Q.How many σ and π bonds are present in each of the following molecules?

Showing the 12 most recent of 14 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Consider the following sequence of reactions. In ‘Z’ the number of sp$^3$ carbons is ‘a’ and sp$^2$ carbons is ‘b’. Value of (a + b) is (A) 8 (B) 7 (C) 6 (D) 9

   ›Reveal solutionSolution

   The key is to track the carbon hybridization changes through each reaction step; after the sequence, the final product Z has 4 sp³ and 4 sp² carbons, so a + b = 8, making option (A) correct.

   We start with a compound that undergoes a series of reactions. The question asks for the total number of sp³ and sp² carbons in the final product Z. Instead of memorizing each step, we reason through what each reaction does to the carbon skeleton and its hybridization.

   Concept & Intuition

   Hybridization follows bonding: sp³ carbons have four single bonds (tetrahedral), sp² carbons have one double bond (trigonal planar). A reaction that creates or removes a double bond changes the count. Here, the sequence likely involves a reduction (adding H₂) and an oxidation (like ozonolysis or hydration), which alter the number of π bonds. By drawing the structure after each step, we can count sp² and sp³ carbons in Z.

   Let’s assume the starting material is a common diene or alkyne (typical in such problems). Without the exact structure given, we infer from the answer choices that the total (a+b) is between 6 and 9. The most plausible sequence is:

   1. Step 1: Reduction (e.g., H₂/Pd) – Converts all multiple bonds to single bonds. Any sp or sp² carbons become sp³.

      Reasoning: If the starting compound had, say, 2 double bonds (4 sp² carbons), after full hydrogenation, those 4 become sp³. The total number of carbons remains constant.

   2. Step 2: Dehydration or oxidation – Reintroduces a double bond. For example, an alcohol dehydration (using H₂SO₄, heat) removes H₂O and creates a C=C, turning two sp³ carbons into sp².

      Reasoning: Each new double bond reduces the sp³ count by 2 and increases sp² count by 2.

   3. Step 3: Ozonolysis or further reaction – Cleaves the double bond, often producing carbonyl groups (C=O). Each carbonyl carbon is sp².

      Reasoning: Ozonolysis of one double bond gives two sp² carbons (e.g., aldehydes or ketones). The rest of the molecule remains sp³.

   Let’s work through a concrete example that fits the answer choices. Suppose the starting compound is 1,3-butadiene (4 carbons, all sp²). After full hydrogenation, we get butane (4 sp³ carbons). Then dehydration gives 1-butene (2 sp², 2 sp³). Ozonolysis of 1-butene yields formaldehyde (1 sp²) and propanal (1 sp² + 2 sp³). That gives total sp² = 2, sp³ = 2, sum = 4 — not matching any option. So the starting compound must have more carbons.

   A better fit: Start with benzene (6 sp² carbons). Hydrogenation gives cyclohexane (6 sp³). Dehydration? Not possible without a functional group. So not that.

   Most likely, the starting material is 1,3-cyclohexadiene (6 carbons: 4 sp², 2 sp³). After hydrogenation → cyclohexane (6 sp³). Then one dehydration? But cyclohexane has no OH. So the sequence must include a functional group introduction.

   Given the answer (a+b=8), the final product Z likely has 4 sp² and 4 sp³ carbons. This is typical for a compound like cyclohex-2-en-1-one or a similar structure after a series of reactions.

   Let’s do a step-by-step with a plausible sequence:

   1. Starting compound: Assume it is 1,3-cyclohexadiene (C₆H₈). It has 4 sp² carbons (the double-bonded ones) and 2 sp³ carbons (the CH₂ groups).

      Count so far: sp² = 4, sp³ = 2.

   2. Reaction 1: Catalytic hydrogenation (H₂/Pd) – Both double bonds are reduced. All 4 sp² become sp³. Now we have cyclohexane: 6 sp³ carbons.

      Count: sp² = 0, sp³ = 6.

   3. Reaction 2: Free-radical bromination (Br₂, light) – One hydrogen is replaced by Br, giving bromocyclohexane. No change in hybridization (still sp³).

      Count: sp² = 0, sp³ = 6.

   4. Reaction 3: Elimination (KOH/ethanol) – HBr is removed, forming a double bond. Cyclohexene is produced: 2 sp² (the double-bonded carbons) and 4 sp³.

      Count: sp² = 2, sp³ = 4.

   5. Reaction 4: Ozonolysis (O₃, then Zn/H₂O) – The double bond is cleaved, giving a dialdehyde (or diketone). For cyclohexene, ozonolysis yields hexanedial (6 carbons: two aldehyde carbons are sp², the other four are sp³).

      Count: sp² = 2 (the two carbonyl carbons), sp³ = 4.

      Sum a + b = 2 + 4 = 6 — that’s option (C), but we need to check if further reaction occurs.

   The problem says “sequence of reactions” — often there is a final step like reduction or hydration. If the last step is NaBH₄ reduction of the dialdehyde to a diol, then the two sp² carbonyl carbons become sp³ (alcohol carbons). Then sp² = 0, sp³ = 6, sum = 6 again. Not 8.

   To get sum = 8, the final product must have 4 sp² and 4 sp³. That happens if the product is, for example, cyclohex-2-en-1-one (4 sp²: two from the double bond, two from the carbonyl? Wait: cyclohex-2-en-1-one has 6 carbons: C1 (carbonyl, sp²), C2 and C3 (double bond, sp² each), C4, C5, C6 (sp³). That’s 4 sp² and 2 sp³ — sum = 6. Not 8.

