Q.Which of the following elements does not show disproportionation tendency?

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.An oxoacid, $X$ is formed by the reaction of $\mathrm{P_4O_6}$ with water. When this acid is heated, it undergoes disproportionation to give a different oxoacid $Y$ and a poisonous gas $Z$. Identify the correct statement regarding $X, Y, Z$ (A) X is a tribasic acid (B) Y has two P-H bonds (C) X forms another tribasic oxoacid on reacting with $\mathrm{PCl_3}$ (D) Z acts as Lewis base

   ›Reveal solutionSolution

   This tests P₄O₆ hydrolysis and the disproportionation of phosphorous acid; the answer is Z (PH₃) acts as a Lewis base.

   Concept and Intuition

   $P_4O_6$ is the acid anhydride of phosphorous acid. On heating, $H_3PO_3$ disproportionates — phosphorus simultaneously oxidises (+3→+5, forming $H_3PO_4$) and reduces (+3→−3, forming $PH_3$). Recognising the exact structures of X, Y, Z (in particular counting P–H bonds and basicity) is the crux of the question.

   Step-by-Step Solution

   1. $P_4O_6 + 6H_2O \to 4H_3PO_3$: X = orthophosphorous acid, $H_3PO_3$, structure $HP(=O)(OH)_2$ — one P–H bond, two P–OH bonds, so it is DIBASIC (only 2 acidic H's), not tribasic. Rules out (A).
   2. Heating: $4H_3PO_3 \to 3H_3PO_4(Y) + PH_3(Z)$ — a disproportionation.
   3. Y = orthophosphoric acid, $H_3PO_4 = OP(OH)_3$: all three H's are on O (P–OH), so it has ZERO P–H bonds, not two. Rules out (B).
   4. (C) claims X reacts with $PCl_3$ to give another tribasic oxoacid — the standard, well-known reaction is X reacting with $PCl_5$ (an oxidising/chlorinating agent) to give $H_3PO_4$ and $POCl_3$; $PCl_3$ is not the correct reagent here, so (C) is false. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Compound X is formed when dry chlorine is passed over heated white phosphorous. X on hydrolysis gives an acid Y. What are the disproportionation products of Y? (A) $H_3PO_4,\ H_3PO_2$ (B) $PH_3,\ H_3PO_4$ (C) $H_3PO_3,\ PH_3$ (D) $H_3PO_2,\ PH_3$

   ›Reveal solutionSolution

   X = $PCl_3$; its hydrolysis product Y = $H_3PO_3$ (phosphorous acid), which disproportionates to $H_3PO_4$ and $PH_3$.

   Concept and Intuition

   Passing dry chlorine over heated white phosphorus is the classic laboratory preparation of phosphorus trichloride: $P_4+6Cl_2 \rightarrow 4PCl_3$. $PCl_3$ hydrolyses readily to phosphorous acid. In phosphorous acid, phosphorus is in the $+3$ oxidation state — neither the highest ($+5$, in $H_3PO_4$) nor the lowest common one ($-3$, in $PH_3$) — so on heating it disproportionates, i.e. simultaneously oxidises and reduces itself.

   Step-by-Step Solution

   1. X: $P_4 + 6Cl_2 \xrightarrow{heat} 4PCl_3$.
   2. Hydrolysis: $PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl$, so Y $=H_3PO_3$.
   3. $H_3PO_3$ has one P–H bond and two P–OH bonds (it is dibasic, not tribasic), with P in the $+3$ state.
   4. On heating, disproportionation: $4H_3PO_3 \rightarrow 3H_3PO_4 + PH_3\uparrow$ ($+3 \rightarrow +5$ and $+3\rightarrow-3$). …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Consider the following $\mathrm{H_3PO_2}$ , $\mathrm{Se_2Cl_2}$ , $\mathrm{HNO_2}$ , $\mathrm{HNO_3}$ , $\mathrm{H_2SO_4}$ , $\mathrm{H_2O_2}$ How many of the above compounds undergo disproportionation reaction? (A) 2 (B) 3 (C) 4 (D) 5

   ›Reveal solutionSolution

   $H_3PO_2$, $Se_2Cl_2$, $HNO_2$ and $H_2O_2$ disproportionate — a count of 4, option (C).

