Master Compound Questions for DAT Exam Success

III. Compound theory questions

Compound questions

This chapter contains 41 questions on chemical compounds and bonding, covering the key concepts in the chapter summary. Allow approximately one minute per question. At the end of each test, select “See Result” to view your score and the answer choices. Incorrect answers are marked in red, and correct answers are marked in green. Explanations or worked solutions appear below each question.

COMPOUND THEORY TEST 1

1 / 20

Which polyatomic ion is correctly matched with its name?

A. PO₄^3- : Phosphorous

B. ClO^2- : Hypochlorate

C. BrO₃^– : Bromate

D. BrO₄^– : Bromine tetraoxide

E. SO₄^2- : Sulfite

Explanation:

PO₄³^– : Phosphate

ClO^– : Hypochlorite

BrO₄^–: Perbromate

SO₄²^– : Sulfate

2 / 20

Which of the following acids is INCORRECTLY named?

A. HCl(aq): Hydrochloric acid

B. HNO₂ : Nitrous acid

C. HNO₃ : Nitric acid

D. HClO₄ : Perchloric acid

E. H₂SO₃ : Sulfuric acid

Explanation:

H₂SO₃ : Sulfurous acid

3 / 20

Which of the following compounds is INCORRECTLY named?

A. CaCl₂ : Calcium chloride

B. Ba₃(PO₄)₂ : Barium phosphate

C. SnF₄ : Tin(IV) fluoride

D. P₂O₅ : Diphosphorus pentoxide

E. CuCrO₄ : Copper(IV) chromate

Explanation:

The chromate ion, CrO₄²⁻, has a charge of −2. A neutral formula unit of CuCrO₄ therefore contains Cu²⁺, so the correct name is copper(II) chromate.

4 / 20

Which formula is correctly matched with its name?

A. PCl₃ : Phosphorus chloride

B. SO₄^2- : Sulfite ion

C. HCN(aq): Hydrocyanic acid

D. Ca(ClO)₂ : Calcium(II) hypochloride

E. SeO₂ : Selenium(II) oxide

Explanation:

HCN(aq) is hydrocyanic acid. The other correct names are phosphorus trichloride for PCl₃, sulfate for SO₄²⁻, calcium hypochlorite for Ca(ClO)₂, and selenium dioxide for SeO₂.

5 / 20

Which of the following is the correct chemical formula for magnesium carbonate?

A. MgCO

B. MgCO₃

C. Mg₂CO₃

D. Mg(CO₃)₂

E. Mg₂(CO₃)₂

Explanation:

Mg²⁺ and CO₃²⁻ combine in a 1:1 ratio to give the neutral formula unit MgCO₃. The subscripts must be in the lowest whole-number ratio.

6 / 20

What is the predominant type of bonding in CaCl₂?

A. Covalent bonding

B. Ionic bonding

C. Hydrogen bonding

D. Metallic bonding

E. B and D

Explanation:

CaCl₂ is an ionic solid. Its Ca²⁺ and Cl⁻ ions are held together by electrostatic attraction between opposite charges.

7 / 20

Which ionic solid is expected to have the greatest lattice energy, defined as the energy required to separate one mole of solid into gaseous ions?

A. MgO

B. MgCl₂

C. KI

D. MgBr₂

E. KBr

Explanation:

Lattice energy is an energy, not a force. It generally increases with larger ionic charge magnitudes and smaller ion sizes. MgO contains Mg²⁺ and O²⁻; its charge product is larger than those of the listed halides, so it has the greatest lattice energy.

8 / 20

Which statements about ionic bonding and ionic substances are correct?

I. Electron transfer can form oppositely charged ions that are held together by electrostatic attraction.

II. Ionic bonding consists of sharing electron pairs between atoms.

III. Ionic compounds commonly contain a metal cation and a nonmetal anion.

IV. Ionic bonds are bonds between neutral nonmetal atoms.

V. Some ionic substances dissolve in polar solvents such as water.

A. II only

B. I and III

C. II and IV

D. II, IV, and V

E. I, III, and V

Explanation:

I, III, and V are correct. Ionic bonding is electrostatic attraction between oppositely charged ions. Metal–nonmetal combinations are common examples, and some ionic compounds dissolve in water; solubility is not universal. Sharing electron pairs describes covalent bonding.

