Q.Which of the following angle corresponds to sp^2 hybridisation?

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Total number of electrons involved in the formation of bridge bonds in the structure of diborane is (A) 6 (B) 2 (C) 4 (D) 3

   ›Reveal solutionSolution

   This tests the electron-deficient bonding in diborane; the two 3c–2e bridge bonds together involve 4 electrons.

   Concept and Intuition

   Diborane is the classic example of electron-deficient, multicentre bonding. Each boron is $sp^3$ hybridized, using 4 terminal B–H bonds (normal 2c–2e bonds, 2 per boron) plus two bridging hydrogens that sit between the two borons. Because there aren't enough electrons for two ordinary 2c–2e bonds to the bridging H atoms, each bridging H instead forms a single 3-centre–2-electron bond spanning B–H–B (a "banana bond"), where 2 electrons are delocalized over 3 nuclei (B, H, B).

   Step-by-Step Solution

   1. Total valence electrons in $\mathrm{B_2H_6}$: $2(3) + 6(1) = 12$ electrons available for bonding.
   2. Four terminal B–H bonds (2 per boron) are normal 2c–2e bonds, using $4\times2 = 8$ electrons.
   3. The remaining $12-8=4$ electrons form the two bridge (3c–2e) bonds — each bridge bond is a single bonding unit holding 2 electrons shared over B–H–B. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The number $\sigma$, $\pi$ – bonds and lone pairs of electrons present in the product $C$ respectively are $C_2H_5NO_2 \xrightarrow{Sn+HCl} A \xrightarrow[\text{Heat}]{CHCl_3+KOH} B \xrightarrow{HgO} C$ (A) $10, 1, 3$ (B) $8, 3, 2$ (C) $9, 2, 3$ (D) $9, 1, 3$

   ›Reveal solutionSolution

   This chains nitro-reduction, the carbylamine reaction, and HgO-oxidation of an isocyanide to an isocyanate; counting bonds/lone pairs in the final ethyl isocyanate gives 9 $\sigma$, 2 $\pi$, 3 lone pairs — option (C).

   Concept and Intuition

   Sn/HCl is a standard reducing system for nitro groups, giving primary amines. Primary amines undergo the carbylamine (isocyanide) test with $CHCl_3$/KOH, forming a foul-smelling isocyanide with one carbon lost from chloroform incorporated as the isocyanide carbon. Isocyanides ($R-N\equiv C$) can be oxidised by mercuric oxide (HgO) to the corresponding isocyanate ($R-N=C=O$), a cumulated-double-bond system analogous to $CO_2$ (the central carbon is sp-hybridised, doubly bonded to both N and O, with no lone pair of its own on that carbon).

   Step-by-Step Solution

   1. $C_2H_5NO_2 \xrightarrow{Sn+HCl} A$: reduction of the nitro group gives the primary amine $\Rightarrow$ A = $C_2H_5NH_2$ (ethylamine).
   2. $A \xrightarrow[\text{Heat}]{CHCl_3+KOH} B$: the carbylamine reaction on a primary amine gives an isocyanide $\Rightarrow$ B = $C_2H_5-NC$ (ethyl isocyanide).
   3. $B \xrightarrow{HgO} C$: mercuric oxide oxidises the isocyanide's terminal carbon, converting $R-N\equiv C$ into the isocyanate $R-N=C=O$ $\Rightarrow$ C = $CH_3CH_2-N=C=O$ (ethyl isocyanate).
   4. Count $\sigma$ bonds in $CH_3-CH_2-N=C=O$: 3 (methyl C–H) + 1 (C–C) + 2 (methylene C–H) + 1 (C–N single bond to the sp carbon system... here it's the alkyl-N single bond) + 1 ($\sigma$ component of N=C) + 1 ($\sigma$ component of C=O) = 9 $\sigma$ bonds.
   5. Count $\pi$ bonds: one in N=C, one in C=O (the central carbon is sp-hybridised with two mutually perpendicular $\pi$ bonds, like in $CO_2$) = 2 $\pi$ bonds. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.The geometry and hybridisation of central atom of the volatile product formed when calamine is heated are respectively (A) $sp^2$, angular (B) $sp^3$, angular (C) $sp^3d$, linear (D) $sp$, linear

   ›Reveal solutionSolution

   Calamine (ZnCO₃) decomposes on heating to give solid ZnO and gaseous CO₂; CO₂'s central carbon is $sp$-hybridised, making the molecule linear.

