Explanation:​
Group 17 melting points increase down the group due to increased strength in intermolecular forces. The molecular structures of each element are held together by London (dispersion) forces. These increase with the number of electrons in the molecule. Therefore statement I is not true, but III is.
Group 1 melting points decrease down a group due to decreasing metallic bond strength (with increase of atomic radius), therefore statement II is also true.

Explanation:
NH3 has a non-bonding pair of electrons it can donate to a central metal ion. BH3 and CH4​ do not have non-bonding electrons.

Explanation:​
Acid rain is formed from sulfur and nitrogen oxides dissolving in the clouds.

Explanation​:
Statement II is incorrect as acids produce solutions with $\mathrm{pH}$ lower than 7

Explanation​:
Recall of transition metal properties.

Explanation:​

The electron configuration of the atom is 1s^2 2s^2 2p^4. There are 8 electrons in total and the electron configuration corresponds to a neutral atom, so the atomic number must be 8 also. The element is oxygen and it lies in group 16 $\text{OR}$ since the atom has 6 valence electrons, its element should be located in group 16.

There are two subshells so the period number is 2. This question is moderately difficult as the student must deduce that the number of electrons is equal to the atomic number or determine which of the orbitals are part of different shells, without the assistance of orbital labels.

Explanation​:
Saturated compounds contain only single bonds.
Butyne is an alkyne, and so, has a triple bond.

Explanation​:
1 and 2 are enantiomers, exhibiting opposite optical activities (rotate plane polarised light in opposite directions by the same amount).
Compound 3 is achiral, so it has a rotation angle of 0 degrees.
This question is in the “hard” category, as the student must be able to determine a molecule's chirality and recall how polarimetry can be used to distinguish between stereoisomers.

Explanation:
Simple recognition of the units used in an IR spectrum. Since the information is given in the syllabus and covered frequently, this question is categorised as easy

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​:
The student must recall the definition of a racemic mixture and understand how the wedge-dash notation varies between enantiomers.
A racemic mixture includes both optical isomers in equal amounts. Option A is the same molecule (possibly a conformational isomer), option C is a diastereomer, and option D is a structural/positional isomer.

Explanation:​

The group of an element is determined by its atomic number, which is found in the lower left corner of the element symbol. Although this conflicts with the representation in the data booklet, students should be able to differentiate between the mass and atomic number based on their relative size. The elements represented by options A, B, C and D are calcium (an alkaline earth metal), iron (a transition metal), gallium (a poor / p block metal) and rubidium (an alkali metal), respectively. This question is moderately difficult as the student must identify the correct value from the given symbol and relate it to the position of an element in the periodic table.

Metalloids lie on the boundary between metals and non-metals in the p block of the periodic table. They are shiny like metals but brittle like non-metals, and can switch between being electrical insulators and electrical conductors. Although this switchable conductivity could be considered as a change from metal-like to non-metal-like behaviour, it does not affect all properties of the element: the element does not switch from being a metal to a non-metal.

Metalloids do not contain incomplete d subshells (this is a property of transition metals) and lie alongside the poor metals in the p block of the periodic table. This question is categorised as moderately difficult as the concept of metalloids is not covered extensively in the syllabus, though the distinction between metals, non-metals and metalloids is highlighted.

Explanation​:
The methyl groups are equivalent so produce one signal with integral 6, while the other protons both produce signals with integrals 1.
The student must recognise that all protons are detectable by NMR, including those attached to $\mathrm{O}$ atoms, and know how to count proton environments (COOH protons are sometimes too broad to detect due to proton exchange and/or hydrogen bonding, but this point is not covered at IB level).
The question is categorised as hard as several features of the $\mathrm{NMR}$ spectrum must be predicted.

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:​
The attraction for a shared pair of electrons in a covalent bond depends on electronegativity. Carbon has the lowest electronegativity value of the given elements across period 2, it has the lowest effective nuclear charge (least number of protons attracting valence electrons in energy level 2). Increasing electronegativity across a period from left to right.

Explanation​:
Each $\mathrm{Cl}$ ligand has a 1-charge.
2+(4x−1)=2−.
As outlined in the Chemistry Guide, [CuCl4]^2− has a tetrahedral shape.

Explanation:
The student must recognise that the described technique is IR spectroscopy, and correctly interpret the diagram to identify a wavelength in the IR range.
The question is of moderate difficulty as it requires both the definition and the diagram to be understood correctly.

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:​
The air is naturally composed of 0.04% carbon dioxide, which dissolves in water to form the weak carbonic acid

Explanation:​
Statement III is not correct as physical properties, including melting point and metallic character, change dramatically across a period due to changes in nuclear charge for the same number of shells.
Elements in the periodic table are arranged in increased atomic number.
Anions are atoms that have gained electrons. This will increase the electron-electron repulsion in the outermost shell, and will cause the electrons to move further apart.Therefore, the radius of the outershell will increase.

Explanation:
The student must recognise that there are three proton environments in a 3:2:3 ratio;
B has only $2$ environments ($3:2$ ratio), while C has 4 (3:2:2:1). D has $3$ environments but they are in a 6:1:1 ratio.
The question is hard as two pieces of information must be extracted from the spectrum and interpreted in the context of the structural formulas given.