Q.Which of the following species, do not show disproportionation reaction and why ? ClO–, ClO2–, ClO3– and ClO4– Also write reaction for each of the species that disproportionates.

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.In the unbalanced reactions given below, the oxidation numbers of oxygen in X, Z and Y are respectively Li + O$_2$ → X K + O$_2$(excess) → Y Na + O$_2$(excess) → Z (A) $-1, -2, -\dfrac{1}{2}$ (B) $-2, -1, -\dfrac{1}{2}$ (C) $-2, -\dfrac{1}{2}, -1$ (D) $-1, -\dfrac{1}{2}, -2$

   ›Reveal solutionSolution

   The key is to recall the specific products formed when alkali metals react with oxygen: Li gives Li₂O (O⁻²), Na gives Na₂O₂ (O⁻¹), and K gives KO₂ (O in −½). The correct order is −2, −1, −½, which corresponds to option (B).

   ---

   The problem asks for the oxidation numbers of oxygen in the products X, Y, and Z formed from the reactions of Li, K, and Na with oxygen. This is not a generic “metal + oxygen → metal oxide” situation — each alkali metal behaves differently due to the size and stability of the resulting anion. The key concept is the nature of the oxide, peroxide, or superoxide formed, which directly determines oxygen’s oxidation state.

   Why this approach works:

   Instead of balancing equations, we recall the characteristic product for each metal:

   • Lithium, being the smallest alkali metal, forms only the normal oxide (O²⁻).
   • Sodium forms the peroxide (O₂²⁻, each O is −1).
   • Potassium (and larger alkali metals) forms the superoxide (O₂⁻, each O is −½).

   Thus, we can directly read off the oxidation numbers.

   ---

   1. Lithium + O₂ → X Lithium is small; the Li⁺ ion strongly stabilizes the small O²⁻ ion. The product is lithium oxide:

   $4\text{Li} + \text{O}_2 \rightarrow 2\text{Li}_2\text{O}$

   In Li₂O, oxygen is in the oxide ion O²⁻, so its oxidation number is −2.

   2. Potassium + excess O₂ → Y Potassium is large; it stabilizes the larger superoxide ion O₂⁻. With excess oxygen, the product is potassium superoxide:

   $\text{K} + \text{O}_2 (\text{excess}) \rightarrow \text{KO}_2$

   In KO₂, the anion is O₂⁻. The overall charge on O₂⁻ is −1, so each oxygen atom has oxidation number −½.

   3. Sodium + excess O₂ → Z Sodium is intermediate in size. With excess oxygen, it forms sodium peroxide:

   $2\text{Na} + \text{O}_2 (\text{excess}) \rightarrow \text{Na}_2\text{O}_2$

   In Na₂O₂, the anion is O₂²⁻ (peroxide). The overall charge is −2, so each oxygen atom has oxidation number −1.

   Thus, the oxidation numbers of oxygen in X, Z, and Y are −2, −1, and −½ respectively.

   Watch out

   A common mistake is to assume all alkali metals form the same oxide (O²⁻). Remember: only Li gives the normal oxide; Na gives peroxide; K, Rb, Cs give superoxides.

   Tip

   You can remember the pattern by the size of the metal ion: small cation (Li⁺) → small anion (O²⁻); medium (Na⁺) → medium (O₂²⁻); large (K⁺) → large (O₂⁻).

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Given below are two statements Statement-I: Reducing property of dioxide increases from SO$_2$ to TeO$_2$. Statement-II: Pb$_3$O$_4$ on heating gives lead dioxide and oxygen The correct answer is (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct

   ›Reveal solutionSolution

   Statement-I is false because reducing power actually decreases down Group 16 (SO₂ > TeO₂), and Statement-II is true because Pb₃O₄ decomposes to PbO and O₂, not PbO₂. So both are not correct.

   Concept & Intuition

   This question tests two separate ideas: (1) the trend in reducing property of dioxides of Group 16 elements (S, Se, Te, Po), and (2) the thermal decomposition behaviour of lead(II,IV) oxide (Pb₃O₄).

   • For Statement-I: Reducing property means the ability to donate electrons (get oxidised). Down Group 16, the +4 oxidation state becomes more stable (the inert‑pair effect), so the dioxide becomes harder to oxidise — hence reducing power decreases.
   • For Statement-II: Pb₃O₄ is a mixed oxide (2PbO·PbO₂). On heating, the PbO₂ part decomposes to PbO and O₂, so the overall product is PbO and O₂, not PbO₂.

   Let’s check each carefully.

   ---

   1. Statement-I: Reducing property of dioxide increases from SO₂ to TeO₂

      • Reducing property means the tendency of the dioxide to get oxidised to a higher oxidation state (e.g., +4 → +6).
      • Down Group 16, the stability of the +4 oxidation state increases due to the inert‑pair effect (the 6s² electrons in Te and Po are reluctant to participate).
      • Therefore, TeO₂ is more stable and less willing to be oxidised than SO₂.
      • In fact, SO₂ is a strong reducing agent (easily oxidised to SO₃ or H₂SO₄), while TeO₂ is quite stable and shows little reducing behaviour.
      • Conclusion: Reducing property decreases down the group. So Statement-I is false.

