Q.Differentiate between the principle of estimation of nitrogen in an organic compound by

Showing the 12 most recent of 28 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.The metal which is refined by Mond process is X and the metal refined by Van Arkel method is Y. What are X and Y respectively? (A) Mn, Ga (B) In, Zr (C) Ti, Ni (D) Ni, Zr

   ›Reveal solutionSolution

   The Mond process refines nickel (Ni) via volatile carbonyl formation, and the Van Arkel method refines zirconium (Zr) via thermal decomposition of its iodide. So X = Ni, Y = Zr, which corresponds to option (D).

   The key to this question is knowing which metals form volatile compounds that can be easily separated from impurities and then decomposed to yield the pure metal. Both the Mond process and the Van Arkel method exploit this principle, but they use different volatile intermediates.

   1. The Mond process is specifically designed for nickel. Crude nickel is heated in a stream of carbon monoxide at around 50–60°C. Nickel reacts to form nickel tetracarbonyl, $\text{Ni}(\text{CO})_4$, which is a volatile liquid (boiling point 43°C). Impurities like iron, copper, and cobalt do not form carbonyls under these mild conditions, so the vapour can be separated. The carbonyl vapour is then heated to about 200–250°C, where it decomposes to deposit pure nickel metal and release CO, which is recycled. This process yields nickel of very high purity (99.9%+). …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Match the following List-1 (Metal) A Na B Ca C Ba D Li List-2 (Flame colour) I Apple green II Yellow III Brick red IV Colourless V Crimson red The correct answer is (A) A – IV, B – III, C – II, D – I (B) A – III, B – I, C – V, D – II (C) A – II, B – III, C – I, D – V (D) A – II, B – III, C – IV, D – V

   ›Reveal solutionSolution

   The flame colours of alkali and alkaline earth metals are characteristic: Na gives yellow, Ca gives brick red, Ba gives apple green, and Li gives crimson red. The correct matching is A–II, B–III, C–I, D–V, which corresponds to option (C).

   The key concept here is flame emission spectroscopy — when metal salts are heated in a flame, their electrons get excited to higher energy levels and then fall back, emitting light of specific wavelengths. Each metal has a unique set of emission lines, giving a characteristic colour to the flame. This is a classic qualitative test for metal ions.

   Let’s match each metal from List-1 with its well-known flame colour from List-2.

   1. Sodium (Na) — Sodium compounds produce an intense, persistent yellow flame. This is due to the strong emission at about 589 nm (the sodium D-line). So Na matches with II (Yellow).

   2. Calcium (Ca) — Calcium salts give a brick red colour. It’s a warm, reddish-orange, not as bright as strontium’s crimson. So Ca matches with III (Brick red).

   3. Barium (Ba) — Barium compounds yield a pale apple green flame. This is a distinctive green, different from copper’s blue-green. So Ba matches with I (Apple green). …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.A sample of water is found to contain $\mathrm{F^-}$ (1 ppm), $\mathrm{SO_4^{2-}}$ (> 500 ppm), $\mathrm{NO_3^-}$ (40 ppm), Which ion/s make (s) the water sample unfit for drinking? (A) $\mathrm{F^-}$ & $\mathrm{NO_3^-}$ (B) $\mathrm{SO_4^{2-}}$ (C) $\mathrm{NO_3^-}$ (D) $\mathrm{F^-}$

   ›Reveal solutionSolution

   Water potability is determined by comparing ion concentrations with permissible limits. In this sample, only sulfate concentration exceeds the maximum permissible limit, making the water unfit for drinking. The correct option is (B).

   The suitability of water for drinking, known as potability, depends on the concentrations of various dissolved substances. Each substance has a specific desirable limit and a maximum permissible limit. Exceeding these limits can lead to adverse health effects, making the water unfit for consumption. The unit 'ppm' (parts per million) is commonly used to express these concentrations, indicating milligrams of a substance per liter of water ($\mathrm{mg/L}$).

   Here, we need to compare the given concentrations of fluoride ($\mathrm{F^-}$), sulfate ($\mathrm{SO_4^{2-}}$), and nitrate ($\mathrm{NO_3^-}$) with the established maximum permissible limits for drinking water, typically set by standards like the Bureau of Indian Standards (BIS).

