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: NH4+,Cl−,SO4^2−,and PO4^3−

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

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:
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:

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

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

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

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:
Na^+, Ne, O^2−have 10 electrons in two energy levels with 11, 10, and 8 protons respectively. The increased number of protons has a greater pull on the 10 electrons and reduces the radius.
Ca^2+ has 18 electrons in 3 energy levels and is therefore, the largest.

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

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

Explanation:
The number of shells and therefore the shielding effect is the same for all of these species.
As the number of protons decreases, the attraction between the nucleus and electrons decreases, resulting in a smaller radius.
K^+ has the most protons (19), whereas S^2− has the least (16).

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

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:
The 6 protons in the nucleus of carbon will exert a stronger attractive force on the second energy level electrons than the 5 protons of boron.
Increasing electron affinity from left to right across a period.

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

Explanation:
The longest carbon chain has 5 carbons and will include the C=C.
The alkene is given priority and is on carbon-1.
The methyl group is bonded to carbon-2.

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

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

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

Explanation:
The sum of the oxidation state of all atoms in a compound is 0 and in an ion is equal to the charge of the ion.
O in the above molecules have an oxidation state of -2.
Group I elements have an oxidation state of +1.
In choices A, B, and C, the oxidation states of Cr on both sides are +6.
In choice D, the oxidation state of Cr changes from +6 to +3.