VSEPR Theory, Valence Bond Theory and Hybridisation — NEET Chemistry MCQs with Solutions

Solution: XeF₂ has the most lone pairs (3) but is linear and symmetric, so its dipole moment is zero. The next highest is ClF₃ with 2 lone pairs; being T-shaped it has a non-zero dipole moment. ClF₃ has 5 electron domains (3 bond pairs + 2 lone pairs), so the central Cl is sp³d hybridised.

Solution: SF₄ is sp³d (see-saw shape) with 1 lone pair, and its axial S–F bonds are longer than the equatorial ones — satisfying all three conditions. XeF₂ is sp³d with lone pairs but both bonds are equal; PF₅ has unequal bonds but no lone pair; XeF₄ is sp³d².

Solution: SO₂: 2 bond pairs + 1 lone pair on S → sp² (bent). NO₂⁻: 2 bond pairs + 1 lone pair on N → sp² (bent). N₃⁻ (azide): central N has 2 bonding domains and no lone pair → sp (linear).

Solution: BrF₅ (5 bond pairs + 1 lone pair, sp³d²) and XeOF₄ (5 bonding domains + 1 lone pair, sp³d²) are both square pyramidal. SbF₅ and PCl₅ have no lone pair and are trigonal bipyramidal.

Solution: NO₃⁻ and BCl₃ are trigonal planar with sp² central atoms. ClO₂⁻ (2 bond pairs + 2 lone pairs) and ClO₃⁻ (3 bond pairs + 1 lone pair) both have 4 electron domains → sp³. So 2 species are sp³.

Solution: BF₃ (3 bond pairs, trigonal planar) and NO₂⁻ (2 bond pairs + 1 lone pair, bent) both have 3 electron domains → sp². NH₂⁻ and H₂O have 4 domains (sp³).

Solution: BF₃ is trigonal planar (120°). PF₃ is pyramidal with one lone pair (≈ 97°). ClF₃ is T-shaped with two lone pairs, compressing the F–Cl–F angle to ≈ 87.5°. Hence ClF₃ < PF₃ < BF₃.

Solution: BrF₅ has 5 bond pairs + 1 lone pair (sp³d²) → square pyramidal. [Ni(CN)₄]²⁻ is square planar (dsp²); PF₅ and PCl₅ are trigonal bipyramidal.

Solution: O in OF₂ has 2 bond pairs + 2 lone pairs → bent (V-shaped), so A and D are correct and E is wrong. Because F is more electronegative than O, the bond pairs are pulled away from O and the FOF angle (≈ 103°) is less than 104.5° — B correct. F being more electronegative, O has oxidation state +2 (not −2) — C wrong. Hence A, B, D.

Solution: SCl₂: S has 2 bond pairs + 2 lone pairs. O₃: central O has 1 lone pair (sp², bent). ClF₃: Cl has 3 bond pairs + 2 lone pairs (T-shaped). SF₆: S has 6 bond pairs, 0 lone pairs. So 2, 1, 2, 0.

Solution: In a trigonal bipyramidal arrangement a lone pair prefers the equatorial position because there it suffers only two 90° lone pair–bond pair repulsions (with the two axial bonds); in an axial position it would face three. Hence SF₄ is see-saw shaped with the lone pair equatorial.

Solution: CH₄, NH₄⁺ and BH₄⁻ each have 10 electrons, so all three are isoelectronic. Each central atom has 4 bond pairs and no lone pair → all are sp³ and tetrahedral.

Solution: Br in BrF₃ has 7 valence electrons: 3 form bond pairs with F, leaving 2 lone pairs. With 5 electron domains (sp³d) and both lone pairs equatorial, the molecule is bent T-shaped.

Solution: NO₂⁻: 2 bond pairs + 1 lone pair on N → sp² (bent). NO₂⁺: 2 bonding domains, no lone pair → sp (linear, isoelectronic with CO₂). NH₄⁺: 4 bond pairs → sp³ (tetrahedral).

