength/power decreases', not just 'reactivity decreases' — the question asks specifically about oxidising agents.

Background Concept

H2 + Cl2 -> 2HCl (gaseous reaction, often with UV light or a flame). Both reactants are coloured (chlorine pale green; hydrogen colourless) while the product HCl is colourless, so the visible change is the loss of the green colour.

Understanding the Question

'State what is observed' — give the visible change only, not the equation.

Approach

Think about which species are coloured before and after reaction.

Step-by-Step Reasoning

Chlorine gas is pale green; as it reacts with hydrogen to form colourless hydrogen chloride gas, the green colour disappears.

Key Takeaways

Observation questions need only what you would see, not chemical explanation.

Common Mistakes

Writing the equation instead of an observation. Saying 'white fumes' — that is seen when HCl meets moist air with ammonia or in some demonstrations, but the credited observation here is the disappearance of the green colour.

Things to Be Careful About

Do not add unobserved details; the mark is for 'green colour disappears'.

Background Concept

Thermal stability of HX depends on the strength of the H–X covalent bond. Down Group 17, the halogen atoms get larger; the H–X bond lengthens and the overlap between the H 1s orbital and the halogen valence orbital becomes poorer, so the bond energy falls (HF strongest, HI weakest).

Understanding the Question

'Explain why' — the observation (decreasing thermal stability) must be linked to a reason (bond strength), not merely stated.

Approach

State the bond-strength trend, then link it to decomposition on heating.

Step-by-Step Reasoning

A weaker H–X bond requires less energy to break, so the molecule splits into its elements more easily when heated. Since bond strength decreases HF > HCl > HBr > HI, thermal stability decreases in the same order.

Key Takeaways

Thermal stability of hydrides tracks bond strength, which decreases with increasing bond length down a group.

Common Mistakes

Explaining via reactivity/oxidising power of the halogen instead of bond strength. Saying 'larger atoms are less stable' without mentioning the bond.

Things to Be Careful About

The mark scheme credits 'H–Hal bond strength decreases'; include that exact idea.

Background Concept

Homolytic fission of a covalent bond splits it evenly, giving each fragment one electron from the shared pair. The resulting species, with an unpaired electron (shown as a dot, e.g. Cl•), is a free radical. Radicals are highly reactive.

Understanding the Question

'Define' — give the precise textbook definition.

Approach

Quote the definition: species + unpaired electron(s).

Step-by-Step Reasoning

The key phrase is 'one or more unpaired electrons'; the species may be an atom or molecule, usually neutral.

Key Takeaways

Radicals arise from homolytic fission and are denoted with a dot.

Common Mistakes

Saying 'an atom with an extra electron' (that is an anion idea) or 'a charged species'. Omitting 'unpaired'.

Things to Be Careful About

The mark scheme wording is 'a species with one or more unpaired electrons' — use it.

Background Concept

The initiation step Cl2 -> 2Cl• requires energy to homolytically cleave the Cl–Cl bond. UV light provides photons with enough energy for homolytic fission, generating the radicals that start the chain reaction.

Understanding the Question

'State the essential condition' for the reaction at room temperature — the answer is the radiation that initiates the radical chain.

Approach

Recall the standard condition for alkane + halogen free-radical substitution.

Step-by-Step Reasoning

Without UV (or high temperature), the Cl–Cl bond does not break homolytically at room temperature and no reaction occurs. UV light is the essential condition.

Key Takeaways

Alkane + halogen requires UV light (or high temperature) to proceed.

Common Mistakes

Writing 'sunlight' (too vague — say UV), or naming a catalyst.

Things to Be Careful About

The mark scheme accepts 'ultraviolet'; 'UV light' is fine.

Background Concept

A radical chain reaction has three stages: initiation (UV splits Cl2 into 2Cl•), propagation (radicals consumed and regenerated — two steps here), and termination (two radicals combine). Propagation steps must show a radical on both sides, keeping the chain going.

Understanding the Question

Write the two propagation equations for chlorination of ethane. Each is worth one mark.

Approach

Step 1: a chlorine radical abstracts H from ethane, forming HCl and the ethyl radical. Step 2: the ethyl radical attacks a Cl2 molecule, forming chloroethane and regenerating Cl•.

Step-by-Step Reasoning

First propagation: Cl• + C2H6 -> C2H5• + HCl — the Cl• takes an H atom; one electron of the C–H bond stays on the carbon, giving the ethyl radical. Second propagation: C2H5• + Cl2 -> C2H5Cl + Cl• — the ethyl radical forms a C–Cl bond, breaking the Cl–Cl bond homolytically so Cl• is regenerated. Adding the two propagation steps gives the overall equation C2H6 + Cl2 -> C2H5Cl + HCl.

Key Takeaways

Propagation steps: radical on both sides; the radical regenerated in step 2 is the one consumed in step 1.

Common Mistakes

Writing termination steps (e.g. Cl• + Cl• -> Cl2) instead of propagation. Forgetting the radical dot. Writing CH3CH2• inconsistently or unbalanced equations.

Things to Be Careful About

Show the radical dot (•) clearly on both equations; each equation is separately marked.

