Q.Isostructural species are those which have the same shape and hybridisation. Among the given species identify the isostructural pairs.

Showing the 12 most recent of 28 on this concept.

1. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.In graphite the C–C bond length with in the layer is X pm and the distance between two adjacent layers is Y pm. X and Y respectively are (A) 340, 141.5 (B) 141.5, 340 (C) 141.5, 154 (D) 143.5, 340

   ›Reveal solutionSolution

   Graphite has strong covalent bonds within layers (short C–C bond length ≈ 141.5 pm) and weak van der Waals forces between layers (large interlayer spacing ≈ 340 pm). The correct pair is (141.5, 340), which is option (B).

   Graphite’s structure is the key: it consists of flat sheets (graphene layers) where each carbon is bonded to three others in a hexagonal honeycomb. The in‑plane C–C bonds are strong and short (like in benzene or graphene), while the layers stack loosely, held only by weak dispersion forces, so the distance between layers is much larger.

   1. Identify the in‑plane bond length (X)

      In a graphene layer, each carbon is sp²‑hybridized, forming σ bonds 120° apart. The C–C bond length in graphite is well‑known to be about 141.5 pm (1.415 Å). This is similar to the bond length in benzene (140 pm) and much shorter than a typical single bond (154 pm). So X = 141.5 pm.

   2. Identify the interlayer distance (Y)

      Adjacent layers are held together by weak van der Waals forces, not covalent bonds. The typical spacing between layers in graphite is about 340 pm (3.4 Å). This is much larger than any covalent bond length, reflecting the weak, non‑bonded interaction.

   3. Match to the options

      • (A) 340, 141.5 → swaps the values (wrong order).
      • (B) 141.5, 340 → correct.
      • (C) 141.5, 154 → 154 pm is a typical C–C single bond length, not the interlayer distance.
      • (D) 143.5, 340 → 143.5 pm is slightly too large for the in‑plane bond.

   Watch out

   A common mistake is to confuse the interlayer distance (≈340 pm) with a bond length, or to think the C–C bond in graphite is as long as a diamond single bond (154 pm). Remember: graphite’s in‑plane bonds have partial double‑bond character, making them shorter.

   Tip

   You can remember the numbers by noting that 141.5 pm is roughly 1.4 Å (the “aromatic” bond length) and 340 pm is about 3.4 Å (the classic van der Waals gap in layered materials).

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.XeF₄ on reaction with O₂F₂ at 143 K gives a Xenon compound A. This on complete hydrolysis gives HF and B. The hybridisation of central atom in B is (A) $sp^3$ (B) $sp^2$ (C) $sp^3d$ (D) $sp^3d^2$

   ›Reveal solutionSolution

   The reaction of XeF₄ with O₂F₂ yields XeF₆ (compound A), which hydrolyzes to XeO₃ (compound B). The central Xe in XeO₃ has $sp^3$ hybridization, so the correct option is (A).

   Concept & Intuition

   This problem tests your understanding of noble gas chemistry and the link between molecular structure and hybridization. The key is to track the xenon atom through two reactions: first, an oxidation/fluorination step that increases the number of fluorine atoms around xenon; second, a hydrolysis that replaces fluorine with oxygen. The final compound’s geometry (and thus hybridization) is determined by counting electron domains around xenon using VSEPR theory.

   Step-by-step reasoning

   1. Identify the first reaction product (compound A). XeF₄ reacts with O₂F₂ at 143 K. O₂F₂ is a strong fluorinating agent (it contains an O–O bond and readily provides fluorine radicals). The reaction is known to produce XeF₆:

   $\text{XeF}_4 + \text{O}_2\text{F}_2 \rightarrow \text{XeF}_6 + \text{O}_2$

   So compound A is XeF₆.

   2. Determine the hydrolysis product (compound B). Complete hydrolysis of XeF₆ gives HF and a xenon oxide. The reaction is:

   $\text{XeF}_6 + 3\text{H}_2\text{O} \rightarrow \text{XeO}_3 + 6\text{HF}$

   Thus compound B is XeO₃ (xenon trioxide).

   3. Find the hybridization of the central atom in XeO₃.

      • Count valence electrons: Xe has 8, each O contributes 0 (as a ligand), and the molecule is neutral.
      • Draw the Lewis structure: Xe forms three double bonds with three oxygen atoms (Xe=O). This uses 6 electrons. Xenon also has one lone pair (the remaining 2 electrons).
      • Electron domain count: 3 double bonds + 1 lone pair = 4 electron domains around Xe.
      • According to VSEPR, 4 domains give a tetrahedral electron‑pair geometry. The lone pair occupies one vertex, so the molecular shape is trigonal pyramidal.
      • Hybridization for 4 domains is $sp^3$.