9 / 20

Which statements about covalent bonding and covalent substances are correct?

I. Covalent bonds form through complete transfer of electrons between atoms.

II. Covalent bonds form by sharing electron pairs between atoms.

III. Covalent bonds occur only between metal and nonmetal atoms.

IV. Covalent bonds commonly form between nonmetal atoms.

V. Covalent substances always dissolve readily in polar solvents.

A. II only

B. I and III

C. II and IV

D. II, IV, and V

E. I, III, and V

Explanation:

II and IV are correct. Covalent bonds involve shared electron pairs and commonly form between nonmetal atoms. They may be polar or nonpolar. Water solubility depends on the substance; it is not a universal property of covalent compounds.

10 / 20

Which statement about metallic bonding is INCORRECT?

A. It involves attraction between metal cations and delocalized electrons.

B. Metals typically conduct heat and electricity well.

C. Many metals have high melting and boiling points.

D. Metals are typically brittle in the solid state.

E. Metallic bonding occurs in pure metals and in many alloys.

Explanation:

D is incorrect. Metallic bonding involves attraction between metal cations and delocalized electrons. Metals typically conduct heat and electricity and can deform without the brittle fracture characteristic of many ionic crystals. Their melting and boiling points vary.

11 / 20

Which molecule contains a triple bond?

A. CO₂

B. HCN

C. Br₂

D. NH₃

E. O₂

Explanation:

HCN contains a C≡N triple bond. Br₂ and NH₃ contain single bonds, while CO₂ and O₂ contain double bonds.

12 / 20

Which statement about HF, LiF, H₂O, and BF₃ is INCORRECT?

A. HF, H₂O, and BF₃ are all polar molecules.

B. HF, H₂O, and BF₃ contain covalent bonds.

C. LiF is an ionic compound.

D. Pure HF and pure H₂O exhibit intermolecular hydrogen bonding.

E. BF₃ is a nonpolar molecule.

Explanation:

A is incorrect. The B–F bonds in BF₃ are polar, but their dipole moments cancel in its trigonal planar geometry, so the molecule is nonpolar. HF and H₂O are polar covalent molecules, and LiF is ionic. Pure HF and water form intermolecular hydrogen bonds.

13 / 20

Which of the following bonds has the highest polarity?

A. C–F

B. C–N

C. C–O

D. C–C

E. F–F

Explanation:

A larger electronegativity difference produces a more polar bond. Among these choices, C–F has the largest electronegativity difference and is the most polar bond.

14 / 20

Which statement about HF and HCl is INCORRECT? If A–D are all correct, select E.

A. The electronegativity difference in H–F is larger than that in H–Cl.

B. The H–F bond is more polar than the H–Cl bond.

C. HF has a higher normal boiling point than HCl.

D. HF has a higher melting point than HCl at standard pressure.

E. None of statements A–D is incorrect.

Explanation:

A–D are all correct, so E is the answer. The H–F bond is more polar because F is more electronegative than Cl. Intermolecular hydrogen bonding in HF contributes to its higher boiling and melting points compared with HCl; these phase-change properties are properties of the substances, not of individual bonds.

15 / 20

Which property difference between two bonded atoms is most useful for predicting whether their covalent bond is polar or nonpolar?

A. Electronegativity

B. Valence electrons

C. Electron affinity

D. Ionization energy

E. Molecular weight

Explanation:

The electronegativity difference between bonded atoms predicts bond polarity. Molecular polarity also depends on geometry and the vector sum of the bond dipoles.

16 / 20

At 25 °C and the same gas partial pressure, which listed gas is most soluble in water?

A. NF₃

B. NH₃

C. CH₄

D. CO₂

E. CF₄

Explanation:

NH₃ is highly soluble in water. It can form hydrogen bonds with water and partially reacts as a weak base to form NH₄⁺ and OH⁻. NF₃, CH₄, CO₂, and CF₄ are less soluble under these conditions. Molecular polarity alone does not determine all solubility behavior.

17 / 20

Which option identifies all the nonpolar molecules listed in A–D?

