201. A metal displacement record gives the reactivity order \( \mathrm{Mg \gt Zn \gt Cu \gt Ag} \). The reaction expected to occur spontaneously in aqueous solution is:
ⓐ. \( \mathrm{Ag + Zn^{2+} \rightarrow Ag^+ + Zn} \)
ⓑ. \( \mathrm{Cu + Mg^{2+} \rightarrow Cu^{2+} + Mg} \)
ⓒ. \( \mathrm{Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn} \)
ⓓ. \( \mathrm{Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag} \)
Correct Answer: \( \mathrm{Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag} \)
Explanation: A more reactive metal can displace the ion of a less reactive metal from solution. In the given order, \( \mathrm{Zn} \) is above \( \mathrm{Ag} \), so zinc can reduce \( \mathrm{Ag^+} \) to silver metal. Zinc itself is oxidised from \(0\) to \(+2\). The other options try to make a less reactive metal displace a more reactive metal ion, which is not supported by the given order. The displacement direction follows the tendency of the free metal to lose electrons.
202. Read the observation note.
A strip of \( \mathrm{Cu} \) is placed in \( \mathrm{AgNO_3} \) solution. A grey metallic coating appears on the copper strip, and the solution gradually becomes blue. The simplified ionic change is \( \mathrm{Cu + 2Ag^+ \rightarrow Cu^{2+} + 2Ag} \).
The blue colour appears mainly because:
ⓐ. \( \mathrm{Cu} \) atoms form \( \mathrm{Cu^{2+}} \) in solution
ⓑ. \( \mathrm{Ag^+} \) ions are oxidised to \( \mathrm{Ag} \)
ⓒ. nitrate ions are reduced to nitrogen gas
ⓓ. copper metal dissolves as neutral \( \mathrm{Cu} \) atoms
Correct Answer: \( \mathrm{Cu} \) atoms form \( \mathrm{Cu^{2+}} \) in solution
Explanation: Copper metal changes from oxidation number \(0\) to \(+2\), so copper is oxidised. The \( \mathrm{Cu^{2+}} \) ions formed enter the solution and are responsible for the blue colour. Silver ions gain electrons and form metallic silver, which appears as the grey coating. Nitrate ions are spectator ions in this simplified displacement reaction. The colour change is therefore linked to oxidation of copper, not to reduction of nitrate.
203. Study the displacement table and choose the row with a supported prediction.
| Row | Given order or condition | Proposed reaction | Prediction |
|---|
| P | \( \mathrm{Zn \gt Cu} \) | \( \mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu} \) | occurs |
| Q | \( \mathrm{Zn \gt Cu} \) | \( \mathrm{Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn} \) | occurs |
| R | \( \mathrm{Mg \gt Ag} \) | \( \mathrm{Ag + Mg^{2+} \rightarrow Ag^+ + Mg} \) | occurs |
| S | \( \mathrm{Fe \gt Cu} \) | \( \mathrm{Cu + Fe^{2+} \rightarrow Cu^{2+} + Fe} \) | occurs |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row R
ⓓ. Row S
Correct Answer: Row P
Explanation: Row P is supported because zinc is more reactive than copper, so zinc can displace copper from \( \mathrm{Cu^{2+}} \) solution. Zinc is oxidised to \( \mathrm{Zn^{2+}} \), while \( \mathrm{Cu^{2+}} \) is reduced to \( \mathrm{Cu} \). Row Q reverses that allowed direction, because copper cannot displace zinc from \( \mathrm{Zn^{2+}} \) solution when zinc is more reactive. Row R is unsupported because silver cannot displace magnesium from \( \mathrm{Mg^{2+}} \) solution when magnesium is more reactive. Row S is unsupported because copper cannot displace iron when iron is more reactive. A free metal displaces the ion of a less reactive metal, not the ion of a more reactive metal.
