Periodic Trends MCQs – 40 Practice Questions with Answers and Explanations

Master the patterns of the Periodic Table with 40 free, original practice questions covering Group 1, Group 2, Group 7, the transition metals and the trends across Period 3. Every question has a full explanation, and the later sections stretch you with the A-Level, IB and AP explanations behind the trends – ionic radius, ionisation energy anomalies and polarising power. Click “Show Answer” on any question to check yourself.

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Section 1: Group 1 & Group 7 Patterns

Q1. The alkali metals become more reactive down Group 1. What is the main reason for this trend?

  • (A) The atoms get bigger, so they collide with water more often
  • (B) The outer electron is further from the nucleus and more shielded, so it is lost more easily
  • (C) The nuclear charge decreases down the group
  • (D) The metals become less dense down the group
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Answer: B – Down the group the outer electron sits in a shell further from the nucleus with more inner shells shielding it, so the attraction is weaker and the electron is lost more easily, even though nuclear charge increases.

Q2. Equal-sized pieces of lithium, sodium, potassium and rubidium are added separately to water. Which metal reacts most vigorously?

  • (A) Lithium
  • (B) Sodium
  • (C) Potassium
  • (D) Rubidium
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Answer: D – Reactivity increases down Group 1, so rubidium, the lowest of the four, reacts most vigorously. Lithium floats and fizzes steadily, sodium melts and darts around, and potassium also ignites with a lilac flame.

Q3. Which observation is correct when a small piece of potassium is placed on water?

  • (A) It melts into a ball, moves across the surface and burns with a lilac flame
  • (B) It sinks and dissolves slowly with no fizzing
  • (C) It burns with a bright orange flame and produces a black solid
  • (D) Nothing happens because potassium is unreactive
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Answer: A – Potassium floats, melts with the heat released, fizzes as hydrogen forms, and the hydrogen ignites with the characteristic lilac flame of potassium.

Q4. Which statement about the physical properties of Group 1 metals is correct?

  • (A) Their melting points increase down the group
  • (B) They are hard metals that are difficult to cut
  • (C) Their melting points decrease down the group
  • (D) They are poor conductors of electricity
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Answer: C – Melting points fall down Group 1 (lithium 180 °C down to caesium 28 °C) because the metallic bonding weakens as the ions get larger. They are soft enough to cut with a knife and conduct electricity well.

Q5. The halogens become less reactive down Group 7. Why?

  • (A) The atoms lose electrons more easily as they get bigger
  • (B) The outer shell is further from the nucleus and more shielded, so an incoming electron is attracted less strongly
  • (C) The number of electron shells decreases down the group
  • (D) The halogens change from gases to solids, and solids cannot react
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Answer: B – Halogens react by gaining an electron into the outer shell. Down the group that shell is further away and more shielded, so the attraction for the extra electron weakens and reactivity falls.

Q6. Chlorine water is added to a solution of potassium bromide. What is observed?

  • (A) The solution stays colourless
  • (B) A purple vapour forms
  • (C) The solution turns orange as bromine is displaced
  • (D) A white precipitate forms
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Answer: C – Chlorine is more reactive than bromine, so it displaces it: Cl2 + 2KBr → 2KCl + Br2. The bromine formed colours the solution orange.

Q7. Which displacement reaction actually takes place?

  • (A) Iodine + potassium chloride → potassium iodide + chlorine
  • (B) Bromine + potassium chloride → potassium bromide + chlorine
  • (C) Iodine + potassium bromide → potassium iodide + bromine
  • (D) Bromine + potassium iodide → potassium bromide + iodine
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Answer: D – A halogen can only displace a less reactive halogen below it in the group. Bromine is above iodine, so it displaces iodine from potassium iodide, giving a brown solution of iodine. Iodine cannot displace bromine or chlorine.

Q8. Which statement correctly describes the trend in the halogens from fluorine to iodine?

  • (A) Melting and boiling points decrease down the group
  • (B) Melting and boiling points increase down the group
  • (C) All the halogens are gases at room temperature
  • (D) The colours become paler down the group
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Answer: B – Melting and boiling points rise down Group 7 as the molecules get larger and the van der Waals forces strengthen: fluorine and chlorine are gases, bromine is a liquid and iodine is a solid at room temperature.

Q9. Which description of iodine at room temperature is correct?