   1. Understand ppm:

      The concentration unit 'ppm' stands for parts per million. For dilute aqueous solutions, 1 ppm is approximately equal to 1 milligram of solute per liter of solution ($\mathrm{1\ ppm \approx 1\ mg/L}$). This unit helps quantify very small amounts of substances present in water.

   2. Recall Permissible Limits for Ions in Drinking Water:

      According to Indian drinking water standards (BIS IS 10500:2012), the desirable and maximum permissible limits for the given ions are:

      • Fluoride ($\mathrm{F^-}$):

        • Desirable limit: $1.0\ \mathrm{ppm}$
        • Maximum permissible limit (in the absence of an alternate source): $1.5\ \mathrm{ppm}$ High concentrations of fluoride can cause fluorosis (dental and skeletal damage).

      • Sulfate ($\mathrm{SO_4^{2-}}$):

        • Desirable limit: $200\ \mathrm{ppm}$
        • Maximum permissible limit (in the absence of an alternate source): $400\ \mathrm{ppm}$ Sulfate in excess of $500\ \mathrm{ppm}$ can cause laxative effects.

      • Nitrate ($\mathrm{NO_3^-}$):

        • Desirable limit: $45\ \mathrm{ppm}$
        • Maximum permissible limit (in the absence of an alternate source): $45\ \mathrm{ppm}$ High concentrations of nitrate (above $45\ \mathrm{ppm}$) can lead to methemoglobinemia, commonly known as "blue baby syndrome," especially in infants.

   3. Compare Given Concentrations with Limits:

      Now, let's compare the concentrations provided in the problem with these standard limits:

      • Fluoride ($\mathrm{F^-}$):
        • Given concentration: $1\ \mathrm{ppm}$
        • Comparison: This concentration is exactly at the desirable limit and well below the maximum permissible limit of $1.5\ \mathrm{ppm}$. Therefore, fluoride at $1\ \mathrm{ppm}$ does not make the water unfit. …

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4. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.German silver is an alloy of (A) Cu, Zn (B) Cu, Zn, Ni (C) Ag, Zn, Ni (D) Cu, Ni

   ›Reveal solutionSolution

   German silver is an alloy of copper, zinc, and nickel — it contains no actual silver. The correct option is (B).

   The name "German silver" is a classic trap. It sounds like it should contain silver, but it doesn't. The trick is to remember that this alloy was developed in Germany as a cheaper, silvery-looking substitute for real silver. The key is its composition: it's primarily copper, with zinc and nickel added to give it that white, corrosion-resistant finish.

   Let's break down the options:

   1. Option (A): Cu, Zn — This is brass. Brass is yellow, not silvery, and lacks the nickel that gives German silver its characteristic white colour and hardness. So this is wrong.

   2. Option (B): Cu, Zn, Ni — This is exactly German silver. The copper provides the base metal, zinc adds strength and workability, and nickel gives the alloy its silvery appearance and resistance to tarnishing. Typical proportions are roughly 60% copper, 20% zinc, and 20% nickel, though these vary. …

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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A solid contains elements A and B. Anions of B form ccp lattice. Cations of A occupy 50% of octahedral voids and 50% of tetrahedral voids. What is the molecular formula of the solid? (A) $AB_3$ (B) $A_3B_2$ (C) $A_2B_3$ (D) $AB$

   ›Reveal solutionSolution

   With $N$ anions B in ccp, A fills half the $N$ octahedral voids and half the $2N$ tetrahedral voids, giving $A:B = 3:2$ — formula $A_3B_2$, option (B).

   Concept

   In a ccp lattice of $N$ anions there are $N$ octahedral voids and $2N$ tetrahedral voids.

   Solution

   Let the number of B anions be $N$.

   • Octahedral voids $= N$; A fills $50\% \Rightarrow \tfrac{1}{2}N$ cations. …

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6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The amine / salt of amine which gives positive test with a mixture of chloroform and alcoholic KOH solution is (A) $C_6H_5-NH-CH_3$ (B) $C_6H_5-N(CH_3)_2$ (C) $C_6H_5-\overset{+}{N}(CH_3)_3 X^-$ (D) $C_6H_5-CH_2-NH_2$

   ›Reveal solutionSolution

   The chloroform + alcoholic KOH test (the carbylamine test) is a diagnostic for primary amines. Of the four species only benzylamine, $\mathrm{C_6H_5CH_2NH_2}$, is primary — option (D).

