Q.Account for the following:

Showing the 12 most recent of 27 on this concept.

1. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.What is the amino acid which does not contain $-\mathrm{NH}_2$ group in its structure? (A) Tyrosine (B) Alanine (C) Proline (D) Methionine

   ›Reveal solutionSolution

   The key idea is that proline is an imino acid, not a standard amino acid — its side chain loops back to form a secondary amine, so it lacks a free $-\mathrm{NH}_2$ group. The correct option is (C).

   Amino acids are the building blocks of proteins. The standard structure of an $\alpha$-amino acid has a central carbon ($\alpha$-carbon) bonded to four groups: an amino group ($-\mathrm{NH}_2$), a carboxyl group ($-\mathrm{COOH}$), a hydrogen atom, and a variable side chain (R). That $-\mathrm{NH}_2$ group is what makes it an "amino" acid.

   But not every molecule classified as an amino acid follows this exact pattern. Proline is the classic exception. Its side chain is not a simple alkyl group — instead, it forms a five-membered ring that connects back to the nitrogen of the amino group. This turns the primary amine ($-\mathrm{NH}_2$) into a secondary amine ($>\mathrm{NH}$), which is part of a ring. Because of this, proline is technically an imino acid, not a true amino acid. It still has a carboxyl group and an $\alpha$-carbon, but the free $-\mathrm{NH}_2$ group is absent.

   Let's check each option:

   1. Tyrosine — has a standard $-\mathrm{NH}_2$ group attached to the $\alpha$-carbon. Its side chain is a phenolic ring. No structural modification to the amino group. So it contains $-\mathrm{NH}_2$.

   2. Alanine — the simplest of the four here, with a methyl group as the side chain. The $\alpha$-carbon carries a normal $-\mathrm{NH}_2$ group. So it contains $-\mathrm{NH}_2$.

   3. Proline — as described, the side chain is a three-carbon chain that loops back and bonds to the nitrogen atom, forming a pyrrolidine ring. The nitrogen now has only one hydrogen (instead of two), and is part of the ring. There is no free $-\mathrm{NH}_2$ group. This is the correct answer.

   4. Methionine — has a standard $-\mathrm{NH}_2$ group. Its side chain contains a sulfur atom in a thioether linkage. No effect on the amino group. So it contains $-\mathrm{NH}_2$.

   Watch out

   A common mistake is to think that all 20 standard amino acids have a free $-\mathrm{NH}_2$ group. Proline is the only one that does not — its nitrogen is part of a ring, making it a secondary amine. In exam questions, "amino acid" is often used loosely to include proline, but the defining feature here is the absence of $-\mathrm{NH}_2$.

   Tip

   If you ever forget, remember the mnemonic: Proline is the "pro" at making rings — its side chain curls back to tie up the amino group. No free $-\mathrm{NH}_2$ means no primary amine.

   ✓Final answer

   The correct option is (C) Proline.

2. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Which of the following compounds will be suitable for estimation of nitrogen by Kjeldahl’s method? (A) I & V only (B) I, II, III only (C) II & V only (D) III & IV only

   ›Reveal solutionSolution

   Kjeldahl's method estimates nitrogen by converting it to ammonium sulfate, then ammonia. It is suitable for nitrogen in amines, amides, and amino acids, but not for nitrogen in nitro, azo, or heterocyclic ring compounds. Based on common examples, compounds like aniline (an amine) and urea (an amide) are suitable.

   Concept and Intuition

   Kjeldahl's method is a quantitative analytical technique used to determine the nitrogen content in organic compounds. The core idea is to convert all the nitrogen present in the sample into a measurable form, specifically ammonia, which can then be quantified.

   The method relies on a crucial chemical transformation:

   1. Digestion: The organic compound is heated with concentrated sulfuric acid in the presence of a catalyst (like $\text{CuSO}_4$) and potassium sulfate ($\text{K}_2\text{SO}_4$). This process oxidizes the organic matter and converts the nitrogen present into ammonium sulfate, $(\text{NH}_4)_2\text{SO}_4$.

   $\text{Organic N-compound} + \text{conc. } \text{H}_2\text{SO}_4 \xrightarrow{\text{catalyst, heat}} (\text{NH}_4)_2\text{SO}_4 + \text{CO}_2 + \text{H}_2\text{O}$

   2. Distillation: The ammonium sulfate solution is then treated with an excess of a strong base, typically sodium hydroxide ($\text{NaOH}$). This liberates ammonia gas ($\text{NH}_3$).

   $(\text{NH}_4)_2\text{SO}_4 + 2\text{NaOH} \longrightarrow \text{Na}_2\text{SO}_4 + 2\text{NH}_3 + 2\text{H}_2\text{O}$

   3. Titration: The liberated ammonia gas is absorbed in a known excess volume of a standard acid (e.g., $\text{H}_2\text{SO}_4$ or $\text{HCl}$). The unreacted acid is then back-titrated with a standard base to determine the amount of ammonia absorbed, and consequently, the amount of nitrogen in the original sample.

   The critical limitation of Kjeldahl's method is that not all forms of nitrogen can be quantitatively converted to ammonium sulfate under the digestion conditions. For the method to be effective, the nitrogen must be in a reduced state or a state that can be easily reduced to ammonia.

