ESAT Chemistry: Atomic structure and the Periodic Table, set 3

• Answer all questions. No calculator.
• Each question has exactly one correct answer.

1. 11 mark

   An atom of potassium-40 has atomic number 19 and mass number 40. Take a proton or a neutron as having relative mass 1, and an electron as having relative mass 1/2000.

   Approximately what percentage of the total mass of this atom is contributed by its electrons?

   1. A About 32%
   2. B About 0.02%
   3. C About 48%
   4. D About 0.95%
2. 21 mark

   An atom has 3 more neutrons than protons, and a mass number of 65.

   What is its atomic number, and how many electrons does a neutral atom of this element contain?

   1. A Atomic number 34; 34 electrons
   2. B Atomic number 33; 33 electrons
   3. C Atomic number 31; 31 electrons
   4. D Atomic number 65; 65 electrons
3. 31 mark

   An ion of chromium, written Cr3+, has mass number 52. Chromium has atomic number 24.

   How many neutrons and how many electrons does this ion contain?

   1. A 28 neutrons and 21 electrons
   2. B 28 neutrons and 27 electrons
   3. C 24 neutrons and 21 electrons
   4. D 76 neutrons and 21 electrons
4. 41 mark

   An ion M2+ has the electron configuration 2,8,8.

   What is the atomic number of element M?

   1. A 18
   2. B 16
   3. C 22
   4. D 20
5. 51 mark

   Moving across Period 3 from sodium to argon, the first ionisation energy of the elements generally increases, even though every atom in the period has the same number of occupied electron shells.

   Which of the following correctly explains this trend, in terms of atomic structure?

   1. A The number of electrons in the complete inner shells increases across the period, so shielding of the outer electrons also increases; this extra shielding is what raises the energy needed to remove an outer electron.
   2. B Nuclear charge increases across the period while the outer electrons occupy the same outer shell, so shielding from the inner shells stays roughly constant and the outer electrons are held more tightly.
   3. C Atomic radius increases across the period, so the outer electrons sit further from the nucleus and are attracted less strongly, but the rising nuclear charge more than compensates, giving an overall increase in ionisation energy.
   4. D First ionisation energy always removes an electron from a complete inner shell rather than the outer shell, and the increasing stability of a complete shell across the period is what raises the energy needed.
6. 61 mark

   A mass spectrometer trace for an element shows two peaks: mass number 24 with relative peak height 4 units, and mass number 26 with relative peak height 1 unit.

   Calculate the relative atomic mass of this element.

   1. A 30.5
   2. B 25.6
   3. C 25.0
   4. D 24.4
7. 71 mark

   Element R exists as three isotopes: 70% with mass number 63, 20% with mass number 65, and 10% with mass number 66.

   Calculate the relative atomic mass of element R.

   1. A 64.7
   2. B 63.0
   3. C 63.7
   4. D 57.1
8. 81 mark

   An element Z is in the same Period as chlorine, and in the same Group as oxygen.

   Which element is Z?

   1. A Sulfur
   2. B Selenium
   3. C Silicon
   4. D Phosphorus
9. 91 mark

   Tellurium has atomic number 52 and a relative atomic mass of about 127.6. Iodine has atomic number 53 and a relative atomic mass of about 126.9.

   In the modern Periodic Table, which of these two elements is placed first, and why?

   1. A Iodine is placed first because it has the lower atomic number of the pair.
   2. B Tellurium is placed first because its atomic number (52) is lower than iodine's (53), even though tellurium's relative atomic mass is slightly higher.
   3. C Iodine is placed first because ordering by relative atomic mass takes priority whenever atomic number and atomic mass disagree.
   4. D Tellurium is placed first because the Periodic Table orders elements by increasing relative atomic mass, and tellurium's mass is the higher of the two.
10. 101 mark

    Each option below lists three Group-to-family pairings, using IUPAC numbering (Groups 1-18).

    Which option is entirely correct?

    1. A Group 1 - alkali metals; Group 2 - alkaline earth metals; Group 17 - halogens.
    2. B Group 1 - alkaline earth metals; Group 17 - halogens; Group 18 - noble gases.
    3. C Group 2 - alkaline earth metals; Group 16 - halogens; Group 18 - noble gases.
    4. D Group 1 - alkali metals; Group 16 - noble gases; Group 17 - halogens.
11. 111 mark

    An atom has the electron configuration 2,8,8,2.

