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MYP 3 · Science

Atomic Structure

60 questions across 3 sub-topics

Use the Sub-Topic filter above to focus on one.

Structure of the Atom Protons, Neutrons and Electrons Electron Arrangement (basic)

Structure of the Atom 20 questions

QUESTION 1 3 marks Criterion A
Easy
AB

The diagram shows a simplified model of an atom, with two regions labelled A and B.

a. Name the region labelled A, and state the two types of particle found there.
[2]
b. Name the particle labelled B, and state its relative charge.
[1]
Show complete worked solution
(a)
Region A is the nucleus, the tiny, dense region at the centre of the atom. It contains protons and neutrons.
(b)
Particle B is an electron. It has a relative charge of $-1$.
QUESTION 2 2 marks Criterion A
Easy

State the names of the three subatomic particles found in an atom. Which two of these are found in the nucleus?

Show complete worked solution

The three subatomic particles are the proton, neutron, and electron. The proton and neutron are found in the nucleus; the electron is found outside the nucleus, in the electron shells.

QUESTION 3 4 marks Criterion A
Medium

A neutral atom always contains an equal number of protons and electrons.

a. Explain why this equal balance of protons and electrons means the atom has no overall electric charge.
[2]
b. An atom loses one electron. Describe what happens to its overall charge, and explain why.
[2]
Show complete worked solution
(a)
A proton has a relative charge of $+1$ and an electron has a relative charge of $-1$. In a neutral atom the number of protons equals the number of electrons, so every $+1$ charge is exactly cancelled out by a $-1$ charge, giving an overall charge of zero.
(b)
The atom now has one more proton than electrons, so the positive charges are no longer fully cancelled out. The atom becomes an ion with an overall charge of $+1$ (it is now positively charged), because it has one uncancelled proton's worth of positive charge.
QUESTION 4 4 marks Criterion A
Medium

Almost all of an atom's mass is concentrated in a region that takes up almost none of its volume.

a. Which part of the atom contains almost all of its mass? Explain your answer by comparing the relative masses of protons, neutrons, and electrons.
[2]
b. Describe, in general terms, how the volume of the nucleus compares with the volume of the whole atom.
[2]
Show complete worked solution
(a)
Almost all of the atom's mass is in the nucleus. Protons and neutrons each have a relative mass of about $1$, while an electron's relative mass is only about $\frac{1}{1836}$ — roughly $2000$ times smaller. Since the nucleus contains all the protons and neutrons, it holds essentially all of the atom's mass.
(b)
The nucleus is extremely small compared with the whole atom — its diameter is roughly $10\,000$ to $100\,000$ times smaller than the diameter of the atom. This means almost all of the atom's volume is empty space occupied only by the tiny, fast-moving electrons.
QUESTION 5 3 marks Criterion A
Medium

State three key ideas of the modern (nuclear) model of the atom.

Show complete worked solution

(1) The atom has a tiny, dense, positively charged nucleus at its centre, containing protons and neutrons. (2) Electrons occupy the space around the nucleus, arranged in shells (energy levels). (3) The atom is mostly empty space — the nucleus takes up only a tiny fraction of the atom's total volume.

QUESTION 6 5 marks Criterion A
Medium

Before the modern nuclear model of the atom was accepted, scientists used a different model called the "plum pudding" model.

a. Name the scientist most associated with the plum pudding model.
[1]
b. Describe the key difference between the plum pudding model and the nuclear model of the atom.
[2]
c. Name the scientist and the experiment that provided evidence against the plum pudding model.
[2]
Show complete worked solution
(a)
J. J. Thomson.
(b)
In the plum pudding model, positive charge was thought to be spread evenly throughout the whole atom, like dough, with electrons ('plums') scattered through it. In the nuclear model, almost all of the positive charge and mass is concentrated in a tiny central nucleus, with the rest of the atom being empty space occupied by orbiting electrons.
(c)
Ernest Rutherford's alpha-particle scattering (gold foil) experiment provided this evidence — a small number of alpha particles bounced back or deflected at large angles, which the plum pudding model could not explain.
QUESTION 7 6 marks Criterion A
Hard

Scientists' model of the atom has changed several times as new evidence was discovered.

a. Describe Dalton's early model of the atom.
[1]
b. Describe Thomson's "plum pudding" model, and state what discovery led to it.
[2]
c. Describe Rutherford's nuclear model, the experiment that led to it, and how it differed from Thomson's model.
[3]
Show complete worked solution
(a)
John Dalton proposed that atoms were tiny, solid, indivisible spheres — with no internal parts at all.
(b)
Thomson discovered the electron — a tiny, negatively charged particle far lighter than a whole atom — which proved atoms were not solid, indivisible spheres after all. He proposed the plum pudding model: a sphere of positive charge, spread evenly throughout the atom, with negative electrons embedded in it like plums in a pudding.
(c)
In the alpha-particle scattering experiment, Rutherford's team fired alpha particles at thin gold foil. Almost all passed straight through, but a small number deflected at large angles and a few bounced straight back — impossible if positive charge were spread evenly through the atom (as Thomson proposed). This showed that positive charge (and almost all the mass) must be concentrated in a tiny, dense central region — the nucleus — with the rest of the atom being mostly empty space. This nuclear model replaced the plum pudding model, since Thomson's evenly-spread positive charge could never have deflected a fast alpha particle back the way it came.
QUESTION 8 6 marks Criterion B
Medium
sourcegold foildetector screenrare bounce-backsmall deflection

A student models Rutherford's alpha-particle scattering experiment using a simulation, firing a beam of particles at a thin sheet of foil and recording where each particle lands on a detector screen behind it, as shown.

a. State the independent variable and the dependent variable in this investigation.
[2]
b. Describe the method and equipment needed to determine what proportion of particles are deflected at each angle.
[3]
c. Suggest one improvement that would make the results more reliable.
[1]
Show complete worked solution
(a)
Independent variable: the angle at which each particle is fired (or which point on the foil it is aimed at). Dependent variable: the angle of deflection recorded on the detector screen for each particle.
(b)
Fire a large number of particles (e.g. several thousand) one at a time from the source, through the same small gap, at the same spot on the thin foil. Use a detector screen curved around the foil to record a flash of light (or a count) at the exact angle each particle lands. Record a tally of how many particles land at each range of angles (e.g. $0$–$1°$, $1$–$10°$, $10$–$90°$, $>90°$), then calculate what percentage of the total falls into each range.
(c)
Fire a much larger total number of particles, since large-angle deflections are rare — a bigger sample makes the measured percentages less affected by chance and gives a more reliable estimate of the true proportions.
QUESTION 9 6 marks Criterion B
Medium

A class builds two physical models to represent competing ideas about atomic structure: Model P, a ball of soft dough with metal beads pushed evenly through it (representing the plum pudding model), and Model Q, a single small, dense metal marble surrounded by empty space (representing the nuclear model). They plan to roll a heavy ball-bearing at each model to see which one better matches Rutherford's real results.

