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.

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:
The two compounds are part of a homologous series because they have the same functional group but different hydrocarbon chains. Both compounds are alcohols, so interact via hydrogen bonding.
The CH2 and CH3 groups of the alkyl chains, however, interact via weaker dispersion forces, as they contain only non-polar bonds. Increasing the length of the alkyl chain does not affect the strength of hydrogen bonding, but the strength of the dispersion forces is increased. More heat energy is needed to separate molecules of Y, so the boiling point is higher.

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

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:
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:
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:
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:
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:
The higher the charge of the ions, the stronger the electrostatic forces of attraction and, therefore, the greater the lattice enthalpy.
The smaller the ion, the greater the charge density and the stronger the electrostatic attraction between oppositely charged ions.
Statement I is incorrect since the smaller the lattice enthalpy, the weaker the electrostatic attraction and the lower the melting point is.

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

Explanation:

Lithium is in group 1 and so forms Li^+ ions. The formula of the nitride ion is N^3−as nitrogen is in group 15. For the positive and negative charges to cancel out, three Li^+ ions are needed for each N^3− ion.

Explanation:
Volatility refers to the ease at which a substance vaporizes. Ionic bonds are very strong and, therefore, a lot of energy is required to turn ionic compounds into gases.

Polar solvents are able to form electrostatic forces of attraction with ions.

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:
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:
A $\sigma$ bonding orbital is produced by end-to-end (direct) overlap of atomic orbitals.
Students should be familiar with orbital overlap diagrams of this type, as they are covered extensively in the course. The question is therefore categorized as easy.

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 carbon atoms in ethyne are sp hybridized, with a bond angle of 180°.
Recognizing the arrangement of hybrid orbitals in ethyne is a key aim of this subtopic. This question is therefore categorized as easy

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:
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:
To successfully answer this question, the student must recall that electrical conductivity requires freely moving electrons.
Metals are able to conduct electricity because the valence electrons are able to move freely through the lattice in the sea of delocalized electrons.
The size and mass of the atom do not factor into the ability of the material to conduct electricity. While transition metal salts can conduct electricity, the ability to form ionic bonds does not explain conductivity.

This question is categorized as easy because it only involves the recall of basic information about metals and conductivity.

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.