    In which Group (IUPAC numbering, Groups 1-18) and Period is this element found?

    1. A Period 4, Group 12
    2. B Period 2, Group 4
    3. C Period 4, Group 8
    4. D Period 4, Group 2
12. 121 mark

    Potassium reacts more vigorously with water than sodium does, even though both are in Group 1.

    Which of the following correctly explains why, in terms of atomic structure?

    1. A Potassium has a smaller atomic radius than sodium, so its nucleus attracts the outer electron more strongly, making it more reactive.
    2. B Potassium has more protons than sodium, so its nucleus attracts electrons from other atoms more strongly, making potassium more reactive.
    3. C Potassium's outer electron is further from the nucleus and more strongly shielded than sodium's, so it is lost more easily.
    4. D Potassium has a full outer shell like sodium, so the extra shielding from a greater number of complete inner shells makes it less stable and more reactive.
13. 131 mark

    Fluorine reacts more vigorously than iodine, even though both are in Group 17.

    Which of the following correctly explains why, in terms of atomic structure?

    1. A Fluorine has more protons than iodine, so its nucleus attracts an incoming electron more strongly, making fluorine more reactive.
    2. B Fluorine's outer shell is closer to the nucleus and less shielded than iodine's, so it attracts an incoming electron more strongly.
    3. C Fluorine has a larger atomic radius than iodine, allowing it to attract an incoming electron from further away, making it more reactive.
    4. D Fluorine has a full outer shell like the noble gases, so it needs only a small energy input to react, making it more reactive.
14. 141 mark

    Four atoms are described: Atom W has mass number 39 and 20 neutrons; Atom X has mass number 41 and 20 neutrons; Atom Y has mass number 40 and 21 neutrons; Atom Z has mass number 39 and 19 neutrons.

    Which two atoms are isotopes of each other?

    1. A Atom W and Atom Z
    2. B Atom X and Atom Z
    3. C Atom W and Atom X
    4. D Atom W and Atom Y
15. 151 mark

    Atom X has 17 protons and 20 neutrons. Atom Y is an isotope of atom X with two fewer neutrons than X. Atom Y then gains one electron to form an ion.

    In which Group (IUPAC numbering, Groups 1-18) and Period of the Periodic Table is element Y found, and what is the charge on the ion it forms?

    1. A Group 17, Period 3, charge 1-
    2. B Group 7, Period 3, charge 1-
    3. C Group 17, Period 3, charge 1+
    4. D Group 18, Period 3, charge 1-

Worked solutions

Every question below carries the reasoning, not just the answer. The official material for this test publishes a correct option letter and nothing else.

1. Question 1Answer: B

   1. The atom's total relative mass is set almost entirely by its 40 nucleons, each of relative mass 1, giving a nucleon mass of 40.
   2. Potassium-40 has atomic number 19, so it has 19 electrons, each of relative mass about 1/2000.
   3. Total relative mass of the electrons = 19 x 1/2000 = 0.0095.
   4. Percentage of the atom's mass contributed by electrons = 0.0095 / 40 x 100% = 0.02375%, which rounds to about 0.02%.
   5. So virtually all of the atom's mass sits in the nucleus, which is option B.
   • Why not A: Treats each electron as contributing as much mass as a nucleon, comparing the electron count (19) with the total particle count (19 protons + 21 neutrons + 19 electrons = 59) rather than weighting by relative mass.
   • Why not C: Divides the number of electrons directly by the mass number (19/40), instead of weighting each electron by its tiny relative mass.
   • Why not D: Correctly totals the electrons' relative mass (19 x 1/2000 = 0.0095) but expresses it as a percentage of a single nucleon's mass rather than of the atom's full mass number of 40.