   Important

   Compounds suitable for Kjeldahl's method:

   • Amines (primary, secondary, tertiary)
   • Amides
   • Amino acids and proteins
   • Ammonium salts

   Compounds NOT suitable for Kjeldahl's method:

   • Nitro compounds (e.g., nitrobenzene, $-\text{NO}_2$)
   • Azo compounds (e.g., azobenzene, $-\text{N}=\text{N}-$)
   • Nitrogen in heterocyclic rings (e.g., pyridine, quinoline)
   • Diazo compounds
   • Nitriles (some sources list them as unsuitable, or requiring modified conditions)

   The reason these unsuitable compounds cannot be estimated is that their nitrogen atoms are either in a highly oxidized state (like nitro compounds) or are part of very stable structures (like heterocyclic rings or azo groups) that do not readily convert to ammonium sulfate under the standard Kjeldahl digestion conditions.

   Let's analyze the given compounds based on these principles. Since the specific structures for I, II, III, IV, V are not provided, we will consider common examples that represent the types of compounds typically tested in such questions.

   Step-by-step Analysis

   We will assume the compounds are:

   • I: Aniline ($\text{C}_6\text{H}_5\text{NH}_2$)
   • II: Nitrobenzene ($\text{C}_6\text{H}_5\text{NO}_2$)
   • III: Pyridine ($\text{C}_5\text{H}_5\text{N}$)
   • IV: Azobenzene ($\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_5$)
   • V: Urea ($\text{H}_2\text{N}-\text{CO}-\text{NH}_2$)

   1. Compound I: Aniline

      • Aniline is a primary aromatic amine. The nitrogen atom is directly bonded to a carbon atom and two hydrogen atoms.
      • This type of nitrogen is readily converted to ammonium sulfate upon digestion with concentrated sulfuric acid.
      • Therefore, aniline is suitable for Kjeldahl's method.

   2. Compound II: Nitrobenzene

      • Nitrobenzene contains a nitro group ($-\text{NO}_2$). The nitrogen in a nitro group is in a highly oxidized state.
      • Under the standard Kjeldahl digestion conditions, this nitrogen is not quantitatively reduced to ammonia.
      • Therefore, nitrobenzene is not suitable for Kjeldahl's method.

   3. Compound III: Pyridine

      • Pyridine is a heterocyclic compound where the nitrogen atom is part of an aromatic ring system.
      • The nitrogen in such a stable ring structure is not quantitatively converted to ammonium sulfate during the digestion process.
      • Therefore, pyridine is not suitable for Kjeldahl's method.

   4. Compound IV: Azobenzene

      • Azobenzene contains an azo group ($-\text{N}=\text{N}-$). In azo compounds, the nitrogen atoms are directly bonded to each other.
      • These nitrogen-nitrogen bonds are not easily broken, and the nitrogen is not quantitatively converted to ammonium sulfate under Kjeldahl conditions.
      • Therefore, azobenzene is not suitable for Kjeldahl's method.

   5. Compound V: Urea

      • Urea is an amide. It contains nitrogen atoms in amide groups ($-\text{CONH}_2$).
      • Nitrogen in amide groups is readily converted to ammonium sulfate upon digestion with concentrated sulfuric acid.
      • Therefore, urea is suitable for Kjeldahl's method.

   Based on this analysis, compounds I (Aniline) and V (Urea) are suitable for estimation of nitrogen by Kjeldahl's method.

   ✓Final answer

   Based on the common examples representing the types of compounds, compounds I (Aniline) and V (Urea) are suitable for estimation of nitrogen by Kjeldahl’s method. The correct option is (A).

3. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.The atomic numbers of elements A, B, D, E respectively are 7, 17, 13, 11. The element which forms basic oxide among them is (A) E (B) B (C) D (D) A

   ›Reveal solutionSolution

   The basic character of an oxide depends on the metallic character of the element. Among the given elements, sodium (atomic number 11) is the most metallic and forms a strongly basic oxide. The correct option is (A) E.

   The key idea here is that oxides of metals are generally basic, while oxides of non-metals are acidic. So to find which element forms a basic oxide, we need to identify which one is a metal — and more specifically, which one is the most metallic.

   Let’s decode the atomic numbers given:

   • A: atomic number 7 → Nitrogen (N) — a non-metal
   • B: atomic number 17 → Chlorine (Cl) — a non-metal
   • D: atomic number 13 → Aluminium (Al) — a metal, but with some amphoteric character
   • E: atomic number 11 → Sodium (Na) — a highly reactive metal

   Now, let’s work through the reasoning step by step.

   1. Metallic character and basicity of oxides

      Metallic elements tend to lose electrons and form positive ions. Their oxides react with water to give hydroxides, which release OH⁻ ions — that’s what makes them basic. Non-metals, on the other hand, form covalent oxides that are typically acidic (they release H⁺ in water).

   2. Identify the metals among the given elements

      From the atomic numbers:

      • 11 (Na) and 13 (Al) are metals.
      • 7 (N) and 17 (Cl) are non-metals. So only D and E are candidates for forming basic oxides.