  • (A) A grey-black solid that gives a purple vapour when warmed
  • (B) A red-brown liquid
  • (C) A greenish-yellow gas
  • (D) A pale yellow gas
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Answer: A – Iodine is a shiny grey-black solid that sublimes on gentle warming to give a purple vapour. The red-brown liquid is bromine, the greenish-yellow gas is chlorine and the pale yellow gas is fluorine.

Q10. The noble gases in Group 0 are almost completely unreactive. What is the reason?

  • (A) They have very low boiling points
  • (B) Their atoms are extremely small
  • (C) They have a full outer shell of electrons, so they do not need to gain, lose or share electrons
  • (D) They exist as single atoms rather than molecules
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Answer: C – A complete outer shell (two electrons for helium, eight for the others) is a very stable arrangement, so noble gas atoms have no tendency to form bonds. Existing as single atoms is a result of this, not the cause.

Section 2: Trends Across a Period

Q11. How does the character of the elements change across Period 3 from sodium to argon?

  • (A) From metals on the left to non-metals on the right
  • (B) From non-metals on the left to metals on the right
  • (C) All the elements in the period are metals
  • (D) Metals and non-metals alternate across the period
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Answer: A – Sodium, magnesium and aluminium are metals, silicon is a metalloid, and phosphorus, sulfur, chlorine and argon are non-metals. This metal-to-non-metal pattern is repeated in every period.

Q12. Which Period 3 element has a giant covalent (macromolecular) structure and a very high melting point of over 1400 °C?

  • (A) Sodium
  • (B) Sulfur
  • (C) Phosphorus
  • (D) Silicon
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Answer: D – Every silicon atom is held in a giant lattice by four strong covalent bonds, so a huge amount of energy is needed to melt it. Sulfur and phosphorus are simple molecular substances with much lower melting points.

Q13. Phosphorus, sulfur, chlorine and argon all have low melting points. Why?

  • (A) Their covalent bonds are very weak
  • (B) They are simple molecular substances held together by weak van der Waals forces
  • (C) They contain no electrons in their outer shells
  • (D) They are gases dissolved in the air
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Answer: B – These elements exist as small molecules (P4, S8, Cl2) or single atoms (Ar). Only the weak forces between the molecules need to be overcome on melting; the strong covalent bonds inside the molecules stay intact.

Q14. Which of these Period 3 elements conducts electricity well at room temperature?

  • (A) Sulfur
  • (B) Phosphorus
  • (C) Aluminium
  • (D) Chlorine
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Answer: C – Aluminium is a metal with delocalised electrons that can carry charge. Sulfur, phosphorus and chlorine are non-metals with no mobile electrons or ions, so they do not conduct. Silicon, in between, is only a semiconductor.

Q15. Sodium oxide, Na2O, is added to water. What type of solution is formed and what does this show about the oxide?

  • (A) An acidic solution, so Na2O is an acidic oxide
  • (B) An alkaline solution, so Na2O is a basic oxide
  • (C) A neutral solution, so Na2O is a neutral oxide
  • (D) No solution forms because Na2O is insoluble
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Answer: B – Sodium oxide reacts with water to form sodium hydroxide, NaOH, giving a strongly alkaline solution. Metal oxides on the left of the period are basic.

Q16. Aluminium oxide, Al2O3, reacts with both hydrochloric acid and sodium hydroxide. How is it classified?

  • (A) Amphoteric
  • (B) Basic
  • (C) Acidic
  • (D) Neutral
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Answer: A – An oxide that reacts with both acids and alkalis is amphoteric. Aluminium sits at the metal/non-metal borderline in Period 3, so its oxide shows both behaviours.

Q17. Sulfur dioxide, SO2, is bubbled into water. What is formed and how is the oxide classified?

  • (A) An alkaline solution; SO2 is a basic oxide
  • (B) A neutral solution; SO2 is a neutral oxide
  • (C) Pure sulfur; SO2 is an amphoteric oxide
  • (D) An acidic solution (sulfurous acid); SO2 is an acidic oxide
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Answer: D – Sulfur dioxide dissolves in water to give sulfurous acid, H2SO3. Non-metal oxides on the right of the period are acidic – this is also why SO2 from burning fossil fuels causes acid rain.

Q18. Atomic radius decreases across a period from left to right. What is the correct explanation?

