Q.Assign oxidation number to the underlined elements in each of the following species:

Showing the 12 most recent of 35 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The reduction products formed when copper and zinc metals are separately oxidised with dilute $\mathrm{HNO_3}$ respectively are (A) NO , $\mathrm{NO_2}$ (B) $\mathrm{N_2O}$ , NO (C) NO , $\mathrm{N_2O}$ (D) $\mathrm{NO_2}$ , NO

   ›Reveal solutionSolution

   Cu with dilute HNO₃ reduces N(+5) only to NO (+2); the more strongly reducing Zn drives the reduction further, to N₂O (+1) with dilute HNO₃. So the pair is NO, N₂O.

   Concept and Intuition

   Nitric acid is a strong oxidising acid, and the extent to which the N is reduced depends on (a) the concentration of the acid and (b) the reducing power/reactivity of the metal:

   • Concentrated HNO₃ is reduced only mildly (to $\mathrm{NO_2}$, N: +5 → +4) because the strong oxidising medium doesn't allow deep reduction and passivates many metals.
   • Dilute HNO₃ is reduced further, typically to NO (N: +5 → +2) with moderately reactive metals like copper.
   • With a strongly electropositive/reactive metal like zinc, the metal is a much better reducing agent, and dilute HNO₃ can be reduced even further — to $\mathrm{N_2O}$ (N: +5 → +1), and with very dilute acid, all the way to $\mathrm{NH_4NO_3}$ (N: +5 → −3).

   This reactivity-dependent depth of reduction is a standard trend taught for metal + HNO₃ reactions.

   Step-by-Step Solution

   1. Copper + dilute HNO₃: $3\mathrm{Cu} + 8\mathrm{HNO_3} \rightarrow 3\mathrm{Cu(NO_3)_2} + 2\mathrm{NO}\uparrow + 4\mathrm{H_2O}$ — reduction product is NO.
   2. Zinc + dilute HNO₃ (zinc being a stronger reducing agent than copper) drives the nitrogen to a lower oxidation state than NO: $4\mathrm{Zn} + 10\mathrm{HNO_3} \rightarrow 4\mathrm{Zn(NO_3)_2} + \mathrm{N_2O}\uparrow + 5\mathrm{H_2O}$ — reduction product is $\mathrm{N_2O}$.
   3. Order asked is (Cu product, Zn product) = (NO, $\mathrm{N_2O}$), matching option (C).

   Common Mistakes

   • Assuming both metals reduce dilute HNO₃ to the same product (NO) — the depth of reduction depends on the metal's reducing strength, not just acid concentration alone.
   • Mixing up concentrated-acid products ($\mathrm{NO_2}$) with dilute-acid products (NO, $\mathrm{N_2O}$).

   ✓Final answer

   The correct option is (C) — NO , $\mathrm{N_2O}$.

   ANSWER: C

2. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.Identify the reaction which occurs in blast furnace at temperature of above 900 K. (A) $\mathrm{3Fe_2O_3 + CO \longrightarrow 2Fe_3O_4 + CO_2}$ (B) $\mathrm{Fe_3O_4 + 4CO \longrightarrow 3Fe + 4CO_2}$ (C) $\mathrm{FeO + CO \longrightarrow Fe + CO_2}$ (D) $\mathrm{Fe_2O_3 + CO \longrightarrow 2FeO + CO_2}$

   ›Reveal solutionSolution

   The blast furnace reduces iron oxide in temperature-dependent stages; the final reduction of FeO to metallic iron by CO happens in the higher-temperature zone, above 900 K.

   Concept and Intuition

   The blast furnace has a temperature gradient from top (cooler) to bottom (hotter). Reduction of iron ore proceeds in stages matched to this gradient:

   • 500–800 K (upper, cooler zone): $\mathrm{Fe_2O_3}$ is progressively reduced — first to $\mathrm{Fe_3O_4}$, then further to $\mathrm{FeO}$.
   • 900–1500 K (lower, hotter zone): the final reduction of $\mathrm{FeO}$ to metallic iron occurs, along with the regeneration of CO from coke and $\mathrm{CO_2}$ ($\mathrm{C+CO_2\to2CO}$).

