Atomic Structure
11 questions, 76 marks, every one with its mark scheme.
Or open them one at a time, as you finish each question.
Question 01
8 marksWhich atom contains two more protons and one more neutron than an atom of 40Ar?
Tick (✓) one box.
- 41K
- 42Ca
- 43Ca
- 43Sc
Mark scheme
- 43Ca1
Complete the table to show the numbers of fundamental particles in each species.
| Species | Number of protons | Number of neutrons | Number of electrons |
|---|---|---|---|
| 65Zn2+ | |||
| 80Br− |
Mark scheme
- 65Zn2+: 30, 35, 281
- 80Br−: 35, 45, 361
In the early nineteenth century, atoms were thought of as tiny solid spheres that could not be divided into smaller particles.
State two ways in which the model of the atom used today differs from this model.
Mark scheme
- A central nucleus containing protons and neutrons1
- Electrons arranged in energy levels around the nucleus1
Neon has two main isotopes, 20Ne and 22Ne.
Considering only the numbers of protons, neutrons and electrons, state one way in which an atom of 20Ne is the same as an atom of 22Ne, and one way in which the two atoms differ.
Mark scheme
- Similarity: same number of protons or electrons1
- Difference: different number of neutrons1
Do not accept: same electron configuration
Explain why 20Ne and 22Ne have the same chemical properties.
Mark scheme
- Same electron configuration1
Do not accept: same number of protons; electrons determine chemical properties
Question 02
4 marksZirconium extracted from a mineral was investigated by time of flight (TOF) mass spectrometry. The table shows the results.
| Mass number | Relative intensity |
|---|---|
| 90 | 25.7 |
| 91 | 5.6 |
| 92 | 8.6 |
| 94 | 8.7 |
| 96 | 1.4 |
State what is meant by relative atomic mass.
Mark scheme
- The average (mean) mass of one atom of the element1
- divided by one twelfth of the mass of one atom of carbon-121
Do not accept: average mass of one atom of the element divided by one twelfth of the mass of one mole of carbon-12
Calculate the relative atomic mass of zirconium from these results.
Give your answer to 1 decimal place.
Mark scheme
- [(90 × 25.7) + (91 × 5.6) + (92 × 8.6) + (94 × 8.7) + (96 × 1.4)] ÷ 50.01
- 91.31
Do not accept: 91.32 ÷ 100 = 45.7; 91
Question 03
8 marksWrite the full electron configuration for an atom of nickel.
Mark scheme
- 1s2 2s2 2p6 3s2 3p6 3d8 4s21
Do not accept: [Ar]3d84s2
Complete the table to show the full electron configuration of each ion.
| Ion | Full electron configuration |
|---|---|
| Ni2+ | |
| S2− |
Mark scheme
- 1s2 2s2 2p6 3s2 3p6 3d81
- 1s2 2s2 2p6 3s2 3p61
Do not accept: 1s2 2s2 2p6 3s2 3p6 3d6 4s2 for Ni2+
Write the full electron configuration for an atom of copper.
Mark scheme
- 1s2 2s2 2p6 3s2 3p6 3d10 4s11
Do not accept: 1s2 2s2 2p6 3s2 3p6 3d9 4s2
An ion X3+ has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d5
Identify element X.
Mark scheme
- Fe1
Do not accept: V
A compound is made of a 1+ ion and a 2− ion. Both ions have the same electron configuration as a neon atom.
Give the formula of the compound.
Mark scheme
- Na2O1
The K+ ion and the Cl− ion are isoelectronic.
Explain why the Cl− ion is larger than the K+ ion.
Mark scheme
- Cl− has fewer protons with the same electron arrangement1
- so a weaker attraction between the nucleus and the outer electrons1
Do not accept: chlorine; the ions have different numbers of electrons
Question 04
10 marksElectron impact is used to turn zinc atoms into ions in a TOF mass spectrometer.
Write an equation, with state symbols, for this ionisation.
Mark scheme
- Zn(g) + e− → Zn+(g) + 2e−1
Do not accept: Zn + e− → Zn+ + 2e− (no state symbols)
Antibiotic A is a large organic molecule. Its relative molecular mass is found by TOF mass spectrometry with electrospray ionisation.
Describe how A is ionised by electrospray ionisation. Give an equation for the ionisation of A.