   3. Compare the basic nature of Na₂O and Al₂O₃

      Sodium oxide (Na₂O) dissolves in water to form sodium hydroxide (NaOH), a strong base. Aluminium oxide (Al₂O₃) is amphoteric — it can react with both acids and bases, but its basic character is much weaker than that of a typical metal oxide. In fact, Al₂O₃ behaves more like an acidic oxide when reacting with strong bases.

   4. Which one is clearly basic?

      Sodium (E) forms a strongly basic oxide. Aluminium (D) forms an amphoteric oxide, not purely basic. So among the options, E is the correct choice.

   Watch out

   A common mistake is to think that all metal oxides are strongly basic. Aluminium oxide is a metal oxide but is amphoteric — it can act as both an acid and a base. So don’t pick D just because it’s a metal.

   Tip

   In periodic table trends, basic character of oxides increases down a group and decreases across a period. Sodium (Group 1) is far more metallic than aluminium (Group 13), so its oxide is more basic.

   ✓Final answer

   The element that forms a basic oxide is E (sodium), so the correct option is (A) E.

4. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.Identify the correct statements from the following I. Conjugate base of chloric acid is $\mathrm{ClO}^-$ II. $\mathrm{AlCl}_3$ is a Lewis acid III. Conjugate base of $\mathrm{NH}_3$ is $\mathrm{NH}_2^-$ (A) I, II, III (B) I, II only (C) II, III only (D) I, III only

   ›Reveal solutionSolution

   Conjugate acid-base pairs differ by one proton; Lewis acids accept electron pairs. Chloric acid is $\ce{HClO3}$, so its conjugate base is $\ce{ClO3-}$, not $\ce{ClO-}$. $\ce{AlCl3}$ is electron-deficient and accepts a lone pair, making it a Lewis acid. $\ce{NH3}$ loses a proton to give $\ce{NH2-}$, so that pair is correct. Only statements II and III are true.

   The core of this question is understanding two distinct definitions: the Brønsted-Lowry concept of conjugate acid-base pairs (where a conjugate base is what remains after the acid donates a proton), and the Lewis concept of acids (electron-pair acceptors). Each statement must be tested against these definitions.

   1. Statement I: Conjugate base of chloric acid is $\ce{ClO-}$

      Chloric acid is $\ce{HClO3}$. When it donates a proton ($\ce{H+}$), it loses one hydrogen atom and the charge adjusts. The species left is $\ce{ClO3-}$ (the chlorate ion). The ion $\ce{ClO-}$ is the conjugate base of hypochlorous acid ($\ce{HClO}$), not chloric acid. So statement I is false.

   2. Statement II: $\ce{AlCl3}$ is a Lewis acid

      A Lewis acid is any species that can accept an electron pair. $\ce{AlCl3}$ has an aluminum atom with only six electrons in its valence shell (three bonds to chlorine, no lone pairs). This electron deficiency means it readily accepts a lone pair from a donor (like $\ce{Cl-}$ in the classic dimerization to $\ce{Al2Cl6}$). Therefore, $\ce{AlCl3}$ is indeed a Lewis acid. Statement II is true.

   3. Statement III: Conjugate base of $\ce{NH3}$ is $\ce{NH2-}$

      Ammonia ($\ce{NH3}$) can act as a Brønsted acid, though weakly. If it donates a proton, it loses one $\ce{H+}$, leaving $\ce{NH2-}$ (the amide ion). This fits the definition: conjugate base = acid minus $\ce{H+}$. Statement III is true.

   Watch out

   A common mistake is confusing the names of oxyacids. Chloric acid ($\ce{HClO3}$) is different from hypochlorous acid ($\ce{HClO}$), chlorous acid ($\ce{HClO2}$), and perchloric acid ($\ce{HClO4}$). Each has a different conjugate base. Always check the formula of the acid first.

   Since only statements II and III are correct, the answer is the option listing those two.

   ✓Final answer

   The correct option is (C) II, III only.

5. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.Arrange the following in the correct order of their boiling points (C$_2$H$_5$)$_2$O \hspace{1cm} CH$_3$(CH$_2$)$_3$OH \hspace{1cm} CH$_3$CH$_2$CH(OH)CH$_3$ \hspace{1cm} CH$_3$–(CH$_2$)$_3$–CH$_3$ I \hspace{2.5cm} II \hspace{2.5cm} III \hspace{2.5cm} IV (A) I > III > II > IV (B) II > I > III > IV (C) III > II > I > IV (D) II > III > IV > I

   ›Reveal solutionSolution

   Hydrogen bonding (alcohols II, III) dominates; among molecules of similar mass the longer-chain alkane IV slightly outboils the weakly polar ether I, and straight-chain 1-butanol (II) beats branched 2-butanol (III). Order: II > III > IV > I — option (D).

   Concept & Intuition

   Boiling point tracks the strength of intermolecular forces: hydrogen bonding > dipole-dipole > London dispersion. Identify each species:

   • I: $(C_2H_5)_2O$ — diethyl ether (polar, no O–H).
   • II: $CH_3(CH_2)_3OH$ — 1-butanol (straight-chain primary alcohol).
   • III: $CH_3CH_2CH(OH)CH_3$ — 2-butanol (branched secondary alcohol).
   • IV: $CH_3(CH_2)_3CH_3$ — n-pentane (nonpolar alkane).