   Step-by-Step Solution

   1. List the reactions and their known temperature zones:
      • $\mathrm{3Fe_2O_3+CO\to2Fe_3O_4+CO_2}$ — occurs at 500–800 K.
      • $\mathrm{Fe_2O_3+CO\to2FeO+CO_2}$ — occurs at 500–800 K.
      • $\mathrm{Fe_3O_4+4CO\to3Fe+4CO_2}$ — this overall step is also associated with the lower-temperature region as an intermediate combination step, not the final >900 K step.
      • $\mathrm{FeO+CO\to Fe+CO_2}$ — occurs at 900–1500 K, the higher-temperature zone.
   2. Since the question asks specifically for the reaction above 900 K, the answer is the reduction of FeO to Fe.

   Common Mistakes

   • Assuming all the iron-oxide-reduction reactions happen at the same temperature; the blast furnace's whole design relies on a staged temperature gradient.
   • Confusing the $\mathrm{Fe_3O_4\to Fe}$ combination step (a summary of the lower-zone chemistry) with the genuinely higher-temperature $\mathrm{FeO\to Fe}$ step.

   ✓Final answer

   The correct option is (C) — $\mathrm{FeO + CO \longrightarrow Fe + CO_2}$.

   ANSWER: C

3. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.Match the following. (g = gas, l = liquid) List-I (reaction) A) $\mathrm{SO_2(g) + Cl_2(g) \rightarrow SO_2Cl_2(l)}$ B) $\mathrm{2SO_2(g) + O_2(g) \rightarrow 2SO_3(g)}$ C) $\mathrm{4HCl + O_2 \rightarrow 2Cl_2 + 2H_2O}$ D) $\mathrm{4NH_3(g) + 5O_2(g) \rightarrow 4NO(g) + 6H_2O(g)}$ List-II (catalyst) I) Pt / Rh gauze II) $\mathrm{CuCl_2}$ III) Charcoal IV) $\mathrm{V_2O_5}$ Correct answer is (A) A-III, B-IV, C-II, D-I (B) A-III, B-II, C-IV, D-I (C) A-IV, B-III, C-I, D-II (D) A-IV, B-I, C-III, D-II

   ›Reveal solutionSolution

   Matching each industrial gas-phase reaction to its standard catalyst: sulfuryl chloride formation uses charcoal, the Contact process uses V₂O₅, the Deacon process uses CuCl₂, and the Ostwald process uses Pt/Rh gauze — giving A-III, B-IV, C-II, D-I.

   Concept and Intuition

   These are four classic catalysed industrial gas reactions, each with a specific, commonly examined catalyst:

   • Charcoal (activated carbon) catalyses the combination of $\mathrm{SO_2}$ and $\mathrm{Cl_2}$ to form sulfuryl chloride.
   • Vanadium pentoxide ($\mathrm{V_2O_5}$) is the classic catalyst of the Contact process for making $\mathrm{SO_3}$ (and hence sulfuric acid).
   • Cupric chloride ($\mathrm{CuCl_2}$) catalyses the Deacon process, oxidising HCl to $\mathrm{Cl_2}$.
   • Platinum-rhodium gauze catalyses the Ostwald process, oxidising ammonia to nitric oxide (the first step of nitric acid manufacture).