2. Question 2Answer: C

   1. Let p be the number of protons; the number of neutrons is then p + 3.
   2. Mass number = protons + neutrons = p + (p + 3) = 2p + 3 = 65.
   3. Solving, 2p = 62, so p = 31: the atomic number is 31.
   4. A neutral atom has equal numbers of protons and electrons, so it contains 31 electrons.
   5. So the atomic number is 31 with 31 electrons, which is option C.
   • Why not A: Solves the relationship with the wrong sign, treating the neutron count as 3 fewer than the proton count (2p - 3 = 65) instead of 3 more, giving p = 34.
   • Why not B: Ignores the stated 3-neutron difference entirely and simply halves the mass number, rounding up (65/2 = 32.5, rounded to 33).
   • Why not D: Uses the mass number itself as if it were the atomic number.

3. Question 3Answer: A

   1. Neutrons = mass number - atomic number = 52 - 24 = 28.
   2. A neutral chromium atom has 24 electrons, equal to its 24 protons.
   3. This ion carries a 3+ charge, meaning it has lost 3 electrons: 24 - 3 = 21 electrons.
   4. So this ion contains 28 neutrons and 21 electrons, which is option A.
   • Why not B: Gets the correct neutron count but treats the 3+ charge as if the ion had gained three electrons (24 + 3 = 27) instead of losing them.
   • Why not C: Gets the correct electron count but mistakes the atomic number (24) for the number of neutrons, instead of subtracting the atomic number from the mass number.
   • Why not D: Gets the correct electron count but adds the mass number and the atomic number (52 + 24 = 76) instead of subtracting one from the other to find the neutrons.

4. Question 4Answer: D

   1. The configuration 2,8,8 sums to 2 + 8 + 8 = 18 electrons in the ion.
   2. Forming a 2+ ion means the atom lost 2 electrons, so the neutral atom had 18 + 2 = 20 electrons.
   3. A neutral atom's electron count equals its atomic number, so M has atomic number 20.
   4. So the atomic number of M is 20, which is option D.
   • Why not A: Uses the ion's electron count directly as the atomic number, forgetting that 2 electrons were removed to form the 2+ ion.
   • Why not B: Subtracts the charge from the ion's electron count instead of adding it back (18 - 2 = 16).
   • Why not C: Adds twice the charge (4) rather than the charge itself (2) back onto the ion's electron count (18 + 4 = 22).

5. Question 5Answer: B

   1. Every element in Period 3, from sodium to argon, has 3 occupied electron shells, with the same fixed 2 + 8 = 10 electrons filling the first two, inner shells.
   2. Moving across the period, each successive element has one more proton, increasing the nuclear charge, while the extra electrons are added to the same outer, third shell rather than to a new shell.
   3. Because the inner-shielding electron count stays fixed at 10, the increasing nuclear charge is not cancelled out, so the effective nuclear charge pulling on the outer electrons rises across the period.
   4. A larger effective nuclear charge holds the outer electron more tightly, so more energy is needed to remove it: first ionisation energy increases across the period.
   5. So the correct explanation is option B.
   • Why not A: For every Period 3 element from sodium to argon, the inner shells hold a fixed 2 + 8 = 10 electrons; only the outer, third shell gains electrons across the period, so shielding from the inner shells does not increase.
   • Why not C: Atomic radius actually decreases across Period 3, not increases, because the rising nuclear charge pulls the same-shielded outer shell in more tightly; a genuinely larger radius would weaken the nuclear attraction rather than leave it still dominant.
   • Why not D: First ionisation energy removes the outermost, most loosely held electron, not one from a complete inner shell; inner-shell electrons are far more tightly bound and are not the ones removed first.

6. Question 6Answer: D

   1. Relative atomic mass is a weighted average of the isotope masses, weighted by relative abundance, here given as peak heights.
   2. The total of the peak heights is 4 + 1 = 5, so mass 24 makes up 4/5 of the sample and mass 26 makes up 1/5.
   3. Weighted sum = (4 x 24) + (1 x 26) = 96 + 26 = 122.
   4. Relative atomic mass = 122 / 5 = 24.4.
   5. So the relative atomic mass of this element is 24.4, which is option D.
   • Why not A: Correctly totals the weighted masses (4 x 24 + 1 x 26 = 122) but then divides by the height of the larger peak (4) instead of by the sum of both peak heights (5).
   • Why not B: Swaps the two peak heights between the masses, weighting mass 24 by 1 unit and mass 26 by 4 units instead of the other way round: (1 x 24 + 4 x 26)/5 = 128/5 = 25.6.
   • Why not C: Takes a simple, unweighted average of the two mass numbers (24 + 26)/2, ignoring the peak heights entirely.

