Which of the following pairs of molecules form hydrogen bonds?

A. C$_{2}$H$_{5}$OH and CH$_{3}$OH

Which of the following statements is correct about ionization energy? It

A. decreases across the periods

B. results in the formation of an anion

C. causes Metalic nuclei to disintegrate

Which of the following pairs of compounds would form a precipitate when their aqueous solutions are mixed?

C. K$_{2}$SO$_{4}$ and BaCl$_{2}$

D. NH$_{4}$NO$_{3}$ and CO$_{3}$

Which of the following statements about elements in Group VII is correct?

C. Cl$_{2}$ will oxidize I$^{-}$

A. Br$_{2}$ will oxidize CI$^{-}$

B. F$_{2}$ has the least tendency to be reduced

C. Cl$_{2}$ will oxidize I$^{-}$

D. l$_{2}$ is a stronger oxidizing agent than F$_{2}$

The reactivity of fluorine is high because of

B. the small size of the fluorine atom

C. the availability of d-orbitals

Electrovalent compounds normally ------------

C. conduct electricity in the solid state

Cathodic protection of metals is based on

D. relative tendencies of oxidation

A. standard electrode potential of hydrogen

D. relative tendencies of oxidation

Student X titrated 25 cm$^{3}$ of Na$_{2}$CO$^{3}$ with 0.1 mol dm$^{-3}$ HCI, using methyl orange as indicator. Student Y carried out the same exercise but used phenolphthalein as an indicator. Which of the following statements about the titration is true?

C. the tirer value obtained by X is twice that of Y

A. hydrogen chloride gas was released in the reactions in both titrations

B. the titer values obtained from the titration are equal

C. the tirer value obtained by X is twice that of Y

D. the titer value obtained by Y is twice that of X

Elements with high ionization energies would-------

C. have high effective nuclear charges

Which of the following statements about Group VII elements is correct?

C. their reactivity decreases down the group

A. they are present in the same physical state

B. they are strong reducing agents

C. their reactivity decreases down the group

D. they exist as monoatomic molecules

Which of the following half reaction equations represent the reaction at the cathode

B. B$^{2+_{(aq)}}$+2e$^{-}$ $\rightarrow$ B$_{(s)}$

A. A$^{3+_{(aq)}}$+3e$^{-}$ $\rightarrow$ A$_{(s)}$

B. B$^{2+_{(aq)}}$+2e$^{-}$ $\rightarrow$ B$_{(s)}$

C. A$_{(s)}$ $\rightarrow$ $^{3+_{(aq)}}$+3e$^{-}$

D. B$_{(s)}$ $\rightarrow$ B$^{2+_{(aq)}}$ + 2e$^{-}$

Under which conditions of pressure (p) and temperature (T) would the volume of an inflated ballon increase? When

C. T is increased and P is decreased

D. T is decreased and P is increased

When CuSO$_{4(aq)}$ is added to Pb(NO$_{3}$)$_{2(aq)}$--------

D. a white precipitate would be formed

A. there would be no visible change

B. a blue precipitate would be formed

C. the resulting solution would become colourless

D. a white precipitate would be formed

What would be observed when aqueous ammonia is added in drops and then in excess to a solution of copper(II) ions?

A. blue precipitate is formed which is soluble in excess ammonia

B. brick red precipitate is produced which is insoluble in excess ammonia

C. white precipitate is formed which is excess in ammonia

D. green precipitate is formed which is insoluble in excess ammonia

The collision between ideal gas molecules are considered to be perfectly elastic because

A. they collide without losing energy

B. they move randomly in a straight line

C. their average kinetic energy is variable

D. the distance between them is large compared to their size

How do I prepare for WAEC (WASSCE) Chemistry?

Practise past questions under timed conditions, focus on your weakest topics, and go back over every wrong answer until you understand why the right option is right. DeeepSolve gives you timed CBT-style practice, and you can ask the AI tutor to explain any question step by step.

WAEC (WASSCE) 2021

2021 WAEC (WASSCE) Chemistry Past Questions and Answers

The 2021 WAEC (WASSCE) Chemistry past questions with answers: 50 objective questions from the 2021 paper in the DeeepSolve question bank, each with its answer behind "Show answer". Try each question before you check, then practise the paper as a timed WAEC-style CBT.