   4. Confirm with known facts.

      XeO₃ is a well‑known explosive solid with a trigonal pyramidal structure (bond angles ~103°). This matches $sp^3$ hybridization.

   Watch out

   A common mistake is to think that because Xe has expanded octet in XeF₆ ($sp^3d^2$), the hydrolysis product also uses d‑orbitals. But in XeO₃, the lone pair and three double bonds give only four domains — no d‑orbitals are needed.

   Tip

   For noble gas compounds, always track the number of electron domains after hydrolysis. Replacing F with O often reduces the domain count because oxygen forms multiple bonds.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

3. TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
   Q.Observe the following reactions XeF$_6$ + H$_2$O $\rightarrow$ A + 2HF XeF$_6$ + 2H$_2$O $\rightarrow$ B + 4HF Hybridization of central atom in A, B respectively is (A) $sp^3d,\, sp^3d$ (B) $sp^3d^2,\, sp^3d$ (C) $sp^3d,\, sp^3$ (D) $sp^3d^2,\, sp^3$

   ›Reveal solutionSolution

   The hydrolysis of XeF$_6$ with one mole of water yields XeOF$_4$ (A) with $sp^3d^2$ hybridization, while hydrolysis with two moles of water yields XeO$_2$F$_2$ (B) with $sp^3d$ hybridization. The correct option is (B).

   The problem asks us to determine the hybridization of the central xenon atom in the products formed from the partial hydrolysis of XeF$_6$. Xenon hexafluoride (XeF$_6$) undergoes hydrolysis in a stepwise manner, where fluorine atoms are progressively replaced by oxygen atoms from water molecules. The extent of hydrolysis depends on the stoichiometry of the reaction, specifically the molar ratio of XeF$_6$ to H$_2$O.

   To determine the hybridization of the central atom (Xenon) in the products, we will use the VSEPR theory, which relies on calculating the steric number. The steric number is the sum of the number of sigma bonds formed by the central atom and the number of lone pairs on the central atom.

   Steric Number = (Number of sigma bonds) + (Number of lone pairs)

   Based on the steric number, the hybridization is:

   • 2: $sp$
   • 3: $sp^2$
   • 4: $sp^3$
   • 5: $sp^3d$
   • 6: $sp^3d^2$
   • 7: $sp^3d^3$

   Let's break down the reactions and determine the hybridization for each product.

   1. Identify Product A and its Hybridization:

      The first reaction given is:

      $\text{XeF}_6 + \text{H}_2\text{O} \rightarrow \text{A} + 2\text{HF}$

      This is a partial hydrolysis reaction where one molecule of water reacts with XeF$_6$. In this process, one oxygen atom from water replaces two fluorine atoms from XeF$_6$.

      Therefore, product A is $\text{XeOF}_4$.

      Now, let's determine the hybridization of Xe in $\text{XeOF}_4$:

      • Central atom: Xenon (Xe)
      • Valence electrons of Xe: 8
      • Atoms bonded to Xe: 4 Fluorine atoms and 1 Oxygen atom.
      • Bonds formed: Each F forms a single bond (1 electron from Xe), and O forms a double bond (2 electrons from Xe).
      • Electrons used in bonding: $(4 \times 1) + (1 \times 2) = 4 + 2 = 6$ electrons.
      • Remaining valence electrons: $8 - 6 = 2$ electrons.
      • Number of lone pairs: $2 / 2 = 1$ lone pair.
      • Number of sigma bonds: 4 (to F atoms) + 1 (to O atom) = 5 sigma bonds.
      • Steric number: (Number of sigma bonds) + (Number of lone pairs) = $5 + 1 = 6$.
      • A steric number of 6 corresponds to $\mathbf{sp^3d^2}$ hybridization.
      • The geometry of XeOF$_4$ is square pyramidal (due to one lone pair distorting the octahedral arrangement).

   2. Identify Product B and its Hybridization:

      The second reaction given is:

      $\text{XeF}_6 + 2\text{H}_2\text{O} \rightarrow \text{B} + 4\text{HF}$

      This is a further hydrolysis reaction where two molecules of water react with XeF$_6$. Here, two oxygen atoms from water replace four fluorine atoms from XeF$_6$.

      Therefore, product B is $\text{XeO}_2\text{F}_2$.