a. Identify one variable that must be controlled for this to be a fair comparison between Model P and Model Q.
[2]
b. Describe how each model would be tested, and what result would provide evidence in favour of the nuclear model over the plum-pudding model.
[3]
c. Suggest why each test should be repeated several times rather than performed only once.
[1]
Show complete worked solution
(a)
The speed (and mass) of the ball-bearing used to test each model must be kept the same, and it must be fired from the same distance and along the same line each time — otherwise any difference in how it bounces could be caused by the test itself rather than by a real difference between the two models.
(b)
Fire the ball-bearing repeatedly at the same point on each model and record whether it passes straight through, is slightly deflected, or bounces straight back. Model P (soft, evenly spread dough) should let the ball-bearing pass through or barely deflect every time, since there is nothing solid to bounce off. Model Q (concentrated dense marble) should let the ball-bearing pass through when it misses the marble, but bounce sharply back on the rare occasions it hits the marble directly. Observing this pattern of mostly straight through, but occasionally a large deflection from Model Q — matching what Rutherford actually observed — would support the nuclear model.
(c)
Repeating the test reduces the effect of chance results and allows an average or overall pattern to be seen, making the conclusion about which model fits the evidence more reliable.
QUESTION 10 8 marks Criterion B
Hard

A student runs a simulation of the gold foil experiment in which a detector screen can only record the deflection angle of each particle to the nearest $10°$.

a. Identify one source of error in this method, and explain how it would affect the results.
[2]
b. Suggest an improvement to the detector that would reduce this error, and explain how it works.
[3]
c. Explain how this improved precision would produce more reliable evidence about the size of the nucleus.
[3]
Show complete worked solution
(a)
The main source of error is the low precision of the detector (only recording to the nearest $10°$). Small but real differences in deflection angle — for example, between $2°$ and $8°$ — would all be rounded into the same "$0$–$10°$" category, so fine detail in the pattern of scattering is lost.
(b)
Replace the coarse detector with one that records deflection angle to a much finer precision, for example to the nearest $1°$, using a more sensitive scintillation screen or an electronic position sensor around the foil. Because each particle's landing position is measured far more precisely, the recorded angle is much closer to the particle's true deflection angle, rather than being rounded into a wide $10°$ band.
(c)
With finer-angle data, the pattern of how the proportion of deflected particles changes with angle can be measured much more precisely, including right at very large angles close to $180°$. Since the fraction of particles that bounce back at large angles is directly related to how small and concentrated the positive nucleus is, a more precise angle measurement gives a more reliable (less rounded, less approximate) estimate of just how tiny the nucleus really is compared with the whole atom.
QUESTION 11 3 marks Criterion C
Easy
ResultNumber of particles
Total alpha particles fired1000
Passed straight through996
Deflected at a small angle3
Bounced back at a large angle1

The table shows the results of a simplified gold-foil scattering experiment.

Show complete worked solution

$$ \text{percentage straight through} = \frac{996}{1000} \times 100 $$

$$ = 99.6\% $$

Answer: $99.6\%$ of the alpha particles passed straight through the foil.

QUESTION 12 2 marks Criterion C
Easy
ResultNumber of particles
Total alpha particles fired500
Passed straight through or nearly straight498
Deflected at a large angle2

Calculate the percentage of particles that were deflected at a large angle.

Show complete worked solution

$$ \frac{2}{500} \times 100 = 0.4\% $$

Answer: $0.4\%$ of the particles were deflected at a large angle.

QUESTION 13 5 marks Criterion C
Medium
Predicted by plum pudding modelActually observed
% deflected at a large angle (>90°)about 0%about 0.1%

The plum pudding model predicted that positive charge was spread evenly through the atom, so no alpha particle should ever be deflected by a large angle.

a. Compare the predicted and observed percentages in the table.
[1]
b. Explain what this difference tells scientists about the structure of the atom.
[2]
c. Evaluate whether these results support the plum pudding model.
[2]
Show complete worked solution
(a)
The plum pudding model predicted almost $0\%$ of particles would deflect at a large angle, but about $0.1\%$ actually did — a small percentage, but clearly not zero.
(b)
Even though $0.1\%$ is small, it means some alpha particles experienced a strong enough force to be deflected sharply backwards. This could only happen if they passed very close to a small region of concentrated positive charge and mass — evidence that positive charge is not spread evenly through the atom, but concentrated in a tiny central nucleus.
(c)
These results do not support the plum pudding model. If positive charge really were spread thinly and evenly through the whole atom, as that model claimed, no single alpha particle would ever encounter a strong enough concentration of charge to bounce back — yet a small but real number of particles did exactly that. The data instead support the nuclear model, in which positive charge is concentrated in a small, dense nucleus.
QUESTION 14 4 marks Criterion C
Medium

A simple scale model uses picometres (pm) to represent the radius of a typical atom and its nucleus.

FeatureApproximate radius
Whole atom100,000 pm
Nucleus5 pm
a. Calculate how many times larger the radius of the atom is than the radius of the nucleus.
[2]
b. Using your answer, explain what this tells you about the structure of an atom.
[2]
Show complete worked solution
(a)
$$ \frac{100\,000}{5} = 20\,000 $$

Answer: the atom's radius is about $20\,000$ times larger than the nucleus's radius.

(b)
Since the nucleus's radius is about $20\,000$ times smaller than the whole atom's radius, the nucleus occupies an incredibly tiny fraction of the atom's total volume. This confirms that an atom is made up almost entirely of empty space, with only electrons (and their motion) filling that space around the minute, dense central nucleus.
QUESTION 15 4 marks Criterion C
Medium
Repeat trial123
Large-angle deflections (out of 8000 fired)8259

A student repeated a gold-foil scattering measurement three times.

a. Calculate the mean number of large-angle deflections, excluding any anomalous trial.
[2]
b. Identify the anomalous trial and suggest a possible reason for it.
[2]
Show complete worked solution
(a)
Trial $2$ ($25$) is anomalous (see part b), so it is excluded: $$ \text{mean} = \frac{8+9}{2} = 8.5 $$

Answer: $8.5$ large-angle deflections (to 1 d.p.).