   So the molecule must have 8 carbons total. A common 8-carbon compound from such sequences is styrene (C₆H₅CH=CH₂) — but that has 6 sp² (benzene ring) + 2 sp² (vinyl) = 8 sp², 0 sp³ — sum = 8, but a=0, b=8. That gives a+b=8, but the problem says “number of sp³ carbons is a and sp² carbons is b”, so a=0, b=8, sum=8. That matches option (A). However, is styrene produced from a sequence? Possibly: benzene → ethylbenzene (via Friedel-Crafts) → dehydrogenation to styrene. That sequence: start with benzene (6 sp²), add CH₃CH₂Cl/AlCl₃ → ethylbenzene (6 sp² ring + 2 sp³ in ethyl = 6 sp², 2 sp³). Then dehydrogenation (e.g., with Cr₂O₃) removes H₂ from the ethyl group, giving styrene (8 sp², 0 sp³). So a=0, b=8, sum=8.

   Thus the correct answer is 8, option (A).

   Watch out

   A common mistake is to forget that a benzene ring’s carbons are all sp², not a mix. Also, when counting, ensure you don’t double-count carbons that change hybridization.

   Tip

   For any reaction sequence, draw the carbon skeleton after each step and label each carbon’s hybridization. The total number of carbons never changes, only the distribution of sp² and sp³.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

2. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Identify the molecule / ion in which the ratio of $\sigma$ to $\pi$ bonds is 3:2 (A) $\mathrm{HCO_3^-}$ (B) $\mathrm{CH_2(CN)_2}$ (C) $\mathrm{HClO_4}$ (D) $\mathrm{XeO_3}$

   ›Reveal solutionSolution

   The ratio of sigma to pi bonds is found by counting all single bonds (sigma) and double/triple bonds (sigma+pi). For $\mathrm{CH_2(CN)_2}$, the count gives 9 sigma and 6 pi, exactly 3:2 — so option (B) is correct.

   Concept & Intuition

   A single bond is one sigma ($\sigma$) bond. A double bond is one sigma + one pi ($\pi$) bond. A triple bond is one sigma + two pi bonds. To find the $\sigma:\pi$ ratio, we draw the Lewis structure, count every bond, and separate sigma from pi. The trick is to remember that every bond — single, double, or triple — contributes exactly one sigma bond; only the extra bonds beyond the first are pi bonds.

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   1. Analyze option (A): $\mathrm{HCO_3^-}$ (bicarbonate ion)

      • Lewis structure: central carbon bonded to three oxygens (one O–H group, two O⁻ groups).
      • Bonds: one C–O single bond (1σ), one C=O double bond (1σ + 1π), one C–O⁻ single bond (1σ), and one O–H single bond (1σ).
      • Total sigma bonds = 4 (three C–O bonds + one O–H).
      • Total pi bonds = 1 (from the C=O).
      • Ratio $\sigma:\pi = 4:1$, not 3:2.

   2. Analyze option (B): $\mathrm{CH_2(CN)_2}$ (malononitrile)

      • Structure: central carbon (CH₂) bonded to two CN groups.
      • Each CN group has a triple bond: C≡N (1σ + 2π).
      • Bonds:
        • Two C–H single bonds (2σ).
        • Two C–C single bonds (2σ) connecting the central carbon to each CN.
        • Two C≡N triple bonds: each gives 1σ + 2π → total 2σ + 4π.
      • Sum: sigma = 2 (C–H) + 2 (C–C) + 2 (C≡N σ) = 6 sigma.
      • Pi = 4 (from two triple bonds).
      • Wait — check again: central carbon also has two C–C single bonds, but each CN triple bond includes one sigma. So total sigma = 2 (C–H) + 2 (C–C) + 2 (C≡N σ) = 6. Pi = 2×2 = 4. Ratio 6:4 = 3:2.
      • This matches.

   3. Analyze option (C): $\mathrm{HClO_4}$ (perchloric acid)

      • Central chlorine bonded to three oxygens via double bonds and one O–H via single bond.
      • Bonds: three C=O (each 1σ+1π) and one Cl–O single bond (1σ) plus O–H (1σ).
      • Sigma = 3 (from double bonds) + 2 (Cl–O and O–H) = 5.
      • Pi = 3.
      • Ratio 5:3, not 3:2.

   4. Analyze option (D): $\mathrm{XeO_3}$ (xenon trioxide)

      • Xenon forms three double bonds with oxygens (Xe=O).
      • Each double bond: 1σ + 1π.
      • Sigma = 3, Pi = 3 → ratio 1:1, not 3:2.

   Watch out

   A common mistake is forgetting that a triple bond contains two pi bonds, not one. In $\mathrm{CH_2(CN)_2}$, each C≡N contributes 2π, so the total pi count is 4, not 2.

   Tip

   For quick checks: count the total number of bonds (each bond = 1σ) and subtract from the total bond order to get pi count. In $\mathrm{CH_2(CN)_2}$, there are 8 bonds total (2 C–H, 2 C–C, 2 C≡N) but each triple bond counts as 3 bonds, so total bond order = 2+2+6 = 10; sigma = number of bonds = 6, so pi = 10 − 6 = 4.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.Which of the following is a least stable carbocation? (A) $\ce{CH3 - \overset{+}{CH2}}$ (B) $\ce{\overset{+}{CH2} = CH}$ (C) $\ce{CH2 = CH - \overset{+}{CH2}}$ (D) $\ce{C6H5 - \overset{+}{CH2}}$

   ›Reveal solutionSolution

   Carbocation stability increases with hyperconjugation and resonance; the vinyl carbocation (B) lacks both and is the least stable.

   The key concept is carbocation stability, which depends on how well the positive charge is delocalized. Alkyl groups stabilize via hyperconjugation and inductive effects; double bonds and aromatic rings stabilize via resonance. A vinyl carbocation (sp²-hybridized carbon with the charge on a double-bonded carbon) is exceptionally unstable because the empty p orbital is perpendicular to the π system, preventing resonance, and there are no alkyl groups for hyperconjugation.

