Compound Structure and Properties: Unit Diagnostic (Chemistry Unit 2)

What this diagnostic is for

Use this page to find which Unit 2 topics to revisit: one question per topic, two for Topics 2.5 and 2.7. Answer every question before opening any answer, then use the table at the end.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your total gives no predicted score. Each wrong answer points to a topic; it is not a grade.

Calculator and data. No calculator is needed. Pauling electronegativities (no unit): H 2.20, C 2.55, N 3.04, O 3.44, Si 1.90, S 2.58. Ionic radii: Na⁺ 102 pm, O²⁻ 140 pm. Use a periodic table for valence electrons.

Question 1 (multiple choice · 2.1)

In which bond does the hydrogen atom carry the partial negative charge (δ−)?

(A) Si–H
(B) C–H
(C) S–H
(D) N–H

Answer and explanation

Answer: (A). The more electronegative atom in a bond carries δ−. Silicon (1.90) is less electronegative than hydrogen (2.20), so the shared pair is pulled slightly towards hydrogen: Si(δ+)–H(δ−).

(B) and (C) Carbon (2.55) and sulfur (2.58) are both more electronegative than hydrogen, so H is δ+ (C–H is usually treated as nonpolar).
(D) N–H is the most polar of the four, but N is δ−. The question asks about direction, not size.

If you missed this: read “Electronegativity and its trends” and “Polar covalent bonds” in the Topic 2.1 study guide.

Question 2 (multiple choice · 2.2)

The potential energy curve for the H–Cl bond has its minimum at a certain distance and depth. Compared with it, where is the minimum of the curve for the H–Br bond?

(A) At a longer internuclear distance, and the well is shallower
(B) At a longer internuclear distance, and the well is deeper
(C) At a shorter internuclear distance, and the well is deeper
(D) At the same distance, because both are single bonds to hydrogen, but the well is shallower

Answer and explanation

Answer: (A). Bromine is one period below chlorine, so it has a larger core. The nuclei cannot get as close, so the bond is longer: the minimum moves right. The shared pair is further from the bromine nucleus and better shielded, so it is held less strongly: the well is shallower.

(B) Larger atoms usually form longer, weaker bonds.
(C) reverses both effects.
(D) Equal bond order does not mean equal length.

If you missed this: read “Factor 1: the size of the atoms’ cores” and Worked example 2 in the Topic 2.2 study guide.

Question 3 (multiple choice · 2.3)

A student draws a two-dimensional section of solid sodium oxide, Na₂O, using 12 ions. Which drawing is consistent with the formula and with the properties of the ions?

(A) 8 Na⁺ and 4 O²⁻ in one continuous array, each O²⁻ touching only Na⁺ ions, with the O²⁻ ions drawn larger
(B) 6 Na⁺ and 6 O²⁻ in one continuous array, alternating in every direction, with the O²⁻ ions drawn larger
(C) 8 Na⁺ and 4 O²⁻ in one continuous array, each O²⁻ touching only Na⁺ ions, with the Na⁺ ions drawn larger
(D) 4 separate Na₂O units, each an O²⁻ ion with two Na⁺ ions attached, with gaps between the units

Answer and explanation

Answer: (A). 2 Na⁺ for every O²⁻, so 12 ions means 8 Na⁺ and 4 O²⁻: 8(+1) + 4(−2) = 0. Opposite charges touch, and O²⁻ (140 pm) is larger than Na⁺ (102 pm).

(B) is 1 : 1, with total charge 6(+1) + 6(−2) = −6.
(C) reverses the sizes. Na⁺ has lost its outer shell, so it is smaller.
(D) shows molecules. An ionic solid is an extended array.

If you missed this: read “Drawing a particulate model of an ionic solid” in the Topic 2.3 study guide.

Question 4 (multiple choice · 2.4)

A particulate model of an alloy shows 20 lattice positions in a regular array. Sixteen hold iron atoms. Four hold chromium atoms, which are almost the same size as iron atoms. Three much smaller carbon atoms sit in the gaps between the atoms. Which statement about this alloy is correct?