Background Concept

Chlorine disproportionates in alkali. In cold dilute NaOH: Cl2 + 2NaOH -> NaCl + NaClO + H2O, giving sodium chlorate(I) (bleach). In hot concentrated NaOH the product is NaClO3 (chlorate(V)) instead, so 'cold' is essential.

Understanding the Question

Identify the reagent and conditions converting Cl2 to NaClO.

Approach

Recall the cold aqueous alkali reaction.

Step-by-Step Reasoning

Cold dilute aqueous sodium hydroxide reacts with chlorine to give chloride and chlorate(I); the cold condition prevents further disproportionation to chlorate(V).

Key Takeaways

Cold NaOH -> NaClO; hot concentrated NaOH -> NaClO3.

Common Mistakes

Omitting 'cold' or 'aqueous'; writing NaOH(s).

Things to Be Careful About

The mark scheme credits 'cold NaOH(aq)' — include both the temperature and state.

Background Concept

In disproportionation one element in one oxidation state simultaneously goes up and down in oxidation number. For chlorine in NaOH: 0 in Cl2 becomes +1 in NaClO (oxidation) and -1 in NaCl (reduction).

Understanding the Question

'Define' — precise wording needed.

Approach

Key elements: same species, oxidised AND reduced, in the same reaction.

Step-by-Step Reasoning

The definition must convey that one species undergoes both processes at the same time; it is a special type of redox reaction.

Key Takeaways

Disproportionation = simultaneous oxidation and reduction of the same species.

Common Mistakes

Saying 'two species react' or omitting 'simultaneously/same reaction'.

Things to Be Careful About

Use 'both oxidised and reduced in the same reaction' — the mark scheme requires the simultaneity.

Background Concept

The haloform reaction: methyl ketones react with NaClO to give a trihalomethane (CHCl3) and the sodium carboxylate. Balancing is done by counting Cl, Na and O atoms on each side.

Understanding the Question

Fill in the coefficients in the given equation.

Approach

Count atoms element by element. CHCl3 needs 3 Cl atoms, so 3 NaClO are required. That supplies 3 Na: one goes to CH3COONa, leaving 2 Na for 2 NaOH. Check O: 3 (from NaClO) = 1 (carboxylate) + 2 (NaOH). Check H: 6 (propanone) = 1 (CHCl3) + 3 (acetate) + 2 (NaOH).

Step-by-Step Reasoning

• Cl balance: 1 Cl in CHCl3 → coefficient of NaClO = 3.
• Na balance: 3 Na from 3 NaClO = 1 (CH3COONa) + 2 (NaOH).
• O balance: 3 = 1 + 2 ✓. H and C balance also check out.

Key Takeaways

Balance by counting the element that appears in the fewest species first (here Cl), then distribute the counter-ion.

Common Mistakes

Putting 2 NaClO and failing the Cl balance; forgetting the NaOH coefficient.

Things to Be Careful About

Verify every element after choosing coefficients — one wrong count invalidates the mark.

Background Concept

The AgNO3/ethanol test for halogenoalkanes: the halogenoalkane is hydrolysed, releasing halide ions which precipitate with Ag+ as AgX. Colour identifies the halogen (AgCl white, AgBr cream, AgI yellow); a precipitate forms faster with weaker C–X bonds. Chloroform, CHCl3, hydrolyses slowly in aqueous conditions but does eventually release Cl-.

Understanding the Question

Two marks: the observation (white ppt) and the explanation (hydrolysis releases Cl-, which reacts with Ag+ to give AgCl).

Approach

State what is seen, then give the chemistry: hydrolysis → Cl- → AgCl(s).

Step-by-Step Reasoning

Ag+ from aqueous silver nitrate meets Cl- released by hydrolysis of CHCl3; AgCl is insoluble, appearing as a white precipitate. The ethanol solvent helps mix the organic and aqueous layers.

Key Takeaways

For halogenoalkane tests, the precipitate only appears after hydrolysis releases the halide ion — the explanation must include this step.

Common Mistakes

Giving only the observation without explanation; saying the C–Cl bond breaks directly with Ag+ without mentioning hydrolysis/Cl- release; wrong colour (cream/yellow are for Br/I).

Things to Be Careful About

The mark scheme M2 accepts: hydrolysis releases Cl-, or Cl- reacts with Ag+, or AgCl is formed — include at least one clearly.

Background Concept

The physical properties of an element in its standard state are direct consequences of its atomic structure and the type of bonding and giant structure it forms. Elements can form giant metallic lattices (delocalised electrons), giant covalent networks (shared electron pairs in a continuous lattice), or simple molecular structures (discrete molecules held by weak intermolecular forces). Graphite is a special case: it has a giant covalent structure where each carbon atom is bonded to three others in layers, leaving one delocalised electron per atom that can move freely, giving it electrical conductivity. Tin (Sn) is a post-transition metal with a giant metallic structure.

Understanding the Question

We are given a table with some properties of carbon (graphite) and tin (Sn) in their standard states. We need to fill in the missing entries for electrical conductivity, type of bonding, and type of structure for both elements.

Approach

Recall the standard state properties of graphite and tin. Graphite is a form of carbon with giant covalent bonding and good electrical conductivity. Tin is a metal with giant metallic bonding and good electrical conductivity.

