This set covers the whole of periodic patterns: group and period from electrons, evidence for reactivity, explaining trends, metals and exceptions, and unfamiliar elements. Questions run from easier to harder.
Attempt each on paper, open the answer, and mark the reasoning as well as the result. Any data tables are invented for practice, not real measurements.
Questions
1. Aluminium has atomic number 13. Write its electron arrangement and give its group and period.
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13 electrons: 2 in shell one, 8 in shell two, 3 left. Arrangement 2,8,3. The last number is 3, so Group III. Three shells, so period 3.
2. An element has electron arrangement 2,8,8,1. Give its group, period and name.
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Last number 1 gives Group I. Four numbers give period 4. Atomic number 2 + 8 + 8 + 1 = 19, which is potassium.
3. An element is in Group VI and period 2. Write its arrangement and atomic number.
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Two shells and 6 outer electrons: 2,6. Atomic number 2 + 6 = 8, oxygen.
4. Magnesium (2,8,2) reacts with chlorine (2,8,7). Describe what happens to the electrons, give the ions formed and write the balanced equation.
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Each magnesium atom loses 2 electrons to form Mg²⁺ (2,8). Each chlorine atom gains 1 electron to form Cl⁻ (2,8,8). One Mg²⁺ needs two Cl⁻ to balance the charge, so the formula is MgCl₂.
Mg + Cl₂ → MgCl₂. Atoms: 1 Mg, 2 Cl on each side. Balanced.
5. Invented data: halogens X₂, Y₂ and Z₂ were each added to solutions of the halides of X, Y and Z. A tick means displacement occurred.
| Halogen added | Halide of X | Halide of Y | Halide of Z |
|---|---|---|---|
| X₂ | not applicable | tick | tick |
| Y₂ | cross | not applicable | tick |
| Z₂ | cross | cross | not applicable |
(a) Put the halogens in order of reactivity. (b) Write the conclusion for Y₂ and the halide of X.
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(a) X₂ displaces both others, Y₂ displaces only Z, and Z₂ displaces neither. Order: X₂ > Y₂ > Z₂.
(b) Y₂ does not displace X from its halide, so Y is less reactive than X.
6. Explain why iodine is less reactive than chlorine.
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Both atoms react by gaining one electron. Iodine’s outer shell is further from the nucleus and is screened by more inner shells. The attraction for an incoming electron is weaker, so iodine gains an electron less easily. Iodine is less reactive.
7. A student writes: “Reactivity increases down every group, so caesium and astatine are the most reactive members of their groups.” Find the error and correct it.
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Caesium (Group I, loses an electron) is indeed the most reactive alkali metal of the two shown. Astatine (Group VII, gains an electron) is the least reactive halogen, because reactivity decreases down Group VII. The rule is not the same for every group.
8. Invented data for three substances:
| Substance | Conducts when solid | Malleable | Melting point / °C |
|---|---|---|---|
| M | yes | yes | 660 |
| N | no | brittle | 115 |
| O | yes | brittle | above 3 500 |
Classify M and N. What can and cannot be said about O?
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M conducts and is malleable: a metal. N does not conduct and is brittle: a non-metal. O conducts, but it is brittle with a very high melting point, which is not the metal pattern. It cannot be classed as a metal from this data. It may be a conducting non-metal like graphite.
9. Invented data: melting points of four metals from one group are J 180 °C, K 100 °C, L missing and M 40 °C. Estimate L and say whether this is interpolation or extrapolation.
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L lies between K and M. (100 + 40) ÷ 2 = 140 ÷ 2 = 70 °C, an estimate by interpolation. The values fall steadily, and 70 sits between 100 and 40, so the estimate is sensible.
10. Element R is in Group II and has more shells than calcium. Predict the ion it forms, the formula of its chloride and whether it is more or less reactive than calcium.
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Group II elements lose 2 electrons, so R forms R²⁺. The chloride is RCl₂. More shells mean the outer electrons are lost more easily, so R is predicted to be more reactive than calcium.
11. Balance: K + H₂O → KOH + H₂
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Balance the hydrogen and oxygen: 2K + 2H₂O → 2KOH + H₂.
Check: left K 2, H 4, O 2; right K 2, O 2, H 2 + 2 = 4. Balanced.
12. In the equation 2Na + 2H₂O → 2NaOH + H₂, a textbook calculation uses 0.23 g of sodium (Ar = 23). Calculate the volume of hydrogen at room temperature and pressure (24 dm³ per mole). This is a theory question only.
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Moles of Na = 0.23 ÷ 23 = 0.010 mol. The ratio Na : H₂ is 2 : 1, so moles of H₂ = 0.010 ÷ 2 = 0.0050 mol. Volume = 0.0050 × 24 = 0.12 dm³, which is 120 cm³.
If you got these wrong
- Questions 1 to 4 wrong: group, period or ions. Go to linking group membership with outer electrons.
- Questions 5 and 9 wrong: reading results or estimating from data. Go to comparing reactivity patterns using evidence and interpreting a table with an unfamiliar element.
- Questions 6, 7 and 10 wrong: explaining a trend or using the wrong direction. Go to explaining a trend without treating all groups alike.
- Question 8 wrong: classification from one property. Go to distinguishing a metal property from an exception.
- Questions 11 and 12 wrong: balancing or amounts. Try the equation balance reasoning trainer and the mole and equation-ratio tutor.
Record each slip in the mistake log and retest queue and retry a similar question after a few days. The module overview is at periodic patterns.
If the same error type keeps returning, online one-to-one Chemistry tuition gives a teacher the chance to watch your working and name the cause.