      Now, let's determine the hybridization of Xe in $\text{XeO}_2\text{F}_2$:

      • Central atom: Xenon (Xe)
      • Valence electrons of Xe: 8
      • Atoms bonded to Xe: 2 Fluorine atoms and 2 Oxygen atoms.
      • Bonds formed: Each F forms a single bond (1 electron from Xe), and each O forms a double bond (2 electrons from Xe).
      • Electrons used in bonding: $(2 \times 1) + (2 \times 2) = 2 + 4 = 6$ electrons.
      • Remaining valence electrons: $8 - 6 = 2$ electrons.
      • Number of lone pairs: $2 / 2 = 1$ lone pair.
      • Number of sigma bonds: 2 (to F atoms) + 2 (to O atoms) = 4 sigma bonds.
      • Steric number: (Number of sigma bonds) + (Number of lone pairs) = $4 + 1 = 5$.
      • A steric number of 5 corresponds to $\mathbf{sp^3d}$ hybridization.
      • The geometry of XeO$_2$F$_2$ is see-saw (due to one lone pair in a trigonal bipyramidal arrangement).

   3. Compare with Options:

      • Hybridization of central atom in A ($\text{XeOF}_4$): $sp^3d^2$
      • Hybridization of central atom in B ($\text{XeO}_2\text{F}_2$): $sp^3d$

      Comparing these with the given options:

      (A) $sp^3d,\, sp^3d$

      (B) $sp^3d^2,\, sp^3d$

      (C) $sp^3d,\, sp^3$

      (D) $sp^3d^2,\, sp^3$

      The calculated hybridizations match option (B).

   ✓Final answer

   The hybridization of the central atom in A ($\text{XeOF}_4$) is $sp^3d^2$, and in B ($\text{XeO}_2\text{F}_2$) is $sp^3d$. Thus, the correct option is (B).

4. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Identify the reaction in which the hybridization of the underlined atom is changed (A) $\mathrm{NH_3(g) + H_2O(l) \rightarrow}$ (B) $\mathrm{POCl_3 + 3H_2O \rightarrow}$ (C) $\mathrm{SO_2(g) + H_2O(l) \rightarrow}$ (D) $\mathrm{PCl_3 + 3H_2O \rightarrow}$

   ›Reveal solutionSolution

   Only in $\text{SO}_2+\text{H}_2\text{O}$ does the central atom change hybridisation, going from $sp^2$ in $\text{SO}_2$ to $sp^3$ in $\text{H}_2\text{SO}_3$ — option (C).

   Check the hybridisation of the central atom on each side of every reaction.

   • (A) $\text{NH}_3\to\text{NH}_4^+$ (with water): N is $sp^3$ in $\text{NH}_3$ and remains $sp^3$ in $\text{NH}_4^+$. No change.
   • (B) $\text{POCl}_3\to\text{H}_3\text{PO}_4$: P is $sp^3$ in $\text{POCl}_3$ and $sp^3$ in $\text{H}_3\text{PO}_4$. No change.
   • (C) $\text{SO}_2\to\text{H}_2\text{SO}_3$: S is $sp^2$ in $\text{SO}_2$ (two bond domains + one lone pair, bent, 3 electron domains) but becomes $sp^3$ in $\text{H}_2\text{SO}_3$ (four electron domains around S). Hybridisation changes $sp^2\!\to\!sp^3$.
   • (D) $\text{PCl}_3\to\text{H}_3\text{PO}_3$: P is $sp^3$ in $\text{PCl}_3$ and $sp^3$ in $\text{H}_3\text{PO}_3$. No change.

   ✓Final answer

   The hybridisation of the central atom changes only in $\text{SO}_2(g)+\text{H}_2\text{O}(l)$ ($sp^2\to sp^3$) — option (C).

5. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.The substituent ‘X’ increases electron density in benzene ring by hyperconjugation effect and substituent ‘Y’ decreases electron density in benzene by resonance effect. ‘X’ and ‘Y’ respectively are (A) $-\mathrm{CH}_3, -\mathrm{NO}_2$ (B) $-\mathrm{OCH}_3, -\mathrm{COCH}_3$ (C) $-\mathrm{OH}, -\mathrm{NO}_2$ (D) $-\mathrm{CH}_3, -\mathrm{NHCH}_3$

   ›Reveal solutionSolution

   The key is to identify one substituent that donates electrons via hyperconjugation (only alkyl groups with $\alpha$-H do this) and another that withdraws electrons via resonance (groups with a $\pi$-system that pulls electron density). The correct pair is $-\mathrm{CH}_3$ (hyperconjugation donor) and $-\mathrm{NO}_2$ (resonance withdrawer), which matches option (A).

   The question tests your ability to distinguish between two common electronic effects: hyperconjugation and resonance. Hyperconjugation is a sigma-bond donation — it requires a $\mathrm{C-H}$ bond adjacent to the ring to overlap with the $\pi$-system. Only alkyl groups like $-\mathrm{CH}_3$ can do this. Resonance, on the other hand, involves pi-electron delocalization; a group that withdraws by resonance must have an electronegative atom or a $\pi$-bond that can pull electron density away from the ring. $-\mathrm{NO}_2$ is the classic example.