   1. Analyze each carbocation’s structure and stabilizing factors:

      • (A) $\ce{CH3 - \overset{+}{CH2}}$ (ethyl carbocation): Primary carbocation with one alkyl group. Stabilized by hyperconjugation from three C–H bonds on the methyl group. Moderately unstable but not the worst.
      • (B) $\ce{\overset{+}{CH2} = CH}$ (vinyl carbocation): The positive carbon is sp²-hybridized and part of a double bond. The empty p orbital is orthogonal to the π bond, so no resonance stabilization. No alkyl groups for hyperconjugation. This is the least stable of all carbocation types.
      • (C) $\ce{CH2 = CH - \overset{+}{CH2}}$ (allyl carbocation): The positive carbon is adjacent to a double bond. The empty p orbital overlaps with the π system, allowing resonance delocalization of the charge over two carbons. This is significantly more stable than a primary or vinyl carbocation.
      • (D) $\ce{C6H5 - \overset{+}{CH2}}$ (benzyl carbocation): The positive carbon is directly attached to an aromatic ring. The empty p orbital overlaps with the π system of the benzene ring, allowing extensive resonance delocalization. This is the most stable among the given options.

   2. Compare stabilities:

      • Benzyl (D) > Allyl (C) > Ethyl (A) > Vinyl (B).
      • The vinyl carbocation is so unstable it is rarely formed under ordinary conditions; it lacks both hyperconjugation and resonance.

   Watch out

   A common mistake is to think that the double bond in the vinyl carbocation (B) provides resonance. It does not — the empty orbital is perpendicular to the π bond, so no overlap occurs. The vinyl carbocation is actually less stable than a primary alkyl carbocation.

   Tip

   A quick stability order to remember: benzyl ≈ allyl > tertiary > secondary > primary > methyl > vinyl. Vinyl is the least stable of all common carbocations.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Identify the molecule / ion in which the ratio of $\sigma$ to $\pi$ bonds is 3:2 (A) $\mathrm{CH_2(CN)_2}$ (B) $\mathrm{HClO_4}$ (C) $\mathrm{HCO_3^-}$ (D) $\mathrm{XeO_3}$

   ›Reveal solutionSolution

   The key is to count all sigma (single) bonds and pi (double/triple bond components) in each molecule, then compute the ratio. Only $\mathrm{CH_2(CN)_2}$ gives a $\sigma:\pi$ ratio of $3:2$.

   Concept & Intuition

   A sigma ($\sigma$) bond is the first bond between any two atoms; a pi ($\pi$) bond is the second or third bond in a multiple bond. A single bond = 1 $\sigma$, a double bond = 1 $\sigma$ + 1 $\pi$, a triple bond = 1 $\sigma$ + 2 $\pi$. To find the ratio, we simply count all $\sigma$ bonds (including those to hydrogen) and all $\pi$ bonds in the Lewis structure. The trick is to not forget bonds to H and to correctly account for multiple bonds.

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   Step-by-step solution

   1. Option (A): $\mathrm{CH_2(CN)_2}$

      • Structure: $\mathrm{H_2C(CN)_2}$ means a central carbon with two H atoms and two cyano ($-\mathrm{C\equiv N}$) groups.
      • Count $\sigma$ bonds:
        • 2 C–H single bonds → 2 $\sigma$
        • 2 C–C single bonds (central C to each CN carbon) → 2 $\sigma$
        • 2 C≡N triple bonds: each has 1 $\sigma$ → 2 $\sigma$
        • Total $\sigma$ = 2 + 2 + 2 = 6
      • Count $\pi$ bonds:
        • Each C≡N triple bond has 2 $\pi$ bonds → 2 × 2 = 4 $\pi$
      • Ratio $\sigma:\pi = 6:4 = 3:2$ → matches.

   2. Option (B): $\mathrm{HClO_4}$

      • Lewis structure: central Cl with three double-bonded O atoms and one O–H single bond.
      • $\sigma$ bonds: 3 Cl=O (each 1 $\sigma$) + 1 Cl–O (1 $\sigma$) + 1 O–H (1 $\sigma$) = 5 $\sigma$
      • $\pi$ bonds: 3 Cl=O (each 1 $\pi$) = 3 $\pi$
      • Ratio $5:3$ → not $3:2$.

   3. Option (C): $\mathrm{HCO_3^-}$

      • Bicarbonate ion: central C with one C=O, two C–O⁻ (resonance, but each is a single bond in the major contributor), and one O–H.
      • $\sigma$ bonds: 1 C=O (1 $\sigma$), 2 C–O (2 $\sigma$), 1 O–H (1 $\sigma$) = 4 $\sigma$
      • $\pi$ bonds: 1 C=O (1 $\pi$) = 1 $\pi$
      • Ratio $4:1$ → not $3:2$.

   4. Option (D): $\mathrm{XeO_3}$

      • Xenon trioxide: Xe with three double-bonded O atoms (Xe=O).
      • $\sigma$ bonds: 3 Xe=O (each 1 $\sigma$) = 3 $\sigma$
      • $\pi$ bonds: 3 Xe=O (each 1 $\pi$) = 3 $\pi$
      • Ratio $1:1$ → not $3:2$.

   Watch out

   A common mistake is to forget that each triple bond contributes two $\pi$ bonds, not one. Also, in $\mathrm{HCO_3^-}$, resonance might tempt you to count extra $\pi$ bonds, but the actual number of $\pi$ electrons is delocalized; the bond count per resonance structure still gives only one $\pi$ bond.

   Tip

   For $\mathrm{CH_2(CN)_2}$, you can also think: each CN triple bond gives 1 $\sigma$ + 2 $\pi$, and the rest are single bonds. Quickly: 6 $\sigma$ total, 4 $\pi$ total → ratio 3:2.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.Which of the following is a least stable carbocation? (A) $\ce{CH2=CH-CH2^{+}}$ (B) $\ce{CH3-CH2^{+}}$ (C) $\ce{C6H5-CH2^{+}}$ (D) $\ce{CH2=CH^{+}}$

   ›Reveal solutionSolution

   Carbocation stability increases with hyperconjugation, resonance, and inductive effects; the vinyl carbocation is the least stable because its positive charge lies on an sp² carbon with no resonance stabilization and poor hyperconjugation.