   The concept first

   Why should the test single out primary amines? Alcoholic KOH pulls a proton off chloroform to give $\mathrm{CCl_3^-}$, which loses $\mathrm{Cl^-}$ to form the electron-deficient carbene $:\!\mathrm{CCl_2}$. The amine nitrogen attacks this carbene. To convert the resulting adduct into an isocyanide, the nitrogen must lose two hydrogens (they leave as HCl, mopped up by KOH) while the carbon loses two chlorines. That is only geometrically possible if the nitrogen started with two N–H bonds, i.e. if the amine is primary. Secondary amines have one N–H, tertiary and quaternary have none — the reaction stops dead.

   Step-by-step

   1. Generate the carbene: $\mathrm{CHCl_3 + KOH \rightarrow :CCl_2 + KCl + H_2O}$.
   2. Nucleophilic attack: $\mathrm{R\!-\!NH_2 + :CCl_2 \rightarrow R\!-\!\overset{+}{N}H_2\!-\!\overset{-}{C}Cl_2}$.
   3. Double dehydrohalogenation: two successive losses of HCl (base-assisted) give $\mathrm{R\!-\!N\!\equiv\!\overset{-}{C}}$. …

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7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.How many of the following metals give oxides and nitrides when burnt in air? Be, Na, Mg, Ba, Sr, Li, K (A) 2 (B) 3 (C) 4 (D) 5

   ›Reveal solutionSolution

   When burnt in air, most metals form oxides, but only the most reactive ones can also overcome the stability of nitrogen to form nitrides. Among the given metals, Beryllium, Magnesium, Barium, Strontium, and Lithium are reactive enough to form both oxides and nitrides, making the count 5.

   The Dance of Metals with Air: Oxides and Nitrides

   When we talk about "burning a metal in air," we're really talking about its reaction with the two main components of air: oxygen ($\text{O}_2$) and nitrogen ($\text{N}_2$). Most metals are eager to react with oxygen, forming various oxides. But nitrogen is a different beast altogether.

   The Concept: Why Nitrogen is Picky

   Air is roughly 78% nitrogen and 21% oxygen. While oxygen is quite reactive, nitrogen gas ($\text{N}_2$) is famously inert. This inertness stems from the incredibly strong triple bond ($\text{N}\equiv\text{N}$) holding the two nitrogen atoms together. Breaking this bond requires a significant amount of energy.

   Watch out

   The stability of the $\text{N}\equiv\text{N}$ triple bond is the primary reason why most elements do not react directly with nitrogen gas under normal conditions. Only very strong reducing agents (highly electropositive metals) can overcome this energy barrier.

   So, for a metal to form a nitride when burnt in air, it must be exceptionally reactive. It needs to be able to:

   1. Break the $\text{N}\equiv\text{N}$ bond: This requires a metal that readily gives up electrons (a strong reducing agent).
   2. Form a stable ionic lattice: The resulting nitride ion ($\text{N}^{3-}$) is quite small and highly charged. For a stable nitride to form, the metal cation and the nitride anion must be able to pack together efficiently, leading to a high lattice energy. This is particularly favored by small, highly charged metal cations.

   Generally, we look to the highly electropositive metals of Group 1 (alkali metals) and Group 2 (alkaline earth metals) for nitride formation. However, even within these groups, there are nuances.

   Step-by-Step Analysis of Each Metal

   Let's examine each metal on our list to see if it forms both an oxide and a nitride when burnt in air.

   1. Lithium (Li)

      • Group: Group 1 (Alkali Metal)
      • Oxide Formation: Lithium readily burns in air to form lithium oxide. $\qquad \text{4Li(s) + O}_2\text{(g)} \rightarrow \text{2Li}_2\text{O(s)}$
      • Nitride Formation: Uniquely among the alkali metals, lithium reacts directly with nitrogen to form lithium nitride. This is due to the exceptionally small size and high charge density of the $\text{Li}^+$ ion, which allows it to form a very stable lattice with the small $\text{N}^{3-}$ ion, leading to high lattice energy. $\qquad \text{6Li(s) + N}_2\text{(g)} \rightarrow \text{2Li}_3\text{N(s)}$
      • Conclusion: Lithium forms both oxides and nitrides.