   Concept and Intuition

   In disproportionation, the same element in an intermediate oxidation state is both oxidised and reduced. An element already in its highest (or lowest) oxidation state cannot disproportionate, though it may still be a plain oxidiser or reductant.

   Step-by-Step Solution

   1. $H_3PO_2$: P is +1 (intermediate between −3 and +5); on heating it gives $PH_3$ and phosphoric acid — disproportionates.
   2. $Se_2Cl_2$: Se is +1 (like $S_2Cl_2$); it disproportionates to Se and $SeCl_4$ — disproportionates.
   3. $HNO_2$: N is +3 (intermediate); $3HNO_2 \to HNO_3 + 2NO + H_2O$ — disproportionates.
   4. $HNO_3$: N is +5, the maximum — cannot disproportionate.
   5. $H_2SO_4$: S is +6, the maximum — cannot disproportionate.
   6. $H_2O_2$: O is −1 (intermediate between 0 and −2); $2H_2O_2 \to 2H_2O + O_2$ — disproportionates.
   7. Total that disproportionate = 4 → option (C).

   Common Mistakes …

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.In which of the following reactions, chlorine undergoes disproportionation? I. Reaction with cold, dilute $\mathrm{NaOH}$ II. Reaction with hot, concentrated $\mathrm{NaOH}$ III. Reaction with $\mathrm{H_2S}$ The correct answer is (A) I, II only (B) I, II, III (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   Cl₂ disproportionates with both cold dilute and hot concentrated NaOH (giving different products), but merely oxidises H₂S without disproportionating itself — so I, II only.

   Concept and Intuition

   Disproportionation is a redox reaction where one element in a single oxidation state splits into two different oxidation states of products (self oxidation-reduction). For chlorine (0 in Cl₂) reacting with alkali, this is exactly what happens — but the products differ depending on temperature/concentration:

   • Cold, dilute: Cl₂ → Cl⁻ (−1) + OCl⁻ (+1) [hypochlorite — used in bleaching powder chemistry]
   • Hot, concentrated: Cl₂ → Cl⁻ (−1) + ClO₃⁻ (+5) [chlorate]

   Both are genuine disproportionations of chlorine, just to different final oxidation states depending on conditions.

   By contrast, $\mathrm{Cl_2 + H_2S \rightarrow 2HCl + S}$ is an ordinary redox reaction: Cl₂ (0) is reduced to Cl⁻ (−1) only, while sulfide S²⁻ (−2) is oxidised to elemental S (0). Chlorine itself ends up in only one oxidation state (−1), so this is not disproportionation of chlorine — it's chlorine acting purely as an oxidising agent.

   Step-by-Step Solution

   1. I: $\mathrm{Cl_2 + 2NaOH(cold,dil) \rightarrow NaCl + NaOCl + H_2O}$ → Cl: 0 → −1 and +1 → disproportionation. ✓ …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Consider the following statements regarding the hydrolysis of $XeF_4$ I. It is a disproportionation reaction II. Xe and $XeO_3$ are formed in 2:1 molar ratio III. $O_2$ gas is evolved in this reaction The correct statements are (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   This tests the balanced hydrolysis equation of $XeF_4$; all three given statements (disproportionation, 2:1 Xe:$XeO_3$ ratio, $O_2$ evolution) are correct, so the answer is (D).

   Concept and Intuition

   $XeF_4$ hydrolyses in water via a disproportionation: xenon in the +4 state is simultaneously reduced to elemental Xe (0) and oxidised to $XeO_3$ (+6), because there is no stable intermediate +4 xenon oxide/oxoacid in water. Tracking the balanced equation lets us check every quantitative claim directly.

   Step-by-Step Solution

   1. Write the balanced hydrolysis: $6XeF_4 + 12H_2O \rightarrow 4Xe + 2XeO_3 + 24HF + 3O_2$.
   2. Statement I — Xe goes from +4 (in $XeF_4$) to 0 (Xe gas, reduction) and to +6 (in $XeO_3$, oxidation) in the same reaction: this is disproportionation. True.
   3. Statement II — from the balanced equation, 4 mol Xe are produced for every 2 mol $XeO_3$, i.e. a 4:2 = 2:1 molar ratio. True. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Chlorine gas reacts with cold, dilute NaOH solution to produce NaCl, $H_2O$ and a sodium salt (X). When it reacts with hot, concentrated NaOH solution, it produces NaCl, $H_2O$ and a sodium salt (Y). Identify the correct statements regarding the oxidation states of chlorine in X and Y. I. The oxidation state of chlorine in Y is same as that of nitrogen in nitric acid II. The oxidation state of chlorine in X is same as that of phosphorus in phosphinic acid III. The sum of the oxidation states of chlorine in X and Y is same as that of iodine in iodic acid The correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   Identifying X = NaOCl (Cl = +1) and Y = NaClO$_3$ (Cl = +5) from the disproportionation of chlorine in cold-dilute vs hot-concentrated alkali shows statements I and II are true but III is false.

