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

Explanation:
First ionization energy is defined as the energy required to remove one mole of electrons from one mole gaseous atoms.

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

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:

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:
Ethers contain C−O single bonds $\text{not}$ the $\mathrm{C}=\mathrm{O}$ carbonyl group.

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

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

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

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:
Structural isomers have the same molecular formula but different structural formulae.
Methoxyethane and propanol both have the molecular formula C3H8O.

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.

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

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:
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.