WAEC Past Questions

WAEC Chemistry 2017
Questions & Answers

20 questions · Correct answers highlighted · 20 with explanations

Explanations reviewed August 2026

This exam paper examines foundational chemical principles including atomic structure, the periodic table, stoichiometry, and chemical equilibrium. The questions frequently emphasize recurring topics such as oxidation numbers and redox reactions, solution concentrations and solubility curves, as well as periodic properties and ionization energy trends. Ensure you work through the mathematical calculations carefully and review the chemical equations step by step rather than rushing to check the answer key.

20 Total Questions
2017 Exam Year
20 With Explanations

Topics in this paper

Solutions and Solubility (6) Acids Bases Salts (6) Atomic Structure (5) Stoichiometric Calculations (5) Periodic Table Trends (5) Redox Reactions (3) Chemical Equilibrium (2) Thermochemistry and Enthalpy (2) Organic Nomenclature (2) Electrochemistry (2) Rates of Reaction (2)
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1
Question 1 of 20
WAEC · Chemistry · 2017

Which of the following species determines the chemical properties of an atom?

A. Electron
B. Neutron
C. Nucleus
D. Proton
Explanation

Options like neutron, nucleus, or proton are incorrect because subatomic particles in the nucleus dictate mass rather than chemical behavior. Valence shell electrons dictate chemical properties because they actively engage in bonding and chemical reactions. Common mistake: Confusing the subatomic particles responsible for atomic mass with those driving chemical reactivity.

2
Question 2 of 20
WAEC · Chemistry · 2017

The following glasswares are used to measure the volume of liquids except

A. graduated beaker
B. pipette
C. test tube
D. burette
Explanation

Students might mistakenly pick graduated beakers, pipettes, or burettes, forgetting that these are precise volumetric tools, whereas test tubes are strictly non-volumetric. A test tube lacks calibration markings and is utilized solely for heating or containing small quantities of substances, unlike true volumetric apparatus. Common mistake: Assuming all standard laboratory glassware is designed for exact volume measurement.

3
Question 3 of 20
WAEC · Chemistry · 2017

Pauli exclusion principle is related to

A. quantity of electrons in the valence shell
B. filling the orbitals with lower energy first
C. the filling of degenerate orbitals
D. quantum numbers of electrons
Explanation

Distractors involving valence electron quantities, lower energy orbital filling, or degenerate orbital filling refer to the Aufbau principle or Hund's rule rather than the Pauli principle. The Pauli exclusion principle dictates the specific quantum numbers of electrons, ensuring no two electrons share an identical set. Common mistake: Confusing the orbital-filling rules of Aufbau with the quantum number restrictions defined by Pauli.

4
Question 4 of 20
WAEC · Chemistry · 2017

The species with no electron in the 3d-orbital is

A. Zn²⁺
B. Sc³⁺
C. Cu
D. Mn
Explanation

Choices like Zn²⁺, Cu, or Mn might seem plausible if their electronic configurations are miscalculated. Scandium, with atomic number 21, loses three electrons to form Sc³⁺ with an argon core configuration [Ar], leaving the 3d orbital completely empty. Common mistake: Failing to subtract electrons correctly from the valence and d-orbitals when determining transition metal ion configurations.

5
Question 5 of 20
WAEC · Chemistry · 2017

The separation of oxygen from nitrogen by fractional distillation of air is possible because

A. nitrogen is less dense than oxygen
B. oxygen is more reactive than nitrogen
C. of the difference in their boiling points
D. they belong to the same period
Explanation

Options referencing gas density, chemical reactivity, or periodic table groups do not explain why industrial air distillation succeeds. Fractional distillation separates gaseous components by exploiting their distinct boiling points, with nitrogen boiling at -196°C and oxygen at -183°C. Common mistake: Attributing gas separation techniques to density or reactivity differences instead of boiling point gaps.

6
Question 6 of 20
WAEC · Chemistry · 2017

The purity of a solid sample can best be determined by its

A. boiling point
B. melting point
C. conductivity
D. solubility
Explanation

Distractors like boiling point, conductivity, or solubility are less sensitive indicators of substance purity than thermal phase changes. A genuinely pure solid exhibits a sharp melting point, whereas impurities cause the melting range to broaden and depress the temperature. Common mistake: Relying on solubility or conductivity tests instead of melting point ranges to check solid purity.

