How to use this review
Use this page after the Unit 2 diagnostic and any topics it flagged. Answer all seven questions on paper before opening any answer.
These are original Marlbridge practice questions, not past exam questions; the alloy in Question 5 is invented. Each rubric is a suggested Marlbridge rubric, not official scoring, and your total gives no predicted score.
Data. Pauling electronegativities (no unit): Li 0.98, Ca 1.00, Be 1.57, Cr 1.66, Fe 1.83, Si 1.90, C 2.55, S 2.58, N 3.04, Cl 3.16, O 3.44, F 3.98. Radii and molar masses are in the calculator note above.
Big ideas of the unit
- Electronegativity decides how a pair is shared. Equal pull: nonpolar; unequal pull: polar; a nonmetal pulling on a metal: ions. Bonding is a continuum (Topic 2.1).
- A bond is a well in a potential energy curve. Position = bond length; depth = bond energy. Smaller atoms and higher bond order give shorter, stronger bonds (Topic 2.2).
- An ionic solid is an extended array. |q₊q₋| / d ranks the attraction; charge usually beats size (Topic 2.3).
- A metal is cores in a sea of delocalized electrons. Added atoms of similar size substitute; much smaller atoms go into gaps (Topic 2.4).
- Properties are evidence. Mobile ions or electrons explain conductivity; like-charge repulsion explains brittleness (2.1, 2.3, 2.4).
- A Lewis diagram accounts for every valence electron. Count, adjust for charge, check octets. Period 2 atoms never exceed 8 (Topic 2.5).
- More than one diagram? Equivalent structures average the bond order; formal charge ranks nonequivalent structures (Topic 2.6).
- Domains decide shape. Count domains, place lone pairs, name geometry and hybridization, then check whether bond dipoles cancel (Topic 2.7).
Key relationships and methods
| Idea | Relationship or method | Topic |
|---|---|---|
| Bond polarity | larger electronegativity difference, larger dipole; δ− on the more electronegative atom | 2.1 |
| Potential energy curve | bond length = distance at the minimum; bond energy = well depth | 2.2 |
| Ionic attraction | ∝ |q₊q₋| / d, with d = r₊ + r₋ for touching ions | 2.2, 2.3 |
| Particulate models | ionic: alternating charges, formula ratio, sensible sizes; metal: identical cores in an electron sea | 2.3, 2.4 |
| Alloy type | % difference = |r(host) − r(added)| ÷ r(host) × 100; similar size substitutes, much smaller fills gaps | 2.4 |
| Lewis count | valence electrons ± charge; bonds = (needed − available) ÷ 2 | 2.5 |
| Formal charge | FC = valence − nonbonding − ½ bonding; sum = overall charge | 2.6 |
| Average bond order | bonds to equivalent atoms ÷ number of equivalent positions | 2.6 |
| VSEPR | domains = bonded atoms + lone pairs; 2, 3, 4 domains → sp, sp², sp³ | 2.7 |
| Dipole moment | polar bonds and a shape in which they do not cancel | 2.1, 2.7 |
Question 1 (multiple choice · mixed)
Which species has a trigonal planar molecular geometry and all its bonds to the central atom identical in length?
- (A) The carbonate ion, CO₃²⁻
- (B) The nitrite ion, NO₂⁻
- (C) Phosgene, COCl₂
- (D) Phosphorus trifluoride, PF₃
Answer and explanation
Answer: (A). CO₃²⁻ has 4 + 3(6) + 2 = 24 electrons (Topic 2.5). Each octet structure has one C=O and two C–O bonds, with the double bond to any of the three oxygens, so the bonds are identical, order 4 ÷ 3 = 1.33 (Topic 2.6). Carbon has 3 domains and no lone pair: trigonal planar (Topic 2.7).
- (B) The two N–O bonds are equal (order 1.5), but N has a lone pair: bent.
- (C) Phosgene is trigonal planar, but C=O and C–Cl bonds differ in length.
- (D) The P–F bonds are identical, but P has a lone pair (26 electrons): trigonal pyramidal.
Question 2 (multiple choice · mixed)
Which molecule has a dipole moment?