A. CH₄

B. Cl₂

C. PH₃

D. CHCl₃

E. A and B

Explanation:

CH₄ and Cl₂ are nonpolar, so E is correct. PH₃ and CHCl₃ have nonzero permanent molecular dipole moments. In CH₄, the tetrahedral bond dipoles cancel; Cl₂ has identical atoms.

18 / 20

Which statement about intermolecular forces is correct?

A. Strong conventional hydrogen bonding often involves H bonded to N, O, or F interacting with a lone pair on a neighboring molecule.

B. London dispersion attractions are always stronger than dipole–dipole attractions between molecules of similar molar mass.

C. Nonpolar halogen molecules and noble-gas atoms have permanent dipole–dipole attractions.

D. The normal boiling points increase in the order HI < HBr < HCl.

E. All covalent bonds are more polar than ionic bonds.

Explanation:

A is correct. Strong conventional hydrogen bonding commonly involves H bonded to N, O, or F interacting with a lone pair on a neighboring molecule. London dispersion forces occur in all atoms and molecules. Nonpolar halogen molecules and noble-gas atoms have no permanent dipole–dipole attractions. The boiling and melting points increase in the order HCl < HBr < HI as polarizability and dispersion attractions increase. These observations do not establish a universal ranking of all intermolecular forces.

19 / 20

Considering only polarizability, which intact group 14 hydride is expected to have the strongest London dispersion attractions?

A. CH₄

B. SiH₄

C. GeH₄

D. SnH₄

E. PbH₄

Explanation:

Polarizability generally increases down group 14, so intact PbH₄ is expected to have the strongest dispersion attractions in this comparison. This is a qualitative intermolecular-force prediction; it does not assume that every listed hydride is stable enough for a normal boiling point to be measured.

20 / 20

Which option identifies all listed pairs that can form a stable coordinate covalent bond by lone-pair donation without breaking an existing bond?

I. BF₃

II. H⁺

III. NH₃

IV. H₂O

A. I with II

B. I with III

C. II with IV

D. III with IV

E. B and C

Explanation:

A coordinate covalent bond forms when one species donates both electrons of a shared pair. NH₃ donates a lone pair to electron-deficient BF₃, forming F₃B←NH₃. H₂O donates a lone pair to H⁺, forming H₃O⁺. Thus both pairings B and C can form coordinate bonds. Hydrogen follows a duet, not an octet.

Your score is

The average score is 57%

COMPOUND THEORY TEST 2

1 / 21

Which statements about London dispersion forces are correct?

I. Instantaneous fluctuations in electron density can induce temporary dipoles in neighboring particles.

II. These forces occur only between nonpolar molecules.

III. For particles with comparable shapes, these forces generally become stronger as polarizability increases.

IV. Larger, more polarizable molecules always have lower boiling points than smaller molecules.

A. I only

B. I and II

C. I and III

D. I, II, and III

E. I, III, and IV

Explanation:

I and III are correct. Dispersion attractions arise from correlated instantaneous and induced dipoles and occur in all atoms and molecules. Their strength depends on polarizability and molecular shape. Boiling points are not directly proportional to molecular size.

2 / 21

Using the types of intermolecular attractions present, which order of normal boiling points is expected from lowest to highest?

I. BF₃

II. H₂O

III. PH₃

A. I, II, III

B. II, I, III

C. I, III, II

D. III, I, II

E. III, II, I

Explanation:

The expected order is I, III, II: BF₃ < PH₃ < H₂O. BF₃ is nonpolar and has dispersion attractions. PH₃ is polar and also has dipole–dipole attractions. Water forms an extensive hydrogen-bond network. All three also have dispersion attractions. This comparison is not a universal force-strength rule for arbitrary molecules.

3 / 21

Which substance does NOT form conventional hydrogen bonds between its own molecules in a pure sample?

A. NH₃

B. NF₃

C. H₂O

D. HF

E. None of the above

Explanation:

NF₃ has no N–H, O–H, or F–H bonds and cannot donate a conventional hydrogen bond to another NF₃ molecule. NH₃, H₂O, and HF can hydrogen-bond to molecules of the same substance.

4 / 21

Which substance forms conventional hydrogen bonds between its own molecules in a pure sample?