Practise this paper as a timed WAEC-style CBT All WAEC (WASSCE) Chemistry years

Questions

1.
The region around the nucleus where electrons can be located is called?
1. A.a spectra
2. B.an orbital
3. C.a quanta
4. D.a field
Show answer
Answer: B. an orbital
AI explanation
Step 1: Analyze the question The question asks for the term used for the region around the nucleus where an electron is most likely to be found.
Step 2: Explain the concept Electrons do not move in fixed circular paths. According to quantum mechanics, we can only describe the probability of finding an electron in a certain region of space. That region is called an orbital.
s orbital: spherical
p orbital: dumbbell-shaped
Each orbital holds a maximum of 2 electrons (with opposite spins).
Step 3: Check each option
A. Spectra: These are the patterns of light (wavelengths) emitted or absorbed by atoms when electrons change energy levels. They are not a region of space.
B. Orbital: This is the region around the nucleus where there is a high probability of finding an electron. ✔
C. Quanta: These are discrete packets of energy (for example, a photon). They are not a region of space.
D. Field: This is a general physics term, such as an electric or magnetic field. It is not the specific name for the electron region in an atom.
Step 4: Answer The correct option is B. an orbital.
Exam tip: Remember "Orbital = Orbiting region." Do not confuse an orbit (a fixed path in Bohr's old model) with an orbital (a probability region in the modern model). In WASSCE, "region/space where an electron is likely to be found" always points to orbital.
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2.
An element, Q, contains 69% of $^{63}$ Q and 31% of $^{65}$ Q. What is the relative atomic mass of Q?
1. A.63.0
2. B.63.6
3. C.65.0
4. D.69.0
Show answer
Answer: B. 63.6
AI explanation
Analyze the question: Q has two isotopes, ⁶³Q (69%) and ⁶⁵Q (31%). The relative atomic mass is the weighted average of the isotopic masses.
Concept: Relative atomic mass = Σ (isotopic mass × % abundance) ÷ 100
Solution:
Contribution from ⁶³Q: 63 × 69 = 4347
Contribution from ⁶⁵Q: 65 × 31 = 2015
Sum = 4347 + 2015 = 6362
Relative atomic mass = 6362 ÷ 100 = 63.62 ≈ 63.6
Check: The answer must lie between 63 and 65. Since most atoms (69%) are ⁶³Q, it should be closer to 63. 63.6 fits this.
Tip: A quick shortcut is to start from the lighter isotope and add the excess weighted by abundance: 63 + (2 × 0.31) = 63.62. This saves you from multiplying big numbers. Also, always check that your answer lies between the two isotopic masses and is closer to the more abundant isotope. That alone eliminates C and D right away.
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3.
The isotopes of neon are represented by the symbols $^{20}$ $_{x}$ Ne, $^{21}$ $_{y}$ Ne, and $^{22}$ $_{z}$ Ne. The relationship between x, y, and z is?
1. A.x > y > z
2. B.x < y < z
3. C.x = y= z
4. D.x < z < y
Show answer
Answer: C. x = y= z
AI explanation
Analyze the question The symbols ²⁰ₓNe, ²¹ᵧNe and ²²_zNe are written with the mass number on top and the atomic number (x, y, z) at the bottom. The question asks how x, y and z are related.
Concept
Isotopes are atoms of the same element with the same atomic number (the same number of protons) but different mass numbers (different numbers of neutrons).
The atomic number identifies the element. All of these atoms are neon, so they must have the same atomic number.
Solution
All three atoms are neon (Ne), so each has atomic number 10.
Therefore x = y = z = 10.
What differs is the number of neutrons: • ²⁰Ne has 20 − 10 = 10 neutrons • ²¹Ne has 21 − 10 = 11 neutrons • ²²Ne has 22 − 10 = 12 neutrons
The mass numbers (20, 21, 22) are different, but the atomic numbers x, y and z are equal.
Answer: C (x = y = z).
Tip For isotopes, remember that the atomic number (Z) stays the same and only the mass number (A) changes. Whenever a question asks you to compare the atomic numbers of isotopes, the answer is "equal".
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4.
What is the empirical formula of a hydrocarbon containing 0.160 moles of carbon and 0.640 moles of hydrogen
1. A.CH $_{2}$
2. B.CH $_{3}$
3. C.CH $_{4}$
4. D.C $_{2}$ H $_{4}$
Show answer
Answer: C. CH $_{4}$
AI explanation
Analyze the question: We are given the moles of carbon and hydrogen in a hydrocarbon and asked for the empirical formula, which is the simplest whole-number ratio of atoms in the compound.
Concept: To find the empirical formula from moles:
Write down the moles of each element.
Divide each by the smallest number of moles.
If needed, round to or scale up to whole numbers.
Solution:
| Element | Moles | Divide by smallest (0.160) | Ratio | |---|---|---|---| | C | 0.160 | 0.160 ÷ 0.160 | 1 | | H | 0.640 | 0.640 ÷ 0.160 | 4 |
The ratio of C : H = 1 : 4, so the empirical formula is CH₄.
Check the other options:
A. CH₂ has a ratio of 1 : 2, which is wrong.
B. CH₃ has a ratio of 1 : 3, which is wrong.
D. C₂H₄ has a ratio of 1 : 2, so it is not in the simplest form and does not match anyway.
Exam tip: When the question gives you moles directly, skip the step of dividing mass by atomic mass. Just divide by the smallest value and read off the ratio. If you see a decimal like 1.5, multiply everything by 2. If you see 1.33, multiply by 3.
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5.
How many molecules of oxygen would occupy a volume of 2.24 cm $^{3}$ at s.t.p? [Molar volume at s.t.p = 22,400 cm $^{3}$ , Avogadro's number = 6.02 x 10 $^{23}$ ]
1. A.3.01 x 10 $^{20}$
2. B.3.01 x 10 $^{27}$
3. C.6.02 x 10 $^{19}$
4. D.6.02 x 10 $^{27}$
Show answer
Answer: C. 6.02 x 10 $^{19}$
AI explanation
Analyze the question: We need the number of O₂ molecules in 2.24 cm³ of gas at s.t.p. Concept: At s.t.p., 1 mole of any gas occupies 22,400 cm³ and contains 6.02 × 10²³ molecules (Avogadro's number). Solution: Number of moles = volume ÷ molar volume n = 2.24 ÷ 22,400 = 1 × 10⁻⁴ mol Number of molecules = n × Avogadro's number = 1 × 10⁻⁴ × 6.02 × 10²³ = 6.02 × 10¹⁹ molecules Tip: Look at the ratio of the given volume to the molar volume first. Here, 2.24 is 1/10,000 of 22,400, so the answer is simply Avogadro's number divided by 10⁴, which gives 10²³ ÷ 10⁴ = 10¹⁹. Always check that the units (cm³ with cm³) match before you divide.
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6.
The energy profile diagram illustrates-------------
1. A.an endothermic reaction
2. B.an exothermic reaction
3. C.a spontaneous reaction
4. D.a redox reaction
Show answer
Answer: A. an endothermic reaction
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7.
Which of the graphs illustrates the variation of the solubility of a salt in water (y-axis), with an increase in temperature (x-axis) if the dissolution process is exothermic?
1. A.I
2. B.II
3. C.III
4. D.IV
Show answer
Answer: D. IV
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8.
Which of the following oxide causes acid rain?
1. A.CO
2. B.NO
3. C.H $_{2}$ O $_{2}$
4. D.NO $_{2}$
Show answer
Answer: D. NO $_{2}$
AI explanation
Analyze the question: We need the oxide among the options that causes acid rain. Acid rain forms when gases dissolve in rainwater to make acidic solutions.
Concept: The main acid-rain-causing oxides are:
Sulphur dioxide (SO₂) and sulphur trioxide (SO₃), which form sulphurous and sulphuric acids
Nitrogen dioxide (NO₂), which forms nitric acid (and nitrous acid)
Key reaction: 4NO₂ + O₂ + 2H₂O → 4HNO₃
Or: 2NO₂ + H₂O → HNO₃ + HNO₂
Checking each option:
A. CO is carbon monoxide. It is a neutral oxide and a toxic gas (it binds to haemoglobin), but it does not form an acid in water. ✗