- (A) SiCl₄
- (B) SCl₂
- (C) BeCl₂ (gas phase)
- (D) CS₂
Answer and explanation
Answer: (B). S–Cl is polar (3.16 − 2.58 = 0.58, Cl δ−; Topic 2.1). SCl₂ has 6 + 2(7) = 20 electrons, leaving two lone pairs on S: 4 domains, bent (Topic 2.7), so the bond dipoles add.
- (A) Si–Cl bonds are very polar (1.26), but four identical tetrahedral bonds cancel.
- (C) Be–Cl bonds are polar (1.59), but Be has only two domains (4 electrons), so the molecule is linear and the dipoles cancel.
- (D) CS₂ is linear, with almost nonpolar C–S bonds (0.03).
Question 3 (multiple choice · mixed)
Potential energy curves are drawn for the N–O bond in NO⁺, NO₂⁻, NO₃⁻ and hydroxylamine, H₂N–OH. Which species has the curve with its minimum at the shortest distance and the deepest well?
- (A) NO⁺
- (B) NO₂⁻
- (C) NO₃⁻
- (D) H₂N–OH
Answer and explanation
Answer: (A). NO⁺ has 5 + 6 − 1 = 10 electrons. Needed: 16, so 6 shared electrons: N≡O with one lone pair on each atom (Topic 2.5). Bond orders: NO⁺ 3, NO₂⁻ 1.5, NO₃⁻ 1.33, H₂N–OH 1 (Topic 2.6). The highest bond order gives the shortest, strongest bond: the well furthest left and deepest (Topic 2.2).
- (B) 3 bonds over 2 positions: order 1.5.
- (C) Nitrate has the most bonds in total, spread over three positions: order 1.33.
- (D) Hydroxylamine (14 electrons) has an N–O single bond: the longest, shallowest curve.
Question 4 (constructed response · mixed)
Calcium sulfide, CaS, and lithium fluoride, LiF, are white solids with high melting points.
(a) Classify the bonding in each solid. Use the electronegativity data and the types of element. (b) Calculate the distance between ion centres, d, and |q₊q₋| / d for each solid, and the ratio of the CaS value to the LiF value. (c) A student claims: “LiF must have the stronger attraction, because its ions are smaller.” Evaluate the claim with numbers from (b), and predict which solid has the higher melting point. (d) Describe a two-dimensional particulate model of CaS that uses 8 ions.
Worked solution and suggested Marlbridge rubric
(a) Both are ionic: each is a metal with a nonmetal. Differences: LiF 3.00; CaS 1.58. CaS’s difference is smaller, but a group 2 metal and a group 16 nonmetal form Ca²⁺ and S²⁻; the difference alone does not decide.
(b) d(CaS) = 100 + 184 = 284 pm; d(LiF) = 76 + 133 = 209 pm. CaS: 4 ÷ 284 = 0.01408 pm⁻¹. LiF: 1 ÷ 209 = 0.00478 pm⁻¹. Ratio CaS : LiF = 2.94.
(c) The claim is not supported. LiF’s ions are closer, but only by a factor of 284 ÷ 209 = 1.36, while CaS’s charge product is 4 times larger. Charge wins (4 ÷ 1.36 = 2.94): CaS has the stronger attraction and is predicted to have the higher melting point.
(d) 4 Ca²⁺ and 4 S²⁻ (charge 4(+2) + 4(−2) = 0) in one continuous array, alternating so each ion touches only opposite charges, S²⁻ drawn larger (184 against 100 pm), charges labelled.
| Point | What earns it |
|---|---|
| 1 | (a) Both ionic, with metal + nonmetal as the reason |
| 1 | (b) Both distances correct |
| 1 | (b) Both values of |q₊q₋| / d and the ratio 2.9 |
| 1 | (c) Rejects the claim: distance factor (about 1.4) against charge factor (4) |
| 1 | (c) CaS has the higher melting point, linked to stronger attraction |
| 1 | (d) 4 : 4 ratio, alternating charges, S²⁻ larger than Ca²⁺ |
Total: 6 points. “CaS, because its charges are larger” without comparing distances earns the prediction point only.
Question 5 (constructed response · mixed)
A stainless steel is 81.6% iron, 18.0% chromium and 0.40% carbon by mass.
