Explanation:
Neither material is electrically conductive. Diamond has no delocalized electrons, while the delocalized electrons in buckminsterfullerene are confined within the molecule. There is no extended lattice of bonds through which the electrons can move in response to an applied potential difference. Diamond is a hard material; a large force is required to break the strong covalent bonds between atoms.

Buckminsterfullerene, however, is soft, as only a small force is needed to overcome the weak intermolecular (London / dispersion) forces between molecules.

This question is moderately difficult, as the student must recall multiple features of the two carbon allotropes or deduce their physical properties from the diagrams given.

Explanation:
Atom 1 has four electron domains and so a tetrahedral shape (109.5$°$).
Atom 2 has three electron domains and so a trigonal planar shape (120$°$).
Atom 3 has four electron domains of which one is a lone pair giving it a trigonal pyramidal shape (107$°$).

Explanation:
2-methylpentane is branched and therefore has a lower surface area. The lower the surface area, the weaker the London (dispersion) forces and the lower the boiling point.

Explanation:
The bonds between N and H are polar due to the difference in electronegativity.
The overall shape of the NH3 is asymmetrical (trigonal pyramidal), resulting in a molecule with a net dipole moment.
CO2 and SO3 both have polar bonds, but the molecules are symmetrical. Therefore, they do not have net dipole moment.
The C-H bonds are not polar due to a small difference in electronegativity.

Explanation:
The sp^2 orbitals in ethene are used to form $\sigma$ bonds, while the non-hybridized p orbitals overlap laterally to form the $C-C\pi$ bonds.
The C=C bond contains one $\sigma$ and one π bond, so it involves both sp^2 and p orbitals.
This question is moderately difficult as the student must recognize the difference between sp^2 and p orbitals and understand how they contribute to both $\sigma$ and $\pi$ bonds without the aid of orbital diagrams.

Explanation:
NH3 has all of the bonds/forces named above, but its high boiling point is due to the presence of hydrogen bonds which are the strongest intermolecular forces. Hydrogen bonding is possible due to the presence of N−H bonds.

Explanation:
Alloys are homogeneous mixtures of metals with other metals or nonmetals.

Explanation:
Ionic bonds form between oppositely charged ions; NH^4+ and SO4^2− in ammonium sulfate and Na+ and CO32−Na+ in sodium carbonate.

Due to a smaller difference in electronegativity and both being nonmetals, H−Cl bonds are polar covalent.

Explanation:
Oppositely charged ions attract, it is an ionic bonding;

Choice A describes metallic bonding;

Choice B describes a simple molecular lattice;

Choice D describes a covalent bond.

Explanation:
In an sp^2 carbon atom, there are three sp^2 orbitals of equal energy and one p orbital of higher energy.
The formation of three σ bonds and one π bond can be understood as being derived from three sp^2 orbitals and one p orbital, respectively, each containing an unpaired electron.
This question is moderately difficult: although the orbital energies of an sp^2 carbon atom is covered explicitly in the syllabus, the student must recognize this distribution without the aid of orbital labels or additional diagrams.

Explanation:
BF3 is trigonal planar, with an unoccupied p orbital perpendicular to the plane of the molecule. The hybridization of B is sp^2.
BF4^− is tetrahedral and the B atom has sp^3 hybridization.
This question is moderately difficult as the student must consider the hybridization of a non-hydrocarbon, recognizing that the B atom in BF3 has only three electron domains and is therefore analogous to the C atom of ethene.

Explanation:
Hybridization is a system for explaining the geometry of molecules.
Hybrid orbitals are therefore a conceptual tool for understanding the formation of molecular orbitals, and do not truly exist within atoms or molecules. They must therefore be equal in number to the original atomic orbitals. Paired electrons in s orbitals may occupy separate sp^3 orbitals, as in the hybrid orbitals of the C atom in methane.
Hybrid orbitals overlap end-on to produce σ bonds; π bonds are formed by the sideways overlap of p orbitals.
This question is categorized as hard as the student must understand several key features of hybrid orbitals and correctly distinguish between σ and π bonding.

Explanation:
SiO2 has a giant covalent structure where each Si atom is singly bonded to four oxygen atoms (while maintaining a Si ratio of 1:2).
As there are four electron domains around each silicon atom, the structure has a tetrahedral arrangement. SiO2 is held together by strong covalent bonds requiring a large amount of energy to overcome, resulting in a high melting point.
The strong bonds, together with the fact that there are no charged particles present, lead to insolubility in polar solvents such as water.

Explanation:
All statements are correct, but only D is a reason for solubility in water

Explanation:
Strontium is in group 2 and so forms Sr^2+ ions. The formula of the sulfide ion is S^2− as sulfur is in group 16. For the positive and negative charges to cancel out, one of each ion is required.

Explanation:
The other three in theory, can form hydrogen bonds due to O−H and F−H bonds present

Explanation:
Magnesium has two electrons in its outermost shell and loses both to achieve a full outer shell (Mg^2+). Fluorine has seven electrons in its outermost shell; it needs one more electron to achieve a full outer shell (F^1−).

Two fluorine atoms are therefore required for each magnesium atom in order to form a neutral ionic compound (where the charges of the positive and negative ions cancel out).

Explanation:
The charges of the positive and negative ions must cancel out to give an overall neutral compound. Group 2 metals form ions with a 2+ charge. The formulae of the counterions above are: $.$

Explanation:
Each carbon atom in ethene forms three σ bonds and one π bond. The s electrons are originally paired but are conceptually unpaired during the formation of the three sp^2 orbitals, allowing the σ bonds to form. The p electrons are unpaired in the original atomic orbitals and remain unpaired in the hybrid orbitals. One p orbital in each carbon atom is used to produce the π bond.
This question is moderately difficult as the student must recall the concept of electron unpairing/promotion in hybridization and differentiate between σ and π orbitals without the aid of a diagram.

Explanation:
N2 has a triple bond, O2 has a double bond, and H2 only has a single bond.
The greater the number of shared pairs of electrons, the stronger the bond.

Explanation:
Metals characteristically have high melting points due to strong metallic bonds.

Explanation:
The central atom is carbon with two C-H single bonds and one C=O bond. This gives 3 electron domains and 8 valence electrons.