A. CH₃CH₂CH₃

B. CH₃CH₂F

C. N(CH₃)₃

D. CH₃Cl

E. CH₃CH₂OH

Explanation:

Ethanol, CH₃CH₂OH, has an O–H group and oxygen lone pairs, so its molecules form hydrogen bonds with one another. The other choices lack a conventional N–H, O–H, or F–H donor. N(CH₃)₃ can accept hydrogen bonds from water, but it cannot donate them to another molecule of itself.

5 / 21

Which statements are correct?

I. Main-group elements in the same group generally have similar valence-electron configurations and similar chemical properties.

II. Each carbon atom in diamond is bonded to four other carbon atoms in a tetrahedral arrangement.

III. In conventional Lewis structures, BF₃ and monomeric AlCl₃ have incomplete octets, whereas KrF₂ has an expanded octet.

IV. Having fewer valid resonance contributors always makes a species more stable.

A. I only

B. I and II

C. I, II, III

D. I, II, IV

E. All of the above

Explanation:

I, II, and III are correct. IV is incorrect: resonance delocalization can stabilize a species, but stability is not determined simply by counting resonance drawings. The quality and relative contributions of the structures matter.

6 / 21

W and X are main-group metals with usual ionic charges +1 and +2; Y and Z are main-group nonmetals with usual ionic charges −2 and −1. Which formula is NOT consistent with charge neutrality or a neutral diatomic halogen molecule?

W: 1 valence electron

X: 2 valence electrons

Y: 6 valence electrons

Z: 7 valence electrons

A. W₂Y

B. WZ

C. XY

D. Z₂

E. X₂Z

Explanation:

W₂Y, WZ, and XY are neutral with the stated ionic charges, and Z₂ is a neutral diatomic halogen molecule. X₂Z would have net charge 2(+2) + (−1) = +3, so E is inconsistent. The neutral ionic formula for X and Z would be XZ₂.

7 / 21

Which pair has the same number of lone pairs on the central atoms in its conventional Lewis structures?

A. CH₂O and H₂O

B. H₂Se and PCl₃

C. NH₃ and IF₄⁺

D. BH₃ and ClF₃

E. BH₄^– and CH₃^–

Explanation:

NH₃ and IF₄⁺ each have one lone pair on the central atom. CH₂O, BH₃, and BH₄⁻ have zero; PCl₃ and CH₃⁻ have one; H₂O, H₂Se, and ClF₃ have two. For IF₄⁺, 7 + 4(7) − 1 = 34 valence electrons; after four bonds and terminal F octets, two electrons remain as one iodine lone pair.

8 / 21

Which species has a central atom that follows the octet rule in its conventional Lewis structure?

A. BCl₃

B. BeBr₂

C. SbCl₅^2-

D. NH₄⁺

E. XeF₄

Explanation:

In NH₄⁺, nitrogen has four bonds and eight electrons around it. BCl₃ and monomeric BeBr₂ have incomplete central-atom octets, while SbCl₅²⁻ and XeF₄ have expanded central-atom octets in conventional Lewis structures.

9 / 21

Which Lewis structure is the major resonance contributor for N₂O?

Explanation:

C is the major Lewis contributor: O⁻–N⁺≡N. It gives each atom an octet and places the negative formal charge on oxygen, the more electronegative terminal atom. A also satisfies the octet rule but places the negative charge on terminal nitrogen, so it contributes less. The other drawings do not give valid octets with the correct electron count.

10 / 21

In the molecule PCl₃, the atom P has __________ lone pair(s) and __________ bond(s). Also, the total number of valence electrons in the molecule is __________.

A. 0, 2, 20

B. 1, 3, 20

C. 1, 2, 24

D. 1, 3, 26

E. 2, 3, 26

Explanation:

In PCl₃, the central phosphorus atom has one lone pair and three P–Cl bonds. The molecule has 5 + 3(7) = 26 valence electrons in total.

11 / 21

According to VSEPR theory, what is the molecular geometry of XeOF₄?

A. Square planar

B. Linear

C. Trigonal planar

D. Trigonal bipyramidal

E. Square pyramidal

Explanation:

The shape of XeOF₄ is square pyramidal.

12 / 21

Which isolated species has the same molecular geometry as AsCl₃ according to VSEPR theory?

