Q.Indicate the steps in the preparation of:

Showing the 12 most recent of 18 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Identify X and Y involved in the extraction of zinc from sphalerite in the sequence given below Sphalerite $\rightarrow$ Froth flotation $\rightarrow$ X $\rightarrow$ reduction with coke $\rightarrow$ Y $\rightarrow$ Pure zinc (A) X= Calcination; Y= Electrolysis (B) X= Roasting; Y= Fractional distillation (C) X= Roasting; Y= Liquation (D) X= Calcination; Y= Fractional distillation

   ›Reveal solutionSolution

   Sphalerite (ZnS) is roasted to ZnO, reduced with coke to impure zinc, then purified by fractional distillation because zinc has a conveniently low boiling point.

   Concept and Intuition

   Zinc's chief ore, sphalerite (ZnS), is a sulfide ore, so the standard pretreatment step for sulfide ores is roasting (heating strongly in air) — this converts the sulfide to the oxide and drives off $\mathrm{SO_2}$, giving $\mathrm{ZnO}$ (calcination, by contrast, is used for carbonate/hydroxide ores to drive off $\mathrm{CO_2}$/$\mathrm{H_2O}$ — not applicable here since sphalerite is a sulfide).

   $\mathrm{ZnO}$ is then reduced with coke (carbon) at high temperature to metallic zinc, which distils off as vapour (zinc boils at ~1180 K, well below iron's melting point, so it separates as vapour directly in the retort/furnace). This crude zinc still carries impurities (Pb, Cd, Fe), and because zinc's boiling point is so much lower than these impurities, fractional distillation cleanly separates and purifies it.

   Step-by-Step Solution

   1. Sphalerite (ZnS) → froth flotation → concentrated ZnS.
   2. X: roasting converts $\mathrm{ZnS}\rightarrow \mathrm{ZnO} + \mathrm{SO_2}$.
   3. $\mathrm{ZnO} + \mathrm{C} \rightarrow \mathrm{Zn} + \mathrm{CO}$ (reduction with coke) gives impure zinc. …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.During the preparation of $\mathrm{K_2Cr_2O_7}$ from chromite ore, in one of the steps, the yellow solution of sodium chromate is converted into orange sodium dichromate crystals. This is achieved by, (A) Increasing the pH (B) Decreasing the pH (C) Maintaining neutral pH (D) Adding $\mathrm{NaCl}$

   ›Reveal solutionSolution

   The chromate–dichromate equilibrium is pH-dependent: chromate (yellow) dominates in alkaline/neutral solution, dichromate (orange) dominates on acidification (decreasing pH).

   Concept and Intuition

   In aqueous solution, chromate and dichromate exist in a pH-sensitive equilibrium:

   $2\mathrm{CrO_4^{2-}} + 2\mathrm{H^+} \rightleftharpoons \mathrm{Cr_2O_7^{2-}} + \mathrm{H_2O}$

   Adding acid (H⁺) pushes this equilibrium to the right (Le Chatelier), converting yellow chromate into orange dichromate. This is exactly the industrial step in the dichromate manufacturing process (chromite ore → sodium chromate solution → acidified with $\mathrm{H_2SO_4}$ → sodium dichromate crystallised out).

   Step-by-Step Solution

   1. Sodium chromate solution (yellow) is obtained after fusing chromite ore with $\mathrm{Na_2CO_3}$ and roasting/leaching.
   2. To convert it to dichromate, the solution is acidified (typically with $\mathrm{H_2SO_4}$), i.e. the pH is decreased. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The fusion of chromite ore with $Na_2CO_3$ in free access of air leads to the formation of yellow coloured solution of compound A and residue B along with the evolution of $CO_2$ gas. Identify the correct statements regarding A and B. I. A contains $Cr-O-Cr$ linkage II. B is $Fe_2O_3$ III. Oxidation state of chromium in A is +6 The correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   Chromite ore fusion is the standard extraction step for chromium compounds; identifying A as sodium chromate (not dichromate) and B as iron(III) oxide resolves which of the three statements are correct. The answer is (B) II, III only.

