The periodic table: arrangement, development, metals & non-metals – GCSE Chemistry exam questions with mark scheme

1Grade 6–74 marks

Mendeleev published a periodic table in 1869.

(a) Mendeleev left gaps in his table. Give the reason why. [1]

(b) In some places Mendeleev did not follow the order of atomic weights. For example, he placed tellurium before iodine even though tellurium has the higher atomic weight. Suggest why. [1]

(a) for elements that had not yet been discovered(1)
(b) so that elements with similar properties were in the same group(1)
(c) isotopes have different numbers of neutrons, so atomic weight does not always follow atomic (proton) number(1)
(c) elements are ordered by atomic number, which puts tellurium before iodine as Mendeleev had placed them allow tellurium has fewer protons than iodine but a greater proportion of heavier isotopes / isotopes with more neutrons(1)
Answer: (a) undiscovered elements (b) to keep elements with similar properties in the same group (c) atomic weight is affected by isotopes/neutrons; ordering by atomic number agrees with Mendeleev
Examiner tip: “Undiscovered elements” is the mark in (a) — just saying “to make it fit” is too vague.

2Grade 1–3Foundation5 marks

The figure shows an outline of the periodic table. The letters A, B, C, D and E show the positions of five elements.

The letters are not the symbols of the elements.

(a) The elements in the periodic table are arranged in order of increasing …

Tick (✓) one box.

☐ atomic number ☐ mass number ☐ number of neutrons ☐ date of discovery [1]

(b) Which letter shows an element in Group 1? [1]

(d) Which two letters show elements with similar chemical properties?

Give a reason for your answer. [2]

(a) atomic number if more than one box ticked, 0 marks(1)
(b) A(1)
(c) A do not accept B (given in the question)(1)
(d) C and E both needed, either order(1)
(d) (they are) in the same group / column allow have the same number of electrons in their outer shell; allow both in Group 7 / both halogens; reason mark can be awarded only if C and E given(1)
Answer: (a) atomic number (b) A (c) A (d) C and E – same group (same number of outer-shell electrons)
Examiner tip: Similar properties come from being in the same group (column), not the same row – C and D are both on the right but are in different groups.

3Grade 1–3Foundation4 marks

Elements can be classified as metals or non-metals.

(a) Which two properties are typical of non-metals?

Tick (✓) two boxes.

☐ good conductor of electricity
☐ low melting point
☐ shiny when freshly cut
☐ brittle when solid
☐ can be hammered into shape [2]

(b) Complete the sentence.

When metals react, their atoms form __________ ions. [1]

(a) low melting point(1)
(a) brittle when solid if more than two boxes ticked, deduct 1 mark for each extra tick(1)
(b) positive allow +; allow cation(s) do not accept negative(1)
(c) (towards the) right (and top) allow right-hand side / top right; do not accept top alone; do not accept bottom(1)
Answer: (a) low melting point; brittle when solid (b) positive (c) on the right-hand side (towards the top)
Examiner tip: In (c) “top” on its own is not enough – non-metals are on the right; hydrogen is the only common exception.

4Grade 4–55 marks

An atom of element X has the electronic structure 2,8,5

(a) Which group of the periodic table is element X in? [1]

(b) Which period of the periodic table is element X in? [1]

Is element Q a metal or a non-metal?

Explain your answer in terms of its electronic structure. [3]

(a) (Group) 5 allow Group 15 / V(1)
(b) (Period) 3(1)
(c) metal(1)
(c) (Q has) one electron in its outer shell / is in Group 1 allow has few outer electrons(1)
(c) (so it) loses (one) electron (when it reacts) to form a positive ion allow forms a 1+ ion / Q⁺; do not accept gains electrons(1)
Answer: (a) Group 5 (b) Period 3 (c) metal – 1 outer electron, which it loses to form a positive (1+) ion
Examiner tip: Group number = number of outer-shell electrons; period number = number of occupied shells. Don’t swap them.

5Grade 4–55 marks

The bar chart shows the melting points of the elements in Period 3 of the periodic table.

(a) Use the bar chart to give the melting point of silicon (Si). [1]

(b) Name the three elements in Period 3 that are metals. [1]

“All elements with a melting point above 500 °C are metals, and all elements with a melting point below 500 °C are non-metals.”