   Concept and Intuition

   Metal carbonates thermally decompose to the metal oxide plus carbon dioxide gas — this is the "volatile product" being asked about, not the zinc compound. For CO₂, carbon forms two σ-bonds (to each oxygen) and has no lone pairs, so VSEPR predicts 2 electron domains ⇒ $sp$ hybridisation ⇒ linear geometry (O=C=O, 180°).

   Step-by-Step Solution

   1. Calamine = ZnCO₃ (zinc carbonate, an ore of zinc).
   2. Heating: $\mathrm{ZnCO_3\xrightarrow{\Delta}ZnO+CO_2\uparrow}$.
   3. The volatile (gaseous) product is $CO_2$.
   4. Central C atom: 2 bonding domains, 0 lone pairs ⇒ $sp$ hybridisation, linear shape. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The ratio of number of $sp^3$ hybrid orbitals to number of $sp^2$ hybrid orbitals in the major product (Z) of the given reaction sequence is $\mathrm{CaC_2 \xrightarrow{H_2O} X \xrightarrow[873K]{\text{Red hot Iron tube}} Y \xrightarrow{\text{Anhy. } AlCl_3} Z}$ where Y is [FIGURE] (a skeletal zigzag structure of a five-carbon chain with one C=C double bond and a terminal $-CH_2Cl$ group, i.e. 4-chloro-2-methylbut-1-ene-type chlorinated alkene chain) (A) 3 : 5 (B) 3 : 2 (C) 2 : 3 (D) 3 : 4

   ›Reveal solutionSolution

   The reaction sequence is $CaC_2 \to$ acetylene $\to$ (cyclic trimerization) benzene $\to$ (Friedel–Crafts alkylation, with carbocation rearrangement) an alkylbenzene. Counting hybrid orbitals in the major product gives $sp^3:sp^2 = 2:3$.

   Concept and Intuition

   This question strings together three classic named reactions: (1) the reaction of calcium carbide with water to give acetylene, (2) the cyclic trimerization of acetylene over a red-hot iron catalyst to give benzene, and (3) Friedel–Crafts alkylation of benzene, in which a primary alkyl-halide-derived carbocation is never used "as drawn" — it rearranges (via a hydride or alkyl shift) to whatever more stable carbocation is accessible, before it ever attacks the aromatic ring. This rearrangement is one of the best-known caveats of Friedel–Crafts alkylation.

   Step-by-Step Solution

   1. $CaC_2 + 2H_2O \to Ca(OH)_2 + C_2H_2$: so $X = C_2H_2$ (acetylene, ethyne).
   2. Passing ethyne through a red-hot iron tube at 873 K causes cyclic polymerization: $3C_2H_2 \to C_6H_6$ (benzene). So $Y = C_6H_6$.
   3. Benzene + a primary alkyl chloride + anhydrous $AlCl_3$ is a Friedel–Crafts alkylation. $AlCl_3$ abstracts $Cl^-$ from the alkyl chloride to generate a carbocation.
   4. A primary carbocation is highly unstable; before it attacks the aromatic ring, it rearranges via a hydride shift to the nearest, more stable secondary carbocation.
   5. This rearranged (secondary) carbocation is what actually attacks the benzene ring, so the major product $Z$ is the branched alkylbenzene (analogous to the textbook case of n-propyl chloride + $AlCl_3$/benzene giving predominantly cumene, isopropylbenzene, rather than n-propylbenzene) — an aromatic ring bearing a branched alkyl substituent whose point of attachment carbon is now secondary. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.In the structure of diborane, the number of 2-centre-2-electron bonds is X and 3-centre-2-electron bonds is Y. The value of (X + Y) is (A) 5 (B) 6 (C) 4 (D) 8

   ›Reveal solutionSolution

   This tests the classic bonding picture of diborane: 4 ordinary terminal B–H (2c-2e) bonds plus 2 bridging B–H–B (3c-2e) "banana" bonds, giving $X+Y=4+2=6$.