(b)
Trial $2$ ($25$ deflections) is anomalous — it is roughly three times higher than trials $1$ and $3$, which are close together. A possible reason is contamination of the gold foil (e.g. with a thicker patch, dirt, or a fold), which would cause more particles than usual to be deflected in that trial.
QUESTION 16 5 marks Criterion C
Medium
Foil thickness (?m)0.10.20.40.8
% of particles deflected at a large angle0.010.020.040.08

A student investigates how gold-foil thickness affects the percentage of alpha particles deflected at a large angle.

a. Describe the pattern shown between foil thickness and the percentage of particles deflected.
[2]
b. Using the pattern, predict the percentage of particles that would be deflected at a large angle for a foil of thickness $1.6\,\mu\text{m}$. Show your working.
[3]
Show complete worked solution
(a)
The percentage deflected is directly proportional to the foil thickness — every time the thickness doubles, the percentage deflected also doubles (e.g. $0.1\to0.2\,\mu\text{m}$ doubles the thickness, and $0.01\%\to0.02\%$ doubles the deflection).
(b)
$$ \text{deflection} \propto \text{thickness} \qquad \Rightarrow \qquad \frac{0.08}{0.8} = 0.1\,\%\ \text{per}\ \mu\text{m} $$

$$ 1.6 \times 0.1 = 0.16 $$

Answer: approximately $0.16\%$ would be deflected at a large angle.

QUESTION 17 7 marks Criterion C
Hard

Imagine a scale model in which the nucleus of an atom is represented by a marble of radius $1\,\text{cm}$. The real ratio of an atom's radius to its nucleus's radius is approximately $10\,000 : 1$.

a. Calculate the radius of the whole atom in this scale model, giving your answer in metres. Show your working.
[3]
b. Using your answer, explain why atoms are described as being "mostly empty space". Give a real-world comparison to illustrate the scale involved.
[4]
Show complete worked solution
(a)
$$ \text{atom radius} = 1\,\text{cm} \times 10\,000 = 10\,000\,\text{cm} $$

Converting to metres: $$ 10\,000\,\text{cm} \div 100 = 100\,\text{m} $$

Answer: the atom's radius would be about $100\,\text{m}$ in this scale model.

(b)
A marble ($1\,\text{cm}$ radius) representing the nucleus, compared with an atom of radius $100\,\text{m}$, is a huge difference in scale — the marble would sit at the centre of a sphere roughly the size of a large sports stadium. Almost the entire volume of that stadium-sized sphere would contain nothing but the marble at its centre and the empty space through which the (much smaller and lighter) electrons move. This shows that even though atoms make up all solid matter, the matter itself — the protons and neutrons — occupies only an infinitesimally small fraction of an atom's total volume; the rest is empty space.
QUESTION 18 5 marks Criterion D
Medium

Positron Emission Tomography (PET) scans use radioactive isotopes injected into a patient's bloodstream. As the unstable nuclei of these isotopes decay, doctors can detect the radiation given off to build a detailed image showing which parts of the body — such as a tumour — are most active, helping to diagnose diseases like cancer.

Discuss one benefit and one drawback of this use of atomic structure in medicine.

Show complete worked solution

Benefit: PET scans allow doctors to see detailed information about what is happening inside the body without invasive surgery, often detecting cancers or other diseases earlier and more precisely than other methods. This can lead to earlier treatment and significantly better outcomes for patients.

Drawback: The patient is exposed to a small dose of radiation from the radioactive isotope, which carries a health risk if used too often or at too high a dose. There is also a cost and access issue — PET scanners are expensive and not available in every hospital, particularly in poorer regions, and radioactive isotopes for the scan must be produced and transported before they decay away, limiting where and how quickly they can be used.

QUESTION 19 6 marks Criterion D
Hard

Many household smoke detectors contain a tiny amount of americium, an element whose unstable nucleus continuously emits alpha particles. These particles ionise the air inside the detector, allowing a small electric current to flow; smoke entering the detector disrupts this current and triggers the alarm.

Evaluate the impact of using this radioactive material in smoke detectors, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: This design gives an extremely reliable, low-cost, and long-lasting way (often over $10$ years without needing replacement) to detect smoke very early, well before a fire becomes dangerous. Because the mechanism does not rely on smoke physically blocking a light beam, it is highly sensitive to the small, fast-burning particles produced by flaming fires, giving people more time to escape and saving lives.

Concern: The americium source is radioactive, so if a detector is broken open, mishandled, or disposed of incorrectly (for example thrown in ordinary household rubbish rather than returned for proper recycling), the radioactive material could pose a small but real environmental and health hazard. This is why regulations in many countries require these detectors to be collected and disposed of through special electronic-waste or radioactive-material recycling schemes rather than general rubbish.

QUESTION 20 6 marks Criterion D
Hard

Understanding the structure of the atom — and that huge amounts of energy can be released by splitting the nucleus of a large atom such as uranium (nuclear fission) — led to the development of nuclear power stations, which now generate a significant share of the world's low-carbon electricity.

Discuss one benefit and one drawback of generating electricity this way.

Show complete worked solution

Benefit: Splitting the nucleus of uranium atoms releases an enormous amount of energy from a very small mass of fuel, and unlike burning fossil fuels, nuclear fission produces electricity without directly releasing carbon dioxide, making it an important tool for reducing greenhouse gas emissions and tackling climate change while still providing a reliable, constant supply of electricity.

Drawback: Nuclear fission produces radioactive waste, some of which remains hazardous for many thousands of years and must be safely stored and isolated from the environment for that entire time, at significant ongoing cost. There is also the risk (though rare) of a serious accident releasing radioactive material, and building and eventually decommissioning nuclear power stations is very expensive and takes many years, meaning decisions about atomic structure and nuclear technology carry consequences that last for generations.

Protons, Neutrons and Electrons 20 questions

QUESTION 1 3 marks Criterion A
Easy

Complete the missing information for the three subatomic particles found in an atom.

Show complete worked solution

ParticleRelative chargeRelative mass
Proton+11
Neutron01
Electron?11/1836 (? 0)

QUESTION 2 2 marks Criterion A
Easy
8p?8nShell 1: 2e?Shell 2: 6e?

The diagram shows a model of an oxygen atom.

Show complete worked solution

Reading the diagram: the nucleus contains 8 protons and 8 neutrons, and the shells contain a total of $2+6=8$ electrons. Since the atom is neutral, the number of electrons is also 8 — matching the number of protons.

QUESTION 3 4 marks Criterion A
Medium

An atom of sodium is represented as $^{23}_{11}\text{Na}$.

a. State the number of protons and the number of electrons in this atom.
[2]
b. Calculate the number of neutrons in this atom.
[2]
Show complete worked solution
(a)
The bottom number ($11$) is the atomic number, which equals the number of protons. Since the atom is neutral: protons $=11$ and electrons $=11$.
(b)
$$ \text{neutrons} = \text{mass number} - \text{atomic number} = 23 - 11 $$

Answer: $12$ neutrons.