   2. Sodium (Na)

      • Group: Group 1 (Alkali Metal)
      • Oxide Formation: Sodium burns in air primarily to form sodium peroxide, but also some sodium oxide. $\qquad \text{2Na(s) + O}_2\text{(g)} \rightarrow \text{Na}_2\text{O}_2\text{(s)}$ $\qquad \text{4Na(s) + O}_2\text{(g)} \rightarrow \text{2Na}_2\text{O(s)}$ (if oxygen is limited)
      • Nitride Formation: Sodium does not react directly with nitrogen to form a stable nitride under normal burning conditions in air. The $\text{Na}^+$ ion is larger than $\text{Li}^+$, leading to lower lattice energy with $\text{N}^{3-}$, which isn't enough to overcome the stability of $\text{N}_2$.
      • Conclusion: Sodium forms oxides but not nitrides.

   3. Potassium (K)

      • Group: Group 1 (Alkali Metal)
      • Oxide Formation: Potassium burns in air primarily to form potassium superoxide. $\qquad \text{K(s) + O}_2\text{(g)} \rightarrow \text{KO}_2\text{(s)}$
      • Nitride Formation: Similar to sodium, potassium does not react directly with nitrogen to form a stable nitride under normal burning conditions in air. The $\text{K}^+$ ion is even larger than $\text{Na}^+$, further reducing the lattice energy with $\text{N}^{3-}$.
      • Conclusion: Potassium forms oxides but not nitrides.

   4. Beryllium (Be)

      • Group: Group 2 (Alkaline Earth Metal)
      • Oxide Formation: Beryllium burns in air to form beryllium oxide. $\qquad \text{2Be(s) + O}_2\text{(g)} \rightarrow \text{2BeO(s)}$
      • Nitride Formation: Beryllium reacts with nitrogen at high temperatures to form beryllium nitride. $\qquad \text{3Be(s) + N}_2\text{(g)} \rightarrow \text{Be}_3\text{N}_2\text{(s)}$
        Tip

        Beryllium exhibits a "diagonal relationship" with aluminum, meaning it shares some properties with aluminum. Like aluminum, beryllium forms a nitride, though it often requires higher temperatures than other Group 2 metals due to its smaller size and more covalent character.

      • Conclusion: Beryllium forms both oxides and nitrides.

   5. Magnesium (Mg)

      • Group: Group 2 (Alkaline Earth Metal)
      • Oxide Formation: Magnesium burns brilliantly in air to form magnesium oxide. $\qquad \text{2Mg(s) + O}_2\text{(g)} \rightarrow \text{2MgO(s)}$ …

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8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.In the estimation of nitrogen by Kjeldahl’s method 0.933 g of an organic compound ‘X’ was analyzed. Ammonia evolved was absorbed in 60 mL of 0.1 M H$_2$SO$_4$. The unreacted acid requires 20 mL of 0.1 M NaOH for complete neutralization. The compound ‘X’ is (A) CH$_3$CH$_2$NH$_2$ (B) C$_6$H$_5$NH$_2$ (C) C$_6$H$_5$CH$_2$NH$_2$ (D) CH$_3$–C–NH$_2$ (with a double bond O above the C)

   ›Reveal solutionSolution

   In Kjeldahl’s method, the nitrogen content is found from the difference between the acid taken and the acid neutralized by the evolved ammonia. Here, the calculated nitrogen percentage (~15.0%) matches aniline (C₆H₅NH₂), so the correct option is (B).

   ---

   Concept & Intuition

   Kjeldahl’s method converts organic nitrogen into ammonia (NH₃), which is then absorbed in a known volume of standard acid. The unreacted acid is titrated with base. The amount of acid consumed by ammonia equals the difference between the initial acid and the acid left after absorption. Since each mole of H₂SO₄ neutralizes two moles of NH₃, we can find moles of nitrogen in the sample. Dividing by the sample mass gives the percentage of nitrogen. Then we compare this percentage to the theoretical nitrogen percentages of the given compounds.