   Step-by-Step Solution

   1. A) $\mathrm{SO_2(g)+Cl_2(g)\to SO_2Cl_2(l)}$: catalysed by charcoal → matches III.
   2. B) $\mathrm{2SO_2(g)+O_2(g)\to2SO_3(g)}$: the Contact process, catalysed by $\mathrm{V_2O_5}$ → matches IV.
   3. C) $\mathrm{4HCl+O_2\to2Cl_2+2H_2O}$: the Deacon process, catalysed by $\mathrm{CuCl_2}$ → matches II.
   4. D) $\mathrm{4NH_3(g)+5O_2(g)\to4NO(g)+6H_2O(g)}$: the Ostwald process, catalysed by Pt/Rh gauze → matches I.
   5. Combined: A-III, B-IV, C-II, D-I.

   Common Mistakes

   • Confusing the Deacon process (HCl oxidation, CuCl₂ catalyst) with the Contact process (SO₂ oxidation, V₂O₅ catalyst) — both are oxidation-by-O₂ reactions but use different catalysts.
   • Forgetting that charcoal (not a metal catalyst) is used for the simple $\mathrm{SO_2Cl_2}$ synthesis.

   ✓Final answer

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

   ANSWER: A

4. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.Which of the following reactions give $\mathrm{H_2}$ as one of the products? (Reactions are not balanced) (only = మాత్రమే) I) $\mathrm{NaBH_4 + I_2 \longrightarrow}$ II) $\mathrm{B_2H_6 + N(CH_3)_3 \longrightarrow}$ III) $\mathrm{Al + NaOH + H_2O \longrightarrow}$ IV) $\mathrm{BF_3 + NaH \longrightarrow}$ (A) I, II & III only (B) II & IV only (C) I & III only (D) II, III & IV only

   ›Reveal solutionSolution

   Checks familiarity with real hydride/boron-hydride reactions from NCERT: only the NaBH4 + I2 diborane-synthesis reaction and the classic Al + NaOH + H2O amphoteric reaction release $\mathrm{H_2}$. Answer: (C).

   Concept and Intuition

   $\mathrm{H_2}$ is liberated when a species carrying hydridic ($\mathrm{H^-}$) or otherwise reducible hydrogen is oxidised, or when a metal that can be oxidised reduces water/hydroxide. Simple adduct formation (a Lewis base donating a lone pair to a Lewis acid, as with boron compounds) and simple substitution/synthesis reactions that just rearrange bonds without any net redox on hydrogen do not release $\mathrm{H_2}$.

   Step-by-Step Solution

   1. I) $\mathrm{NaBH_4}$ (hydridic $\mathrm{H^-}$) reacts with $\mathrm{I_2}$ (oxidiser): $2\mathrm{NaBH_4}+\mathrm{I_2}\rightarrow \mathrm{B_2H_6}+2\mathrm{NaI}+\mathrm{H_2}$ — a standard laboratory preparation of diborane. $\mathrm{H_2}$ is released. TRUE.
   2. II) $\mathrm{B_2H_6}$ is a Lewis acid; trimethylamine is a Lewis base. They simply combine: $\mathrm{B_2H_6}+2\mathrm{N(CH_3)_3}\rightarrow 2\mathrm{H_3B\!\cdot\!N(CH_3)_3}$, an acid–base adduct with no redox and no gas evolved. FALSE.
   3. III) Aluminium is amphoteric and reacts with hot concentrated NaOH: $2\mathrm{Al}+2\mathrm{NaOH}+2\mathrm{H_2O}\rightarrow 2\mathrm{NaAlO_2}+3\mathrm{H_2}$ — a textbook source of hydrogen gas. TRUE.
   4. IV) $\mathrm{BF_3}$ (Lewis acid, electron deficient boron) reacts with hydridic NaH: $4\mathrm{NaH}+\mathrm{BF_3}\xrightarrow{450\,K, \text{diglyme}}\mathrm{NaBH_4}+3\mathrm{NaF}$. The hydride ions are transferred onto boron to make $\mathrm{BH_4^-}$, not released as $\mathrm{H_2}$. FALSE.
   5. So only I and III liberate $\mathrm{H_2}$ → option (C).