Step-by-Step Reasoning

• Electrical conductivity of C (graphite): Graphite conducts electricity because each carbon atom is bonded to three others, leaving one delocalised electron per atom that is free to move through the layers. Answer: good or conductor.
• Type of bonding in C (graphature): The carbon atoms are held together by strong covalent bonds. Answer: covalent.
• Type of structure of Sn: Tin is a metal, so it forms a continuous lattice of positive ions in a sea of delocalised electrons. Answer: giant (specifically, giant metallic).

Key Takeaways

Graphite is a giant covalent structure that conducts electricity due to delocalised electrons. Metals like tin have giant metallic structures and are good conductors.

Common Mistakes

• Writing "ionic" for the bonding in graphite.
• Forgetting that graphite has a "giant" structure, not "simple molecular".

Things to Be Careful About

Ensure the terminology matches the mark scheme: "covalent" for bonding, "giant" for structure, and "good" or "conductor" for conductivity. Do not write "metallic" for carbon.

Background Concept

Lattice structures are classified by the type of particle and the forces holding them together: giant ionic (ions), giant metallic (metal cations and delocalised electrons), giant covalent (atoms in a continuous network), and simple molecular (discrete molecules). Graphite consists of layers of carbon atoms. Within each layer, the atoms are bonded covalently in a hexagonal lattice, but the layers themselves are held together by weak van der Waals forces. Because the structure is built from discrete molecular-like layers (often described as giant molecular or giant covalent with intermolecular forces between layers), it is classified as a giant molecular structure in many Cambridge mark schemes, though "giant covalent" is also accepted in broader contexts. The specific mark scheme here rewards "giant molecular".

Understanding the Question

We need to identify the specific lattice structure classification for graphite.

Approach

Recall that graphite is made of layers of carbon atoms. The layers are giant 2D networks, but the overall structure is often termed giant molecular due to the weak forces between layers.

Step-by-Step Reasoning

Graphite consists of layers of carbon atoms bonded covalently in a hexagonal arrangement. These layers are held together by weak intermolecular forces. The standard classification for this in this context is giant molecular.

Key Takeaways

Graphite is classified as having a giant molecular structure (or giant covalent with layers).

Common Mistakes

• Calling it "simple molecular" (it is not, as the layers are continuous).
• Calling it "giant ionic" or "giant metallic".

Things to Be Careful About

Use the exact terminology expected: "giant molecular".

Background Concept

In a metallic lattice, the metal atoms donate their outer electrons to a "sea" of delocalised electrons that are free to move throughout the structure. When a potential difference is applied, these delocalised electrons can drift towards the positive terminal, carrying charge and thus conducting electricity.

Understanding the Question

We need to explain why tin (Sn), a metal, has good electrical conductivity.

Approach

State the feature of metallic bonding that allows charge to flow: the presence of mobile charge carriers.

Step-by-Step Reasoning

Tin has a giant metallic structure. The metal atoms lose their outer electrons to form positive ions, and these electrons become delocalised. These delocalised electrons are free to move through the lattice, allowing tin to conduct electricity.

Key Takeaways

Metallic conductivity is due to the mobility of delocalised electrons.

Common Mistakes

• Saying "electrons can move" without specifying they are delocalised.
• Saying "ions move" (ions are fixed in the lattice and do not move to conduct electricity in solids).

Things to Be Careful About

Ensure you mention "delocalised electrons" and that they are "free to move".

Background Concept

Metal carbonates react with dilute acids to produce a salt, carbon dioxide gas, and water. This is a neutralisation reaction where the carbonate ion ($\text{CO}_3^{2-}$) acts as a base, accepting protons from the acid.

Understanding the Question

Write the balanced chemical equation for the reaction between magnesium carbonate and dilute hydrochloric acid.

Approach

Reactants: $\text{MgCO}_3$ and $\text{HCl}$. Products: $\text{MgCl}_2$, $\text{CO}_2$, and $\text{H}_2\text{O}$. Balance the equation.

Step-by-Step Reasoning

Magnesium carbonate ($\text{MgCO}_3$) reacts with hydrochloric acid ($\text{HCl}$). The magnesium ion ($\text{Mg}^{2+}$) combines with chloride ions ($\text{Cl}^-$) to form magnesium chloride ($\text{MgCl}_2$). The carbonate ion reacts with hydrogen ions to form carbon dioxide and water. To balance the chlorides, we need 2 $\text{HCl}$ molecules.

Key Takeaways

Carbonates react with acids to give salt, water, and carbon dioxide. Always balance the equation, especially the acid coefficient.

Common Mistakes

• Forgetting to balance the $\text{HCl}$ (writing $\text{MgCO}_3 + \text{HCl} \rightarrow \dots$).
• Writing $\text{H}_2\text{CO}_3$ as a product instead of $\text{CO}_2 + \text{H}_2\text{O}$ (carbonic acid decomposes).

Things to Be Careful About

Ensure the equation is fully balanced with correct formulae for all species.

Background Concept

The thermal stability of Group 2 carbonates increases down the group. This is because the larger the metal ion, the lower its charge density, and the less it polarises the large carbonate ion. Less polarisation means the carbonate ion is less distorted and less easily decomposed by heat.

Understanding the Question

Describe the trend in thermal stability of Group 2 carbonates as you go down the group.

Approach

State the trend directly: stability increases down the group.

