ESAT Chemistry Mock Module 4 Worked Solutions

Correct Answer: E. $\text{CaCl}_2$

Fastest Approach (🚀):
$\text{Ca}^{2+}$ and $\text{Cl}^-$.
Two $\text{Cl}^-$ balance one $\text{Ca}^{2+}$.

Matches Option E.

Step-by-Step Breakdown:

1. Find the charges from the Group

Calcium is in Group 2, so it loses two electrons: $\text{Ca}^{2+}$. Chlorine is in Group 17, so it gains one: $\text{Cl}^-$.

2. Balance the total charge

The compound must be electrically neutral. One $2+$ needs two $1-$ ions:
$(+2) + 2(-1) = 0$

So the formula is $\text{CaCl}_2$.

3. The cross-over shortcut

Write the charges and swap them as subscripts: $\text{Ca}^{2+}$ and $\text{Cl}^{1-}$ give $\text{Ca}_1\text{Cl}_2$. Then cancel any common factor - which is what turns $\text{Mg}_2\text{O}_2$ into $\text{MgO}$.

4. What the formula means here

$\text{CaCl}_2$ is not a molecule. It is a giant ionic lattice, and the formula gives the ratio of ions in it: two chloride ions for every calcium ion.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $\text{CaCl}$ — Charge Unbalanced
  Total charge $+1$, not zero. One chloride ion cannot balance a $2+$ calcium ion.
• B. $\text{Ca}_2\text{Cl}$ — Ratio Inverted
  Total charge $+3$. This has the ratio inverted as well - it is the chloride that is needed in excess.
• C. $\text{CaCl}_3$ — Charge Unbalanced
  Total charge $-1$. Three chloride ions would balance a $3+$ ion, such as aluminium.
• D. $\text{Ca}_2\text{Cl}_3$ — Charge Unbalanced
  Total charge $+1$, and not a simplest ratio in any case.

Common Mistake (⚠️):
Assuming a one-to-one ratio because there are two elements. The ratio is set by the charges, and only equal charges give a $1:1$ formula.

Takeaway (📌):
Charges from the Group, cross them over, cancel. An ionic formula always totals zero charge.

Correct Answer: A. In order of increasing atomic number

Fastest Approach (🚀):
Atomic number $=$ number of protons $=$ what fixes the electron arrangement.

Matches Option A.

Step-by-Step Breakdown:

1. What atomic number is

The number of protons in the nucleus. It defines the element, and in a neutral atom it also fixes the number of electrons and therefore the electron configuration.

2. Why that ordering makes the table work

Chemical behaviour comes from the outer electrons. Ordering by atomic number means elements with the same outer configuration line up in vertical Groups, and the periodicity appears on its own.

3. Why mass ordering fails

Mendeleev did order by relative atomic mass, and it works nearly everywhere - but not quite. Argon ($A_r \approx 40$) comes before potassium ($A_r \approx 39$) despite being heavier, because argon has $18$ protons and potassium $19$. Ordering by mass would put a noble gas among the alkali metals. Tellurium and iodine are a second such pair.

4. Why mass ordering nearly works

Mass number is protons plus neutrons, and the two rise roughly together, so the two orderings mostly agree. Isotopic abundance is what causes the exceptions.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. In order of increasing relative atomic mass — Historical Scheme
  Mendeleev's original scheme. It fails at argon and potassium, where the heavier element has the lower atomic number.
• C. Alphabetically, by chemical symbol — Arbitrary Order
  Symbols come from several languages and reflect nothing about chemistry - alphabetical order would destroy the Groups.
• D. In order of decreasing chemical reactivity — No Single Order
  Reactivity varies in different directions for metals and non-metals, so no single reactivity sequence exists to order by.
• E. By the date on which each element was discovered — Arbitrary Order
  Discovery order is historical accident: several elements known since antiquity sit in quite different parts of the table.

Common Mistake (⚠️):
Answering 'relative atomic mass'. It is very nearly right and was the original scheme, but the argon-potassium inversion is exactly why the modern table does not use it.

Takeaway (📌):
Increasing atomic number. Mass ordering agrees almost everywhere and fails at argon-potassium and tellurium-iodine.

Correct Answer: B. $0.50\ \text{mol}$

Fastest Approach (🚀):
$M_r = 12 + 4 = 16$.
$n = \dfrac{8.0}{16} = 0.50\ \text{mol}$.

Matches Option B.

Step-by-Step Breakdown:

1. Relative formula mass

$\text{CH}_4$ has one carbon and four hydrogens:
$M_r = 12 + (4 \times 1) = 16$

So one mole of methane has a mass of $16\ \text{g}$.

2. Convert the mass

$n = \dfrac{\text{mass}}{M_r} = \dfrac{8.0}{16} = 0.50\ \text{mol}$

3. Predict the direction first

$8.0\ \text{g}$ is less than the mass of one mole, so the answer must be below $1\ \text{mol}$. That eliminates three options before any division is done, and it is the check that catches an inverted fraction.

4. What half a mole contains

$0.5 \times 6.02\times10^{23} = 3.01\times10^{23}$ molecules - and five times that many atoms, since each molecule holds five.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $0.25\ \text{mol}$ — Wrong Molar Mass
  Uses $M_r = 32$, which is the value for oxygen gas or for methanol, not for methane.
• C. $1.0\ \text{mol}$ — Wrong Molar Mass
  Would need $M_r = 8$. Methane's is $16$, so $8.0\ \text{g}$ is half a mole.
• D. $2.0\ \text{mol}$ — Fraction Inverted
  $\tfrac{16}{8.0}$ - the fraction inverted. Less than a mole's mass cannot give more than a mole.
• E. $8.0\ \text{mol}$ — Mass As Moles
  The mass in grams reported as a number of moles. The two are different quantities.

Common Mistake (⚠️):
Inverting the fraction and calculating $\tfrac{16}{8}$. The check is direction: less than one mole's mass must give less than one mole.

Takeaway (📌):
$n = \dfrac{m}{M_r}$. Work out $M_r$ first and predict whether the answer is above or below $1$.

Correct Answer: E. Run X had the faster initial rate, but both runs used the same amounts of reactant, because the two curves level off at the same volume

Fastest Approach (🚀):
Steeper $\Rightarrow$ faster (X).
Same plateau $\Rightarrow$ same amount of reactant.

Matches Option E.

Step-by-Step Breakdown:

1. What the gradient tells you

The rate of reaction is the rate at which product appears, which is the gradient of the curve. X is steeper at the start, so its initial rate is higher. Something has been changed - a higher temperature, a greater concentration, a finer powder or a catalyst.

2. What the plateau tells you

Each curve flattens when the limiting reactant runs out. The height at which it flattens is the total gas produced, and that depends only on how much limiting reactant there was - not at all on how fast it reacted.

3. Putting the two together

Equal plateaus mean equal amounts of reactant. Different gradients mean different rates. So the runs differ in a condition that changes the rate but not the quantity.