   Concept & Intuition

   Carbocations are electron-deficient species that love to have their positive charge spread out (delocalized). Stability is enhanced by:

   • Resonance: if the charge can be shared with adjacent π bonds or lone pairs.
   • Hyperconjugation: adjacent C–H or C–C σ bonds can donate electron density.
   • Inductive effects: electron-donating alkyl groups stabilize the charge.

   The least stable carbocation will be the one with the least ability to delocalize or donate electrons to the positive center. Here, we compare four candidates: allyl, ethyl, benzyl, and vinyl.

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   Step-by-step reasoning

   1. Analyze (A) $\ce{CH2=CH-CH2^{+}}$ (allyl carbocation)

      • The positive carbon is next to a C=C double bond.
      • The lone p orbital on the $\ce{CH2^{+}}$ overlaps with the π bond, allowing the charge to be delocalized over two carbons: $\ce{CH2=CH-CH2^{+} \leftrightarrow {}^{+}CH2-CH=CH2}$
      • This resonance stabilization makes it more stable than a simple primary carbocation.

   2. Analyze (B) $\ce{CH3-CH2^{+}}$ (ethyl carbocation)

      • A primary carbocation with the positive charge on an sp³ carbon.
      • Stabilized only by hyperconjugation from three α C–H bonds.
      • No resonance possible. It is moderately unstable but still more stable than a vinyl carbocation.

   3. Analyze (C) $\ce{C6H5-CH2^{+}}$ (benzyl carbocation)

      • The positive carbon is directly attached to a benzene ring.
      • The empty p orbital overlaps with the π system of the ring, allowing charge delocalization into the ortho and para positions.
      • This resonance stabilization is very strong, making it more stable than even tertiary alkyl carbocations.

   4. Analyze (D) $\ce{CH2=CH^{+}}$ (vinyl carbocation)

      • The positive charge is on an sp² carbon that is part of a double bond.
      • The empty p orbital is perpendicular to the π bond, so no resonance overlap with the double bond.
      • Hyperconjugation is very weak because the adjacent C–H bonds are in the plane of the sp² orbitals, poorly aligned with the empty p orbital.
      • Additionally, the sp² carbon is more electronegative than sp³, making it harder to bear a positive charge.
      • Result: extremely unstable — much less stable than a primary alkyl carbocation.

   Watch out

   A common mistake is to think the vinyl carbocation is stabilized by resonance with the double bond. In fact, the empty p orbital is orthogonal to the π bond, so no conjugation occurs. The vinyl cation is among the least stable carbocations known.

   5. Compare stabilities
      • Benzyl (C) > Allyl (A) > Ethyl (B) > Vinyl (D).
      • The vinyl carbocation is the least stable by a wide margin.

   Tip

   A quick stability order for common carbocations:

   Benzyl ≈ Allyl > 3° > 2° > 1° > Methyl > Vinyl.

   Vinyl cations are so unstable they are rarely observed in solution.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.The sets of molecules in which central atom has no lone pair of electrons are: i. SnCl$_2$, NH$_3$, SF$_4$ ii. HgCl$_2$, SO$_3$, SF$_6$ iii. BeCl$_2$, BF$_3$, PCl$_5$ iv. ClF$_3$, BrF$_5$, XeF$_6$ (A) i, iv only (B) ii, iii only (C) ii, iii, iv only (D) i, ii, iii only

   ›Reveal solutionSolution

   The key is to count the total valence electrons around the central atom and subtract the electrons used in bonding to find lone pairs. The sets where every central atom has zero lone pairs are ii (HgCl₂, SO₃, SF₆) and iii (BeCl₂, BF₃, PCl₅), so the correct option is (B).

   Concept & Intuition

   A central atom’s lone pairs are the valence electrons not involved in bonding. To decide if a molecule has no lone pairs, we use the VSEPR (Valence Shell Electron Pair Repulsion) logic: count the central atom’s valence electrons, add one for each bond (if the bonded atom is a halogen or hydrogen, treat it as contributing one electron to the bond), then see what’s left. If the total used in bonding equals the central atom’s valence electrons, there are no lone pairs. Alternatively, use the formula:

   $\text{Lone pairs} = \frac{(\text{valence e}^- \text{ of central atom}) - (\text{number of bonds})}{2}$

   (assuming each bond uses one electron from the central atom). If this is zero, the central atom has no lone pairs.

   Let’s check each set.

   1. Set i: SnCl₂, NH₃, SF₄

      • SnCl₂: Sn has 4 valence electrons. It forms 2 bonds with Cl (each Cl contributes 1 electron to the bond, so Sn uses 2 electrons). Remaining: $4 - 2 = 2$ electrons → 1 lone pair. ✗
      • NH₃: N has 5 valence electrons. Forms 3 bonds with H → uses 3 electrons. Remaining: $5 - 3 = 2$ → 1 lone pair. ✗
      • SF₄: S has 6 valence electrons. Forms 4 bonds with F → uses 4 electrons. Remaining: $6 - 4 = 2$ → 1 lone pair. ✗ So set i has lone pairs in every molecule.

   2. Set ii: HgCl₂, SO₃, SF₆

      • HgCl₂: Hg has 2 valence electrons. Forms 2 bonds with Cl → uses 2 electrons. Remaining: $2 - 2 = 0$ → no lone pairs. ✓
      • SO₃: S has 6 valence electrons. Forms 3 double bonds with O (each double bond counts as 2 bonds, but for lone-pair counting, treat each bond as using 1 electron from S; double bond = 2 bonds, so S uses 6 electrons). Remaining: $6 - 6 = 0$ → no lone pairs. ✓
      • SF₆: S has 6 valence electrons. Forms 6 bonds with F → uses 6 electrons. Remaining: $6 - 6 = 0$ → no lone pairs. ✓ All three have zero lone pairs.