Step-by-Step Reasoning

As you move down Group 2 from Mg to Ba, the ionic radius of the metal cation increases. The charge density decreases, so the polarising power on the carbonate ion decreases. This makes the carbonate more stable to heat. Therefore, thermal stability increases down the group.

Key Takeaways

Group 2 carbonate thermal stability increases down the group.

Common Mistakes

• Saying "decreases down the group" (confusing with solubility of hydroxides or something else).
• Not specifying "thermal stability" if the question asks for it.

Things to Be Careful About

The question asks to "describe" the trend, so a simple statement of the trend is sufficient. Do not explain it unless asked.

Background Concept

According to the Brønsted-Lowry theory, an acid is a proton ($\text{H}^+$) donor and a base is a proton acceptor. The ammonium ion ($\text{NH}_4^+$) can donate a proton to become ammonia ($\text{NH}_3$), so it acts as a weak acid. Hydroxide ions ($\text{OH}^-$) accept protons to form water, so they act as a base.

Understanding the Question

Explain why the reaction between ammonium carbonate and sodium hydroxide is an acid-base reaction.

Approach

Identify the acidic species ($\text{NH}_4^+$) and the basic species ($\text{OH}^-$) and apply the Brønsted-Lowry definitions.

Step-by-Step Reasoning

Ammonium carbonate contains the ammonium ion ($\text{NH}_4^+$). In water, $\text{NH}_4^+$ can donate a proton to a base, making it a Brønsted-Lowry acid. Sodium hydroxide provides hydroxide ions ($\text{OH}^-$), which accept protons to form water, making them a Brønsted-Lowry base. Thus, the reaction is an acid-base reaction where $\text{NH}_4^+$ donates a proton to $\text{OH}^-$.
Alternatively, one can state that $\text{NaOH}$ is a stronger base than $\text{NH}_3$, so the equilibrium lies to the right.

Key Takeaways

$\text{NH}_4^+$ acts as a weak acid (proton donor) and reacts with strong bases like $\text{OH}^-$.

Common Mistakes

• Calling $\text{NH}_4^+$ a base.
• Not mentioning proton donation/acceptance explicitly.

Things to Be Careful About

Use the exact terms "proton donor" and "proton acceptor" or "H+ donor/acceptor" as these are the mark scheme keywords.

Background Concept

Successive ionisation energies of an element increase because each successive electron is removed from an increasingly positive ion. Large jumps in ionisation energy occur when an electron is removed from a new, closer principal quantum shell (core electron) or a more stable subshell (e.g., from a full or half-full subshell). Silicon (Group 14) has 4 valence electrons ($3\text{s}^2 3\text{p}^2$). Removing these 4 electrons (1st to 4th IE) shows a gradual increase. Removing the 5th electron requires breaking into the $n=2$ shell ($2\text{p}^6$), causing a large jump.

Understanding the Question

We are given a graph of the first 2 and 7th to 14th ionisation energies of silicon. We need to plot the missing 3rd, 4th, 5th, and 6th points.

Approach

Use the electron configuration of Si to determine the relative magnitudes of the missing ionisation energies. Ensure a general upward trend and a large jump between the 4th and 5th IE.

Step-by-Step Reasoning

1. Electron configuration: $\text{Si}: 1\text{s}^2 2\text{s}^2 2\text{p}^6 3\text{s}^2 3\text{p}^2$.
2. 1st-4th IE: Remove $3\text{p}^2$ and $3\text{s}^2$. These are valence electrons. The IE increases gradually from 1st to 4th. Plot 3rd and 4th points higher than 2nd but lower than 7th.
3. 5th-6th IE: Remove $2\text{p}$ electrons. The 5th electron is removed from the $n=2$ shell, which is much closer to the nucleus and more strongly attracted. Therefore, the 5th IE is noticeably higher than the 4th IE (the largest jump on this part of the graph). Plot the 5th point significantly higher than the 4th, roughly near the level of the 7th point. Plot the 6th point higher than the 5th.
4. General trend: All points from 1st to 14th should show a general increase.

Key Takeaways

Large jumps in IE indicate a change in principal quantum shell. For Group 14, the jump is between the 4th and 5th IE.

Common Mistakes

• Plotting the 5th IE point too low (not showing the large jump from the 4th).
• Plotting the 3rd and 4th IE points higher than the 7th IE.
• Not showing a general increasing trend.

Things to Be Careful About

The exact y-values are not required, but the relative positions must show a general increase and a distinct large jump between the 4th and 5th ionisation energies.

Background Concept

The second ionisation energy is the energy required to remove one mole of electrons from one mole of gaseous 1+ ions to form one mole of gaseous 2+ ions. State symbols (g) are essential.

Understanding the Question

Write the equation representing the second ionisation energy of silicon.

Approach

Start with the gaseous 1+ ion, remove one electron, and form the gaseous 2+ ion.

Step-by-Step Reasoning

Reactant: $\text{Si}^+(\text{g})$. Product: $\text{Si}^{2+}(\text{g})$ and $\text{e}^-$. Equation: $\text{Si}^+(\text{g}) \rightarrow \text{Si}^{2+}(\text{g}) + \text{e}^-$.

Key Takeaways

Always include state symbols (g) and show the correct ion charges for successive ionisation energies.