Evaluate the student’s statement. Use the bar chart and your answer to part (b). [3]

(a) any value from 1400 to 1430 (°C)(1)
(b) sodium, magnesium and aluminium all three needed; allow Na, Mg, Al(1)
(c) magnesium and aluminium (are metals and) have melting points above 500 °C, so these support the statement allow either one named(1)
(c) but silicon has a melting point above 500 °C / of about 1400 °C and is a non-metal allow silicon is not a metal or but sodium has a melting point below 500 °C / of about 100 °C and is a metal(1)
(c) so the statement is not correct / melting point alone cannot be used to classify an element as a metal or non-metal allow conclusion only if at least one counter-example (Si or Na) is given; allow suggestion of a better test e.g. electrical conductivity or type of ion formed(1)
Answer: (a) about 1414 °C (b) Na, Mg, Al (c) Mg and Al fit, but Si (high mp, non-metal) and Na (low mp, metal) do not, so the statement is wrong
Examiner tip: ‘Evaluate’ needs evidence on both sides and a conclusion – quote the data for at least one element that breaks the rule.

6Grade 4–55 marks

Before atomic numbers were known, chemists arranged elements in order of relative atomic mass.

The table shows information about four elements.

Element	Atomic number	Relative atomic mass
chlorine	17	35.5
argon	18	39.9
potassium	19	39.1
calcium	20	40.1

(a) Write the four elements in order of increasing relative atomic mass. [1]

(b) Arranging these elements by relative atomic mass would place potassium in Group 0 and argon in Group 1.

Explain why this arrangement is wrong. [2]

(a) chlorine, potassium, argon, calcium allow symbols Cl, K, Ar, Ca(1)
(b) potassium is a (reactive) metal / reacts like lithium and sodium, so it should be in Group 1 (with the alkali metals)(1)
(b) argon is unreactive / is a (noble) gas, so it does not have similar properties to Group 1 elements allow argon should be with the noble gases / in Group 0; allow elements would not be in groups with elements that have similar properties for 1 mark max(1)
(c) (modern table) is arranged in order of atomic number / proton number, so argon (18) comes before potassium (19)(1)
(c) potassium has one electron in its outer shell (2,8,8,1) and argon has a full outer shell (2,8,8) allow elements in the same group have the same number of outer-shell electrons(1)
Answer: (a) Cl, K, Ar, Ca (b) K is a reactive metal like the Group 1 metals; Ar is an unreactive noble gas – each would be in a group with elements of different properties (c) ordered by atomic number (Ar 18, K 19); K has 1 outer electron, Ar a full outer shell
Examiner tip: In (b) the mark is for linking position to properties – saying “it’s in the wrong place” without saying how the elements behave scores nothing.

7Grade 6–7Higher5 marks

In 1871 Mendeleev left a gap below silicon in his periodic table. He predicted the properties of the missing element, which he called eka-silicon.

The element germanium was discovered in 1886. The table shows some of Mendeleev’s predictions and the measured properties of germanium.

Property	Predicted for eka-silicon	Measured for germanium
Relative atomic mass	72	72.6
Density in g/cm³	5.5	5.32
Formula of oxide	EsO₂	GeO₂

(a) Calculate the difference between the predicted density and the measured density as a percentage of the measured density.

Give your answer to 2 significant figures. [2]

(b) Suggest how Mendeleev was able to predict the properties of eka-silicon. [1]

(a) (5.5 − 5.32 =) 0.18 allow 0.18 / 5.32 (× 100) seen(1)
(a) 3.4 (%) an answer of 3.4 scores 2 marks; 3.38… or 3.383 scores 1 mark (not to 2 s.f.); 3.3 (from 0.18 ÷ 5.5) scores 1 mark(1)
(b) (from the properties of) the elements around the gap / in the same group as eka-silicon (e.g. silicon and tin) allow elements in the same group have similar properties(1)
(c) the properties of germanium matched / were very close to Mendeleev’s predictions allow reference to data, e.g. relative atomic mass 72.6 vs 72, same oxide formula(1)
(c) so this showed that the gaps were for undiscovered elements / that his table could be used to make correct predictions allow showed his table was correct / reliable(1)
Answer: (a) 3.4% (b) from the properties of neighbouring elements in the same group (e.g. Si, Sn) (c) germanium’s properties closely matched the predictions, showing the gaps were for undiscovered elements and the table could predict correctly
Examiner tip: Divide by the measured value as the question asks – dividing by 5.5 gives 3.3% and loses the accuracy mark.

8Grade 6–75 marks

Oxygen, sulfur and selenium are in Group 6 of the periodic table.

The table shows the formulae of two compounds of oxygen and two compounds of sulfur.

Element	Compound with sodium	Compound with hydrogen
oxygen	Na₂O	H₂O
sulfur	Na₂S	H₂S
selenium	?	?

(a) How many electrons are in the outer shell of a selenium atom? [1]

(b) Selenium forms an ion when it reacts with sodium.