B. NO is nitric oxide. It is also a neutral oxide and does not react directly with water to form an acid. It can be oxidised in air to NO₂, which then causes acid rain, but NO itself is not the direct cause. ✗
C. H₂O₂ is hydrogen peroxide. It is not an oxide gas pollutant that causes acid rain (it is a peroxide). ✗
D. NO₂ is nitrogen dioxide. It is an acidic oxide that dissolves in rainwater to form nitric acid, so it directly causes acid rain. ✓
Answer: D (NO₂)
Tip: Remember that acid rain comes from the acidic oxides of sulphur and nitrogen (SO₂, SO₃, NO₂). Neutral oxides such as CO, NO, and N₂O do not make acids with water. Think "Sulphur and Nitrogen = Sour Nature."
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9.
Which of the following pairs of molecules form hydrogen bonds?
1. A.C $_{2}$ H $_{5}$ OH and CH $_{3}$ OH
2. B.CH $_{3}$ OH and H $_{2}$
3. C.H $_{2}$ S and CH $_{4}$
4. D.NH $_{3}$ and SO $_{2}$
Show answer
Answer: A. C $_{2}$ H $_{5}$ OH and CH $_{3}$ OH
AI explanation
Step 1: Analyze the question We need the pair in which the molecules can form hydrogen bonds with each other.
Step 2: The concept A hydrogen bond forms when a hydrogen atom covalently bonded to a very electronegative atom (F, O or N) is attracted to a lone pair on another F, O or N atom. So, for a pair to hydrogen bond, at least one molecule must have an H attached to F/O/N, and the other must have a lone pair on F/O/N.
Step 3: Check each option
A. C₂H₅OH and CH₃OH: Both are alcohols with an –OH group. Each has an O–H (donor) and lone pairs on O (acceptor). Hydrogen bonds form between them. ✔
B. CH₃OH and H₂: H₂ has no electronegative atom and no lone pairs, so it cannot take part. ✘
C. H₂S and CH₄: Sulphur is not electronegative enough, and C–H in methane is nonpolar. No hydrogen bonding. ✘
D. NH₃ and SO₂: NH₃ has N–H, but SO₂ has no H and its oxygen atoms are bonded to S, not H. NH₃ could in principle be attracted to SO₂ by dipole interactions, but this is not the classic hydrogen bonding pair expected here. ✘
Step 4: Answer The correct option is A, since both molecules contain the –OH group, which allows hydrogen bonding between them (this is also why alcohols have relatively high boiling points and dissolve well in water).
Exam tip: Remember "FON" + H. Look for H directly attached to F, O or N. If the molecule pair has an –OH or –NH group, hydrogen bonding is likely. Molecules like H₂S and CH₄ are common distractors.
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10.
The enthalpy change of the reaction is------------
1. A.E $_{A}$
2. B.E $_{B}$
3. C.E $_{C}$
4. D.E $_{D}$
Show answer
Answer: D. E $_{D}$
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11.
The following ions have the same electron configuration except $_{8}$ O, $_{12}$ Mg, $_{13}$ AI, $_{17}$ CI
1. A.CI $^{-}$
2. B.O $^{2-}$
3. C.Mg $^{2+}$
4. D.AI $^{3+}$
Show answer
Answer: A. CI $^{-}$
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12.
Which of the following statements is correct about ionization energy? It
1. A.decreases across the periods
2. B.results in the formation of an anion
3. C.causes Metalic nuclei to disintegrate
4. D.decreases down the group
Show answer
Answer: D. decreases down the group
AI explanation
Analyze the question: We need the correct statement about ionization energy, which is the energy required to remove one mole of electrons from one mole of gaseous atoms, forming a positive ion (cation).
Concept: Ionization energy trends in the Periodic Table:
Across a period (left to right): it generally increases, because nuclear charge increases while the electrons are added to the same shell. The nucleus holds the outer electrons more tightly and atomic radius decreases.
Down a group: it decreases, because atomic radius increases, there are more inner shells shielding the outer electron, and the outer electron is held less tightly and is easier to remove.
Evaluating each option:
A. Decreases across the periods: Wrong. It generally increases across a period.
B. Results in the formation of an anion: Wrong. Losing electrons forms a cation (positive ion). Gaining electrons forms an anion.
C. Causes metallic nuclei to disintegrate: Wrong. Ionization removes outer electrons only, and the nucleus is unaffected. Nuclear disintegration is radioactivity.
D. Decreases down the group: Correct. For example, in Group 1: Li (520 kJ/mol) > Na (496) > K (419), so the energy needed falls as you go down.
Tip: Remember "Across → Up, Down → Down" for ionization energy: it goes up across a period and down a group. Also remember that ionization makes a cation (think "caTion = posiTive"), so you can eliminate option B quickly.
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13.
What number of moles of oxygen would exert a pressure of 10 atm at 320 K in an 8.2 dm $^{2}$ cylinder? [R = 0.082 atm dm $^{3}$ mol $^{-1}$ K $^{-1}$ ]
1. A.0.32
2. B.1.56
3. C.3.13
4. D.31.25
Show answer
Answer: C. 3.13
AI explanation
Analyze the question: We're given pressure, temperature and volume of oxygen gas and asked for the number of moles. This is a direct use of the ideal gas equation. (The "8.2 dm²" in the question is a typo for 8.2 dm³, since volume is measured in dm³ and R is given in dm³ units.)
Concept: The ideal gas law is
PV = nRT, so n = PV / RT
Solution:
P = 10 atm
V = 8.2 dm³
R = 0.082 atm dm³ mol⁻¹ K⁻¹
T = 320 K
n = (10 × 8.2) / (0.082 × 320)
Numerator: 10 × 8.2 = 82
Denominator: 0.082 × 320 = 26.24
n = 82 / 26.24 = 3.125 ≈ 3.13 mol
Answer: The number of moles is about 3.13, which is option C.
Tip: Notice that 8.2 and 0.082 differ by a factor of 100, so you can simplify before calculating: n = (10 × 100) / 320 = 1000/320 = 3.125. Cancelling like this saves time in the exam. Also always check that your units match those of R (atm, dm³, K) before substituting.
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14.
The ratio of carbon atoms of hydrogen atoms in a hydrocarbon is 1:2. If its molecular mass is 56, what is its molecular formula?
1. A.C $_{3}$ H $_{6}$
2. B.C $_{4}$ H $_{8}$
3. C.C $_{2}$ H $_{4}$
4. D.CH $_{2}$
Show answer
Answer: B. C $_{4}$ H $_{8}$
AI explanation
Analyze the question: The ratio of C atoms to H atoms is 1:2, so the empirical formula is CH₂. The molecular mass is 56, and we need the molecular formula.
Concept: Molecular formula = (empirical formula)ₙ, where
n = molecular mass ÷ empirical formula mass
Solution:
Empirical formula = CH₂
Empirical formula mass = 12 + (2 × 1) = 14
n = 56 ÷ 14 = 4
Molecular formula = (CH₂)₄ = C₄H₈
Check: C₄H₈ has a mass of (4 × 12) + (8 × 1) = 48 + 8 = 56 ✓
Checking the other options:
A. C₃H₆ = 42 ✗
C. C₂H₄ = 28 ✗
D. CH₂ = 14 ✗
Tip: For these questions, find the empirical formula mass first, then divide the molecular mass by it to get n. You can also test the options quickly by adding up their masses. Only the option that gives 56 is correct, and it must also keep the 1:2 ratio.
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15.
The basic property of salts used as drying agents is by?
1. A.efforescence
2. B.high melting point
3. C.hygroscopy
4. D.low solubility
Show answer
Answer: C. hygroscopy
AI explanation
Analyze the question: The question asks which property lets certain salts work as drying agents (desiccants), meaning they remove moisture from substances or from the air.
Explain the concept:
Efflorescence: a hydrated salt loses its water of crystallization to the air and turns into a powder (e.g. Na₂CO₃·10H₂O). The salt gives out water, so it cannot dry anything.
Hygroscopy: a substance absorbs moisture from the atmosphere but does not necessarily dissolve (e.g. anhydrous CaCl₂, CaO, conc. H₂SO₄, silica gel). This is the principle behind drying agents.