7. Question 7Answer: C

   1. Relative atomic mass is the percentage-weighted average of all the isotopes' mass numbers.
   2. Mass 63 contributes 70%: 0.7 x 63 = 44.1.
   3. Mass 65 contributes 20%: 0.2 x 65 = 13.0.
   4. Mass 66 contributes 10%: 0.1 x 66 = 6.6.
   5. Adding all three contributions: 44.1 + 13.0 + 6.6 = 63.7.
   6. So the relative atomic mass of element R is 63.7, which is option C.
   • Why not A: Takes a simple, unweighted average of the three mass numbers (63 + 65 + 66)/3 = 194/3 = 64.7, ignoring the stated abundances.
   • Why not B: Uses only the mass number of the most abundant isotope (63), ignoring the other two isotopes entirely.
   • Why not D: Weights only the first two isotopes (0.7 x 63 + 0.2 x 65 = 44.1 + 13 = 57.1) and forgets that the remaining 10% (mass 66) must also be included.

8. Question 8Answer: A

   1. Chlorine's configuration 2,8,7 fills three shells, so chlorine is in Period 3: the unknown element Z must also be in Period 3.
   2. Oxygen's configuration 2,6 gives an outer-shell count of 6, and 6 + 10 = 16, so oxygen is in Group 16: Z must also be in Group 16.
   3. The Period-3 element in Group 16 is sulfur, with configuration 2,8,6 (outer-shell count 6, giving Group 6 + 10 = 16).
   4. So element Z is sulfur, which is option A.
   • Why not B: Shares oxygen's Group (16), but sits in Period 4, one Period below chlorine, rather than in chlorine's Period 3.
   • Why not C: Shares chlorine's Period (3), but sits in Group 14, not Group 16.
   • Why not D: Shares chlorine's Period (3), but sits in the adjacent Group 15, not Group 16.

9. Question 9Answer: B

   1. The modern Periodic Table orders elements by increasing atomic number, not by relative atomic mass.
   2. Tellurium has atomic number 52 and iodine has atomic number 53, so tellurium is placed before iodine.
   3. This holds despite tellurium's relative atomic mass (about 127.6) being slightly higher than iodine's (about 126.9), exactly the kind of pair that shows atomic number, not mass, is what fixes the order.
   4. So tellurium is placed first because of its lower atomic number, which is option B.
   • Why not A: Misremembers which element has the smaller atomic number: iodine's atomic number (53) is actually higher than tellurium's (52), not lower.
   • Why not C: Reverses which property is the actual ordering principle; the modern table always orders by atomic number, and relative atomic mass never overrides it.
   • Why not D: Reaches the right element for the wrong reason, and additionally misstates the ordering rule as increasing atomic mass; under such a rule the higher-mass element would in any case come later, not first.

10. Question 10Answer: A

    1. Using IUPAC numbering, Group 1 holds the alkali metals and Group 2 the alkaline earth metals.
    2. Group 16 holds common non-metals such as oxygen and sulfur, Group 17 holds the halogens, and Group 18 holds the noble gases.
    3. Checking option A against these facts: Group 1 alkali metals, Group 2 alkaline earth metals, and Group 17 halogens are all correctly paired.
    4. So the option that is entirely correct is A.
    • Why not B: Swaps Group 1's alkali metals for Group 2's alkaline earth metals, mislabelling Group 1.
    • Why not C: Mislabels Group 16 as the halogens; the halogens are Group 17, while Group 16 holds common non-metals such as oxygen and sulfur.
    • Why not D: Mislabels Group 16 as the noble gases; the noble gases are Group 18, not Group 16.

11. Question 11Answer: D

    1. The number of occupied shells gives the Period, so four shells (2,8,8,2) place this element in Period 4.
    2. For Groups 1 and 2, the Group number equals the outer-shell electron count directly, with no offset; the +10 offset applies only to outer-shell counts of 3 to 8, for Groups 13 to 18.
    3. The outer shell here holds 2 electrons, so the Group number is 2.
    4. So this element is in Period 4, Group 2, which is option D.
    • Why not A: Wrongly applies the +10 offset, used only for outer-shell counts of 3 to 8, to an outer-shell count of 2.
    • Why not B: Transposes the Period and Group values.
    • Why not C: Uses the second shell's electron count (8) instead of the outer shell's (2) to fix the Group number.