4. Why this is the classic catalyst signature

A catalyst raises the gradient and leaves the plateau untouched - exactly the pattern here. Adding more reactant, by contrast, would raise the plateau as well, which is how the two are told apart on a graph.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Run X used a greater volume of acid than run Y — Plateau Ignored
  More acid would produce more gas and a higher plateau, unless the acid was in excess in both - and then it would not be the limiting reactant setting the plateau.
• B. Run X produced more product in total than run Y — Rate As Amount
  The plateaus are equal, so the totals are equal. Only the rate differs.
• C. Run X was carried out at a lower temperature than run Y — Direction Reversed
  A lower temperature would make X slower, not faster. The steeper curve is the faster one.
• D. Run Y must have used a catalyst — Assigned To Wrong Run
  A catalyst raises the rate, so it would be in X - the steeper curve - not in Y.

Common Mistake (⚠️):
Reading a steeper curve as 'more product'. Steeper means faster; the amount of product is the height it finishes at, and those are independent.

Takeaway (📌):
Gradient is rate; plateau is amount. A change that lifts the gradient but not the plateau does not add reactant.

Correct Answer: C. Relative mass $1$, relative charge $0$

Fastest Approach (🚀):
Neutron: same mass as a proton, no charge.
Mass $1$, charge $0$.

Matches Option C.

Step-by-Step Breakdown:

1. The three rows

| Particle | Relative mass | Relative charge |
|----------|---------------|-----------------|
| proton | $1$ | $+1$ |
| neutron | $1$ | $0$ |
| electron | $\tfrac{1}{1836}$ | $-1$ |

2. Why the neutron's mass is $1$ and not $0$

The neutron is very slightly heavier than the proton, but on the relative scale both are taken as $1$. Its mass being $1$ is the whole reason the mass number counts protons and neutrons.

3. Why the electron's mass is not quite zero

It is often quoted as negligible, and for mass-number purposes it is - but it is a small fraction, about $\tfrac{1}{1836}$, not exactly nothing.

4. The check that catches a swapped row

An atom is neutral because the numbers of protons and electrons are equal and their charges are $+1$ and $-1$. If the row you have chosen would not let that work, it belongs to a different particle.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Relative mass $1$, relative charge $+1$ — Proton Row
  The proton's row. The neutron has the same mass but no charge.
• B. Relative mass $\tfrac{1}{1836}$, relative charge $-1$ — Electron Row
  The electron's row - a tiny mass and a charge of $-1$.
• D. Relative mass $0$, relative charge $0$ — Mass Zeroed
  The charge is right but the mass is not. A neutron carries a full unit of relative mass, which is why it counts towards the mass number.
• E. Relative mass $\tfrac{1}{1836}$, relative charge $0$ — Rows Mixed
  Takes the electron's mass and pairs it with the neutron's charge.

Common Mistake (⚠️):
Giving the neutron a relative mass of zero because it has no charge. Mass and charge are independent, and the neutron carries a full unit of mass.

Takeaway (📌):
Proton $(1, +1)$, neutron $(1, 0)$, electron $(\tfrac{1}{1836}, -1)$. Learn it as a table, not as three separate facts.

Correct Answer: B. With alternating current the electrodes would reverse polarity many times a second, so the ions would move back and forth and no product would accumulate at either electrode

Fastest Approach (🚀):
Ions must travel consistently to one electrode.
A.c. reverses the polarity constantly $\Rightarrow$ ions oscillate $\Rightarrow$ no product collects.

Matches Option B.

Step-by-Step Breakdown:

1. How electrolysis works

In a molten or dissolved ionic compound the ions are free to move. Positive ions (cations) are attracted to the negative cathode, where they gain electrons; negative ions (anions) go to the positive anode, where they lose them. The products then collect at their own electrode.

2. What alternating current would do

A.c. reverses the polarity of both electrodes fifty times a second in the UK. An ion drifting towards one electrode is pulled back before it has travelled far, then pushed forward again. Over any useful period it goes nowhere.

3. Even if an ion did arrive

Any product discharged during one half-cycle would be immediately re-formed into ions in the next, when that electrode's polarity reversed. Nothing would accumulate at either end.

4. What the wrong answers get wrong

A.c. does pass through an electrolyte - that is why an electrolyte conducts at all. And direct current does not carry more energy than alternating current: it is the fixed polarity, not the amount of energy, that matters.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Alternating current cannot pass through a liquid at all — Conduction Denied
  Alternating current passes through an electrolyte perfectly well - that is why an ionic solution conducts.
• C. Alternating current would heat the electrolyte too strongly and boil it away — Heating Invoked
  Heating depends on current and resistance, not on whether the current alternates. Electrolysis cells are often deliberately run hot.
• D. Direct current carries more energy than alternating current at the same voltage — Energy Invoked
  At the same voltage the energy delivered is comparable; the difference that matters is direction, not quantity.
• E. Alternating current would react chemically with the electrolyte and contaminate the products — Category Error
  A current is a flow of charge, not a chemical substance. It cannot react with anything.

Common Mistake (⚠️):
Reaching for an energy or heating explanation. Electrolysis is about the direction the ions travel, and only a steady polarity provides one.

Takeaway (📌):
Electrolysis needs a fixed cathode and anode. Alternating current swaps them constantly, so the ions oscillate and nothing is collected.

Correct Answer: C. It is endothermic, $\Delta H$ is positive, and energy is taken in from the surroundings

Fastest Approach (🚀):
Products above reactants $\Rightarrow \Delta H > 0 \Rightarrow$ endothermic, energy taken in.

Matches Option C.

Step-by-Step Breakdown:

1. Read the step

$\Delta H = E_{\text{products}} - E_{\text{reactants}}$. Products higher means the difference is positive.

2. Name it and say where the energy came from

A positive $\Delta H$ is endothermic: the products hold more energy than the reactants, and the extra has been absorbed from the surroundings. That is why an endothermic reaction makes its container feel cold.

3. The activation energy is a separate feature

The curve still rises to a peak between reactants and products - that hump is $E_a$, the energy needed to get started. It exists whichever way $\Delta H$ points, and for an endothermic reaction it is always larger than $\Delta H$, never equal to it.

4. Endothermic reactions are perfectly ordinary

Thermal decomposition, photosynthesis and the reaction of citric acid with sodium hydrogencarbonate in a cold pack are all endothermic. Nothing forbids them; they simply need a supply of energy.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. It is exothermic, and $\Delta H$ is negative — Sign Inverted
  Exothermic reactions release energy, so their products sit lower than the reactants and $\Delta H$ is negative.
• B. It has no activation energy, because the products are already higher — Activation Energy Denied
  Every reaction has an activation energy - the hump between reactants and products - regardless of which is higher.
• D. The reaction cannot occur, because products cannot be at a higher energy than reactants — Reaction Denied
  Endothermic reactions are common: thermal decomposition and photosynthesis are both examples. They simply absorb energy.
• E. The activation energy is equal to $\Delta H$ — Quantities Confused
  The activation energy is measured from the reactants to the peak, and in an endothermic reaction it always exceeds $\Delta H$.

Common Mistake (⚠️):
Inverting the sign convention. Exothermic gives out energy, so the products end up lower and $\Delta H$ is negative - the opposite of the diagram described.

Takeaway (📌):
Products above: endothermic, $\Delta H$ positive, surroundings cool. Products below: exothermic, $\Delta H$ negative, surroundings warm.