   Concept and Intuition

   Diborane ($B_2H_6$) is electron-deficient — there are not enough valence electrons for every B–H linkage to be a normal shared electron-pair bond. Its structure resolves this with two different bond types: four TERMINAL B–H bonds (two on each boron) that are perfectly normal 2-centre-2-electron bonds, and two BRIDGING hydrogens that each simultaneously bond to both borons using just one pair of electrons spread over three atoms — the famous 3-centre-2-electron "banana" bonds that hold the two $BH_2$ units together.

   Step-by-Step Solution

   1. Diborane's structure: each boron is bonded to 2 terminal H atoms (4 terminal H total) and shares 2 bridging H atoms with the other boron.
   2. Terminal B–H bonds: there are 4 of these (2 per boron), each a normal 2-centre-2-electron (2c-2e) bond. So $X=4$.
   3. Bridging B–H–B bonds: there are 2 bridge hydrogens, each forming a 3-centre-2-electron (3c-2e) bond spanning both borons and the one hydrogen. So $Y=2$.
   4. $X+Y = 4+2 = 6$.

   Common Mistakes …

   🔒 Start your 14-day free trial to see the full solution →
6. AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
   Q.Match the following List-I (complex) — List-II (hybridisation) A) $[\mathrm{CoF_6}]^{3-}$ — I) $dsp^2$ B) $[\mathrm{NiCl_4}]^{2-}$ — II) $d^2sp^3$ C) $[\mathrm{Ni(CN)_4}]^{2-}$ — III) $sp^3$ D) $[\mathrm{Co(NH_3)_6}]^{3+}$ — IV) $sp^3d^2$ The correct answer is (A) A-IV, B-I, C-III, D-II (B) A-II, B-I, C-III, D-IV (C) A-IV, B-III, C-I, D-II (D) A-II, B-III, C-I, D-IV

   ›Reveal solutionSolution

   Matching each complex to its hybridisation using ligand field strength and metal $d$-electron count gives A-IV, B-III, C-I, D-II.

   Concept and Intuition

   Valence bond theory predicts hybridisation from whether a ligand is weak-field (high spin, uses outer $d$-orbitals or simple $sp^3$/$sp^3d^2$ geometry) or strong-field (low spin, uses inner $(n-1)d$ orbitals, e.g. $dsp^2$/$d^2sp^3$). F⁻ and Cl⁻ are weak-field ligands; CN⁻ and NH₃ (for Co³⁺ specifically) are strong-field ligands.

   Step-by-Step Solution

   1. A) $[\mathrm{CoF_6}]^{3-}$: $\mathrm{Co^{3+}}$ is $d^6$; F⁻ is weak field $\Rightarrow$ high-spin, outer-orbital octahedral complex, hybridisation $sp^3d^2$ = IV.
   2. B) $[\mathrm{NiCl_4}]^{2-}$: $\mathrm{Ni^{2+}}$ is $d^8$; Cl⁻ is weak field, and with 4 ligands (no strong field to pair electrons) gives tetrahedral geometry, hybridisation $sp^3$ = III.
   3. C) $[\mathrm{Ni(CN)_4}]^{2-}$: $\mathrm{Ni^{2+}}$ is $d^8$; CN⁻ is a strong-field ligand, forcing pairing of the two unpaired $d$-electrons, giving a square planar geometry, hybridisation $dsp^2$ = I. …

   🔒 Start your 14-day free trial to see the full solution →
7. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.Identify the correctly matched pair from the following (A) $\mathrm{C-C}$ --------120 pm (B) $\mathrm{C=C}$ --------110 pm (C) $\mathrm{C \equiv C}$ --------154 pm (D) C–C Bond length in Benzene --------139 pm

   ›Reveal solutionSolution

   Of the four bond-length claims, only benzene's C–C bond length of 139 pm (intermediate between single and double bonds due to aromatic delocalization) is correct.