12. Question 12Answer: C

    1. Potassium and sodium are both in Group 1, but potassium has one more occupied shell than sodium.
    2. This means potassium's outer electron is further from the nucleus and shielded by an additional complete inner shell.
    3. A more distant, more shielded outer electron is held less tightly, so it is lost more easily.
    4. Losing its single outer electron more easily is what makes potassium react more vigorously with water than sodium does.
    5. So the correct explanation is option C.
    • Why not A: Reverses the atomic radius trend: potassium has a larger atomic radius than sodium, not a smaller one, because it has an extra occupied shell.
    • Why not B: Misapplies a stronger nuclear attraction to electrons on OTHER atoms; the relevant mechanism here is how easily potassium loses its OWN outer electron, and a stronger pull on that electron would make it less reactive, not more.
    • Why not D: Wrongly claims Group 1 metals have a full outer shell; they have exactly one outer electron, not eight.

13. Question 13Answer: B

    1. Fluorine and iodine are both in Group 17, but fluorine has far fewer occupied shells than iodine.
    2. This means fluorine's outer shell sits closer to the nucleus and is shielded by fewer inner shells.
    3. A closer, less-shielded outer shell attracts an incoming electron more strongly than a more distant, more-shielded one.
    4. This stronger attraction for an extra electron is what makes fluorine react more vigorously than iodine.
    5. So the correct explanation is option B.
    • Why not A: Fluorine actually has far fewer protons than iodine (9 compared with 53), so this option misstates the comparison it relies on.
    • Why not C: Reverses the atomic radius trend: fluorine has a smaller atomic radius than iodine, and a larger radius would weaken, not strengthen, the attraction for an incoming electron.
    • Why not D: Confuses the halogens with the noble gases; Group 17 atoms have 7 outer-shell electrons, not a full outer shell.

14. Question 14Answer: D

    1. Isotopes must share the same number of protons (mass number minus neutrons), differing only in neutron number and so mass number.
    2. Atom W: protons = 39 - 20 = 19. Atom X: protons = 41 - 20 = 21. Atom Y: protons = 40 - 21 = 19. Atom Z: protons = 39 - 19 = 20.
    3. Atom W and Atom Y both have 19 protons, with different mass numbers (39 and 40), so they are isotopes of each other.
    4. So the pair that are isotopes of each other is Atom W and Atom Y, which is option D.
    • Why not A: Both have mass number 39, but different numbers of protons (19 and 20), so this is a coincidental mass-number match between different elements, not an isotope pair.
    • Why not B: Neither the proton counts (21 and 20) nor any coincidental mass number links these two atoms.
    • Why not C: The proton counts (19 and 21) differ between these atoms, so there is no basis for calling them isotopes.

15. Question 15Answer: A

    1. Atom X has 17 protons, so it is chlorine (atomic number 17); with 20 neutrons its mass number is 37.
    2. Atom Y is an isotope of X, sharing the same 17 protons, with two fewer neutrons (18), giving mass number 35: still chlorine.
    3. Chlorine's electron configuration is 2,8,7, occupying 3 shells, so it is in Period 3; its outer shell holds 7 electrons, and for Groups 13 to 18 the Group number is the outer-shell count plus 10, giving Group 7 + 10 = 17.
    4. Atom Y gains one electron to form an ion: gaining a negatively charged electron gives the ion an overall charge of 1-.
    5. So element Y is in Group 17, Period 3, and forms a 1- ion, which is option A.
    • Why not B: Gets the Period and the ion's charge right, but uses the outer-shell electron count (7) directly as the Group number, omitting the +10 offset that applies to outer-shell counts of 3 to 8.
    • Why not C: Gets the Group and Period right, but assumes gaining an electron makes the ion positively charged, when gaining an extra, negatively charged electron gives the ion an overall negative charge.
    • Why not D: Gets the Period and the charge right, but treats element Y's outer shell as full at 8 electrons, as if it were a noble gas, instead of the correct value of 7, giving Group 18 instead of Group 17.