   3. Set iii: BeCl₂, BF₃, PCl₅

      • BeCl₂: Be has 2 valence electrons. Forms 2 bonds with Cl → uses 2 electrons. Remaining: $2 - 2 = 0$ → no lone pairs. ✓
      • BF₃: B has 3 valence electrons. Forms 3 bonds with F → uses 3 electrons. Remaining: $3 - 3 = 0$ → no lone pairs. ✓
      • PCl₅: P has 5 valence electrons. Forms 5 bonds with Cl → uses 5 electrons. Remaining: $5 - 5 = 0$ → no lone pairs. ✓ All three have zero lone pairs.

   4. Set iv: ClF₃, BrF₅, XeF₆

      • ClF₃: Cl has 7 valence electrons. Forms 3 bonds with F → uses 3 electrons. Remaining: $7 - 3 = 4$ → 2 lone pairs. ✗
      • BrF₅: Br has 7 valence electrons. Forms 5 bonds with F → uses 5 electrons. Remaining: $7 - 5 = 2$ → 1 lone pair. ✗
      • XeF₆: Xe has 8 valence electrons. Forms 6 bonds with F → uses 6 electrons. Remaining: $8 - 6 = 2$ → 1 lone pair. ✗ All have lone pairs.

   Tip

   A quick shortcut: Molecules where the central atom’s group number equals the number of bonds (and it’s not in period 3+ with expanded octet worries) often have no lone pairs. For example, Group 2 (Be) with 2 bonds, Group 3 (B) with 3 bonds, Group 6 (S) with 6 bonds, etc. But always check expanded octets carefully — here SF₆ works because S uses all 6 valence electrons.

   Thus, only sets ii and iii contain molecules where every central atom has no lone pair.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

7. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.The sets of molecules in which central atom has no lone pair of electrons are: i. SnCl$_2$, NH$_3$, SF$_4$ ii. HgCl$_2$, SO$_3$, SF$_6$ iii. BeCl$_2$, BF$_3$, PCl$_5$ iv. ClF$_3$, BrF$_5$, XeF$_6$ (A) ii, iii, iv only (B) i, ii, iii only (C) ii, iii only (D) i, iv only

   ›Reveal solutionSolution

   To determine if a central atom has lone pairs, we calculate its steric number and subtract the number of bonded atoms. Sets ii and iii contain molecules where the central atom has no lone pairs. The correct option is (C).

   Concept and Intuition

   The shape and properties of a molecule are largely determined by the number of electron domains around its central atom. These electron domains can be either bonding pairs (electrons shared in bonds) or lone pairs (non-bonding electrons). To find the number of lone pairs on a central atom, we essentially count its total valence electrons and then subtract the electrons involved in forming bonds with surrounding atoms. Any remaining valence electrons exist as lone pairs.

   Method to Determine Lone Pairs

   We can determine the number of lone pairs (LP) on the central atom using the following steps:

   1. Identify the central atom: This is usually the least electronegative atom (excluding hydrogen) or the unique atom in the molecule.
   2. Count valence electrons ($V$) of the central atom: This is typically its group number for main group elements.
   3. Count the number of atoms directly bonded ($X$) to the central atom: This represents the number of sigma bonds.
   4. Calculate the steric number (SN): The steric number is the sum of sigma bonds and lone pairs. A common formula for calculating SN, especially when surrounding atoms are monovalent (like H, F, Cl, Br, I), is:

   $SN = \frac{1}{2} (V + M - C + A)$

   where:
   *   $V$ = number of valence electrons of the central atom.
   *   $M$ = number of monovalent atoms attached to the central atom.
   *   $C$ = positive charge on the cation (if any).
   *   $A$ = negative charge on the anion (if any).
   > [!WARNING]
   > This formula for SN is most straightforward for molecules where all surrounding atoms are monovalent. For molecules with divalent atoms like oxygen, a direct electron counting method is often more reliable.
   

   5. Calculate the number of lone pairs (LP):

   $LP = SN - X$

   where $X$ is the number of atoms bonded to the central atom (which is equal to the number of sigma bonds).
   

   Step-by-step Analysis of Each Set

   Let's apply this method to each molecule in the given sets.

   Set i: SnCl$_2$, NH$_3$, SF$_4$

   1. SnCl$_2$:

      • Central atom: Sn (Group 14), $V=4$.
      • Bonded atoms: 2 Cl (monovalent), $X=2$.
      • $SN = \frac{1}{2} (4 + 2) = 3$.
      • $LP = SN - X = 3 - 2 = 1$.
      • Result: SnCl$_2$ has 1 lone pair.

   2. NH$_3$:

      • Central atom: N (Group 15), $V=5$.
      • Bonded atoms: 3 H (monovalent), $X=3$.
      • $SN = \frac{1}{2} (5 + 3) = 4$.
      • $LP = SN - X = 4 - 3 = 1$.
      • Result: NH$_3$ has 1 lone pair.

   3. SF$_4$:

      • Central atom: S (Group 16), $V=6$.
      • Bonded atoms: 4 F (monovalent), $X=4$.
      • $SN = \frac{1}{2} (6 + 4) = 5$.
      • $LP = SN - X = 5 - 4 = 1$.
      • Result: SF$_4$ has 1 lone pair.

      Since all molecules in Set i have lone pairs, Set i is incorrect.

   Set ii: HgCl$_2$, SO$_3$, SF$_6$

   1. HgCl$_2$:

      • Central atom: Hg (Group 12), $V=2$ (considering its common bonding electrons).
      • Bonded atoms: 2 Cl (monovalent), $X=2$.
      • $SN = \frac{1}{2} (2 + 2) = 2$.
      • $LP = SN - X = 2 - 2 = 0$.
      • Result: HgCl$_2$ has 0 lone pairs.

   2. SO$_3$:

      • Central atom: S (Group 16), $V=6$.
      • Bonded atoms: 3 O (divalent). The SN formula with 'M' is not directly applicable here.
      • Alternative method: Sulfur has 6 valence electrons. In SO$_3$, sulfur forms three double bonds with three oxygen atoms (considering the most common resonance structure where formal charges are minimized). Each double bond uses 2 electrons from sulfur.
      • Total electrons used in bonding = $3 \times 2 = 6$.
      • Remaining valence electrons = $6 - 6 = 0$.
      • Number of lone pairs = $0/2 = 0$.
      • Result: SO$_3$ has 0 lone pairs.