   2. Statement-II: Pb₃O₄ on heating gives lead dioxide and oxygen

      • Pb₃O₄ (red lead) is actually 2PbO·PbO₂.
      • When heated strongly, the PbO₂ component decomposes:

   $\text{PbO}_2 \xrightarrow{\Delta} \text{PbO} + \frac{1}{2}\text{O}_2$

   • So the overall reaction is:

   $\text{Pb}_3\text{O}_4 \xrightarrow{\Delta} 3\text{PbO} + \frac{1}{2}\text{O}_2$

   • It does not give PbO₂ as a product; PbO₂ itself decomposes.
   • Conclusion: Statement-II is also false.

   Watch out

   A common mistake is to think Pb₃O₄ simply loses oxygen to give PbO₂. In reality, PbO₂ is less stable than PbO at high temperature, so the final product is PbO.

   3. Final evaluation
      • Statement-I: Incorrect
      • Statement-II: Incorrect
      • Therefore, both statements are not correct.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Match the following List – 1 (Chemical) List – 2 (Use) A. KOH I. Coolant B. Na($l$) II. Antacid C. Li III. Electrochemical cells D. $\mathrm{Mg(OH)_2}$ IV. Absorbent for $\mathrm{CO_2}$ The correct answer is (A) A – II, B – III, C – IV, D – I (B) A – IV, B – I, C – III, D – II (C) A – IV, B – III, C – II, D – I (D) A – III, B – IV, C – I, D – II

   ›Reveal solutionSolution

   This matching question tests your knowledge of the industrial and everyday uses of common chemicals. The correct pairing is: KOH as a CO₂ absorbent, liquid sodium as a coolant, lithium in electrochemical cells, and magnesium hydroxide as an antacid — which corresponds to option (B).

   The key to solving this is not memorising random facts, but understanding why each substance is used for its specific purpose. The properties of the chemical — its reactivity, physical state, and chemical behaviour — directly determine its application.

   Let’s go through each one.

   1. KOH (Potassium hydroxide) — This is a strong base, similar to NaOH but more soluble and more reactive. It readily reacts with acidic gases like carbon dioxide to form potassium carbonate ($K_2CO_3$). This makes it an excellent absorbent for CO₂, especially in gas purification systems or in closed environments (like submarines or spacecraft). So A matches with IV.

   2. Na(l) (Liquid sodium) — Sodium metal has a low melting point (about 98°C) and an exceptionally high thermal conductivity. As a liquid, it can flow through pipes and carry heat away very efficiently. It is also chemically stable at high temperatures in the absence of air/water. For these reasons, liquid sodium is used as a coolant in fast-breeder nuclear reactors. So B matches with I.

   3. Li (Lithium) — Lithium is the lightest metal and has the highest electrochemical potential (standard reduction potential of -3.04 V). This means it can produce a high voltage per cell. Combined with its low density, lithium is ideal for electrochemical cells (batteries) — especially rechargeable lithium-ion batteries, where it provides high energy density. So C matches with III.

   4. Mg(OH)₂ (Magnesium hydroxide) — This is a sparingly soluble base. It does not dissociate completely in water, so it gives a mild, buffered alkaline solution. This makes it safe to ingest in small amounts — it neutralises excess stomach acid (HCl) without causing a sudden, harsh reaction. That is why it is used as an antacid (often sold as milk of magnesia). So D matches with II.

   Watch out

   A common mistake is to confuse KOH with an antacid. But KOH is far too caustic — it would burn the stomach lining. Antacids must be mild bases like Mg(OH)₂ or Al(OH)₃. Similarly, liquid sodium is not used in batteries (that’s solid sodium or lithium), and lithium is not used as a coolant (its melting point is too high for that purpose).

   Tip

   If you ever forget, think of the physical state: liquids (like Na(l)) are great for carrying heat; solids (like Li) are used in solid electrodes; and mild, insoluble bases (like Mg(OH)₂) are safe for the stomach.

   ✓Final answer

   The correct option is (B) — A–IV, B–I, C–III, D–II.

4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Which of the following reactions is an example of Clemmensen reduction? (A) $\begin{array}{c} \mathrm{O} \\ \vert \\ \mathrm{R-C-OC_2H_5} \xrightarrow{\text{1. DIBAL-H}} \xrightarrow{\text{2. H}_2\mathrm{O}} \mathrm{R-C-H} + \mathrm{C_2H_5OH} \\ \phantom{\mathrm{R-C-OC_2H_5}} \end{array}$ (B) $\begin{array}{c} \mathrm{O} \\ \vert \\ \mathrm{R-C-H} \xrightarrow{\text{(i) NH}_2\text{-NH}_2} \xrightarrow{\text{(ii) KOH} \vert \mathrm{CH}_2\mathrm{CH}_2(\mathrm{OH})_2} \mathrm{R-CH_3} \\ \phantom{\mathrm{R-C-H}} \end{array}$ (C) $\begin{array}{c} \mathrm{O} \\ \vert \\ \mathrm{R-C-CH_3} \xrightarrow{\text{Zn-Hg}} \xrightarrow{\text{HCl}} \mathrm{R-CH_2-CH_3} \\ \phantom{\mathrm{R-C-CH_3}} \end{array}$ (D) $\begin{array}{c} \mathrm{O} \\ \vert \\ \mathrm{R-C-Cl} + \mathrm{H}_2 \xrightarrow{\text{Pd}} \xrightarrow{\text{BaSO}_4} \mathrm{R-C-H} + \mathrm{HCl} \\ \phantom{\mathrm{R-C-Cl}} \end{array}$

   ›Reveal solutionSolution

   Clemmensen reduction uses Zn‑Hg / HCl to reduce a carbonyl group (C=O) directly to a methylene group (CH₂). The reaction in option (C) matches this exactly, giving the final product R–CH₂–CH₃ from a ketone.