   All have similar molar mass (~72–74 g/mol), so hydrogen bonding is the deciding factor.

   Step-by-step reasoning.

   1. Alcohols come first. II and III both have –OH and hydrogen-bond, so they boil highest (1-butanol ~117 °C, 2-butanol ~99 °C).

   2. Straight vs branched alcohol. Branching lowers surface area and weakens dispersion, so straight-chain II boils higher than III.

   3. Ether vs alkane. Diethyl ether (I, ~35 °C) is only weakly polar; n-pentane (IV, ~36 °C) is nonpolar but has a slightly longer carbon chain and marginally stronger dispersion, so IV boils just above I.

   4. Full order:

   $\text{II}\,(\sim117\,^\circ\text{C}) > \text{III}\,(\sim99\,^\circ\text{C}) > \text{IV}\,(\sim36\,^\circ\text{C}) > \text{I}\,(\sim35\,^\circ\text{C}).$

   Watch out

   Do not assume a polar molecule always outboils a nonpolar one of similar mass. Diethyl ether's dipole is weak, and n-pentane's slightly larger dispersion gives it a marginally higher boiling point.

   Tip

   For nearly equal molar masses, the molecule with the longer, less-branched carbon chain (more surface contact) tends to boil higher — here n-pentane edges out diethyl ether.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.Identify the set of molecules which are not in the correct order of their dipole moments (A) $\mathrm{HF > HCl > HBr}$ (B) $\mathrm{H_2O > H_2S > CO_2}$ (C) $\mathrm{H_2S > HCl > HF}$ (D) $\mathrm{NH_3 > NF_3 > BF_3}$

   ›Reveal solutionSolution

   Dipole moment depends on both bond polarity and molecular geometry. Checking each option against known values shows that (A), (B), and (D) all list the correct order, but (C) reverses the true order — the actual order is HF > HCl > H₂S, not H₂S > HCl > HF. So the set that is not in the correct order is (C).

   Concept & Intuition

   A common trap is to assume greater electronegativity difference always means a larger dipole moment — but bond length and molecular geometry matter too. Even though HF has the largest electronegativity difference among the hydrogen halides, its dipole moment still comes out largest once you compare actual measured values, and H₂S (despite having two polar S–H bonds) has a smaller net dipole than either HCl or HF.

   Step-by-step reasoning

   1. Option (A): HF > HCl > HBr

      Measured dipole moments: HF ≈ 1.91 D, HCl ≈ 1.08 D, HBr ≈ 0.82 D.

      → Order HF > HCl > HBr is correct.

   2. Option (B): H₂O > H₂S > CO₂

      H₂O is bent (~104.5°) with a large dipole (≈1.85 D) because oxygen is highly electronegative. H₂S is also bent but sulfur is less electronegative, giving a smaller dipole (≈0.97 D). CO₂ is linear and symmetric, so its two C=O bond dipoles cancel exactly, giving zero net dipole.

      → Order H₂O > H₂S > CO₂ is correct.

   3. Option (C): H₂S > HCl > HF

      Using the same measured values as above: H₂S ≈ 0.97 D, HCl ≈ 1.08 D, HF ≈ 1.91 D. The actual order is HF > HCl > H₂S — the complete reverse of what this option claims.

      → This order is incorrect.

   4. Option (D): NH₃ > NF₃ > BF₃

      NH₃ is pyramidal with a lone pair reinforcing the bond dipoles, giving a net dipole of ≈1.47 D. NF₃ is also pyramidal, but because fluorine is more electronegative than nitrogen, the N–F bond dipoles point toward F, opposing the lone-pair contribution — this largely cancels out and leaves a much smaller net dipole (≈0.24 D). BF₃ is trigonal planar and symmetric, giving zero net dipole.

      → Order NH₃ > NF₃ > BF₃ is correct.

   Watch out

   Don't assume more electronegative atoms always mean a larger dipole moment. In NF₃, the bond dipoles point away from the lone pair and partially cancel it, making its net dipole much smaller than NH₃'s — the opposite of what a naive electronegativity comparison with NH₃ would suggest.

   Tip

   For molecules with lone pairs, always take the vector sum: the lone pair contributes its own dipole, and the bond dipoles may add to or subtract from it depending on their direction.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

7. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.Observe the following species

   (i) $\mathrm{NH_3}$

   (ii) $\mathrm{AlCl_3}$

   (iii) $\mathrm{SnCl_4}$

   (iv) $\mathrm{CO_2}$

   (v) $\mathrm{Ag^+}$

   (vi) $\mathrm{HSO_4^-}$ How many of the above species act as Lewis acids? (A) 5 (B) 3 (C) 4 (D) 2

   ›Reveal solutionSolution

   A Lewis acid is an electron-pair acceptor. Checking each species: AlCl₃, SnCl₄, CO₂, and Ag⁺ can all accept an electron pair, while NH₃ (a lone-pair donor) and HSO₄⁻ (a Brønsted acid/base with no accessible empty orbital) cannot. That gives 4 Lewis acids, so the correct option is (C).

