Chemistry 2025 Jamb Examination
Chemistry 2025 Jamb Examination
Question 1
The molecule with the highest number of lone pair of electrons is
To find the molecule with the highest number of lone pairs, we examine the Lewis structures of each option.
In \(H_2O\), Oxygen has 6 valence electrons; 2 are used for bonding with Hydrogen, leaving 4 electrons or 2 lone pairs. In \(NH_3\), Nitrogen has 5 valence electrons; 3 are used for bonding, leaving 1 lone pair. In \(CH_4\), Carbon has 4 valence electrons and all 4 are used for bonding, leaving 0 lone pairs. In \(CO_2\), Carbon forms two double bonds with two Oxygen atoms. Each Oxygen atom retains 2 lone pairs, resulting in a total of 4 lone pairs for the entire \(CO_2\) molecule. Therefore, carbon dioxide has the highest count.
In \(H_2O\), Oxygen has 6 valence electrons; 2 are used for bonding with Hydrogen, leaving 4 electrons or 2 lone pairs. In \(NH_3\), Nitrogen has 5 valence electrons; 3 are used for bonding, leaving 1 lone pair. In \(CH_4\), Carbon has 4 valence electrons and all 4 are used for bonding, leaving 0 lone pairs. In \(CO_2\), Carbon forms two double bonds with two Oxygen atoms. Each Oxygen atom retains 2 lone pairs, resulting in a total of 4 lone pairs for the entire \(CO_2\) molecule. Therefore, carbon dioxide has the highest count.
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Question 2
The composition of petroleum varies because it is a
Petroleum is a complex mixture of various hydrocarbons, including alkanes, cycloalkanes, and aromatic hydrocarbons, along with other organic compounds.
Because it is a mixture and not a single pure chemical compound, its exact composition (the ratio of different hydrocarbon molecules) varies significantly depending on the geographical location and the geological conditions under which it was formed.
Because it is a mixture and not a single pure chemical compound, its exact composition (the ratio of different hydrocarbon molecules) varies significantly depending on the geographical location and the geological conditions under which it was formed.
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Question 3
In the table provided, the two compounds that will combine in the presence of an acid catalyst to produce compound V are:
\begin{array}{|l|l|l|l|l|l|} \hline \text{COMPOUND} & \text{I} & \text{II} & \text{III} & \text{IV} & \text{V} \\ \hline \text{FORMULA} & \text{ROH} & \text{RCOR'} & \text{ROR'} & \text{RCOOH} & \text{RCOOR'} \\ \hline \end{array}
\begin{array}{|l|l|l|l|l|l|} \hline \text{COMPOUND} & \text{I} & \text{II} & \text{III} & \text{IV} & \text{V} \\ \hline \text{FORMULA} & \text{ROH} & \text{RCOR'} & \text{ROR'} & \text{RCOOH} & \text{RCOOR'} \\ \hline \end{array}
Compound V is an ester (\(RCOOR'\)). Esters are produced through a process called esterification, which involves the reaction between an alkanol (alcohol) and an alkanoic acid (carboxylic acid) in the presence of an acid catalyst.
Looking at the table:
Compound I (\(ROH\)) is an alkanol.
Compound IV (\(RCOOH\)) is an alkanoic acid.
Therefore, combining I and IV will yield the ester (Compound V).
Looking at the table:
Compound I (\(ROH\)) is an alkanol.
Compound IV (\(RCOOH\)) is an alkanoic acid.
Therefore, combining I and IV will yield the ester (Compound V).
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Question 4
An example of an alkaline gas is
An alkaline gas is one that dissolves in water to form a basic solution or reacts with acids to form salts. Ammonia (\(NH_3\)) is the only common alkaline gas. When it dissolves in water, it forms ammonium hydroxide (\(NH_4OH\)), which turns red litmus paper blue. In contrast, \(HCl\) and \(NO_2\) are acidic gases, while \(N_2\) is neutral.
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Question 5
The dusty and sand particles present in the air is an example of
A suspension is a heterogeneous mixture containing solid particles large enough for sedimentation.
Dust and sand particles in the air are solid particles dispersed in a gas. They are visible to the naked eye and will eventually settle out due to gravity if the air is still. This fits the definition of a suspension. An emulsion involves two liquids, and a saturated solution is a homogeneous mixture.