   Concept & Intuition

   The Clemmensen reduction is a classic method for converting a carbonyl group (in an aldehyde or ketone) into a methylene (CH₂) group. The key reagents are zinc amalgam (Zn‑Hg) and concentrated hydrochloric acid (HCl). The reaction works under strongly acidic conditions, where the carbonyl oxygen is protonated and then reduced by the zinc metal, ultimately replacing the C=O with two hydrogen atoms. This is the go‑to method when you want to remove a carbonyl group from a molecule without using basic conditions (which would be needed for the alternative Wolff‑Kishner reduction).

   Now let’s examine each option.

   1. Option (A) – This shows an ester (R–COOEt) being reduced to an aldehyde (R–CHO) using DIBAL‑H, followed by hydrolysis. That is a partial reduction of an ester to an aldehyde, not a Clemmensen reduction. The reagents are completely different.

   2. Option (B) – This shows an aldehyde (R–CHO) being treated with hydrazine (NH₂–NH₂) and then KOH in ethylene glycol. That is the Wolff‑Kishner reduction, which also converts a carbonyl to a methylene group, but under basic conditions. The Clemmensen reduction uses acidic conditions (Zn‑Hg / HCl), not basic.

   3. Option (C) – Here we have a ketone (R–CO–CH₃) treated with Zn‑Hg / HCl. The product is R–CH₂–CH₃, meaning the carbonyl oxygen has been replaced by two hydrogens. This is the textbook definition of a Clemmensen reduction. The reagents and transformation match perfectly.

   4. Option (D) – This shows an acid chloride (R–COCl) being hydrogenated over Pd/BaSO₄ (the Rosenmund reduction) to give an aldehyde. That is a catalytic hydrogenation that stops at the aldehyde stage, not a Clemmensen reduction.

   Watch out

   A common mistake is confusing Clemmensen (acidic, Zn‑Hg/HCl) with Wolff‑Kishner (basic, NH₂‑NH₂/KOH). Both reduce C=O to CH₂, but the conditions are opposite. Option (B) is Wolff‑Kishner, not Clemmensen.

   Tip

   To quickly identify a Clemmensen reduction in a multiple‑choice question, look for the phrase “Zn‑Hg / HCl” or “zinc amalgam / hydrochloric acid” next to a carbonyl compound. No other common reduction uses that exact reagent pair.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
   Q.Identify the correct statements I. Lithium halides are somewhat covalent in nature II. NaNO$_3$ on heating gives NO$_2$ gas III. LiHCO$_3$ is a solid IV. All alkali metals form ethynides on reaction with ethyne The correct option is (A) I, II, III only (B) II, IV only (C) I only (D) I, III only

   ›Reveal solutionSolution

   Only statement I is correct. Lithium halides exhibit some covalent character due to the small size and high polarizing power of the Li$^+$ ion. The final answer is (C).

   Let's analyze each statement to determine its correctness.

   Concept and Intuition

   This question tests your understanding of the anomalous properties of lithium and general trends in alkali metal chemistry, including Fajan's rules, thermal decomposition of nitrates, stability of bicarbonates, and reactivity with ethyne. Lithium, being the first element in Group 1, often shows properties that differ significantly from the rest of the group due to its exceptionally small size and high charge density.

   Step-by-step Analysis

   1. Statement I: Lithium halides are somewhat covalent in nature.

      • Reasoning: This statement relates to Fajan's rules, which describe the factors influencing the covalent character in ionic compounds. A small cation with a high positive charge density has a greater ability to polarize the electron cloud of an anion. Lithium (Li$^+$) is the smallest cation among the alkali metals. Its high charge density allows it to significantly distort the electron cloud of halide anions (F$^-$, Cl$^-$, Br$^-$, I$^-$). This distortion leads to a sharing of electron density between the lithium ion and the halide ion, imparting a noticeable covalent character to lithium halides, even though they are predominantly ionic. This effect is more pronounced with larger anions (e.g., LiI is more covalent than LiF).
      • Verdict: This statement is correct.

   2. Statement II: NaNO$_3$ on heating gives NO$_2$ gas.

      • Reasoning: The thermal decomposition of alkali metal nitrates follows different pathways depending on the metal.
        • Lithium nitrate (LiNO$_3$) decomposes to lithium oxide, nitrogen dioxide gas, and oxygen gas, similar to the nitrates of Group 2 elements:

   $4\text{LiNO}_3(s) \xrightarrow{\Delta} 2\text{Li}_2\text{O}(s) + 4\text{NO}_2(g) + \text{O}_2(g)$

       *   However, other alkali metal nitrates (NaNO$_3$, KNO$_3$, RbNO$_3$, CsNO$_3$) decompose to form the corresponding nitrite and oxygen gas, but *not* nitrogen dioxide gas:
   

   $2\text{NaNO}_3(s) \xrightarrow{\Delta} 2\text{NaNO}_2(s) + \text{O}_2(g)$

   *   Since NaNO$_3$ produces NaNO$_2$ and O$_2$, and not NO$_2$ gas, this statement is incorrect.
   *   **Verdict:** This statement is **incorrect**.
   