(a) Use the radii to classify how chromium and carbon atoms fit into the iron lattice. (b) Calculate the number of Fe, Cr and C atoms per 100 atoms. Explain why carbon’s atom percent is much larger than its mass percent. (c) State how many atoms of each element a 50-atom particulate model should show, and where the carbon atoms go. (d) A student claims: “Chromium transfers electrons to iron, so this alloy is ionic.” Evaluate the claim using electronegativity and one property of the alloy.
Worked solution and suggested Marlbridge rubric
(a) Chromium: |126 − 128| ÷ 126 × 100 = 1.6%, a comparable size, so Cr replaces Fe at lattice positions: substitutional. Carbon: |126 − 75| ÷ 126 × 100 = 40.5%, much smaller, so C fits in the gaps: interstitial. The alloy is both.
(b) Take 100 g.
- n(Fe) = 81.6 ÷ 55.85 = 1.4611 mol
- n(Cr) = 18.0 ÷ 52.00 = 0.3462 mol
- n(C) = 0.40 ÷ 12.01 = 0.0333 mol
- Total = 1.8405 mol
Per 100 atoms: 79.4 Fe, 18.8 Cr, 1.8 C. A carbon atom is much lighter than an iron atom (12.01 against 55.85), so a small mass of carbon contains many atoms: 0.40% by mass is 1.8% of the atoms.
(c) 50 × 0.794 = 39.7, 50 × 0.188 = 9.4, 50 × 0.018 = 0.9: about 40 Fe and 9 Cr at lattice positions and 1 C in a gap, not a carbon atom in every gap.
(d) The claim is incorrect. Chromium (1.66) and iron (1.83) are both low-electronegativity metals (difference 0.17); neither takes electrons from the other. Both release valence electrons into a delocalized sea: metallic bonding. Evidence: the alloy conducts electricity as a solid, which an ionic solid cannot.
| Point | What earns it |
|---|---|
| 1 | (a) Cr substitutional (small size difference) |
| 1 | (a) C interstitial (large size difference) |
| 1 | (b) Moles of all three elements from 100 g |
| 1 | (b) 79.4 : 18.8 : 1.8 per 100 atoms, with the light carbon atom as the reason |
| 1 | (c) About 40 Fe, 9 Cr and 1 C, with C in a gap |
| 1 | (d) Rejects the claim: both low-electronegativity metals, delocalized electrons, a metallic property as evidence |
Total: 6 points. Accept carried-forward values in (c).
Question 6 (constructed response · mixed)
The cyanate ion, OCN⁻, has carbon in the centre. The fulminate ion, CNO⁻, has nitrogen in the centre. Both are linear.
(a) Calculate the number of valence electrons in each ion and the number of bonds an octet diagram needs. (b) Describe the three octet resonance structures of OCN⁻ with the formal charge on each atom, and identify the main contributor. (c) The best structure of CNO⁻ is [C≡N–O]⁻. Give its formal charges, and use formal charge to predict which arrangement, cyanate or fulminate, is more stable. (d) Give the number of electron domains on the central atom and its hybridization in each ion. (e) Is the C–O bond in cyanate longer or shorter than a C=O double bond? Justify your answer.
Worked solution and suggested Marlbridge rubric
(a) 6 + 4 + 5 + 1 = 16 electrons each. Needed: 24, so 8 shared electrons = 4 bonds.
(b)
| Structure | FC on O | FC on C | FC on N | Sum |
|---|---|---|---|---|
| [O=C=N]⁻ | 6 − 4 − 2 = 0 | 4 − 0 − 4 = 0 | 5 − 4 − 2 = −1 | −1 ✓ |
| [O≡C–N]⁻ | 6 − 2 − 3 = +1 | 0 | 5 − 6 − 1 = −2 | −1 ✓ |
| [O–C≡N]⁻ | 6 − 6 − 1 = −1 | 0 | 5 − 2 − 3 = 0 | −1 ✓ |
[O–C≡N]⁻ contributes most. It and [O=C=N]⁻ each have just one −1, but it puts the −1 on oxygen, the most electronegative atom. [O≡C–N]⁻ has −2 and +1 on oxygen.