   Concept and Intuition

   Chlorine disproportionates in alkali; the products depend on temperature/concentration. In cold, dilute NaOH, Cl$_2$ disproportionates mildly to Cl$^-$ (−1) and OCl$^-$ (+1) — giving NaOCl. In hot, concentrated NaOH, the disproportionation goes further, to Cl$^-$ (−1) and ClO$_3^-$ (+5) — giving NaClO$_3$. Comparing these oxidation states to those of other well-known central atoms (N in HNO$_3$, P in phosphinic acid, I in iodic acid) tests familiarity with common oxoacid oxidation states.

   Step-by-Step Solution

   1. Cold dilute: $Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O$. In NaOCl, Cl is $+1$ (X).
   2. Hot concentrated: $3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O$. In NaClO$_3$, Cl is $+5$ (Y).
   3. Statement I: Oxidation state of Cl in Y is $+5$. In $HNO_3$, N is $+5$ (H is $+1$, three O's are $-2$ each: $1 + x - 6 = 0 \Rightarrow x=+5$). Matches — True. …

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7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Chlorine reacts separately with cold, dilute NaOH and hot, concentrated NaOH. Which of the following statements regarding these reactions is not correct? (A) Both are disproportionation reactions (B) NaCl is a common product in both (C) Hypochlorite ion is formed in cold, dilute conditions (D) Perchlorate ion is formed in hot, concentrated conditions

   ›Reveal solutionSolution

   Chlorine disproportionates differently depending on NaOH concentration/temperature — cold dilute gives hypochlorite, hot concentrated gives chlorate (never perchlorate). The false statement is (D).

   Concept and Intuition

   Chlorine (oxidation state 0) is unstable in strongly basic solution and disproportionates — part of it is reduced to $Cl^-$ (−1) and part is oxidised, with the oxidation product depending on conditions. Cold, dilute base traps the intermediate +1 oxidation state (hypochlorite) before it can disproportionate further. Hot, concentrated base pushes the oxidised chlorine all the way to +5 (chlorate) via further disproportionation of hypochlorite itself. Perchlorate (+7) is never reached by direct reaction of $Cl_2$ with aqueous NaOH under any standard condition — it requires a completely different route (electrolytic oxidation of chlorate).

   Step-by-Step Solution

   1. Cold, dilute NaOH: $Cl_2 + 2NaOH \xrightarrow{\text{cold, dilute}} NaCl + NaOCl + H_2O$. Chlorine disproportionates: one atom $0 \to -1$ (chloride), the other $0 \to +1$ (hypochlorite). This confirms statements (A) 'both are disproportionation' and (C) 'hypochlorite formed in cold dilute' are TRUE.
   2. Hot, concentrated NaOH: $3Cl_2 + 6NaOH \xrightarrow{\text{hot, concentrated}} 5NaCl + NaClO_3 + 3H_2O$. Here chlorine disproportionates $0 \to -1$ (5 parts) and $0 \to +5$ (1 part, chlorate ion $ClO_3^-$), NOT perchlorate ($ClO_4^-$, oxidation state +7). …

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8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.White phosphorus on heating with concentrated $\mathrm{NaOH}$ solution in an inert atmosphere of $\mathrm{CO_2}$ gives a salt 'X' and gas 'Y'. The oxidation state of central atom in X and Y is respectively (A) $-3, +1$ (B) $+1, -3$ (C) $0, -3$ (D) $+1, +2$

   ›Reveal solutionSolution

   This is the classic disproportionation of white phosphorus with hot NaOH, giving sodium hypophosphite (P = +1) and phosphine gas (P = −3). The answer is (B) +1, −3.