   3. Statement III: LiHCO$_3$ is a solid.

   * Reasoning: Lithium bicarbonate (LiHCO$_3$) is known to exist only in aqueous solution. It is unstable and cannot be isolated as a stable solid. When attempts are made to crystallize it, it decomposes into lithium carbonate (Li$_2$CO$_3$), water, and carbon dioxide:

   $\text{LiHCO}_3(aq) \xrightarrow{\text{evaporation}} \text{Li}_2\text{CO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)$

   *   In contrast, bicarbonates of other alkali metals (like NaHCO$_3$ and KHCO$_3$) are stable solids and can be readily isolated. This is another anomalous property of lithium.
   *   **Verdict:** This statement is **incorrect**.
   

   4. Statement IV: All alkali metals form ethynides on reaction with ethyne.

   * Reasoning: Ethyne (acetylene, H-C≡C-H) has acidic hydrogens. Alkali metals can react with ethyne to form ethynides (acetylides).

   * Lithium reacts directly with ethyne gas to form lithium ethynide:

   $2\text{Li}(s) + \text{C}_2\text{H}_2(g) \rightarrow \text{Li}_2\text{C}_2(s) + \text{H}_2(g)$

       *   However, for other alkali metals like sodium and potassium, the direct reaction with ethyne gas is not as facile. They typically require specific conditions, such as reaction in liquid ammonia, to form their ethynides:
   

   $2\text{Na}(s) + \text{C}_2\text{H}_2(g) \xrightarrow{\text{liq. NH}_3} \text{Na}_2\text{C}_2(s) + \text{H}_2(g)$

   *   Since the statement implies a general ability for *all* alkali metals to form ethynides *on reaction with ethyne* under similar or direct conditions, and this is not true for all (e.g., Na and K require specific solvents like liquid ammonia), the statement is considered incorrect in the context of typical direct reactions.
   *   **Verdict:** This statement is **incorrect**.
   

   Based on the analysis, only statement I is correct.

   ✓Final answer

   The correct option is (C).

6. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Match the following List – 1 (Compound) A $\mathrm{O_3}$ B $\mathrm{SO_2}$ C $\mathrm{H_2SO_4}$ D $\mathrm{PH_3}$ List – 2 (Use) I anti-chlor II Storage batteries III germicide IV smoke screens (A) A – III, B – II, C – I, D – IV (B) A – IV, B – II, C – I, D – III (C) A – III, B – I, C – II, D – IV (D) A – IV, B – I, C – II, D – III

   ›Reveal solutionSolution

   This question requires matching common chemical compounds with their primary applications based on their characteristic properties. We will identify the key use for Ozone, Sulfur Dioxide, Sulfuric Acid, and Phosphine, leading to the correct option (C).

   The utility of a chemical compound is directly linked to its unique physical and chemical properties. Understanding these properties allows us to predict and explain why certain substances are used for specific purposes in industry, medicine, or everyday life. For instance, strong oxidizing agents are often used as disinfectants, while strong acids have roles in batteries or industrial processes.

   Let's match each compound from List-1 with its appropriate use from List-2:

   1. Compound A: $\mathrm{O_3}$ (Ozone)

      Ozone is a powerful oxidizing agent. This property makes it highly effective in killing bacteria, viruses, and other microorganisms.

      • Use: Due to its strong oxidizing and germicidal properties, ozone is widely used as a disinfectant and sterilizing agent, for example, in water purification and air treatment.
      • Match: Therefore, $\mathrm{O_3}$ matches with III germicide.

   2. Compound B: $\mathrm{SO_2}$ (Sulfur Dioxide)

      Sulfur dioxide acts as a reducing agent and also has bleaching properties (though often temporary). One of its important applications is in removing excess chlorine.

      • Use: After fabrics are bleached with chlorine, residual chlorine can damage the fabric or interfere with subsequent dyeing processes. Sulfur dioxide is used to neutralize this excess chlorine, acting as an "anti-chlor."
      • Match: Therefore, $\mathrm{SO_2}$ matches with I anti-chlor.

   3. Compound C: $\mathrm{H_2SO_4}$ (Sulfuric Acid)

      Sulfuric acid is a strong mineral acid and is highly corrosive. It is one of the most widely produced industrial chemicals.

      • Use: A significant application of sulfuric acid is as an electrolyte in lead-acid storage batteries, commonly found in automobiles. Its ability to conduct electricity and participate in reversible chemical reactions makes it ideal for this purpose.
      • Match: Therefore, $\mathrm{H_2SO_4}$ matches with II Storage batteries.

   4. Compound D: $\mathrm{PH_3}$ (Phosphine)

      Phosphine is a highly toxic gas. When impure (containing traces of $\mathrm{P_2H_4}$), it is spontaneously flammable in air.

      • Use: This property of spontaneous combustion, producing dense white fumes of phosphorus pentoxide ($\mathrm{P_4O_{10}}$), makes it useful in applications requiring smoke generation. For example, it is used in Holme's signals, which are used as smoke screens or flares at sea.
      • Match: Therefore, $\mathrm{PH_3}$ matches with IV smoke screens.

   Combining these matches:

   • A - III
   • B - I
   • C - II
   • D - IV

   This combination corresponds to option (C).

   ✓Final answer

   The correct match is A – III, B – I, C – II, D – IV, which corresponds to option (C).

7. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Match the following List I (Ore) A Siderite B Malachite C Sphalerite D Zincite List II (Composition) I ZnS II ZnO III FeCO$_3$ IV Cu$_2$S V CuCO$_3$.Cu(OH)$_2$ Options : (A) A-III B-I C-V D-II (B) A-III B-IV C-I D-II (C) A-III B-V C-I D-II (D) A-IV B-V C-II D-I

   ›Reveal solutionSolution

   This question tests knowledge of common ores and their chemical compositions. The correct matches are: Siderite → FeCO₃, Malachite → CuCO₃·Cu(OH)₂, Sphalerite → ZnS, Zincite → ZnO, which corresponds to option (C).

   The key here is to recall the standard chemical formulas for these well-known ores. Each ore is a naturally occurring mineral from which a metal is extracted. The composition is often given as a simple carbonate, sulfide, oxide, or a basic carbonate. Let’s match them one by one.

   1. Siderite (A) is an iron ore. Its name comes from the Greek sideros (iron). It is iron(II) carbonate, so its composition is FeCO₃. This matches III in List II.

      → So A → III.

   2. Malachite (B) is a green copper ore, often used as a gemstone. It is a basic copper carbonate, with the formula CuCO₃·Cu(OH)₂. This matches V in List II.

      → So B → V.

   3. Sphalerite (C) is the chief ore of zinc. It is zinc sulfide, ZnS. This matches I in List II.

      → So C → I.

   4. Zincite (D) is another zinc ore, but it is zinc oxide, ZnO. This matches II in List II.

      → So D → II.

   Now, looking at the options:

   • (A) A-III, B-I, C-V, D-II → Incorrect (B should be V, not I).
   • (B) A-III, B-IV, C-I, D-II → Incorrect (B should be V, not IV; IV is Cu₂S, which is chalcocite).
   • (C) A-III, B-V, C-I, D-II → Correct.
   • (D) A-IV, B-V, C-II, D-I → Incorrect (A should be III, not IV; C should be I, not II; D should be II, not I).

   Tip

   A common pitfall is confusing Malachite (basic carbonate) with Chalcocite (Cu₂S) or Cuprite (Cu₂O). Remember: Malachite is green and has the distinctive formula CuCO₃·Cu(OH)₂.

   Watch out

   Do not mix up Sphalerite (ZnS) and Zincite (ZnO). Sphalerite is the most common zinc ore; Zincite is rarer and often red or orange.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. TG EAPCET 2024Set ap-2024-05-07-FN1 markMCQ
   Q.Which of the following does not evolve O$_2$, when made to react with water? (A) F$_2$ (B) XeF$_2$ (C) XeF$_4$ (D) XeF$_6$

   ›Reveal solutionSolution

   When reacted with water, F$_2$, XeF$_2$, and XeF$_4$ undergo redox reactions that oxidize water to O$_2$, but XeF$_6$ undergoes a non-redox hydrolysis to form XeO$_3$, meaning it does not evolve O$_2$. The correct option is (D).

   The Concept: Hydrolysis and Oxidation States

   At the heart of this question lies the concept of hydrolysis, which is a reaction with water. Specifically, we need to determine if these hydrolysis reactions are also redox reactions (reduction-oxidation reactions) where oxygen from water is oxidized to molecular oxygen (O$_2$).

   Water (H$_2$O) contains oxygen in the -2 oxidation state. For O$_2$ to be evolved, the oxygen in water must be oxidized from -2 to 0. This means that another species in the reaction must be reduced.

   Let's consider the general behavior:

   • Strong Oxidizing Agents: If a compound is a very strong oxidizing agent, it will readily oxidize water to O$_2$.
   • Xenon Fluorides (XeF$_n$): These compounds feature xenon in positive oxidation states. When they react with water, two main scenarios can occur:
     1. Redox Hydrolysis: Xenon is reduced (e.g., to Xe gas, oxidation state 0) or disproportionates (changes to both higher and lower oxidation states), and in doing so, it oxidizes water to O$_2$. This typically happens when xenon is in a lower positive oxidation state.
     2. Non-Redox Hydrolysis: Xenon's oxidation state remains unchanged. Water molecules simply replace fluoride ligands, forming xenon oxides or oxyfluorides. In this case, oxygen's oxidation state also remains -2, and no O$_2$ is evolved. This is more likely when xenon is already in a high, stable oxidation state.

   We'll examine each option by looking at the oxidation states of xenon and oxygen before and after the reaction with water.

   Step-by-Step Analysis

   Let's break down the reaction of each compound with water and track the oxidation states.

   1. (A) F$_2$ with H$_2$O

   Fluorine (F$_2$) is the most electronegative element and the strongest oxidizing agent known. It readily oxidizes water.

   • Initial Oxidation States:
     • F in F$_2$: 0
     • O in H$_2$O: -2
   • Reaction: The reaction is vigorous, even explosive, and produces oxygen gas. $\qquad 2\text{F}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{HF}(\text{aq}) + \text{O}_2(\text{g})$
   • Final Oxidation States:
     • F in HF: -1 (Fluorine is reduced from 0 to -1)
     • O in O$_2$: 0 (Oxygen is oxidized from -2 to 0)

   Since oxygen's oxidation state changes from -2 to 0, O$_2$ is evolved.

   2. (B) XeF$_2$ with H$_2$O

   Xenon difluoride (XeF$_2$) reacts with water, and this is a redox process.