   Concept and Intuition

   Chromite ore ($FeCr_2O_4$) is fused with sodium carbonate in the presence of air (an oxidative fusion). The air oxidises chromium(III) in the ore to chromium(VI), while iron ends up as iron(III) oxide, an insoluble residue. The soluble product is sodium chromate, $Na_2CrO_4$, which is yellow in colour — this is compound A. Only on acidifying this yellow chromate solution does it convert to the orange dichromate ion ($Cr_2O_7^{2-}$), which does have the Cr–O–Cr bridging linkage; the chromate ion itself is a simple monomeric tetrahedral $CrO_4^{2-}$ ion with no such linkage.

   Step-by-Step Solution

   1. Reaction: $4FeCr_2O_4 + 8Na_2CO_3 + 7O_2 \rightarrow 8Na_2CrO_4 + 2Fe_2O_3 + 8CO_2$.
   2. Compound A (yellow solution) = sodium chromate, $Na_2CrO_4$; residue B = iron(III) oxide, $Fe_2O_3$.
   3. Statement I: "A contains Cr–O–Cr linkage." The chromate ion $CrO_4^{2-}$ is a single tetrahedral unit with no bridging oxygen between two chromium atoms — that bridging (Cr–O–Cr) only appears in the dichromate ion $Cr_2O_7^{2-}$, formed later on acidification. So statement I is false. …

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4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Wrought iron is prepared from cast iron in a reverberatory furnace. The substance commonly used to line the furnace and the chemical process involved in it are respectively (A) Magnetite, reduction (B) Magnetite, oxidation (C) Haematite, oxidation (D) Haematite, reduction

   ›Reveal solutionSolution

   The puddling process converts cast iron to wrought iron in a haematite-lined reverberatory furnace, where the haematite oxidises the impurities out of the molten iron.

   Concept and Intuition

   Cast iron contains several % of carbon plus other impurities (Si, S, P, Mn) that make it brittle. Wrought iron is the purest commercial form of iron (almost no carbon). To go from cast iron to wrought iron, impurities must be removed by oxidation — this is done in a reverberatory furnace whose lining ($Fe_2O_3$, haematite) itself acts as an oxidising agent, converting impurities to their oxides which float off as slag, in the classic "puddling" process.

   Step-by-Step Solution

   1. Molten cast iron is melted in a reverberatory furnace lined with haematite ($Fe_2O_3$).
   2. Haematite oxidises carbon (to CO/CO$_2$), silicon (to $SiO_2$), sulphur, phosphorus and manganese in the melt.
   3. These oxidised impurities combine with the furnace lining/flux to form a fusible slag, which is skimmed off. …

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5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The method by which very pure nitrogen can be obtained is (A) Thermal decomposition of ammonium dichromate (B) Thermal decomposition of barium azide (C) Reaction of aqueous solution of ammonium chloride with sodium nitrite (D) Thermal decomposition of ammonium nitrate

   ›Reveal solutionSolution

   The routine lab preparation of N$_2$ (from NH$_4$Cl + NaNO$_2$) carries trace NO/HNO$_3$ impurities; genuinely pure, dry nitrogen is instead obtained via thermal decomposition of an azide such as barium (or sodium) azide.

   Concept and Intuition

   Several routes give nitrogen gas, but not all give it in a pure state. The everyday laboratory method (heating an aqueous mixture of ammonium chloride and sodium nitrite) is convenient but contaminated by small amounts of NO and HNO$_3$ as side products. To get genuinely pure nitrogen, chemists instead thermally decompose an ionic azide, which cleanly releases only N$_2$ gas (plus the metal), with no such side products.