   Let’s check each option carefully.

   1. Option (A): $-\mathrm{CH}_3, -\mathrm{NO}_2$

      $-\mathrm{CH}_3$ has three $\mathrm{C-H}$ bonds whose $\sigma$ electrons can delocalize into the ring — this is hyperconjugation, which increases electron density. $-\mathrm{NO}_2$ has a nitrogen with a positive formal charge and two oxygen atoms; its $\pi$-system pulls electrons from the ring via resonance, decreasing electron density. This fits perfectly.

   2. Option (B): $-\mathrm{OCH}_3, -\mathrm{COCH}_3$

      $-\mathrm{OCH}_3$ donates electrons by resonance (lone pairs on oxygen), not by hyperconjugation. $-\mathrm{COCH}_3$ withdraws by resonance (carbonyl group), so the second part is correct, but the first is not.

   3. Option (C): $-\mathrm{OH}, -\mathrm{NO}_2$

      $-\mathrm{OH}$ donates by resonance (lone pairs), not hyperconjugation. $-\mathrm{NO}_2$ is correct for the second, but again the first fails.

   4. Option (D): $-\mathrm{CH}_3, -\mathrm{NHCH}_3$

      $-\mathrm{CH}_3$ is correct for hyperconjugation, but $-\mathrm{NHCH}_3$ donates electrons by resonance (lone pair on nitrogen), not withdraws. So the second part is wrong.

   Watch out

   A common mistake is to think that any group with lone pairs (like $-\mathrm{OH}$ or $-\mathrm{OCH}_3$) donates by hyperconjugation. They don’t — they donate by resonance. Hyperconjugation is exclusive to alkyl groups with at least one $\alpha$-hydrogen.

   Tip

   To quickly recall: hyperconjugation = alkyl groups only (e.g., $-\mathrm{CH}_3$, $-\mathrm{C}_2\mathrm{H}_5$). Resonance donation = groups with lone pairs on atoms directly attached to the ring (e.g., $-\mathrm{OH}$, $-\mathrm{NH}_2$, $-\mathrm{OCH}_3$). Resonance withdrawal = groups with a $\pi$-bond to an electronegative atom (e.g., $-\mathrm{NO}_2$, $-\mathrm{CN}$, $-\mathrm{CHO}$, $-\mathrm{COOH}$).

   ✓Final answer

   The correct option is (A).

6. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Dehydration of an organic acid X with concentrated H$_2$SO$_4$ at 373K gives H$_2$O and gas Y. The hybridisation of the carbon in Y and nature of Y are respectively (A) sp$^2$, Neutral (B) sp, Neutral (C) sp$^2$, acidic (D) sp, acidic

   ›Reveal solutionSolution

   The dehydration of an organic acid with concentrated H₂SO₄ at 373 K typically produces carbon monoxide (CO), where the carbon is sp‑hybridized and the gas is neutral. The correct option is (B).

   Concept & Intuition

   When an organic acid (like formic acid, HCOOH) is heated with concentrated sulfuric acid, it undergoes dehydration — the acid loses a water molecule. The sulfuric acid acts as a powerful dehydrating agent. The product gas is not CO₂ (which would be acidic) but CO, which is neutral and has a triple bond between carbon and oxygen. The carbon in CO is sp‑hybridized because it forms two sigma bonds (one to oxygen, one lone pair) and two pi bonds, giving a linear geometry.

   Step‑by‑Step Reasoning

   1. Identify the reaction type The problem says “dehydration of an organic acid X with conc. H₂SO₄ at 373 K”. The classic example is formic acid (HCOOH):

   $\text{HCOOH} \xrightarrow{\text{conc. H}_2\text{SO}_4,\ 373\text{ K}} \text{H}_2\text{O} + \text{CO}$

   This is a standard laboratory preparation of carbon monoxide.

   2. Determine the gas Y

      The gas produced is carbon monoxide (CO). It is not CO₂ because dehydration of a carboxylic acid at this temperature (with conc. H₂SO₄) removes water from the –COOH group, leaving CO. (Oxalic acid would give CO₂ + CO, but a simple mono‑carboxylic acid gives CO.)

   3. Hybridisation of carbon in CO

      In CO, the carbon is bonded to oxygen by a triple bond (one σ, two π). The carbon has two regions of electron density: the triple bond counts as one region, and the lone pair on carbon counts as another. Two regions → sp hybridisation.

   $\text{Hybridisation: sp}$

   4. Nature of CO (acidic or neutral?)

      CO is a neutral gas. It does not turn litmus red (unlike CO₂, which forms carbonic acid in water). It is neither acidic nor basic — it is a neutral oxide.