Correct Answer: A. Magnesium is oxidised because it loses electrons, and the copper(II) ion is reduced because it gains them

Fastest Approach (🚀):
Mg: $0 \to +2$, loses $2e^-$ $\Rightarrow$ oxidised.
Cu: $+2 \to 0$, gains $2e^-$ $\Rightarrow$ reduced.

Matches Option A.

Step-by-Step Breakdown:

1. Track the charges

Magnesium starts as a neutral atom and ends as $\text{Mg}^{2+}$: its charge rises by $2$. The copper ion starts at $2+$ and ends as a neutral atom: its charge falls by $2$.

2. Convert charge changes into electron transfers

A rise in charge means electrons have been lost; a fall means they have been gained. So magnesium loses two electrons and the copper ion gains them - the same two.

3. Name the processes

Oxidation Is Loss, Reduction Is Gain. Magnesium is oxidised, the copper(II) ion is reduced.

4. Why they cannot be separated

Electrons do not float free in solution: any that one species loses another must gain. Oxidation and reduction always occur together, which is why the combined process is called a redox reaction and why option C is impossible.

5. The half-equations

$\text{Mg} \rightarrow \text{Mg}^{2+} + 2e^-$ and $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$. Adding them cancels the electrons and returns the overall equation.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Magnesium is reduced because it gains electrons — Direction Reversed
  Magnesium's charge rises from $0$ to $+2$, which is a loss of electrons, not a gain.
• C. Both species are oxidised, because both change their charge — Both Oxidised
  Impossible: electrons lost by one species must be gained by another. Their charges also change in opposite directions.
• D. Copper is oxidised because it gains electrons — Terms Swapped
  Gaining electrons is reduction, not oxidation - the definition is applied to the wrong word.
• E. No electrons are transferred, because no oxygen takes part in the reaction — Oxygen Required
  Oxygen is not required. The general definition is electron transfer, and here two electrons plainly move from magnesium to the copper ion.

Common Mistake (⚠️):
Requiring oxygen before calling something oxidation. Gain of oxygen is only the elementary definition; the general one is loss of electrons, and it covers reactions with no oxygen at all.

Takeaway (📌):
Charge up means electrons lost means oxidation. The two halves always occur together.

Correct Answer: E. Sodium chloride and water

Fastest Approach (🚀):
$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$.

Matches Option E.

Step-by-Step Breakdown:

1. The general equation

acid $+$ alkali $\rightarrow$ salt $+$ water

2. Naming the salt

The first part of the salt's name comes from the metal in the alkali - sodium. The second comes from the acid: hydrochloric acid gives chlorides, sulfuric acid gives sulfates, nitric acid gives nitrates. So the salt is sodium chloride.

3. The balanced equation

$\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)$

4. What is really happening

The ionic equation is $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$. The sodium and chloride ions are spectators: they are in solution before and after. That is why every acid-alkali neutralisation produces water, whichever pair is used.

5. Where the other gases come from

Hydrogen appears with an acid and a metal; carbon dioxide with an acid and a carbonate. Neither reactant here is either of those.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Sodium chloride and hydrogen — Wrong Reaction Type
  Hydrogen is produced when an acid reacts with a metal, not with an alkali.
• B. Sodium chloride and carbon dioxide — Wrong Reaction Type
  Carbon dioxide comes from an acid with a carbonate. There is no carbonate here.
• C. Chlorine and water — Wrong Product
  Chlorine gas is not a product of neutralisation. The chloride ion stays in solution as part of the salt.
• D. Sodium hydroxide and hydrogen chloride — Reactants Repeated
  The reactants restated, so nothing has reacted. Neutralisation forms new substances.

Common Mistake (⚠️):
Adding a gas to the products. Acid with a metal gives hydrogen and acid with a carbonate gives carbon dioxide, but acid with an alkali gives only a salt and water.

Takeaway (📌):
Acid + alkali → salt + water. Metal from the alkali, the rest from the acid, water every time.

Correct Answer: E. Sulfur dioxide, produced by burning fuels that contain sulfur impurities, dissolves in atmospheric water to produce acid rain

Fastest Approach (🚀):
Sulfur impurity burns $\Rightarrow \text{SO}_2 \Rightarrow$ dissolves in water $\Rightarrow$ acid rain.

Matches Option E.

Step-by-Step Breakdown:

1. The four pollutants, sorted

• Carbon dioxide - complete combustion of any carbon fuel. Effect: it is a greenhouse gas contributing to global warming. It is not a cause of acid rain.
• Carbon monoxide - incomplete combustion, where oxygen is limited. Effect: toxic, because it binds to haemoglobin in place of oxygen.
• Sulfur dioxide - sulfur impurities in the fuel. Effect: acid rain.
• Nitrogen oxides - nitrogen and oxygen from the air combining in the high temperatures of an engine. Effect: acid rain and photochemical smog.

2. The sulfur dioxide chain

$\text{S} + \text{O}_2 \rightarrow \text{SO}_2$, then the gas dissolves in cloud droplets and is oxidised to sulfuric acid. The rain that falls damages trees, acidifies lakes and corrodes limestone buildings.

3. The fix

Sulfur is removed from fuels before sale, and flue gases at power stations are scrubbed with calcium oxide or calcium hydroxide, which neutralise the $\text{SO}_2$.

4. Where nitrogen oxides differ

They do not come from the fuel at all. Nitrogen and oxygen already in the air combine only at the temperatures inside an engine, which is why catalytic converters exist to reverse the reaction.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. Carbon dioxide, produced by incomplete combustion, causes acid rain — Origin and Effect Wrong
  Carbon dioxide comes from complete combustion and acts as a greenhouse gas. Acid rain is caused by sulfur dioxide and nitrogen oxides.
• B. Sulfur dioxide, produced by the complete combustion of pure hydrocarbons, is toxic because it binds to haemoglobin — Effect Swapped
  Sulfur dioxide comes from sulfur impurities, not from pure hydrocarbons, and it is carbon monoxide that binds to haemoglobin.
• C. Carbon monoxide, produced from sulfur impurities in fuel, causes acid rain — Origin and Effect Wrong
  Carbon monoxide comes from incomplete combustion of carbon fuels, and its effect is toxicity, not acid rain.
• D. Nitrogen oxides, produced when sulfur burns, deplete the ozone layer — Origin and Effect Wrong
  Nitrogen oxides come from nitrogen and oxygen in the air at high temperature, and they cause acid rain and smog rather than ozone depletion.

Common Mistake (⚠️):
Attributing acid rain to carbon dioxide. It does form a very weakly acidic solution, but acid rain in the environmental sense means sulfur dioxide and nitrogen oxides.

Takeaway (📌):
$\text{CO}$: incomplete combustion, toxic. $\text{SO}_2$: sulfur impurities, acid rain. $\text{NO}_x$: air at high temperature, acid rain and smog. $\text{CO}_2$: complete combustion, greenhouse gas.

Correct Answer: C. Potassium

Fastest Approach (🚀):
Lilac $\Rightarrow$ potassium.

Matches Option C.