   Step-by-Step Solution

   1. $NH_4Cl(aq) + NaNO_2(aq) \xrightarrow{\Delta} N_2 + NaCl + 2H_2O$ — this is the standard lab method but yields nitrogen contaminated with traces of NO and HNO$_3$.
   2. To obtain very pure, dry $N_2$, azides are thermally decomposed instead: $Ba(N_3)_2 \xrightarrow{\Delta} Ba + 3N_2\uparrow$ (similarly $2NaN_3 \xrightarrow{\Delta} 2Na + 3N_2$).
   3. This decomposition gives clean N$_2$ gas without the nitrogen-oxide contaminants of the aqueous nitrite method. …

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6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Identify the incorrect statement from the following (A) $K_2CO_3$ can be prepared by solvay process (B) In solvay process $CaCl_2$ is the byproduct (C) Aqueous solution of $Na_2CO_3$ is basic in nature due to hydrolysis of $CO_3^{2-}$ ion (D) Sodium hydrogen carbonate on heating gives $Na_2CO_3, CO_2$ and $H_2O$

   ›Reveal solutionSolution

   The classic limitation of the Solvay process is that it cannot be used to make $K_2CO_3$, because $KHCO_3$ is too soluble to precipitate — this makes statement (A) the incorrect one.

   Concept and Intuition

   The Solvay (ammonia–soda) process manufactures sodium carbonate by passing $CO_2$ and $NH_3$ into brine, precipitating relatively insoluble $NaHCO_3$, which is then calcined to $Na_2CO_3$. The entire process depends on $NaHCO_3$ being sufficiently insoluble to crystallize out. The analogous potassium salt, $KHCO_3$, is much more soluble in water and does not precipitate under the same conditions — so the Solvay process simply cannot be adapted to prepare $K_2CO_3$, which is why potassium carbonate is manufactured by a different industrial route.

   Step-by-Step Solution

   1. Check (A): 'K2CO3 can be prepared by solvay process' — false, as explained above (textbook exception of the Solvay process).
   2. Check (B): $CaCl_2$ is indeed a byproduct of the Solvay process ($CaCO_3 \rightarrow CaO+CO_2$; $CaO+2NH_4Cl \rightarrow CaCl_2+2NH_3+H_2O$) — true.
   3. Check (C): $Na_2CO_3$ solution is basic because $CO_3^{2-}$, being the conjugate base of a weak acid ($HCO_3^-$), hydrolyses to give $OH^-$ — true. …

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7. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.Identify the reaction in which diborane is produced on industrial scale ? (A) Reaction of $\mathrm{BF_3}$ with $\mathrm{LiAlH_4}$ in diethyl ether (B) Oxidation of $\mathrm{NaBH_4}$ with $\mathrm{I_2}$ (C) Reaction of $\mathrm{BF_3}$ with NaH at 450 K (D) By heating $\mathrm{H_3BO_3}$ to above 370 K temperature

   ›Reveal solutionSolution

   This tests the industrial vs laboratory preparation of diborane; the industrial route uses $\mathrm{BF_3}$ + NaH at 450 K.

   Concept and Intuition

   Diborane, $\mathrm{B_2H_6}$, can be made by more than one route, but exam questions on boron hydrides specifically distinguish the laboratory-scale method from the industrial-scale method because only one is economical at bulk production.

   Step-by-Step Solution

   1. Laboratory method: $4\mathrm{BF_3} + 3\mathrm{LiAlH_4} \rightarrow 2\mathrm{B_2H_6} + 3\mathrm{LiF} + 3\mathrm{AlF_3}$, carried out in diethyl ether — this is a small-scale/lab preparation, not an industrial one.
   2. Industrial method: sodium hydride is reacted with $\mathrm{BF_3}$ at 450 K: $2\mathrm{BF_3} + 6\mathrm{NaH} \xrightarrow{450\,K} \mathrm{B_2H_6} + 6\mathrm{NaF}$. This is the route used on an industrial scale because NaH is far cheaper and easier to handle in bulk than $\mathrm{LiAlH_4}$.
   3. Oxidation of $\mathrm{NaBH_4}$ with $\mathrm{I_2}$ gives diborane too, but it is used as a convenient small-scale/lab method, not the industrial process. …

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8. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.$\mathrm{NaBH_4 + I_2 \longrightarrow A + NaI + H_2}$ $\mathrm{A + LiH \longrightarrow B}$ In these reactions, A and B respectively are (Reactions are not balanced) (A) $\mathrm{B_2H_6,\ LiBH_4}$ (B) $\mathrm{B_2H_6,\ (BN)_x}$ (C) $\mathrm{BH_3.CO,\ LiBH_4}$ (D) $\mathrm{BH_3.CO,\ B_3N_3H_6}$

   ›Reveal solutionSolution

   This is a standard boron-hydride synthesis sequence: sodium borohydride with iodine gives diborane, which then reacts with lithium hydride to give lithium borohydride.