   5. Match with options

      • (A) sp², Neutral → wrong hybridisation
      • (B) sp, Neutral → matches
      • (C) sp², acidic → wrong both
      • (D) sp, acidic → wrong nature

   Watch out

   A common mistake is to think the gas is CO₂ (which is acidic and has sp² carbon). But dehydration of a simple organic acid like formic acid gives CO, not CO₂. CO₂ would come from decarboxylation (different conditions).

   Tip

   Remember: “Formic acid + conc. H₂SO₄ → CO + H₂O” is a classic. The carbon in CO is sp‑hybridized (linear, triple bond). CO is neutral — it does not react with water to form an acid.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Choose the correct statements about allotropes of carbon I. Graphite has layered structure II. Buckminster fullerene is not aromatic in nature III. The distance between two adjacent layers in graphite is 141.5 pm IV. The hybridization of carbons in graphite and Buckminster fullerene is same (A) I & IV (B) I & II (C) II & III (D) III & IV

   ›Reveal solutionSolution

   Graphite has a layered structure (I true), buckminsterfullerene is aromatic (II false), the interlayer distance in graphite is 335 pm not 141.5 pm (III false), and both allotropes use sp² hybridization (IV true). So statements I and IV are correct.

   Concept & Intuition

   Carbon allotropes differ in bonding geometry. Graphite’s layers are held by weak forces, while buckminsterfullerene (C₆₀) is a closed cage with delocalized π‑electrons — that makes it aromatic. The key is to recall the actual interlayer spacing in graphite and the hybridization in each structure.

   Step‑by‑step reasoning

   1. Statement I: Graphite has a layered structure

      Graphite consists of flat sheets of carbon atoms arranged in hexagonal rings. Within each sheet, strong covalent bonds hold atoms; between sheets, only weak van der Waals forces exist. This layered structure is why graphite is a lubricant and conducts electricity only along the sheets.

      → True

   2. Statement II: Buckminster fullerene is not aromatic in nature

      C₆₀ has 60 carbon atoms, each sp² hybridized, forming a soccer‑ball shape. The π‑electrons are delocalized over the entire surface, satisfying Hückel’s rule for spherical aromaticity (2(N+1)² π‑electrons with N = 2 gives 18 π‑electrons, but C₆₀ actually has 60 π‑electrons; however, it is considered aromatic due to three‑dimensional delocalization). Chemically, it undergoes addition reactions typical of aromatic systems.

      → False (it is aromatic)

   3. Statement III: The distance between two adjacent layers in graphite is 141.5 pm

      The in‑plane C–C bond length in graphite is about 141.5 pm. The distance between layers (interlayer spacing) is much larger, approximately 335 pm. The given value confuses the bond length with the interlayer distance.

      → False

   4. Statement IV: The hybridization of carbons in graphite and Buckminster fullerene is same

      In graphite, each carbon is sp² hybridized (three σ‑bonds, one p‑orbital for π‑bonding). In C₆₀, each carbon is also sp² hybridized (three σ‑bonds, one p‑orbital for delocalized π‑system). The geometry is trigonal planar in both, though curved in fullerene.

      → True

   Watch out

   A common mistake is to confuse the in‑plane C–C bond length (141.5 pm) with the interlayer distance (335 pm). Always check units and context.

   Tip

   Remember: sp² hybridization in carbon always gives a planar or nearly planar arrangement with one p‑orbital for π‑bonding — true for both graphite and fullerenes.

   Conclusion: Statements I and IV are correct.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Choose the correct statements about allotropes of carbon I. Graphite has layered structure II. Buckminster fullerene is not aromatic in nature III. The distance between two adjacent layers in graphite is 141.5 pm IV. The hybridization of carbons in graphite and Buckminster fullerene is same (A) I & II (B) I & IV (C) II & III (D) III & IV

   ›Reveal solutionSolution

   Graphite has a layered structure (true), Buckminster fullerene is aromatic (false), the interlayer distance in graphite is 335 pm (not 141.5 pm), and both graphite and fullerene use sp² hybridization (true). So only statements I and IV are correct.

   Concept & Intuition

   Allotropes of carbon differ in how carbon atoms bond and arrange themselves. Graphite forms flat sheets of sp²-hybridized carbons in a hexagonal lattice, with weak forces between layers. Buckminster fullerene (C₆₀) is a soccer-ball-shaped molecule where each carbon is also sp²-hybridized, but the curved surface creates a delocalized π-system that is aromatic. The key is to recall the exact interlayer spacing in graphite (not the in-plane bond length) and the hybridization common to both.