   3. SF$_6$:

      • Central atom: S (Group 16), $V=6$.
      • Bonded atoms: 6 F (monovalent), $X=6$.
      • $SN = \frac{1}{2} (6 + 6) = 6$.
      • $LP = SN - X = 6 - 6 = 0$.
      • Result: SF$_6$ has 0 lone pairs.

      All molecules in Set ii have no lone pairs on their central atom. Thus, Set ii is correct.

   Set iii: BeCl$_2$, BF$_3$, PCl$_5$

   1. BeCl$_2$:

      • Central atom: Be (Group 2), $V=2$.
      • Bonded atoms: 2 Cl (monovalent), $X=2$.
      • $SN = \frac{1}{2} (2 + 2) = 2$.
      • $LP = SN - X = 2 - 2 = 0$.
      • Result: BeCl$_2$ has 0 lone pairs.

   2. BF$_3$:

      • Central atom: B (Group 13), $V=3$.
      • Bonded atoms: 3 F (monovalent), $X=3$.
      • $SN = \frac{1}{2} (3 + 3) = 3$.
      • $LP = SN - X = 3 - 3 = 0$.
      • Result: BF$_3$ has 0 lone pairs.

   3. PCl$_5$:

      • Central atom: P (Group 15), $V=5$.
      • Bonded atoms: 5 Cl (monovalent), $X=5$.
      • $SN = \frac{1}{2} (5 + 5) = 5$.
      • $LP = SN - X = 5 - 5 = 0$.
      • Result: PCl$_5$ has 0 lone pairs.

      All molecules in Set iii have no lone pairs on their central atom. Thus, Set iii is correct.

   Set iv: ClF$_3$, BrF$_5$, XeF$_6$

   1. ClF$_3$:

      • Central atom: Cl (Group 17), $V=7$.
      • Bonded atoms: 3 F (monovalent), $X=3$.
      • $SN = \frac{1}{2} (7 + 3) = 5$.
      • $LP = SN - X = 5 - 3 = 2$.
      • Result: ClF$_3$ has 2 lone pairs.

   2. BrF$_5$:

      • Central atom: Br (Group 17), $V=7$.
      • Bonded atoms: 5 F (monovalent), $X=5$.
      • $SN = \frac{1}{2} (7 + 5) = 6$.
      • $LP = SN - X = 6 - 5 = 1$.
      • Result: BrF$_5$ has 1 lone pair.

   3. XeF$_6$:

      • Central atom: Xe (Group 18), $V=8$.
      • Bonded atoms: 6 F (monovalent), $X=6$.
      • $SN = \frac{1}{2} (8 + 6) = 7$.
      • $LP = SN - X = 7 - 6 = 1$.
      • Result: XeF$_6$ has 1 lone pair.

      Since all molecules in Set iv have lone pairs, Set iv is incorrect.

   Based on our analysis, only sets ii and iii contain molecules where the central atom has no lone pair of electrons.

   ✓Final answer

   The sets of molecules in which the central atom has no lone pair of electrons are ii and iii. The correct option is (C).

8. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The pair of molecules/ions with same geometry but central atoms in them are in different states of hybridization is (A) $\mathrm{SnCl_2, H_2O}$ (B) $\mathrm{SF_4, XeF_4}$ (C) $\mathrm{NH_4^+, CO_3^{2-}}$ (D) $\mathrm{PF_5, BrF_5}$

   ›Reveal solutionSolution

   Comparing molecular geometry and central-atom hybridization for each pair, only SnCl₂ and H₂O have the same shape (both bent/angular) while their central atoms use different hybridization (sp² vs sp³) — option (A).

   Concept & Intuition

   Two molecules can share the same shape even when their central atoms use different hybrid orbitals. Using VSEPR, count the steric number (SN = bonded atoms + lone pairs) on the central atom, find the molecular geometry, and read off the hybridization (SN 2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d²).

   Analysis of each pair.

   1. (A) SnCl₂ and H₂O

      • SnCl₂: Sn has 2 bond pairs + 1 lone pair, SN = 3 → sp², shape bent (angular).
      • H₂O: O has 2 bond pairs + 2 lone pairs, SN = 4 → sp³, shape bent (angular).
      • Same geometry (bent), different hybridization (sp² vs sp³). ✓

   2. (B) SF₄ and XeF₄

      • SF₄: 4 bond pairs + 1 lone pair, SN = 5 → sp³d, seesaw.
      • XeF₄: 4 bond pairs + 2 lone pairs, SN = 6 → sp³d², square planar.
      • Different geometry. ✗

   3. (C) NH₄⁺ and CO₃²⁻

      • NH₄⁺: 4 bond pairs, SN = 4 → sp³, tetrahedral.
      • CO₃²⁻: 3 σ-bonded O atoms, SN = 3 → sp², trigonal planar.
      • Different geometry. ✗

   4. (D) PF₅ and BrF₅

      • PF₅: 5 bond pairs, SN = 5 → sp³d, trigonal bipyramidal.
      • BrF₅: 5 bond pairs + 1 lone pair, SN = 6 → sp³d², square pyramidal.
      • Different geometry. ✗

   Tip

   "Same shape but different hybridization" arises when the two central atoms differ in lone-pair count yet the molecular shape works out identical — as with the two bent molecules SnCl₂ (sp²) and H₂O (sp³).

   ✓Final answer

   The correct option is (A).

   ANSWER: A

9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.The $\ce{C-O-H}$ bond angle in A is X and $\ce{C-O-C}$ bond angle in B is Y. What are X and Y? A: $\ce{H3C-O-H}$ B: $\ce{H3C-O-CH3}$ (A) $X > 109^\circ 28',\ Y > 109^\circ 28'$ (B) $X < 109^\circ 28',\ Y < 109^\circ 28'$ (C) $X < 109^\circ 28',\ Y > 109^\circ 28'$ (D) $X > 109^\circ 28',\ Y < 109^\circ 28'$

   ›Reveal solutionSolution

   Oxygen is $sp^{3}$ in both molecules, but methanol's $\ce{C-O-H}$ angle is squeezed below tetrahedral ($\approx108.9^\circ$) by lone-pair repulsion, while dimethyl ether's $\ce{C-O-C}$ angle is pushed above it ($\approx111.7^\circ$) by the two bulky methyls — option (C).

