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 (Pauling electronegativities, no unit): H 2.20, B 2.04, C 2.55, N 3.04, O 3.44, F 3.98, Si 1.90, P 2.19, S 2.58, Cl 3.16, Br 2.96, I 2.66, Ca 1.00. Experimental data in Questions 6 and 7 are invented for practice.
Question 1 (multiple choice · foundation)
Which single bond has the largest bond dipole?
- (A) P–Cl
- (B) S–Cl
- (C) Si–Cl
- (D) Cl–Cl
Answer and explanation
Answer: (C). For single bonds, the bigger the electronegativity difference, the bigger the bond dipole. Si–Cl: 3.16 − 1.90 = 1.26, the largest.
- (A) P–Cl has a difference of 0.97: polar, but less than Si–Cl.
- (B) S–Cl has 0.58. Sulfur is further right in period 3 than silicon, so it is closer to chlorine in electronegativity.
- (D) Cl–Cl has a difference of 0: identical atoms share equally, so there is no bond dipole.
Shortcut without numbers: all four partners of Cl are in period 3, and electronegativity rises across a period, so the partner furthest left (Si) differs most from Cl.
Question 2 (multiple choice · core)
Which statement best describes the bond in O–F?
- (A) O carries a partial negative charge and F a partial positive charge.
- (B) O carries a partial positive charge and F a partial negative charge.
- (C) The bond is nonpolar because both atoms are nonmetals.
- (D) The bond is ionic, made of O²⁺ and F²⁻ ions.
Answer and explanation
Answer: (B). Fluorine (3.98) is more electronegative than oxygen (3.44), so the shared pair is pulled towards F. F is δ− and O is δ+. The difference, 0.54, makes this a polar covalent bond.
- (A) is the trap: oxygen is usually the δ− atom in its bonds, but not when its partner is fluorine, the most electronegative element.
- (C) Two nonmetals form a covalent bond, but it is only nonpolar if their electronegativities are equal or very similar.
- (D) Two nonmetals do not form ions with each other, and “F²⁻” is not an ion fluorine forms.
Question 3 (multiple choice · core)
Which description of the bonding in solid calcium metal is best?
- (A) Each Ca atom shares one pair of electrons with each neighbouring atom in fixed covalent bonds.
- (B) Half the atoms become Ca²⁺ and half become Ca²⁻, and these ions attract each other.
- (C) Ca²⁺ cations are held in an array by a sea of valence electrons that belong to no single atom.
- (D) Each Ca atom keeps its own two valence electrons, and neutral atoms are held only by weak attractions.
Answer and explanation
Answer: (C). In a metal, each atom’s valence electrons are delocalized across the whole solid. The cations left behind are attracted to this shared electron sea.
- (A) describes covalent bonding. Metals hold their valence electrons too weakly for fixed shared pairs.
- (B) needs electron transfer between identical atoms. With equal electronegativity, neither atom can take electrons from the other.
- (D) ignores the strong metallic bonding; calcium is a solid with a high melting point, not a collection of weakly attracted atoms.
Question 4 (classification · core)
For each pair, state the type of bonding expected (ionic, polar covalent, nonpolar or effectively nonpolar covalent, or metallic). Where a bond is polar, name the atom with the partial negative charge.
(a) calcium and oxygen in CaO (b) the B–H bond in a boron hydride (c) the N–H bond in NH₃ (d) the atoms in a piece of pure aluminium
Worked solution and suggested Marlbridge rubric
(a) Ionic. Ca is a metal and O a nonmetal; the difference is 3.44 − 1.00 = 2.44. Ions Ca²⁺ and O²⁻.
(b) Effectively nonpolar covalent. Both are nonmetals; the difference is only 2.20 − 2.04 = 0.16. If you comment on direction, H is very slightly δ−, because hydrogen is the more electronegative of the two.
(c) Polar covalent, N is δ−. Difference 3.04 − 2.20 = 0.84.
(d) Metallic. Only aluminium atoms; the valence electrons are delocalized.
| Point | What earns it |
|---|---|
| 1 | (a) ionic, with a reason (metal + nonmetal or large difference) |
| 1 | (b) nonpolar or only very slightly polar, with the small difference as the reason |
| 1 | (c) polar covalent with N as δ− |
| 1 | (d) metallic |
Accept “polar covalent, H slightly δ−” for (b) if the answer says the polarity is very small.
Question 5 (explanation · core)
The hydrogen halides are HF, HCl, HBr and HI.
(a) Rank the H–X bonds in order of decreasing bond dipole. (b) Explain the trend using the structure of the halogen atoms.