  • (A) Electrons are removed one by one across the period
  • (B) The number of electron shells increases across the period
  • (C) Nuclear charge increases while the electrons are added to the same shell, so they are pulled closer to the nucleus
  • (D) The atoms become less dense across the period
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Answer: C – Across a period each element has one more proton, but the extra electron goes into the same outer shell, so shielding barely changes. The stronger nuclear pull draws the whole electron cloud inwards.

Q19. Which of the Period 3 metals sodium, magnesium and aluminium has the highest melting point?

  • (A) Sodium
  • (B) Magnesium
  • (C) All three are the same
  • (D) Aluminium
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Answer: D – Aluminium (660 °C) is just above magnesium (650 °C) and far above sodium (98 °C). Aluminium contributes three delocalised electrons per atom to the metallic bonding, compared with two for magnesium and one for sodium, so its lattice is strongest.

Q20. Which sequence correctly summarises the acid-base character of the Period 3 oxides from sodium to sulfur?

  • (A) Basic → amphoteric → acidic
  • (B) Acidic → amphoteric → basic
  • (C) All of them are basic
  • (D) All of them are acidic
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Answer: A – Na2O and MgO are basic, Al2O3 is amphoteric, SiO2 is weakly acidic, and the oxides of phosphorus and sulfur (such as SO2 and SO3) are acidic. The oxides mirror the metal-to-non-metal trend.


Section 3: Group 2 & Transition Metals

Q21. Barium is more reactive than magnesium. Which statement explains the trend in Group 2 reactivity?

  • (A) Ionisation energy increases down the group, so electrons are lost faster
  • (B) The atoms gain electrons more easily down the group
  • (C) Ionisation energy decreases down the group because the outer electrons are further away and more shielded, so they are lost more easily
  • (D) Barium has fewer electron shells than magnesium
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Answer: C – Group 2 metals react by losing two outer electrons. Down the group those electrons are further from the nucleus and more shielded, so the total ionisation energy falls and reactivity rises.

Q22. The Group 2 carbonates are heated strongly. Which carbonate decomposes most easily (at the lowest temperature)?

  • (A) Magnesium carbonate
  • (B) Calcium carbonate
  • (C) Strontium carbonate
  • (D) Barium carbonate
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Answer: A – Thermal stability of Group 2 carbonates (and nitrates) increases down the group. The small Mg2+ ion polarises the carbonate ion most strongly and weakens its bonds, so MgCO3 breaks down to MgO and CO2 most readily.

Q23. Which Group 2 hydroxide is the most soluble in water?

  • (A) Magnesium hydroxide
  • (B) Calcium hydroxide
  • (C) Strontium hydroxide
  • (D) Barium hydroxide
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Answer: D – The solubility of Group 2 hydroxides increases down the group. Mg(OH)2 is only sparingly soluble (it is used as “milk of magnesia”), while Ba(OH)2 is soluble enough to be used as an alkali in titrations.

Q24. The solubility of Group 2 sulfates shows the opposite trend to the hydroxides. Which statement is correct?

  • (A) All Group 2 sulfates are very soluble
  • (B) Barium sulfate is insoluble, and sulfate solubility decreases down the group
  • (C) Magnesium sulfate is insoluble and is used in barium meals
  • (D) Sulfate solubility increases down the group
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Answer: B – Sulfate solubility decreases down Group 2: MgSO4 is soluble, CaSO4 is sparingly soluble and BaSO4 is insoluble. Barium sulfate’s insolubility (and opacity to X-rays) is why it is used in barium meals – and why adding a soluble sulfate is a test for barium ions.

Q25. Which transition metal compound or element is the catalyst in the Haber process for making ammonia?

  • (A) Iron
  • (B) Vanadium(V) oxide
  • (C) Manganese(IV) oxide
  • (D) Nickel
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Answer: A – Finely divided iron catalyses N2 + 3H2 ⇌ 2NH3. Vanadium(V) oxide is the Contact process catalyst, manganese(IV) oxide catalyses hydrogen peroxide decomposition, and nickel is used in hydrogenating alkenes.

Q26. In the Contact process, sulfur dioxide is converted to sulfur trioxide: 2SO2 + O2 ⇌ 2SO3. Which catalyst is used?

  • (A) Iron
  • (B) Concentrated sulfuric acid
  • (C) Vanadium(V) oxide, V2O5
  • (D) Platinum gauze
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Answer: C – Vanadium(V) oxide, V2O5, catalyses the Contact process at about 450 °C. It works by cycling between oxidation states – a typical transition metal behaviour.