Deliquescence: a related property where a substance absorbs so much moisture that it dissolves in it (e.g. NaOH, anhydrous CaCl₂, MgCl₂).
Melting point and solubility: neither of these has any direct link to absorbing water.
Step-by-step solution:
A drying agent must take water from its surroundings.
Efflorescence releases water, so A is ruled out.
High melting point (B) describes thermal stability, not water absorption.
Low solubility (D) has nothing to do with attracting water. Many drying agents are actually very soluble.
Hygroscopy (C) is the ability to absorb water vapour from the air, which is exactly what a drying agent does.
Answer: C. hygroscopy
Exam tip: Remember that efflorescence = emits water and hygroscopy = holds/absorbs water. If a question mentions drying agents or desiccators, think "hygroscopic."
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16.
The 25°C evaporation of a 100 cm $^{3}$ solution of K $_{2}$ CO $_{3}$ to dryness gave 14g of the salt. What is the solubility of K $_{2}$ CO $_{3}$ at 25°C? [K $_{2}$ CO $_{3}$ = 138]
1. A.0.01 mol dm $^{-3}$
2. B.0.101 mol dm $^{-3}$
3. C.1.01 mol dm $^{-3}$
4. D.10.0 mol dm $^{-3}$
Show answer
Answer: C. 1.01 mol dm $^{-3}$
AI explanation
Analyze the question: A 100 cm³ saturated solution of K₂CO₃ contains 14 g of the salt after evaporation. We need the solubility in mol dm⁻³.
Concept: Solubility (mol dm⁻³) = moles of solute ÷ volume of solution in dm³. Moles = mass ÷ molar mass.
Solution:
Moles of K₂CO₃ = 14 g ÷ 138 g/mol = 0.1014 mol
Volume = 100 cm³ = 0.1 dm³ (since 1000 cm³ = 1 dm³)
Solubility = 0.1014 mol ÷ 0.1 dm³ = 1.014 mol dm⁻³ ≈ 1.01 mol dm⁻³
Common mistake: Option B (0.101) comes from forgetting to convert 100 cm³ to 0.1 dm³. It is just the number of moles, not the concentration.
Tip: Always convert cm³ to dm³ by dividing by 1000 before calculating concentration. A quick shortcut is that for 100 cm³, you multiply moles by 10 to get mol dm⁻³.
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17.
Which of the following pairs of compounds would form a precipitate when their aqueous solutions are mixed?
1. A.NaCl and KNO $_{3}$
2. B.KCL and NaNO $_{3}$
3. C.K $_{2}$ SO $_{4}$ and BaCl $_{2}$
4. D.NH $_{4}$ NO $_{3}$ and CO $_{3}$
Show answer
Answer: C. K $_{2}$ SO $_{4}$ and BaCl $_{2}$
AI explanation
Step 1: Analyze the question A precipitate forms when two aqueous solutions are mixed and the ions combine to make an insoluble compound. To solve this, swap the partners (double displacement) and check the solubility of the new products.
Step 2: Concept: solubility rules
All nitrates (NO₃⁻) are soluble.
All sodium, potassium and ammonium salts are soluble.
Chlorides are soluble, except AgCl, PbCl₂ and Hg₂Cl₂.
Sulphates are soluble, except BaSO₄, PbSO₄ and CaSO₄ (slightly soluble).
Step 3: Check each option
A. NaCl + KNO₃ → NaNO₃ + KCl. Both products are soluble, so there is no precipitate.
B. KCl + NaNO₃ → KNO₃ + NaCl. Both products are soluble, so there is no precipitate. (This is the same ion mixture as A.)
C. K₂SO₄ + BaCl₂ → BaSO₄ + 2KCl. • KCl is soluble. • BaSO₄ is insoluble, so it forms a white precipitate.
Ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
D. NH₄NO₃ and CO₃ is not a proper pair of compounds, since "CO₃" is only an ion fragment and not a compound. Even if it meant (NH₄)₂CO₃, ammonium salts and nitrates are all soluble, so there would be no precipitate.
Step 4: Answer The pair that forms a precipitate is K₂SO₄ and BaCl₂, which gives white BaSO₄.
Exam tip: Whenever you see Ba²⁺ with a sulphate, think "white precipitate". This is also the standard test for sulphate ions: add BaCl₂ solution with dilute HCl, and a white precipitate that stays in the acid confirms SO₄²⁻. Also remember that Na⁺, K⁺, NH₄⁺ and NO₃⁻ salts are always soluble, so any option with only these ions can be eliminated quickly.
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18.
The formation of ethene from dehydration of ethanol can be described as-------
1. A.an addition reaction
2. B.an elimination reaction
3. C.an oxidation reaction
4. D.a substitution reaction
Show answer
Answer: B. an elimination reaction
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19.
The hydrolysis of proteins by diluting mineral acids produces
1. A.sucrose
2. B.glucose
3. C.amino acids
4. D.fatty acids
Show answer
Answer: C. amino acids
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20.
Which of the following statements about elements in Group VII is correct?
1. A.Br $_{2}$ will oxidize CI $^{-}$
2. B.F $_{2}$ has the least tendency to be reduced
3. C.Cl $_{2}$ will oxidize I $^{-}$
4. D.l $_{2}$ is a stronger oxidizing agent than F $_{2}$
Show answer
Answer: C. Cl $_{2}$ will oxidize I $^{-}$
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21.
The reactivity of fluorine is high because of
1. A.its high electronegativity
2. B.the small size of the fluorine atom
3. C.the availability of d-orbitals
4. D.the strong F-F bond
Show answer
Answer: A. its high electronegativity
AI explanation
Step 1: Analyze the question We need the main reason fluorine is so reactive. Fluorine is the most reactive non-metal, so we look for the property that best explains this.
Step 2: Explain the concept Fluorine is in Group 17 (halogens). Halogens react by gaining one electron to complete their octet. Reactivity here depends on how strongly an atom attracts electrons (electronegativity) and how easily the F-F bond breaks.
Step 3: Evaluate each option
A. High electronegativity: Fluorine has the highest electronegativity of all elements (4.0 on the Pauling scale). It attracts electrons very strongly, so it readily forms fluoride ions or polar bonds. This is the main reason for its reactivity. ✔
B. Small atomic size: This contributes to fluorine's high electronegativity, since the nucleus is close to the incoming electron. However, it is the underlying cause, not the direct explanation. It is a weaker choice than A.
C. Availability of d-orbitals: Fluorine is in Period 2 and has no d-orbitals. ✘
D. Strong F-F bond: The F-F bond is actually weak (about 159 kJ/mol), because lone pair repulsion between the small atoms destabilises it. A weak bond makes fluorine reactive, so "strong" is incorrect. ✘
Step 4: Answer The best answer is A, its high electronegativity.
Note that a weak F-F bond also helps fluorine's reactivity, but that is not what option D says, and the examiners' focus is electronegativity.
Tip: For halogen reactivity questions, remember: "Up the group, electronegativity and oxidising power go up." Fluorine is at the top, so it is the strongest oxidising agent. Also, never link fluorine with d-orbitals, because Period 2 elements have none.
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22.
The activation energy of the reaction is
1. A.E $_{A}$
2. B.E $_{B}$
3. C.E $_{C}$
4. D.E $_{D}$
Show answer
Answer: C. E $_{C}$
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23.
The most common method of preparing insoluble salts is by
1. A.filtration
2. B.decomposition
3. C.neutralization
4. D.double decomposition
Show answer
Answer: D. double decomposition
AI explanation
Analyze the question: It asks for the usual method of preparing insoluble salts, not separating them.
Concept: The method for making a salt depends on its solubility:
Soluble salts are made by neutralization (acid + base/alkali), acid + metal, or acid + carbonate.
Insoluble salts are made by double decomposition, also called precipitation. Two soluble salt solutions are mixed, and their ions swap partners to form one insoluble product (the precipitate) and one soluble product.
Example: Preparing barium sulphate:
BaCl₂(aq) + Na₂SO₄(aq) → BaSO₄(s) + 2NaCl(aq)