   Concept and Intuition

   Bond length decreases as bond order increases (more shared electron pairs pull the nuclei closer): a C–C single bond (~154 pm) is longer than a C=C double bond (~134 pm), which is longer than a C≡C triple bond (~120 pm). Benzene's bonds are all equal and intermediate (~139 pm) because of resonance delocalization of the π electrons around the ring — it's neither a pure single nor double bond.

   Step-by-Step Solution

   1. (A) claims C–C = 120 pm — this is actually close to the triple bond length, not the single bond length (~154 pm). Incorrect.
   2. (B) claims C=C = 110 pm — the real C=C double bond length is about 134 pm. Incorrect.
   3. (C) claims C≡C = 154 pm — the real triple bond length is about 120 pm (this is actually the single-bond value, swapped). Incorrect. …

   🔒 Start your 14-day free trial to see the full solution →
8. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Hybridization of positively charged and negatively charged carbons of the following respectively are $\mathrm{(CH_3)_3C^+}$ $\mathrm{CH_3C \equiv C^-}$ (A) $sp^2$, $sp$ (B) $sp^2$, $sp^2$ (C) $sp^3$, $sp^3$ (D) $sp^3$, $sp^2$

   ›Reveal solutionSolution

   Tests hybridization of a carbocation vs. a carbanion in a triple-bond system. Answer: $sp^2$ for the cation, $sp$ for the anion (option A).

   Concept and Intuition

   Hybridization of a charged carbon is decided by the number of sigma-bonded/lone-pair "electron domains" around it. A carbocation with 3 sigma bonds and an empty orbital is $sp^2$ (trigonal planar, like a carbon in an alkene minus one substituent). A carbanion that is part of a triple bond (as in an acetylide) already uses two of its hybrid orbitals for the two sigma bonds that make up part of the triple-bond framework — its geometry stays linear, so it remains $sp$ hybridized, with the negative charge (lone pair) occupying an $sp$ orbital, giving high s-character and hence high electronegative character (this is why terminal alkynes are acidic).

   Step-by-Step Solution

   1. $(CH_3)_3C^+$: central carbon bonded to 3 methyl groups via 3 sigma bonds, with an empty p orbital — 3 regions of electron density around it → $sp^2$ hybridization, trigonal planar geometry. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.Match the following List-I (Molecule/ion) : List II (Number of lone pairs of electrons on the central atom) A) $\mathrm{XeF_2}$ : I) 2 B) $\mathrm{XeO_3}$ : II) 0 C) $\mathrm{XeF_4}$ : III) 3 D) $\mathrm{PF_6^-}$ : IV) 1 The correct answer is (A) A-III, B-IV, C-I, D-II (B) A-I, B-II, C-IV, D-III (C) A-II, B-I, C-III, D-IV (D) A-III, B-IV, C-II, D-I

   ›Reveal solutionSolution

   Counting lone pairs via VSEPR gives XeF2=3, XeO3=1, XeF4=2, PF6-=0, matching List-II items III, IV, I, II respectively.

   Concept and Intuition

   For a central atom, total electron domains = bond pairs + lone pairs. Xenon starts with 8 valence electrons; each bond to F or O (Xe=O counted as one domain) uses one domain. Electrons not used in bonding remain as lone pairs. For $\mathrm{PF_6^-}$, phosphorus's 5 valence electrons plus 1 extra from the negative charge give 6 pairs, all used in 6 P-F bonds.

   Step-by-Step Solution

   1. $\mathrm{XeF_2}$: Xe (8 e-) makes 2 bond pairs, leaving 6 e- = 3 lone pairs (AX2E3, linear).
   2. $\mathrm{XeO_3}$: 3 Xe=O bonds use 3 domains, leaving 2 electrons = 1 lone pair (AX3E1, pyramidal).
   3. $\mathrm{XeF_4}$: 4 Xe-F bonds, leaving 4 electrons = 2 lone pairs (AX4E2, square planar). …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.The hybridization of central atom of $\mathrm{ClF_3}$, $\mathrm{NH_3}$, $\mathrm{SO_3}$ are respectively (A) $sp^2, sp^2, sp^2$ (B) $sp^3d, sp^3, sp^2$ (C) $sp^2, sp^3, sp^3d$ (D) $sp^3d, sp^3, sp^3$

    ›Reveal solutionSolution

    Counting electron domains on the central atom of each molecule gives hybridisations $sp^3d$ (ClF₃), $sp^3$ (NH₃), $sp^2$ (SO₃).