7
Question 7 of 20
WAEC · Chemistry · 2017

The mass of one mole of (NH₄)₂CO₃ is [O=16.0, N=14.0, C=12.0, H=1.0]

A. 66.0 g
B. 76.0 g
C. 80.0 g
D. 96.0 g
Explanation

Options like 66.0 g, 76.0 g, or 80.0 g arise from arithmetic errors in atomic mass summation. The molar mass is computed by adding the atomic weights: two nitrogens (2×14), eight hydrogens (8×1), one carbon (12), and three oxygens (3×16), totaling 96 g/mol. Common mistake: Miscounting the number of constituent atoms inside parentheses when calculating formula mass.

8
Question 8 of 20
WAEC · Chemistry · 2017

Atomic orbital is

A. the circular path through which electrons revolve
B. a region around the nucleus where electrons are most likely to be found
C. the path around the nucleus through which protons move
D. the path around the nucleus through which electrons move
Explanation

Choices describing circular or proton paths are fundamentally flawed because protons reside in the nucleus and do not orbit freely. Quantum mechanics defines an atomic orbital as a probabilistic region where locating an electron is most likely, rather than a fixed physical track. Common mistake: Visualizing electron orbitals as fixed planetary orbits instead of probability clouds.

9
Question 9 of 20
WAEC · Chemistry · 2017

The property of elements which increases down a group of the Periodic Table is

A. electronegativity
B. electron affinity
C. ionic radius
D. ionization energy
Explanation

Students might mistakenly select electronegativity, electron affinity, or ionization energy, which actually decrease down a periodic group. Ionic radius expands down a group because additional electron shells are progressively added. Common mistake: Mixing up the downward trend of atomic and ionic size with the trends of ionization energy and electronegativity.

10
Question 10 of 20
WAEC · Chemistry · 2017

The reason for the decrease in the atomic size of elements across a period is that

A. nuclear charge increases while the outermost electrons are drawn closer to the nucleus
B. nuclear charge decreases while the outermost electrons are drawn closer to the nucleus
C. valence electrons increase across the period while the valence shell remains constant
D. nuclear charge decreases while the distance of the valence shell from the nucleus is increasing
Explanation

Distractors suggesting decreasing nuclear charge or changing valence shell distances misrepresent periodic trends. Moving across a period, increasing protons elevate the nuclear charge, pulling the outermost electrons closer despite the principal energy shell remaining constant. Common mistake: Assuming atomic size increases across a period due to a higher count of valence electrons.

11
Question 11 of 20
WAEC · Chemistry · 2017

The oxidation number of nitrogen in Pb(NO₃)₂ is

A. +2
B. +3
C. +4
D. +5
Explanation

Options such as +2, +3, or +4 are incorrect calculations derived from improper assignments of ion charges. Within the nitrate ion NO₃⁻, oxygen carries a -2 oxidation state, meaning three oxygens total -6, requiring nitrogen to be +5 to yield an overall -1 charge. Common mistake: Forgetting to account for the outer subscript of polyatomic ions when balancing oxidation numbers.

12
Question 12 of 20
WAEC · Chemistry · 2017

Which of the following arrangements indicates the correct decreasing order of ionization energy of period 3 elements?

A. Na, Al, P, Cl
B. Cl, P, Al, Na
C. Na, P, Cl, Al
D. Al, Na, Cl
Explanation

Options featuring incorrect periodic sequences violate the general rule of ionization energy progression across period 3. Ionization energy generally rises across a period, resulting in the decreasing order Cl > P > Al > Na, with a slight anomaly at aluminum due to p-subshell stability. Common mistake: Overlooking periodic anomalies when ranking elements by ionization energy.

13
Question 13 of 20
WAEC · Chemistry · 2017

Which of the following metals has the strongest metallic bond?

A. ¹⁹K
B. ¹³Al
C. ¹²Mg
D. ¹¹Na
Explanation

Choices like potassium, magnesium, or sodium possess fewer delocalized valence electrons or larger atomic volumes. Aluminum features three valence electrons available for delocalization alongside a smaller atomic size, yielding much stronger metallic bonding than Na, Mg, or K. Common mistake: Assuming alkali metals have stronger metallic bonds than group 13 elements due to higher reactivity.