   Concept and Intuition

   White phosphorus ($P_4$) is highly reactive because of strained P–P bonds. With hot, concentrated NaOH (in an inert atmosphere, so the phosphine produced doesn't spontaneously catch fire in air), it undergoes a disproportionation reaction: some phosphorus atoms are oxidised (to hypophosphite) while others are reduced (to phosphine gas):

   $\mathrm{P_4 + 3NaOH + 3H_2O \longrightarrow 3NaH_2PO_2 + PH_3\uparrow}$

   Here X = $\mathrm{NaH_2PO_2}$ (sodium hypophosphite, the salt), Y = $\mathrm{PH_3}$ (phosphine gas).

   Step-by-Step Solution

   1. Elemental $P_4$ has oxidation state 0.
   2. Find P's oxidation state in $\mathrm{NaH_2PO_2}$: assign Na = +1, O = −2, H = +1 (standard convention, consistent with known values like +5 in $\mathrm{H_3PO_4}$ and +3 in $\mathrm{H_3PO_3}$). The anion $\mathrm{H_2PO_2^-}$ has 2 H and 2 O: $x + 2(+1) + 2(-2) = -1 \Rightarrow x + 2 - 4 = -1 \Rightarrow x = +1$. So P is +1 in X.
   3. Find P's oxidation state in $\mathrm{PH_3}$: $x + 3(+1) = 0 \Rightarrow x = -3$. So P is −3 in Y. …

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9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Which of the following is not correct ? (A) Potassium permanganate on heating gives potassium manganate and manganese dioxide only (B) Phosphine is used in smoke screens (C) Bleaching action of chlorine is due to oxidation (D) Noble gases have very low boiling points

   ›Reveal solutionSolution

   Tests recall of p-block/d-block facts; the flawed statement hides an incomplete product list for KMnO₄ decomposition.

   Concept and Intuition

   "Not correct" questions reward checking each option against a precise reaction/fact rather than a vague impression. KMnO₄'s thermal decomposition is a textbook reaction whose complete balanced equation includes oxygen gas as a product — omitting it changes "only two products" from true to false.

   Step-by-Step Solution

   1. (A): $2\mathrm{KMnO_4} \xrightarrow{\Delta} \mathrm{K_2MnO_4} + \mathrm{MnO_2} + \mathrm{O_2}\uparrow$. Three products form, so "gives potassium manganate and manganese dioxide only" is FALSE — this is the incorrect statement being asked for.
   2. (B): Phosphine (with traces of $\mathrm{P_2H_4}$) is used in Holme's signal and in smoke screens on ships — a genuine, correct fact.
   3. (C): Chlorine bleaches by oxidation — $\mathrm{Cl_2}+\mathrm{H_2O}\rightarrow \mathrm{HCl}+\mathrm{HOCl}$, HOCl gives nascent oxygen that oxidises (and hence bleaches) coloured matter — correct. …

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10. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.Alkaline oxidative fusion of $\mathrm{MnO_2}$ gives 'X'. The products formed when 'X' undergoes disproportionation in acid medium are (only = only) I) $\mathrm{KMnO_4}$ II) $\mathrm{MnO_2}$ III) $\mathrm{O_2}$ IV) $\mathrm{H_2O}$ (A) I, II only (B) I, II, III, IV (C) I, II, III only (D) I, II, IV only

    ›Reveal solutionSolution

    This tests the classic $\mathrm{MnO_2} \to \mathrm{K_2MnO_4} \to$ disproportionation sequence; the acid-medium disproportionation of manganate gives $\mathrm{KMnO_4}$, $\mathrm{MnO_2}$ and $\mathrm{H_2O}$ only — never $\mathrm{O_2}$.

    Concept and Intuition

    Manganese shows a rich range of oxidation states ($+2$ to $+7$). Pyrolusite ($\mathrm{MnO_2}$, Mn in $+4$) is oxidised under strongly alkaline, oxidising fusion conditions to the $+6$ manganate ion, which is stable only in strongly alkaline solution. The moment you acidify it, the medium can no longer stabilise $\mathrm{Mn^{6+}}$: it undergoes disproportionation, i.e. it simultaneously oxidises itself (to $+7$, permanganate) and reduces itself (to $+4$, back to $\mathrm{MnO_2}$). This self-redox is a direct consequence of the instability of the intermediate $+6$ state in acid.

