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:​
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.

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:​

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

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:​
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:
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:
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:
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:​
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:​
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:​
C3H6​ belongs to the alkene class with the general formula CnH2n​, and has a three carbon chain.

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 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