   Common Mistakes

   • Assuming every boron-hydride reaction releases $\mathrm{H_2}$ just because hydrogen is present — adduct/substitution reactions (II, IV) do not.
   • Confusing the NaH + BF3 preparation of NaBH4 (IV) with the NaBH4 + I2 oxidative decomposition (I) — they look similar but go in opposite directions.

   ✓Final answer

   The correct option is (C) — I & III only give $\mathrm{H_2}$.

   ANSWER: C

5. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
   Q.In the reaction of sodium borohydride with 'X' gives diborane as one product. Identify X and type of reaction. (A) $\mathrm{I_2}$, neutralisation reaction (B) $\mathrm{O_2}$, addition reaction (C) $\mathrm{O_2}$, substitution reaction (D) $\mathrm{I_2}$, redox reaction

   ›Reveal solutionSolution

   Diborane is prepared in the laboratory by treating sodium borohydride with iodine; iodine is reduced while hydride hydrogen is oxidised to H$_2$, making it a redox reaction.

   Concept and Intuition

   A standard laboratory route to diborane is the reaction of sodium borohydride with iodine:

   $2\mathrm{NaBH_4} + \mathrm{I_2} \xrightarrow{\Delta} \mathrm{B_2H_6} + 2\mathrm{NaI} + \mathrm{H_2}$

   Here, the oxidation state of boron stays $+3$ throughout (in both $\mathrm{BH_4^-}$ and $\mathrm{B_2H_6}$), but iodine goes from $0$ (in $I_2$) to $-1$ (in $I^-$) — it is reduced — while hydridic hydrogen goes from $-1$ (in $\mathrm{BH_4^-}$) to $0$ (in $\mathrm{H_2}$) — it is oxidised. Since both oxidation and reduction occur simultaneously, this is a redox reaction.

   Step-by-Step Solution

   1. Identify the reagent X that reacts with $\mathrm{NaBH_4}$ to give diborane: it is $\mathrm{I_2}$.
   2. Track oxidation states: iodine $0\to-1$ (reduced); hydride hydrogen $-1\to0$ (oxidised, forming $\mathrm{H_2}$ gas as a by-product).
   3. Since both oxidation and reduction happen together, classify this as a redox reaction.
   4. This matches option (D): $\mathrm{I_2}$, redox reaction.

   Common Mistakes

   • Assuming boron's oxidation state changes (it doesn't — it remains +3 in both reactant and product).
   • Misclassifying the reaction as simple substitution or addition rather than recognising the electron transfer between iodine and hydride hydrogen.

   ✓Final answer

   The correct option is (D) — $\mathrm{I_2}$, redox reaction.

   ANSWER: D

6. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.Which of the following reactions is not a metal displacement reaction? (A) $2Mg(s) + TiCl_4(l) \rightarrow 2MgCl_2(s) + Ti(s)$ (B) $2Al(s) + Cr_2O_3(s) \rightarrow Al_2O_3(s) + 2Cr(s)$ (C) $5Ca(s) + V_2O_5(s) \rightarrow 5CaO(s) + 2V(s)$ (D) $2Fe(s) + 3H_2O(l) \rightarrow Fe_2O_3(s) + 3H_2(g)$

   ›Reveal solutionSolution

   A metal displacement reaction is one metal displacing another (less reactive) metal from its compound. Options A, B, C are all such reactions (Mg/Al/Ca displacing Ti/Cr/V); option D is iron reacting with water/steam to liberate hydrogen — a metal-water reaction, not a metal-displacing-metal reaction.

   Concept and Intuition

   A metal displacement (single displacement) reaction occurs when a more reactive metal displaces a less reactive metal from a compound such as its halide or oxide, typically because the more reactive metal is a stronger reducing agent. Reactions of a metal with water/steam producing the metal oxide/hydroxide plus hydrogen gas are a different reaction category entirely — the metal is displacing hydrogen from water, not displacing another metal.