    Step-by-Step Solution

    1. Alkaline oxidative fusion: $\mathrm{MnO_2} + \mathrm{O_2} + 4\mathrm{KOH} \xrightarrow{\Delta} 2\mathrm{K_2MnO_4} + 2\mathrm{H_2O}$ (or using $\mathrm{KNO_3}$ as the oxidant). This is 'X' = $\mathrm{K_2MnO_4}$, potassium manganate (green).
    2. Disproportionation in acid medium: manganate ($\mathrm{Mn^{6+}}$) disproportionates to permanganate ($\mathrm{Mn^{7+}}$) and manganese dioxide ($\mathrm{Mn^{4+}}$): $3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2}\downarrow + 2\mathrm{H_2O}$ …

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11. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
    Q.In the disproportionation reaction of nitrous acid, X is formed along with nitric acid and water. X can also be obtained by the reaction of (dil. = dilute, conc. = concentrated) (A) $\mathrm{Zn + dil.\ HNO_3}$ (B) $\mathrm{Zn + Conc.\ HNO_3}$ (C) $\mathrm{Cu + dil.\ HNO_3}$ (D) $\mathrm{Cu + Conc.\ HNO_3}$

    ›Reveal solutionSolution

    The disproportionation of $\mathrm{HNO_2}$ produces NO gas as X; among the options,

    Cu with dilute $\mathrm{HNO_3}$ is the classic reaction that also produces NO.

    Concept and Intuition

    Nitrous acid is unstable and disproportionates — the same nitrogen (oxidation state +3)

    is both oxidized to +5 (in $\mathrm{HNO_3}$) and reduced to +2 (in NO):

    $3\mathrm{HNO_2}\rightarrow \mathrm{HNO_3}+2\mathrm{NO}+\mathrm{H_2O}$

    So X = nitric oxide (NO). The reactivity of nitric acid with metals depends on both the

    metal's reactivity and the acid's concentration: dilute $\mathrm{HNO_3}$ is a weaker

    oxidant that gets reduced only to NO (not further to $\mathrm{N_2O}$ typically, except

    with very reactive metals and very dilute acid), while concentrated $\mathrm{HNO_3}$ is

    reduced to $\mathrm{NO_2}$.

    Step-by-Step Solution

    1. Write the disproportionation: $3\mathrm{HNO_2}\rightarrow\mathrm{HNO_3}+2\mathrm{NO}+\mathrm{H_2O}$, confirming X = NO.
    2. Recall standard metal + nitric acid reactions:
       • Cu + dilute $\mathrm{HNO_3}$ → $\mathrm{Cu(NO_3)_2}$ + NO + $\mathrm{H_2O}$.
       • Cu + concentrated $\mathrm{HNO_3}$ → $\mathrm{Cu(NO_3)_2}$ + $\mathrm{NO_2}$ + $\mathrm{H_2O}$. …

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12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The disproportionation products of ortho phosphorous acid are (A) $\mathrm{H_3PO_4, PH_3}$ (B) $\mathrm{H_3PO_2, H_3PO_3}$ (C) $\mathrm{H_3PO_4, HPO_3}$ (D) $\mathrm{H_3PO_2, P_2H_4}$

    ›Reveal solutionSolution

    On heating, orthophosphorous acid ($\mathrm{H_3PO_3}$, P in +3) disproportionates to orthophosphoric acid ($\mathrm{H_3PO_4}$, P in +5) and phosphine ($\mathrm{PH_3}$, P in −3).

    Concept and Intuition

    Disproportionation is a redox reaction where the same element is simultaneously oxidised and reduced. In $\mathrm{H_3PO_3}$, phosphorus is in the +3 state, which is intermediate; on heating it disproportionates into a higher (+5, in $\mathrm{H_3PO_4}$) and a lower (−3, in $\mathrm{PH_3}$) oxidation state.

    Step-by-Step Solution

    1. Write the reaction: $4\mathrm{H_3PO_3}\xrightarrow{\Delta}3\mathrm{H_3PO_4}+\mathrm{PH_3}$.
    2. Check oxidation states: P is +3 in $\mathrm{H_3PO_3}$ on the left; on the right, P is +5 in $\mathrm{H_3PO_4}$ (oxidation) and −3 in $\mathrm{PH_3}$ (reduction). …

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