Common Mistakes

• Starting with the neutral atom $\text{Si}(\text{g})$ (that's the first IE).
• Forgetting state symbols.
• Writing $\text{e}^-$ on the left side.

Things to Be Careful About

Ensure the equation is balanced and state symbols are present.

Background Concept

The boiling points of simple molecular substances depend on the strength of the intermolecular forces (IMFs) between the molecules. For non-polar molecules like the Group 14 hydrides ($\text{EH}_4$), the only IMFs present 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, as larger electron clouds are more polarisable.

Understanding the Question

Explain the trend in boiling points shown in the bar chart for $\text{CH}_4$, $\text{SiH}_4$, $\text{GeH}_4$, $\text{SnH}_4$, and $\text{PbH}_4$.

Approach

Identify the trend (boiling point increases down the group). Link this to the increasing size of the molecules, specifically the increasing number of electrons, and how this affects the intermolecular forces.

Step-by-Step Reasoning

1. Trend: The boiling point increases from $\text{CH}_4$ to $\text{PbH}_4$.
2. Reason: As you go down the group from C to Pb, the central atom becomes larger and has more electrons. Therefore, the molecules $\text{CH}_4$ to $\text{PbH}_4$ have a greater number of electrons.
3. Effect on IMFs: More electrons mean the electron cloud is more polarisable, leading to stronger instantaneous dipole–induced dipole forces (London dispersion forces) between the molecules.
4. Effect on boiling point: Stronger IMFs mean more energy is required to overcome them, resulting in a higher boiling point.

Key Takeaways

Boiling points of non-polar hydrides increase down a group due to increasing London dispersion forces from more electrons.

Common Mistakes

• Saying "more energy to break bonds" (boiling overcomes IMFs, not covalent bonds).
• Not mentioning "electrons" or "polarisability".
• Using incorrect terminology like "dipole-dipole" (these molecules are non-polar).

Things to Be Careful About

Use the correct terms: "instantaneous dipole–induced dipole forces" or "London forces" or "dispersion forces". Do not say "covalent bonds break".

Background Concept

The shape of a molecule is determined by the number of bonding pairs and lone pairs of electrons around the central atom (VSEPR theory). $\text{SiH}_4$ has a central silicon atom bonded to four hydrogen atoms with no lone pairs. This is an $\text{AX}_4$ system.

Understanding the Question

Deduce the shape of a molecule of $\text{SiH}_4$.

Approach

Count the electron domains around the central Si atom.

Step-by-Step Reasoning

Silicon has 4 valence electrons. It forms 4 single bonds with 4 hydrogen atoms. There are 4 bonding pairs and 0 lone pairs. The electron pairs repel each other to be as far apart as possible, resulting in a tetrahedral shape with bond angles of $109.5^\circ$.

Key Takeaways

$\text{AX}_4$ molecules with no lone pairs are tetrahedral.

Common Mistakes

• Saying "square planar" (requires lone pairs, e.g., $\text{XeF}_4$).
• Forgetting to specify "tetrahedral".

Things to Be Careful About

Just state the shape: "tetrahedral".

Background Concept

The formation of a compound from its constituent elements in their standard states is a synthesis reaction. Silicon is a solid, and chlorine is a diatomic gas in its standard state.

Understanding the Question

Write the balanced equation for the formation of $\text{SiCl}_4$ from silicon and chlorine.

Approach

Reactants: $\text{Si}$ and $\text{Cl}_2$. Product: $\text{SiCl}_4$. Balance the chlorine atoms.

Step-by-Step Reasoning

Silicon reacts with chlorine gas to form silicon tetrachloride. One silicon atom reacts with two chlorine molecules ($\text{Cl}_2$) to give one $\text{SiCl}_4$ molecule.

Key Takeaways

Formation equations use elements in their standard states. Remember chlorine is $\text{Cl}_2$.

Common Mistakes

• Writing $\text{Cl}$ instead of $\text{Cl}_2$.
• Not balancing the equation.

Things to Be Careful About

Ensure the equation is balanced and uses correct molecular formulae for elements.

Background Concept

$\text{SiCl}_4$ undergoes vigorous hydrolysis with water. The silicon atom has empty d-orbitals and can accept a lone pair from water, leading to the breakdown of Si-Cl bonds and formation of Si-O-Si bonds (or $\text{SiO}_2$) and $\text{HCl}$ gas.

Understanding the Question

Describe what is observed when $\text{SiCl}_4$ is added to water.

Approach

Recall the products of $\text{SiCl}_4$ hydrolysis: a white solid/liquid (silicic acid/$\text{SiO}_2$) and $\text{HCl}$ gas. Describe the visual observations.

Step-by-Step Reasoning

When $\text{SiCl}_4$ reacts with water, it produces $\text{HCl}$ gas and a white precipitate/solution of silicic acid or silicon dioxide. The $\text{HCl}$ gas reacts with moisture in the air to form misty or steamy fumes. The reaction is vigorous, so effervescence (bubbling) may also be observed.

Key Takeaways

$\text{SiCl}_4$ hydrolyses vigorously to give misty fumes of $\text{HCl}$ and a white solid.

Common Mistakes

• Saying "no reaction" (that's for $\text{CCl}_4$).
• Not mentioning the fumes or effervescence.

Things to Be Careful About

Use the terms "misty fumes" or "steamy fumes" and/or "effervescence".