   • Initial Oxidation States:
     • Xe in XeF$_2$: +2 (since F is -1)
     • O in H$_2$O: -2
   • Reaction: $\qquad 2\text{XeF}_2(\text{s}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{Xe}(\text{g}) + 4\text{HF}(\text{aq}) + \text{O}_2(\text{g})$
   • Final Oxidation States:
     • Xe in Xe: 0 (Xenon is reduced from +2 to 0)
     • O in O$_2$: 0 (Oxygen is oxidized from -2 to 0)

   Here, xenon is reduced, and water is oxidized. Therefore, O$_2$ is evolved.

   3. (C) XeF$_4$ with H$_2$O

   Xenon tetrafluoride (XeF$_4$) also undergoes a redox hydrolysis, which is a disproportionation reaction for xenon.

   • Initial Oxidation States:
     • Xe in XeF$_4$: +4
     • O in H$_2$O: -2
   • Reaction: The hydrolysis of XeF$_4$ is complex, involving disproportionation of xenon. $\qquad 6\text{XeF}_4(\text{s}) + 12\text{H}_2\text{O}(\text{l}) \rightarrow 2\text{XeO}_3(\text{aq}) + 4\text{Xe}(\text{g}) + 24\text{HF}(\text{aq}) + 3\text{O}_2(\text{g})$
   • Final Oxidation States:
     • Xe in XeO$_3$: +6
     • Xe in Xe: 0
     • (Xenon disproportionates from +4 to +6 and 0)
     • O in O$_2$: 0 (Oxygen is oxidized from -2 to 0)

   In this reaction, xenon disproportionates, and water is oxidized. Thus, O$_2$ is evolved.

   4. (D) XeF$_6$ with H$_2$O

   Xenon hexafluoride (XeF$_6$) reacts with water in a different manner. Depending on the amount of water, it can undergo partial or complete hydrolysis. For complete hydrolysis, the reaction is:

   • Initial Oxidation States:
     • Xe in XeF$_6$: +6
     • O in H$_2$O: -2
   • Reaction (Complete Hydrolysis): $\qquad \text{XeF}_6(\text{s}) + 3\text{H}_2\text{O}(\text{l}) \rightarrow \text{XeO}_3(\text{aq}) + 6\text{HF}(\text{aq})$ This reaction forms xenon trioxide (XeO$_3$), which is a highly explosive compound.
   • Final Oxidation States:
     • Xe in XeO$_3$: +6 (since O is -2, $x + 3(-2) = 0 \Rightarrow x = +6$)
     • O in XeO$_3$: -2

   Notice that the oxidation state of xenon remains +6, and the oxidation state of oxygen remains -2. There is no change in the oxidation state of oxygen from water to O$_2$. This is a non-redox hydrolysis where fluoride ligands are replaced by oxide ligands from water.

   Therefore, O$_2$ is NOT evolved when XeF$_6$ reacts with water.

   Tip

   The key difference for XeF$_6$ is that xenon is already in a very high oxidation state (+6). It doesn't need to be reduced to Xe(0) or disproportionate in a way that forces water to be oxidized to O$_2$. Instead, it forms XeO$_3$, where xenon retains its +6 oxidation state, and oxygen from water simply becomes part of the XeO$_3$ molecule without changing its own oxidation state.

   Watch out

   While XeF$_6$ can undergo partial hydrolysis to form XeOF$_4$ and XeO$_2$F$_2$, these reactions also do not involve a change in the oxidation state of xenon or the evolution of O$_2$. The complete hydrolysis to XeO$_3$ is the most relevant for this question.

   The compound that does not evolve O$_2$ when made to react with water is XeF$_6$.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Which of the following is the strongest reducing agent? (A) $\mathrm{TeO_3}$ (B) $\mathrm{SO_3}$ (C) $\mathrm{TeO_2}$ (D) $\mathrm{SO_2}$

   ›Reveal solutionSolution

   The strongest reducing agent is the species most easily oxidized (loses electrons). Among the given oxides, $\mathrm{SO_2}$ is the most readily oxidized to a higher oxidation state, making it the strongest reducing agent. The correct option is (D).

   Concept and Intuition

   A reducing agent donates electrons and gets oxidized itself. To compare reducing strength among oxides, we look at the oxidation state of the central atom (S or Te) and its tendency to increase further. Lower oxidation states are generally easier to oxidize, but stability trends in the periodic table also matter: sulfur in +4 is less stable than tellurium in +4, so $\mathrm{SO_2}$ is more eager to lose electrons than $\mathrm{TeO_2}$. Higher oxides like $\mathrm{SO_3}$ and $\mathrm{TeO_3}$ are already in their maximum oxidation states (+6) and cannot be easily oxidized—they are oxidizing agents, not reducing ones.

   Step-by-step reasoning

   1. Determine oxidation states

      • In $\mathrm{SO_2}$: S is +4 (since O is -2 each, total -4, so S = +4).
      • In $\mathrm{SO_3}$: S is +6.
      • In $\mathrm{TeO_2}$: Te is +4.
      • In $\mathrm{TeO_3}$: Te is +6. A reducing agent must be able to increase its oxidation state. Species already at +6 (the maximum for group 16) cannot be oxidized further under normal conditions—they are at the top of their oxidation ladder. So (A) and (B) are out.