   Concept & Intuition

   A Lewis acid is any species that can accept a pair of electrons — typically because it has an incomplete octet, a vacant low-energy orbital, or an electron-deficient centre that can coordinate a lone pair. Go through each species and ask: can it accept an electron pair?

   Step-by-step reasoning

   1. NH₃ — Nitrogen has a lone pair and a complete octet; it donates electrons rather than accepting them.

      → Not a Lewis acid (it is a Lewis base).

   2. AlCl₃ — Aluminium has only six electrons around it (an incomplete octet) and readily accepts a lone pair (e.g., from Cl⁻ to form AlCl₄⁻).

      → Lewis acid.

   3. SnCl₄ — Tin can expand its octet using empty d-orbitals and accepts a lone pair to form species like SnCl₆²⁻.

      → Lewis acid.

   4. CO₂ — The carbon in CO₂ is strongly electron-deficient (δ⁺) and readily accepts a lone pair from a nucleophile, as in its reaction with OH⁻/H₂O to form HCO₃⁻/H₂CO₃. This is a standard textbook example of a Lewis acid, alongside BF₃, SO₂, and SO₃.

      → Lewis acid.

   5. Ag⁺ — A silver cation has vacant orbitals and readily accepts lone pairs from ligands (e.g., NH₃, forming [Ag(NH₃)₂]⁺).

      → Lewis acid.

   6. HSO₄⁻ — Sulfur here is already tetrahedrally coordinated with no accessible empty orbital, and the species behaves as a Brønsted acid/base (donating or accepting a proton), not as an electron-pair acceptor.

      → Not a Lewis acid.

   Counting the Lewis acids: AlCl₃, SnCl₄, CO₂, Ag⁺ — that's 4 species.

   Watch out

   Don't overlook CO₂ as a Lewis acid just because it has no formal incomplete octet — its electron-deficient carbon still readily accepts a lone pair from a nucleophile, which is exactly the Lewis-acid criterion.

   Tip

   Electron-deficient molecules such as BF₃, AlCl₃, CO₂, SO₂, and SO₃ are classic textbook examples of Lewis acids — keep this list in mind for quick recall.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.In the given reaction sequence, Z is

   [!FORMULA] $\mathrm{C_6H_5CO_2H} \xrightarrow[\text{(2) } \Delta]{\text{(1) } \mathrm{NH_3}} \mathrm{X} \xrightarrow[\mathrm{NaOH}]{\mathrm{Br_2}\ /} \mathrm{Y} \xrightarrow[\mathrm{KOH}]{\mathrm{CHCl_3}\ /} \mathrm{Z}$

   (A) $\mathrm{C_6H_5{-}Cl}$ (B) $\mathrm{C_6H_5{-}OH}$ (C) $\mathrm{C_6H_5{-}NC}$ (D) $\mathrm{C_6H_5{-}CN}$

   ›Reveal solutionSolution

   Benzoic acid $\to$ benzamide $\to$ (Hofmann) aniline $\to$ (carbylamine reaction) phenyl isocyanide $\ce{C6H5NC}$ — option (C).

   The concept first. Three named reactions, each with a give-away signature:

   1. Acid + ammonia + heat: forms the ammonium salt, which on heating dehydrates to the amide. (Acid $\to$ amide is the standard way in.)
   2. Hofmann bromamide degradation ($\ce{Br2}/\ce{NaOH}$): converts an amide $\ce{RCONH2}$ into a primary amine $\ce{RNH2}$ with the loss of one carbon. The mechanism goes through an N-bromoamide, then a nitrene-like migration of R from carbon to nitrogen, giving an isocyanate that is hydrolysed.
   3. Carbylamine (isocyanide) test ($\ce{CHCl3} + \ce{KOH}$, heat): the diagnostic test for primary amines. $\ce{KOH}$ deprotonates chloroform to give dichlorocarbene $\ce{:CCl2}$, which the amine attacks; loss of $2\ce{HCl}$ gives the isocyanide $\ce{R-N#C}$, recognisable by its very unpleasant smell.

   Step 1 — X.

   $\ce{C6H5COOH + NH3 -> C6H5COO^- NH4+ ->[\Delta, -H2O] C6H5CONH2}$

   $X$ = benzamide.

   Step 2 — Y.

   $\ce{C6H5CONH2 ->[Br2/NaOH] C6H5NH2 + Na2CO3 + NaBr}$

   Seven carbons become six, and $\ce{-NH2}$ is now attached directly to the ring. $Y$ = aniline.

   Step 3 — Z.

   $\ce{C6H5NH2 + CHCl3 + 3KOH ->[\Delta] C6H5NC + 3KCl + 3H2O}$

   $Z$ = phenyl isocyanide, $\ce{C6H5-N#C}$.

   Step 4 — Beware the look-alike. Option (D), $\ce{C6H5CN}$ (benzonitrile), is bonded through carbon and would come from a diazonium salt with $\ce{CuCN}$ (Sandmeyer), not from the carbylamine reaction. Options (A) and (B) would need diazotisation followed by $\ce{CuCl}$ or by warm water.

   ✓Final answer

   $Z$ is phenyl isocyanide, $\ce{C6H5NC}$, so the correct option is (C).