204. In the reaction \( \mathrm{3Mg + 2Fe^{3+} \rightarrow 3Mg^{2+} + 2Fe} \), the electron transfer count is:
ⓐ. \(3\) electrons lost and \(2\) electrons gained
ⓑ. \(6\) electrons lost and \(6\) electrons gained
ⓒ. \(2\) electrons lost and \(3\) electrons gained
ⓓ. \(5\) electrons lost and \(5\) electrons gained
Correct Answer: \(6\) electrons lost and \(6\) electrons gained
Explanation: \( \textbf{Oxidation of magnesium:} \)
\[
\mathrm{Mg \rightarrow Mg^{2+}+2e^-}
\]
\( \textbf{For three magnesium atoms:} \)
\[
3\times2=6\,\text{electrons lost}
\]
\( \textbf{Reduction of iron ion:} \)
\[
\mathrm{Fe^{3+}+3e^- \rightarrow Fe}
\]
\( \textbf{For two iron ions:} \)
\[
2\times3=6\,\text{electrons gained}
\]
\( \textbf{Electron balance:} \) The number of electrons lost equals the number gained.
\( \textbf{Final answer:} \) \(6\) electrons are transferred in the balanced equation. The coefficients \(3\) and \(2\) are selected from the electron counts \(2\) and \(3\).
205. Assertion: In a metal displacement reaction, the spectator ion may be absent from the net ionic equation.
Reason: Spectator ions do not undergo oxidation-number change and cancel from the complete ionic equation.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Assertion is false, but Reason is true
ⓓ. Both Assertion and Reason are true, and Reason explains Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: In reactions such as \( \mathrm{Zn + CuSO_4 \rightarrow ZnSO_4 + Cu} \), the sulphate ion remains unchanged. The net ionic equation is written as \( \mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu} \), because \( \mathrm{SO_4^{2-}} \) appears unchanged on both sides. Such ions do not take part in electron transfer. The Reason explains why they are removed when writing the net ionic equation. The redox change is carried by the metal and metal ion.
206. A halogen displacement reaction follows the order \( \mathrm{Cl_2 \gt Br_2 \gt I_2} \) in oxidising strength. Bromine water is added to a solution containing chloride ions. The most suitable prediction is:
ⓐ. \( \mathrm{Br_2} \) displaces \( \mathrm{Cl^-} \) to form \( \mathrm{Cl_2} \)
ⓑ. \( \mathrm{Cl^-} \) reduces \( \mathrm{Br_2} \) and chlorine gas forms
ⓒ. no displacement occurs because bromine is weaker than chlorine
ⓓ. bromide ions are oxidised to bromine by chloride ions
Correct Answer: no displacement occurs because bromine is weaker than chlorine
Explanation: A more reactive halogen displaces the halide ion of a less reactive halogen. Since \( \mathrm{Cl_2} \) is above \( \mathrm{Br_2} \), bromine cannot oxidise chloride ions to chlorine under the stated order. Chloride ions are already the reduced form of the stronger halogen. A reaction such as \( \mathrm{Cl_2 + 2Br^- \rightarrow 2Cl^- + Br_2} \) is favoured, not its reverse. The direction depends on relative oxidising strength of the halogen molecule.
207. A graph compares displacement ability of halogens.
The y-axis represents oxidising strength of halogen molecules. The plotted values decrease in the order \( \mathrm{Cl_2} \), \( \mathrm{Br_2} \), \( \mathrm{I_2} \).
Based on the graph, the reaction most consistent with the trend is:
ⓐ. \( \mathrm{I_2 + 2Cl^- \rightarrow 2I^- + Cl_2} \)
ⓑ. \( \mathrm{I_2 + 2Br^- \rightarrow 2I^- + Br_2} \)
ⓒ. \( \mathrm{Br_2 + 2I^- \rightarrow 2Br^- + I_2} \)
ⓓ. \( \mathrm{Cl^- + I_2 \rightarrow Cl_2 + I^-} \)
Correct Answer: \( \mathrm{Br_2 + 2I^- \rightarrow 2Br^- + I_2} \)
Explanation: The graph shows \( \mathrm{Br_2} \) has greater oxidising strength than \( \mathrm{I_2} \). Therefore \( \mathrm{Br_2} \) can accept electrons from \( \mathrm{I^-} \), reducing itself to \( \mathrm{Br^-} \). Iodide is oxidised to \( \mathrm{I_2} \). Iodine cannot displace chloride or bromide ions when it is lower in oxidising strength. The halogen molecule higher on the graph can oxidise the halide ion of a lower halogen.