   2. Compare the +4 oxides: $\mathrm{SO_2}$ vs $\mathrm{TeO_2}$

      Both can be oxidized to +6 (forming $\mathrm{SO_3}$ or $\mathrm{TeO_3}$). Which does so more readily?

      • Sulfur is a smaller, more electronegative atom than tellurium. The +4 state for sulfur is less stable because the small size leads to greater electron-electron repulsion and a higher tendency to lose electrons to reach the more stable +6 state.
      • Tellurium, being larger and more metallic, is more comfortable in the +4 state; it is harder to oxidize further.
      • Experimentally, $\mathrm{SO_2}$ is a well-known reducing agent (e.g., it decolorizes $\mathrm{KMnO_4}$, reduces $\mathrm{Fe^{3+}}$ to $\mathrm{Fe^{2+}}$), while $\mathrm{TeO_2}$ is much less reactive as a reductant.

   3. Eliminate the weaker candidates

      • $\mathrm{TeO_3}$ and $\mathrm{SO_3}$ are already fully oxidized—they can only be reduced, not oxidized. They are oxidizing agents.
      • $\mathrm{TeO_2}$ is a weak reducing agent compared to $\mathrm{SO_2}$.

   4. Conclusion

      $\mathrm{SO_2}$ is the strongest reducing agent among the four.

   Tip

   A quick mnemonic: In group 16, the lower the period, the more stable the +4 state becomes. So sulfur's +4 is "unstable" and eager to become +6, making $\mathrm{SO_2}$ a strong reductant. Tellurium's +4 is more stable, so $\mathrm{TeO_2}$ is a weaker reductant.

   Watch out

   A common mistake is to think that a lower oxidation state always means a stronger reducing agent. While true in a general sense, you must also consider the element's position in the periodic table. For example, $\mathrm{H_2O}$ (O is -2) is not a strong reducing agent because oxygen is very electronegative and stable in that state. Always compare within the same group or similar contexts.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.Match the following List–I (Substance) A) $\mathrm{Na_2O_2}$ B) $\mathrm{D_2O}$ C) Cs D) $\mathrm{Mg(OH)_2}$ List–II (Use) I. Photoelectric cells II. Antacid III. Oxidising agent IV. Moderator The correct answer is (A) A - III, B - IV, C - I, D - II (B) A - IV, B - III, C - II, D - I (C) A - III, B - IV, C - II, D - I (D) A - II, B - IV, C - I, D - III

    ›Reveal solutionSolution

    This is a matching problem linking four substances to their primary uses. The correct pairing is: Na₂O₂ as an oxidising agent, D₂O as a moderator, Cs in photoelectric cells, and Mg(OH)₂ as an antacid — which corresponds to option (A).

    The key is to recall the characteristic chemical or physical property of each substance and then match it to the most prominent application listed. Let’s go through each one carefully.

    1. A) Na₂O₂ (Sodium peroxide)

       • Sodium peroxide contains the peroxide ion (O₂²⁻), which is a strong oxidising agent. It readily releases oxygen or accepts electrons in reactions (e.g., with water or CO₂).
       • Its common use is as an oxidising agent in bleaching, in oxygen generators, and in chemical synthesis.
       • Therefore, A matches with III (Oxidising agent).

    2. B) D₂O (Deuterium oxide, heavy water)

       • D₂O is water where hydrogen is replaced by deuterium (²H). Its key property is a much lower neutron absorption cross-section than ordinary water, making it ideal for slowing down (moderating) neutrons in nuclear reactors without capturing them.
       • It is not used as an oxidising agent, antacid, or in photoelectric cells.
       • Hence, B matches with IV (Moderator).

    3. C) Cs (Caesium)

       • Caesium is an alkali metal with the lowest ionisation energy among stable elements. This makes it extremely sensitive to light — electrons are easily ejected when photons strike its surface (the photoelectric effect).
       • It is widely used in photoelectric cells (e.g., in light sensors, night-vision devices).
       • So, C matches with I (Photoelectric cells).

    4. D) Mg(OH)₂ (Magnesium hydroxide)

       • Magnesium hydroxide is a mild base, poorly soluble in water. It neutralises stomach acid without being too harsh, and is the active ingredient in many antacids (e.g., milk of magnesia).
       • It is not used as an oxidising agent, moderator, or in photoelectric cells.
       • Thus, D matches with II (Antacid).

    Watch out

    A common mistake is to confuse D₂O with an oxidising agent because of the “oxide” in its name, or to think Cs is used as a moderator because it’s a metal. Always focus on the specific property: D₂O’s neutron moderation and Cs’s photoelectric sensitivity.

    Putting the matches together:

    A → III, B → IV, C → I, D → II.

    This corresponds exactly to option (A).

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.Consider the reaction $P_4 + 3NaOH + 3H_2O \rightarrow Q + 3NaH_2PO_2$ Identify the reaction in which Q is not the product. (Equations are not balanced.) (A) $Ca_3P_2 + H_2O \rightarrow$ (B) $H_3PO_3 \xrightarrow{\Delta}$ (C) $PH_4I + KOH \rightarrow$ (D) $PCl_3 + H_2O \rightarrow$

    ›Reveal solutionSolution

    The reaction given produces phosphine ($PH_3$) as Q; we must find which option does not yield $PH_3$. The correct option is (B) because $H_3PO_3$ on heating disproportionates into $H_3PO_4$ and $PH_3$, but the question asks for the reaction where Q is not a product — and in (B) Q is a product, so it is not the answer; actually we need the one where Q is absent. Re‑checking: (B) does produce $PH_3$, so it is not the answer. The reaction that does not give $PH_3$ is (D) $PCl_3 + H_2O \rightarrow H_3PO_3 + HCl$, so Q is absent. Final result: (D).

