Chemical Bonding and Molecular Structure: JEE Advanced PYQ, 2007–2026

Answer: (A)

Nitrogen has 5 valence electrons, no lone pair and 8 bonding electrons (four bonds): 5 − 0 − 4 = +1. With −1 on each of the two singly bonded oxygens, the total is −1, the charge on the ion.

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: (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)

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: (C)

For effective LCAO the combining atomic orbitals must have comparable energies, the same symmetry about the bond axis and adequate overlap. All four pairs have non-zero overlap. The 2p_x–2p_x, 2s–2s and 2s–2p_z pairs are close in energy, so they combine well or at least significantly. The 1s–2s pair is the exception: 1s lies much lower in energy than 2s, so the two combine poorly. Inner 1s orbitals therefore form their own σ1s and σ*1s MOs.

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: (A)

Boiling point and viscosity depend on the strength of intermolecular forces. Ethylene glycol and propan-1-ol have almost identical molar masses, so London forces are comparable. Each –OH group is both a hydrogen-bond donor and an acceptor. Glycol has two such groups per molecule, so each molecule can link to several neighbours and form a three-dimensional network. That network resists flow (high viscosity) and needs much more energy to break apart (high boiling point). The explanations based on molar mass, a stronger individual bond, or an absence of hydrogen bonding in propan-1-ol are not supported.

Answer: (A)

Step 1: Both NH₃ and NF₃ have 4 electron pairs around N (3 bond pairs and 1 lone pair), so the electron geometry is tetrahedral. The lone pair repels the bond pairs more than they repel each other, and this compresses the bond angle below 109.5°. Step 2: How much the angle closes depends on how strongly the bond pairs repel one another. In NH₃, N is more electronegative than H, so the bond pairs sit nearer N and repel each other strongly. This holds the angle at 107°. Step 3: In NF₃, F is far more electronegative than N, so the bond pairs are displaced towards the fluorine atoms. They are farther from N and repel one another less. The lone pair can then push them closer together, giving about 102°. Step 4: The other explanations either predict the wrong direction (larger F atoms would widen the angle) or attribute the effect to F lone pairs or to displacement of the N lone pair, none of which VSEPR uses.

Answer: (D)

Write the configuration of each molecule and count bonding and antibonding electrons. Li₂ has σ1s² σ*1s² σ2s², so bond order = (4−2)/2 = 1 with no unpaired electrons. B₂ has bond order (6−4)/2 = 1, but its last two electrons occupy separate degenerate π2p orbitals, so it is paramagnetic. C₂ has bond order 2 and N₂ has bond order 3. Only Li₂ satisfies both conditions.