   Step-by-step reasoning

   1. Statement I: Graphite has a layered structure

      Graphite consists of stacked, two-dimensional sheets of carbon atoms arranged in a honeycomb lattice. Within each layer, strong covalent bonds hold atoms together; between layers, only weak van der Waals forces exist. This is a defining property of graphite.

      → True

   2. Statement II: Buckminster fullerene is not aromatic in nature

      C₆₀ has 60 π-electrons, which satisfies Hückel’s rule for spherical aromaticity (2(N+1)² for N=2 gives 18, but C₆₀’s π-system is delocalized over the entire sphere and exhibits aromatic character). It undergoes reactions typical of aromatic compounds and is considered aromatic.

      → False (it is aromatic)

   3. Statement III: The distance between two adjacent layers in graphite is 141.5 pm

      The in-plane C–C bond length in graphite is about 141.5 pm, but the distance between layers (interlayer spacing) is much larger — approximately 335 pm (3.35 Å). The given value confuses the bond length with the interlayer distance.

      → False

   4. Statement IV: The hybridization of carbons in graphite and Buckminster fullerene is same

      In graphite, each carbon is sp²-hybridized (three σ bonds, one p orbital for π bonding). In C₆₀, each carbon also forms three σ bonds (two single bonds and one double bond in the alternating pattern) and uses sp² hybridization. The curvature does not change the hybridization.

      → True

   Since only I and IV are correct, the answer is option (B).

   Watch out

   A common mistake is to confuse the in-plane C–C bond length (~141.5 pm) with the interlayer spacing (~335 pm). Also, some think fullerene is non-aromatic because of its curved shape, but it is indeed aromatic.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The number of lone pairs of electrons on the central atom of XeO$_3$, XeOF$_4$ and XeF$_6$ respectively is (A) 3, 2, 1 (B) 2, 1, 0 (C) 1, 2, 1 (D) 1, 1, 1

   ›Reveal solutionSolution

   The number of lone pairs on the central Xe atom is determined by counting total valence electrons, subtracting those used in bonding and in the terminal atoms' own lone pairs, and dividing the remainder by two. For XeO$_3$, XeOF$_4$, and XeF$_6$, the lone pairs are 1, 1, and 1 respectively — option (D).

   The key to this problem is understanding that xenon is a noble gas with 8 valence electrons. In its compounds, it forms bonds by expanding its octet, using d-orbitals. The number of lone pairs on Xe is simply the leftover electrons after accounting for all bonds and the terminal atoms' own octets.

   Let's work through each molecule step by step.

   1. XeO$_3$

      Xenon has 8 valence electrons; each oxygen contributes 6, and there are 3 oxygens.

      Total valence electrons available: $8 + 3 \times 6 = 26$.

      In the Lewis structure, each Xe=O double bond uses 4 electrons, so the three double bonds use $3 \times 4 = 12$ electrons. Each double-bonded oxygen already has 4 electrons from the bond and needs 4 more to complete its octet — that's 2 lone pairs (4 electrons) per oxygen, or $3 \times 4 = 12$ electrons total for the three oxygens. Adding the bonding electrons and the oxygens' own lone-pair electrons: $12 + 12 = 24$. The remaining $26 - 24 = 2$ electrons go on xenon as 1 lone pair.

   2. XeOF$_4$

      Xenon: 8 valence electrons. Oxygen: 6. Each fluorine: 7, and there are 4 fluorines.

      Total valence electrons: $8 + 6 + 4 \times 7 = 8 + 6 + 28 = 42$.

      The structure: Xe forms a double bond with O (4 electrons) and single bonds with each F (2 electrons each, total 8). So bonding uses $4 + 8 = 12$ electrons.

      Each fluorine gets 3 lone pairs (6 electrons each, total $4 \times 6 = 24$). Oxygen gets 2 lone pairs (4 electrons). That accounts for $24 + 4 = 28$ electrons in lone pairs on terminal atoms.

      Total electrons used so far: $12 + 28 = 40$. Remaining: $42 - 40 = 2$ electrons → 1 lone pair on Xe.

   3. XeF$_6$

      Xenon: 8. Each fluorine: 7, total $6 \times 7 = 42$.

      Total: $8 + 42 = 50$ electrons.

      Six Xe–F single bonds use $6 \times 2 = 12$ electrons.

      Each fluorine gets 3 lone pairs (6 electrons each), total $6 \times 6 = 36$ electrons.

      Used: $12 + 36 = 48$. Remaining: $50 - 48 = 2$ electrons → 1 lone pair on Xe.

   Watch out

   A common mistake is to forget that oxygen in XeO$_3$ forms double bonds, not single bonds. If you treat it as single-bonded, you'll get 3 lone pairs on Xe — which is wrong because oxygen would then have a formal charge of -1, and the molecule would not be neutral.