The white precipitate of BaSO₄ forms immediately because it is insoluble. It is then separated by filtration, washed, and dried.
Checking the options:
A. Filtration: This only separates the precipitate after it forms. It is not the preparation reaction itself.
B. Decomposition: This is the breakdown of one compound into simpler substances, usually by heat. It is not a general way to make insoluble salts.
C. Neutralization: This is used for soluble salts, such as NaCl from HCl + NaOH.
D. Double decomposition: This is the correct method, because it produces the insoluble salt as a precipitate.
Answer: D. double decomposition
Exam tip: Remember "Insoluble = Double decomposition (precipitation)". Filtration is only the follow-up step. If the question says "preparing a soluble salt", think neutralization, or the acid with metal or carbonate.
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24.
Which of the following cell notations represent the diagram
1. A.B $^{2}$ +/B/ /A/A $^{3+}$
2. B.A $^{3+}$ /A/ /B/B $^{2+}$
3. C.B/B $^{2+}$ / /A/ $^{3+}$
4. D.A/A $^{3+}$ / /B $^{2+}$ /B
Show answer
Answer: B. A $^{3+}$ /A/ /B/B $^{2+}$
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25.
Which of the graphs illustrates the variation of the pH of a given volume of strong acid solution (y-axis) with the volume of strong base titrated against its (x-axis)?
1. A.I
2. B.II
3. C.III
4. D.IV
Show answer
Answer: A. I
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26.
Electrovalent compounds normally ------------
1. A.have low boiling points
2. B.have mobile electrons
3. C.conduct electricity in the solid state
4. D.dissolve in polar solvent
Show answer
Answer: D. dissolve in polar solvent
AI explanation
Analyze the question: It asks about a general property of electrovalent (ionic) compounds, such as NaCl and MgO.
Concept: Electrovalent compounds form when electrons are transferred from a metal to a non-metal, producing positive and negative ions. These ions are held in a crystal lattice by strong electrostatic forces of attraction.
Evaluate each option:
A. Have low boiling points. False. The electrostatic forces between ions are strong and need a lot of energy to overcome, so ionic compounds have high melting and boiling points. (Low boiling points are typical of covalent molecular compounds.)
B. Have mobile electrons. False. Mobile electrons are a feature of metals. In ionic compounds the electrons are held by the ions, and none are free to move.
C. Conduct electricity in the solid state. False. In the solid, the ions are fixed in the lattice and cannot move. They conduct only when molten or dissolved in water, when the ions become free to move.
D. Dissolve in polar solvent. True. Polar solvents like water have partial positive and negative ends. These attract the ions and pull them out of the lattice (hydration), so ionic compounds typically dissolve in water.
Answer: D
Tip: Remember "ionic = high melting point, soluble in water, conducts only when molten or in solution." Also remember "like dissolves like": ionic and polar substances dissolve in polar solvents, while non-polar substances dissolve in non-polar solvents such as kerosene or benzene.
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27.
Cathodic protection of metals is based on
1. A.standard electrode potential of hydrogen
2. B.its electrical conductivity
3. C.nature of oxides formed
4. D.relative tendencies of oxidation
Show answer
Answer: D. relative tendencies of oxidation
AI explanation
Analyze the question: Cathodic protection is a way of preventing corrosion (rusting) of a metal such as iron or steel pipes, ship hulls, and underground tanks. We need to identify the principle behind it.
Explain the concept: Corrosion is an oxidation process: the metal loses electrons.
Fe → Fe²⁺ + 2e⁻
Cathodic protection makes the metal we want to protect act as the cathode (the site of reduction), so it does not lose electrons. There are two main methods:
Sacrificial protection: Iron is connected to a more reactive metal such as zinc or magnesium. Zinc is more easily oxidised than iron (it has a more negative electrode potential), so zinc corrodes instead and the iron is protected. This is the principle behind galvanising and the magnesium blocks fixed on ship hulls.
Impressed current: An external power source supplies electrons to the metal, making it the cathode.
Solution:
A. Standard electrode potential of hydrogen: Hydrogen is only the reference electrode (0.00 V), so this is not the basis of the protection.
B. Electrical conductivity: Conductivity is needed for the circuit, but it is not the principle.
C. Nature of oxides formed: This applies to protection by a passive oxide layer, as in aluminium, which is a different concept.
D. Relative tendencies of oxidation: The more reactive metal (the one with the greater tendency to be oxidised) is oxidised in preference to the metal being protected. This is the principle of cathodic protection.
Answer: D
Tip: Remember "Cathodic = Cathode is safe." The metal that is more easily oxidised (higher in the reactivity series) is sacrificed to protect the less reactive one. Whenever you see "sacrificial", "galvanising" or "zinc/magnesium bars", think of the relative tendency to lose electrons.
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28.
Potassium trioxonitrate (V) can be obtained from its solution by----------
1. A.distillation
2. B.evapoartion
3. C.crystallization
4. D.filtration
Show answer
Answer: C. crystallization
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29.
Which of the following gasses is highly soluble in water at room temperature
1. A.ammonia
2. B.carbon (IV) oxide
3. C.chlorine
4. D.nitrogen
Show answer
Answer: A. ammonia
AI explanation
Analyze the question: You are asked which gas dissolves most readily (is highly soluble) in water at room temperature.
Concept: Gas solubility in water depends mainly on polarity and on whether the gas reacts with water or forms hydrogen bonds with it.
Ammonia (NH₃): It is polar and forms hydrogen bonds with water. It also reacts reversibly to form ammonium and hydroxide ions: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ It is extremely soluble, about 700 volumes of NH₃ per 1 volume of water at room temperature. This is why it is used in the fountain experiment.
Carbon (IV) oxide (CO₂): It is only slightly soluble, about 1 volume per 1 volume of water. It forms a weak carbonic acid.
Chlorine (Cl₂): It is moderately soluble, about 2 volumes per 1 volume of water. It partly reacts to form HCl and HOCl.
Nitrogen (N₂): It is non-polar and almost insoluble in water.
Comparing them: NH₃ ≫ Cl₂ > CO₂ > N₂ in terms of solubility (roughly). Ammonia is by far the most soluble, so it is the only one that fits "highly soluble".
Answer: A. ammonia
Tip: Remember that ammonia and hydrogen chloride are the two classic "very soluble" gases in WASSCE and JAMB. They are collected by upward or downward delivery, never over water. Gases like N₂, O₂, H₂ and CO are almost insoluble, so they can be collected over water.
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30.
Student X titrated 25 cm $^{3}$ of Na $_{2}$ CO $^{3}$ with 0.1 mol dm $^{-3}$ HCI, using methyl orange as indicator. Student Y carried out the same exercise but used phenolphthalein as an indicator. Which of the following statements about the titration is true?
1. A.hydrogen chloride gas was released in the reactions in both titrations
2. B.the titer values obtained from the titration are equal
3. C.the tirer value obtained by X is twice that of Y
4. D.the titer value obtained by Y is twice that of X
Show answer
Answer: C. the tirer value obtained by X is twice that of Y
AI explanation
Analyze the question
