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:​
All three central atoms (N, C, and O) have four electron domains. An increasing number of lone pairs reduces the angle between the bonding pairs.
In CH4​ there are no lone pairs and it has a tetrahedral shape with bond angles of 109.5$°$.
In NH3​ there is one lone pair and it has a trigonal pyramidal shape with bond angles of 107$°$.
In H2​O, there are two lone pairs, and it has a bent shape with bond angles of 104.5°.

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:
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​:
Fluorine is the most electronegative element on the periodic table. Therefore, the greatest difference in electronegativity between $\mathrm{C}$ and $\mathrm{F}$ results in the most polar bond.

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

Explanation:​
Recall of description of metallic bonding as positive ions surrounded by delocalized electrons.

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:​
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:​
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:
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:​
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:​
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 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:​
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:​
Students must have knowledge of metallic bonding to answer this question. Specifically, they must recall that transition elements bond via metallic bonding which is a result of strong interactions between delocalized $s$ and $d$ electrons and metal ions.
Stronger bonds result in higher melting points and so the number of delocalized electrons correlates with melting point.
Comparatively, an $s$ block metal would only have 1 or 2 delocalized electrons and therefore a lower melting point.
Hybridization commonly occurs in organic molecules involving $s$ and $p$ orbitals and can also occur in the formation of metal-ligand complexes. Hybridization is not present in metallic bonds.
Finally, the sharing of electrons takes place in covalent bonds and is between 2 non-metals.

This question is categorized as medium level because there are many concepts that the student must recall and apply to successfully answer this question.

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:​
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​:
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 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:​
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:​
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.