   Thus, all three molecules have exactly one lone pair on the central xenon atom.

   ✓Final answer

   The correct option is (D) 1, 1, 1.

10. TG EAPCET 2025Set ap-2025-04-29-AN1 markMCQ
    Q.The pair of molecules having same number of lone pair of electrons on central atom is (A) $\mathrm{SF_4},\ \mathrm{XeF_4}$ (B) $\mathrm{ClF_3},\ \mathrm{BrF_5}$ (C) $\mathrm{ClF_3},\ \mathrm{XeF_4}$ (D) $\mathrm{SF_4},\ \mathrm{ClF_3}$

    ›Reveal solutionSolution

    The key is to count the lone pairs on the central atom using the VSEPR formula (total valence electrons minus bonding electrons, divided by 2). The pair with the same lone-pair count is $\mathrm{ClF_3}$ and $\mathrm{XeF_4}$, each having 2 lone pairs on the central atom.

    The question asks which two molecules have the same number of lone pairs on their central atom. This is a classic VSEPR (Valence Shell Electron Pair Repulsion) problem. The central idea is simple: count the total valence electrons around the central atom, subtract the electrons used in bonding (each bond uses 2 electrons), and the remainder, divided by 2, gives the number of lone pairs.

    Let’s work through each option systematically.

    1. Option (A): $\mathrm{SF_4}$ and $\mathrm{XeF_4}$

       • For $\mathrm{SF_4}$: Sulfur has 6 valence electrons. It forms 4 S–F bonds, using $4 \times 2 = 8$ electrons. But wait — sulfur only has 6 valence electrons, so it uses 4 of its own for bonding and gets 4 more from fluorine atoms? Actually, the correct method: total valence electrons = $6 + 4 \times 7 = 34$. Bonding pairs = 4 bonds × 2 electrons = 8 electrons used in bonds. Remaining electrons = $34 - 8 = 26$. These are distributed as lone pairs on the central atom and on fluorines. Each fluorine gets 3 lone pairs (6 electrons), so 4 fluorines take $4 \times 6 = 24$ electrons. That leaves $26 - 24 = 2$ electrons on sulfur, which is 1 lone pair.
       • For $\mathrm{XeF_4}$: Xenon has 8 valence electrons. Total valence = $8 + 4 \times 7 = 36$. Bonding uses 8 electrons. Remaining = 28. Each fluorine takes 6 electrons (3 lone pairs), so 4 fluorines take 24. Leftover = $28 - 24 = 4$ electrons on xenon, which is 2 lone pairs.
       • So $\mathrm{SF_4}$ has 1 lone pair, $\mathrm{XeF_4}$ has 2 — not the same.

    2. Option (B): $\mathrm{ClF_3}$ and $\mathrm{BrF_5}$

       • $\mathrm{ClF_3}$: Chlorine has 7 valence electrons. Total valence = $7 + 3 \times 7 = 28$. Bonding uses 6 electrons. Remaining = 22. Each fluorine takes 6 electrons, so 3 fluorines take 18. Leftover = $22 - 18 = 4$ electrons on chlorine, which is 2 lone pairs.
       • $\mathrm{BrF_5}$: Bromine has 7 valence electrons. Total valence = $7 + 5 \times 7 = 42$. Bonding uses 10 electrons. Remaining = 32. Each fluorine takes 6 electrons, so 5 fluorines take 30. Leftover = $32 - 30 = 2$ electrons on bromine, which is 1 lone pair.
       • Not the same (2 vs. 1).

    3. Option (C): $\mathrm{ClF_3}$ and $\mathrm{XeF_4}$

       • We already found $\mathrm{ClF_3}$ has 2 lone pairs and $\mathrm{XeF_4}$ has 2 lone pairs.
       • So this pair matches.

    4. Option (D): $\mathrm{SF_4}$ and $\mathrm{ClF_3}$

       • $\mathrm{SF_4}$ has 1 lone pair, $\mathrm{ClF_3}$ has 2 — not the same.

    Tip

    A faster way: use the formula $\text{Lone pairs} = \frac{1}{2} \left( V - 8n \right)$, where $V$ is total valence electrons and $n$ is the number of atoms bonded to the central atom (assuming each bond uses 2 electrons and each terminal atom gets an octet). For $\mathrm{ClF_3}$: $V = 28$, $n = 3$, so lone pairs = $\frac{1}{2}(28 - 24) = 2$. For $\mathrm{XeF_4}$: $V = 36$, $n = 4$, so lone pairs = $\frac{1}{2}(36 - 32) = 2$. Works like a charm.

