Atomic Structure
5 questions, 47 marks, every one with its mark scheme.
Or open them one at a time, as you finish each question.
Question 01
10 marksComplete the table to show the number of protons, neutrons and electrons in each species.
| Species | Protons | Neutrons | Electrons |
|---|---|---|---|
| 79Br− | |||
| 58Ni2+ | |||
| 32S2− |
Mark scheme
- 79Br−: 35, 44, 361
- 58Ni2+: 28, 30, 261
- 32S2−: 16, 16, 181
Give the relative mass of an electron.
Mark scheme
- 1/18361
Do not accept: 0
Define the term mass number.
Mark scheme
- number of protons plus neutrons in the nucleus of an atom1
Selenium-78 and selenium-80 are isotopes of selenium.
State what is meant by the term isotopes.
Mark scheme
- atoms of one element with the same number of protons1
- different number of neutrons1
Explain why selenium-78 and selenium-80 react in the same way.
Mark scheme
- same electron configuration1
Explain why an atom of selenium-78 and an atom of selenium-80 have the same atomic radius.
Mark scheme
- same number of protons1
- same number of electrons1
Do not accept: the isotope with more neutrons has the larger atomic radius
Question 02
7 marksA sample of gallium recovered from an old electronic component was analysed in a time of flight mass spectrometer. The table shows the results.
| m/z | Relative intensity |
|---|---|
| 69 | 12.0 |
| 71 | 8.4 |
Define the term relative atomic mass.
Mark scheme
- 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: the mass of one atom over one twelfth of the mass of one mole of carbon-12
Work out, to one decimal place, the relative atomic mass of the gallium in this sample.
Mark scheme
- (69 × 12.0 + 71 × 8.4) ÷ 20.41
- 69.81
The relative atomic mass of gallium in the Periodic Table is 69.7
Suggest why the value for this sample is different.
Mark scheme
- sample has different isotopes or different abundances of isotopes1
Iridium has two isotopes, 191Ir and 193Ir. A sample of iridium has a relative atomic mass of 192.2
Calculate the percentage abundance of 191Ir in this sample.
Mark scheme
- 192.2 = (191x + 193(100 − x)) ÷ 1001
- 40.0 %1
Question 03
17 marksA sample of rubidium is analysed in a time of flight mass spectrometer. Electron impact is used to form the ions.
Give an equation, with state symbols, showing what happens to a rubidium atom when it is hit by a high-energy electron.
Mark scheme
- Rb(g) → Rb+(g) + e−1
Give two reasons why the rubidium atoms must be ionised.
Mark scheme
- ions, not atoms, will interact with and be accelerated by an electric field1
- only ions will create a current when hitting the detector1
Do not accept: a magnetic field
Explain why the spectrometer is kept under a vacuum.
Mark scheme
Any 2 from:
- ions would collide with particles in the air1
- ions would not move in a straight line1
- particles in the air would be ionised and produce a peak1
Do not accept: to remove the air
The sample contains 85Rb and 87Rb.
State which of the two ions reaches the detector first. Explain your answer.
Mark scheme
- 85Rb+ has a shorter time of flight1
- both ions have the same kinetic energy1
- 87Rb+ has a lower velocity because it is heavier1
Do not accept: the same speed
Explain how the spectrometer measures the abundance of each ion.
Mark scheme
- ions hit the detector and accept electrons causing current to flow1
- bigger current = higher abundance of that ion1
Each 85Rb+ ion is given a kinetic energy of 4.00 × 10−16 J. The flight tube is 1.50 m long.
KE = ½mv2 where m is in kg. The Avogadro constant L = 6.022 × 1023 mol−1.
Calculate the time, in s, taken for a 85Rb+ ion to travel along the flight tube.
Give your answer to 3 significant figures.
Mark scheme
- m = 85 ÷ (1000 × 6.022 × 1023) = 1.41 × 10−25 kg1
- v = √(2 × 4.00 × 10−16 ÷ 1.41 × 10−25) = 7.53 × 104 m s−11
- t = d / v = 1.50 ÷ 7.53 × 1041
- 1.99 × 10−5 s1
The spectrometer is then used with a sample of a different element. A 1+ ion of one isotope, given the same kinetic energy, takes 2.03 × 10−5 s to travel along the same flight tube.
Calculate the mass number of this isotope and identify the element.
Mark scheme
- m = 2KEt2 / d2 = 2 × 4.00 × 10−16 × (2.03 × 10−5)2 ÷ 1.502 = 1.47 × 10−25 kg1
- 1.47 × 10−25 × 1000 × 6.022 × 1023 = 88.21
- mass number 88, strontium1
Question 04
6 marksGive the full electron configuration of a vanadium atom.
Mark scheme
- 1s2 2s2 2p6 3s2 3p6 4s2 3d31
Give the full electron configuration of a Ni2+ ion.
Mark scheme
- 1s2 2s2 2p6 3s2 3p6 3d81
Give the full electron configuration of a P3− ion.
Mark scheme
- 1s2 2s2 2p6 3s2 3p61
Copper does not follow the usual filling order. Give the full electron configuration of an atom of copper.
Mark scheme
- 1s2 2s2 2p6 3s2 3p6 4s1 3d101
Do not accept: 1s2 2s2 2p6 3s2 3p6 4s2 3d9
An ion X3+ has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d6
Identify element X.
Mark scheme
- cobalt1
Calcium and phosphorus react to form an ionic compound containing Ca2+ and P3− ions.
Give the formula of this compound.
Mark scheme
- Ca3P21
Question 05
7 marksElement Z is in Period 3. The table shows its first seven successive ionisation energies.
| Ionisation | 1st | 2nd | 3rd | 4th | 5th | 6th | 7th |
|---|---|---|---|---|---|---|---|
| Energy / kJ mol−1 | 1012 | 1907 | 2914 | 4964 | 6274 | 21268 | 25431 |
Use the data to deduce which group Z is in. Explain your answer.
Mark scheme
- Group 51
- large jump after the fifth electron is removed, as the sixth electron is taken from a shell closer to the nucleus1
Write an equation, including state symbols, to represent the third ionisation energy of element Z.
Mark scheme
- Z2+(g) → Z3+(g) + e−1
Do not accept: 2Z2+(g) → 2Z3+(g) + 2e−
Explain why the second ionisation energy of Z is greater than its first ionisation energy.
Mark scheme
- more energy is required to remove an electron from a positively charged ion than from an atom1
Do not accept: the second electron feels different shielding from the first
Which element has the highest fourth ionisation energy?
Tick (✓) one box.
- Sodium
- Magnesium
- Aluminium
- Silicon
Mark scheme
- Aluminium1
The third ionisation energy of magnesium is 7733 kJ mol−1. The third ionisation energy of aluminium is 2745 kJ mol−1.
Explain why the third ionisation energy of magnesium is greater than that of aluminium.
Mark scheme
- the electron is removed from the 2p sub-shell in Mg and from the 3s sub-shell in Al1
- the electron removed from Mg is less shielded than a 3s electron1
Do not accept: molecule