QUESTION 4 4 marks Criterion A
Medium
6p?6nC-126p?8nC-14same element

Carbon-12 and carbon-14 are both forms of the element carbon.

a. Define the term "isotope".
[2]
b. State what is the same and what is different between an atom of carbon-12 and an atom of carbon-14, in terms of protons, neutrons, and electrons.
[2]
Show complete worked solution
(a)
Isotopes are atoms of the same element (so they have the same number of protons) but with a different number of neutrons (and therefore a different mass number).
(b)
Both have the same number of protons ($6$) and electrons ($6$), since they are both carbon. Carbon-12 has $12-6=6$ neutrons, while carbon-14 has $14-6=8$ neutrons — a different number of neutrons.
QUESTION 5 4 marks Criterion A
Medium

A neutral sodium atom has $11$ protons and $11$ electrons. The atom then loses one electron.

a. State what happens to the atom's overall charge, and explain why.
[2]
b. State the resulting number of protons and electrons, and the name given to a charged atom like this.
[2]
Show complete worked solution
(a)
The atom becomes positively charged (overall charge $+1$). It still has $11$ protons (giving $11$ positive charges) but now only $10$ electrons (giving $10$ negative charges), so one positive charge is left uncancelled.
(b)
The resulting particle has $11$ protons and $10$ electrons. A charged atom formed this way is called an ion (specifically, a positive ion, or cation).
QUESTION 6 5 marks Criterion A
Medium

A chlorine ion, $\text{Cl}^-$, is formed when a neutral chlorine atom (which has $17$ protons and $18$ neutrons) gains one extra electron.

a. State the atomic number and the mass number of this chlorine ion.
[2]
b. State the overall charge of this ion, and explain why.
[2]
c. Explain why this ion is still considered chlorine, and not a different element.
[1]
Show complete worked solution
(a)
The atomic number is the number of protons, which does not change when an ion forms: atomic number $=17$. The mass number is protons $+$ neutrons $=17+18$: mass number $=35$.
(b)
The overall charge is $-1$. The ion has $17$ protons ($17$ positive charges) but $17+1=18$ electrons ($18$ negative charges), leaving one negative charge uncancelled.
(c)
The number of protons defines which element an atom (or ion) is, and this has not changed — it is still $17$. Only the number of electrons changed, which affects charge but not the identity of the element.
QUESTION 7 6 marks Criterion A
Hard
ParticleProtonsNeutronsElectrons
Atom X121212
Atom Y121312
Ion Z121210

The table shows the composition of three particles, all based on the element magnesium.

a. State the mass number of atom X.
[1]
b. Explain how atom Y differs from atom X, and state the term that describes this relationship.
[2]
c. Calculate the overall charge of ion Z, and explain your reasoning.
[3]
Show complete worked solution
(a)
Mass number $= 12+12 = \mathbf{24}$.
(b)
Atom Y has the same number of protons ($12$) as atom X, but one extra neutron ($13$ instead of $12$), giving it a different mass number ($25$ instead of $24$). Atoms of the same element with different numbers of neutrons are called isotopes of each other.
(c)
Ion Z has $12$ protons ($12$ positive charges) but only $10$ electrons ($10$ negative charges). $$ \text{charge} = (+12) + (-10) = +2 $$

Answer: ion Z has an overall charge of $+2$, because it has $2$ more protons than electrons, leaving $2$ positive charges uncancelled.

QUESTION 8 6 marks Criterion B
Medium

In a simulation, a beam of unknown charged particles passes between two oppositely charged metal plates, similar to J. J. Thomson's experiment that led to the discovery of the electron. A student uses this setup to investigate whether a mystery particle is positively or negatively charged.

a. State the independent variable and the dependent variable in this investigation.
[2]
b. Explain how the direction the beam bends would show whether the mystery particle is positively or negatively charged.
[2]
c. Suggest one variable that should be controlled for this to be a fair test.
[2]
Show complete worked solution
(a)
Independent variable: which plate is made positive and which is made negative (or, testing with the plates on/off). Dependent variable: the direction the particle beam bends towards.
(b)
Opposite charges attract. If the beam bends towards the positive plate, the particles must be negatively charged (attracted to positive charge, repelled by negative). If it bends towards the negative plate, the particles must be positively charged.
(c)
The voltage (strength of charge) applied to the plates should be kept the same in every trial, along with the speed of the particle beam entering the plates — changing either of these would change how much the beam bends regardless of the particle's charge, making comparisons unfair.
QUESTION 9 6 marks Criterion B
Medium

A science class wants to investigate whether two samples of chlorine gas, collected from different sources, might contain different proportions of the chlorine-35 and chlorine-37 isotopes, which would give them slightly different average masses per atom.

a. State the independent variable and the dependent variable for this investigation.
[2]
b. Describe the method and equipment needed to measure the average mass accurately.
[2]
c. State one variable that must be controlled for this to be a fair comparison.
[2]
Show complete worked solution
(a)
Independent variable: which sample of chlorine gas is tested (Sample 1 or Sample 2). Dependent variable: the measured average mass per atom (or per mole) of the gas sample.
(b)
Use a precise mass balance to weigh a carefully measured volume (or a known number of moles) of each gas sample under the same temperature and pressure, then calculate the mass per atom for each. In a real laboratory, a mass spectrometer would be used, since it can measure the mass and relative abundance of each isotope directly and very precisely.
(c)
The temperature and pressure of the gas must be kept the same for both samples when measuring volume or mass, since both affect how much gas (and therefore mass) occupies a given space.
QUESTION 10 8 marks Criterion B
Hard
Trial12345
% abundance of chlorine-35 measured74.978.275.673.476.0

A student uses a simplified mass-spectrometer model to measure the percentage abundance of chlorine-35 in a chlorine sample five times.

a. Identify a likely source of error that would explain the spread in these repeated readings.
[2]
b. Suggest an improvement to the method that would reduce this error, and explain how it works.
[3]
c. Explain how this improvement would make a calculated relative atomic mass for chlorine more reliable.
[3]
Show complete worked solution
(a)
The spread ($73.4\%$ to $78.2\%$, a range of nearly $5$ percentage points) is likely due to random error in reading or calibrating the instrument each time (for example, small variations in how the detector counts each isotope, or in the exact size of the sample used in each trial).
(b)
Take the mean of a larger number of repeat readings (e.g. $10$ or more trials instead of $5$), and ensure the instrument is carefully recalibrated with a known reference sample before each run. Averaging more readings reduces the effect of random fluctuations in any single measurement, since high and low errors tend to cancel out, giving a result closer to the true abundance.
(c)
The relative atomic mass of chlorine is calculated as a weighted average using the percentage abundance of each isotope. If the abundance percentage itself is unreliable (as shown by the scatter in this data), the resulting relative atomic mass will also be unreliable. A more precise, well-calibrated, averaged abundance measurement feeds directly into a more accurate and trustworthy relative atomic mass calculation.
QUESTION 11 3 marks Criterion C
Easy
AtomAtomic numberMass number
Lithium37
Fluorine919
Aluminium1327

Calculate the number of neutrons in an atom of aluminium.

Show complete worked solution

$$ \text{neutrons} = \text{mass number} - \text{atomic number} = 27 - 13 $$

Answer: $14$ neutrons.

QUESTION 12 2 marks Criterion C
Easy
SpeciesProtonsElectrons
W99
X1110
Y1717

Identify which species is not a neutral atom, and explain your reasoning.