Step-by-Step Breakdown:

1. The table

| Ion | Flame colour |
|-----|--------------|
| lithium, $\text{Li}^+$ | crimson |
| sodium, $\text{Na}^+$ | yellow |
| potassium, $\text{K}^+$ | lilac |
| calcium, $\text{Ca}^{2+}$ | orange-red |
| copper, $\text{Cu}^{2+}$ | blue-green |

2. Why the wire must be clean

It is dipped in concentrated hydrochloric acid and heated until it gives no colour of its own. Sodium contamination is the usual problem: its yellow is so intense that a trace masks every other colour, and the lilac of potassium is the one most easily lost that way.

3. Where the colour comes from

Heat promotes electrons to higher energy levels; as they fall back they emit light of definite wavelengths. The energy levels are characteristic of the element, which is what makes the colour diagnostic.

4. Its limits

A flame test identifies the metal ion only. The anion needs a separate test - silver nitrate for halides, barium chloride for sulfate, dilute acid for carbonate.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Sodium — Wrong Colour
  Sodium gives an intense yellow. It is the contaminant that most often hides a lilac flame.
• B. Calcium — Wrong Colour
  Calcium gives orange-red.
• D. Copper — Wrong Colour
  Copper gives blue-green.
• E. Lithium — Wrong Colour
  Lithium gives crimson - a clear red rather than the pale purple of potassium.

Common Mistake (⚠️):
Confusing lilac with crimson. Lithium is a clear red; potassium is the pale purple one, and it is easily masked by any sodium present.

Takeaway (📌):
Li crimson, Na yellow, K lilac, Ca orange-red, Cu blue-green. Flame tests identify the cation only.

Correct Answer: C. $5$

Fastest Approach (🚀):
C: $3 \Rightarrow 3\text{CO}_2$. H: $8 \Rightarrow 4\text{H}_2\text{O}$.
O on the right: $6+4 = 10 \Rightarrow 5\text{O}_2$.

Matches Option C.

Step-by-Step Breakdown:

1. Carbon first

Three carbon atoms on the left, and carbon appears only in $\text{CO}_2$, so the coefficient there is $3$: $3\text{CO}_2$.

2. Hydrogen next

Eight hydrogen atoms on the left, and hydrogen appears only in $\text{H}_2\text{O}$, which carries two each. So $4\text{H}_2\text{O}$.

3. Oxygen last

Now count the oxygen the products need: $3 \times 2 = 6$ from the carbon dioxide and $4 \times 1 = 4$ from the water, giving $10$ atoms. Oxygen arrives as $\text{O}_2$, so $\dfrac{10}{2} = 5$ molecules.

$\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$

4. Check every element

C: $3 = 3$. H: $8 = 8$. O: $10 = 6 + 4$. All balanced, with whole numbers and no common factor.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. $3$ — Water Ignored
  Balances the carbon dioxide's oxygen alone, ignoring the four atoms in the water.
• B. $4$ — Under-counted
  Gives $8$ oxygen atoms where the products need $10$.
• D. $6$ — Over-counted
  Gives $12$ oxygen atoms, two more than the products require.
• E. $7$ — Wrong Alkane
  The coefficient for butane, $\text{C}_4\text{H}_{10}$, at $6.5$ doubled - not for propane.

Common Mistake (⚠️):
Balancing oxygen early. Because it appears in both products, its count keeps changing as the others are adjusted - leave it until the products are fixed.

Takeaway (📌):
Carbon, hydrogen, oxygen, in that order. For $\text{C}_n\text{H}_m$ the oxygen coefficient is $n + \tfrac{m}{4}$.

Correct Answer: A. Zinc is more reactive than copper, because a more reactive metal displaces a less reactive one from its compound

Fastest Approach (🚀):
Zinc displaces copper $\Rightarrow$ zinc more reactive.
Copper cannot displace zinc $\Rightarrow$ order confirmed.

Matches Option A.

Step-by-Step Breakdown:

1. What displacement means

A more reactive metal has the stronger tendency to form positive ions. Placed in a solution of a less reactive metal's salt, it gives up electrons and forces the less reactive metal out of solution as the metal:
$\text{Zn}(s) + \text{CuSO}_4(aq) \rightarrow \text{ZnSO}_4(aq) + \text{Cu}(s)$

2. Reading the observations

The copper coating is copper metal deposited on the zinc. The fading blue is $\text{Cu}^{2+}$ leaving the solution - zinc sulfate solution is colourless. Both observations say the same thing.

3. Why the second experiment matters

Copper in zinc sulfate does nothing. If it had reacted, the conclusion would be contradictory. The negative result is what turns 'zinc appears more reactive' into a firm ordering.

4. As a redox process

$\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^-$ and $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$. The more reactive metal is the better reducing agent, which is the same statement in redox language.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Copper is more reactive than zinc, because copper was the metal that was deposited — Reversed
  The deposited metal is the one displaced, so it is the less reactive. Zinc went into solution, which is what the more reactive metal does.
• C. The two metals are equally reactive, since a reaction occurred in one direction — Order Denied
  Equal reactivity would produce no reaction in either direction. One reaction occurring settles the order.
• D. Zinc sulfate is more soluble in water than copper(II) sulfate — Solubility Invoked
  Both sulfates are soluble. Solubility plays no part in which metal displaces which.
• E. Nothing about reactivity, because displacement reactions depend only on solubility — Method Denied
  Displacement is precisely the standard method for establishing a reactivity series.

Common Mistake (⚠️):
Concluding that the metal deposited is the more reactive one. It is the opposite: the deposited metal is the one that has been pushed out of solution.

Takeaway (📌):
The metal that dissolves is the more reactive; the one deposited is the less reactive. A negative control confirms the order.

Correct Answer: B. They have the general formula $\text{C}_n\text{H}_{2n+2}$, are saturated, and do not decolourise bromine water

Fastest Approach (🚀):
Saturated $\Rightarrow \text{C}_n\text{H}_{2n+2}$, no C=C $\Rightarrow$ bromine water stays orange.

Matches Option B.

Step-by-Step Breakdown:

1. The general formula

$\text{C}_n\text{H}_{2n+2}$: methane $\text{CH}_4$, ethane $\text{C}_2\text{H}_6$, propane $\text{C}_3\text{H}_8$. The extra two hydrogens are the ones capping the ends of the chain, and they are what mark the series as saturated.

2. What saturated means

Every carbon-carbon bond is a single bond, so each carbon already holds as much hydrogen as it can. There is no double bond to open.

3. Why bromine water is unaffected

The test works by addition across a $\text{C}=\text{C}$ double bond, which removes the orange colour. An alkane has none, so the bromine water stays orange - which is exactly how the test distinguishes the two series. Alkanes do react with bromine by substitution, but only in ultraviolet light, not under the conditions of the test.

4. What alkanes do instead

They burn, completely to carbon dioxide and water and incompletely to carbon monoxide, and they can be cracked into shorter alkanes and alkenes. Polymerisation, though, needs the double bond - so it belongs to the alkenes.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. They have the general formula $\text{C}_n\text{H}_{2n}$ and readily decolourise bromine water — Alkene Described
  Both halves belong to the alkenes: $\text{C}_n\text{H}_{2n}$ and a positive bromine water test.
• C. They contain at least one carbon-carbon double bond — Alkene Described
  A carbon-carbon double bond makes a compound an alkene. Alkanes have only single bonds.
• D. They undergo addition polymerisation to form poly(alkanes) — Alkene Reaction
  Addition polymerisation needs a double bond to open, so it is an alkene reaction.
• E. They are unsaturated and react readily with bromine at room temperature in the dark — Alkene Described
  Alkanes are saturated, and they react with bromine only by substitution in ultraviolet light - not in the dark.

