Q.In both water and dimethyl ether (CH3—O—CH3), oxygen atom is central atom, and has the same hybridisation, yet they have different bond angles. Which one has greater bond angle? Give reason.

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.In gas phase structure of $\mathrm{H_2O_2}$, the dihedral angle is (A) $94.5°$ (B) $101.9°$ (C) $111.5°$ (D) $90.2°$

   ›Reveal solutionSolution

   This tests the gas-vs-solid structural difference of $\mathrm{H_2O_2}$; the gas-phase dihedral (twist) angle is $111.5°$.

   Concept and Intuition

   $\mathrm{H_2O_2}$ has a non-planar, "open book" structure: two O–H bonds are twisted out of the O–O plane by a dihedral angle. This angle is not fixed — it changes with the environment because the molecule is fairly flexible about the O–O axis, and intermolecular hydrogen bonding (present in the solid/liquid, absent in isolated gas-phase molecules) changes the preferred twist to minimize lattice energy.

   Step-by-Step Solution

   1. In the gas phase, $\mathrm{H_2O_2}$ molecules are essentially isolated (no significant H-bonding to neighbours), so the molecule adopts its intrinsic minimum-energy skew conformation.
   2. This intrinsic (gas-phase) dihedral angle is experimentally found to be $111.5°$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Following is the structural representation of $N_2O_3$ molecule in which x, y, z are bond angles and p, q, r are bond lengths [FIGURE: structural diagram of N2O3 showing two N atoms bonded to each other (bond length q), left N bonded to two O atoms (bond lengths p, with angle x between the two O-N bonds), right N bonded to two O atoms (one via angle y, bond length r), and z labelling the angle at the N-N bond region] The correct orders of bond angles and bond lengths respectively are (A) $x<z<y$ & $p<q<r$ (B) $x<y<z$ & $p<r<q$ (C) $x<z<y$ & $p<r<q$ (D) $x<y<z$ & $p<q<r$

   ›Reveal solutionSolution

   $N_2O_3$ ($O{=}N{-}NO_2$) has a famously weak, long N–N bond (the reason it dissociates into NO + $NO_2$), a short true N=O double bond on the nitroso end, and an intermediate resonance-delocalised N–O bond on the nitro end — giving $p<r<q$ for lengths, and $x<z<y$ for the angles once lone-pair and steric effects are accounted for.

   Concept and Intuition

   $N_2O_3$ is best understood as two fragments joined by a single N–N bond: a nitrosyl-type nitrogen ($-N{=}O$, one oxygen, one lone pair) bonded to a nitro-type nitrogen ($-NO_2$, two oxygens, delocalised over both N–O bonds). Bond length tracks bond order: a genuine double bond (the terminal N=O on the nitrosyl end) is shorter than a bond with partial (resonance-averaged, ~1.5) double-bond character (the N–O bonds on the nitro end), which in turn is much shorter than the N–N bond joining the two halves. That N–N bond is a classic textbook weak point — it is this unusually long, weak single bond that makes $N_2O_3$ thermally unstable and prone to falling apart into NO and $NO_2$.

   Bond angle is governed by electron-domain geometry (VSEPR) and steric crowding. At the nitrosyl nitrogen, one of its three electron domains is a lone pair, which compresses the O–N–N angle below the ideal trigonal value. At the nitro nitrogen, all three domains are bonding pairs (to N and to two O's), so its angles stay closer to trigonal (~120°), but they're not identical to each other — whichever oxygen sits on the same side as (syn/closer to) the nitrosyl oxygen across the N–N bond experiences more steric repulsion and is pushed to a wider angle, while the oxygen on the far side keeps a more moderate angle.

   Step-by-Step Solution

   1. Identify the two nitrogens: the nitrosyl N (one O, one lone pair, bonded to the other N) and the nitro N (two O's, bonded to the other N). …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Which of the following statement is not correct about the structures of ozone and sulphur dioxide? (A) Both have angular shape (B) Both have resonance structures (C) Both have same hybridization (D) Both have same bond angles

   ›Reveal solutionSolution

   $O_3$ and $SO_2$ share angular shape, $sp^2$ hybridization and resonance, but NOT the same bond angle — so (D) is the incorrect statement.