   Step-by-Step Solution

   1. (A) $2Mg + TiCl_4 \rightarrow 2MgCl_2 + Ti$: magnesium, being more reactive, displaces titanium from its chloride — a metal displacement reaction (used industrially in the Kroll-type reduction of titanium).
   2. (B) $2Al + Cr_2O_3 \rightarrow Al_2O_3 + 2Cr$: aluminium displaces chromium from its oxide — a classic aluminothermic (thermite-type) metal displacement reaction.
   3. (C) $5Ca + V_2O_5 \rightarrow 5CaO + 2V$: calcium displaces vanadium from its oxide — again a metal displacement reaction, used to extract vanadium.
   4. (D) $2Fe + 3H_2O \rightarrow Fe_2O_3 + 3H_2$: here iron reacts with water (steam) to form iron(III) oxide and hydrogen gas. No other metal is being displaced from a compound — instead, iron displaces hydrogen from water. This is categorised as a reaction of a metal with water, not a metal displacement reaction.
   5. Hence (D) is the one that is NOT a metal displacement reaction.

   Common Mistakes

   • Treating any redox reaction where a metal ends up in elemental form as automatically a 'metal displacement reaction' — the defining feature is specifically one metal displacing another metal from a compound.
   • Overlooking that reaction (D) displaces hydrogen (a non-metal, in the sense of the reactivity series placement), which is a separate reaction class from metal-metal displacement.

   ✓Final answer

   The correct option is (D) — $2Fe(s) + 3H_2O(l) \rightarrow Fe_2O_3(s) + 3H_2(g)$.

   ANSWER: D

7. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.What happens when $H_2O_2$ is added to an aqueous solution of ferrous sulphate in acid medium? (A) Only $H_2$ is evolved (B) $Fe^{2+}$ is reduced (C) $Fe^{2+}$ is oxidized (D) $H_2$ is evolved and $Fe^{2+}$ is oxidized

   ›Reveal solutionSolution

   $H_2O_2$ acts as an oxidizing agent towards ferrous ion in acid medium, converting $Fe^{2+}$ (ferrous) to $Fe^{3+}$ (ferric); no hydrogen gas is produced.

   Concept and Intuition

   Hydrogen peroxide is a versatile species that can act as either an oxidizing or a reducing agent depending on what it reacts with. Towards a good reducing agent like $Fe^{2+}$, $H_2O_2$ behaves as an oxidant, itself being reduced to water while oxidizing $Fe^{2+}$ to $Fe^{3+}$.

   Step-by-Step Solution

   1. Half reaction (oxidation): $Fe^{2+} \rightarrow Fe^{3+} + e^-$
   2. Half reaction (reduction): $H_2O_2 + 2H^+ + 2e^- \rightarrow 2H_2O$
   3. Combine (balancing electrons, ×2 on the Fe half-reaction): $2Fe^{2+} + H_2O_2 + 2H^+ \rightarrow 2Fe^{3+} + 2H_2O$
   4. No $H_2$ gas is evolved in this reaction — that would require reduction of $H^+$ to $H_2$, which is not what happens here.

   Common Mistakes

   • Assuming $H_2O_2$ always acts as an oxidizer that liberates $H_2$ gas — it does not here; it is reduced all the way to water, not to hydrogen gas.

   ✓Final answer

   The correct option is (C) — $Fe^{2+}$ is oxidized.

   ANSWER: C

8. AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
   Q.Which of the following reactions is not correct with respect to products formed? (dilute = dilute; conc. = concentrated) (A) $3Cu + 8HNO_3(dilute) \rightarrow 3Cu(NO_3)_2 + N_2O + 4H_2O$ (B) $4Zn + 10HNO_3(dilute) \rightarrow 4Zn(NO_3)_2 + N_2O + 5H_2O$ (C) $Cu + 4HNO_3(conc) \rightarrow Cu(NO_3)_2 + 2NO_2 + 2H_2O$ (D) $Zn + 4HNO_3(conc) \rightarrow Zn(NO_3)_2 + 2NO_2 + 2H_2O$

   ›Reveal solutionSolution

   Copper with dilute nitric acid produces NO gas (not $N_2O$); the equation given in option (A) is both chemically wrong (wrong product) and fails to atom-balance, unlike the other three standard reactions.