Show complete worked solution

Species X is not a neutral atom, because its number of protons ($11$) does not equal its number of electrons ($10$) — an unequal number of protons and electrons means the particle carries an overall charge, making it an ion.

QUESTION 13 4 marks Criterion C
Medium
IsotopeMass number% abundance
Chlorine-353575.77
Chlorine-373724.23

Natural chlorine is a mixture of two isotopes, shown in the table.

a. Explain what makes chlorine-35 and chlorine-37 isotopes of each other.
[1]
b. Calculate the relative atomic mass of chlorine, using the abundances given. Give your answer to 1 decimal place.
[3]
Show complete worked solution
(a)
They both have the same number of protons (chlorine's atomic number, $17$) but a different number of neutrons — $18$ for chlorine-35 and $20$ for chlorine-37 — giving them different mass numbers.
(b)
$$ A_r = (35 \times 0.7577) + (37 \times 0.2423) $$

$$ = 26.5195 + 8.9651 = 35.4846 $$

Answer: $A_r = 35.5$ (1 d.p.) — this matches the accepted relative atomic mass of chlorine shown on the periodic table.

QUESTION 14 4 marks Criterion C
Medium
SampleProtonsNeutronsElectrons
1666
2676
3777

A student was asked to record data for three samples, all supposedly isotopes of carbon.

a. Identify which sample is not actually an isotope of carbon.
[1]
b. Explain your reasoning.
[3]
Show complete worked solution
(a)
Sample 3 is not an isotope of carbon.
(b)
Isotopes of the same element must all have the same number of protons, since the number of protons (atomic number) defines which element an atom is. Samples 1 and 2 both have $6$ protons, matching carbon, and differ only in neutron number ($6$ and $7$) — making them true isotopes of each other. Sample 3 has $7$ protons, not $6$, meaning it is actually an atom of a different element (nitrogen), not an isotope of carbon at all.
QUESTION 15 5 marks Criterion C
Medium
IsotopeMass number% abundance
Boron-101019.9
Boron-111180.1

Natural boron is a mixture of two isotopes, shown in the table. The accepted relative atomic mass of boron on the periodic table is $10.81$.

a. Calculate the relative atomic mass of boron using the data in the table. Give your answer to 2 decimal places.
[3]
b. Comment on how well your calculated value agrees with the accepted value of $10.81$.
[1]
c. Explain why the relative atomic mass of boron is not a whole number.
[1]
Show complete worked solution
(a)
$$ A_r = (10 \times 0.199) + (11 \times 0.801) $$

$$ = 1.99 + 8.811 = 10.801 $$

Answer: $A_r = 10.80$ (2 d.p.).

(b)
The calculated value ($10.80$) is extremely close to the accepted value ($10.81$), differing by only $0.01$ — a very good level of agreement, well within the precision expected from the rounded abundance data given.
(c)
Relative atomic mass is a weighted average of all of an element's naturally occurring isotopes, each of which does have a whole-number mass number. Since boron exists as a mixture of two different isotopes in specific proportions, the weighted average falls between $10$ and $11$, giving a non-whole-number result.
QUESTION 16 4 marks Criterion C
Medium
Unknown particleRelative massRelative charge
Particle 110
Particle 21/1836?1
Particle 31+1

A beam experiment measured the relative mass and relative charge of three unknown particles.

a. Match each unknown particle to a proton, neutron, or electron.
[2]
b. Explain how the data allowed you to distinguish Particle 1 from Particle 3, since they have the same relative mass.
[2]
Show complete worked solution
(a)
Particle 1 (mass $1$, charge $0$) is a neutron. Particle 2 (mass $\approx0$, charge $-1$) is an electron. Particle 3 (mass $1$, charge $+1$) is a proton.
(b)
Particle 1 and Particle 3 have the same relative mass ($1$), so mass alone cannot tell them apart. However, their relative charge is different — Particle 1 has no charge ($0$) while Particle 3 has a positive charge ($+1$) — and it is this difference in charge that identifies Particle 1 as a neutron and Particle 3 as a proton.
QUESTION 17 7 marks Criterion C
Hard
IsotopeMass number% abundance
Magnesium-242478.99
Magnesium-252510.00
Magnesium-262611.01

Natural magnesium is a mixture of three isotopes, shown in the table.

a. Calculate the relative atomic mass of magnesium, using all three isotopes. Give your answer to 2 decimal places.
[4]
b. State which isotope is most abundant, and explain why the relative atomic mass calculated is closest to that isotope's mass number.
[2]
c. Suggest why it is important to measure these abundance percentages very precisely.
[1]
Show complete worked solution
(a)
$$ A_r = (24 \times 0.7899) + (25 \times 0.1000) + (26 \times 0.1101) $$

$$ = 18.9576 + 2.5000 + 2.8626 $$

$$ = 24.3202 $$

Answer: $A_r = 24.32$ (2 d.p.).

(b)
Magnesium-24 is by far the most abundant isotope ($78.99\%$, almost four-fifths of all magnesium atoms). Because a weighted average is dominated by whichever value has the largest weighting, the overall relative atomic mass ($24.32$) ends up very close to magnesium-24's mass number ($24$), only slightly higher due to the smaller contributions from the heavier isotopes.
(c)
Relative atomic mass values are used throughout chemistry for accurate calculations (for example, working out the mass of reactants and products in a reaction), so even a small error in the measured abundances would carry through and cause errors in every calculation that relies on that relative atomic mass.
QUESTION 18 5 marks Criterion D
Medium

Carbon dating uses the fact that living things constantly absorb a tiny, steady proportion of the radioactive isotope carbon-14 alongside the much more common, stable carbon-12. After an organism dies, its carbon-14 gradually decays at a known rate, allowing scientists to estimate the age of ancient remains, wooden artefacts, and other organic archaeological finds.

Discuss one benefit and one drawback of using this isotope-based dating method.

Show complete worked solution

Benefit: Carbon dating gives archaeologists and historians a scientific, fairly accurate way to determine the age of organic remains going back tens of thousands of years, without needing any written records. This has been essential for building an accurate timeline of human history and for confirming or correcting the dates of major archaeological discoveries.

Drawback: The method only works for objects that were once living (containing carbon) and becomes unreliable beyond about $50\,000$ years, since by then almost all of the carbon-14 has decayed away, leaving too little to measure accurately. Results can also be thrown off by contamination of the sample with more recent carbon, so careful sample handling and cross-checking against other dating methods is needed to trust the result.

QUESTION 19 6 marks Criterion D
Hard

Iodine-131 is a radioactive isotope of iodine. Because the body naturally absorbs iodine into the thyroid gland (using its chemical properties, which depend on its number of protons, not neutrons), doctors can give patients iodine-131 to treat certain thyroid conditions, including some thyroid cancers — the radiation it emits destroys targeted thyroid cells from within.