Mark scheme
- A is dissolved in a volatile polar solvent such as methanol1
- injected through a needle at high voltage1
- each molecule gains a proton, H+1
- A + H+ → AH+1
Do not accept: atoms gain a proton; molecules lose an electron
The spectrum of A shows a single peak at m/z = 734
Give the relative molecular mass of A.
Mark scheme
- 7331
Do not accept: 734
Suggest why electrospray ionisation is used for A, rather than electron impact.
Mark scheme
- A does not break up or fragment1
Give two reasons why the particles in a sample must be turned into ions before TOF mass spectrometry can be carried out on them.
Mark scheme
- Ions, not molecules, are accelerated by an electric field1
- Only ions create a current when they hit the detector1
Do not accept: accelerated by a magnetic field
Four samples are investigated in one TOF mass spectrometer. Every ion formed has a charge of 1+ and the electric field gives every ion an identical kinetic energy.
Which sample forms ions that reach the detector at the same time as 32S+ ions?
Tick (✓) one box.
- HCOOH ionised by electrospray
- CH3OH ionised by electrospray
- H2S ionised by electron impact
- N2H4 ionised by electron impact
Mark scheme
- N2H4 ionised by electron impact1
Question 05
7 marksA sample of gallium containing the isotopes 69Ga and 71Ga is investigated by TOF mass spectrometry, with electron impact ionisation.
Describe how the gallium ions are accelerated in the spectrometer.
Mark scheme
- Positive ions are accelerated by an electric field1
- to a constant kinetic energy1
Do not accept: magnetic field
Identify the ion that reaches the detector first. Explain your answer.
Mark scheme
- 69Ga+1
- Same kinetic energy1
- lower m/z moves faster, so it arrives at the detector first1
Do not accept: Ga
Explain how the ions are detected and how their relative abundance is measured.
Mark scheme
- Each ion hits the detector, a negative plate, and gains an electron, so a current is generated1
- The current is proportional to the abundance1
Do not accept: positive plate; the detector counts the ions
Question 06
7 marksA sample of silicon used to make solar cells contains the isotopes 28Si, 29Si and 30Si.
The sample contains 5.0% 29Si. The relative atomic mass of silicon in the sample is 28.11
Calculate the percentage abundances of 28Si and 30Si in the sample.
Mark scheme
- Let % of 30Si = x and % of 28Si = (95.0 − x)1
- 28.11 = [28(95.0 − x) + (29 × 5.0) + 30x] ÷ 1001
- x = % of 30Si = 3.0%1
- % of 28Si = 92.0%1
Element E has two isotopes. In a sample of E, 69.2% of the atoms have a mass number of 63
The relative atomic mass of E is 63.6
Calculate the mass number of the other isotope of E.
Mark scheme
- Abundance of the other isotope = 30.8%1
- 63.6 = [(63 × 69.2) + (m × 30.8)] ÷ 1001
- m = 64.95, so mass number = 651
Do not accept: 64.95; 64.9
Question 07
4 marksA sample of chlorine gas, Cl2, has been enriched so that it contains 35Cl and 37Cl atoms in the ratio 1 : 3
Electron impact ionisation is used to record a TOF mass spectrum of the sample. The spectrum shows peaks for Cl2+ ions only.
Complete the table to show the relative abundance of each peak.
| m/z | Relative abundance |
|---|---|
| 70 | |
| 72 | |
| 74 | 9 |
Mark scheme
- 11
- 61
Do not accept: 3 for m/z = 72
Calculate the relative atomic mass of the enriched chlorine.
Give your answer to 1 decimal place.
Mark scheme
- [(35 × 1) + (37 × 3)] ÷ 41
- 36.51
Question 08
8 marksElement Q is investigated by TOF mass spectrometry. Electron impact converts the atoms into 1+ ions.
After acceleration, ions of one isotope of Q each carry 4.10 × 10−16 J of kinetic energy. These ions take 1.214 × 10−5 s to travel through a flight tube of length 82.5 cm.
Use KE = ½mv2, and take 6.022 × 1023 mol−1 for L, the Avogadro constant.
Calculate the mass number of this isotope of Q.
Show your working.