(c) [C≡N–O]⁻: C = 4 − 2 − 3 = −1, N = 5 − 0 − 4 = +1, O = 6 − 6 − 1 = −1 (sum −1). Fulminate’s best structure has three nonzero formal charges, including −1 on carbon, the least electronegative atom. Cyanate’s has a single −1, on oxygen. Formal charge predicts the cyanate arrangement is more stable.
(d) Both central atoms: 2 bonded atoms, no lone pair: 2 domains, sp, linear.
(e) Longer. The main structure has a C–O single bond; only the minor structure [O=C=N]⁻ has C=O. The average C–O bond order is between 1 and 2, so the bond is longer and weaker than a C=O double bond.
| Point | What earns it |
|---|---|
| 1 | (a) 16 electrons and 4 bonds |
| 1 | (b) Three valid structures with correct formal charges |
| 1 | (b) [O–C≡N]⁻ chosen: −1 on the most electronegative atom |
| 1 | (c) −1, +1, −1, and cyanate preferred: fewer, better-placed charges |
| 1 | (d) 2 domains and sp for both central atoms |
| 1 | (e) Longer: average bond order between 1 and 2 |
Total: 6 points.
Question 7 (constructed response · mixed)
Boron trifluoride, BF₃, and ammonia, NH₃, combine to form one molecule, F₃B–NH₃, with a new B–N bond.
(a) Describe the Lewis diagrams of BF₃ and NH₃, with the number of electrons around B and N. (b) Before they combine, give the shape, approximate bond angle and hybridization around B and N. (c) State which of BF₃ and NH₃ has a dipole moment, with a reason for each. (d) The B–N bond is made from nitrogen’s lone pair. Calculate the formal charges on B and N in F₃B–NH₃. (e) Predict the shape, F–B–F angle and hybridization around boron in F₃B–NH₃. (f) Use your answers to (a) to explain why boron, and not nitrogen, accepts the electron pair.
Worked solution and suggested Marlbridge rubric
(a) BF₃ (24 electrons): three B–F single bonds, three lone pairs on each F. B has 6 electrons (incomplete octet; all formal charges zero). NH₃ (8 electrons): three N–H bonds and one lone pair on N. N has 8.
(b) BF₃: 3 domains, trigonal planar, 120°, sp². NH₃: 4 domains, one a lone pair, trigonal pyramidal, just under 109.5°, sp³.
(c) NH₃ has a dipole moment: polar N–H bonds (N 3.04, H 2.20) in a pyramid, so the dipoles add. BF₃ has none: three identical polar B–F dipoles 120° apart cancel.
(d) In the combined molecule, B and N each have 4 bonds and no lone pair. B: 3 − 0 − ½(8) = −1. N: 5 − 0 − ½(8) = +1. Every F and H is 0, so the sum is 0, as it must be for a neutral molecule.
(e) Boron now has 4 domains: tetrahedral around B, F–B–F about 109.5° (down from 120°), sp³.
(f) Boron has only 6 electrons, so it has room for one more pair, which completes its octet. Nitrogen already has an octet and, as a period 2 atom, cannot hold more than 8.
| Point | What earns it |
|---|---|
| 1 | (a) Both diagrams: 6 electrons on B, 8 on N |
| 1 | (b) BF₃ trigonal planar, 120°, sp² |
| 1 | (b) NH₃ trigonal pyramidal, under 109.5°, sp³ |
| 1 | (c) NH₃ polar, BF₃ not, each with a shape reason |
| 1 | (d) B −1, N +1, sum checked |
| 1 | (e) Tetrahedral around B, about 109.5°, sp³ |
| 1 | (f) B has an incomplete octet; N is full and cannot exceed 8 |
Total: 7 points.
How did you do?
Total: 28 points (3 multiple choice, 25 constructed response). Where you lost points matters more than the total.
- Bond polarity or type (Q2, 4(a), 5(d)): Topic 2.1.
- Curves, Coulomb’s law, ionic solids (Q3, 4): Topics 2.2 and 2.3.
- Alloys and metallic models (Q5): Topic 2.4.
- Electron counts, formal charge, resonance (Q1, 3, 6, 7(a), 7(d)): Topics 2.5 and 2.6.
- Shapes, hybridization, dipoles (Q1, 2, 6(d), 7(b), 7(c), 7(e)): Topic 2.7.
If several topics were weak, return to the Unit 2 diagnostic, then tick off each topic checklist: