Q.In which of the following molecules, σ2p_z molecular orbital is filled after π2p_x and π2p_y molecular orbitals?

Showing the 12 most recent of 60 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\mathrm{BCl_3}$ on hydrolysis gives a complex ion $X$. $\mathrm{AlCl_3}$ in acidified aqueous solution forms a complex ion $Y$. The hybridization of central atoms in $X$ and $Y$ are respectively (A) $sp^3, sp^3$ (B) $sp^3, sp^3d$ (C) $sp^2, sp^3d^2$ (D) $sp^3, sp^3d^2$

   ›Reveal solutionSolution

   $X=[\mathrm{B(OH)_4}]^-$ is tetrahedral ($sp^3$ boron); $Y=[\mathrm{Al(H_2O)_6}]^{3+}$ is octahedral ($sp^3d^2$ aluminium). Answer: $sp^3, sp^3d^2$.

   Concept and Intuition

   Both boron and aluminium halides are electron-deficient Lewis acids that hydrolyse in water and then continue to act as Lewis acids toward additional water/hydroxide, because the central atom still has an empty low-lying orbital (or in Al's case, can expand its coordination number using d-orbitals, unlike period-2 boron). This is why boric acid's "acidity" in water is not proton donation but hydroxide-ion acceptance, and why $\mathrm{Al^{3+}}$ readily forms a hexa-aqua octahedral complex rather than stopping at a simple hydroxide.

   Step-by-Step Solution

   1. $\mathrm{BCl_3} + 3\mathrm{H_2O} \rightarrow \mathrm{B(OH)_3} + 3\mathrm{HCl}$ gives boric acid, $\mathrm{B(OH)_3}$.
   2. Boric acid is a Lewis acid: it doesn't ionise to release $\mathrm{H^+}$ itself; instead it accepts a hydroxide ion from a water molecule: $\mathrm{B(OH)_3} + \mathrm{H_2O} \rightleftharpoons [\mathrm{B(OH)_4}]^- + \mathrm{H^+}$.
   3. In $X=[\mathrm{B(OH)_4}]^-$, boron is surrounded by 4 $\mathrm{OH}$ groups tetrahedrally — boron uses $sp^3$ hybrid orbitals (boron is a period-2 element, no d-orbitals available, so it cannot exceed 4-coordination).
   4. In acidified aqueous solution, $\mathrm{Al^{3+}}$ (from hydrolysed $\mathrm{AlCl_3}$) is strongly hydrated, forming the octahedral complex $Y=[\mathrm{Al(H_2O)_6}]^{3+}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.The pair of molecules with same type of hybridisation is (A) $\mathrm{H_3BO_3,\ H_3PO_3}$ (B) $\mathrm{NO_2,\ SO_3}$ (C) $\mathrm{XeO_3,\ BF_3}$ (D) $\mathrm{PCl_3,\ ClF_3}$

   ›Reveal solutionSolution

   This tests assigning hybridisation from electron-domain count (bond pairs + lone pairs); only the $NO_2$/$SO_3$ pair shares the same (sp²) hybridisation.

   Concept and Intuition

   Hybridisation of the central atom is fixed by the number of $\sigma$-bonds plus lone pairs around it (VSEPR electron-domain count): 2→sp, 3→sp², 4→sp³, 5→sp³d, 6→sp³d². Matching hybridisation across two different molecules means checking each central atom's true electron-domain count, not just guessing from formula similarity.

   Step-by-Step Solution

   1. $H_3BO_3$: boron is bonded to 3 –OH groups, no lone pair on B → 3 domains → sp² (trigonal planar). $H_3PO_3$: phosphorous acid is $HP(=O)(OH)_2$; P is bonded to 1 H, 2 –OH, and a double-bonded O → 4 $\sigma$-domains → sp³. Different from B. ✗
   2. $NO_2$: N has 2 $\sigma$-bonds to O plus one unpaired (non-bonding) electron → effectively 3 domains → sp² (bent, ~134°). $SO_3$ (gaseous, monomeric): S has 3 $\sigma$-bonds, no lone pair → 3 domains → sp² (trigonal planar). Same as $NO_2$. ✓
   3. $XeO_3$: Xe has 3 $\sigma$-bonds + 1 lone pair → 4 domains → sp³ (pyramidal). …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.How many of the following molecules are linear with no lone pairs of electrons on the central atom? $BeCl_2, O_3, SCl_2, XeF_2, SnCl_2, PbCl_2, HgCl_2$ (A) 2 (B) 3 (C) 4 (D) 1

   ›Reveal solutionSolution

   Only molecules that are BOTH linear AND have zero lone pairs on the central atom count. That's just $BeCl_2$ and $HgCl_2$ — answer 2.

   Concept and Intuition

   Linear geometry can arise two different ways: either the central atom has exactly 2 bonding domains and no lone pairs (sp hybridisation, e.g. $BeCl_2$, $HgCl_2$), or it has 2 bonding domains plus lone pairs arranged so the lone pairs occupy equatorial positions and the bonds end up 180° apart (e.g. $XeF_2$, sp$^3$d with 3 lone pairs). The question specifically asks for the first kind — linear and lone-pair-free — so a molecule like $XeF_2$, though geometrically linear, must be excluded because Xe does carry lone pairs.

   Step-by-Step Solution

   1. $BeCl_2$: Be has 2 bond pairs, 0 lone pairs → sp, linear. Qualifies.
   2. $O_3$: central O has 1 lone pair → bent (~117°). Excluded.
   3. $SCl_2$: S has 2 lone pairs → bent (~103°), like water. Excluded.
   4. $XeF_2$: Xe has 3 lone pairs (sp$^3$d) → shape is linear, but lone pairs are present on the central atom, so it fails the "no lone pairs" condition. Excluded.
   5. $SnCl_2$: Sn(II) has a lone pair (inert pair effect) → bent, like $SO_2$. Excluded.
   6. $PbCl_2$: similarly, Pb(II) has a lone pair → bent as a discrete molecule. Excluded. …

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Observe the following set of reactions $Y \xleftarrow[\Delta]{Na/NH_3(l)} C_2H_2 \xrightarrow{H_2/Pt} X$ Hybridization of carbon in X and Y respectively is (A) $sp^2, sp^2$ (B) $sp^3, sp^3$ (C) $sp^2, sp^3$ (D) $sp^3, sp^2$

   ›Reveal solutionSolution

   Acetylene reduced by $\mathrm{Na/NH_3(l)}$ gives the alkene ethylene ($Y$, $sp^2$ carbon), while full catalytic hydrogenation with $\mathrm{H_2/Pt}$ gives the alkane ethane ($X$, $sp^3$ carbon).

   Concept and Intuition

   Alkynes can be partially reduced to alkenes using dissolving-metal conditions (Na in liquid ammonia), which adds only one equivalent of $\mathrm{H_2}$ across the triple bond (typically giving the trans alkene for internal alkynes). Alternatively, catalytic hydrogenation over a metal catalyst like Pt with excess $\mathrm{H_2}$ adds two equivalents of $\mathrm{H_2}$, fully saturating the triple bond all the way to the alkane. The degree of saturation directly determines the carbon hybridisation: triple bond carbon is $sp$, double bond carbon is $sp^2$, and fully single-bonded (saturated) carbon is $sp^3$.

   Step-by-Step Solution

   1. Start: $\mathrm{C_2H_2}$ (acetylene, $HC\equiv CH$), carbon here is $sp$-hybridised (not asked, but the starting point).
   2. Path to $Y$: $\mathrm{Na/NH_3(l)}, \Delta$ performs a partial (dissolving-metal) reduction, adding one $\mathrm{H_2}$ equivalent across the triple bond to give the alkene $\mathrm{CH_2=CH_2}$ (ethylene) as $Y$. Each carbon here has 3 σ-bonds (to 2 H and 1 C) plus one π-bond, i.e., $sp^2$ hybridised. …

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5. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The number of chlorine atoms attached to $sp^2$ carbon in the structure of DDT is X. The number of chlorine atoms attached to $sp^3$ carbon in chloramphenicol is Y. What are X and Y respectively? (A) 3, 2 (B) 2, 3 (C) 5, 2 (D) 2, 2

   ›Reveal solutionSolution

   DDT has 2 chlorines on aromatic ($sp^2$) carbons (one on each chlorophenyl ring) and 3 on the $sp^3$ $-CCl_3$ carbon; chloramphenicol has exactly 2 chlorines, both on its $sp^3$ dichloromethyl carbon. So X = 2, Y = 2.

   Concept and Intuition

   DDT (dichlorodiphenyltrichloroethane) has the structure $(4\text{-}ClC_6H_4)_2CH-CCl_3$ — two para-chlorophenyl rings joined to a central $CH$ carbon, which is itself joined to a $-CCl_3$ group. Its five chlorine atoms split into two structurally distinct sets: the two ring-chlorines (on aromatic $sp^2$ carbons) and the three trichloromethyl chlorines (on one aliphatic $sp^3$ carbon).

   Chloramphenicol's structure is a p-nitrophenyl group attached to $-CH(OH)-CH(NHCOCHCl_2)-CH_2OH$; the antibacterial-active dichloroacetyl group $-COCHCl_2$ carries both of the molecule's chlorine atoms on one $sp^3$ carbon.

   Step-by-Step Solution

   1. Draw DDT: $Cl_3C-CH(C_6H_4Cl)_2$.
   2. Each chlorophenyl ring contributes 1 Cl bonded to a ring ($sp^2$) carbon → 2 such chlorines total (X = 2).
   3. The $-CCl_3$ carbon (sp3) carries the other 3 chlorines — not what's being asked (that's for a different sub-question), so it doesn't affect X. …

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6. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The conjugate base of hydrogen carbonate ion is $\underline{X}$. The hybridization of central atom in X is (A) $sp$ (B) $sp^2$ (C) $sp^3$ (D) $sp^3d$

   ›Reveal solutionSolution

   This tests conjugate acid-base pairs and hybridisation from molecular geometry.

   Concept and Intuition

   A conjugate base is formed when a species donates (loses) a proton ($H^+$). $HCO_3^-$ losing a proton gives $CO_3^{2-}$. The carbonate ion is a classic trigonal-planar, resonance-stabilised oxyanion (three equivalent C–O bonds, bond order $4/3$ each), which is only possible if the central carbon is $sp^2$ hybridised, leaving one unhybridised $p$ orbital for the delocalised $\pi$ system.

   Step-by-Step Solution

   1. $HCO_3^-$ (hydrogen carbonate/bicarbonate ion) loses $H^+$: $HCO_3^- \to CO_3^{2-} + H^+$.
   2. So $X = CO_3^{2-}$ (carbonate ion), the central atom being carbon.
   3. $CO_3^{2-}$ has three C–O bonds arranged symmetrically at $120°$ (trigonal planar), with delocalised $\pi$ bonding shown by three resonance structures. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Identify the correct set of molecules / ions in which hybridization of central atom is not same (A) $H_2O,\ NH_3,\ CH_4$ (B) $H_3O^+,\ NH_4^+,\ ClO_4^-$ (C) $CH_3^+,\ BF_3,\ SO_2$ (D) $SF_4,\ XeF_4,\ SO_3$

   ›Reveal solutionSolution

   This tests hybridization assignment via steric number (bond pairs + lone pairs) for several common species. Only the set $SF_4, XeF_4, SO_3$ has three different hybridizations at the central atom.

   Concept and Intuition

   Hybridization of a central atom is determined by its steric number = (number of sigma bonds) + (number of lone pairs). Matching steric numbers across a molecular family (VSEPR) tells us whether their central-atom hybridizations agree, even though the molecules look chemically different.

   Step-by-Step Solution

   1. (A) $H_2O$: O has 2 bond pairs + 2 lone pairs = steric number 4 → $sp^3$. $NH_3$: N has 3 bond pairs + 1 lone pair = 4 → $sp^3$. $CH_4$: C has 4 bond pairs = 4 → $sp^3$. All same.
   2. (B) $H_3O^+$: O has 3 bond pairs + 1 lone pair = 4 → $sp^3$. $NH_4^+$: N has 4 bond pairs = 4 → $sp^3$. $ClO_4^-$: Cl has 4 sigma bonds = 4 → $sp^3$. All same.
   3. (C) $CH_3^+$: C has 3 bond pairs + 0 lone pair (empty p-orbital) = 3 → $sp^2$. $BF_3$: B has 3 bond pairs = 3 → $sp^2$. $SO_2$: S has 2 bond pairs + 1 lone pair = 3 → $sp^2$. All same. …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Observe the following reaction $\mathrm{Na_2B_4O_7 \xrightarrow{H_2O} Acid + alkali}$ The hybridisation of central atom of acid is (A) $sp^3$ (B) $sp^2$ (C) $dsp^2$ (D) $sp^3d^2$

   ›Reveal solutionSolution

   Borax hydrolyses to give sodium hydroxide (alkali) and orthoboric acid $H_3BO_3$ (acid); boron in $H_3BO_3$ is $sp^2$ hybridised.

   Concept and Intuition

   Borax, $Na_2B_4O_7\cdot10H_2O$, is the sodium salt of a weak acid (boric acid) and behaves as a salt of a strong base + weak acid, so in water it hydrolyses to release both an alkali and the free weak acid. The key structural fact you need is the shape of boric acid: it is a planar molecule, $B(OH)_3$, with boron surrounded by three $-OH$ groups arranged in a trigonal plane and an empty $p$-orbital perpendicular to that plane (which is why boric acid is a Lewis acid, not a Bronsted acid — it accepts $OH^-$ from water rather than donating a proton).

   Step-by-Step Solution

   1. Write the hydrolysis of borax: $Na_2B_4O_7 + 7H_2O \rightarrow 2NaOH + 4H_3BO_3$.
   2. Identify the acid: $H_3BO_3$ (orthoboric acid); the alkali is $NaOH$.
   3. Determine boron's hybridisation in $H_3BO_3$: boron forms 3 $\sigma$-bonds to oxygen (of the 3 $-OH$ groups) and has no lone pair on boron itself — 3 bonding regions, planar arrangement. …

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9. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Which of the following represent the most stable structures of $SF_4$ and $ClF_3$ respectively? [FIGURE] (Lewis structures pairing $SF_4$ (S bonded to 4 F atoms) with $ClF_3$ (Cl bonded to 3 F atoms); the four options differ only in where the lone pair(s) -- one on S, two on Cl -- are drawn relative to the fluorine atoms, testing the correct VSEPR seesaw ($SF_4$) / T-shape ($ClF_3$) lone-pair placement) (A) [FIGURE] ($SF_4$: lone pair drawn on the left side of S, with F atoms up, right, lower-right and down; $ClF_3$: one F drawn up via a double bond line, one F right, one F down, with both lone pairs drawn on the left side, upper-left and lower-left) (B) [FIGURE] ($SF_4$: lone pair drawn at the top of S (axial), with F atoms left, upper-right, right and down; $ClF_3$: lone pair drawn near the top F (upper area) and the other lone pair drawn near the bottom F (lower area)) (C) [FIGURE] ($SF_4$: lone pair drawn at the top of S (axial), with F atoms left, upper-right, right and down -- same skeleton as option B; $ClF_3$: both lone pairs drawn on the same (left) side, upper-left and lower-left) (D) [FIGURE] ($SF_4$: lone pair drawn upper-left of S, with F atoms up, right, lower-right and down; $ClF_3$: lone pairs drawn upper-left and lower-left, with F atoms upper-left, right and lower-left)

   ›Reveal solutionSolution

   This tests VSEPR lone-pair placement in trigonal-bipyramidal-derived shapes: the most stable $SF_4$ (seesaw) has its lone pair equatorial, and the most stable $ClF_3$ (T-shape) has both lone pairs equatorial and adjacent to each other — only one option draws both correctly.

   Concept and Intuition

   Both $SF_4$ and $ClF_3$ derive their geometry from an underlying trigonal bipyramidal (5 electron-domain) arrangement, which has two distinct kinds of positions: two axial positions (180° apart, perpendicular to the equatorial plane) and three equatorial positions (120° apart from each other, 90° from the axial positions). Lone pairs are more diffuse and repel more strongly than bonding pairs, so the most stable arrangement is the one that minimises the number of 90° (strongest) repulsions involving lone pairs.

   For $SF_4$: sulfur has 4 bonding pairs (to F) and 1 lone pair. If the lone pair were axial, it would have three 90° interactions with the three equatorial bonding pairs — high repulsion. If the lone pair is equatorial instead, it has only two 90° interactions (with the two axial bonding pairs) and two 120° interactions (with the other two equatorial bonding pairs) — much lower net repulsion. So the stable/observed "seesaw" geometry places the lone pair equatorially.

   For $ClF_3$: chlorine has 3 bonding pairs and 2 lone pairs. To minimise 90° lone-pair interactions, both lone pairs should occupy equatorial positions (there are 3 equatorial slots; 2 are taken by lone pairs, 1 by an F atom, and the remaining 2 F atoms occupy the two axial positions). Because the two lone pairs are both equatorial, they sit adjacent to each other (120° apart) rather than being split between an axial and an equatorial slot (which would create additional 90° lone-pair/bond-pair repulsions) or being placed to bracket the axial fluorines. This equatorial-equatorial adjacent placement is exactly what gives the familiar bent "T-shape" molecular geometry (bond angles slightly less than 90° and 180°).

   Step-by-Step Solution

   1. Recognise $SF_4$ has 5 electron domains (4 bond pairs + 1 lone pair) → trigonal bipyramidal electron geometry, seesaw molecular geometry.
   2. Apply the rule: lone pairs prefer equatorial positions in TBP geometry to minimise 90° repulsions → the correct $SF_4$ picture has its lone pair drawn to the side (equatorial), not at the "top"/axial position.
   3. Recognise $ClF_3$ has 5 electron domains (3 bond pairs + 2 lone pairs) → trigonal bipyramidal electron geometry, T-shaped molecular geometry.
   4. Apply the same equatorial-preference rule twice: both lone pairs should be equatorial, and since there are only 3 equatorial slots, they end up adjacent to each other on the same side, with the two axial slots taken by F atoms. …

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10. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
    Q.Observe the following statements Statement – I: The correct order of O – O bond length in $\mathrm{O_2, H_2O_2}$ and $\mathrm{O_3}$ is $\mathrm{H_2O_2 > O_3 > O_2}$ Statement – II: Hybridisation of carbon in graphite and pyridine is same The correct answer is (A) Both statements I and II are correct (B) Statement I is correct, but statement II is not correct (C) Statement I is not correct, but statement II is correct (D) Both statements I and II are not correct

    ›Reveal solutionSolution

    Both statements check out: O–O bond length really is $H_2O_2>O_3>O_2$ (bond order 1, 1.5, 2 respectively — longer bond ↔ lower bond order), and carbon is $sp^2$ in both graphite and pyridine.

    Concept and Intuition

    Bond length is inversely related to bond order: more shared electron pairs (higher bond order) pull the nuclei closer together, shortening the bond. In $O_2$ the O–O bond order is 2 (double bond); in $O_3$, resonance delocalizes one $\pi$ bond over two O–O linkages giving each an effective bond order of 1.5; in $H_2O_2$ the O–O bond is a pure single bond (order 1). So bond length should increase as bond order decreases: $O_2 < O_3 < H_2O_2$, i.e. $H_2O_2 > O_3 > O_2$ — exactly Statement I.

    For hybridization, both graphite and pyridine are built from planar, hexagonally-arranged $sp^2$ carbons with a delocalized $\pi$-electron system: graphite is stacked sheets of fused six-membered all-carbon aromatic-like rings, and pyridine is a six-membered aromatic ring (isoelectronic with benzene, N replacing one CH). Both give trigonal-planar, $sp^2$-hybridized ring carbons.

    Step-by-Step Solution

    1. Identify O–O bond order in each species: $O_2\to2$, $O_3\to1.5$ (per O–O bond, due to resonance), $H_2O_2\to1$.
    2. Apply bond-order–bond-length inverse relation: lower bond order → longer bond. So length ranks $H_2O_2 (1) > O_3 (1.5) > O_2 (2)$ — matches Statement I exactly.
    3. Recall graphite's structure: fused six-membered rings of $sp^2$ carbon, each carbon $\sigma$-bonded to three neighbours plus one delocalized $p$-orbital electron in the $\pi$ system — hybridization $sp^2$. …

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11. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.Aluminium carbide on reaction with heavy water gives a carbon compound $X$. The hybridization in $X$ is (A) $sp$ (B) $sp^2$ (C) $sp^3$ (D) $dsp^2$

    ›Reveal solutionSolution

    $Al_4C_3$ reacts with $D_2O$ to give deutero-methane ($CD_4$), whose carbon is $sp^3$ hybridised.

    Concept and Intuition

    Aluminium carbide is the carbide of a highly electropositive metal and hydrolyses readily in water (or heavy water) to release a hydrocarbon — methane (or its deuterated analogue) — along with the metal hydroxide. Isotopic substitution (H → D) changes mass and bond vibrational frequency but not the electronic structure or geometry, so hybridisation is unaffected.

    Step-by-Step Solution

    1. Write the hydrolysis reaction with heavy water: $Al_4C_3 + 12D_2O \rightarrow 4Al(OD)_3 + 3CD_4$.
    2. The carbon compound $X$ formed is $CD_4$ (methane-$d_4$).
    3. Carbon in $CD_4$ forms four equivalent $\sigma$ bonds to D atoms arranged tetrahedrally, exactly as in $CH_4$. …

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12. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
    Q.Acrolein, formaldehyde and peroxy acetyl nitrate are the main compounds present in photochemical smog. The total number of $sp^2$ carbons present in these three compounds is (A) 2 (B) 3 (C) 4 (D) 5

    ›Reveal solutionSolution

    Counting $sp^2$ carbons in acrolein, formaldehyde and PAN — the three characteristic photochemical-smog compounds — gives a total of 5.

    Concept and Intuition

    Any carbon involved in a C=C or C=O double bond is $sp^2$ hybridised (trigonal planar, one unhybridised p orbital for the $\pi$ bond). Carbons bonded only by single bonds (like a methyl group) remain $sp^3$.

    Step-by-Step Solution

    1. Acrolein, $CH_2\!=\!CH\!-\!CHO$: three carbons — $C1(=CH_2)$, $C2(=CH-)$ both part of the C=C double bond ($sp^2$), and $C3$ (the aldehyde carbon, C=O) also $sp^2$. Total = 3.
    2. Formaldehyde, $HCHO$: one carbon, the carbonyl carbon, $sp^2$. Total = 1. …

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