Background Concept

Melting points depend on the strength of the forces that must be overcome to separate the particles in the solid state. In giant covalent structures (like $\text{SiO}_2$, diamond, silicon), atoms are bonded in a continuous network by strong covalent bonds. Melting requires breaking these bonds, which takes a lot of energy. In simple molecular structures (like $\text{SiCl}_4$, $\text{CO}_2$), molecules are held together by weak intermolecular forces (van der Waals/London forces). Melting only requires overcoming these weak forces, not the covalent bonds within the molecules.

Understanding the Question

Explain the large difference in melting points between $\text{SiO}_2$ (>1700 °C) and $\text{SiCl}_4$ (liquid at room temp) using structure and bonding.

Approach

Identify the structure and bonding of each compound. Explain that melting $\text{SiO}_2$ breaks strong covalent bonds, while melting $\text{SiCl}_4$ only overcomes weak intermolecular forces.

Step-by-Step Reasoning

1. $\text{SiO}_2$ structure: Each silicon is bonded to four oxygens in a giant covalent (tetrahedral) lattice. To melt it, strong covalent bonds must be broken. This requires a very high temperature (>1700 °C).
2. $\text{SiCl}_4$ structure: It consists of discrete $\text{SiCl}_4$ molecules held together by weak intermolecular forces (instantaneous dipole–induced dipole / London forces). The covalent Si-Cl bonds within the molecule are not broken during melting.
3. Comparison: Much less energy is required to overcome the weak intermolecular forces in $\text{SiCl}_4$ compared to the strong covalent bonds in the giant lattice of $\text{SiO}_2$. Hence, $\text{SiCl}_4$ has a much lower melting point.

Key Takeaways

Giant covalent structures have high melting points due to strong bonds; simple molecular structures have low melting points due to weak IMFs.

Common Mistakes

• Saying "bonds are broken in $\text{SiCl}_4$" (only IMFs are broken).
• Calling $\text{SiO}_2$ "ionic" or "molecular".
• Not mentioning the specific types of forces (covalent bonds vs intermolecular forces).

Things to Be Careful About

Clearly distinguish between the bonds within molecules ($\text{SiCl}_4$) and the forces between molecules. Use "giant covalent" for $\text{SiO}_2$ and "simple molecular" for $\text{SiCl}_4$.

Background Concept

Nucleophilic addition is the characteristic reaction mechanism of carbonyl compounds (aldehydes and ketones). The C=O bond is strongly polarised because oxygen is more electronegative than carbon, giving carbon a partial positive charge (δ⁺) and oxygen a partial negative charge (δ⁻). This makes the carbonyl carbon electrophilic and susceptible to attack by nucleophiles.

In the case of propanone reacting with HCN/KCN, the cyanide ion (CN⁻) acts as the nucleophile. The reaction proceeds in two steps: (1) the nucleophile attacks the electrophilic carbon, breaking the π bond of C=O and pushing the electron pair onto oxygen to form a tetrahedral alkoxide intermediate; (2) the alkoxide is protonated by HCN (or H⁺) to give the neutral hydroxynitrile product.

Understanding the Question

The question asks you to complete Fig. 4.2, which already shows propanone with its C=O double bond and two methyl groups, and the cyanide ion (:CN⁻) below it, with the final product A shown on the right. You need to fill in the mechanism: show the curly arrows, the bond dipole, the intermediate structure, and the protonation step.

The command word is "Complete" — this is a drawing task where the mechanism IS the answer.

Approach

Draw the mechanism in two stages:

1. Show the nucleophilic attack: a curly arrow from the lone pair on carbon of CN⁻ to the carbonyl carbon, the δ⁺/δ⁻ dipole on C=O, and a curly arrow from the C=O π bond to oxygen. Draw the resulting tetrahedral intermediate with O⁻.
2. Show protonation: a curly arrow from a lone pair on O⁻ to H⁺ (representing the proton source, HCN).

Step-by-Step Reasoning

Step 1 — Nucleophilic attack:

• The cyanide ion has a lone pair on its carbon atom (shown as :CN⁻). Carbon is the nucleophilic end because it bears the negative charge.
• Draw a curly arrow from this lone pair to the carbonyl carbon (which is δ⁺ due to the polarised C=O bond).
• The C=O double bond must break heterolytically: draw a curly arrow from the middle of the C=O π bond to the oxygen atom, giving oxygen a negative formal charge.
• The intermediate is a tetrahedral alkoxide: the central carbon now has four single bonds (to two CH₃ groups, to CN, and to O⁻).

Step 2 — Protonation:

• The O⁻ in the intermediate is a strong base. It abstracts a proton from HCN (or equivalently from H⁺).
• Draw a curly arrow from a lone pair on O⁻ to H⁺.
• This gives the neutral product A (2-hydroxy-2-methylpropanenitrile).

Mark allocation (3 marks):

• M1: curly arrow from lone pair on C of CN⁻ to carbonyl C, AND correct dipole on C=O, AND curly arrow from C=O bond to O, AND correct intermediate structure.
• M2: curly arrow from lone pair on O⁻ to H⁺.
• (The marks are distributed across these features as listed in the mark scheme.)