   Concept and Intuition

   Nitric acid's reduction product depends on its concentration and the reducing power of the metal. With dilute $HNO_3$, copper (a moderately reactive metal) reduces $HNO_3$ to NO, the standard textbook reaction. Very dilute acid with a more reactive metal like zinc can go further, to $N_2O$. With concentrated $HNO_3$, both Cu and Zn reduce it only to $NO_2$ (less reduction, since concentrated acid is a weaker oxidizer per mole in some contexts, but conventionally taught to give $NO_2$).

   Step-by-Step Solution

   1. Standard known reaction: $3Cu + 8HNO_3(\text{dilute}) \to 3Cu(NO_3)_2 + 2NO + 4H_2O$ — the real product is NO, not $N_2O$ as option (A) states.
   2. Checking atom balance for option (A) as written (with $N_2O$): LHS has 24 O atoms (from 8 $HNO_3$); RHS has $3Cu(NO_3)_2$ (18 O) + $N_2O$ (1 O) + $4H_2O$ (4 O) = 23 O — the equation doesn't even balance, confirming it's not a valid/correct equation.
   3. Option (B): $4Zn+10HNO_3(\text{dil.})\to4Zn(NO_3)_2+N_2O+5H_2O$ — this is the standard reaction of zinc with very dilute nitric acid, and it balances (N: 10=8+2; O: 30=24+1+5; H: 10=10). TRUE.
   4. Option (C): $Cu+4HNO_3(\text{conc.})\to Cu(NO_3)_2+2NO_2+2H_2O$ — standard, balances. TRUE.
   5. Option (D): $Zn+4HNO_3(\text{conc.})\to Zn(NO_3)_2+2NO_2+2H_2O$ — standard, balances. TRUE.
   6. So (A) is the reaction that is NOT correct.

   Common Mistakes

   • Assuming Cu with dilute $HNO_3$ gives $N_2O$ (that's more characteristic of very reactive metals like Zn with very dilute acid); Cu with dilute acid gives NO.

   ✓Final answer

   The correct option is (A) — $3Cu + 8HNO_3(dilute) \rightarrow 3Cu(NO_3)_2 + N_2O + 4H_2O$.

   ANSWER: A

9. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.Observe the following reaction $\mathrm{H_2O(s) + F_2(g) \rightarrow HF(g) + HOF(g)}$ In this reaction (A) Hydrogen is reduced and fluorine is oxidized (B) Oxygen is reduced and fluorine is oxidized (C) Oxygen is oxidized and fluorine is reduced (D) Hydrogen is oxidized and fluorine is reduced

   ›Reveal solutionSolution

   Because fluorine is the most electronegative element (even more than oxygen), oxygen — not fluorine — is the one oxidized in this reaction; fluorine is reduced.

   Concept and Intuition

   When assigning oxidation states, the more electronegative atom in a bond is assigned the negative number. Since $F$ is more electronegative than $O$, in a compound like HOF the $O$–$F$ bond gives $F$ the $-1$ state, which can force $O$ to a positive or zero oxidation state — the reverse of oxygen's usual $-2$.

   Step-by-Step Solution

   1. In $\mathrm{H_2O}$: $H=+1$ (×2), so $O = -2$ (sum $=0$).
   2. In $\mathrm{HF}$: $H=+1$, $F=-1$.
   3. In $\mathrm{HOF}$: assign $H=+1$ and $F=-1$ (F is more electronegative than O); for the neutral molecule, $O = -(1-1) = 0$.
   4. Compare oxidation states: Oxygen goes from $-2$ (in $H_2O$) to $0$ (in HOF) → oxidized (loses 2 electrons).
   5. Fluorine goes from $0$ (in $F_2$) to $-1$ in both HF and HOF → reduced (each F atom gains 1 electron).
   6. Electron balance: O loses 2 e⁻; the two F atoms together gain 2 e⁻ (1 each) — balanced.
   7. Hydrogen stays at $+1$ throughout — no redox change.