Na₂CO₃ is a dibasic salt of a weak acid. It reacts with HCl in two stages, and each indicator detects a different stage.
Explain the concept
Stage 1: Na₂CO₃ + HCl → NaHCO₃ + NaCl. The pH at the end of this stage is about 8.3, and phenolphthalein changes colour here (range 8.3–10).
Stage 2: NaHCO₃ + HCl → NaCl + H₂O + CO₂. The pH at the end of this stage is about 3.5–4, and methyl orange changes colour here (range 3.1–4.4).
Solution
Student Y (phenolphthalein) only reaches the end of stage 1. This needs 1 mole of HCl per mole of Na₂CO₃, so the titre is V.
Student X (methyl orange) reaches the end of stage 2. This needs 2 moles of HCl per mole of Na₂CO₃, so the titre is 2V.
So X's titre is twice Y's titre.
Checking the other options
A is wrong because HCl gas is not released. CO₂ is released, and only in the second stage.
B is wrong because the titres are different (2V and V).
D is wrong because it reverses the relationship. Y's titre is half of X's, not double.
Answer: C
Tip: Phenolphthalein "sees" only half of the carbonate neutralisation, while methyl orange sees all of it. So methyl orange titre = 2 × phenolphthalein titre. Remember "Methyl orange = Most acid = Maximum volume."
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31.
What is the concentration of a solution which contains 0.28g of potassium hydroxide in 100 cm $^{3}$ of solution [KOH = 56]
1. A.0.01 mol dm $^{-3}$
2. B.0.05 mol dm $^{-3}$
3. C.0.10 mol dm $^{-3}$
4. D.0.50 mol dm $^{-3}$
Show answer
Answer: B. 0.05 mol dm $^{-3}$
AI explanation
Step 1: Analyze the question We need the molar concentration (mol dm⁻³) of KOH, given the mass dissolved and the volume of solution.
Step 2: Concept Concentration (mol dm⁻³) = number of moles ÷ volume in dm³ Number of moles = mass ÷ molar mass
Step 3: Solution
Moles of KOH = 0.28 g ÷ 56 g/mol = 0.005 mol
Volume = 100 cm³ = 100 ÷ 1000 = 0.1 dm³
Concentration = 0.005 ÷ 0.1 = 0.05 mol dm⁻³
Alternatively, using the mass concentration route:
Mass concentration = 0.28 g ÷ 0.1 dm³ = 2.8 g/dm³
Molar concentration = 2.8 ÷ 56 = 0.05 mol dm⁻³
Step 4: Answer The correct option is **B. 0.
Tip: Always convert cm³ to dm³ first by dividing by 1000. Many students get option A or D by forgetting this conversion or mixing up the steps. Remember the formula triangle: moles = concentration × volume (in dm³).
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32.
A molecular of phosphorus is
1. A.diatomic
2. B.triatomic
3. C.tetraatomic
4. D.monoatomic
Show answer
Answer: C. tetraatomic
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33.
The Bohr model of the atom proposed the existence of?
1. A.the nucleus
2. B.electron shells
3. C.nucleons
4. D.neutrons
Show answer
Answer: B. electron shells
AI explanation
Step 1: Analyze the question We are asked what the Bohr model specifically introduced or proposed about the atom.
Step 2: Explain the concept Different atomic models added different ideas:
Dalton: atoms are indivisible solid spheres.
Thomson: the "plum pudding" model, with electrons embedded in a positive sphere.
Rutherford: a small, dense, positively charged nucleus at the center, with electrons around it.
Bohr (1913): electrons move in fixed circular paths called orbits, or energy levels (shells), around the nucleus. Electrons do not radiate energy while in these orbits. They absorb or emit energy only when they jump between levels.
Chadwick (1932): discovered the neutron.
Step 3: Check the options
A. The nucleus: This was Rutherford's discovery. Bohr built on it, but did not propose it.
B. Electron shells: This is Bohr's key contribution, since electrons occupy specific quantized energy levels.
C. Nucleons: This is a general term for protons and neutrons, and it is not part of Bohr's proposal.
D. Neutrons: These were discovered by Chadwick after Bohr's model.
Step 4: Answer The correct option is B. electron shells.
Tip: Link each scientist to one keyword:
Thomson → electron
Rutherford → nucleus
Bohr → shells (energy levels)
Chadwick → neutron
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34.
Elements with high ionization energies would-------
1. A.lose electrons easily
2. B.have large atomic radii
3. C.have high effective nuclear charges
4. D.have low atomic numbers
Show answer
Answer: C. have high effective nuclear charges
AI explanation
Analyze the question: Ionization energy is the energy needed to remove the outermost electron from a gaseous atom. A high ionization energy means the electron is held tightly and is hard to remove. We need to pick the statement that goes with this.
Concept: The ionization energy depends on:
Effective nuclear charge (Zeff): the higher the Zeff, the stronger the pull on the outer electron, so the ionization energy is higher.
Atomic radius: the larger the radius, the weaker the pull, so the ionization energy is lower.
Shielding: more inner shells mean more shielding and a lower ionization energy.
Evaluate each option:
A. Lose electrons easily: Wrong. High ionization energy means electrons are lost with difficulty.
B. Have large atomic radii: Wrong. Large atoms have outer electrons far from the nucleus, which gives low ionization energy (e.g. Cs, Fr).
C. Have high effective nuclear charges: Correct. A strong net pull from the nucleus holds the outer electrons firmly, so more energy is needed to remove one. This is why ionization energy increases across a period, as Zeff rises and the radius shrinks.
D. Have low atomic numbers: Not a reliable rule. Ionization energy generally decreases down a group, so low atomic number can go with high ionization energy (e.g. He), but atomic number alone does not decide it. Sodium (Z = 11) has a much lower ionization energy than neon (Z = 10) only because of the shell and nuclear charge effects. The direct cause is Zeff, not atomic number.
Answer: C. have high effective nuclear charges
Tip: Remember "High IE = tight grip." Anything that strengthens the nucleus's grip (high Zeff, small radius) raises the ionization energy. Anything that loosens it (large radius, more shielding) lowers it.
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35.
If humid air is polluted by chlorine discharged, the air can be restored by sprinkling
1. A.solid MnO $_{2}$
2. B.acidified KMnO $_{4}$
3. C.acidified FeSO $_{4}$
4. D.saturated NaCl(aq)
Show answer
Answer: C. acidified FeSO $_{4}$
AI explanation
Analyze the question: Chlorine gas has polluted humid air. We need a substance that, when sprinkled, removes the chlorine by reacting with it. Chlorine is an oxidizing agent, so we need a reducing agent to convert Cl₂ into harmless chloride ions (Cl⁻).
Concept: Cl₂ + 2e⁻ → 2Cl⁻. Any substance that readily gives up electrons will remove it.
Checking each option:
A. Solid MnO₂: This is an oxidizing agent. It is used to produce chlorine from concentrated HCl, so it cannot remove it.
B. Acidified KMnO₄: This is also a strong oxidizing agent. It reacts with HCl to liberate chlorine, so it would make things worse.
C. Acidified FeSO₄: Fe²⁺ is a reducing agent. It reduces chlorine to chloride: 2Fe²⁺ + Cl₂ → 2Fe³⁺ + 2Cl⁻ This removes the chlorine from the air.
D. Saturated NaCl(aq): Chlorine is much less soluble in brine because of the common-ion effect, which is why chlorine is collected over brine. It does not chemically remove chlorine, so it is not a way to restore the air.
Tip: Gas is an oxidizer, so you want a reducing agent. Remember that Fe²⁺ (green) → Fe³⁺ (yellow-brown) is a classic test of oxidizing agents. When you see "remove an oxidizing gas," look for the reducing agent among the options. Both MnO₂ and KMnO₄ are oxidizers, so you can eliminate them quickly.
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36.
