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

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

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 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:
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:

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:
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:
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:
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 H2O, there are two lone pairs, and it has a bent shape with bond angles of 104.5°.

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:
Metals and non-metals have the greatest differences in electronegativity and so ionic bonds are formed.

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

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:
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:
Non-metal elements bond to each other covalently. Ca3N2 is made up of a metal and a nonmetal (with a large difference in electronegativity) and has ionic bonding.

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.