   Concept and Intuition

   Both molecules have a central atom with one lone pair and two bonded oxygens in a bent (angular) $AX_2E$ arrangement, both delocalize their central atom's p-electron density across resonance structures, and both use $sp^2$ hybrid orbitals. But the exact bond angle depends on subtle differences in lone-pair/bond-pair repulsion and the electronegativity/size of the central atom (O vs S), so the specific angles are not identical.

   Step-by-Step Solution

   1. (A) Both angular — TRUE for both $O_3$ and $SO_2$.
   2. (B) Both have resonance structures — TRUE; each is a resonance hybrid of two equivalent canonical forms.
   3. (C) Both have same hybridization — TRUE; central atom is $sp^2$ in both cases. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Which of the following sets contain isostructural molecules? I. $H_2O$, $OF_2$, $SCl_2$ II. $CO_2$, $BeCl_2$, $HgCl_2$ III. $SiCl_4$, $SF_4$, $XeF_4$ Correct answer is (A) I, II, III (B) I, III only (C) II, III only (D) I, II only

   ›Reveal solutionSolution

   Isostructural means same shape/geometry class; sets I ($H_2O$/$OF_2$/$SCl_2$, all bent) and II ($CO_2$/$BeCl_2$/$HgCl_2$, all linear) qualify, but set III's members have three different shapes.

   Concept and Intuition

   Two species are isostructural when they share the same molecular geometry — which, by VSEPR, is decided by the total number of electron domains (bond pairs + lone pairs) around the central atom, not just its identity. Same central-atom electron-domain "formula" (like AX$_2$E$_2$ or AX$_2$) means same shape, even across different elements.

   Step-by-Step Solution

   1. Set I: $H_2O$ (O: 2 bond pairs + 2 lone pairs), $OF_2$ (O: 2 bond pairs + 2 lone pairs), $SCl_2$ (S: 2 bond pairs + 2 lone pairs). All are AX$_2$E$_2$, sp$^3$, bent/angular. Isostructural.
   2. Set II: $CO_2$ (C: sp, 2 $\sigma$ + no lone pair), $BeCl_2$ (Be: sp, 2 $\sigma$, no lone pair, gas-phase monomer), $HgCl_2$ (Hg: sp, 2 $\sigma$, no lone pair). All linear AX$_2$. Isostructural. …

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5. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.Match the following. List – I (Molecule) : (A) $\text{SO}_2$ (B) $\text{NO}_2$ (C) $\text{O}_3$ (D) $\text{S}_8$ List – II (Bond angle in degrees) : (I) 134 (II) 117 (III) 90 (IV) 119.5 (V) 107 The correct answer is: (A) A-IV, B-I, C-II, D-V (B) A-V, B-IV, C-II, D-III (C) A-IV, B-I, C-III, D-V (D) A-V, B-I, C-IV, D-III

   ›Reveal solutionSolution

   Matching each molecule to its known bond angle — SO₂ 119.5°, NO₂ 134°, O₃ 117°, S₈ 107° — gives A-IV, B-I, C-II, D-V.

   Concept and Intuition

   Bond angle in a bent/angular molecule is set by the balance of lone-pair and bonding-pair repulsions (VSEPR) around the central atom, and for radicals, by the presence of a single unpaired electron instead of a full lone pair (which repels less than a lone pair, widening the angle).

   Step-by-Step Solution

   1. $\text{NO}_2$: nitrogen has one unpaired electron (not a full lone pair) besides two bonding regions; the reduced repulsion from a single electron (vs. a lone pair) widens the angle to about $134^\circ$ — matches (I).
   2. $\text{O}_3$ (ozone): central O has one lone pair plus resonance-delocalised bonding — bond angle is close to $117^\circ$ — matches (II).
   3. $\text{SO}_2$: sulfur has one lone pair, similar bent geometry to ozone but slightly more open — bond angle $\approx 119.5^\circ$ — matches (IV). …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The structure of $\mathrm{Al_2Cl_6}$ is given below [FIGURE] (a dimeric Al2Cl6 structure: two Al atoms each bonded to two terminal Cl atoms, bridged by two Cl atoms between the Al centres, with curved arrows showing the dative bonding of the bridging chlorines; angle X is the Cl-Al-Cl angle at the bridging chlorines on the left Al, angle Y is the Al-Cl-Al bridging bond angle, and angle Z is the terminal Cl-Al-Cl angle on the right Al) The correct order of bond angles X, Y and Z is (A) $X > Y > Z$ (B) $Z > X > Y$ (C) $Y > X > Z$ (D) $Z > Y > X$

   ›Reveal solutionSolution

   Terminal angle (Z) is largest, ring angle at Al (X) next, bridge angle at Cl (Y) smallest: $Z > X > Y$.

   Concept and Intuition

   $\mathrm{Al_2Cl_6}$ is a dimer in which two $\mathrm{AlCl_3}$ units join through two bridging chlorines, forming a puckered 4-membered $\mathrm{Al\text{-}Cl\text{-}Al\text{-}Cl}$ ring. Each Al is roughly $sp^3$. The terminal $\mathrm{Cl}$ atoms are held by ordinary covalent bonds and spread apart to reduce repulsion, so the terminal $\mathrm{Cl\text{-}Al\text{-}Cl}$ angle opens out to about $120^\circ$. The angles inside the strained ring are forced to be much smaller.

   Step-by-Step Solution

   1. Identify the three angles: X = $\mathrm{Cl\text{-}Al\text{-}Cl}$ at Al between the two bridging chlorines (ring angle at Al); Y = $\mathrm{Al\text{-}Cl\text{-}Al}$ at the bridging chlorine; Z = terminal $\mathrm{Cl\text{-}Al\text{-}Cl}$ at Al. …

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7. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Consider the following molecules $PF_5,\ H_2O,\ NH_3,\ XeF_2,\ BF_3,\ SF_6,\ IF_7$ In how many of the above molecules, the ratio between the number of bond pairs of electrons and lone pairs of electrons is 1 : 3? (A) 4 (B) 3 (C) 2 (D) 5

   ›Reveal solutionSolution

   Counting ALL bond pairs and ALL lone pairs in each whole molecule (not just on the central atom) reveals a pattern: whenever the central atom has zero lone pairs and is surrounded only by fluorine atoms (each contributing 3 lone pairs), the bond-pair:lone-pair ratio is always exactly 1:3.

   Concept and Intuition

   Each terminal fluorine atom in these molecules is singly bonded and carries 3 lone pairs (since F has 7 valence electrons, 1 used in the bond, 6 left = 3 lone pairs). So if a central atom has $n$ bonds to fluorine and 0 lone pairs of its own, the whole molecule has $n$ bond pairs and $3n$ lone pairs — an automatic 1:3 ratio, regardless of $n$. Molecules with lone pairs on the central atom, or with non-F terminal atoms (like H, which has no lone pairs), break this neat pattern.

   Step-by-Step Solution

   1. PF5: P has 5 bonds to F, 0 lone pairs on P. Bond pairs = 5; lone pairs (on the 5 F atoms) = $5\times3=15$. Ratio $5:15=1:3$. ✓
   2. H2O: O has 2 bonds to H, 2 lone pairs on O; H atoms have no lone pairs. Bond pairs = 2; lone pairs = 2. Ratio $2:2=1:1$. ✗
   3. NH3: N has 3 bonds to H, 1 lone pair on N. Bond pairs = 3; lone pairs = 1. Ratio $3:1$. ✗
   4. XeF2: Xe has 2 bonds to F, 3 lone pairs on Xe; each F has 3 lone pairs (2×3=6). Bond pairs = 2; lone pairs = $3+6=9$. Ratio $2:9$. ✗
   5. BF3: B has 3 bonds to F, 0 lone pairs on B. Bond pairs = 3; lone pairs = $3\times3=9$. Ratio $3:9=1:3$. ✓ …

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8. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.White phosphorus reacts with sulphuryl chloride and forms a solid substance A and a gas D. When A is heated at high temperature, it decomposes to give a colorless oily liquid 'X' and a gas C. The shape of X is (A) Pyramidal (B) Trigonal bipyramidal (C) Square planar (D) Tetrahedral

   ›Reveal solutionSolution

   The sequence identifies A as solid $PCl_5$ and X as $PCl_3$, whose shape (like $NH_3$) is trigonal pyramidal due to phosphorus's lone pair.

   Concept and Intuition

   White phosphorus reacts with excess sulphuryl chloride to fully chlorinate to $PCl_5$, a solid, releasing $SO_2$ gas as a byproduct. $PCl_5$ is thermally unstable and readily dissociates on heating into $PCl_3$ (a colourless, fuming, oily liquid at room temperature) and chlorine gas. The shape of any $AX_3E$ species (3 bond pairs + 1 lone pair around the central atom, $sp^3$ hybridised) is trigonal pyramidal — the classic example being $NH_3$, and $PCl_3$ is directly analogous since P has 5 valence electrons, uses 3 to bond to Cl and retains one lone pair.

   Step-by-Step Solution

   1. White phosphorus + excess $SO_2Cl_2$ (a chlorinating agent) $\rightarrow$ solid A $= PCl_5$ + gas D $= SO_2$.
   2. Heating A ($PCl_5$) at high temperature causes thermal dissociation: $PCl_5 \rightarrow PCl_3 + Cl_2$.
   3. X $= PCl_3$ (the 'colourless oily liquid'), C $= Cl_2$ (the gas). …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The increasing order of number of lone pair of electrons on the central atom of the following molecules is I) $\mathrm{ClF_3}$ II) $\mathrm{XeF_2}$ III) $\mathrm{SF_4}$ IV) $\mathrm{SiH_4}$ (A) IV < III < II < I (B) I < II < III < IV (C) II < I < III < IV (D) IV < III < I < II

   ›Reveal solutionSolution

   This tests VSEPR lone-pair counting on the central atom of four molecules. The increasing order of lone pairs is $SiH_4 (0) < SF_4 (1) < ClF_3 (2) < XeF_2 (3)$, i.e. IV < III < I < II.

   Concept and Intuition

   For a central atom, total electron domains = bonding pairs + lone pairs. Central-atom valence electrons plus contributions from surrounding atoms (minus charge adjustments) give the steric number; subtracting the number of $\sigma$-bonds (= number of surrounding atoms, since none of these molecules has multiple bonds to the central atom) leaves the lone pairs. This is a standard VSEPR bookkeeping exercise.

   Step-by-Step Solution

   1. $SiH_4$: Si (group 14, 4 valence e$^-$) bonds to 4 H atoms using all 4 electrons in 4 bond pairs. Lone pairs $= 0$. Shape: tetrahedral.
   2. $SF_4$: S (group 16, 6 valence e$^-$) forms 4 S–F bonds (using 4 electrons, one each) leaving $6-4=2$ electrons $=1$ lone pair. Shape: see-saw.
   3. $ClF_3$: Cl (group 17, 7 valence e$^-$) forms 3 Cl–F bonds (using 3 electrons) leaving $7-3=4$ electrons $=2$ lone pairs. Shape: T-shaped.
   4. $XeF_2$: Xe (group 18, 8 valence e$^-$) forms 2 Xe–F bonds (using 2 electrons) leaving $8-2=6$ electrons $=3$ lone pairs. Shape: linear. …

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10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.How many of the following molecules have two lone pairs of electrons on central atom? $\mathrm{SF_6, BF_3, ClF_3, PCl_5, BrF_5, XeF_4, H_2O, SF_4}$ (A) 5 (B) 4 (C) 3 (D) 2

    ›Reveal solutionSolution

    This tests VSEPR-based lone-pair counting on the central atom of eight species; three of them (ClF3, XeF4, H2O) carry exactly two lone pairs.