Evaluate the impact of this use of isotopes in medicine, discussing both a benefit and a concern.

Show complete worked solution

Benefit: Because iodine-131 behaves chemically just like ordinary iodine, it travels naturally and specifically to the thyroid gland, allowing doctors to deliver a targeted radioactive treatment directly to diseased thyroid tissue while affecting far less of the rest of the body than external radiotherapy would. This has made it a highly effective, relatively non-invasive treatment for certain thyroid cancers and other thyroid conditions, saving many lives.

Concern: The patient becomes radioactive for a period after treatment, meaning they must follow strict precautions (such as limiting close contact with others, especially children and pregnant women) to avoid exposing other people to radiation. There are also risks associated with producing, transporting, storing, and disposing of radioactive isotopes safely, and access to this treatment can be limited in areas without the specialist facilities and trained staff required to handle radioactive medicine safely.

QUESTION 20 6 marks Criterion D
Hard

Natural uranium contains mostly uranium-238, with only a small percentage of uranium-235 — the isotope needed for nuclear power stations (and nuclear weapons) to work. "Enrichment" is the process of increasing the proportion of uranium-235 in a sample by separating the two isotopes, which differ very slightly in mass due to their different neutron numbers.

Evaluate the impact of uranium enrichment technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: Enrichment makes it possible to produce nuclear fuel with enough uranium-235 to sustain the chain reaction needed in a nuclear power station, generating large amounts of reliable, low-carbon electricity from a relatively small amount of fuel — an important tool in reducing dependence on fossil fuels and tackling climate change.

Concern: The same enrichment technology and expertise used to make low-enriched fuel for power stations can, if enrichment is taken much further, be used to produce highly enriched uranium suitable for nuclear weapons. This "dual-use" nature means uranium enrichment is tightly controlled and monitored internationally, since the spread of enrichment technology raises serious concerns about nuclear weapons proliferation, alongside the ongoing challenge of safely storing the radioactive waste that nuclear power production creates.

Electron Arrangement (basic) 20 questions

QUESTION 1 3 marks Criterion A
Easy
11p?12n

The diagram shows the electron arrangement of an atom of sodium.

a. Write the electron arrangement shown, as a series of numbers (e.g. "2, 8, ...").
[2]
b. State the atomic number of this atom.
[1]
Show complete worked solution
(a)
Counting the electrons in each shell from the diagram, working outward: $2, 8, 1$.
(b)
The atomic number equals the total number of electrons: $2+8+1=\mathbf{11}$.
QUESTION 2 2 marks Criterion A
Easy

State the maximum number of electrons that can occupy the first, second, and third electron shells of an atom (for the first 20 elements).

Show complete worked solution

First shell: maximum $2$ electrons. Second shell: maximum $8$ electrons. Third shell: maximum $8$ electrons (for the first $20$ elements).

QUESTION 3 4 marks Criterion A
Medium

Chlorine has an atomic number of $17$.

a. Write the electron arrangement of a chlorine atom.
[2]
b. State how many more electrons chlorine would need to completely fill its outer shell.
[2]
Show complete worked solution
(a)
$17$ electrons fill the shells: $2$ in the first shell, $8$ in the second shell, leaving $17-2-8=7$ for the third shell. Electron arrangement: $2, 8, 7$.
(b)
The third (outer) shell can hold up to $8$ electrons, but chlorine has only $7$ in it. It would need $1$ more electron to fill its outer shell.
QUESTION 4 4 marks Criterion A
Medium

An atom has the electron arrangement $2, 8, 2$.

a. Calculate the atomic number of this atom.
[1]
b. State the number of electron shells this atom has, and explain what this tells you about the atom's position in the periodic table (its period).
[3]
Show complete worked solution
(a)
Atomic number $=$ total electrons $= 2+8+2 = \mathbf{12}$ (this is magnesium).
(b)
This atom has $3$ occupied electron shells. The number of occupied shells matches the period number in the periodic table, so this element is in Period 3.
QUESTION 5 4 marks Criterion A
Medium
13p?14n

The diagram shows the electron arrangement of an atom of aluminium.

a. State the atomic number of this atom, using the diagram.
[1]
b. State the number of electron shells shown.
[1]
c. Predict which group of the periodic table this element belongs to, giving a reason.
[2]
Show complete worked solution
(a)
Total electrons $=2+8+3=\mathbf{13}$.
(b)
$3$ electron shells.
(c)
The number of electrons in the outer shell usually matches the group number for main-group elements. This atom has $3$ outer-shell electrons, so it is predicted to be in Group 13 (this is aluminium).
QUESTION 6 5 marks Criterion A
Medium
10p?10n

Noble gases, such as helium, neon, and argon, are described as very unreactive elements.

a. State the electron arrangement of a neon atom (atomic number $10$).
[1]
b. Explain, using the idea of a "full outer shell", why noble gases are unreactive.
[2]
c. Argon has the electron arrangement $2, 8, 8$. Predict, with a reason, whether argon would also be unreactive.
[2]
Show complete worked solution
(a)
$2, 8$.
(b)
Neon's outer (second) shell contains $8$ electrons, which is the maximum it can hold — a full outer shell. Atoms with a full outer shell are very stable and have little tendency to lose, gain, or share electrons, which is why they rarely react with other elements.
(c)
Argon would also be predicted to be unreactive. Its outer (third) shell contains $8$ electrons, which is the maximum for that shell at this level — a full outer shell, just like neon's — giving argon the same kind of stability.
QUESTION 7 6 marks Criterion A
Hard
19p?20n

Potassium has an atomic number of $19$.

a. Determine the electron arrangement of a potassium atom, showing how you worked it out.
[2]
b. Explain why the third shell holds only $8$ electrons here, rather than continuing to fill with the $9$th electron.
[2]
c. Using this electron arrangement, predict potassium's group and comment on its expected reactivity.
[2]
Show complete worked solution
(a)
Fill the first shell: $2$ electrons ($19-2=17$ remaining). Fill the second shell: $8$ electrons ($17-8=9$ remaining). Fill the third shell up to its maximum: $8$ electrons ($9-8=1$ remaining). The final electron goes into a new, fourth shell: $1$ electron. Electron arrangement: $2, 8, 8, 1$.
(b)
At this basic level, the third shell is treated as holding a maximum of $8$ electrons (for the first $20$ elements), the same as the second shell. Once it reaches $8$, any additional electrons must start occupying a new, outer shell rather than continuing to add to the third shell — this is why potassium's $19$th electron starts a new, fourth shell instead of making the third shell hold $9$.
(c)
With $1$ electron in its outer shell, potassium is predicted to be in Group 1. Group 1 elements are highly reactive metals, since they only need to lose that single outer electron to achieve a full outer shell — potassium is expected to be very reactive, more so than lighter Group 1 metals, since its outer electron is further from the nucleus and less strongly attracted to it.
QUESTION 8 6 marks Criterion B
Medium

A class investigates the flame test colours produced by different metal ions, in order to identify unknown metal salts, using a nichrome wire dipped into each sample and held in a blue Bunsen flame.

a. State the independent variable and the dependent variable in this investigation.
[2]
b. Describe the method and equipment needed to collect reliable flame-test data for several different metal salts.
[3]
c. State one variable that should be controlled for a fair comparison between samples.
[1]
Show complete worked solution
(a)
Independent variable: the metal salt (sample) being tested. Dependent variable: the colour of the flame produced.
(b)
Clean a nichrome wire loop by dipping it in dilute hydrochloric acid and holding it in a roaring blue Bunsen flame until no colour shows. Dip the clean wire into the first metal salt sample, then hold it at the edge of the blue flame and record the colour observed. Clean the wire thoroughly between each test (repeating the acid-and-flame cleaning step) before testing the next sample, to avoid contamination carrying colour over from one test to the next.
(c)
The amount (mass) of each metal salt sample used, and the position/distance in the flame where the wire is held, should be kept the same for every test.
QUESTION 9 6 marks Criterion B
Medium

A student wants to investigate how reliably flame test colours can be used to identify an unknown metal salt, by comparing an unknown sample's flame colour against flame colours from several known reference salts.

a. State the independent and dependent variables when comparing the unknown sample to the known reference salts.
[2]
b. Explain how testing the known reference salts alongside the unknown sample, rather than relying on memory or a textbook picture, would improve the reliability of the identification.
[2]
c. Suggest an improvement using more precise equipment that would make the identification more objective.
[2]
Show complete worked solution
(a)
Independent variable: which sample is tested (the unknown, or each known reference salt). Dependent variable: the flame colour produced, compared against the unknown's colour.
(b)
Flame colours can look different depending on lighting conditions, the exact Bunsen flame used, and how a person perceives colour. Testing known reference salts side by side, under the exact same conditions, as the unknown sample gives a direct, fair comparison, removing the guesswork of matching a live flame colour to a memory or a printed picture that may look different.
(c)
Use a flame photometer or spectroscope to view or measure the exact wavelengths of light produced, rather than relying on the human eye to judge colour. This gives an objective, numerical reading of the light emitted, removing the uncertainty of visually comparing similar-looking colours.
QUESTION 10 8 marks Criterion B
Hard
Sample testedStudent's flame colour IDActual metal ion
1Red (identified as strontium)Strontium
2Crimson red (identified as strontium)Lithium
3Lilac (identified as potassium)Potassium
4Yellow (identified as sodium)Sodium

A student carried out flame tests on four unknown metal salts, judging each colour by eye. Sample $2$ was misidentified — lithium produces a crimson-red flame, which the student confused with strontium's similar red flame.

a. Identify the source of error that led to sample 2 being misidentified.
[2]
b. Suggest an improvement to the method that would reduce this kind of error, and explain why it works.
[3]
c. Explain, using ideas about electron arrangement, why each metal produces its own distinctive set of coloured lines in its spectrum.
[3]
Show complete worked solution
(a)
The error is subjective judgement of very similar colours by eye — lithium (crimson-red) and strontium (red) produce flame colours that look extremely similar to a human observer, making them easy to confuse without a more precise method.
(b)
Use a spectroscope to examine the flame's light spectrum, or a flame photometer to measure the exact wavelengths of light emitted, instead of judging colour by eye. Different metals produce light at specific, distinct wavelengths even when the overall colours look similar to the eye, so an instrument that measures wavelength precisely can objectively distinguish lithium from strontium in a way the human eye cannot.
(c)
In a flame, energy from the heat causes electrons in a metal atom to jump up to higher energy levels (further-out shells). When these electrons fall back down to their original, lower energy level, they release the extra energy as light of a very specific colour (wavelength). Because each element has its own unique arrangement of electrons and energy levels, the exact "jumps" possible — and so the exact colours of light produced — are different for every element, giving each metal its own characteristic set of spectral lines, even when the overall flame colour looks similar to another element by eye.
QUESTION 11 3 marks Criterion C
Easy
ElementAtomic number
Beryllium4
Fluorine9
Silicon14

Write the electron arrangement of an atom of fluorine.

Show complete worked solution

Fill the first shell: $2$ electrons ($9-2=7$ remaining), all of which go into the second shell (maximum $8$, and $7\le8$).

Answer: $2, 7$.

QUESTION 12 2 marks Criterion C
Easy
ElementElectron arrangement
Nitrogen2, 5
Neon2, 8
Sodium2, 8, 1

Identify which of these elements is a noble gas, and explain how you can tell from its electron arrangement.

Show complete worked solution

Neon is the noble gas. Its outer shell ($8$ electrons) is completely full — the maximum the second shell can hold — which is the defining feature of a noble gas's electron arrangement.

QUESTION 13 4 marks Criterion C
Medium
ElementSodiumMagnesiumAluminiumArgon
Electron arrangement2, 8, 12, 8, 22, 8, 32, 8, 8
Group121318

The table shows the electron arrangement and periodic table group of four Period 3 elements.

a. Describe the pattern between the number of outer-shell electrons and the group number.
[2]
b. Using this pattern, predict the electron arrangement of phosphorus (atomic number $15$, Group $15$).
[2]
Show complete worked solution
(a)
For these main-group elements, the number of electrons in the outer shell matches the group number exactly for groups $1$ and $2$ (e.g. $1$ outer electron $\to$ Group $1$, $2$ outer electrons $\to$ Group $2$). For Group $13$ and Group $18$, the pattern is the outer-shell electron count plus $10$ (e.g. $3$ outer electrons $\to$ Group $13$; $8$ outer electrons $\to$ Group $18$).
(b)
Following the pattern, Group $15$ should have $5$ outer-shell electrons ($15-10=5$). Filling the shells: $2$ in the first, $8$ in the second, leaving $15-2-8=5$ for the third (outer) shell. Predicted electron arrangement: $2, 8, 5$ — matching the Group $15$ prediction.
QUESTION 14 4 marks Criterion C
Medium
Metal ionFlame test colour
LithiumCrimson red
SodiumYellow/orange
PotassiumLilac
CalciumBrick red
CopperBlue-green

An unknown sample produces a lilac flame when tested.

a. Identify the metal most likely present in the unknown sample.
[2]
b. Explain, in terms of electron arrangement, why different metals produce different flame colours.
[2]
Show complete worked solution
(a)
Potassium — its flame test colour, lilac, matches the table exactly.
(b)
Heat energy from the flame makes electrons jump up to a higher shell (energy level). When they fall back down, they release energy as light of a specific colour. Since each metal has a different electron arrangement, the specific "jumps" its electrons can make are different, so each metal releases light of a different, characteristic colour.
QUESTION 15 5 marks Criterion C
Medium
ElementZRecorded arrangement
Lithium32, 1
Beryllium42, 2
Boron52, 3
Carbon62, 4
Nitrogen72, 5
Oxygen82, 5, 1
Fluorine92, 7
Neon102, 8