Common Mistake (⚠️):
Attributing the bromine water reaction to alkanes. Decolourising bromine water is the standard positive test for unsaturation, so a substance that decolourises it is not an alkane.

Takeaway (📌):
Alkanes: $\text{C}_n\text{H}_{2n+2}$, saturated, bromine water unchanged. Alkenes: $\text{C}_n\text{H}_{2n}$, unsaturated, bromine water decolourised.

Correct Answer: D. They gain enough energy to break free of their fixed positions and of the forces holding them, going straight from a regular, closely packed, vibrating arrangement to a random, widely spaced, fast-moving one

Fastest Approach (🚀):
Solid $\to$ gas directly: energy in, spacing up, order lost, no liquid stage.

Matches Option D.

Step-by-Step Breakdown:

1. The starting arrangement

In a solid the particles are close together in a regular lattice, vibrating about fixed positions. They cannot move past one another, which is why a solid keeps its shape.

2. What sublimation supplies

Energy. It must be enough to overcome the forces between the particles completely, not merely enough to let them slide past one another - that lesser amount would produce a liquid.

3. The finishing arrangement

Widely spaced, random, and moving rapidly in all directions. The gas fills its container because nothing holds the particles together at all.

4. The point that is always worth stating

The particles themselves are unchanged - same size, same mass, same chemical identity. What changes is their spacing, their arrangement and their energy. A substance that has sublimed is chemically the same substance, which is why solid carbon dioxide gives carbon dioxide gas.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. They move closer together and lose energy — Reverse Process
  Both halves reversed: particles move much further apart and gain energy. This describes deposition, the reverse process.
• B. They gain energy and take up a more regular arrangement — Order Reversed
  Energy is right, but a gas is far less ordered than a solid. Arrangement becomes random, not more regular.
• C. They pass briefly through the liquid state before becoming a gas — Liquid Stage Added
  Sublimation is defined by the absence of a liquid stage - that is what makes it sublimation rather than melting followed by boiling.
• E. The particles themselves expand in size as the solid becomes a gas — Particles Expand
  The particles never change size. It is the spacing between them that increases enormously.

Common Mistake (⚠️):
Saying the particles expand. Nothing about a particle's size changes with state - it is the space between particles that grows, and the difference matters in every state-change question.

Takeaway (📌):
Sublimation: solid straight to gas. Energy in, spacing up, regularity lost, particles themselves unchanged.

Correct Answer: D. Each sodium atom transfers one electron to a chlorine atom, forming $\text{Na}^+$ and $\text{Cl}^-$ ions that are held together by electrostatic attraction

Fastest Approach (🚀):
Metal loses, non-metal gains $\Rightarrow \text{Na}^+$ and $\text{Cl}^-$.
Opposite charges attract $\Rightarrow$ ionic lattice.

Matches Option D.

Step-by-Step Breakdown:

1. Which atom loses and which gains

Sodium is in Group 1 with one outer electron; losing it leaves a full shell beneath. Chlorine is in Group 17 with seven; gaining one completes its shell. So the electron moves from sodium to chlorine, never the other way.

2. The ions formed

$\text{Na} \rightarrow \text{Na}^+ + e^-$ and $\text{Cl} + e^- \rightarrow \text{Cl}^-$. Note the charges: losing a negative electron leaves a positive ion.

3. What holds the compound together

Electrostatic attraction between the oppositely charged ions, acting in every direction. Sodium chloride is therefore a giant lattice, not a molecule, with each ion surrounded by six of the other kind.

4. Why it is not the other two bond types

Sharing is covalent, and happens between non-metals. A sea of delocalised electrons is metallic, and happens between metals. A metal with a non-metal gives ionic bonding, which is the quickest way to classify any binary compound.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. Sodium and chlorine share a pair of electrons between them — Covalent
  Sharing is covalent bonding, which occurs between non-metals. Sodium is a metal.
• B. Chlorine transfers one electron to sodium, forming $\text{Na}^-$ and $\text{Cl}^+$ — Direction Reversed
  The transfer runs the wrong way, and the charges are inverted: losing an electron makes an ion positive, not negative.
• C. Both atoms release electrons into a sea of delocalised electrons that holds them together — Metallic
  A sea of delocalised electrons is metallic bonding, which needs two metals.
• E. The two atoms exchange protons until both have full outer shells — Protons Moved
  Protons are locked in the nucleus and never move in a chemical reaction - moving one would change the element.

Common Mistake (⚠️):
Getting the direction of transfer backwards. The metal loses and becomes positive; a Group 1 atom with one outer electron has nothing to gain by taking another.

Takeaway (📌):
Metal + non-metal: transfer, ions, electrostatic attraction, giant lattice. Non-metal + non-metal: sharing. Metal + metal: delocalised sea.

Correct Answer: D. They have the same number of electrons in their outermost shell, and it is those electrons that take part in bonding

Fastest Approach (🚀):
Same outer-shell electron count $\Rightarrow$ same bonding behaviour $\Rightarrow$ similar chemistry.

Matches Option D.

Step-by-Step Breakdown:

1. What a Group is

A vertical column. Every element in it has the same number of electrons in its outermost shell - one for Group 1, two for Group 2, seven for Group 17, and so on.

2. Why that determines the chemistry

Reactions involve outer electrons only: losing them, gaining them or sharing them to reach a full outer shell. Sodium and potassium each have one outer electron to lose, so both form $+1$ ions and both react vigorously with water.

3. Why 'similar' rather than 'identical'

Going down a Group the outer electrons sit further from the nucleus and are screened by more inner shells, so they are held less tightly. That is why Group 1 metals get more reactive down the group and Group 17 non-metals get less so - a trend superimposed on the shared behaviour.

4. What Periods do instead

A Period is a horizontal row, across which the number of outer electrons changes at every step. Elements in a Period therefore differ sharply in their chemistry, which is why option E gets the direction wrong.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. They have the same number of complete electron shells — Period Confused
  The number of complete shells is what a Period shares. Down a Group it increases at every step.
• B. They have the same relative atomic mass — Mass Irrelevant
  Relative atomic masses differ widely within a Group - lithium is $7$ and caesium $133$.
• C. They contain the same number of neutrons — Neutrons Irrelevant
  Neutron number varies within a Group and even between isotopes of one element, and it has no effect on chemistry.
• E. They lie in the same Period, so their atoms are the same size — Period Confused
  Group and Period swapped, and atoms in a Group differ greatly in size - which is exactly why reactivity changes down it.

Common Mistake (⚠️):
Confusing Groups with Periods. Groups are vertical and share outer-electron count; Periods are horizontal and share the number of shells.

Takeaway (📌):
Same Group means same number of outer electrons, which is what fixes the chemistry. Trends down the Group come from distance and shielding.

Correct Answer: D. $8.0\ \text{g}$

Fastest Approach (🚀):
$n(\text{Mg}) = \dfrac{4.8}{24} = 0.20$.
Ratio $2:2 \Rightarrow n(\text{MgO}) = 0.20$.
$m = 0.20 \times 40 = 8.0\ \text{g}$.