   ---

   Step-by-step reasoning

   1. Find moles of H₂SO₄ initially taken Volume = 60 mL = 0.060 L, concentration = 0.1 M

   $\text{moles H}_2\text{SO}_4 = 0.060 \times 0.1 = 0.0060 \text{ mol}$

   2. Find moles of NaOH used to neutralize the unreacted acid Volume = 20 mL = 0.020 L, concentration = 0.1 M

   $\text{moles NaOH} = 0.020 \times 0.1 = 0.0020 \text{ mol}$

   The reaction is:

   $\text{H}_2\text{SO}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$

   So moles of unreacted H₂SO₄ = half the moles of NaOH:

   $\text{unreacted H}_2\text{SO}_4 = \frac{0.0020}{2} = 0.0010 \text{ mol}$

   3. Find moles of H₂SO₄ that reacted with NH₃

   $\text{reacted H}_2\text{SO}_4 = 0.0060 - 0.0010 = 0.0050 \text{ mol}$

   4. Relate to moles of NH₃ (and hence nitrogen) The reaction of ammonia with sulfuric acid:

   $2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$

   So 1 mol H₂SO₄ reacts with 2 mol NH₃.

   $\text{moles NH}_3 = 2 \times 0.0050 = 0.0100 \text{ mol}$

   Each mole of NH₃ contains one mole of nitrogen, so:

   $\text{moles N} = 0.0100 \text{ mol}$

   5. Calculate mass of nitrogen Atomic mass of N = 14 g/mol

   $\text{mass of N} = 0.0100 \times 14 = 0.140 \text{ g}$

   6. Calculate percentage of nitrogen in the sample Sample mass = 0.933 g $$$$

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9. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
   Q.Identify the crystal system in which body centered lattice is not present (A) Cubic (B) Hexagonal (C) Tetragonal (D) Orthorhombic

   ›Reveal solutionSolution

   The body-centered lattice is absent in the hexagonal crystal system because the required symmetry of a hexagonal lattice cannot be maintained with a body-centered arrangement. The correct option is (B).

   The key idea here is that not every crystal system can support a body-centered lattice. A body-centered lattice has an extra lattice point at the body center of the unit cell, and this point must be exactly equivalent (by symmetry) to the corner points. The crystal system’s symmetry determines whether such an arrangement is possible.

   Let’s walk through each option systematically.

   1. Cubic system – This has the highest symmetry: four three-fold rotation axes along the body diagonals. A body-centered cubic (BCC) lattice is perfectly valid because the body center is symmetrically equivalent to the corners under these rotations. So BCC exists here.

   2. Hexagonal system – This is the critical one. The hexagonal system has a six-fold rotation axis (or a three-fold axis with a perpendicular mirror). In a simple hexagonal lattice, the unit cell has atoms only at the corners. If you try to add a body center, that point would lie exactly halfway along the vertical axis. But the six-fold symmetry requires that any point at the body center would have to be repeated by rotation, creating additional points that break the hexagonal symmetry. In fact, the body-centered hexagonal lattice is mathematically identical to a simple hexagonal lattice with a different choice of axes — it does not form a distinct Bravais lattice. The hexagonal system only allows primitive (P) lattices. So body-centered is not present here. …

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10. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.A crystal is formed by X (cations) and Y (anions). Atoms of Y form ccp and atoms of X occupy half of octahedral voids and half of tetrahedral voids. What is the molecular formula of crystal? (A) $X_2Y_3$ (B) $XY_3$ (C) $X_3Y$ (D) $X_3Y_2$

    ›Reveal solutionSolution

    In a ccp arrangement of anions, the number of octahedral voids equals the number of anions and tetrahedral voids are twice that. X occupies half of each type of void, so the ratio of X to Y is 3:2, giving the formula $X_3Y_2$.

    The key to this problem is understanding void geometry in a cubic close-packed (ccp) lattice. In ccp, the anions (Y) are arranged in a face-centered cubic pattern. For every sphere in a ccp arrangement, there is exactly one octahedral void and two tetrahedral voids. This is a fixed geometric fact — it comes from the stacking of close-packed layers.

    So if we let the number of Y atoms be $n$, then:

    • Number of octahedral voids = $n$
    • Number of tetrahedral voids = $2n$

    Now the problem says X occupies half of the octahedral voids and half of the tetrahedral voids. That means:

    1. Octahedral voids filled = $\frac{1}{2} \times n = \frac{n}{2}$
    2. Tetrahedral voids filled = $\frac{1}{2} \times 2n = n$ …

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11. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.Consider the following reactions $X + O_2 \rightarrow Cu_2O + SO_2$ $Cu_2O + X \rightarrow Cu + Y \uparrow$ The shape of the molecule Y is (A) Linear (B) Tetrahedral (C) Pyramidal (D) Angular

    ›Reveal solutionSolution

    The problem is about identifying the product Y from two consecutive reactions involving copper and a sulfur-containing compound; Y is SO₂, which has an angular (bent) shape, so the correct option is (D).