(A) It is a substitutional alloy only. The carbon atoms must have replaced iron atoms.
(B) It is an interstitial alloy only. The chromium atoms sit in the gaps.
(C) It is both: chromium atoms substitute for some iron atoms, and carbon atoms occupy gaps.
(D) It cannot be an alloy, because carbon is a nonmetal and alloys contain only metals.

Answer and explanation

Answer: (C). Size decides where an added atom goes. Chromium is about the same size as iron, so it replaces iron atoms at lattice positions (substitutional). Carbon is much smaller, so it fits in the gaps (interstitial). One alloy can show both.

(A) All 20 lattice positions hold metal atoms; carbon is in the gaps.
(B) Atoms of similar size cannot fit into the gaps.
(D) An alloy can contain nonmetal atoms; steel contains carbon.

If you missed this: read “Alloys: adding a second element” and Worked example 1 in the Topic 2.4 study guide.

Question 5 (multiple choice · 2.5)

Nitrosyl chloride, ONCl, has nitrogen as the central atom. Which describes its best Lewis diagram?

(A) An N=O double bond, an N–Cl single bond and one lone pair on nitrogen
(B) Two single bonds and two lone pairs on nitrogen
(C) An N=Cl double bond, an N–O single bond and one lone pair on nitrogen
(D) An N≡O triple bond, an N–Cl single bond and no lone pair on nitrogen

Answer and explanation

Answer: (A). Total: 6 + 5 + 7 = 18 electrons. Two single bonds (4) and three lone pairs on each outer atom (12) leave 2 for a lone pair on N. Nitrogen has only 6, so an oxygen lone pair becomes a bond: N=O. Recount: 6 + 4 + 6 + 2 = 18.

(B) contains 20 electrons, two more than the molecule has.
(C) has 18 electrons, but an outer halogen keeps one single bond; oxygen forms the double bond.
(D) contains only 16 electrons. A triple bond is not needed.

If you missed this: read “A step-by-step method” and Worked example 1 in the Topic 2.5 study guide.

Question 6 (short answer · 2.5)

The sulfite ion, SO₃²⁻, and sulfur trioxide, SO₃, both have sulfur bonded to three oxygen atoms.

(a) Calculate the number of valence electrons in each species. (b) Describe the Lewis diagram of SO₃²⁻ in which every atom has an octet, including how the charge is shown. (c) Use the “needed minus available” check to explain why SO₃²⁻ needs no double bond, but an octet diagram of SO₃ does.

Worked answer

(a) SO₃²⁻: 6 + 3(6) + 2 = 26. SO₃: 6 + 3(6) = 24. The 2− charge adds two electrons.

(b) Three S–O single bonds (6) and three lone pairs on each O (18) use 24. The last 2 go on sulfur as a lone pair, giving it 8. Draw square brackets around the diagram with 2− outside the top right. Recount: 6 + 18 + 2 = 26.

(c) Needed for four octets: 4 × 8 = 32.

SO₃²⁻: 32 − 26 = 6 shared electrons = 3 bonds, exactly the three single bonds of the skeleton.
SO₃: 32 − 24 = 8 shared electrons = 4 bonds. The skeleton has only 3 lines, so one must be a double bond (S=O).

Common slips: subtracting the charge (22); forgetting the lone pair on sulfur.

If you missed this: read “A quick check for multiple bonds” and Worked example 2 in the Topic 2.5 study guide.

Question 7 (multiple choice · 2.6)

In the hydrogen carbonate ion, HCO₃⁻, carbon is bonded to three oxygen atoms, and the hydrogen is bonded to one of those oxygen atoms. Which statement about its C–O bonds is best?

(A) All three C–O bonds are identical, each with bond order 1.33, as in the carbonate ion.
(B) The C–OH bond is a single bond; the other two C–O bonds are identical, each with a bond order of about 1.5.
(C) One C–O bond is a fixed double bond and the other two are single bonds. Resonance cannot occur because the ion contains hydrogen.
(D) The ion switches rapidly between structures, so at any instant one bond is double and two are single.