   Concept and Intuition

   Diborane ($B_2H_6$) is classically prepared in the laboratory by the reaction of sodium borohydride with iodine. It also reacts with electron-rich hydrides like LiH to form borohydride salts by accepting a hydride ion.

   Step-by-Step Solution

   1. $NaBH_4+I_2\rightarrow A+NaI+H_2$: this is the known laboratory preparation of diborane, so $A=B_2H_6$: $2NaBH_4+I_2\rightarrow B_2H_6+2NaI+H_2$.
   2. $A+LiH\rightarrow B$: diborane reacts with lithium hydride (a hydride donor) to form lithium borohydride: $B_2H_6+2LiH\rightarrow2LiBH_4$, so $B=LiBH_4$.

   Common Mistakes …

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9. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.Which of the following source materials generate $SO_2$ that is used in contact process? (A) S, $FeS_2$ (B) S, FeS (C) $H_2S$, $FeS_2$ (D) $Na_2S$, $FeS_2$

   ›Reveal solutionSolution

   The Contact Process sources its $SO_2$ feedstock from burning sulfur and roasting iron pyrites ($FeS_2$) — the standard raw materials taught for industrial $H_2SO_4$ manufacture.

   Concept and Intuition

   The Contact Process converts $SO_2$ to $SO_3$ (catalytically, over $V_2O_5$) and then to $H_2SO_4$. The very first step is generating $SO_2$ itself, which industrially comes from two main sources:

   1. Burning elemental sulfur in air/oxygen: $S + O_2 \rightarrow SO_2$.
   2. Roasting (oxidative roasting) of iron pyrites (fool's gold, $FeS_2$): $4FeS_2 + 11O_2 \rightarrow 2Fe_2O_3 + 8SO_2$.

   Other sulfur-containing species like $FeS$, $H_2S$, or $Na_2S$ are not the standard industrial feedstocks for this process (in fact roasting $FeS$ is not the conventional pyrites-roasting reaction taught, and $H_2S$/$Na_2S$ are not the textbook sources).

   Step-by-Step Solution

   1. Recall that the Contact Process needs a supply of $SO_2$ as its starting material. …

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10. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.Small quantities of NO and $\mathrm{HNO_3}$ are formed as impurities, when $\mathrm{N_2}$ is prepared from $\mathrm{NH_4Cl}$ (aq) and $\mathrm{NaNO_2}$ (aq), these impurities can be removed by passing the $\mathrm{N_2}$ gas through which of the following? (A) $\mathrm{H_2SO_4}$ (aq) containing $\mathrm{SO_3}$ (B) $\mathrm{H_2SO_4}$ (aq) containing $\mathrm{K_2Cr_2O_7}$ (C) $\mathrm{H_2SO_4}$ (aq) containing $\mathrm{KMnO_4}$ (D) HCl (aq) containing $\mathrm{KMnO_4}$

    ›Reveal solutionSolution

    The standard lab preparation of $N_2$ from $NH_4Cl$ and $NaNO_2$ gives a gas contaminated with NO and $HNO_3$; this is purified by passing through acidified potassium dichromate solution, i.e. $H_2SO_4$(aq) containing $K_2Cr_2O_7$ — option (B).