Matches Option D.

Step-by-Step Breakdown:

1. Mass to moles

$n(\text{Mg}) = \dfrac{4.8}{24} = 0.20\ \text{mol}$

2. Moles to moles, through the equation

The coefficients give $2\text{Mg} : 2\text{MgO}$, a one-to-one ratio, so $0.20\ \text{mol}$ of magnesium produces $0.20\ \text{mol}$ of magnesium oxide. This is the only step the balanced equation is needed for, and it is the step that is skipped when people scale the masses directly.

3. Moles back to mass

$M_r(\text{MgO}) = 24 + 16 = 40$

$m = n \times M_r = 0.20 \times 40 = 8.0\ \text{g}$

4. Check by conservation of mass

The oxygen absorbed is $8.0 - 4.8 = 3.2\ \text{g}$, which is $0.10\ \text{mol}$ of $\text{O}_2$ - exactly half the magnesium, as the $2:1$ coefficients require. The whole calculation is consistent.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $4.8\ \text{g}$ — Mass Unchanged
  The starting mass unchanged, as though the oxygen added nothing. Magnesium gains mass when it burns.
• B. $6.4\ \text{g}$ — Wrong Molar Mass
  Would need $M_r(\text{MgO}) = 32$. Adding one oxygen to magnesium gives $24 + 16 = 40$.
• C. $7.2\ \text{g}$ — Masses Scaled
  $4.8 \times 1.5$ - a plausible-looking scaling with no basis in the mole ratio.
• E. $9.6\ \text{g}$ — Coefficient As Mass
  $4.8 \times 2$, taking the coefficient $2$ as a mass factor. Coefficients scale moles, not grams.

Common Mistake (⚠️):
Scaling the masses in the ratio of the coefficients. Coefficients are ratios of moles, never of grams, because the two substances have different molar masses.

Takeaway (📌):
Grams to moles, moles to moles, moles to grams. Only the middle step uses the equation.

Correct Answer: A. The particles must collide with at least the activation energy, and in a suitable orientation

Fastest Approach (🚀):
Enough energy ($\ge E_a$) and the right orientation.

Matches Option A.

Step-by-Step Breakdown:

1. The energy condition

Existing bonds must be broken before new ones form, and breaking bonds costs energy. The minimum a collision must supply is the activation energy, $E_a$. Below it, the particles simply bounce apart unchanged.

2. The orientation condition

Even an energetic collision fails if the molecules meet the wrong way round, so that the parts which need to react are not brought into contact. For all but the simplest particles, most collisions fail on this count alone.

3. Why the rate is so much lower than the collision rate

Particles in a gas or solution collide enormously often. Only a small fraction of those collisions satisfies both conditions, which is why reactions take seconds or hours rather than being instantaneous.

4. How this explains every rate factor

Raising the temperature increases both the frequency of collisions and the fraction with enough energy. Raising the concentration or the surface area increases the frequency. A catalyst lowers $E_a$, so a larger fraction of the existing collisions succeeds. All four act through the same two conditions.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. The particles need only collide; any collision at any energy leads to reaction — No Conditions
  If every collision reacted, all reactions would be instantaneous. Most collisions fail on energy or orientation.
• C. The two particles must be at exactly the same temperature as each other — Temperature Misapplied
  Temperature is a property of a large collection of particles, not of a single one. Individual particles have a spread of energies.
• D. The collision must take place in the presence of a catalyst — Catalyst Required
  A catalyst makes reaction easier but is not required - most reactions proceed without one.
• E. The particles must collide with less than the activation energy, so that they are not broken apart — Energy Reversed
  Inverted. Below the activation energy the particles bounce apart unchanged; the energy is what makes reaction possible.

Common Mistake (⚠️):
Giving only the energy condition. Orientation is the second requirement and is worth a mark of its own in most mark schemes.

Takeaway (📌):
Successful collision = at least $E_a$, in a suitable orientation. Every rate factor works by changing the frequency or the success fraction.

Correct Answer: B. $35.5$

Fastest Approach (🚀):
$\dfrac{35\times75 + 37\times25}{100} = \dfrac{3550}{100} = 35.5$.

Matches Option B.

Step-by-Step Breakdown:

1. Why a weighted mean

Relative atomic mass is the average mass of the atoms as they actually occur, so each isotope counts in proportion to how common it is. A plain average of $35$ and $37$ would give $36$ and would only be right if the two were equally abundant.

2. Substitute

$A_r = \dfrac{(35 \times 75) + (37 \times 25)}{100}$

$= \dfrac{2625 + 925}{100} = \dfrac{3550}{100} = 35.5$

3. A quicker route for a 3:1 split

The isotopes differ by $2$, and the heavier one holds a quarter of the total, so the mean sits $\tfrac14 \times 2 = 0.5$ above the lighter: $35 + 0.5 = 35.5$.

4. The check that catches most errors

The answer must lie between $35$ and $37$, and closer to $35$ because that isotope is three times as common. Options C and D are outside or at the edge of that range and can be discarded on sight.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. $36.0$ — Unweighted Mean
  The plain average of $35$ and $37$, which would be correct only for a $50:50$ mixture.
• C. $35.0$ — Abundance Ignored
  The lighter isotope's mass alone. The $25\%$ of heavier atoms must pull the mean up.
• D. $37.0$ — Abundance Ignored
  The heavier isotope's mass alone, and the further of the two from the true weighted mean.
• E. $36.5$ — Weights Swapped
  Weights the mean towards the heavier isotope. The $75\%$ share belongs to the lighter one.

Common Mistake (⚠️):
Taking a plain average of $35$ and $37$. That treats the isotopes as equally abundant, which the $75:25$ split explicitly rules out.

Takeaway (📌):
Weighted mean: multiply each isotopic mass by its percentage, add, divide by $100$. The answer always sits nearer the more abundant isotope.

Correct Answer: D. $\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$

Fastest Approach (🚀):
Cathode negative $\Rightarrow$ attracts $\text{Pb}^{2+}$.
Gains electrons (reduction) $\Rightarrow$ electrons on the left.

Matches Option D.

Step-by-Step Breakdown:

1. Which ion arrives at the cathode

The cathode is the negative electrode, so it attracts positive ions. In molten $\text{PbBr}_2$ that is $\text{Pb}^{2+}$.

2. What happens there

The ion gains electrons from the electrode and is discharged as lead metal, which collects as a molten pool. Gaining electrons is reduction, so reduction always happens at the cathode.

3. Which side the electrons go

Electrons are gained, so they are a reactant and belong on the left:
$\text{Pb}^{2+} + 2e^- \rightarrow \text{Pb}$

Two electrons, because the charge must balance: $(+2) + 2(-1) = 0$ on the left, matching the neutral atom on the right.

4. The other electrode

At the anode, bromide ions lose electrons: $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$, giving brown bromine vapour. That is oxidation, and it is option B - the right equation at the wrong electrode.

Matches Option D.

