Molecular Structure and Bonding Theory Questions and Answers

Answer: B

P₄ and CH₄ share the same Geometry: both are tetrahedral.

In white phosphorus the four P atoms sit at the corners of a tetrahedron, each bonded to the other three (P-P-P angles of 60°). In methane the four H atoms sit at the corners of a tetrahedron around carbon (H-C-H angles of 109.5°). The arrangements differ in detail, but both molecules have tetrahedral shape.

They do not share the other properties. White phosphorus is a pale yellow waxy solid at 300 K with a boiling point near 280 °C, while methane is a colourless gas boiling at about -162 °C.

Answer: C

The order is (I) \(\lt\) (IV) \(\lt\) (III) \(\lt\) (II): toluene \(\lt\) 4-methylphenol \(\lt\) nitrobenzene \(\lt\) 4-nitrotoluene.

Toluene has a small moment (about 0.4 D) from the weakly donating methyl group. 4-Methylphenol (p-cresol) is around 1.5 D, dominated by the OH dipole. Nitrobenzene is strongly polar (about 4.0 D) because the nitro group withdraws electrons from the ring.

In 4-nitrotoluene the electron-donating methyl and electron-withdrawing nitro group are para to each other, so their moments point the same way along the axis and add, giving about 4.4 D, the largest of the four. Orders placing nitrobenzene or 4-nitrotoluene low ignore the size of the nitro dipole.

Answer: D

[BrF₄]⁻ has sp³d² hybridization.

Bromine has 7 valence electrons, plus 1 for the negative charge, giving 8. Four are used in bonds to F, leaving two lone pairs. With four bonding pairs and two lone pairs, there are six electron pairs around Br, arranged octahedrally (sp³d²). The two lone pairs occupy trans positions, so the ion is square planar, like XeF₄, with which it is isoelectronic.

SF₄ has five electron pairs (one lone pair), sp³d, and a see-saw shape, so (A) and (B) describe a different electron count. sp³ would allow only four electron pairs.

Answer: D

The weakest binding is van der Waals forces.

Van der Waals forces are weak, short-range attractions between molecules arising from temporary or permanent dipoles, typically under 10 kJ/mol. They are intermolecular forces, not true chemical bonds.

Covalent, coordinate and ionic bonds are true chemical bonds with energies of the order of hundreds of kJ/mol, tens of times stronger.

Answer: B

\([\text{NO}_{2}]^{+}\) has a shape similar to \(\text{SiO}_{2}\), taken as the linear O=Si=O unit.

The nitronium ion is isoelectronic with CO₂: nitrogen has two double bonds to oxygen and no lone pair, so it is linear (sp, AX₂). A monomeric SiO₂ unit, with silicon in the same group as carbon, would likewise be linear. (Solid silica is a network of SiO₄ tetrahedra, but the question compares the triatomic formula.)

SO₂ and O₃ have a lone pair on the central atom (AX₂E) and are bent at about 119° and 117°; water has two lone pairs and is bent at 104.5°. None of them is linear.

Answer: B

The hardness of diamond is due to covalent bonding, option B. Each carbon atom is bonded to four others by strong covalent bonds in a continuous three-dimensional network, so that there are no weak planes in the crystal, and a great deal of energy is needed to break it.

Coordinate bonding (option A) is a covalent bond in which one atom donates both electrons and is not the bond in diamond. Electrovalent bonding (option C) is ionic and needs two different elements. Van der Waals forces (option D) are the weak forces between the layers of graphite, which makes graphite soft. The contrast with graphite is the usual way this question is set.

Answer: C, D

The molecules with a non-zero dipole moment are (C) NO and (D) SO₂. The final key accepts C and D.

A dipole needs a net displacement of charge. N₂ is homonuclear, so no polarity. CO₂ is linear and its two C=O dipoles cancel, so the net moment is zero. NO is heteronuclear, with a small but real dipole (about 0.16 D) from the difference in electronegativity of N and O. SO₂ is bent, so its two S–O dipoles add up to a net dipole (about 1.6 D).

Answer: A, B, D

The occupied orbitals are A, B and D; (C) is an empty antibonding orbital.

Carbon monoxide has 10 valence electrons, which fill five molecular orbitals: two sigma orbitals built mainly from the 2s orbitals, the pair of pi bonding orbitals, and the sigma orbital that is the HOMO (largely a carbon lone pair). The lowest unoccupied orbitals are the pi* pair (the LUMO).

Reading the drawings, (A) and (D) are s-based sigma orbitals, and (B) is a pi bonding orbital: the two p orbitals overlap side by side with the same sign (both upper lobes shaded), with more weight on the more electronegative oxygen. In (C) the p orbitals have opposite signs on the two atoms, which puts a node between them: that is a pi* antibonding orbital, empty in the ground state. I am identifying (A) and (D) from the drawing's shading and size, which the booklet does not label; the accepted set A, B, D agrees with the official key.

Answer: A, B, D

Hydrogen bonding is central to (A), (B) and (D).

• (A) In ice, each water molecule is hydrogen-bonded to four others in an open tetrahedral network, which makes ice less dense than liquid water, so it floats.
• (B) In water, the ammonium ion from a primary amine, RNH₃⁺, has three N-H bonds to hydrogen-bond with water, while that from a tertiary amine has only one. The better-solvated ion is more stable, so primary amines are the stronger bases in water.
• (D) In benzene, acetic acid exists largely as a dimer held together by two O-H...O hydrogen bonds, which is why its measured molar mass there is about double.

(C) is an inductive effect, not hydrogen bonding: the methyl group of acetic acid pushes electron density towards the carboxylate and destabilises it, so formic acid, with no such group, is stronger.