   ANSWER: C

9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.Consider the following carbocations

   [!FORMULA] $\begin{array}{|c|c|c|c|c|} \hline \ce{C6H5CH2^+} & \ce{CH2=CH^+} & \ce{CH3 - C^+ - H} \\ & & | \\ & & \ce{CH3} \\ \hline \text{I} & \text{II} & \text{III} & \ce{CH3 - CH2^+} & \ce{HC \equiv C^+} \\ & & & \text{IV} & \text{V} \\ \hline \end{array}$

   Arrange the above carbocations in the order of decreasing stability (A) I > III > IV > II > V (B) V > II > IV > III > I (C) V > II > III > I > IV (D) I > III > IV > V > II

   ›Reveal solutionSolution

   Carbocation stability increases with greater delocalization/donation of electron density onto the electron-deficient carbon. Ranking these five: I (benzylic, resonance-stabilized) > III (secondary, two alkyl groups) > IV (primary, one alkyl group) > II (vinyl, no effective stabilization) > V (alkynyl, least stable of all). This is option (A).

   Concept and intuition:

   Carbocations are electron-deficient species; anything that spreads out (delocalizes) or donates into the positive charge makes them more stable. The most powerful stabilizer here is resonance (as in the benzyl cation), followed by hyperconjugation and the inductive effect from alkyl groups (more alkyl substitution = more stability). Vinyl and alkynyl carbocations are exceptionally unstable because the positive charge sits on an sp²/sp carbon, which holds electrons more tightly (higher s-character) and offers essentially no resonance stabilization from the adjacent π system, since the empty orbital does not align with it.

   Step-by-step reasoning:

   1. Identify each carbocation's structure and stabilization mode.

      • I: $\ce{C6H5CH2^+}$ — the benzyl carbocation. The positive charge is delocalized into the aromatic ring through resonance (several resonance structures place the charge on the ring carbons). This is the most stable of the five.
      • II: $\ce{CH2=CH^+}$ — the vinyl carbocation. The cationic carbon is essentially sp-hybridized with the empty orbital lying in the plane of the molecule, orthogonal to the π system, so the adjacent double bond cannot donate into it by resonance. Very unstable.
      • III: $\ce{CH3-CH^+-CH3}$ — the isopropyl carbocation. This carbon bears two methyl groups and one hydrogen, so it is a secondary carbocation, stabilized by hyperconjugation and induction from its two alkyl groups.
      • IV: $\ce{CH3-CH2^+}$ — the ethyl carbocation. This carbon bears one methyl group and two hydrogens, so it is a primary carbocation, stabilized by only one alkyl group.
      • V: $\ce{HC\equiv C^+}$ — the ethynyl (alkynyl) carbocation. The cationic carbon is sp-hybridized (50% s-character), holding the bonding electrons closest to the nucleus and offering no resonance or hyperconjugative stabilization. This is the least stable of the five.

   2. Rank by known stability order.

      General order: benzylic/allylic > tertiary > secondary > primary > vinyl > alkynyl. With only benzylic, secondary, primary, vinyl and alkynyl represented here:

      I (benzylic) > III (secondary) > IV (primary) > II (vinyl) > V (alkynyl).

   3. Match with the options.

      • (A) I > III > IV > II > V — matches exactly.
      • (B) V > II > IV > III > I — reverses the order; wrong, since V is the least stable, not the most.
      • (C) V > II > III > I > IV — also has V first; wrong.
      • (D) I > III > IV > V > II — places V ahead of II, but V (alkynyl) is less stable than II (vinyl); wrong.

   Watch out

   A common mistake is to think a triple bond's π electrons can stabilize an adjacent positive charge the way a benzene ring does. In fact, an sp carbon is more electron-withdrawing (more s-character) and destabilizes a cation further rather than stabilizing it. Likewise, the vinyl cation's empty orbital lies in the plane of the molecule, perpendicular to the π bond, so no resonance donation from the double bond is possible.

   Tip

   Remember the order: benzylic/allylic > tertiary > secondary > primary > vinyl > alkynyl — the more s-character at the cationic carbon (and the less alkyl/resonance support it has), the less stable the carbocation.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2023Set ap-2023-05-11-AN1 markMCQ
    Q.In which of the following, ions are correctly arranged with respect to their bond orders? (A) $\mathrm{N_2^{2-} \rightarrow O_2^- \rightarrow O_2^{2-} \rightarrow C_2^{2-}}$ (B) $\mathrm{O_2^{2-} \rightarrow O_2^- \rightarrow N_2^{2-} \rightarrow C_2^{2-}}$ (C) $\mathrm{C_2^{2-} > O_2 > O_2^2 > N_2^2}$ (D) $\mathrm{C_2^{2-} \rightarrow N_2^{2-} \rightarrow O_2^- \rightarrow O_2^{2-}}$

    ›Reveal solutionSolution

    Bond order decreases as antibonding orbitals are progressively filled. Using molecular orbital theory, $\mathrm{C_2^{2-}}$ has bond order 3, $\mathrm{N_2^{2-}}$ has 2, $\mathrm{O_2^-}$ has 1.5, and $\mathrm{O_2^{2-}}$ has 1. The correct arrangement is (D).