208. The net ionic equation \( \mathrm{Cl_2 + 2I^- \rightarrow 2Cl^- + I_2} \) shows:
ⓐ. \( \mathrm{Cl_2} \) is the reducing agent and \( \mathrm{I^-} \) is the oxidising agent
ⓑ. \( \mathrm{Cl_2} \) is the oxidising agent and \( \mathrm{I^-} \) is the reducing agent
ⓒ. both \( \mathrm{Cl_2} \) and \( \mathrm{I^-} \) are oxidising agents
ⓓ. both \( \mathrm{Cl_2} \) and \( \mathrm{I^-} \) are reducing agents
Correct Answer: \( \mathrm{Cl_2} \) is the oxidising agent and \( \mathrm{I^-} \) is the reducing agent
Explanation: Chlorine changes from \(0\) in \( \mathrm{Cl_2} \) to \(-1\) in \( \mathrm{Cl^-} \), so chlorine is reduced. The species reduced acts as the oxidising agent, making \( \mathrm{Cl_2} \) the oxidising agent. Iodine changes from \(-1\) in \( \mathrm{I^-} \) to \(0\) in \( \mathrm{I_2} \), so iodide ions are oxidised. The species oxidised acts as the reducing agent. This reaction is a non-metal displacement redox reaction.
209. Read the reaction record.
Chlorine gas reacts with cold dilute alkali to form chloride and hypochlorite ions. The simplified ionic change may be written as \( \mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O} \).
This reaction is called disproportionation because:
ⓐ. chlorine is both reduced to \(-1\) and oxidised to \(+1\)
ⓑ. chlorine is only oxidised from \(0\) to \(+1\) in the products
ⓒ. chlorine is only reduced from \(0\) to \(-1\) in the products
ⓓ. hydroxide ion is both oxidised and reduced
Correct Answer: chlorine is both reduced to \(-1\) and oxidised to \(+1\)
Explanation: In \( \mathrm{Cl_2} \), chlorine has oxidation number \(0\). In \( \mathrm{Cl^-} \), chlorine has oxidation number \(-1\), so part of chlorine is reduced. In \( \mathrm{ClO^-} \), oxygen is \(-2\), and the ion charge is \(-1\), so chlorine is \(+1\), meaning part of chlorine is oxidised. Disproportionation occurs when the same element from one oxidation state forms products in higher and lower oxidation states. Here chlorine moves in both directions from \(0\).
210. A reaction can be recognised as disproportionation when:
ⓐ. two different elements exchange ions without oxidation-number change
ⓑ. one metal displaces another metal from solution
ⓒ. an acid and a base form salt and water
ⓓ. one oxidation state forms two different oxidation states
Correct Answer: one oxidation state forms two different oxidation states
Explanation: Disproportionation is a special redox reaction. The same element undergoes oxidation and reduction in the same overall reaction. This means one part of the element's atoms move to a higher oxidation number, while another part moves to a lower oxidation number. The starting oxidation state is usually intermediate, so both increase and decrease are possible. Ordinary ion exchange or neutralisation does not satisfy this condition.
211. In \( \mathrm{2H_2O_2 \rightarrow 2H_2O + O_2} \), oxygen undergoes disproportionation. The oxidation-number changes of oxygen are:
ⓐ. \(-1\rightarrow-2\) and \(-1\rightarrow0\)
ⓑ. \(-2\rightarrow-1\) and \(-2\rightarrow0\)
ⓒ. \(0\rightarrow-1\) and \(0\rightarrow-2\)
ⓓ. \(+1\rightarrow0\) and \(+1\rightarrow-2\)
Correct Answer: \(-1\rightarrow-2\) and \(-1\rightarrow0\)
Explanation: \( \textbf{Oxygen in } \mathrm{H_2O_2}: \) It is a peroxide, so oxygen is \(-1\).
\( \textbf{Oxygen in } \mathrm{H_2O}: \) Oxygen has its usual oxidation number \(-2\).
\( \textbf{Oxygen in } \mathrm{O_2}: \) Elemental oxygen has oxidation number \(0\).
\( \textbf{Reduction path:} \)
\[
-1\rightarrow-2
\]
\( \textbf{Oxidation path:} \)
\[
-1\rightarrow0
\]
\( \textbf{Final answer:} \) Oxygen is both reduced and oxidised. The same starting oxidation number \(-1\) splits into lower and higher values.
