Q.Display electron distribution around the oxygen atom in the water molecule and state the shape of the molecule. Also write the H-O-H bond angle.

Showing the 12 most recent of 20 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following is mismatch regarding shape?

   (a) XeF4 — Octahedral

   (b) PCl5 — Trigonal bipyramidal

   (c) SO2 — Bent

   (d) CH4 — Tetrahedral

   ›Reveal solutionSolution

   Check each pair against VSEPR theory. XeF4 has 4 bond pairs + 2 lone pairs on Xe; the electron-pair arrangement is octahedral, but the MOLECULAR SHAPE (what the atoms actually trace out) is square planar, not octahedral.

   Go through each option:

   1. XeF4: Xe has 8 valence electrons; 4 are used in Xe-F bonds, leaving 2 lone pairs. Total electron domains = 6, so the electron-pair geometry is octahedral -- but the two lone pairs occupy opposite (axial) positions to minimise repulsion, leaving the 4 fluorine atoms in a plane. The resulting MOLECULAR shape is square planar, not octahedral. Labelling XeF4's shape as 'Octahedral' is a mismatch.
   2. PCl5: 5 bond pairs, no lone pairs on P -- trigonal bipyramidal. Correct match.
   3. SO2: S has 1 lone pair plus 2 bond regions (with the double-bond character); the molecular shape is bent/angular. Correct match.
   4. CH4: 4 bond pairs, no lone pairs on C -- tetrahedral. Correct match. Only XeF4's labelled shape is wrong.

   ✓Final answer

   (a) XeF4 -- Octahedral (its real molecular shape is square planar).

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Match the correct pair: Square planner

   (a) XeF4

   (b) Stock notation

   (c) Lassaigne's test

   (d) Size of orbital

   (e) Exothermic reaction

   ›Reveal solutionSolution

   XeF4 has a square planar shape — its central Xe atom has 6 electron domains (4 bond pairs + 2 lone pairs) arranged octahedrally, with the two lone pairs on opposite (axial) sides.

   In XeF4, the central xenon atom has 8 valence electrons: 4 are used to form 4 Xe-F sigma bonds, leaving 2 lone pairs. This gives 6 electron domains total, which arrange themselves octahedrally to minimize repulsion. The two lone pairs occupy positions opposite each other (axial positions) to minimize lone pair-lone pair repulsion, leaving the four fluorine atoms in a plane around the xenon atom — resulting in a square planar molecular geometry (the printed option 'Square planner' on the paper is the intended term 'square planar').

   ✓Final answer

   Square planar matches with (a) XeF4.

3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following is not a linear molecule ?

   (a) CO2

   (b) BeCl2

   (c) C2H2

   (d) H2S

   ›Reveal solutionSolution

   H2S is not linear; it is bent.

   CO2 (sp, O=C=O), BeCl2 (sp, Cl–Be–Cl) and C2H2 (sp, H–C≡C–H) are all linear (180°). H2S has a central sulphur with two bond pairs and two lone pairs, so its shape is bent/angular (bond angle about 92°). Hence H2S is not linear.

   ✓Final answer

   (D) H2S.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following has the largest bond angle ?

   (a) H2O

   (b) CO2

   (c) NH3

   (d) CH4

   ›Reveal solutionSolution

   CO2 has the largest bond angle (180°).

   Bond angles: H2O ≈ 104.5° (two lone pairs), NH3 ≈ 107° (one lone pair), CH4 = 109.5° (tetrahedral), CO2 = 180° (linear, sp carbon, no lone pairs). The linear CO2 has the largest bond angle.

   ✓Final answer

   (B) CO2.

5. CBSE 2025Set ANNUAL1 markMCQ
   Q.What is the shape of the molecule NH3?

   (a) Square pyramidal

   (b) V-shape

   (c) Trigonal pyramidal

   (d) Tetrahedral

   ›Reveal solutionSolution

   NH3 has 4 electron domains (3 bond pairs + 1 lone pair) arranged tetrahedrally, but since molecular SHAPE only counts the positions of ATOMS (not lone pairs), the visible shape is trigonal pyramidal.

   Step 1 — Electron domains on N: Nitrogen has 5 valence electrons. In NH3, 3 of these electrons pair up with 3 hydrogen atoms (3 sigma bonds), and the remaining 2 electrons form 1 lone pair. Total electron domains = 3 bonding + 1 lone pair = 4.