    Bond order measures the net number of bonding electron pairs holding two atoms together. In molecular orbital theory, we calculate it as:

    $\text{Bond order} = \frac{1}{2}(\text{electrons in bonding MOs} - \text{electrons in antibonding MOs})$

    Higher bond order means a stronger, shorter bond. When electrons are added to a molecule (forming anions), they occupy the next available orbital according to the aufbau principle. If that orbital is antibonding, the bond order drops.

    Let me work through each ion systematically, counting electrons and applying the MO filling sequence.

    Determining each bond order

    1. $\mathrm{C_2^{2-}}$: 12 + 2 = 14 electrons

    The neutral $\mathrm{C_2}$ molecule has 12 electrons. Adding two more gives 14 total.

    For homonuclear diatomics of C and N (where $2s$-$2p$ mixing is significant), the order is:

    $\sigma_{2s} < \sigma_{2s}^* < \pi_{2p_x} = \pi_{2p_y} < \sigma_{2p_z} < \pi_{2p_x}^* = \pi_{2p_y}^* < \sigma_{2p_z}^*$

    Filling 14 electrons:

    • $\sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^2, \pi_{2p_y}^2, \sigma_{2p_z}^2$

    Bonding: 2 + 4 + 2 = 8

    Antibonding: 2

    Bond order = $\frac{1}{2}(8 - 2) = 3$

    2. $\mathrm{N_2^{2-}}$: 14 + 2 = 16 electrons

    Adding two electrons to the 14 in $\mathrm{N_2}$, they enter the degenerate $\pi_{2p}^*$ orbitals:

    • $\sigma_{2s}^2, \sigma_{2s}^{*2}, \pi_{2p_x}^2, \pi_{2p_y}^2, \sigma_{2p_z}^2, \pi_{2p_x}^{*1}, \pi_{2p_y}^{*1}$

    Bonding: 8

    Antibonding: 2 + 2 = 4

    Bond order = $\frac{1}{2}(8 - 4) = 2$

    3. $\mathrm{O_2^-}$: 16 + 1 = 17 electrons

    For oxygen (no $s$-$p$ mixing), the $\sigma_{2p_z}$ comes before the $\pi_{2p}^*$ orbitals. Neutral $\mathrm{O_2}$ has 16 electrons with configuration ending in $\pi_{2p_x}^{*1}, \pi_{2p_y}^{*1}$ (two unpaired). Adding one electron:

    • Configuration ends: $\pi_{2p_x}^{*2}, \pi_{2p_y}^{*1}$

    Bonding: 8

    Antibonding: 2 + 3 = 5

    Bond order = $\frac{1}{2}(8 - 5) = 1.5$

    4. $\mathrm{O_2^{2-}}$: 16 + 2 = 18 electrons

    Adding two electrons to $\mathrm{O_2}$:

    • Configuration ends: $\pi_{2p_x}^{*2}, \pi_{2p_y}^{*2}$

    Bonding: 8

    Antibonding: 2 + 4 = 6

    Bond order = $\frac{1}{2}(8 - 6) = 1$

    Tip

    For oxygen-family species, each electron added goes into antibonding $\pi^*$ orbitals, decreasing bond order by 0.5 per electron.

    Arranging by bond order

    From highest to lowest:

    $\mathrm{C_2^{2-}}(3) > \mathrm{N_2^{2-}}(2) > \mathrm{O_2^-}(1.5) > \mathrm{O_2^{2-}}(1)$

    This matches option (D) exactly.

    ✓Final answer

    The correct option is (D): $\mathrm{C_2^{2-} \rightarrow N_2^{2-} \rightarrow O_2^- \rightarrow O_2^{2-}}$ with bond orders 3, 2, 1.5, and 1 respectively.

11. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.Match the following List-I (Molecule/ion) A) SnCl$_2$ B) NH$_3$ C) I$_3^-$ D) SO$_3$ List-II (Shape) I. Trigonal planar II. Linear III. Angular IV. Trigonal pyramidal The correct answer is (A) A - III, B - I, C - II, D - IV (B) A - IV, B - III, C - I, D - II (C) A - III, B - I, C - IV, D - II (D) A - III, B - IV, C - II, D - I

    ›Reveal solutionSolution

    The key idea is to determine the molecular shape of each species using VSEPR theory (steric number and lone pairs). The correct matches are: SnCl₂ → Angular, NH₃ → Trigonal pyramidal, I₃⁻ → Linear, SO₃ → Trigonal planar, so the answer is option (D).

    Concept & Intuition

    Molecular geometry is predicted by the Valence Shell Electron Pair Repulsion (VSEPR) theory. The central atom’s steric number (number of bonded atoms + lone pairs) determines the electron-pair geometry, while the number of lone pairs modifies the final molecular shape. For each species, we count valence electrons, assign the central atom, and then apply VSEPR.

    ---

    1. SnCl₂ (A)

       • Tin (Sn) has 4 valence electrons; each Cl contributes 1 for bonding, total = 4 + 2×1 = 6 electrons (3 pairs).
       • Sn is central, forms two single bonds with Cl, leaving one lone pair on Sn.
       • Steric number = 2 bonds + 1 lone pair = 3 → electron-pair geometry is trigonal planar.
       • With one lone pair, the molecular shape is angular (bent, ~120°).
       • So A matches III.