Dust and sand particles in the air are solid particles dispersed in a gas. They are visible to the naked eye and will eventually settle out due to gravity if the air is still. This fits the definition of a suspension. An emulsion involves two liquids, and a saturated solution is a homogeneous mixture.
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Question 6
The property of metal that makes it suitable as a catalyst is
Transition metals are widely used as catalysts because of their ability to adopt multiple oxidation states and form complexes. This is primarily due to the presence of partially filled d-orbitals. These incomplete d-orbitals allow the metal to accept electrons from or donate electrons to reactant molecules, lowering the activation energy of the reaction and providing an alternative reaction pathway.
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Question 7
When \(\Delta H\) is positive and small, and \(\Delta S\) is positive and large, the reaction will be
The spontaneity of a reaction is determined by the Gibbs free energy equation:\[\Delta G = \Delta H - T\Delta S\]For a reaction to be spontaneous, \(\Delta G\) must be negative.
If \(\Delta H\) is positive (endothermic) and \(\Delta S\) is positive (increase in disorder), the term \(-T\Delta S\) becomes increasingly negative as the temperature \(T\) increases. If \(\Delta S\) is 'large' and \(\Delta H\) is 'small', the \(T\Delta S\) term will easily outweigh \(\Delta H\) even at moderate temperatures, making \(\Delta G\) negative and the reaction spontaneous.
If \(\Delta H\) is positive (endothermic) and \(\Delta S\) is positive (increase in disorder), the term \(-T\Delta S\) becomes increasingly negative as the temperature \(T\) increases. If \(\Delta S\) is 'large' and \(\Delta H\) is 'small', the \(T\Delta S\) term will easily outweigh \(\Delta H\) even at moderate temperatures, making \(\Delta G\) negative and the reaction spontaneous.
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Question 8
Calculate the pH of 0.001M KOH solution.
Potassium hydroxide (\(KOH\)) is a strong base that dissociates completely in water. Therefore, the concentration of hydroxide ions \([OH^-]\) is \(0.001\) M, which can be written as \(10^{-3}\) M.
First, calculate the \(pOH\):\[pOH = -\log[OH^-] = -\log(10^{-3}) = 3\]Using the relationship between \(pH\) and \(pOH\) at 25°C:\[pH + pOH = 14\]\[pH = 14 - 3 = 11\]The \(pH\) of the solution is 11, indicating it is strongly basic.
First, calculate the \(pOH\):\[pOH = -\log[OH^-] = -\log(10^{-3}) = 3\]Using the relationship between \(pH\) and \(pOH\) at 25°C:\[pH + pOH = 14\]\[pH = 14 - 3 = 11\]The \(pH\) of the solution is 11, indicating it is strongly basic.
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Question 9
The acid used in making baking soda and soft drink is
Tartaric acid is often used in combination with sodium bicarbonate (baking soda) to create baking powder; it acts as the acid source that reacts with the soda to release carbon dioxide. However, for soft drinks, the characteristic sharp taste is often provided by citric acid (found in citrus-flavoured drinks). Since the question asks for the acid used in 'making' both (referring to the acidic component of baking powder and a common beverage acidifier), tartaric acid is the standard academic answer in this context for the leavening component, though citric acid is also used in many effervescent formulations.
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Question 10
NH3(g) + HCl(g) \(\rightarrow\) NH4Cl(s)
In the reaction above, an increase in pressure will
In the reaction above, an increase in pressure will
According to Le Chatelier's Principle, an increase in pressure favours the side of the reaction with the fewer number of moles of gas. [Image showing the effect of pressure on gaseous equilibrium systems]
In the given reaction:
Reactants = 1 mole of \(NH_3\) (gas) + 1 mole of \(HCl\) (gas) = 2 moles of gas.
Products = 1 mole of \(NH_4Cl\) (solid) = 0 moles of gas.
Since the product side has fewer moles of gas (zero), an increase in pressure will shift the equilibrium to the right, favouring the formation of the product.
In the given reaction:
Reactants = 1 mole of \(NH_3\) (gas) + 1 mole of \(HCl\) (gas) = 2 moles of gas.
Products = 1 mole of \(NH_4Cl\) (solid) = 0 moles of gas.
Since the product side has fewer moles of gas (zero), an increase in pressure will shift the equilibrium to the right, favouring the formation of the product.