Why the Other Options Are Wrong (❌):

• A. $\text{Pb} \rightarrow \text{Pb}^{2+} + 2e^-$ — Oxidation At Cathode
  Lead losing electrons - oxidation of the metal, which is not what happens at a cathode.
• B. $2\text{Br}^- \rightarrow \text{Br}_2 + 2e^-$ — Wrong Electrode
  The correct anode half-equation. Bromide ions are negative and travel to the positive electrode.
• C. $\text{Br}_2 + 2e^- \rightarrow 2\text{Br}^-$ — Reaction Reversed
  The reverse of the anode reaction. There is no bromine present at the start to be reduced.
• E. $\text{Pb}^{2+} \rightarrow \text{Pb} + 2e^-$ — Charge Unbalanced
  The charges do not balance: $+2$ on the left against $-2$ on the right. A positive ion becoming a neutral atom must gain electrons.

Common Mistake (⚠️):
Putting the electrons on the wrong side. Check the charge balance: both sides of a half-equation must carry the same total charge.

Takeaway (📌):
Cathode: positive ions, reduction, electrons on the left. Anode: negative ions, oxidation, electrons on the right.

Correct Answer: E. $-150\ \text{kJ}\,\text{mol}^{-1}$, exothermic

Fastest Approach (🚀):
$\Delta H = 2750 - 2900 = -150\ \text{kJ}\,\text{mol}^{-1}$.
Negative $\Rightarrow$ exothermic.

Matches Option E.

Step-by-Step Breakdown:

1. Which process is which

Breaking bonds requires energy - endothermic. Forming bonds releases energy - exothermic. Every bond-energy calculation is the balance between the two.

2. Subtract

$\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds formed})$

$= 2750 - 2900 = -150\ \text{kJ}\,\text{mol}^{-1}$

3. Interpret the sign

More energy came out than went in, so the surplus is released to the surroundings and the mixture warms. A negative $\Delta H$ is exothermic - the sign and the word must always agree, which is what rules out option D.

4. Why adding is wrong

The two figures point in opposite directions: one is energy in, the other energy out. Adding them gives $5650$, which is the total energy that changed hands, not the net change - and no reaction has a $\Delta H$ of that size from these numbers.

Matches Option E.

Why the Other Options Are Wrong (❌):

• A. $+150\ \text{kJ}\,\text{mol}^{-1}$, endothermic — Sign Inverted
  Right size, but both the sign and the word are wrong. Releasing more energy than is absorbed gives a negative $\Delta H$.
• B. $+5650\ \text{kJ}\,\text{mol}^{-1}$, endothermic — Added Not Subtracted
  Adds the two totals instead of subtracting. They point in opposite directions, so the net change is their difference.
• C. $-5650\ \text{kJ}\,\text{mol}^{-1}$, exothermic — Added Not Subtracted
  The same addition, with the sign corrected but the magnitude still wrong.
• D. $+150\ \text{kJ}\,\text{mol}^{-1}$, exothermic — Sign and Word Disagree
  The sign and the word contradict each other: a positive $\Delta H$ is endothermic by definition.

Common Mistake (⚠️):
Pairing a positive $\Delta H$ with the word exothermic. The two halves of the answer must agree: negative and exothermic, or positive and endothermic.

Takeaway (📌):
$\Delta H = $ broken $-$ formed. Negative means exothermic; positive means endothermic. Check that sign and word match before answering.

Correct Answer: B. The copper(II) ion, because it oxidises the magnesium while itself being reduced

Fastest Approach (🚀):
Oxidising agent $=$ the species reduced.
$\text{Cu}^{2+}$ is reduced $\Rightarrow$ it is the oxidising agent.

Matches Option B.

Step-by-Step Breakdown:

1. What the word 'agent' means here

An agent is named for the change it brings about in its partner. An oxidising agent oxidises something else - so it must take the electrons, and taking electrons is being reduced.

2. Apply it

$\text{Cu}^{2+}$ takes two electrons from magnesium. Magnesium is therefore oxidised, so $\text{Cu}^{2+}$ is the oxidising agent - and it is itself reduced in the process.

3. The other label

Magnesium supplies the electrons, so it causes the reduction of the copper ion: magnesium is the reducing agent, and it is itself oxidised. Every redox reaction has one of each.

4. The rule in one line

The oxidising agent is the species that is reduced; the reducing agent is the species that is oxidised. It reads backwards, which is precisely why it has to be learned as a sentence rather than reasoned out under time pressure.

Matches Option B.

Why the Other Options Are Wrong (❌):

• A. Magnesium, because it is the species that is oxidised — Labels Swapped
  Magnesium is oxidised, which makes it the reducing agent. The labels are named for the effect on the partner.
• C. Magnesium, because it is the reducing agent, and reducing agents cause oxidation — Wrong Label Chosen
  The first half is right - magnesium is the reducing agent - but that makes it the reducing agent, not the oxidising one.
• D. The copper atom formed in the products — Product Chosen
  Copper metal is a product, formed after the electron transfer. The agent is the species present at the start that accepts the electrons.
• E. Neither, because no oxygen takes part in the reaction — Oxygen Required
  Redox does not require oxygen. Electrons transfer here, so both agents are present.

Common Mistake (⚠️):
Assuming the oxidising agent is the species that is oxidised. It is the reverse: the agent is named for what it does to its partner.

Takeaway (📌):
Oxidising agent is reduced; reducing agent is oxidised. One of each in every redox reaction.

Correct Answer: C. Sodium chloride is a giant ionic lattice, so melting requires many strong electrostatic attractions to be overcome, while solid chlorine is a molecular solid held together only by weak intermolecular forces

Fastest Approach (🚀):
NaCl: giant ionic lattice, strong electrostatic attractions everywhere $\Rightarrow$ high m.p.
$\text{Cl}_2$: molecular solid, weak intermolecular forces $\Rightarrow$ very low m.p.

Matches Option C.

Step-by-Step Breakdown:

1. What melting actually breaks

Melting separates particles from one another. It does not break the bonds inside a molecule - chlorine gas is still $\text{Cl}_2$ well above its melting point, and its covalent bond is strong.

2. Sodium chloride

A giant ionic lattice: every $\text{Na}^+$ is surrounded by six $\text{Cl}^-$ and vice versa, throughout the crystal. Melting means overcoming an enormous number of strong electrostatic attractions at once, so the temperature required is very high.

3. Chlorine

A simple molecular substance. Within each $\text{Cl}_2$ the covalent bond is strong, but between molecules there are only weak intermolecular forces. Very little energy is needed to separate them, so the melting point is far below room temperature.

4. The general principle

Giant structures - ionic, metallic or giant covalent such as diamond - melt high. Simple molecular substances melt low, whatever the strength of the bonds inside their molecules. That single distinction predicts most melting-point comparisons at this level.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. Chlorine molecules are heavier than sodium chloride formula units — Mass Invoked
  Chlorine's $M_r$ of $71$ is in fact higher than sodium chloride's $58.5$, so mass points the wrong way. Structure, not mass, is what decides.
• B. Sodium chloride is covalently bonded and chlorine is ionically bonded — Bonding Swapped
  The two bonding types are swapped: sodium chloride is ionic and chlorine is covalent.
• D. The covalent bonds inside chlorine molecules are unusually weak — Wrong Bond
  The covalent bond in $\text{Cl}_2$ is strong, and melting never breaks it - only the weak forces between molecules.
• E. Sodium chloride conducts electricity when solid, and conduction raises the melting point — Factually Wrong
  Solid sodium chloride does not conduct, because its ions are fixed in the lattice. It conducts only when molten or dissolved.