    Concept and Intuition

    The reaction $NH_4Cl(aq) + NaNO_2(aq) \xrightarrow{\Delta} N_2(g) + NaCl(aq) + 2H_2O(l)$ is the classic laboratory route to dinitrogen gas. However, due to side reactions during the heating, the evolved $N_2$ carries small amounts of oxides of nitrogen (NO) and nitric acid vapor ($HNO_3$) as impurities. Since $N_2$ itself is chemically very inert (strong $N\equiv N$ triple bond), it passes through an oxidizing acidic scrubbing solution unaffected, while the reactive NO/$HNO_3$ impurities get oxidized/absorbed and stripped out. Acidified potassium dichromate ($K_2Cr_2O_7$ in $H_2SO_4$) is precisely such an oxidizing scrubber and is the standard purification step taught for this preparation.

    Step-by-Step Solution

    1. Recall the preparation: heating aqueous $NH_4Cl$ and $NaNO_2$ gives $N_2$ gas, along with trace NO and $HNO_3$ as side-product impurities.
    2. To purify, the gas stream is bubbled through a scrubbing solution that will react with/absorb the impurities but not the inert $N_2$.
    3. Acidified $K_2Cr_2O_7$ (an oxidizing agent) reacts with and removes the NO/$HNO_3$ impurities, letting pure $N_2$ pass through. …

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11. AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
    Q.Beryllium fluoride can be prepared from the decomposition of ______ (A) $(NH_4)_2BeF_4$ (B) $(NH_4)_4BeF_4$ (C) $(NH_3)_2BeF_2$ (D) $(NH_3)_4BeF_2$

    ›Reveal solutionSolution

    This tests a specific preparation method of beryllium fluoride; $BeF_2$ comes from thermal decomposition of $(NH_4)_2BeF_4$.

    Concept and Intuition

    Beryllium, being the smallest and most electronegative of the alkaline earth metals, forms compounds through routes distinct from its heavier congeners; $BeF_2$'s standard preparation exploits the thermal instability of the ammonium fluoroberyllate salt.

    Step-by-Step Solution

    1. $BeF_2$ is prepared by first forming the complex ammonium salt $(NH_4)_2BeF_4$ (ammonium tetrafluoroberyllate).
    2. On heating, this salt decomposes:

    $(NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_4F$

    3. This is the standard textbook preparation of anhydrous $BeF_2$, distinguishing Be from the other Group 2 metals (whose fluorides are prepared differently, e.g. via direct reaction with $F_2$/HF). …

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12. AP EAPCET 2021Set ap-2021-09-06-FN1 markMCQ
    Q.Pure nitrogen gas is prepared in the laboratory by heating a mixture of ________ (A) $\mathrm{NH_4OH}$ & $\mathrm{NaNH_2}$ (B) $\mathrm{NH_4Cl}$ & $\mathrm{NaNO_2}$ (C) $\mathrm{NH_4NO_2}$ & $\mathrm{KNH_2}$ (D) $\mathrm{NH_4F}$ & $\mathrm{NaNO_3}$

    ›Reveal solutionSolution

    The laboratory preparation of pure $N_2$ is by heating a mixture of ammonium chloride and sodium nitrite. Answer: (B).

    Concept and Intuition

    Ammonium salts contain $N$ in the $-3$ oxidation state, while nitrites contain $N$ in the $+3$ state. When heated together in solution, these comproportionate: the ammonium ion is oxidised and the nitrite ion is reduced, both converging to $N_2$ (oxidation state $0$), which escapes as a gas along with water. This is the standard lab-scale route to pure, dry nitrogen (avoiding the traces of oxides of nitrogen/other impurities that atmospheric nitrogen or other industrial routes carry).

    Step-by-Step Solution

    1. Write the mixture: aqueous $NH_4Cl$ + aqueous $NaNO_2$.
    2. On gentle heating: $NH_4Cl + NaNO_2 \xrightarrow{\Delta} NaCl + 2H_2O + N_2\uparrow$.
    3. Check the redox: $N$ in $NH_4^+$ is $-3 \to 0$ (oxidised, loses $3e^-$); $N$ in $NO_2^-$ is $+3 \to 0$ (reduced, gains $3e^-$) — electrons balance, confirming this comproportionation is valid. …

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