Mark scheme
- v = d / t = 0.825 ÷ 1.214 × 10−5 = 6.796 × 104 m s−11
- m = 2KE / v2 = 2 × 4.10 × 10−16 ÷ (6.796 × 104)2 = 1.776 × 10−25 kg1
- mass of one ion = 1.776 × 10−22 g1
- mass of one mole of ions = 1.776 × 10−22 × 6.022 × 1023 = 106.9 g1
- mass number = 1071
Do not accept: 106.9 as the mass number
Another isotope of Q has a mass number of 109
Calculate the time taken for ions of this isotope to travel through the same flight tube when accelerated to the same kinetic energy.
Give your answer to 3 significant figures.
Mark scheme
- m1 / t12 = m2 / t221
- t = 1.214 × 10−5 × √(109 / 107)1
- t = 1.23 × 10−51
Question 09
7 marksGive an equation, including state symbols, to represent the third ionisation energy of aluminium.
Mark scheme
- Al2+(g) → Al3+(g) + e−1
Do not accept: Al(g) → Al3+(g) + 3e−
Explain why the second ionisation energy of aluminium is greater than its first ionisation energy.
Mark scheme
- The electron is removed from a positive ion, so the attraction holding it is stronger1
Magnesium and calcium are both in Group 2.
Explain why the first ionisation energy of calcium is smaller than the first ionisation energy of magnesium.
Mark scheme
- Calcium has more shells, so the outer electron is further from the nucleus1
- The outer electron is more shielded1
- Weaker attraction between the nucleus and the outer electron1
Do not accept: calcium has a bigger atomic radius
Sodium has a higher second ionisation energy than magnesium.
Explain why.
Mark scheme
- Mg+ loses an electron from a 3s orbital and Na+ loses an electron from a 2p orbital1
- More shielding (in Mg+)1
Do not accept: Mg loses an electron from a 3s orbital (with no comparison)
Question 10
7 marksWhich one of these elements has the highest first ionisation energy?
Tick (✓) one box.
- Magnesium
- Aluminium
- Silicon
- Sulfur
Mark scheme
- Sulfur1
Beryllium and boron are neighbours in Period 2. State which of the two needs less energy to remove its first electron, and explain why.
Mark scheme
- B1
- Its outer electron is in a 2p sub-shell1
- 2p is higher in energy than 2s, so the electron is easier to remove1
Do not accept: Be; the electron is removed from a higher energy level
An element X is in Period 3. The table gives its first eight successive ionisation energies.
| Ionisation energy | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th | 8th |
|---|---|---|---|---|---|---|---|---|
| Value / kJ mol−1 | 1 251 | 2 298 | 3 822 | 5 159 | 6 542 | 9 362 | 11 018 | 33 604 |
Identify the element. Explain how the data show this.
Mark scheme
- Chlorine (Cl)1
- A large increase from the 7th to the 8th ionisation energy1
- The 8th electron is removed from a lower energy level, closer to the nucleus1
Question 11
6 marksDescribe and explain the trend in first ionisation energy of the elements across Period 3 from sodium to argon.
In your answer, explain any elements that do not follow the general trend.
Mark scheme
| Level | Marks | Description |
|---|---|---|
| 3 | 5-6 | All three stages are covered and the explanation of each stage is generally correct and virtually complete. The answer is communicated coherently and shows a logical progression through stages 1, 2 and 3. |
| 2 | 3-4 | Two stages are covered with matching justifications that are generally correct and virtually complete, or all three stages are covered but the justifications are incomplete or contain inaccuracies. |
| 1 | 1-2 | One stage is covered with a matching justification that is generally correct and virtually complete, or two stages are covered but the justifications are incomplete or contain inaccuracies. |
| 0 | No relevant content |
Indicative content
- Stage 1: general trend
- 1a. First ionisation energy increases across the period
- 1b. More protons, an increased nuclear charge
- 1c. Electrons in the same shell
- 1d. No extra shielding
- 1e. Stronger attraction between the nucleus and the outer electron
- Stage 2: Mg to Al
- 2a. Al has a lower first ionisation energy than Mg
- 2b. Outer electron of Al is in a 3p sub-shell
- 2c. 3p is higher in energy than 3s, so the electron is easier to remove
- Stage 3: P to S
- 3a. S has a lower first ionisation energy than P
- 3b. Outer electrons in the 3p sub-shell begin to pair
- 3c. Paired electrons repel, so less energy is needed to remove one