    Concept and Intuition

    The number of lone pairs on a central atom = (its group valence electrons − electrons used in bonds actually shown, since AP EAPCET options here are for simple single-bonded species) worked out via the standard VSEPR electron-pair count: total electron pairs around the central atom = bond pairs + lone pairs, and total electron pairs = (central atom valence electrons + electrons contributed by bonded halogens counted as one shared pair each)/2 for these AX_n type halides.

    Step-by-Step Solution

    1. $\mathrm{SF_6}$: S is $sp^3d^2$ hybridised, forms 6 S–F bonds, uses all 6 valence electrons of S in bonding → 0 lone pairs (octahedral).
    2. $\mathrm{BF_3}$: B has 3 valence electrons, all used in 3 B–F bonds → 0 lone pairs (trigonal planar).
    3. $\mathrm{ClF_3}$: Cl has 7 valence electrons; 3 are used in the 3 Cl–F bonds, leaving 4 electrons = 2 lone pairs (T-shaped, $sp^3d$).
    4. $\mathrm{PCl_5}$: P has 5 valence electrons, all used in 5 P–Cl bonds → 0 lone pairs (trigonal bipyramidal).
    5. $\mathrm{BrF_5}$: Br has 7 valence electrons; 5 used in the 5 Br–F bonds, leaving 2 electrons = 1 lone pair (square pyramidal, $sp^3d^2$).
    6. $\mathrm{XeF_4}$: Xe has 8 valence electrons; 4 used in the 4 Xe–F bonds, leaving 4 electrons = 2 lone pairs (square planar, $sp^3d^2$). …

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11. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.How many of the following molecules / ions have trigonal planar structure ? $\mathrm{BO_3^{3-}, NH_3, PCl_3, BCl_3, ClF_3, XeO_3}$ (A) 5 (B) 2 (C) 4 (D) 3

    ›Reveal solutionSolution

    Tests VSEPR shape prediction; only $BO_3^{3-}$ and $BCl_3$ (both central atoms with 3 bond pairs, 0 lone pairs) are trigonal planar — a count of 2.

    Concept and Intuition

    VSEPR theory says molecular shape depends on the total number of electron domains (bond pairs + lone pairs) around the central atom. A species is trigonal planar only when the central atom has exactly 3 bonding domains and zero lone pairs (sp² hybridization with no lone pair distorting the geometry). If a lone pair is present alongside 3 bond pairs, the shape becomes pyramidal instead.

    Step-by-Step Solution

    1. $BO_3^{3-}$: B has 3 bonds to O and no lone pair (B contributes 3 valence electrons, all used in bonding) → trigonal planar.
    2. $NH_3$: N has 3 bond pairs + 1 lone pair → trigonal pyramidal, not planar.
    3. $PCl_3$: P has 3 bond pairs + 1 lone pair → trigonal pyramidal, not planar.
    4. $BCl_3$: B has 3 bonds, no lone pair → trigonal planar.
    5. $ClF_3$: Cl has 3 bond pairs + 2 lone pairs → T-shaped, not planar. …

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12. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.A molecule has T-shape. The total number of electron pairs in the valence shell of central atom of it is (A) 4 (B) 5 (C) 6 (D) 3

    ›Reveal solutionSolution

    This tests VSEPR geometry-to-electron-pair-count mapping; a T-shape needs 5 electron pairs on the central atom, option (B).

    Concept and Intuition

    VSEPR theory says molecular shape is decided by the total number of electron pairs (bonding + lone) around the central atom, arranged to minimize repulsion. A T-shape arises from a trigonal bipyramidal electron-pair arrangement ($AX_5$-type, 5 pairs) where 2 of the 5 positions (the equatorial ones, which have least repulsion) are occupied by lone pairs instead of bonds, leaving 3 bonds arranged in a T (classic example: $\mathrm{ClF_3}$).

    Step-by-Step Solution

    1. A T-shaped molecule has 3 bonding pairs (giving the 3 visible bonds) with the shape distorted from trigonal planar/pyramidal by lone-pair repulsion. …

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