Solution: AX₃E₂: 5 electron domains → trigonal bipyramidal electron geometry with both lone pairs equatorial, leaving the three bonds in a T-shape (e.g. ClF₃).

Solution: In the azide ion N₃⁻ the central N has two bonding domains and no lone pair (sp) → linear. O₃ and Cl₂O are bent (lone pairs on the central atom), and NO₂ is bent with an odd electron on N.

Solution: SO₄²⁻ and CrO₄²⁻ are both tetrahedral; SiCl₄ and TiCl₄ are both tetrahedral — so A and B are isostructural. NH₃ is pyramidal while NO₃⁻ is trigonal planar; BCl₃ is trigonal planar while BrCl₃ is T-shaped (2 lone pairs on Br).

Solution: In I₃⁻ the central I has 2 bond pairs + 3 lone pairs (sp³d). The three lone pairs occupy the equatorial positions of the trigonal bipyramid, leaving the two bonds axial → linear, 180°.

Solution: SF₄ is see-saw shaped (sp³d, 1 lone pair): its two axial S–F bonds are longer than the two equatorial ones. BF₄⁻ and SiF₄ are tetrahedral and XeF₄ is square planar — all with equal bond lengths.

Solution: Tetrahedral, square planar and rectangular planar AB₄ are all symmetric, so the four A–B bond moments cancel and the molecule is non-polar. In a square pyramidal arrangement (A above the plane of the four B atoms) the bond moments do not cancel, giving a net dipole — a polar AB₄ can have this geometry.

Solution: SF₄ has 4 bond pairs + 1 lone pair = 5 electron domains. Including the lone pair, the electron-domain geometry is trigonal bipyramidal (the molecular shape, ignoring the lone pair, is see-saw).

Solution: XeF₄ + SbF₅ → [XeF₃]⁺[SbF₆]⁻: Xe changes from sp³d² (in XeF₄) to sp³d (in XeF₃⁺). In the other reactions the underlined atom keeps its hybridisation: S stays sp³ (H₂SO₄ → NaHSO₄), N stays sp³ (NH₃ → NH₄⁺), and P stays sp³ (H₃PO₂ → PH₃ + H₃PO₃).

Solution: CH₄ has no lone pair, so its angle is the full tetrahedral 109.5°. Lone pairs compress the angles in NH₃ (107°) and H₂O (104.5°); H₂S is even smaller (≈ 92°) because S uses nearly pure p orbitals.

Solution: The central I of I₃⁻ carries 3 lone pairs (sp³d, linear) and each of the two terminal I atoms carries 3 lone pairs: 3 + 3 + 3 = 9.

Solution: I₃⁻ is linear (180°) > BF₃ trigonal planar (120°) > NH₃ (107°) > PF₃ (≈ 97°, angle smaller than NH₃ because the electronegative F atoms pull bond pairs away from P).

Solution: XeOF₂: Xe has 3 bonding domains + 2 lone pairs (sp³d) → T-shaped. XeOF₄: 5 bonding domains + 1 lone pair (sp³d²) → square pyramidal. IO₃⁻ is pyramidal, ICl₂⁻ is linear, and IO₄⁻ is tetrahedral.

Solution: NF₃ has a lone pair on N (like NH₃), so it is trigonal pyramidal, not trigonal planar. The other three geometries are correct.

Solution: BF₃ (sp², trigonal planar) → BF₄⁻ (sp³, tetrahedral): both hybridisation and shape change. H₂O → H₃O⁺ and NH₃ → NH₄⁺ change shape but stay sp³; CH₄ → C₂H₆ keeps sp³ carbon and tetrahedral geometry.

Solution: PF₅ has only 5 electron domains and is sp³d (trigonal bipyramidal). SF₆ (6 bond pairs) and BrF₅ (5 bond pairs + 1 lone pair) are sp³d², and [CrF₆]³⁻ is an octahedral complex.

Solution: NO₂⁺ is isoelectronic with CO₂: N has two bonding domains and no lone pair → sp, linear (O=N=O)⁺. NO₂⁻ and NO₂ are bent (sp²) and NO₃⁻ is trigonal planar (sp²).