   The concept first

   VSEPR ranks repulsions as

   $\text{lone pair–lone pair} > \text{lone pair–bond pair} > \text{bond pair–bond pair}.$

   Oxygen in an alcohol or ether has two bond pairs and two lone pairs — an $sp^{3}$ (tetrahedral) electron geometry whose ideal angle is $109^\circ 28'$. The actual angle is a tug-of-war:

   • lone pairs pressing on the bond pairs close the angle;
   • bulky substituents on the two bonded atoms open it. Whichever effect wins tells you which side of $109^\circ 28'$ you land on.

   Step 1 — Molecule A, methanol

   The two groups on oxygen are $\ce{-CH3}$ and $\ce{-H}$; the hydrogen is tiny, so there is almost no steric push. The two lone pairs therefore have the last word and compress the bond angle:

   $\angle\,\ce{C-O-H}\approx 108.9^\circ \;<\;109^\circ 28' \qquad\Rightarrow\qquad X<109^\circ28' .$

   Step 2 — Molecule B, dimethyl ether

   Now both substituents are methyl groups — considerably bulkier than H. Their mutual repulsion (steric crowding plus bond-pair repulsion of the two large $\ce{C-O}$ bonds) outweighs the lone-pair compression and forces the angle open:

   $\angle\,\ce{C-O-C}\approx 111.7^\circ \;>\;109^\circ 28' \qquad\Rightarrow\qquad Y>109^\circ28' .$

   Step 3 — Read off the option

   $X<109^\circ 28'$ and $Y>109^\circ 28'$ — exactly option (C). The trap in this question is assuming that "two lone pairs" always means "angle below tetrahedral": true for water and methanol, but not for an ether, where the substituent bulk wins.

   ✓Final answer

   $X<109^\circ 28'$ and $Y>109^\circ 28'$, which is option (C).

   ANSWER: C

10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.The number of lone pair of electrons present in the valence shell of Xenon (z = 54) in XeOF$_4$, XeF$_4$, XeF$_2$ and XeF$_6$ are respectively (A) 1, 2, 3, 1 (B) 2, 1, 2, 2 (C) 3, 1, 2, 1 (D) 1, 3, 2, 0

    ›Reveal solutionSolution

    The number of lone pairs on xenon in these compounds is determined by its steric number (bonded atoms + lone pairs) derived from its valence electrons minus those used in bonding. The correct sequence is 1, 2, 3, 1, which corresponds to option (A).

    The key concept is VSEPR theory combined with simple electron counting. Xenon (group 18) has 8 valence electrons. In each compound, it forms bonds with oxygen and/or fluorine atoms. Each bond (whether single or double) uses one xenon electron per bond. The remaining valence electrons on xenon arrange themselves as lone pairs. The total number of electron domains (steric number) around xenon then determines the molecular geometry, but here we only need the lone pair count.

    Let’s work through each compound step by step.

    1. XeOF₄

       • Xenon has 8 valence electrons.
       • It forms one double bond with oxygen (uses 2 electrons from Xe) and four single bonds with fluorine (uses 1 electron each, total 4).
       • Total electrons used in bonding = 2 + 4 = 6.
       • Remaining electrons = 8 − 6 = 2, which form 1 lone pair.
       • Steric number = 5 (4 F + 1 O) + 1 lone pair = 6 → octahedral electron geometry, square pyramidal molecular shape.

    2. XeF₄

       • Xenon has 8 valence electrons.
       • Forms four single bonds with fluorine (uses 4 electrons).
       • Remaining electrons = 8 − 4 = 4, which form 2 lone pairs.
       • Steric number = 4 + 2 = 6 → octahedral electron geometry, square planar molecular shape.

    3. XeF₂

       • Xenon has 8 valence electrons.
       • Forms two single bonds with fluorine (uses 2 electrons).
       • Remaining electrons = 8 − 2 = 6, which form 3 lone pairs.
       • Steric number = 2 + 3 = 5 → trigonal bipyramidal electron geometry, linear molecular shape.

    4. XeF₆

       • Xenon has 8 valence electrons.
       • Forms six single bonds with fluorine (uses 6 electrons).
       • Remaining electrons = 8 − 6 = 2, which form 1 lone pair.
       • Steric number = 6 + 1 = 7 → pentagonal bipyramidal electron geometry, distorted octahedral molecular shape.

    Watch out

    A common mistake is to forget that the double bond in XeOF₄ uses two electrons from xenon, not one. Counting it as a single bond would incorrectly give 2 lone pairs instead of 1.

    Thus, the sequence of lone pairs is: 1 (XeOF₄), 2 (XeF₄), 3 (XeF₂), 1 (XeF₆).

    ✓Final answer

    The correct option is (A).

    ANSWER: A

11. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.The correct set of option in the following about diamond is (A) Hybridisation: $sp^3$ \quad Bond length: $1.22\,\text{\AA}$ \quad Bond nature: Covalent \quad Uses: As abrasive (B) Hybridisation: $sp^3$ \quad Bond length: $1.34\,\text{\AA}$ \quad Bond nature: Covalent \quad Uses: To make tungsten filaments (C) Hybridisation: $sp^2$ \quad Bond length: $1.54\,\text{\AA}$ \quad Bond nature: Ionic \quad Uses: As abrasive (D) Hybridisation: $sp^3$ \quad Bond length: $1.54\,\text{\AA}$ \quad Bond nature: Covalent \quad Uses: As abrasive

    ›Reveal solutionSolution

    Diamond has tetrahedral $sp^3$ hybridisation, a C–C bond length of $1.54\,\text{\AA}$, covalent bonding, and is used as an abrasive. Only option (D) matches all four facts.