Answer and explanation

Answer: (B). Total: 1 + 4 + 3(6) + 1 = 24 electrons. Carbon needs one double bond, to either oxygen not bonded to hydrogen: two equivalent resonance structures, 3 bonds over 2 positions, bond order 1.5. A double bond to the OH oxygen gives that oxygen +1 and both others −1, so that structure contributes very little.

(A) The hydrogen makes one oxygen different, so 4 ÷ 3 = 1.33 applies to carbonate, not here.
(C) Hydrogen does not stop resonance; two positions are equivalent.
(D) A resonance hybrid is one structure all the time; it does not flip.

If you missed this: read “The resonance hybrid” and “Average bond order” in the Topic 2.6 study guide.

Question 8 (multiple choice · 2.7)

Which species has a square pyramidal molecular geometry?

(A) AsF₅
(B) ClF₅
(C) SeF₄
(D) BrF₄⁻

Answer and explanation

Answer: (B). ClF₅: 7 + 5(7) = 42 electrons. Bonds use 10 and fluorine lone pairs 30, leaving one lone pair on Cl. Five bonded atoms + 1 lone pair = 6 domains; with one position taken by the lone pair, the atoms form a square pyramid.

(A) AsF₅: 40 electrons, no lone pair on As: trigonal bipyramidal.
(C) SeF₄: 34 electrons, one lone pair on Se. 5 domains with the lone pair equatorial: seesaw.
(D) BrF₄⁻: 36 electrons, two lone pairs on Br. 6 domains with the lone pairs opposite each other: square planar.

If you missed this: read “Five and six domains” and Worked example 2 in the Topic 2.7 study guide.

Question 9 (short answer · 2.7)

Ketene has the structure H₂C=C=O: the first carbon is bonded to two H atoms, and the second carbon is bonded to the first carbon and to oxygen. Oxygen has two lone pairs.

(a) Show that this diagram uses every valence electron. (b) State the number of σ bonds and π bonds. (c) Give the hybridization of each carbon atom and of the oxygen atom. (d) Predict the H–C–H and C–C–O bond angles. (e) State whether ketene has a dipole moment, with a reason.

Worked answer

(a) Total: 2(4) + 2(1) + 6 = 16. Drawn: two C–H (4) + C=C (4) + C=O (4) + two lone pairs on O (4) = 16.

(b) Each bond has one σ bond: two C–H, C=C and C=O give 4 σ. Each double bond adds one π bond: 2 π.

(c) CH₂ carbon: 3 domains, sp². Middle carbon: 2 domains, sp. Oxygen: 1 bonded atom + 2 lone pairs = 3 domains, sp².

(d) H–C–H: about 120° (sp² carbon). C–C–O: 180° (sp carbon, no lone pair): the C, C and O atoms lie in a line.

(e) Yes. The C=O bond is polar (O 3.44, C 2.55) and the C–H bonds are close to nonpolar, so nothing balances the C=O dipole.

Common slips: counting a double bond as two domains; calling oxygen “sp” because it has one bonded atom.

If you missed this: read “Hybridization: naming the atom”, “Sigma and pi bonds” and Worked example 3 in the Topic 2.7 study guide.

Your next step

Tick each question you got wrong. Count a short-answer question as missed if any part was wrong.

Topic	Question(s)	If you missed it, read
2.1 Types of Chemical Bonds	1	Topic 2.1 study guide
2.2 Intramolecular Force and Potential Energy	2	Topic 2.2 study guide
2.3 Structure of Ionic Solids	3	Topic 2.3 study guide
2.4 Structure of Metals and Alloys	4	Topic 2.4 study guide
2.5 Lewis Diagrams	5, 6	Topic 2.5 study guide
2.6 Resonance and Formal Charge	7	Topic 2.6 study guide
2.7 VSEPR and Hybridization	8, 9	Topic 2.7 study guide

How to use your result

Look at topics, not the total. If you guessed, count the question as missed.
Fix gaps in order. Topics 2.5 to 2.7 build on each other: a wrong electron count (2.5) spoils formal charges (2.6) and shapes (2.7). Start with the earliest topic you missed.
Then practise. After each study guide, try that topic’s practice questions, then the Unit 2 mixed review, where each question combines topics.