Worked solution and suggested Marlbridge rubric
(a) Differences: H–F 1.78, H–Cl 0.96, H–Br 0.76, H–I 0.46. Order: H–F > H–Cl > H–Br > H–I. In every bond the halogen is δ−.
(b) Going down group 17, each halogen has one more occupied shell. The shared electron pair in the H–X bond is therefore further from the halogen nucleus and more shielded by core electrons. By Coulomb’s law the attraction for the shared pair is weaker, so electronegativity falls from F to I. The difference from hydrogen shrinks, so the shared pair is shifted less and the bond dipole is smaller.
| Point | What earns it |
|---|---|
| 1 | Correct order |
| 1 | Bonding electrons further from the halogen nucleus and/or more shielded down the group |
| 1 | Links weaker attraction (lower electronegativity) to a smaller difference and a smaller bond dipole |
Do not award the second point for “the atom is bigger” with no link to the attraction for the bonding electrons.
Question 6 (constructed response · core)
A student has three unlabelled solids. One is sucrose (table sugar, C₁₂H₂₂O₁₁), one is potassium chloride (KCl) and one is tin (Sn). Her results:
| Solid | Conducts as a solid? | Conducts when molten? | Dissolves in water? | Solution conducts? | When hammered |
|---|---|---|---|---|---|
| X | no | yes | yes | yes | shatters |
| Y | yes | yes | no | not tested | flattens |
| Z | no | no | yes | no | crumbles |
(a) Identify X, Y and Z. (b) Make a claim about the type of bonding in Z and justify it using two pieces of evidence from the table. (c) Use a particle-level model to explain why Y conducts electricity as a solid.
Worked solution and suggested Marlbridge rubric
(a) X is potassium chloride, Y is tin, Z is sucrose.
(b) Z contains covalent bonds within molecules (C, H and O are nonmetals). Evidence: it does not conduct as a solid or when molten, and its solution does not conduct. There are no ions and no delocalized electrons, so there are no charged particles free to move. It also crumbles instead of flattening, unlike a metal.
(c) Tin is a metal. Its valence electrons are delocalized: they are not attached to any one atom and can move through the array of tin cations. When a voltage is applied, these mobile electrons carry charge through the solid.
| Point | What earns it |
|---|---|
| 1 | All three identified correctly |
| 1 | Claim: covalent (molecular) bonding in Z |
| 1 | Two pieces of evidence, linked to the absence of mobile charged particles |
| 1 | Delocalized valence electrons in tin |
| 1 | Those electrons are free to move and carry charge |
Accept any two of: no conduction as solid, no conduction when molten, solution does not conduct, not malleable.
Question 7 (argumentation · stretch)
A textbook offers the rule: “If the electronegativity difference is greater than 1.7, the bonding is ionic; otherwise it is covalent.” Consider two compounds (data invented for practice):
- Compound A is formed by a metal and a nonmetal; difference 1.5. Melting point 990 °C; conducts when molten, not as a solid; brittle.
- Compound B is formed by two nonmetals; difference 1.8. Boiling point 15 °C; the pure liquid is a very poor conductor.
(a) State what the rule predicts for each compound, and whether the properties agree. (b) Explain why a single cut-off value cannot decide the type of bonding.
Model answer and suggested Marlbridge rubric
(a) The rule predicts A is covalent (1.5 < 1.7) and B is ionic (1.8 > 1.7). The properties say the opposite. A behaves as an ionic compound: high melting point, conducts only when molten (ions become mobile), brittle. B behaves as a molecular covalent substance: very low boiling point and almost no mobile ions in the liquid. The rule fails for both.
(b) Bond character changes gradually: every polar bond has some ionic character, so there is no point on the scale where bonding suddenly switches type. Electronegativity difference is only one factor; whether the elements are a metal and a nonmetal also matters, and the compound’s properties are the best evidence of the type of bonding.
| Point | What earns it |
|---|---|
| 1 | Correct predictions from the rule for both A and B |
| 1 | Uses the properties to show A behaves as ionic and B as covalent |
| 1 | Bonding is a continuum: polar bonds have partial ionic character, so no sharp boundary |
| 1 | Properties (or metal/nonmetal identity) must be considered, not the difference alone |
How did you do?
- Q1, Q2 or Q5 wrong: re-read “Polar covalent bonds” and Worked example 1 in the study guide.
- Q3 or Q4 wrong: look again at Figure 2 and Worked example 3.
- Q6 or Q7 incomplete: link every piece of evidence to a particle-level reason (mobile ions, mobile electrons or neither), as in Worked example 2.
Then tick off the topic checklist.