Predict the charge on the selenium ion. Explain your prediction. [2]

(d) Predict the formula of the compound formed between hydrogen and selenium. [1]

(a) 6 / six(1)
(b) 2− allow −2 / Se²⁻; do not accept 2+(1)
(b) (each selenium atom) gains two electrons (to get a full outer shell / 8 outer electrons) allow same group as oxygen / sulfur, which form 2− ions (O²⁻, S²⁻)(1)
(c) Na₂Se allow ecf from the charge in (b) only if formula is consistent with Na⁺; do not accept NA2SE or Na2se(1)
(d) H₂Se allow SeH₂; do not accept H2SE or HSe(1)
Answer: (a) 6 (b) Se²⁻ – gains 2 electrons, like O and S in the same group (c) Na₂Se (d) H₂Se
Examiner tip: Use the pattern in the group: elements in the same group form ions with the same charge and compounds with the same type of formula.

9Grade 6–76 marks

The table shows information about two elements, J and L.

Property	Element J	Element L
Appearance	shiny grey solid	dull yellow solid
Melting point in °C	1538	115
Conducts electricity when solid?	yes	no
When hit with a hammer	changes shape	shatters
Ion formed in its compounds	J²⁺	L²⁻
Oxide	reacts with acids	dissolves in water to form an acidic solution

Compare the properties of elements J and L.

In your answer you should:

identify which element is a metal and which is a non-metal, using both physical and chemical properties
link the ions formed to what happens to the electrons in each atom
suggest where in the periodic table each element is found.

[6]

Physical: J is shiny but L is dull(1)
Physical: J has a high melting point but L has a low melting point(1)
Physical: J conducts electricity but L does not (L is an insulator)(1)
Physical: J is malleable / can be hammered into shape but L is brittle(1)
Chemical: J forms a positive ion but L forms a negative ion(1)
J atoms lose (two) electrons when they react; L atoms gain (two) electrons(1)
Chemical: oxide of J reacts with acids / is basic but oxide of L is acidic(1)
Conclusion: J is a metal and L is a non-metal(1)
J is on the left / in the centre / towards the bottom of the table (e.g. Group 2 or the transition metals)(1)
L is on the right / towards the top of the table(1)
L forms a 2− ion so it has 6 outer electrons / is in Group 6(1)
J forms a 2+ ion so it has (few, e.g.) 2 outer electrons to lose / could be in Group 2(1)

Level	Marks	What it looks like
Level 3	5–6	A coherent, linked comparison using both physical and chemical properties, correctly identifies J as the metal and L as the non-metal, explains the ions in terms of electrons lost and gained, and gives sensible positions for both elements.
Level 2	3–4	Correctly identifies J as the metal and L as the non-metal with some direct comparisons of properties; makes at least one link to electrons or to position in the periodic table. Reasoning may be incomplete.
Level 1	1–2	Simple statements about the properties of J and/or L, or identifies the metal with little supporting comparison.
No relevant content	0	Nothing worthy of credit.

Answer: J: shiny, high mp, conducts, malleable, forms J²⁺ by losing electrons, basic oxide → metal, on the left/centre of the table. L: dull, low mp, insulator, brittle, forms L²⁻ by gaining 2 electrons, acidic oxide → non-metal, on the right (Group 6).
Examiner tip: Write linked comparisons (“J conducts but L does not”) rather than two separate lists – and remember the chemical properties (ions, oxides), not just the physical ones.

10Grade 8–9Higher8 marks

Element E has the atomic number 20.

(a) Give the electronic structure of an atom of E. [1]

(b) Give the group number and the period number of element E. [2]

Predict the formula of this compound. Explain your answer. [2]

(d) Magnesium is directly above element E in the periodic table.

Predict whether element E is more reactive or less reactive than magnesium. Explain your prediction in terms of electrons. [3]

(a) 2,8,8,2(1)
(b) Group 2 allow ecf from (a): group = number of outer electrons(1)
(b) Period 4 allow ecf from (a): period = number of shells(1)
(c) E forms a 2+ ion / loses two electrons and chlorine forms a 1− ion / Cl⁻ allow two chloride ions needed to balance the charge of one E ion(1)
(c) ECl₂ allow CaCl₂; correct formula with no explanation scores 1 mark(1)
(d) more reactive no further marks if ‘less reactive’(1)
(d) (in E) the outer electrons are further from the nucleus / E has more (occupied) shells allow larger atom(1)
(d) so there is a weaker (electrostatic) attraction between the nucleus and the outer electrons, so the outer electrons are lost more easily allow more shielding (by inner shells) for ‘weaker attraction’; do not accept ‘harder to lose’; ignore references to gaining electrons(1)
Answer: (a) 2,8,8,2 (b) Group 2, Period 4 (c) ECl₂ – E²⁺ and two Cl⁻ (d) more reactive: outer electrons further from nucleus, weaker attraction, lost more easily
Examiner tip: Reactivity of metals depends on how easily electrons are lost – say the outer electrons are further from the nucleus and less strongly attracted.

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