    2. NH₃ (B)

       • Nitrogen has 5 valence electrons; each H contributes 1, total = 5 + 3×1 = 8 electrons (4 pairs).
       • N forms three N–H bonds, leaving one lone pair.
       • Steric number = 3 bonds + 1 lone pair = 4 → electron-pair geometry is tetrahedral.
       • With one lone pair, the molecular shape is trigonal pyramidal.
       • So B matches IV.

    3. I₃⁻ (C)

       • Iodine has 7 valence electrons; three I atoms give 21, plus 1 for the negative charge = 22 electrons (11 pairs).
       • The central I is bonded to two terminal I atoms. The central I has 7 valence electrons, uses 2 for bonds, leaving 5 electrons (2 lone pairs + 1 electron in a 3-center-4-electron bond). Actually, the standard VSEPR treatment: central I has 2 bonds and 3 lone pairs (steric number = 5), but the molecular shape is determined by the arrangement of atoms only.
       • With 2 bonded atoms and 3 lone pairs, the electron-pair geometry is trigonal bipyramidal, but the lone pairs occupy equatorial positions, leaving the two I atoms axial → linear shape.
       • So C matches II.

    4. SO₃ (D)

       • Sulfur has 6 valence electrons; each O contributes 6, but for bonding we consider S as central with three double bonds (or resonance). Total valence = 6 + 3×6 = 24 electrons.
       • S forms three bonds (each double bond counts as one region of electron density) and has no lone pairs.
       • Steric number = 3 bonds + 0 lone pairs = 3 → electron-pair geometry is trigonal planar.
       • With no lone pairs, the molecular shape is trigonal planar.
       • So D matches I.

    ---

    Watch out

    A common mistake is to think NH₃ is “trigonal planar” because it has three bonds, but the lone pair pushes the hydrogens down, making it pyramidal. Also, I₃⁻ is often misidentified as bent — but the three lone pairs on the central iodine force the two iodines into a straight line.

    Tip

    For I₃⁻, remember that the central iodine has 3 lone pairs; the VSEPR “AX₂E₃” designation always gives a linear shape (the two atoms are axial, lone pairs equatorial).

    ---

    Now match the lists:

    A → III, B → IV, C → II, D → I.

    This corresponds to option (D).

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.What is the correct order of melting temperature of C, Si, Ge? (A) C $>$ Ge $>$ Si (B) Si $>$ C $>$ Ge (C) C $>$ Si $>$ Ge (D) Si $>$ Ge $>$ C

    ›Reveal solutionSolution

    Melting points of Group 14 elements decrease down the group due to decreasing bond strength, but carbon is anomalously high because of its small size and strong covalent bonding. The correct order is C > Si > Ge, which corresponds to option (C).

    The key concept here is trends in melting points for Group 14 elements (carbon, silicon, germanium, tin, lead). All these elements form giant covalent (diamond-like) structures in their solid state, except for tin and lead which are metallic. The melting point depends on the strength of the covalent bonds holding the atoms together.

    Why this approach works:

    Melting a solid means breaking the bonds between atoms. For diamond (carbon), the bonds are extremely strong because carbon atoms are very small, allowing close overlap of orbitals and very short, strong bonds. As we go down the group, atoms get larger, bond lengths increase, and bond strength decreases. So melting points should generally decrease. But carbon is an outlier — its melting point is far higher than the others. Silicon and germanium follow the expected decreasing trend.

    ---

    1. Recognize the structure type

       Carbon (as diamond), silicon, and germanium all crystallize in the diamond cubic structure — a giant covalent network. Melting requires breaking many strong covalent bonds simultaneously, so melting points are very high.

    2. Understand the bond strength trend

       Bond strength in a covalent network depends on orbital overlap. Smaller atoms have shorter bonds and better overlap. Carbon’s 2p orbitals are much smaller than silicon’s 3p or germanium’s 4p. Thus, the C–C bond is the strongest, Si–Si is weaker, and Ge–Ge is weaker still.

    3. Apply the trend to the given elements

       • Carbon (diamond): melting point ~3550°C (actually sublimes, but effectively the highest).
       • Silicon: melting point ~1414°C.
       • Germanium: melting point ~938°C. So the order is clearly C > Si > Ge.

    4. Check the options

       (A) C > Ge > Si — incorrect, because Si melts higher than Ge.

       (B) Si > C > Ge — incorrect, because C is far higher than Si.

       (C) C > Si > Ge — correct.

       (D) Si > Ge > C — incorrect, because C is highest.

    Watch out

    A common mistake is to think that since silicon is a semiconductor and carbon is a nonmetal, silicon might have a higher melting point. But the small size of carbon gives it uniquely strong bonds — diamond is the hardest natural material and has the highest melting point of any element.

    Tip

    A quick memory aid: In Group 14, melting points decrease down the group, but carbon is in a league of its own. So the order is always C > Si > Ge > Sn > Pb (though Sn and Pb are metallic and have much lower melting points).

    ✓Final answer

    The correct option is (C).

    ANSWER: C