   Common Mistakes

   • Assuming fluorine, being a strong oxidizer, must always be the "oxidizing" (i.e., itself-reduced) partner and mislabeling it as "oxidized" by default.
   • Forgetting that fluorine's electronegativity exceeds oxygen's, which is exactly what forces oxygen's unusual $0$ oxidation state in HOF.

   ✓Final answer

   The correct option is (C) — Oxygen is oxidized and fluorine is reduced.

   ANSWER: C

10. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.What are X and Y respectively in the following reactions? $\mathrm{Cu + HNO_3\ (dilute) \rightarrow Cu(NO_3)_2 + H_2O + X}$ $\mathrm{Zn + HNO_3\ (dilute) \rightarrow Zn(NO_3)_2 + H_2O + Y}$ (Note: Equations are not balanced) (A) NO, $\mathrm{N_2O}$ (B) NO, $\mathrm{NO_2}$ (C) $\mathrm{NO_2}$, $\mathrm{N_2O}$ (D) $\mathrm{N_2O}$, $\mathrm{NO_2}$

    ›Reveal solutionSolution

    This tests the standard reduction products of dilute nitric acid with a less reactive metal (Cu) versus a more reactive metal (Zn).

    Concept and Intuition

    The nitrogen reduction product of $\mathrm{HNO_3}$ depends on the acid's concentration and on how strong a reducing agent the reacting metal is. Less reactive metals reduce dilute $\mathrm{HNO_3}$ only mildly (to NO); more reactive/active metals are stronger reducing agents and can push the reduction further (to $\mathrm{N_2O}$, or even $\mathrm{NH_4NO_3}$ with very active metals and very dilute acid).

    Step-by-Step Solution

    1. $Cu + \text{dilute } HNO_3$: the balanced reaction is $3Cu + 8HNO_3 \rightarrow 3Cu(NO_3)_2 + 4H_2O + 2NO\uparrow$. Copper, being less reactive, reduces dilute nitric acid only to nitric oxide. So $X = NO$.
    2. $Zn + \text{dilute } HNO_3$: the standard textbook reaction is $4Zn + 10HNO_3 \rightarrow 4Zn(NO_3)_2 + 5H_2O + N_2O\uparrow$. Zinc is a much more active metal and reduces dilute nitric acid further, to nitrous oxide. So $Y = N_2O$.
    3. Hence $X = NO$, $Y = N_2O$.

    Common Mistakes

    • Assuming both metals give an identical product with the same dilute acid — the degree of reduction of nitrogen depends strongly on how reactive the metal is, not just on acid concentration alone.

    ✓Final answer

    The correct option is (A) — NO, $\mathrm{N_2O}$.

    ANSWER: A

11. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
    Q.Which of the following reducing agents liberates hydrogen from dilute acid? (A) $\mathrm{Mn^{2+}}$ (B) $\mathrm{Fe^{2+}}$ (C) $\mathrm{Cr^{2+}}$ (D) $\mathrm{Co^{2+}}$

    ›Reveal solutionSolution

    This tests which 3d-metal $M^{2+}$ ion is a strong enough reducing agent to liberate hydrogen gas from a dilute acid.

    Concept and Intuition

    A species can reduce $\mathrm{H^+}$ to $\mathrm{H_2}$ only if it is thermodynamically a stronger reducing agent than hydrogen, i.e. the standard reduction potential of its own couple is more negative than $0\,\mathrm{V}$ (the reference $\mathrm{H^+/H_2}$ potential).