212. Study the table and identify the disproportionation reaction.
| Row | Reaction | Key oxidation-number pattern |
|---|
| P | \( \mathrm{Zn + Cu^{2+} \rightarrow Zn^{2+} + Cu} \) | \( \mathrm{Zn}:0\rightarrow+2 \), \( \mathrm{Cu}:+2\rightarrow0 \) |
| Q | \( \mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O} \) | \( \mathrm{Cl}:0\rightarrow-1 \) and \(0\rightarrow+1\) |
| R | \( \mathrm{NaCl + AgNO_3 \rightarrow AgCl + NaNO_3} \) | no oxidation-number change |
| S | \( \mathrm{HCl + NaOH \rightarrow NaCl + H_2O} \) | no oxidation-number change |
ⓐ. Row P
ⓑ. Row R
ⓒ. Row S
ⓓ. Row Q
Correct Answer: Row Q
Explanation: Row Q shows chlorine beginning at oxidation number \(0\) in \( \mathrm{Cl_2} \). In the products, chlorine appears as \(-1\) in \( \mathrm{Cl^-} \) and \(+1\) in \( \mathrm{ClO^-} \). The same element is therefore both reduced and oxidised. Row P is an ordinary displacement redox reaction involving two different elements changing in opposite directions. Rows R and S are non-redox reactions because oxidation numbers remain unchanged.
213. The oxidation state of an element in a disproportionation reaction is usually intermediate because:
ⓐ. it must already be the maximum possible value
ⓑ. it must already be the minimum possible value
ⓒ. it must be able to increase and also decrease
ⓓ. it must be zero in every disproportionation reaction
Correct Answer: it must be able to increase and also decrease
Explanation: Disproportionation requires the same element to undergo oxidation and reduction. Oxidation means an increase in oxidation number, while reduction means a decrease. A starting oxidation state that is already at an extreme may not easily move in both directions. An intermediate oxidation state can form one product with a higher value and another with a lower value. The starting value may be \(0\), as in \( \mathrm{Cl_2} \), but it does not have to be zero in every case.
214. For \( \mathrm{3NO_2 + H_2O \rightarrow 2HNO_3 + NO} \), nitrogen undergoes disproportionation. What are the oxidation numbers of nitrogen in \( \mathrm{NO_2} \), \( \mathrm{HNO_3} \), and \( \mathrm{NO} \), respectively?
ⓐ. \(+4, +5, +2\)
ⓑ. \(+5, +4, +2\)
ⓒ. \(+4, +2, +5\)
ⓓ. \(+2, +5, +4\)
Correct Answer: \(+4, +5, +2\)
Explanation: \( \textbf{Nitrogen in } \mathrm{NO_2}: \)
\[
x+2(-2)=0
\]
\[
x=+4
\]
\( \textbf{Nitrogen in } \mathrm{HNO_3}: \)
\[
(+1)+x+3(-2)=0
\]
\[
x=+5
\]
\( \textbf{Nitrogen in } \mathrm{NO}: \)
\[
x+(-2)=0
\]
\[
x=+2
\]
\( \textbf{Interpretation:} \) Nitrogen in \( \mathrm{NO_2} \) changes from \(+4\) to \(+5\) in one product and to \(+2\) in another product.
\( \textbf{Final answer:} \) The sequence is \(+4, +5, +2\). The same nitrogen oxidation state is split into one higher and one lower value.
215. The reaction \( \mathrm{2Cu^+ \rightarrow Cu^{2+} + Cu} \) is a disproportionation reaction because:
ⓐ. copper changes from \(0\) to \(+1\) only
ⓑ. copper is only reduced from \(+2\) to \(0\)
ⓒ. copper changes from \(+1\) to \(+2\) and \(0\)
ⓓ. copper remains at \(+1\) throughout the reaction
Correct Answer: copper changes from \(+1\) to \(+2\) and \(0\)
Explanation: In \( \mathrm{Cu^+} \), copper has oxidation number \(+1\). In \( \mathrm{Cu^{2+}} \), copper is \(+2\), so part of copper is oxidised. In elemental \( \mathrm{Cu} \), copper has oxidation number \(0\), so part of copper is reduced. The same element copper begins in one oxidation state and forms two products with different oxidation states. This is the defining oxidation-number pattern of disproportionation.
216. A graph shows oxidation number of one element during a reaction.
The reactant point is at \(+4\). Two product branches are drawn: one branch rises to \(+5\), and another branch falls to \(+2\).
