JAMB Chemistry 2019 past questions

Checked by the Skolra question team · Last updated 23 October 2026

Ten questions from the 2019 JAMB Chemistry paper, with answers and explanations. We hold 74 questions from this paper, and the Skolra app has all of them.

JAMB Chemistry 2019 questions and answers

1. 1.

   Which two gases can be used for the demonstration of the fountain experiment?

   1. A
      Sulphur(IV) oxide and carbon(IV) oxide
   2. B
      Nitrogen and hydrogen
   3. C
      Carbon(IV) oxide and oxygen
   4. D
      Ammonia and hydrogen chloride
      Correct
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   Answer: D

   The fountain experiment demonstrates the very high solubility of a gas in water. A round-bottomed flask is filled with the dry gas, fitted with a jet tube dipped in water; when a little water is squirted in, the gas dissolves so fast that the pressure inside drops sharply and water is forced up the jet in a fountain.

   Ammonia and hydrogen chloride are the two gases classically used for this demonstration, because both are extremely soluble in water (about 700 volumes of gas dissolve in 1 volume of water for ammonia, and even more for hydrogen chloride), so the pressure drop is fast and dramatic.

2. 2.

   The molecular shape and bond angle of water are respectively

   1. A
      Linear and 180°
   2. B
      Trigonal pyramidal and 107°
   3. C
      Bent (V-shaped/angular) and 104.5°
      Correct
   4. D
      Tetrahedral and 109.5°
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   Answer: C

   Oxygen in water has two bonding pairs (to each hydrogen) and two lone pairs, so its electron-pair arrangement is tetrahedral, but since lone pairs are not counted when naming the molecular shape, water is described as bent (V-shaped/angular).

   The two lone pairs on oxygen repel the bonding pairs more strongly than bonding pairs repel each other, squeezing the H-O-H angle down from the ideal tetrahedral value of 109.5° to about 104.5°.

3. 3.

   The IUPAC name of the compound CF$_3$CHBrCl is

   1. A
      2-bromo-2-chloro-1,1,1-trifluoroethane
      Correct
   2. B
      1,2-dibromo-1-chloro-2,2-difluoroethane
   3. C
      2-bromo-1-chloro-1,1,2-trifluoroethane
   4. D
      1-bromo-1-chloro-2,2,2-trifluoroethane
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   Answer: A

   The molecule CF₃CHBrCl has only two carbons, so it is named as an ethane. One carbon holds three fluorine atoms; the other holds one bromine, one chlorine and one hydrogen.

   Numbering: IUPAC rules require the chain to be numbered so that the substituents get the lowest locants at the first point of difference. Starting from the CF₃ carbon gives the set 1,1,1,2,2. Starting from the other carbon gives 1,1,2,2,2. Comparing these, 1,1,1,2,2 is lower, so the CF₃ carbon is C1 and the CHBrCl carbon is C2.

4. 4.

   Which of the following conditions will most enhance the spontaneity of a reaction?

   1. A
      An increase in the activation energy of the reaction
   2. B
      An increase in the entropy of the system
      Correct
   3. C
      A positive value of the free energy change
   4. D
      A decrease in the temperature of the system
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   Answer: B

   Spontaneity of a reaction is governed by the Gibbs free energy equation: ΔG = ΔH − TΔS. A reaction is spontaneous when ΔG is negative.

   An increase in the entropy of the system means ΔS becomes more positive. Since ΔS is subtracted after being multiplied by T (a positive absolute temperature), a more positive ΔS makes the −TΔS term more negative, pulling ΔG down and favouring spontaneity. This is why processes that increase disorder (more particles, more randomness, gases forming from solids/liquids) tend to proceed on their own.

5. 5.

   Which of the following metals is the most essential in the regulation of blood volume, blood pressure and osmotic equilibrium?

   1. A
      Potassium
   2. B
      Sodium
      Correct
   3. C
      Iron
   4. D
      Calcium
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   Answer: B

   Sodium is the metal ion most essential in regulating blood volume, blood pressure and osmotic equilibrium.

   Because sodium is the ion whose concentration the body actively balances to control fluid volume and pressure, it is the correct answer.

6. 6.

   Which of the following properties increases from left to right along the period but decreases down the group in the Periodic Table?