The alkanol represented by the structure above is
1. A.primary and dihydric
2. B.secondary and monohydric
3. C.tertiary and dihydric
4. D.secondary and dihydric
Show answer
Answer: D. secondary and dihydric
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37.
Which of the following statements about Group VII elements is correct?
1. A.they are present in the same physical state
2. B.they are strong reducing agents
3. C.their reactivity decreases down the group
4. D.they exist as monoatomic molecules
Show answer
Answer: C. their reactivity decreases down the group
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38.
What is the relative molecular mass of the compound below? [H = 1.0; C = 12.0; O = 16.0]
1. A.137
2. B.136
3. C.64
4. D.59
Show answer
Answer: B. 136
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39.
Which of the following half reaction equations represent the reaction at the cathode
1. A.A $^{3+_{(aq)}}$ +3e $^{-}$ $\rightarrow$ A $_{(s)}$
2. B.B $^{2+_{(aq)}}$ +2e $^{-}$ $\rightarrow$ B $_{(s)}$
3. C.A $_{(s)}$ $\rightarrow$ $^{3+_{(aq)}}$ +3e $^{-}$
4. D.B $_{(s)}$ $\rightarrow$ B $^{2+_{(aq)}}$ + 2e $^{-}$
Show answer
Answer: B. B $^{2+_{(aq)}}$ +2e $^{-}$ $\rightarrow$ B $_{(s)}$
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40.
A coordinate covalent bond could be formed between?
1. A.NH $_{3}$ and PCl $_{3}$
2. B.BCl $_{3}$ and AlCl $_{3}$
3. C.BCl $_{3}$ and NH $_{3}$
4. D.H $^{+}$ and AlCl $_{3}$
Show answer
Answer: C. BCl $_{3}$ and NH $_{3}$
AI explanation
Analyze the question A coordinate (dative) covalent bond forms when one atom supplies both electrons of the shared pair. So we need one species with a lone pair (donor) and another with an empty orbital (acceptor).
Concept
Donor: has a lone pair, e.g. NH₃, H₂O, Cl⁻, PCl₃ (P has a lone pair).
Acceptor: has an electron-deficient centre, e.g. BCl₃, AlCl₃ (only 6 electrons around the central atom), H⁺ (empty 1s orbital).
Check each option
A. NH₃ and PCl₃: Both are donors (each has a lone pair). Neither has an empty orbital to accept, so no coordinate bond forms.
B. BCl₃ and AlCl₃: Both are electron-deficient acceptors. Neither has a lone pair on the central atom to donate, so no coordinate bond forms between them.
C. BCl₃ and NH₃: NH₃ has a lone pair on nitrogen, and BCl₃ has an empty p-orbital on boron. The nitrogen donates its lone pair to boron, forming H₃N→BCl₃, a classic coordinate covalent bond.
D. H⁺ and AlCl₃: H⁺ is an acceptor and AlCl₃ is also an acceptor. Neither can donate a pair of electrons from the central atom, so no coordinate bond forms this way.
Tip: Look for a donor + acceptor pair. Whenever you see NH₃ (or H₂O) paired with BCl₃, AlCl₃ or H⁺, think coordinate bond. If both species are donors or both are acceptors, no dative bond forms. Common examples: NH₄⁺ (NH₃ + H⁺), H₃O⁺ (H₂O + H⁺), and H₃N→BF₃.
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41.
Consider the structure above How many carbon atoms does the parent chain contain?
1. A.5
2. B.4
3. C.3
4. D.2
Show answer
Answer: A. 5
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42.
How many coulombs of electricity would liberate 1.08g of Ag from a solution of silver salt? [Ag = 108.0; 1F = 96500 C]
1. A.96500 C
2. B.9650 C
3. C.965 C
4. D.9.65 C
Show answer
Answer: C. 965 C
AI explanation
Analyze the question: We need the quantity of electricity (in coulombs) that deposits 1.08 g of silver from a silver salt solution. Concept: Faraday's laws of electrolysis. One mole of electrons (1 Faraday = 96500 C) deposits one mole of a metal if the metal ion carries a +1 charge. For silver: Ag⁺ + e⁻ → Ag So 1 F (96500 C) liberates 1 mole of Ag = 108 g. Solution: Step 1: Find the moles of Ag liberated. Moles = mass ÷ molar mass = 1.08 ÷ 108 = 0.01 mol Step 2: Each mole of Ag needs 1 mole of electrons, so the charge needed is: Q = 0.01 × 96500 = 965 C Alternatively, by proportion: 108 g of Ag needs 96500 C 1.08 g of Ag needs (1.08 ÷ 108) × 96500 = 965 C Answer: C. 965 C Tip: First check the ion's charge (Ag⁺ means 1 F per mole). Then use the ratio: Charge = (mass given ÷ molar mass) × charge on ion × 96500. Here 1.08 is exactly 1/100 of 108, so the answer is 96500 ÷ 100 = 965 C. Spotting such simple ratios saves time in the exam.
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43.
Which of the following illustrates the variation of the rate of evolution of gas from a given length of magnesium ribbon (y-axis) with an increase in the concentration of the added (x-axis)?
1. A.I
2. B.II
3. C.III
4. D.IV
Show answer
Answer: D. IV
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44.
Under which conditions of pressure (p) and temperature (T) would the volume of an inflated ballon increase? When
1. A.P and T are increased
2. B.both T and P are decreased
3. C.T is increased and P is decreased
4. D.T is decreased and P is increased
Show answer
Answer: C. T is increased and P is decreased
AI explanation
Analyze the question: We want the conditions of pressure (P) and temperature (T) that make the volume of a fixed amount of gas in a balloon increase.
Concept (Gas Laws):
Boyle's Law: At constant T, V ∝ 1/P. Lowering the pressure increases the volume.
Charles's Law: At constant P, V ∝ T (in kelvin). Raising the temperature increases the volume.
Combined gas law:
$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} \quad\Rightarrow\quad V_2 = V_1 \cdot \frac{P_1}{P_2}\cdot\frac{T_2}{T_1}$
So V is directly proportional to T and inversely proportional to P.
Step-by-step check of the options:
A. P and T both increased: Higher T increases V, but higher P decreases V. The effects oppose each other, so the result is uncertain. ✗
B. P and T both decreased: Lower P increases V, but lower T decreases V. Again the effects oppose, so the result is uncertain. ✗
C. T increased, P decreased: Higher T increases V and lower P increases V. Both effects work together, so V definitely increases. ✓
D. T decreased, P increased: Both effects decrease V. ✗
Answer: The volume definitely increases when temperature is raised and pressure is lowered, which is option C.
Exam tip: Remember "T goes with V, P goes against V." For V to rise with certainty, the two changes must push the same way: T up and P down (or the reverse for a decrease in V). When both factors point in the same direction, that is usually the intended answer. Also remember to convert °C to kelvin (K = °C + 273) in calculations.
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45.
When CuSO $_{4(aq)}$ is added to Pb(NO $_{3}$ ) $_{2(aq)}$ --------
1. A.there would be no visible change
2. B.a blue precipitate would be formed
3. C.the resulting solution would become colourless
4. D.a white precipitate would be formed
Show answer
Answer: D. a white precipitate would be formed
AI explanation
Analyze the question: We are mixing two aqueous salt solutions, copper(II) sulphate and lead(II) nitrate. We must predict what is observed.
Concept: When two soluble salts are mixed, a double displacement (precipitation) reaction may occur if one of the possible products is insoluble. Recall the solubility rules:
All nitrates are soluble.
Most sulphates are soluble, except those of lead (PbSO₄), barium (BaSO₄) and calcium (CaSO₄, sparingly soluble).
Solution:
The ions swap partners:
CuSO₄(aq) + Pb(NO₃)₂(aq) → PbSO₄(s) + Cu(NO₃)₂(aq)
Cu(NO₃)₂ is a nitrate, so it is soluble and stays in solution (it keeps the blue colour of Cu²⁺).
PbSO₄ is insoluble and appears as a white precipitate.
Ionic equation:
Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s)
Checking the options:
A. No visible change: wrong, a precipitate forms.
B. Blue precipitate: wrong. Blue Cu(OH)₂ would need hydroxide ions (e.g., NaOH or NH₃), and the Cu²⁺ stays dissolved here.