Key Takeaways

• Nucleophilic addition to a carbonyl always involves the nucleophile attacking the δ⁺ carbon and the π electrons moving to oxygen.
• The intermediate is always an alkoxide (O⁻) before protonation.
• Curly arrows show electron movement: from a lone pair or bond TO an atom or bond.
• The C=O dipole must be shown to explain WHY the carbon is attacked.

Common Mistakes

• Drawing the curly arrow from CN⁻ to oxygen instead of to carbon.
• Forgetting the dipole on C=O (this is a specific mark).
• Not showing the π bond breaking (curly arrow from C=O bond to O).
• Drawing the intermediate incorrectly (e.g. still showing a double bond, or wrong charge).
• Forgetting the protonation step entirely.
• Drawing the curly arrow from H⁺ to O⁻ instead of from the lone pair on O⁻ to H⁺ (arrows show electron flow, so they must start at the electron source).

Things to Be Careful About

• The lone pairs on the intermediate oxygen are not required for the mark (the mark scheme explicitly states this), but charges ARE required.
• The dipole must be shown on the starting propanone, not on the intermediate.
• State that the curly arrow goes from the lone pair on the carbon of CN⁻ specifically (not nitrogen).

Background Concept

Optical isomerism (enantiomerism) occurs when a molecule contains a chiral centre — a carbon atom bonded to four different atoms or groups of atoms. The two enantiomers are non-superimposable mirror images that rotate plane-polarised light in opposite directions. If any two of the four groups attached to a carbon are identical, that carbon is not chiral and the molecule cannot show optical isomerism.

Understanding the Question

Compound A is 2-hydroxy-2-methylpropanenitrile. The question asks why this molecule does not exhibit optical isomerism. The central carbon (C-2) has four groups attached: OH, CN, CH₃, and CH₃.

Approach

Examine the four groups attached to the central carbon and check whether they are all different. If two are the same, the carbon is not chiral.

Step-by-Step Reasoning

The central carbon in A is bonded to:

1. —OH
2. —CN
3. —CH₃
4. —CH₃

Groups 3 and 4 are identical (both methyl groups). Therefore the central carbon does not have four different groups attached, and is not a chiral centre. Without a chiral centre, the molecule cannot show optical isomerism.

Key Takeaways

• A chiral centre requires four DIFFERENT groups on a single carbon.
• Two identical groups (even if the other two are different) means no chirality.
• Always check all four substituents systematically.

Common Mistakes

• Saying "it has no chiral centre" without explaining WHY (the mark requires the reasoning about four different groups).
• Confusing this with geometric isomerism.
• Saying the molecule is symmetrical without specifying which groups are identical.

Things to Be Careful About

• The mark scheme requires the specific point: "not bonded to four different atoms/groups." A bare statement like "no chiral carbon" without the reason may not score.

Background Concept

Reduction of a ketone to a secondary alcohol involves the addition of two hydrogen atoms across the C=O double bond: one H adds to the carbon and one H adds to the oxygen. In simplified notation, [H] represents one atom of hydrogen from the reducing agent. Since two hydrogen atoms are needed (one for C, one for O), we write 2[H].

Understanding the Question

Reaction 3 converts propanone to compound C (propan-2-ol) using [H] as the reducing agent. The question asks for a balanced equation using [H] notation.

Approach

Write the molecular formula of propanone on the left, add 2[H], and write the formula of propan-2-ol on the right. Check that atoms balance.

Step-by-Step Reasoning

Propanone: $\text{CH}_3\text{COCH}_3$ (or $\text{C}_3\text{H}_6\text{O}$)
Product C: $\text{CH}_3\text{CH(OH)CH}_3$ (or $\text{C}_3\text{H}_8\text{O}$)

Difference: the product has 2 more H atoms than the reactant. Therefore 2[H] are needed.

$\text{CH}_3\text{COCH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_3$

Check: Left side has C₃H₆O + 2H = C₃H₈O. Right side has C₃H₈O. Balanced.

Key Takeaways

• Reduction of C=O to CH—OH requires 2[H].
• The [H] notation is a simplified way of representing reduction without specifying the actual reducing agent (e.g. NaBH₄ or LiAlH₄).

Common Mistakes

• Writing [H] instead of 2[H] (not balanced).
• Writing the molecular formula incorrectly.
• Adding state symbols that are not required.

Things to Be Careful About

• Use the structural formula format as shown in the question (CH₃COCH₃) rather than molecular formula (C₃H₆O) for clarity, though either may be accepted.

Background Concept

Alcohols are named by identifying the longest carbon chain containing the —OH group, numbering from the end nearest the —OH, and using the suffix -ol with a locant number. Propan-2-ol is a three-carbon chain with the hydroxyl group on carbon 2, making it a secondary alcohol.

Understanding the Question

Compound C is $\text{CH}_3\text{CH(OH)CH}_3$. The question asks for its name.

Approach

Identify the parent chain (3 carbons = propane), locate the —OH group (on carbon 2), and apply the suffix -ol with the locant.

Step-by-Step Reasoning

• Longest chain containing —OH: 3 carbons → "propan"
• Position of —OH: carbon 2 → "-2-"
• Suffix: "-ol"
• Full name: propan-2-ol

Note: "isopropanol" or "2-propanol" are common names but the IUPAC name is propan-2-ol.