    Step-by-Step Solution

    1. $\mathrm{Mn^{2+}}$: manganese(II) has a very stable half-filled $d^5$ configuration, making it a poor reducing agent — it does not liberate $\mathrm{H_2}$ from acid.
    2. $\mathrm{Fe^{2+}}$: $E^\circ(Fe^{3+}/Fe^{2+}) = +0.77\,\mathrm{V}$, which is positive, so $\mathrm{Fe^{2+}}$ is not a strong enough reducing agent to reduce $\mathrm{H^+}$.
    3. $\mathrm{Cr^{2+}}$: $E^\circ(Cr^{3+}/Cr^{2+}) = -0.41\,\mathrm{V}$, clearly more negative than $0\,\mathrm{V}$, so $\mathrm{Cr^{2+}}$ is a strong reducing agent capable of reducing $\mathrm{H^+}$ ions in dilute acid, being oxidised itself to $\mathrm{Cr^{3+}}$ while liberating $\mathrm{H_2}$ gas.
    4. $\mathrm{Co^{2+}}$: not a significant reducing agent under these conditions.
    5. So the correct answer is $\mathrm{Cr^{2+}}$.

    Common Mistakes

    • Assuming any $M^{2+}$ ion from the middle of the 3d series behaves similarly — only ions with strongly negative $M^{3+}/M^{2+}$ potentials (like $\mathrm{Cr^{2+}}$ and $\mathrm{V^{2+}}$) actually liberate hydrogen from dilute acid.

    ✓Final answer

    The correct option is (C) — $\mathrm{Cr^{2+}}$.

    ANSWER: C

12. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
    Q.In which of the following reactions, hydrogen is one of the products formed? i. Reaction of $\mathrm{BF_3}$ with $\mathrm{LiAlH_4}$ in diethyl ether ii. Hydrolysis of diborane iii. oxidation of sodium borohydride with iodine iv. combustion of diborane (A) ii, iv only (B) ii, iii only (C) i, iii only (D) i, iv only

    ›Reveal solutionSolution

    Among the four listed boron-hydride reactions, only the hydrolysis of diborane and the iodine oxidation of sodium borohydride release free H₂ gas.

    Concept and Intuition

    Diborane and borohydride chemistry is a favourite exam ground for spotting when hydridic hydrogen (H⁻-like, on B) reacts with something that can accept it to release H₂ gas — this typically happens with water (protic H⁺ + hydridic H⁻ → H₂) or with mild oxidants. Combustion, by contrast, fully oxidizes everything to the most stable oxides (B₂O₃ and H₂O), leaving no free hydrogen; and the ether-based reduction of BF₃ by LiAlH₄ is simply a hydride-transfer synthesis of diborane, not a hydrogen-releasing reaction.

    Step-by-Step Solution

    1. i. $\mathrm{4BF_3 + 3LiAlH_4 \xrightarrow{ether} 2B_2H_6 + 3LiF + 3AlF_3}$ — this is the classic diborane synthesis; hydride is transferred to boron, no H₂ evolved.
    2. ii. $\mathrm{B_2H_6 + 6H_2O \to 2H_3BO_3 + 6H_2}$ — water hydrolyzes diborane, and H₂ gas is a product.
    3. iii. $\mathrm{2NaBH_4 + I_2 \to B_2H_6 + 2NaI + H_2}$ — a known preparative route to diborane that also releases H₂.
    4. iv. $\mathrm{B_2H_6 + 3O_2 \to B_2O_3 + 3H_2O}$ — complete combustion gives water, not free hydrogen.
    5. So H₂ is a product only in ii and iii.

    Common Mistakes

    • Assuming any reaction "involving hydrogen" in the reagents must release H₂ gas.
    • Confusing the BF₃/LiAlH₄ synthesis (which makes B–H bonds) with a reaction that releases H₂.

    ✓Final answer

    The correct option is (B) — ii, iii only.

    ANSWER: B