C. Colourless solution: wrong, Cu²⁺ still remains in solution, so the liquid stays blue.
D. White precipitate: correct, this is PbSO₄.
Answer: D
Tip: Remember that sulphates of Pb, Ba and Ca are the exceptions among sulphates. Whenever Pb²⁺ meets SO₄²⁻ or Cl⁻ (PbCl₂, which dissolves in hot water), expect a white precipitate.
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46.
Which of the following species has the largest ionic radius
1. A.S $^{2-}$
2. B.K $^{+}$
3. C.CI $^{-}$
4. D.Ca $^{2+}$
Show answer
Answer: A. S $^{2-}$
AI explanation
Analyze the question: We are asked to compare the ionic radii of four ions: S²⁻, Cl⁻, K⁺ and Ca²⁺.
Concept: These ions are isoelectronic. Each has 18 electrons, the same configuration as argon: 1s² 2s² 2p⁶ 3s² 3p⁶.
For isoelectronic species, the electrons are the same in number, so the size depends on the nuclear charge (number of protons):
More protons pull the electron cloud in more strongly, so the radius is smaller.
Fewer protons pull less strongly, so the radius is larger.
Step-by-step solution:
| Ion | Protons | Electrons | |---|---|---| | S²⁻ | 16 | 18 | | Cl⁻ | 17 | 18 | | K⁺ | 19 | 18 | | Ca²⁺ | 20 | 18 |
Order of nuclear charge: S (16) < Cl (17) < K (19) < Ca (20)
Order of ionic radius is the reverse: S²⁻ > Cl⁻ > K⁺ > Ca²⁺
(Approximate values: S²⁻ ≈ 184 pm, Cl⁻ ≈ 181 pm, K⁺ ≈ 138 pm, Ca²⁺ ≈ 100 pm.)
S²⁻ has the fewest protons holding the same 18 electrons, so it has the largest radius.
Answer: A. S²⁻
Exam tip: When you see ions with the same number of electrons, count the protons. The more protons, the smaller the ion. Remember: "Isoelectronic: highest atomic number = smallest size."
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47.
Protons and electrons are called fundamental particles because they----------
1. A.are indivisible
2. B.have different charges
3. C.are the lightest particles
4. D.are found in all matter
Show answer
Answer: D. are found in all matter
AI explanation
Analyze the question We need to find why protons and electrons are called "fundamental particles" of an atom. In chemistry at WASSCE level, fundamental (subatomic) particles are the basic building blocks that make up atoms: protons, neutrons and electrons.
Explain the concept Fundamental particles of the atom are the particles common to the atoms of all elements. Every atom (apart from ordinary hydrogen, which has no neutron) is built from protons, neutrons and electrons. So they are "fundamental" because they are the basic constituents present in all matter.
Check each option
A. are indivisible: At the WASSCE level this is not the reason. Protons are in fact made up of quarks, so this is not strictly true.
B. have different charges: This describes a property, not why they are called fundamental. Neutrons are also fundamental but have no charge.
C. are the lightest particles: Not true. A proton is about 1836 times heavier than an electron, so a proton is not "the lightest".
D. are found in all matter: Correct. All atoms contain them, and since all matter is made of atoms, these particles are the basic building blocks of all matter.
Answer The correct option is D.
Tip When you see "fundamental particles", think "basic building blocks present in every atom". Be careful with "indivisible", as it is a tempting distractor, since modern physics shows protons can be divided into quarks.
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48.
What would be observed when aqueous ammonia is added in drops and then in excess to a solution of copper(II) ions?
1. A.blue precipitate is formed which is soluble in excess ammonia
2. B.brick red precipitate is produced which is insoluble in excess ammonia
3. C.white precipitate is formed which is excess in ammonia
4. D.green precipitate is formed which is insoluble in excess ammonia
Show answer
Answer: A. blue precipitate is formed which is soluble in excess ammonia
AI explanation
1. Analyze the question You are asked what happens when NH₄OH (aqueous ammonia) is added first in drops, then in excess, to Cu²⁺ solution. This is a qualitative analysis (cation test) question.
2. The concept Aqueous ammonia supplies OH⁻ ions, which precipitate many metal hydroxides. Some hydroxides then dissolve in excess ammonia because the metal ion forms a soluble complex ion with NH₃.
3. Step-by-step solution
Step 1: A few drops of ammonia Cu²⁺ reacts with OH⁻ to form copper(II) hydroxide:
Cu²⁺(aq) + 2OH⁻(aq) → Cu(OH)₂(s)
Observation: a pale blue precipitate.
Step 2: Excess ammonia The precipitate dissolves because the tetraamminecopper(II) complex forms:
Cu(OH)₂(s) + 4NH₃(aq) → [Cu(NH₃)₄]²⁺(aq) + 2OH⁻(aq)
Observation: a deep blue (inky blue) solution.
Checking the other options:
B: Brick red is not the colour of any copper(II) hydroxide precipitate. (Brick-red is the colour of calcium flame, or of Cu₂O in Fehling's test.)
C: White precipitates are formed by Zn²⁺, Pb²⁺, or Al³⁺. With Cu²⁺ the precipitate is blue.
D: A green precipitate is typical of Fe²⁺, and it does not dissolve in excess ammonia.
4. Answer The blue precipitate dissolves in excess ammonia to give a deep blue solution, so the answer is A.
5. Tip Remember: Cu²⁺ = blue ppt → deep blue solution in excess NH₃. The cations whose precipitates dissolve in excess ammonia are Cu²⁺, Zn²⁺, and Ag⁺ (in dilute solution). A quick way to recall them is "Come Zee Ammonia". Fe²⁺, Fe³⁺, Pb²⁺, and Al³⁺ do not dissolve in excess ammonia.
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49.
The collision between ideal gas molecules are considered to be perfectly elastic because
1. A.they collide without losing energy
2. B.they move randomly in a straight line
3. C.their average kinetic energy is variable
4. D.the distance between them is large compared to their size
Show answer
Answer: A. they collide without losing energy
AI explanation
Analyze the question: It asks why collisions between ideal gas molecules are described as perfectly elastic.
Concept: In the kinetic theory of gases, an elastic collision is one in which no kinetic energy is lost. Total kinetic energy before the collision equals total kinetic energy after it. The molecules may exchange energy with one another, but none is converted to heat, sound, or deformation. This is why a gas at constant temperature does not "run down" and the molecules keep moving indefinitely.
Evaluating the options:
A. They collide without losing energy. This is the definition of a perfectly elastic collision. ✔
B. They move randomly in a straight line. This is a separate postulate (random, straight-line motion between collisions). It describes how they move, not why collisions are elastic.
C. Their average kinetic energy is variable. This is wrong. The average kinetic energy is proportional to the absolute temperature, so it is constant at a fixed temperature. Variable energy would not explain elasticity.
D. The distance between them is large compared to their size. This is also a postulate. It explains why intermolecular forces and molecular volume are negligible, but not why collisions are elastic.
Answer: Option A.
Exam tip: Whenever you see "perfectly elastic", think "no loss of kinetic energy". Link each kinetic theory postulate to its purpose:
Elastic collisions → no energy loss
Large distances → negligible volume and forces
Random straight-line motion → gas fills its container
Average KE ∝ T → energy depends on temperature
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50.
Which of the following scientists discovered the electrons?
1. A.Joseph J. Thompson
2. B.James Chadwick
3. C.Amedeo Avogadro
4. D.Ernest Rutherford
Show answer
Answer: A. Joseph J. Thompson
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