   Step 2 — Electron geometry vs molecular shape: 4 electron domains arrange themselves tetrahedrally (electron-pair geometry = tetrahedral) to minimise repulsion. However, molecular SHAPE is described only by the positions of the atoms, ignoring the lone pair's position (even though the lone pair does occupy space and affect bond angles).

   Step 3 — Resulting shape: With 3 atoms (H) visible and 1 lone pair 'hidden' at one vertex of the tetrahedron, the 3 N-H bonds form a pyramid with a triangular base — this shape is called trigonal pyramidal. The H-N-H bond angle is compressed slightly to about 107 degrees (from the ideal 109.5 degrees) because the lone pair repels the bonding pairs more strongly than bonding pairs repel each other.

   ✓Final answer

   (c) Trigonal pyramidal.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.The repulsive interaction of electron pairs decrease in the order (Here lp = lone pair, bp = bond pair):

   (a) bp – bp > lp – bp > lp – lp

   (b) lp – lp > lp – bp > bp – bp

   (c) lp – lp > bp – bp > lp – bp

   (d) bp – bp > lp – lp > lp – bp

   ›Reveal solutionSolution

   VSEPR theory ranks repulsions as lone pair-lone pair strongest, then lone pair-bond pair, then bond pair-bond pair weakest.

   According to the Valence Shell Electron Pair Repulsion (VSEPR) theory, a lone pair (lp) is held by only one nucleus and so occupies more space around the central atom than a bond pair (bp), which is shared (and pulled) by two nuclei and is more confined.

   Because lone pairs spread out more, they repel neighbouring electron pairs more strongly. This gives the order of repulsive interaction strength:

   $\text{lp-lp} > \text{lp-bp} > \text{bp-bp}$

   This is why, for example, in NH$_3$ and H$_2$O the bond angles are compressed below the ideal tetrahedral angle — the lone pair(s) push the bonding pairs closer together.

   ✓Final answer

   lp-lp > lp-bp > bp-bp — option (b).

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Which of the following molecules have trigonal bipyramidal shape?

   (a) [Ni(CN)4]2-

   (b) SF6

   (c) [CrF6]3-

   (d) PCl5

   ›Reveal solutionSolution

   PCl5 (sp3d, 5 bond pairs, 0 lone pairs) is trigonal bipyramidal; the others are square planar/octahedral.

   Shape is decided by the number of electron pairs (VSEPR) around the central atom:

   • $[Ni(CN)_4]^{2-}$: Ni is $dsp^2$ hybridized, 4 bond pairs → square planar.
   • $SF_6$: S is $sp^3d^2$ hybridized, 6 bond pairs → octahedral.
   • $[CrF_6]^{3-}$: Cr is $d^2sp^3/sp^3d^2$ hybridized, 6 bond pairs → octahedral.
   • $PCl_5$: P has 5 bond pairs and no lone pair, $sp^3d$ hybridized → trigonal bipyramidal (3 equatorial + 2 axial Cl atoms, bond angles 120° and 90°).

   Only $PCl_5$ has exactly 5 electron pairs with no lone pairs, giving the trigonal bipyramidal geometry.

   ✓Final answer

   (d) PCl5

8. CBSE 2025Set ANNUAL1 markMCQ
   Q.Shape of molecule of CH4 is:

   (a) Trigonal

   (b) Bent structure

   (c) Tetrahedral

   (d) Linear

   ›Reveal solutionSolution

   CH4 is tetrahedral because carbon is sp3 hybridised with 4 bond pairs and no lone pairs.

   Carbon in methane has 4 valence electrons, each forming a single bond with one hydrogen atom (4 bond pairs, 0 lone pairs on carbon). To minimise repulsion between these 4 electron pairs (VSEPR theory), carbon undergoes sp3 hybridisation, and the four sp3 orbitals point towards the corners of a regular tetrahedron with H-C-H bond angles of 109.5 degrees.

   ✓Final answer

   The correct option is (c) Tetrahedral -- CH4 has a tetrahedral shape with bond angle 109.5 degrees.

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Which of the following is linear?

   (a) CO2

   (b) NO2

   (c) SO2

   (d) ClO2

   ›Reveal solutionSolution

   CO2 (O=C=O) is linear; NO2, SO2, ClO2 are all bent due to a lone pair/odd electron on the central atom.