    We start by recognizing that the first reaction involves a compound X reacting with oxygen to produce copper(I) oxide (Cu₂O) and sulfur dioxide (SO₂). This tells us X must contain copper and sulfur. The most common copper sulfide is Cu₂S (copper(I) sulfide). Let’s check:

    1. Identify X from the first reaction The reaction is:

    $X + O_2 \rightarrow Cu_2O + SO_2$

    The common copper(I) sulfide is Cu₂S. Taking X = Cu₂S:

    $Cu_2S + O_2 \rightarrow Cu_2O + SO_2$

    Count atoms: Left: 2 Cu, 1 S, 2 O. Right: 2 Cu, 1 S, 1+2=3 O. Not balanced. We need more O on left. Try:

    $2Cu_2S + 3O_2 \rightarrow 2Cu_2O + 2SO_2$

    Check: Left: 4 Cu, 2 S, 6 O. Right: 4 Cu, 2 S, 2+4=6 O. Balanced! So X is indeed Cu₂S (copper(I) sulfide).

    2. Use the second reaction to find Y The second reaction is the self-reduction step of copper smelting, in which cuprous oxide and cuprous sulfide react to give copper metal:

    $2Cu_2O + Cu_2S \rightarrow 6Cu + SO_2 \uparrow$

    Atom check — Left: 6 Cu, 2 O, 1 S; Right: 6 Cu, 2 O, 1 S. Balanced. So Y = SO₂.

    3. Shape of SO₂ …

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12. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.In contact process of manufacture of $H_2SO_4$, the arsenic purifier used in the industrial plant contains (A) $Al_2O_3 \cdot xH_2O$ (B) $FeO \cdot xH_2O$ (C) $Cr_2O_3 \cdot xH_2O$ (D) $Fe_2O_3 \cdot xH_2O$

    ›Reveal solutionSolution

    The arsenic purifier in the contact process uses hydrated ferric oxide ($Fe_2O_3 \cdot xH_2O$) to adsorb arsenic impurities from sulfur dioxide gas. The correct option is (D).

    The contact process for manufacturing sulfuric acid involves burning sulfur or roasting sulfide ores to produce sulfur dioxide ($SO_2$). This gas must be purified before catalytic oxidation to $SO_3$, because impurities—especially arsenic compounds—poison the vanadium(V) oxide catalyst. The arsenic purifier is a specific hydrated metal oxide that selectively adsorbs arsenic oxides (like $As_2O_3$) from the gas stream.

    Why this approach works:

    Hydrated oxides of certain metals act as excellent adsorbents for polar impurities. Among the options, hydrated ferric oxide ($Fe_2O_3 \cdot xH_2O$) is well-known for its high surface area and affinity for arsenic species. It forms a gel-like mass that traps arsenic compounds physically and chemically, while other hydrated oxides (alumina, chromia, ferrous oxide) are either less effective or not used industrially for this purpose.

    Step-by-step reasoning:

    1. Identify the impurity to be removed.

       In the contact process, $SO_2$ gas from roasting pyrites ($FeS_2$) or other ores often contains arsenic trioxide ($As_2O_3$) vapor. This compound, if not removed, would deactivate the $V_2O_5$ catalyst used in the next step.

    2. Recall the industrial purification train.

       The gas is first cooled, then passed through a series of scrubbers and purifiers: a dust chamber, a water scrubber, and then a special "arsenic purifier" containing a solid adsorbent. The material in this purifier must be cheap, stable, and highly selective for arsenic.

    3. Evaluate each option based on known industrial practice.

       • (A) $Al_2O_3 \cdot xH_2O$ (hydrated alumina): Used as a desiccant or catalyst support, but not specifically for arsenic removal in this process.
       • (B) $FeO \cdot xH_2O$ (hydrated ferrous oxide): Unstable; ferrous iron oxidizes readily to ferric, so it is not used as a stable purifier.
       • (C) $Cr_2O_3 \cdot xH_2O$ (hydrated chromia): Chromium compounds are toxic and expensive; not employed for bulk gas purification. …

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