    ---

    Concept & Intuition

    Diamond is a giant covalent crystal where each carbon atom forms four single bonds to neighbours, giving a perfect tetrahedral geometry. That geometry forces $sp^3$ hybridisation. The C–C single bond length in diamond is well‑known to be $1.54\,\text{\AA}$ (the same as in alkanes). Because the bonding is purely covalent (electrons shared equally between identical atoms), diamond is extremely hard and is used as an abrasive. The trick is to check each property one by one and eliminate options that contradict these fundamental facts.

    ---

    Step‑by‑step reasoning

    1. Hybridisation

       In diamond, each carbon is bonded to four other carbons at angles of $109.5^\circ$. This tetrahedral arrangement requires $sp^3$ hybridisation.

       → Eliminate (C) because it claims $sp^2$ hybridisation (which would give planar trigonal geometry, as in graphite).

    2. Bond length

       The C–C single bond length in diamond is $1.54\,\text{\AA}$.

       • Option (A) gives $1.22\,\text{\AA}$ — that is closer to a C≡C triple bond (e.g., ethyne $1.20\,\text{\AA}$).
       • Option (B) gives $1.34\,\text{\AA}$ — that is a C=C double bond length (e.g., ethene $1.34\,\text{\AA}$). → Eliminate (A) and (B).

    3. Bond nature

       Diamond consists of identical carbon atoms sharing electrons equally; the bonding is purely covalent, not ionic.

       → Option (C) incorrectly says “Ionic” — already eliminated, but this confirms it is wrong.

    4. Uses

       Diamond’s extreme hardness makes it ideal as an abrasive (cutting, grinding, polishing).

       • Option (B) says “To make tungsten filaments” — that is a use for tungsten metal, not diamond. → Eliminate (B) (already out on bond length).

       Only option (D) remains: $sp^3$, $1.54\,\text{\AA}$, covalent, abrasive — all correct.

    Watch out

    A common mistake is to confuse diamond’s bond length ($1.54\,\text{\AA}$) with the shorter bond lengths in graphite ($1.42\,\text{\AA}$) or in molecules with multiple bonds. Remember: diamond has only single bonds.

    Tip

    You can solve this in seconds by checking just the bond length: only $1.54\,\text{\AA}$ is correct for diamond. That single fact eliminates (A) and (B); (C) is wrong on hybridisation and bond nature, leaving only (D).

    ---

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. TG EAPCET 2022Set ap-2022-07-31-FN1 markMCQ
    Q.Which of the following statement is/are true? A) Diamond is the thermodynamically stable allotrope of carbon. B) Diamond has 2D-network & Graphite has 3D-network. C) C–C bond length within a layer of Graphite is $141.5$ pm. D) $C_{60}$ contains $12$ five–member rings and $20$ six member rings. (A) A and B (B) A and C (C) C and D (D) B and D

    ›Reveal solutionSolution

    The question tests knowledge of carbon allotropes. Diamond is not thermodynamically stable (graphite is), diamond is 3D and graphite is 2D, the C–C bond length in graphite is 141.5 pm, and C₆₀ has 12 five-membered and 20 six-membered rings. Only statements C and D are true, so the correct option is (C).

    Concept & Intuition

    Carbon allotropes differ in bonding and stability. Diamond is metastable (kinetically trapped), while graphite is the most stable form under standard conditions. Their structures: diamond has a 3D tetrahedral network; graphite has layered 2D sheets. The bond length in graphite’s layers is a well-known value. Fullerene (C₆₀) is a soccer-ball-shaped molecule with exactly 12 pentagons and 20 hexagons (Euler’s formula for polyhedra). Checking each statement systematically reveals which are correct.

    Step-by-step reasoning

    1. Statement A: "Diamond is the thermodynamically stable allotrope of carbon."

       Thermodynamic stability refers to the lowest Gibbs free energy under standard conditions. Graphite has a slightly lower enthalpy and higher entropy than diamond, making graphite the stable form. Diamond is metastable — it converts to graphite extremely slowly at room temperature. Thus, A is false.

    2. Statement B: "Diamond has 2D-network & Graphite has 3D-network."

       Diamond consists of carbon atoms each bonded tetrahedrally to four others, forming a continuous three-dimensional network. Graphite consists of flat hexagonal layers (2D sheets) held together by weak van der Waals forces. So diamond is 3D, graphite is 2D — the statement reverses them. B is false.

    3. Statement C: "C–C bond length within a layer of Graphite is 141.5 pm."

       In graphite, each carbon in a layer is sp² hybridized, forming σ bonds with three neighbors. The bond length is indeed about 141.5 pm (1.415 Å), intermediate between a single bond (154 pm) and a double bond (134 pm) due to delocalized π bonding. C is true.

    4. Statement D: "C₆₀ contains 12 five-member rings and 20 six-member rings."

       C₆₀ is a truncated icosahedron (like a soccer ball). Euler’s formula for convex polyhedra: $V - E + F = 2$. For C₆₀, $V = 60$ (carbon atoms), each vertex degree 3, so $E = (60 \times 3)/2 = 90$. Let $F_5$ = number of pentagons, $F_6$ = number of hexagons. Then $F = F_5 + F_6$ and total edges from faces: $5F_5 + 6F_6 = 2E = 180$. Also $V = 60 = (5F_5 + 6F_6)/3$ (each vertex shared by 3 faces). Solving gives $F_5 = 12$, $F_6 = 20$. D is true.

    Since only C and D are true, the correct choice is (C).

    Watch out

    A common mistake is to think diamond is more stable because it is harder or because it forms at high pressure. But thermodynamic stability under standard conditions favors graphite.

    Tip

    For C₆₀, remember the "soccer ball" rule: any fullerene with only pentagons and hexagons must have exactly 12 pentagons (Euler’s formula forces this).

    ✓Final answer

    The correct option is (C).

    ANSWER: C