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.

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

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:
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:
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:
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:
The bonds between N and H are polar due to the difference in electronegativity.
The overall shape of the NH3 is asymmetrical (trigonal pyramidal), resulting in a molecule with a net dipole moment.
CO2 and SO3 both have polar bonds, but the molecules are symmetrical. Therefore, they do not have net dipole moment.
The C-H bonds are not polar due to a small difference in electronegativity.

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

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:
The other three in theory, can form hydrogen bonds due to O−H and F−H bonds present

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

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 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