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​:
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:​
Electrostatic forces of attraction between the positive ions and delocalized electrons (metallic bonds) must be broken for metal to melt. The stronger the metallic bond, the higher the melting point.

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

Explanation:​
Graphite is made up of individual graphene layers. Both have carbon atoms with three electron domains and trigonal planar shapes. Each carbon atom in diamond has four electron domains resulting in tetrahedral arrangements.

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:​
Across the same period, metals have lower ionization energies than non-metals due to lower nuclear charges with the same number of shells (shielding effect).

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:​
The student must recall that the delocalized valence electrons in transition elements result in strong metallic bonds and therefore high melting points. The delocalized $s$ and $d$ orbital electrons responsible for the strong metallic bonds also make the metals hard and electrically conductive.

This question is categorized as easy since the student only needs to recall the basic attributes of transition metals

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

Explanation:
Simple molecular covalent structures cannot conduct electricity due to an absence of charged particles. NaCl is an ionic compound that has ions that are free to move when in a molten state.

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:
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 student must recall the appearance of a π bond and the way in which it is formed. This information is stated clearly in the syllabus, so the question is categorized as easy.
Option A describes the formation of a $\sigma$ bond.

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​:
There is a non-bonding (lone) pair of electrons on the nitrogen atom and three on each chlorine. There are three N−Cl bonding pairs of electrons. Only the outermost electrons are shown for each element

Explanation:
There is no polyatomic ion with the formula of CO4^2−. The correct formula for the carbonate ion is CO3^2−. The formulas and charges of the other ions are correct.

Explanation:
Carboxylic acids have hydrogen bonding due to O−H bonds.

Explanation:​
To answer this question, the student must use the elements’ positions on the periodic table and the number of valence electrons to conclude on the melting point and hardness.
Metals with high melting points are those with stronger metallic bonds. The strongest metallic bonds are those that involve the largest number of delocalized electrons.
Calcium has only 2 delocalized electrons compared to nickel’s 10 valence electrons, which are capable of delocalization. As a result, nickel has a stronger metallic bond than calcium, resulting in a higher melting point.
Hardness is also related to metallic bond strength, where stronger bonds equal a harder material.

This question is categorized as medium difficulty because there are many factors to consider in arriving at the final answer. Importantly, the student must recall and apply many definitions such as metallic bonding, melting point, conductivity, and delocalized electrons.