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​:
The electron configuration of Zn2+ is [Ar] 3d^10. There are no vacancies in the $d$ subshell so a transition between the split energy levels of the octahedral complex is not possible.
This question is categorized as easy as the colourless nature of d10 ions is covered explicitly in the syllabus.

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 overall charge on the complex ion is $1-$, water is neutral and the chloride ion has a charge of $1-$ = $+$ Cr.
Cr has an oxidation state of 3+.

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

Explanation​:
Increasing atomic radius results in a weaker electrostatic force of attraction between the nucleus and electrons in the covalent bond. An increased number of shells down the group increases the shielding between the nucleus and electrons in the covalent bond.

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:​
Choices III and IV are ethene and ethyne with a double bond and a triple bond, respectively. They belong to the classes of alkenes and alkynes, which can be deduced by the general formulae, CnH2n and CnH2n−2​​ .
Choice I is the alkane methane, and choice II is a halogenoalkane (chloromethane).

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.

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:
NH2 is the amine functional group, RCOOR′ is the ester functional group.

Explanation:​
In HBrO, the oxidation states of both H and Br are +1.

Explanation:
The masses of options A, B, C and D are 29, 45, 31 and 28, respectively.
The mass of the fragment ion is 148 - 117 = 31, making option C correct.
This question is of moderate difficulty as two simple calculations are required

Explanation​:
There are 4 proton environments as the CH(CH3)2 protons are equivalent. Integration gives the relative number of protons in each environment. There are 3 on C1, 2 on C2, 1 of C3 and 6 on C4 and 5.

Explanation:​

The atomic number of Cr is higher than that of V by 1, and its relative atomic mass is also higher by approximately 1. The atomic number is unique for each element and dictates the position of the element in the periodic table; thus, Cr always has an atomic number of 24 and vanadium an atomic number of 23. However, isotopes of an element may have different atomic masses, so an atom of Cr may have a lower atomic mass than an atom of V. This question is categorised as easy since the difference between atomic number and atomic mass and the variation in mass between isotopes of an element are covered extensively in the syllabus.

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:
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:
First ionization energy decreases down a group due to increased shielding (number of shells). Beryllium must therefore have a higher first IE than magnesium. And similarly, oxygen higher than sulfur.
First ionization energy increases across a period due to increased nuclear charge for the same number of shells. Hence, oxygen has higher first IE than beryllium.

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​:
This question tests the ability of the student to convert a 3D model to wedge-dash notation.
The difficulty is low as the shape of the model is close to that of the 2D representation – no rotation around the single bonds is needed to determine the relative orientations of the substituent groups.

Carbon-1 is attached to an H-atom coming out of the page (wedge bond), an NH2​ group going into the page (dashed bond), and a CH3​ group in the plane of the page.
Carbon-2 is attached to a $\mathrm{Cl}$ atom coming out of the page (wedge bond), an H atom going into the page (dashed bond), and a CH3​ group in the plane of the page.

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:
Ligands are Lewis bases with a lone pair of electrons that they can donate to a central metal ion.

Explanation​:
Amides have a C=O and C−N on the same central carbon atom.