Chemical Bonding and Molecular Structure: NEET PYQ, 2018–2026

Answer: (A)

Step 1: Sulphur has 6 valence electrons. Each of its three S–O bonds uses one of them, leaving no lone pair on S in the resonance structures. Step 2: Electron domains around S = 3, all bonding. Step 3: Three bonding domains with no lone pairs give a trigonal planar molecular shape. Lone pairs on the terminal oxygen atoms do not count for the central atom.

Answer: (D)

By Fajans' rules, covalent character in an ionic bond increases when the cation is small and highly charged, and when the anion is large. All four cations here carry a charge of +1, so the size decides. Li⁺ is the smallest, giving it the highest charge density and the greatest polarising power over the Cl⁻ cloud, so the electron density is pulled partly between the nuclei. The order of covalent character is LiCl > NaCl > KCl > CsCl. Choosing CsCl confuses a large ion being easily polarised with it being a strong polariser, and assuming equal charge means equal character ignores the effect of size.

Answer: (D)

The hybridisation of the central atom depends on its steric number, which is the number of σ-bonds plus the number of lone pairs. Double bonds such as Xe=O contribute only one σ-bond to this count. XeO₃: 3σ + 1 lone pair = 4, so sp³ (pyramidal). SF₄: 4σ + 1 lone pair = 5, so sp³d (see-saw). XeOF₄: 4 Xe–F σ + 1 Xe–O σ + 1 lone pair = 6, so sp³d² (square pyramidal). The answer is sp³, sp³d, sp³d². The usual errors are forgetting the lone pair on the central atom or counting the π-bonds.

Answer: (A)

Principle: a hydrogen bond needs an H atom covalently bonded to a small, highly electronegative atom (F, O or N) acting as donor, and a lone pair on F, O or N acting as acceptor. Apply to each option: methanol has O–H (donor) and O lone pairs (acceptor), so it qualifies. Methane has only C–H bonds, which are not polar enough. Chloromethane has no H on Cl, and C–H bonds do not donate. Trimethylamine has lone pairs but no N–H, so it can accept but not donate, and therefore cannot form H-bonds with its own kind.

Answer: (D)

Carbon has 4 valence electrons, 2 electrons in its lone pair and 6 shared electrons in the triple bond. Formal charge = 4 − 2 − ½(6) = −1. Oxygen, by the same count, is 6 − 2 − 3 = +1, and the two add up to zero for the neutral molecule.

Answer: (A)

For each species, write the configuration, count the bond order and locate the unpaired electron. C₂⁻ has 9 valence electrons: σ2s², σ*2s², π2p⁴, σ2p_z¹. Nb = 7 and Na = 2, so bond order = 2.5, and the odd electron is in the bonding σ2p_z. O₂⁺ also has bond order 2.5, but its odd electron is in π*2p, so it fails the orbital condition. F₂⁺ has bond order 1.5 with an odd electron in π*2p. Be₂⁺ has bond order 0.5 with an odd electron in σ*2s. Only C₂⁻ meets both conditions.

Answer: (A)

Cation formation removes electrons. Na⁺ has 10 electrons and 11 protons, so the remaining electrons feel a stronger effective nuclear pull and the ion is smaller. Anion formation adds electrons to the same shell. Cl⁻ has 18 electrons and 17 protons, so repulsion increases and the ion is larger. The ionic radius trends are the basis for the size effects seen in ionic solids.

Answer: (D)

Count the electron pairs around carbon. In CH₃⁺ there are three bond pairs and no lone pair. Carbon is sp² hybridised, the shape is trigonal planar, and an empty unhybridised p orbital lies perpendicular to the plane. In CH₃⁻ there are three bond pairs and one lone pair, four pairs in all. Carbon is sp³ hybridised, and the lone pair occupies one hybrid orbital, so the shape is trigonal pyramidal.

Answer: (A)

Step 1: Count electron pairs on S. Six valence electrons plus four from F gives 10 electrons, i.e. 5 pairs: 4 bond pairs and 1 lone pair. The electron geometry is trigonal bipyramidal. Step 2: Choose the lone pair site. Repulsions at 90° matter most. Axial site: the lone pair has 3 bond pairs at 90°. Equatorial site: the lone pair has 2 bond pairs at 90° (the 120° interactions are weaker). The equatorial site is therefore preferred. Step 3: Name the shape. With the lone pair equatorial, the four F atoms form a see-saw (distorted tetrahedral) arrangement. Step 4: Lone pair–bond pair repulsion is greater than bond pair–bond pair repulsion, so the axial S–F bonds are pushed back from the lone pair and the axial F–S–F angle is about 173°, below 180°. Only the first option has the correct shape, reason and angle.

Answer: (B)

Principle: the steric number (number of bond pairs plus lone pairs on the central atom) determines the hybridisation, and the steric number gives the geometry. For BeCl₂, Be has 2 bond pairs and 0 lone pairs, so the steric number is 2. Two orbitals mean sp hybridisation. Two sp orbitals are 180° apart, so the molecule is linear with a Cl–Be–Cl angle of 180°.