The graph best represents:
ⓐ. simple oxidation only
ⓑ. simple reduction only
ⓒ. disproportionation
ⓓ. precipitation without redox
Correct Answer: disproportionation
Explanation: The same starting oxidation number \(+4\) gives two product oxidation numbers. One branch rises from \(+4\) to \(+5\), which represents oxidation. The other branch falls from \(+4\) to \(+2\), which represents reduction. Since the same element undergoes both processes, the graph represents disproportionation. A single rising branch would show only oxidation, while a single falling branch would show only reduction.
217. Assertion: \( \mathrm{2H_2O_2 \rightarrow 2H_2O + O_2} \) is a disproportionation reaction.
Reason: Oxygen in \( \mathrm{H_2O_2} \) has oxidation number \(-1\), and it changes to both \(-2\) and \(0\).
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Both Assertion and Reason are true, and Reason explains Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: \( \mathrm{H_2O_2} \) is a peroxide, so oxygen has oxidation number \(-1\). In \( \mathrm{H_2O} \), oxygen is \(-2\), meaning part of oxygen is reduced. In \( \mathrm{O_2} \), oxygen is \(0\), meaning part of oxygen is oxidised. The same element starts from one oxidation state and moves in both directions. The Reason directly gives the oxidation-number evidence for disproportionation.
218. The reaction \( \mathrm{Cl_2 + H_2O \rightarrow HCl + HClO} \) shows chlorine:
ⓐ. only reduced from \(0\) to \(-1\) in both chlorine products
ⓑ. only oxidised from \(0\) to \(+1\) in both chlorine products
ⓒ. unchanged at \(0\) in both chlorine-containing products
ⓓ. reduced in \( \mathrm{HCl} \) and oxidised in \( \mathrm{HClO} \)
Correct Answer: reduced in \( \mathrm{HCl} \) and oxidised in \( \mathrm{HClO} \)
Explanation: Chlorine in \( \mathrm{Cl_2} \) has oxidation number \(0\). In \( \mathrm{HCl} \), hydrogen is \(+1\), so chlorine is \(-1\). In \( \mathrm{HClO} \), hydrogen is \(+1\), oxygen is \(-2\), so chlorine must be \(+1\). Chlorine therefore forms one product with a lower oxidation number and another with a higher oxidation number. This is a disproportionation pattern.
219. In \( \mathrm{Cl_2 + 2OH^- \rightarrow Cl^- + ClO^- + H_2O} \), what fraction of chlorine atoms is oxidised?
ⓐ. \( \frac{1}{4} \)
ⓑ. \( \frac{1}{2} \)
ⓒ. \( \frac{1}{3} \)
ⓓ. \(1\)
Correct Answer: \( \frac{1}{2} \)
Explanation: \( \textbf{Reactant chlorine:} \) One molecule of \( \mathrm{Cl_2} \) contains \(2\) chlorine atoms at oxidation number \(0\).
\( \textbf{Reduced product:} \) One chlorine atom becomes \( \mathrm{Cl^-} \), where chlorine is \(-1\).
\( \textbf{Oxidised product:} \) One chlorine atom becomes \( \mathrm{ClO^-} \), where chlorine is \(+1\).
\( \textbf{Number oxidised:} \) Out of \(2\) chlorine atoms, \(1\) is oxidised.
\[
\text{fraction oxidised}=\frac{1}{2}
\]
\( \textbf{Final answer:} \) The fraction of chlorine atoms oxidised is \( \frac{1}{2} \). The other half is reduced, which completes the disproportionation pair.
220. A reaction note says:
An element \( \mathrm{E} \) in oxidation state \(+3\) forms two products. In one product, \( \mathrm{E} \) has oxidation state \(+5\); in the other, \( \mathrm{E} \) has oxidation state \(+1\).
The note describes:
ⓐ. acid-base neutralisation
ⓑ. disproportionation
ⓒ. single displacement
ⓓ. ionic precipitation
Correct Answer: disproportionation
Explanation: The same element begins in one oxidation state, \(+3\). One product has a higher oxidation state, \(+5\), so part of the element is oxidised. The other product has a lower oxidation state, \(+1\), so part of the element is reduced. This simultaneous increase and decrease for the same element is disproportionation. The description does not depend on precipitate formation or acid-base neutralisation.