    Tip

    The key is to identify Q first. From the given equation $P_4 + 3NaOH + 3H_2O \rightarrow Q + 3NaH_2PO_2$, note that $NaH_2PO_2$ is sodium hypophosphite. This is a classic disproportionation of white phosphorus in warm alkali: $P_4$ is both oxidised (to hypophosphite) and reduced (to phosphine). Hence Q is phosphine, $PH_3$.

    Now we examine each option to see which one does not produce $PH_3$.

    1. Option (A): $Ca_3P_2 + H_2O \rightarrow$ Calcium phosphide reacts with water to give phosphine:

    $Ca_3P_2 + 6H_2O \rightarrow 3Ca(OH)_2 + 2PH_3$

    So Q ($PH_3$) is produced. Not the answer.

    2. Option (B): $H_3PO_3 \xrightarrow{\Delta}$ Phosphorous acid on heating disproportionates:

    $4H_3PO_3 \xrightarrow{\Delta} 3H_3PO_4 + PH_3$

    Again, $PH_3$ is formed. So Q is produced. Not the answer.

    3. Option (C): $PH_4I + KOH \rightarrow$ Phosphonium iodide reacts with a base to give phosphine:

    $PH_4I + KOH \rightarrow KI + H_2O + PH_3$

    Q is produced. Not the answer.

    4. Option (D): $PCl_3 + H_2O \rightarrow$ Phosphorus trichloride hydrolyses to phosphorous acid and hydrochloric acid:

    $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$

    No phosphine is formed. Hence Q is not a product here.

    Watch out

    A common mistake is to think that because $PCl_3$ contains phosphorus in the +3 state, it might disproportionate like $H_3PO_3$. But $PCl_3$ simply hydrolyses; it does not undergo redox under these conditions. Only the acid $H_3PO_3$ disproportionates on heating.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.In +2 oxidation state, which of the following lanthanoids act as reducing agents? (A) Ce, Pr (B) Eu, Gd (C) Eu, Yb (D) Lu, Er

    ›Reveal solutionSolution

    The key is that lanthanoids in +2 oxidation states act as reducing agents when they can easily lose an electron to reach a stable +3 state, typically due to a half-filled or fully filled f-subshell. The correct pair is Eu and Yb, so option (C) is correct.

    Concept and Intuition

    Lanthanoids are most stable in the +3 oxidation state, which corresponds to the loss of two 6s electrons and one 4f electron (or sometimes one 5d electron). A +2 state is relatively rare and occurs only when it leads to a particularly stable electronic configuration — either a half-filled (4f⁷) or fully filled (4f¹⁴) f-subshell.

    A lanthanoid in the +2 state will act as a reducing agent if it can easily lose another electron to revert to the stable +3 state. That ease is greatest when the +2 configuration is already close to a noble-gas-like f-shell (f⁷ or f¹⁴), so the +3 state is even more stable. The classic examples are Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴).

    Watch out

    A common mistake is to think that any lanthanoid with a +2 state is a reducing agent. But Ce²⁺, for instance, has a 4f² configuration — not especially stable — and actually tends to be oxidized to Ce⁴⁺ (a stable f⁰ state) rather than to Ce³⁺. So Ce²⁺ is not a reducing agent in the same sense.

    Step-by-Step Reasoning

    1. Recall the stable +2 configurations

       The +2 oxidation state is most stable when it yields either 4f⁷ (half-filled) or 4f¹⁴ (fully filled). Among lanthanoids, these occur for:

       • Eu²⁺: [Xe]4f⁷ (half-filled)
       • Yb²⁺: [Xe]4f¹⁴ (fully filled)

    2. Why these act as reducing agents

       A reducing agent donates electrons. For Eu²⁺ and Yb²⁺, the +2 state is already quite stable, but the +3 state is even more stable (Eu³⁺ is 4f⁶, Yb³⁺ is 4f¹³). The energy difference is small, so they readily lose one more electron to become +3. This makes them strong reducing agents.

    3. Check the other options

       • (A) Ce, Pr: Ce²⁺ (4f²) is not particularly stable; Ce is more known for Ce⁴⁺ (4f⁰). Pr²⁺ is also unstable and not a typical reducing agent.
       • (B) Eu, Gd: Eu²⁺ works, but Gd²⁺ (4f⁷? No — Gd³⁺ is 4f⁷, so Gd²⁺ would be 4f⁸, not half-filled) is not stable; Gd is not a reducing agent in +2.
       • (D) Lu, Er: Lu²⁺ (4f¹⁴? Actually Lu³⁺ is 4f¹⁴, so Lu²⁺ would be 4f¹⁵ — impossible) and Er²⁺ (4f¹²) are not stable reducing agents.

    4. Conclusion

       Only Eu and Yb among the listed pairs reliably act as reducing agents in the +2 state.

    Tip

    A quick memory aid: "Europium and Ytterbium are the two that go from +2 to +3 easily" — both have special f-shell stability in the +2 state.

    ✓Final answer

    The correct option is (C).

    ANSWER: C