Q27. A few drops of hydrogen peroxide solution are added to a black powder and oxygen is produced rapidly. The black powder, which is unchanged at the end, is most likely:

  • (A) Carbon (soot)
  • (B) Iron filings
  • (C) Copper metal
  • (D) Manganese(IV) oxide, MnO2
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Answer: D – MnO2 is the classic catalyst for decomposing hydrogen peroxide, 2H2O2 → 2H2O + O2. It speeds up the reaction but is chemically unchanged and can be recovered at the end.

Q28. Iron is found in compounds as both Fe2+ and Fe3+. Which transition metal property does this illustrate?

  • (A) Forming colourless compounds
  • (B) Having variable oxidation states
  • (C) Having low densities
  • (D) Forming only one type of ion
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Answer: B – Transition metals characteristically form ions with different charges because their 3d and 4s electrons are at similar energies. Group 1 and Group 2 metals, by contrast, form only +1 and +2 ions.

Q29. Compared with the Group 1 metals, the transition metals generally have:

  • (A) Higher densities, greater hardness and higher melting points
  • (B) Lower densities and lower melting points
  • (C) Softer structures that can be cut with a knife
  • (D) Only one oxidation state each
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Answer: A – Transition metals such as iron and copper are dense, hard and strong with high melting points, which is why they are used for structures and machinery. The Group 1 metals are soft, light and low-melting by comparison (chromium and manganese are partial exceptions to the hardness pattern, but the general trend holds).

Q30. When ammonia solution is added to copper(II) sulfate solution, a deep blue solution eventually forms containing [Cu(NH3)4(H2O)2]2+. What type of species is this?

  • (A) A free radical
  • (B) An allotrope of copper
  • (C) A complex ion, with ligands bonded to a central transition metal ion
  • (D) A giant covalent structure
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Answer: C – Transition metal ions form complex ions in which ligands (here NH3 and H2O) donate lone pairs into empty metal orbitals. Complex formation is also why transition metal compounds are often strongly coloured – the pale blue of [Cu(H2O)6]2+ deepens as ammonia replaces water ligands.

Section 4: Explaining Trends (A-Level / IB / AP Stretch)

Q31. A sodium ion, Na+, is much smaller than a sodium atom. Why?

  • (A) The nucleus loses a proton when the ion forms
  • (B) The outer electron shell is lost completely, and the remaining electrons are pulled closer by the same nuclear charge
  • (C) The electrons repel each other more strongly in the ion
  • (D) Ions always shrink because they carry a charge
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Answer: B – Sodium loses its entire third shell when it forms Na+. The same 11 protons now attract only 10 electrons in two shells, so the ion is far smaller than the atom. Cations are always smaller than their parent atoms.

Q32. A chloride ion, Cl–, is larger than a chlorine atom. What is the best explanation?

  • (A) The ion contains an extra electron shell
  • (B) The nuclear charge increases when the electron is gained
  • (C) Chlorine atoms contain fewer protons than chloride ions
  • (D) The same nuclear charge now attracts more electrons, so each electron is held less tightly and the cloud expands
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Answer: D – Gaining an electron means 17 protons attract 18 electrons. The pull per electron falls and electron-electron repulsion spreads the outer shell out, so anions are always larger than their parent atoms.

Q33. The ions N3-, O2-, F–, Na+, Mg2+ and Al3+ are isoelectronic (they all contain 10 electrons). Which is the smallest?

  • (A) Al3+
  • (B) N3-
  • (C) F–
  • (D) They are all exactly the same size
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Answer: A – In an isoelectronic series the number of electrons is fixed, so the ion with the most protons pulls them in tightest. Sizes fall steadily from N3- (7 protons) to Al3+ (13 protons), which is the smallest.

Q34. Which statement about electronegativity is correct?

  • (A) It decreases across a period because the atoms get bigger
  • (B) Fluorine is the least electronegative element
  • (C) It increases across a period and decreases down a group, with fluorine the most electronegative element
  • (D) Noble gases have the highest electronegativities of all
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Answer: C – Across a period, rising nuclear charge and shrinking radius pull bonding electrons more strongly. Down a group, extra shielding and distance weaken the pull despite the higher charge. Fluorine, top right of the table, has the highest value.

