These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. Data: Be 9.012, Mg 24.31, Ca 40.08, Sr 87.62, Ba 137.33, Cl 35.45 g mol⁻¹. Use a periodic table for group numbers.
Question 1 (multiple choice · foundation)
What is the formula of barium nitride, the ionic compound of barium and nitrogen?
- (A) BaN
- (B) Ba₂N₃
- (C) Ba₃N₂
- (D) BaN₂
Answer and explanation
Answer: (C). Barium is in group 2 and forms Ba²⁺. Nitrogen is in group 15 with 5 valence electrons and gains 3 to form N³⁻. The lowest common multiple of 2 and 3 is 6, so three Ba²⁺ (+6) balance two N³⁻ (−6). Check: 3(+2) + 2(−3) = 0.
- (A) would need the charges to be equal in size, for example Ba²⁺ and N²⁻.
- (B) swaps the numbers: 2(+2) + 3(−3) = −5, so it is not neutral.
- (D) treats nitrogen as N⁻: 1(+2) + 2(−3) = −4 with the real charges.
Question 2 (multiple choice · core)
Lithium and oxygen form the ionic compound Li₂O. What is the most likely formula of the compound formed by rubidium and selenium?
- (A) RbSe
- (B) Rb₂Se
- (C) RbSe₂
- (D) Rb₂Se₃
Answer and explanation
Answer: (B). Rubidium is in group 1, like lithium. Selenium is in group 16, like oxygen. Elements in the same groups form analogous compounds, so the formula type of Li₂O carries over: Rb₂Se (Rb⁺ and Se²⁻; 2(+1) + 1(−2) = 0).
- (A) would need Rb²⁺ or Se⁻; neither is typical.
- (C) reverses the ratio, as if selenium were the metal.
- (D) uses the subscripts of a group 13/group 16 compound such as Al₂O₃; rubidium is not in group 13.
Question 3 (multiple choice · core)
Chlorine reacts more vigorously with hot iron wool than bromine does. Which statement best explains this?
- (A) A chlorine atom has more valence electrons than a bromine atom.
- (B) An electron gained by a chlorine atom enters a shell closer to the nucleus, so it is attracted more strongly.
- (C) A chlorine nucleus has a larger charge than a bromine nucleus.
- (D) Bromine atoms are smaller than chlorine atoms, so their electrons are more shielded.
Answer and explanation
Answer: (B). Halogens react by gaining an electron. In chlorine the incoming electron joins the third shell; in bromine it joins the fourth shell, further from the nucleus and more shielded by core electrons. By Coulomb’s law the attraction is weaker for bromine, so bromine is less reactive.
- (A) is false: both are in group 17 and have 7 valence electrons.
- (C) is false: bromine has 35 protons and chlorine 17. The extra charge is mostly cancelled by shielding.
- (D) is false: bromine atoms are larger than chlorine atoms.
Question 4 (calculation · core)
The first four ionization energies of a main-group element X are 600, 1200, 2000 and 9800 kJ mol⁻¹. (These values are invented for this question, but they follow the real pattern.)
(a) How many valence electrons does X have? Justify your answer with the data. (b) Write the formula of the typical ion of X, and the formulas of its oxide and chloride.
Worked solution and suggested Marlbridge rubric
(a) Compare each ionization energy with the previous one:
- IE₂ ÷ IE₁ = 1200 ÷ 600 = 2.00
- IE₃ ÷ IE₂ = 2000 ÷ 1200 = 1.67
- IE₄ ÷ IE₃ = 9800 ÷ 2000 = 4.90
The steady rises come from removing electrons from an increasingly positive ion. The large jump comes between IE₃ and IE₄, so the fourth electron is a core electron. X has 3 valence electrons (it is in group 13).
(b) X loses its 3 valence electrons: X³⁺. With O²⁻: lowest common multiple 6, so X₂O₃ (2(+3) + 3(−2) = 0). With Cl⁻: XCl₃.
| Point | What earns it |
|---|---|
| 1 | Identifies the largest jump as between IE₃ and IE₄, with a ratio or difference as evidence |
| 1 | 3 valence electrons, linked to the fourth electron coming from an inner shell |
| 1 | X³⁺ |
| 1 | X₂O₃ and XCl₃ (both needed) |
Common error: reading “IE₂ is double IE₁” as the big jump and answering X⁺. A doubling is normal; look for the ratio that is clearly larger than the others.
Question 5 (calculation · core)
A white solid is the chloride of a group 2 metal, M. A sample of mass 3.12 g contains 0.0150 mol of formula units.
(a) Magnesium chloride is MgCl₂. Predict the formula of the chloride of M. (b) Identify M. Show your working. (c) Another student assumed the formula was MCl. Explain how the calculation shows that this assumption is wrong.
Worked solution and suggested Marlbridge rubric
(a) M is in group 2, like magnesium, so it forms M²⁺ and an analogous chloride: MCl₂.
(b) Molar mass of the compound: 3.12 g ÷ 0.0150 mol = 208.0 g mol⁻¹. Mass of two Cl: 2 × 35.45 = 70.90 g mol⁻¹. Molar mass of M: 208.0 − 70.90 = 137.1 g mol⁻¹, which matches barium (137.33 g mol⁻¹). The compound is BaCl₂.
(c) With MCl, the metal would have a molar mass of 208.0 − 35.45 = 172.6 g mol⁻¹. No group 2 metal in the data has a molar mass near this value, and a group 2 metal forms a 2+ ion, so a 1 : 1 ratio with Cl⁻ cannot be neutral.
| Point | What earns it |
|---|---|
| 1 | MCl₂, linked to group 2 or to the analogy with MgCl₂ |
| 1 | 208 g mol⁻¹ for the compound |
| 1 | Subtracts 2 × 35.45 and identifies barium |
| 1 | Explains why MCl fails: no matching group 2 metal or charges would not balance |
Accept a check in the other direction: 0.0150 mol × 208.23 g mol⁻¹ = 3.12 g for BaCl₂.
Question 6 (constructed response · core)
Calcium reacts steadily with cold water: Ca + 2H₂O → Ca(OH)₂ + H₂. Magnesium reacts only very slowly with cold water.
(a) Write a balanced equation for the reaction of barium with water. (b) Predict whether barium reacts more or less vigorously with water than calcium does. Explain your prediction at the level of atoms and electrons. (c) A student writes: “Barium is more reactive because it has more protons, so its nucleus attracts its valence electrons more strongly.” Evaluate the student’s reasoning.
Worked solution and suggested Marlbridge rubric
(a) Barium is in group 2, like calcium, so the reaction is analogous: Ba + 2H₂O → Ba(OH)₂ + H₂. Check: 1 Ba, 4 H and 2 O on each side.
(b) More vigorously. Group 2 metals react by losing their two valence electrons. Barium’s valence electrons are in the sixth shell (6s²), much further from the nucleus than calcium’s 4s² electrons, and they are shielded by more core electrons. By Coulomb’s law they are attracted less strongly and are removed more easily, so barium reacts faster. The trend Mg (very slow) → Ca (steady) fits the same pattern down the group.
(c) The reasoning is wrong. Barium does have many more protons, but almost all of the extra nuclear charge is cancelled by the extra core electrons (shielding). The effective nuclear charge on the valence electrons is about the same, while the distance is much larger, so the attraction is weaker, not stronger. The student has the conclusion right but the explanation backwards.
| Point | What earns it |
|---|---|
| 1 | Correct balanced equation with Ba(OH)₂ |
| 1 | Predicts barium is more reactive, linked to the trend down the group or the Mg/Ca data |
| 1 | Explains with distance of valence electrons from the nucleus and shielding (Coulomb’s law) |
| 1 | Identifies that shielding cancels the extra protons, so the student’s claim of stronger attraction is wrong |
Do not award the third point for “barium is bigger” without linking size to the attraction on the valence electrons.
Question 7 (explanation · stretch)
Sodium and chlorine react to form the ionic compound NaCl. Neon forms no stable compounds with chlorine. Explain this difference in terms of the interactions between valence electrons and nuclei.
Model answer and suggested Marlbridge rubric
Model answer. Sodium has a single 3s valence electron. It is far from the nucleus and shielded by ten core electrons, so it is weakly held and easily removed. Chlorine has seven valence electrons in the third shell with a high effective nuclear charge, so its nucleus strongly attracts one more electron into that shell. The electron transfers from sodium to chlorine, giving Na⁺ and Cl⁻, which attract each other in an ionic solid.
Neon’s eight valence electrons are in the second shell, close to a nucleus with a high effective nuclear charge, so they are held very strongly; neon will not lose an electron to chlorine. Neon also has no space in its valence shell: an extra electron would have to enter the third shell, much further out, so neon does not gain an electron either. Neither transfer nor sharing is favourable, so no bond forms.
| Point | What earns it |
|---|---|
| 1 | Sodium’s valence electron is weakly attracted (distance and/or shielding) and chlorine’s nucleus strongly attracts an extra electron |
| 1 | Neon’s valence electrons are strongly attracted (close to the nucleus, high effective nuclear charge), so it does not lose them |
| 1 | Neon’s valence shell is full, so a gained electron would enter a new, more distant shell |
“Neon has a full shell so it is stable” alone earns no point: the answer must say why, in terms of attraction to the nucleus.
How did you do?
- Q1, Q2 or Q4(b) wrong: re-read “Writing ionic formulas” and Figure 1 in the study guide.
- Q4(a) wrong: compare the ratios in Worked example 2 and Figure 2.
- Q5 wrong: revisit molar mass from Topic 1.1, then the section “Same group, same kind of compound”.
- Q3, Q6 or Q7 incomplete: your explanation needs both distance and shielding, and must say whether the element loses or gains electrons.
Then tick off the topic checklist.