14
Question 14 of 20
WAEC · Chemistry · 2017

Which of the following substances does not have van der Waals' forces?

A. Copper
B. Graphite
C. Neon
D. Iodine crystals
Explanation

Distractors like graphite, neon, and iodine crystals rely on intermolecular or atomic dispersion forces, whereas copper does not. Copper is held together by metallic bonds, whereas neon, graphite, and iodine all rely on van der Waals' forces. Common mistake: Confusing metallic bonding in transition elements with the intermolecular forces found in non-metals.

15
Question 15 of 20
WAEC · Chemistry · 2017

The oxidation number of X in XF₄ is

A. +1
B. +3
C. +4
D. +5
Explanation

Options like +1, +3, or +5 are common miscalculations resulting from improper fluorine valency assumptions. Since each fluorine atom exhibits an oxidation state of -1, four fluorine atoms total -4, forcing element X to assume a +4 oxidation state for neutrality. Common mistake: Misapplying halogen oxidation states when determining the charge of a central atom.

16
Question 16 of 20
WAEC · Chemistry · 2017

The ionization energy is affected by the following factors except

A. distance of the outermost electron(s) from the nucleus
B. size of the positive nuclear charge
C. ability to attract shared electron(s)
D. screening effect of the inner electron(s)
Explanation

Options concerning electron distance, nuclear charge, or screening effects correctly influence ionization energy as they dictate how tightly electrons are held. The ability to attract shared electrons pertains to electronegativity in bonded molecules rather than the energy required to remove an electron from an isolated atom. Common mistake: Confusing the energy required to remove an electron with the tendency to attract shared bonding electrons.

17
Question 17 of 20
WAEC · Chemistry · 2017

Which of the following statements is correct?

A. Covalent compounds would readily ionize in solution
B. covalent compounds consist of ions
C. Hydrogen bond is formed between a metal and a non-metal
D. Ionic compounds in solution would conduct electricity
Explanation

1) Distractor check: You might lean towards options A or B if you mistakenly assume covalent molecules dissociate or break apart into ions like ionic grids do. 2) Reasoning to the answer: Electrical conductivity in liquid media depends on mobile charge carriers. Because ionic structures dissociate into free ions in aqueous solution, they readily permit conduction, whereas the alternative descriptions fail. 3) Common mistake: Assuming covalent molecules behave like ionic salts in water.

18
Question 18 of 20
WAEC · Chemistry · 2017

The bond angle in tetrahedral molecules such as tetrachloromethane is

A. 180°
B. 109.5°
C. 105.5°
D. 90°
Explanation

1) Distractor check: Option A (180°) or D (90°) might look tempting if you mix up tetrahedral spatial arrangements with linear or octahedral layouts. 2) Reasoning to the answer: When a molecule features sp³ hybridization yielding tetrahedral geometry, the resulting symmetrical spacing pushes the spatial intervals to approximately 109.5°. 3) Common mistake: Confusing tetrahedral bond angles with planar or linear values.

19
Question 19 of 20
WAEC · Chemistry · 2017

The properties of a good primary standard include the following except

A. high molecular mass
B. low molecular mass
C. high degree of purity
D. readily available
Explanation

1) Distractor check: You might accidentally choose option A, forgetting that high molecular mass is actually a desired trait rather than something to avoid. 2) Reasoning to the answer: High molar mass is preferred for primary standards because it minimizes weighing and purity errors from impurities or hygroscopic absorption, meaning a low molecular mass is undesirable. 3) Common mistake: Forgetting whether high or low molar mass benefits accurate titration preparation.

20
Question 20 of 20
WAEC · Chemistry · 2017

The use of diamond in abrasives is due to its

A. high melting point
B. hardness
C. octahedral shape
D. durability
Explanation

1) Distractor check: Option A might catch your eye since a high melting point sounds like a general indicator of toughness, but it is not the direct cause of abrasive efficiency. 2) Reasoning to the answer: Strong covalent bonds operating inside a rigid tetrahedral structure give diamond extreme hardness, making it exceptionally well-suited for use in abrasives. 3) Common mistake: Selecting melting point instead of hardness when considering mechanical cutting applications.

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