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

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

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:
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:
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:
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:
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 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:
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:
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:
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:
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:
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:
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 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