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

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

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

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​:
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:
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:​
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:​
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:
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.