Key Takeaways

• The locant number is essential for alcohols with more than 2 carbons (to distinguish propan-1-ol from propan-2-ol).
• Secondary alcohols have the —OH on a carbon bonded to two other carbons.

Common Mistakes

• Writing "propanol" without the locant (ambiguous).
• Writing "2-propanol" (older IUPAC style, may be accepted but propan-2-ol is preferred).

Things to Be Careful About

• Ensure the hyphen placement is correct: propan-2-ol (not propan2ol or propan-2ol).

Background Concept

In IR spectroscopy, a particular absorption is useful for identifying a compound only if it is present in some candidate compounds and absent in others. If all possible compounds share the same bond type, the corresponding absorption cannot distinguish between them. The C—H stretch (2850–2950 cm⁻¹) is present in virtually all organic compounds containing alkyl groups.

Understanding the Question

Compounds A, B, and C all contain methyl groups (CH₃), and therefore all have C—H bonds. The question asks why the absorption at 2850–2950 cm⁻¹ is not useful for determining which compound produced the spectrum.

Approach

Check the structures: do all three compounds contain the bond responsible for this absorption? If yes, it cannot be diagnostic.

Step-by-Step Reasoning

• Compound A: has two CH₃ groups → contains C—H bonds
• Compound B: has two CH₃ groups → contains C—H bonds
• Compound C: has two CH₃ groups → contains C—H bonds

All three compounds contain C—H bonds (from their methyl groups), so all three will show an absorption at 2850–2950 cm⁻¹. This absorption is therefore present in all candidates and cannot be used to distinguish between them.

Key Takeaways

• For an IR absorption to be diagnostic, it must be present in some candidates and absent in others.
• C—H stretches are ubiquitous in organic compounds and rarely help distinguish between similar molecules.

Common Mistakes

• Saying "the peak is too small" or "it overlaps with other peaks" — the reason is that ALL three compounds have it.
• Not referencing the structures of A, B, and C.

Things to Be Careful About

• The mark scheme requires the specific point: "All three have a C—H bond." Simply saying "it's not useful" without the reason scores zero.

Background Concept

IR spectroscopy identifies functional groups by their characteristic bond vibrations. Each bond type absorbs at a specific wavenumber range. By matching observed absorptions (and noting absent ones) to the expected functional groups in candidate compounds, the correct compound can be identified.

The key diagnostic regions here are:

• C≡N: 2200–2250 cm⁻¹ (present in A only)
• C=O: 1670–1740 cm⁻¹ (present in B only, as carboxyl C=O)
• O—H (carboxyl): 2500–3000 cm⁻¹, very broad (present in B only)
• O—H (hydroxy): 3200–3600 cm⁻¹, broad (present in A, B, and C)

Understanding the Question

Fig. 4.3 shows an IR spectrum. The candidate compounds are A (has OH and CN), B (has OH and COOH), and C (has OH only). The spectrum shows a broad peak around 3350 cm⁻¹ (O—H hydroxy), a peak just below 3000 cm⁻¹ (C—H), a peak around 1100–1200 cm⁻¹ (C—O), and critically a peak at 2200–2250 cm⁻¹ (C≡N). There is NO peak in the C=O region (1670–1740 cm⁻¹) and NO very broad absorption spanning 2500–3000 cm⁻¹.

Approach

1. Note the presence of C≡N absorption (2200–2250 cm⁻¹) → only compound A has this.
2. Confirm by noting the absence of C=O (rules out B) and the absence of the very broad carboxyl O—H (also rules out B).
3. Compound C has only OH and CH — it would not show the C≡N peak.

Step-by-Step Reasoning

Present absorptions:

• ~3350 cm⁻¹ (broad): O—H (hydroxy) — present in all three, not diagnostic
• ~2900 cm⁻¹: C—H — present in all three, not diagnostic
• ~2200–2250 cm⁻¹: C≡N — only compound A has a nitrile group
• ~1100–1200 cm⁻¹: C—O — present in all three (all have C—O from the OH)

Absent absorptions:

• No peak at 1670–1740 cm⁻¹: no C=O → rules out compound B (which has a carboxyl C=O)
• No very broad absorption at 2500–3000 cm⁻¹: no carboxyl O—H → also rules out B

The presence of the C≡N absorption at 2200–2250 cm⁻¹ is the decisive evidence for compound A.

Key Takeaways

• IR identification works by both presence AND absence of characteristic peaks.
• The C≡N stretch at 2200–2250 cm⁻¹ is a very distinctive, sharp peak that is hard to miss.
• Carboxylic acids show BOTH a C=O peak AND a very broad O—H spanning 2500–3000 cm⁻¹ — the absence of either rules them out.

Common Mistakes

• Identifying compound C (propan-2-ol) because it has an OH — but C has no C≡N, so it cannot produce the peak at 2200–2250 cm⁻¹.
• Identifying compound B and not noticing the absence of C=O.
• Confusing the broad hydroxy O—H (3200–3600) with the very broad carboxyl O—H (2500–3000). The spectrum shows the former (centred around 3350), not the latter.

Things to Be Careful About

• The mark scheme requires BOTH the identification (compound A) AND the reasoning (C≡N absorption at 2200–2250 cm⁻¹) for the 1 mark.
• Be precise about which wavenumber range you cite — use the values from Table 4.1.