A student recorded the electron arrangement of eight elements. One entry contains an error.

a. Identify which entry is inconsistent with the rest of the data.
[1]
b. Explain how you identified it, using the shell-filling rule.
[2]
c. State the correct electron arrangement for this element.
[2]
Show complete worked solution
(a)
Oxygen's recorded arrangement, "$2, 5, 1$", is inconsistent.
(b)
The first shell should always fill to its maximum of $2$ before the second shell starts filling, and the second shell should fill (up to $8$) before any electrons go into a third shell. Oxygen has $8$ electrons in total, which should all fit within the first two shells ($2+6=8$), so a third shell should not appear at all — the recorded "$2,5,1$" incorrectly starts a third shell far too early, and its total ($2+5+1=8$) reveals the numbers were split incorrectly between shells.
(c)
$2, 6$.
QUESTION 16 4 marks Criterion C
Medium
MetalLithiumSodiumPotassium
Electron arrangement2, 12, 8, 12, 8, 8, 1
Approx. time to fully react with water30 s5 s~1 s (almost instant)

The table shows data for three Group 1 metals reacting with water.

a. Describe the trend shown in reaction time going down the group, from lithium to potassium.
[2]
b. Using the electron arrangements given, suggest a reason for this trend.
[2]
Show complete worked solution
(a)
Reaction time decreases (reactions get faster) going down the group, from $30\,\text{s}$ for lithium, to $5\,\text{s}$ for sodium, to almost instant for potassium — meaning reactivity increases down the group.
(b)
Each metal has just $1$ electron in its outer shell, but the number of shells increases going down the group ($2$ shells for lithium, $3$ for sodium, $4$ for potassium). This means the outer electron is further from the nucleus in the heavier metals, so it is held less tightly and is easier to lose — which is why reactivity increases down the group.
QUESTION 17 7 marks Criterion C
Hard
MetalLithium (Z=3)Sodium (Z=11)Potassium (Z=19)
Electron arrangement2, 12, 8, 12, 8, 8, 1
Time to fully react with water30 s5 s~1 s

The table shows electron arrangement and reactivity data for three Group 1 metals.

a. State what is the same about the electron arrangement of all three metals, and describe how the number of shells changes going down the group.
[2]
b. Describe the trend in reactivity shown by the reaction-time data.
[2]
c. Explain this trend fully, using ideas about electron arrangement and the attraction between the nucleus and the outer electron.
[3]
Show complete worked solution
(a)
All three metals have exactly $1$ electron in their outer shell. Going down the group, the number of occupied electron shells increases by one each time — $2$ shells for lithium, $3$ for sodium, $4$ for potassium.
(b)
Reactivity increases down the group — the reaction time drops sharply from $30\,\text{s}$ (lithium) to $5\,\text{s}$ (sodium) to about $1\,\text{s}$ (potassium), meaning each metal down the group reacts faster (and therefore more vigorously) than the one above it.
(c)
In each case, reactivity depends on how easily the single outer electron can be lost. As the number of shells increases down the group, the outer electron sits further from the positively charged nucleus, and there are more inner shells of electrons between it and the nucleus (partly "shielding" it from the nucleus's pull). Both effects weaken the attraction between the nucleus and the outer electron, making it easier to remove — which is exactly why potassium's outer electron is lost fastest (most reactive) and lithium's is held on to longest (least reactive) among the three.
QUESTION 18 5 marks Criterion D
Medium

Argon, a noble gas with a full outer shell of electrons, is used to fill the space inside many light bulbs, including LED and incandescent bulbs. Its lack of reactivity (due to its stable, full outer shell) means it does not react with the hot metal filament or other components inside the bulb.

Discuss one benefit and one drawback of using argon in this way.

Show complete worked solution

Benefit: Because argon is so unreactive, it does not react with or oxidise the hot filament inside the bulb, which would otherwise burn out much faster in ordinary air (which contains reactive oxygen). This significantly extends the working life of light bulbs, meaning fewer bulbs need to be manufactured and thrown away, saving resources, energy, and money for consumers over time.

Drawback: Argon must be extracted from the air (through an energy-intensive industrial process called fractional distillation of liquid air), which uses a significant amount of energy and has its own environmental footprint. Manufacturing bulbs that are sealed to contain the argon gas also adds cost and complexity compared with simpler designs, and once a bulb reaches the end of its life, it must be disposed of or recycled correctly, since simply smashing it releases the gas and any other bulb components.

QUESTION 19 6 marks Criterion D
Hard

Lithium has the electron arrangement $2, 1$ — a single, loosely held outer electron that it readily loses. This property makes lithium extremely useful in the rechargeable lithium-ion batteries that power phones, laptops, and increasingly, electric cars.

Evaluate the impact of lithium-ion battery technology, discussing both a benefit and a concern it raises.

Show complete worked solution

Benefit: Lithium's electron arrangement makes lithium-ion batteries able to store and release a large amount of energy for their weight, compared with older battery types. This has enabled the portable electronics that most people now rely on daily, and is a key technology supporting the shift from petrol and diesel cars to electric vehicles, helping to reduce transport-related greenhouse gas emissions.

Concern: Mining the lithium (and other metals) needed for these batteries can cause significant environmental damage, including large water use in some lithium-rich regions where water is already scarce, and habitat disruption. Lithium's high reactivity (the same property that makes it useful in batteries) also means damaged or poorly made lithium-ion batteries carry a real fire risk, and disposing of old batteries safely and recycling their materials remains a growing challenge as demand for this technology increases.

QUESTION 20 6 marks Criterion D
Hard

Helium has a full outer shell of electrons ($2$), making it extremely unreactive (inert) and giving it an unusually low boiling point. Liquid helium's extreme coldness and inertness make it essential for cooling the powerful superconducting magnets inside MRI (Magnetic Resonance Imaging) scanners, used in hospitals worldwide for non-invasive medical diagnosis.

Evaluate the impact of relying on helium for this technology, discussing both a benefit and a concern.

Show complete worked solution

Benefit: Helium's full outer shell means it does not react with the sensitive magnet materials it cools, and its extremely low boiling point allows it to cool the magnets enough to become superconducting, which is essential for producing the strong, stable magnetic fields MRI scanners need to create detailed images inside the human body. This has made MRI a hugely valuable, non-invasive diagnostic tool, helping doctors detect and treat conditions such as tumours, injuries, and diseases without surgery.

Concern: Helium is a finite, non-renewable resource on Earth — once released into the air, its light atoms escape Earth's atmosphere into space and are lost forever, unlike most other resources that can eventually be recycled. Global helium shortages have already affected hospitals and scientific research, raising concerns about the long-term availability and rising cost of helium for essential medical technology like MRI scanners, unless new sources are found or usage becomes more efficient.