   I. Atomic Number ii. Ionization energy iii. Metallic character iv. Electron affinity

   1. A
      I and iii only
   2. B
      I and ii only
   3. C
      ii and iv only
      Correct
   4. D
      iii and iv only
   Show answerHide answer

   Answer: C

   Ionization energy and electron affinity are the two properties that increase from left to right along a period but decrease down a group.

   Moving across a period, the nuclear charge increases while electrons are added to the same shell, so the nucleus pulls electrons in more tightly. This makes it harder to remove an electron (ionization energy rises) and gives atoms a stronger pull on an extra electron (electron affinity also rises). Moving down a group, atoms gain extra electron shells, so the outermost electrons sit farther from the nucleus and are more shielded — both ionization energy and electron affinity fall.

7. 7.

   The hybridization in the compound $CH_3 - CH_2 - C \equiv H$ is

   1. A
      sp³ only
   2. B
      sp only
   3. C
      sp³ and sp only
      Correct
   4. D
      sp² and sp³ only
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   Answer: C

   In an alkyne such as CH₃-CH₂-C≡CH (the stem's "C≡H" is a printing slip for C≡CH, a triple bond), the molecule is not uniformly hybridized because the carbon atoms are in different bonding environments.

   The CH3 and CH2 carbons are each joined to their neighbours by single (sigma) bonds only, with four separate groups attached, so they are sp3 hybridized, just like in an alkane.

   So the correct description is that the compound shows sp3 and sp hybridization together — not sp3 alone, which ignores the triple bond, not sp alone, which ignores the ethyl carbons, and not sp2/sp3, which wrongly assumes a double bond (sp2) instead of a triple bond.

8. 8.

   The combustion of carbon(ii)oxide in oxygen can be represented by equation.

   2CO + O$_2$ → 2CO$_2$

   Calculate the volume of the resulting mixture at the end of the reaction if 50cm$_3$ of carbon(ii)oxide was exploded in 100cm$_3$ of oxygen

   1. A
      100cm³
   2. B
      125cm³
      Correct
   3. C
      75cm³
   4. D
      150cm³
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   Answer: B

   From the equation 2CO + O₂ → 2CO₂, 2 volumes of CO react with 1 volume of O₂ to give 2 volumes of CO₂.

   50cm³ of CO therefore reacts with 25cm³ of O₂ (half of the CO volume) and produces 50cm³ of CO₂.

   Since 100cm³ of O₂ was supplied but only 25cm³ was used up, the O₂ left over (excess/unreacted) is 100 − 25 = 75cm³.

   The resulting mixture at the end of the reaction is the CO₂ formed plus the unreacted O₂: 50cm³ + 75cm³ = 125cm³.

9. 9.

   Which of the following sets of operation will completely separate a mixture of sodium chloride, sand and iodine?

   1. A
      Sublimation, filtration, evaporation
      Correct
   2. B
      Sublimation, evaporation, filtration
   3. C
      Filtration, sublimation, evaporation
   4. D
      Evaporation, sublimation, filtration
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   Answer: A

   The mixture contains iodine (which sublimes on heating), sand (insoluble in water) and sodium chloride (soluble in water).

   The correct separation is: sublimation first, heating the mixture gently so the iodine turns straight to vapour and is driven off, leaving sand and salt behind. Next, the sand and salt are stirred into water; the salt dissolves but the sand does not, so filtration removes the sand as residue, leaving a sodium chloride solution as filtrate. Finally, evaporation of the filtrate drives off the water and leaves solid sodium chloride crystals.

10. 10.

    When the end alkyl groups of ethyl ethanoate are interchanged, the compound formed is

    1. A
       Methyl propanoate
       Correct
    2. B
       Propyl ethanoate
    3. C
       Ethyl methanoate
    4. D
       Methyl ethanoate
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    Answer: A

    Ethyl ethanoate has the structure CH₃-CO-O-CH₂CH₃. It is built from two alkyl parts: the methyl group attached to the carbonyl carbon (contributed by ethanoic acid) and the ethyl group attached to the oxygen (contributed by ethanol).

    When these two end alkyl groups are interchanged, the ethyl group moves to the carbonyl side and the methyl group moves to the oxygen side, giving CH₃CH₂-CO-O-CH₃. This is the ester of propanoic acid and methanol, so methyl propanoate is correct.

The questions come from past JAMB, WAEC and NECO papers. We use software to help write the answer options and explanations, and every answer is checked twice before it goes on the site. How we check

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