Q35. The first ionisation energy of aluminium (577 kJ mol-1) is lower than that of magnesium (736 kJ mol-1), even though aluminium has a higher nuclear charge. Why?

  • (A) Aluminium has one more electron shell than magnesium
  • (B) Aluminium’s outer electron is in a 3p sub-shell, which is higher in energy and further from the nucleus than magnesium’s 3s electron
  • (C) Magnesium’s outer electrons are unpaired
  • (D) Aluminium is a metal but magnesium is not
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Answer: B – Magnesium loses a 3s electron; aluminium loses its single 3p electron. The 3p sub-shell is slightly higher in energy, further out and more shielded by the filled 3s sub-shell, so the electron comes off more easily – one of the two famous dips in the Period 3 ionisation energy pattern.

Q36. The first ionisation energy of sulfur (1000 kJ mol-1) is slightly lower than that of phosphorus (1012 kJ mol-1). What causes this second dip?

  • (A) Sulfur has a smaller nuclear charge than phosphorus
  • (B) Sulfur’s outer electron is in a new shell
  • (C) Sulfur is a non-metal and phosphorus is a metal
  • (D) In sulfur, one 3p orbital contains a pair of electrons, and their mutual repulsion makes one of them easier to remove
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Answer: D – Phosphorus has three unpaired 3p electrons (one in each orbital). Sulfur’s fourth 3p electron is forced to pair up, and the repulsion between the paired electrons raises its energy, so it is removed slightly more easily despite the extra proton.

Q37. Sodium chloride melts at 801 °C, but aluminium chloride sublimes at about 180 °C. What does this tell us about aluminium chloride?

  • (A) It has substantial covalent character and exists as Al2Cl6 molecules rather than a giant ionic lattice
  • (B) It is more ionic than sodium chloride
  • (C) Its ions carry bigger charges, so its lattice is stronger
  • (D) It contains no chlorine
Show Answer

Answer: A – The small, highly charged Al3+ ion distorts the electron clouds of the large Cl– ions so much that the bonding is mainly covalent. AlCl3 forms simple Al2Cl6 dimers, so only weak forces between molecules must be broken on heating.

Q38. In the Al2Cl6 dimer, two of the chlorine atoms form bridges between the aluminium atoms. How are these bridging bonds formed?

  • (A) Each bridging chlorine shares two electrons equally, one from each metal
  • (B) The aluminium atoms donate electrons to each other
  • (C) Each bridging chlorine donates a lone pair into an empty orbital on an aluminium atom, forming coordinate (dative) bonds
  • (D) The bridges are hydrogen bonds
Show Answer

Answer: C – Each Al atom in AlCl3 has only six outer electrons and an empty orbital. A chlorine lone pair from the neighbouring molecule fills it, giving coordinate bonds and the dimer Al2Cl6, in which each aluminium ends up with a share in eight electrons.

Q39. Silicon tetrachloride, SiCl4, reacts rapidly with cold water, giving steamy fumes and a white solid. Which statement explains this behaviour?

  • (A) SiCl4 is ionic and simply dissolves
  • (B) Water molecules can attack the silicon atom (it has accessible empty 3d orbitals and is large enough), leading to hydrolysis to SiO2 and HCl
  • (C) SiCl4 contains free chloride ions
  • (D) Silicon is a metal, so its chloride behaves like NaCl
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Answer: B – SiCl4 is hydrolysed: SiCl4 + 2H2O → SiO2 + 4HCl. A water lone pair attacks silicon, which can expand its octet using empty 3d orbitals – unlike carbon in CCl4, which is too small and has no available d orbitals, so CCl4 is inert to water. PCl5 hydrolyses vigorously for similar reasons.

Q40. Fajans’ rules explain why some “ionic” compounds show covalent character. Which cation would cause the greatest polarisation of a large anion such as I–?

  • (A) A large cation with a +1 charge, such as Cs+
  • (B) A large cation with a +2 charge, such as Ba2+
  • (C) Any cation, because polarisation does not depend on the cation
  • (D) A small cation with a high charge, such as Al3+, because it has the highest charge density
Show Answer

Answer: D – Polarising power increases with charge density: small size and high charge concentrate the positive charge, so Al3+ distorts (polarises) the electron cloud of a large, easily distorted anion the most. This is why aluminium iodide is covalent while caesium iodide is ionic.

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