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 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:

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 N3− ion.

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

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 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:
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:
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 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:

Elements in the same group (column) of the periodic table have the same number of outer shell (valence) electrons, represented by the group number; the number of electron shells increases down the group. Elements in the same row (period) have the same number of electron shells (or principal quantum number), equal to the period number; the number of valence electrons increases across the period. This question is moderately difficult as the student must recall multiple pieces of information and correctly interpret a schematic illustration of the periodic table.

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

Explanation:
Atomic radius and metallic character decrease across a period. Ionization energy decreases down a group due to increased shielding.

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

Explanation:

The names of the labelled groups are covered explicitly in the syllabus.

Explanation:
Group 2 metal oxides produce alkaline metallic hydroxides when dissolved in water.

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

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.