A. NH₄⁺

B. PF₅

C. ClF₃

D. CCl₃^–

E. AlH₃

Explanation:

AsCl₃ and CCl₃⁻ are trigonal pyramidal, so D is correct. NH₄⁺ is tetrahedral, PF₅ is trigonal bipyramidal, ClF₃ is T-shaped, and isolated monomeric AlH₃ is trigonal planar.

13 / 21

According to VSEPR theory, which of these molecules has a shape that is different from the others?

A. H₂O

B. SO₂

C. CS₂

D. Br₂O

E. H₂Te

Explanation:

All have a bent shape except CS₂. CS₂ has a linear shape.

14 / 21

Which of the following is an INCORRECT statement?

A. H-O-H bond angle in H₂O is 109.5°.

B. O-C-O bond angle in CO₂ is 180°.

C. F-B-F bond angle in BF₃ is 120°.

D. Adjacent F–S–F bond angles in octahedral SF₆ are 90°.

E. H-Si-H bond angle in SiH₄ is 109.5°.

Explanation:

A is incorrect. The H–O–H angle in water is about 104.5°, smaller than the ideal tetrahedral angle of 109.5°. Lone-pair–bonding-pair repulsions compress the bond angle. CO₂ is linear, BF₃ is trigonal planar, SF₆ is octahedral, and SiH₄ is tetrahedral.

15 / 21

In the conventional valence-bond model, which molecule is described as having an sp³-hybridized central atom?

A. ClF₃

B. KrF₂

C. PCl₃

D. BeCl₂

E. AlCl₃

Explanation:

In the conventional valence-bond model, PCl₃ has three P–Cl bonds and one lone pair on phosphorus, corresponding to four electron domains and an sp³ description. ClF₃ and KrF₂ have five electron domains, BeCl₂ has two, and monomeric AlCl₃ has three. VSEPR domain counting predicts geometry; expanded-octet bonding need not be described as actual d-orbital hybridization.

16 / 21

A. 2, 7

B. 3, 3

C. 3, 6

D. 6, 3

E. 7, 2

Explanation:

17 / 21

Nitrogen has two single bonds and two lone pairs in the displayed Lewis structure.

A. –2

B. –1

C. 0

D. +1

E. +2

Explanation:

Formal charge = valence electrons of the neutral atom − nonbonding electrons − ½(bonding electrons). The nitrogen shown has four nonbonding electrons and two single bonds, so its formal charge is 5 − 4 − ½(4) = −1.

18 / 21

Which statement does NOT describe ordinary paramagnetism?

A. It can arise from unpaired electrons.

B. It has positive magnetic susceptibility.

C. It is generally weaker than ferromagnetism.

D. Magnetization generally disappears when the external magnetic field is removed.

E. Bulk Fe, Co, and Ni are typical examples at room temperature.

Explanation:

E describes ferromagnetism, not ordinary paramagnetism. Paramagnetic species are weakly attracted to an applied field and have positive magnetic susceptibility; their magnetization generally disappears when the field is removed. Molecular O₂ is paramagnetic because it has two unpaired electrons. Bulk Fe, Co, and Ni are ferromagnetic at room temperature.

19 / 21

Which listed substance is paramagnetic under ordinary room-temperature conditions? For metals, consider bulk samples.

A. O₂

B. F₂

C. Cu

D. Ag

E. Co

Explanation:

O₂ is paramagnetic because its molecular-orbital configuration has two unpaired electrons. F₂ is diamagnetic. Bulk copper and silver are diamagnetic, while bulk cobalt is ferromagnetic at room temperature.

20 / 21

How many atoms are in a body-centered cubic unit cell?

A. 1

B. 2

C. 3

D. 4

E. 5

Explanation:

The eight corner atoms contribute 8 × ⅛ = 1 atom, and the atom at the body center contributes 1 whole atom. A body-centered cubic unit cell therefore contains 2 atoms.

21 / 21

Two crystal lattice structures have one atom and four atoms per unit cell. These structures are called _______________ and _______________.

A. Simple cubic, face-centered cubic

B. Body-centered cubic, face-centered cubic

C. Face-centered cubic, simple cubic

D. Body-centered cubic, simple cubic

E. Simple cubic, body-centered cubic

Explanation:

Simple cubic: 1 atom

Body-centered cubic: 2 atoms

Face-centered cubic: 4 atoms

Your score is

The average score is 55%

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