   In CO2, the central carbon forms two σ (and two π) bonds to oxygen with no lone pairs, so by VSEPR the two bonding domains arrange at 180° — a linear molecule. In NO2, SO2, and ClO2, the central atom (N, S, Cl respectively) carries a lone pair or an unpaired electron in addition to its bonding pairs, which bends the molecule (angle < 180°).

   ✓Final answer

   (A) CO2.

10. CBSE 2024Set ANNUAL1 markMCQ
    Q.Which of the following molecules has a linear shape?

    (a) BeCl2

    (b) BCl3

    (c) CCl4

    (d) PCl5

    ›Reveal solutionSolution

    BeCl2's central Be atom has only 2 electron domains (2 bond pairs, no lone pairs), so VSEPR predicts a linear geometry.

    Applying VSEPR to each option's central atom:

    1. BeCl2: Be has 2 valence electrons, forms 2 bonds to Cl, no lone pairs -> sp hybridisation -> linear, bond angle 180 degrees.
    2. BCl3: B has 3 bond pairs, no lone pairs -> sp2 hybridisation -> trigonal planar (120 degrees).
    3. CCl4: C has 4 bond pairs, no lone pairs -> sp3 hybridisation -> tetrahedral (109.5 degrees).
    4. PCl5: P has 5 bond pairs -> sp3d hybridisation -> trigonal bipyramidal. Only BeCl2 has the 2-electron-domain arrangement that produces a linear molecule.

    ✓Final answer

    (a) BeCl2.

11. CBSE 2024Set ANNUAL1 markMCQ
    Q.The shape of IF5 molecule is __________.

    (a) Square pyramidal

    (b) Trigonal bipyramidal

    (c) Octahedral

    (d) Square planar

    ›Reveal solutionSolution

    IF5 has 5 bond pairs and 1 lone pair on iodine; the underlying electron geometry is octahedral, but with a lone pair occupying one vertex, the actual molecular shape is square pyramidal.

    Iodine in IF5 has 7 valence electrons; it forms 5 sigma bonds to the 5 fluorine atoms (using 5 electrons) and retains 1 lone pair (the remaining 2 electrons). By VSEPR (Valence Shell Electron Pair Repulsion) theory:

    Total electron domains around I = 5 bond pairs + 1 lone pair = 6 domains.

    6 electron domains arrange themselves to minimise repulsion in an OCTAHEDRAL electron-pair geometry. However, the MOLECULAR SHAPE describes only the positions of the atoms, not the lone pair. With one of the six octahedral positions occupied by a lone pair (which pushes the four equatorial F atoms slightly away from it and pulls the axial F atom in), the five fluorine atoms end up arranged as a square pyramid: four F atoms forming a square base around iodine, with the fifth F atom at the apex, and the lone pair taking the remaining position opposite that apex (below the central atom).

    This is directly analogous to how NH3 (electron geometry tetrahedral, 1 lone pair) has molecular shape trigonal pyramidal, or how H2O (tetrahedral electron geometry, 2 lone pairs) has molecular shape bent.

    ✓Final answer

    The correct option is (a) Square pyramidal — IF5 has 5 bonding pairs + 1 lone pair (octahedral electron geometry), and removing one position for the lone pair gives a square pyramidal molecular shape.

12. CBSE 2024Set ANNUAL1 mark
    Q.Arrange $NH_3$, $H_2O$ and $CH_4$ in the increasing order of bond angle.

    ›Reveal solutionSolution

    More lone pairs on the central atom compress the bond angle more, because lone pair-bond pair repulsion exceeds bond pair-bond pair repulsion.

    $CH_4$ has 4 bond pairs and 0 lone pairs on carbon — a perfect tetrahedral angle of 109.5°.

    $NH_3$ has 3 bond pairs and 1 lone pair on nitrogen; the extra lone pair-bond pair repulsion compresses the H-N-H angle to about 107°.

    $H_2O$ has 2 bond pairs and 2 lone pairs on oxygen; two lone pairs push the bond pairs even closer together, giving the smallest angle, about 104.5°.

    ✓Final answer

    Increasing order: $H_2O\ (104.5°) < NH_3\ (107°) < CH_4\ (109.5°)$