    Concept and Intuition

    Hybridisation of a central atom is determined by the total number of electron domains (sigma bonds + lone pairs) around it, per VSEPR theory — not by the number of atoms bonded alone. Lone pairs count just as much as bonding pairs toward the domain total.

    Step-by-Step Solution

    1. $ClF_3$: Cl (7 valence electrons) forms 3 Cl–F sigma bonds and retains 2 lone pairs $\Rightarrow 3+2=5$ domains $\Rightarrow sp^3d$ (T-shaped molecule).
    2. $NH_3$: N forms 3 N–H sigma bonds and retains 1 lone pair $\Rightarrow 3+1=4$ domains $\Rightarrow sp^3$ (pyramidal).
    3. $SO_3$: S forms 3 S=O bonds (with no lone pair left on S in the resonance-averaged structure) $\Rightarrow 3$ domains $\Rightarrow sp^2$ (trigonal planar). …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
    Q.Assertion (A): Carbon can form $p\pi-p\pi$ bonds with itself and with other small sized and high electronegative atoms. Reason (R): Other heavier elements of group 14 can also form $p\pi-p\pi$ bonds. (A) Both (A) and (R) are correct and R is the correct explanation of A (B) Both (A) and (R) are correct, but R is not the correct explanation of A (C) (A) is correct but (R) is not correct (D) (A) is incorrect but (R) is correct

    ›Reveal solutionSolution

    Carbon uniquely forms strong pπ-pπ bonds due to its small size and good orbital overlap, but heavier Group 14 elements cannot — so the Assertion is true and the Reason is false.

    Concept and Intuition

    Effective $p\pi-p\pi$ (sideways) orbital overlap requires the two overlapping p-orbitals to be of comparable, small size, so their lobes can overlap efficiently at bonding distance. Carbon's small atomic radius makes this overlap strong (as in C=C, C=O, C=N). Descending group 14 (Si, Ge, Sn, Pb), atomic and orbital sizes increase substantially, degrading this overlap, so these elements essentially cannot form stable $\pi$ bonds with themselves the way carbon can — they instead tend to catenate via single bonds or adopt higher coordination numbers.

    Step-by-Step Solution

    1. Assess Assertion: carbon (2p orbitals, small size) readily forms $\pi$ bonds with itself (C=C) and with other small electronegative atoms like N, O (C=N, C=O) via good $p\pi$-$p\pi$ overlap. This is a well-established fact. TRUE. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Calcium carbide + $\mathrm{D_2O} \rightarrow \underline{X} + \mathrm{Ca(OD)_2}$. The hybridization of carbon atom/s in X (A) $sp^2$ (B) $sp$ (C) $sp^3$ (D) $dsp^2$

    ›Reveal solutionSolution

    Calcium carbide's hydrolysis (here with heavy water) gives (deutero)acetylene, whose carbons are triple-bonded and therefore $sp$ hybridized — the isotope substitution (D for H) doesn't change the hybridization at all.

    Concept and Intuition

    Calcium carbide contains the carbide ion $C_2^{2-}$ ($:C\equiv C:^{2-}$), which on hydrolysis picks up two protons (or deuterons) to form acetylene ($HC\equiv CH$) or its deuterated analogue ($DC\equiv CD$). Isotopic substitution (H → D) changes mass but not electronic structure or bonding, so the hybridization of carbon is unaffected.

    Step-by-Step Solution

    1. Balanced reaction: $\mathrm{CaC_2 + 2D_2O \rightarrow C_2D_2 + Ca(OD)_2}$ (mirrors $\mathrm{CaC_2+2H_2O\rightarrow C_2H_2+Ca(OH)_2}$).
    2. X $=C_2D_2$, structurally $D-C\equiv C-D$.
    3. Each carbon forms one $\sigma$ bond to D, one $\sigma$ bond to the other C, and two $\pi$ bonds to the other C (triple bond) — 2 $\sigma$ regions of electron density around each C. …

    🔒 Start your 14-day free trial to see the full solution →