Common Mistake (⚠️):
Explaining a low melting point by 'weak covalent bonds'. The covalent bond in $\text{Cl}_2$ is strong; it is the forces between molecules that are weak, and only those are broken on melting.

Takeaway (📌):
Melting breaks the forces between particles. Giant structures melt high; simple molecular substances melt low regardless of their internal bonds.

Correct Answer: A. Reactivity decreases and boiling point increases

Fastest Approach (🚀):
Gaining an electron gets harder down the group $\Rightarrow$ less reactive.
Bigger molecules, stronger intermolecular forces $\Rightarrow$ higher boiling point.

Matches Option A.

Step-by-Step Breakdown:

1. The reactivity trend and its cause

A halogen reacts by gaining one electron to complete its outer shell. Going down the group, that shell lies further from the nucleus and is screened by more inner shells, so the attraction on an incoming electron weakens. Fluorine captures one most easily and is the most reactive; iodine the least.

2. Why this is the opposite of Group 1

Alkali metals react by losing an electron, and the same increase in distance and shielding makes losing one easier down the group. Same cause, opposite effect - which is why the two trends run in opposite directions and are so often confused.

3. The boiling point trend

Physical, not chemical. Down the group the molecules have more electrons, so the intermolecular forces between them are stronger and more energy is needed to separate them. Fluorine and chlorine are gases at room temperature, bromine is a liquid and iodine a solid.

4. The evidence for the reactivity order

Displacement: chlorine displaces bromine from potassium bromide solution, and bromine displaces iodine from potassium iodide - but never the reverse.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. Reactivity increases and boiling point increases — Reactivity Reversed
  The boiling point trend is right, but reactivity falls: gaining an electron becomes harder down the group.
• C. Reactivity decreases and boiling point decreases — Boiling Point Reversed
  Reactivity is right, but boiling point rises - iodine is a solid at room temperature while fluorine is a gas.
• D. Reactivity increases and boiling point decreases — Both Reversed
  Both trends reversed. This is the Group 1 reactivity trend paired with an incorrect physical trend.
• E. Both reactivity and boiling point remain essentially unchanged — Trends Denied
  Both change markedly: fluorine is a highly reactive gas and iodine a much less reactive solid.

Common Mistake (⚠️):
Carrying the Group 1 reactivity trend across to Group 17. Both are explained by increasing distance and shielding, but because one group loses electrons and the other gains them, the trends point opposite ways.

Takeaway (📌):
Halogens: reactivity falls down the group, boiling point rises. Alkali metals: reactivity rises. Distance and shielding explain both.

Correct Answer: A. $0.50\ \text{mol}\,\text{dm}^{-3}$

Fastest Approach (🚀):
$500\ \text{cm}^3 = 0.5\ \text{dm}^3$.
$c = \dfrac{0.25}{0.5} = 0.50\ \text{mol}\,\text{dm}^{-3}$.

Matches Option A.

Step-by-Step Breakdown:

1. Convert the volume

$1\ \text{dm}^3 = 1000\ \text{cm}^3$ (a litre), so
$500\ \text{cm}^3 = \dfrac{500}{1000} = 0.500\ \text{dm}^3$

2. Divide

$c = \dfrac{n}{V} = \dfrac{0.25}{0.500} = 0.50\ \text{mol}\,\text{dm}^{-3}$

3. Check the direction

Half a cubic decimetre holds $0.25\ \text{mol}$, so a full one would hold twice as much. Dividing by a number below one must increase the value, which is why the answer is above $0.25$ rather than below it.

4. The conversion that catches people out

Dividing by $500$ instead of $0.5$ makes the answer a thousand times too small. Convert to $\text{dm}^3$ as a separate written line, before the division, and it cannot happen.

Matches Option A.

Why the Other Options Are Wrong (❌):

• B. $0.125\ \text{mol}\,\text{dm}^{-3}$ — Operation Inverted
  Multiplies by $0.5$ instead of dividing. Halving the volume doubles the concentration.
• C. $2.0\ \text{mol}\,\text{dm}^{-3}$ — Fraction Inverted
  $\dfrac{0.5}{0.25}$ - moles and volume the wrong way round.
• D. $5.0\ \text{mol}\,\text{dm}^{-3}$ — Conversion Slip
  Uses a volume of $0.05\ \text{dm}^3$; $500\ \text{cm}^3$ is $0.5$.
• E. $0.25\ \text{mol}\,\text{dm}^{-3}$ — Volume Ignored
  The number of moles restated as a concentration, as though the volume were $1\ \text{dm}^3$.

Common Mistake (⚠️):
Dividing by the volume in $\text{cm}^3$. The unit $\text{mol}\,\text{dm}^{-3}$ names the volume unit required, and it is not cubic centimetres.

Takeaway (📌):
$c = \dfrac{n}{V}$ with $V$ in $\text{dm}^3$. Convert first: divide the $\text{cm}^3$ figure by $1000$.

Correct Answer: C. It increases the rate of reaction by providing an alternative route of lower activation energy, and is chemically unchanged at the end

Fastest Approach (🚀):
Lower $E_a$ $\Rightarrow$ faster. Not consumed. Yield unchanged.

Matches Option C.

Step-by-Step Breakdown:

1. How it speeds the reaction up

The catalyst provides an alternative route with a lower activation energy. The reactant particles have not changed, but a larger fraction of them now has enough energy for a collision to succeed - so more collisions are effective and the rate rises.

2. It is not consumed

The catalyst may take part - forming an intermediate, or providing a surface for adsorption - but it is regenerated by the end, chemically unchanged. That is why a small mass can process a very large quantity of reactant, and why industrial catalysts are replaced only when contaminated.

3. What it does not change

It does not alter how much product is obtainable, because it does not change the energies of the reactants or the products - only the barrier between them. $\Delta H$ is unaffected.

4. And at equilibrium

It speeds the forward and reverse reactions equally, so equilibrium is reached sooner but its position is unmoved. That is a common misconception and the reason option E is on the list.

Matches Option C.

Why the Other Options Are Wrong (❌):

• A. It increases the total yield of product obtainable from the reaction — Yield Changed
  A catalyst changes how quickly the final amount is reached, not how much there is. The yield is fixed by the amounts of reactant.
• B. It is used up during the reaction and must be replaced after each run — Catalyst Consumed
  It is regenerated and chemically unchanged - which is why a small quantity lasts for a very long time.
• D. It raises the activation energy, which is why the reaction proceeds more quickly — Direction Reversed
  Backwards: raising the barrier would slow the reaction. A catalyst lowers the activation energy.
• E. It shifts the position of an equilibrium towards the products — Equilibrium Shifted
  It speeds the forward and reverse reactions equally, so equilibrium arrives sooner but sits in exactly the same place.

Common Mistake (⚠️):
Assuming a catalyst increases the yield. It changes only the speed at which the final amount is reached; it cannot make more product exist.

Takeaway (📌):
Lower activation energy, faster rate, unchanged catalyst, unchanged yield, unchanged equilibrium position.