    Watch out

    A common mistake is to forget that terminal atoms (like F) also have lone pairs. The leftover electrons after bonding are not all on the central atom — you must account for the octet on each terminal atom first.

    ✓Final answer

    The correct option is (C).

11. TG EAPCET 2025Set ap-2025-04-29-FN1 markMCQ
    Q.Which of the following is used as stabilizer for H$_2$O$_2$? (A) CO(CH$_3$)$_2$ (B) H$_2$NCONH$_2$ (C) MnO$_2$ (D) CH$_3$CHO

    ›Reveal solutionSolution

    Hydrogen peroxide decomposes in the presence of catalysts or impurities; stabilizers slow this decomposition. The correct stabilizer among the options is acetanilide-like urea (H₂NCONH₂), which acts as a negative catalyst. The answer is (B).

    Concept & Intuition

    Hydrogen peroxide (H₂O₂) is inherently unstable — it slowly decomposes into water and oxygen:

    $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$

    This decomposition is accelerated by heat, light, and especially by catalysts like metal ions (e.g., Fe³⁺, Cu²⁺) or solid surfaces (e.g., MnO₂). To make H₂O₂ safe for storage and transport, we add stabilizers — substances that inhibit decomposition. A stabilizer works by either:

    • Chelating (binding) trace metal ions that catalyze decomposition, or
    • Acting as a negative catalyst (inhibitor) that slows the reaction.

    Now, let’s examine each option:

    1. Option (A): CO(CH₃)₂ (acetone)

       Acetone is a common organic solvent. It does not chelate metal ions effectively nor does it inhibit H₂O₂ decomposition. In fact, acetone can react with H₂O₂ under certain conditions to form explosive peroxides (e.g., acetone peroxide). So it is not a stabilizer — it’s a hazard.

    2. Option (B): H₂NCONH₂ (urea)

       Urea is a well-known stabilizer for hydrogen peroxide. It works by forming weak hydrogen bonds with H₂O₂ molecules and by sequestering trace metal impurities. Urea is often added to commercial H₂O₂ solutions to slow decomposition. This is the correct choice.

    3. Option (C): MnO₂ (manganese dioxide)

       MnO₂ is a catalyst for H₂O₂ decomposition — it speeds up the reaction dramatically (you see rapid bubbling of oxygen). It is the opposite of a stabilizer. A classic demonstration: dropping MnO₂ into H₂O₂ produces a vigorous fizz.

    4. Option (D): CH₃CHO (acetaldehyde)

       Acetaldehyde can react with H₂O₂ to form peracetic acid (an oxidizer), but it does not stabilize H₂O₂. Like acetone, it may even promote decomposition or side reactions. Not a stabilizer.

    Watch out

    A common mistake is to think that any organic compound will stabilize H₂O₂. In reality, many organics (like aldehydes and ketones) react with H₂O₂, while solids like MnO₂ catalyze its decomposition. Only specific inhibitors like urea or phosphoric acid are used.

    Tip

    Urea is cheap, non-toxic, and highly effective. It’s also used in some hair-lightening products to stabilize the peroxide. Remember: stabilizers = negative catalysts.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
    Q.Identify the number of molecules having permanent dipole moment from the following $\mathrm{CCl_4, NF_3, H_2S, HBr, SF_4, SiF_4, XeF_4, BeCl_2, SnCl_2, BrF_5, SO_2}$ (A) $5$ (B) $7$ (C) $4$ (D) $6$

    ›Reveal solutionSolution

    Seven of the eleven molecules are polar: $\text{NF}_3,\ \text{H}_2\text{S},\ \text{HBr},\ \text{SF}_4,\ \text{SnCl}_2,\ \text{BrF}_5,\ \text{SO}_2$.

    A molecule has a permanent dipole moment only if its bond dipoles do not cancel by symmetry:

    • $\text{CCl}_4$ — regular tetrahedral, symmetric → non-polar
    • $\text{NF}_3$ — pyramidal (lone pair) → polar ✓
    • $\text{H}_2\text{S}$ — bent → polar ✓
    • $\text{HBr}$ — polar diatomic → polar ✓
    • $\text{SF}_4$ — see-saw (lone pair) → polar ✓
    • $\text{SiF}_4$ — regular tetrahedral, symmetric → non-polar
    • $\text{XeF}_4$ — square planar, symmetric → non-polar
    • $\text{BeCl}_2$ — linear, symmetric → non-polar
    • $\text{SnCl}_2$ — bent (lone pair) → polar ✓
    • $\text{BrF}_5$ — square pyramidal (lone pair) → polar ✓
    • $\text{SO}_2$ — bent → polar ✓

    Counting the polar molecules gives $7$.

    ✓Final answer

    Number of molecules with a permanent dipole moment $= 7$ → option (B).