First ionisation enthalpy is a definition that needs to be learned:

First ionisation enthalpy is the enthalpy change when one electron is removed from each atom in one mole of gaseous atoms.

Thus 1 only is the correct answer.

One mole of 1+ ions are formed, but first ionisation enthalpy involves gaseous atoms (not the element in its standard state)

First ionisation enthalpy is a definition that needs to be learned:

First ionisation enthalpy is the enthalpy change when one electron is removed from each atom in one mole of gaseous atoms.

Across period 3 there is a general increase in 1st ionization energy (IE) due to increasing nuclear charge and electrons filling the same energy level; that is filling the 3s and 3p sub-levels. The electrons being in the same energy level are 'initially' no further form the nucleus, but the nuclear charge increases, increasing the attraction between the nucleus and the electrons. Overall the attraction between the outer electrons and the nucleus outweighs the increasing electron-electron repulsion, so statement 3 is incorrect.

Thus 1 and 2 only is the correct answer.

The electron removed from oxygen is removed from the 2p sub-level. The electron removed from nitrogen is also in the 2p sub-level. Thus explanations 1 and 3 do not contribute. Oxygen's outer electron is however, paired-up, unlike the electrons of nitrogen which occupy orbitals singly (on their own) so explanation 2 does apply.

The electron removed from potassium is removed from the 4s sub-level. The electron removed from argon is from the 3p sub-level. Explanations 1 and 3 both contribute to the ionisation energy of potassium being lower than argon; the 4s sub-level is in the n=4 energy level, further from the nucleus with less nuclear attraction and the inner complete levels of electrons will shield (repel) the outer electron to a greater extent (than in argon). Potassium's outer electron in 4s is, however, not paired-up, so explanation 2 does not apply.

First electron affinity and second electron affinity are definitions that need to be learned:

First electron affinity is the enthalpy change when one electron is gained by each atom in one mole of gaseous atoms.

Second electron affinity is the enthalpy change when one electron is gained by each 1− ion in one mole of gaseous ions.

Thus O_(g) + e⁻ → O⁻_(g) is the correct answer, as this is the first electron affinity of oxygen. Equations involving oxygen molecules (O₂) are incorrect.

Across period 3 there is a general decrease in atomic radii due to increasing nuclear charge and electrons filling the same energy level; that is filling the 3s and 3p sub-levels.

The electrons are in the same energy level, but the nuclear charge increases, increasing the attraction between the nucleus and the electrons and pulling them in more tighly/closer to the nucleus. Overall the attraction between the outer electrons and the nucleus outweighs the increasing electron-electron repulsion.

The number of electrons increasing is incidental; increased electron-electron repulsion would cause an increase in atomic size, but here it is outweighed by the increased attraction.

Thus 1 and 3 only is the correct answer.

Down any group of the periodic table there is a general increase in atomic radii due to the electrons filling new energy levels.

The nuclear charge increases, but because the electrons are being added to energy levels that are further form the nucleus, there is increased shielding (repulsion by inner electrons) decreasing the attraction between the nucleus and the electrons and pulling them in less tighly to the nucleus. Overall the attraction between the outer electrons and the nucleus is weaker than the increasing electron-electron repulsion due to shielding.

Statements 1 and 2 here are therefore false as they state the opposite trend. Statement 3 is true.

Thus 3 only is the correct answer.

Down any group of the periodic table there is a general increase in atomic radii due to the electrons filling new energy levels.

The nuclear charge increases, but because the electrons are being added to energy levels that are further from the nucleus, there is increased shielding (repulsion by inner electrons) decreasing the attraction between the nucleus and the electrons and pulling them in less tighly to the nucleus. Overall the attraction between the outer electrons and the nucleus is weaker than the increasing electron-electron repulsion due to shielding. Statements 1 is therefore true.

Anions (negative ions) like Cl^– are always larger than the corresponding atom (and cations are always smaller) as increased number of electrons increases electron-electron repulsion, so statement 2 is also true.

Statement 3 is also true as electronegativity increases to the right and upwards on the periodic table (increasing up a group), so chlorine does have a greater electronegativity than bromine.

Thus 1, 2 and 3 is the correct answer.

Metals tend to form basic (alkaline) oxides, not acidic oxides, so statement 1 is false.

Metals' outer electrons are delocalised and they have giant structures. Metals are often: good conductors of electricity and heat; high in melting and boiling point; malleable; ductile; sonorous; lustrous (shiny). Therefore statement 2 is true.

Metals have low electronegativity and tend to lose electrons when they react. Therefore statement 3 is true (as non-metals tend to gain electrons).

Thus 2 and 3 only is the correct answer.

Metals (Na, Mg , Al) have high mp/bp; strong metallic bonds in giant structures are broken. Higher ionic charge and smaller ions tend to have higher mp/bp due to stronger metallic bonds. Statement 1 is true.

Non-metals or metalloids with giant structures (Si) have high mp/bp; strong covalent bonds are broken. So statement 2 is false as silicon does not have a low boiling point or a simple molecular structure.

Non-metals with simple molecular structures (S₈, Cl₂) or gaseous atoms (Ar) have low mp/bp; only intermolecular forces need to be broken. Different intermolecular forces vary in strength, but are all relatively weak. Statement 3 is true - the molecules get smaller and so have weaker London dispersion forces between them.

Thus 1 and 3 only is the correct answer.

Group 1 metals have metallic bonding and therefore giant structures. Melting points are relatively low for metals as the atoms are large and form 1+ ions so the attraction between the nucleus and ‘sea of electrons’ is not as strong as in other metals. Nonetheless strong metallic bonds must be broken. The melting points (values in data book) become lower down the group because of the increasing atomic/ion size leads to less attraction..

Group 17 non-metals; the halogens, have covalent bonding and simple molecular structures. The atoms get bigger down the group, so the diatomic halogen molecules get bigger down the group. These non-polar molecules have weak London dispersion forces that must be broken. The melting points (data book) become higher down the group as London dispersion forces get stronger as the number of electrons/molar mass of a molecule increases.

Thus, The melting points decrease down group 1 and increase down group 17 is the correct answer.

Metals have high mp/bp; strong metallic bonds in giant structures are broken. Statement 1 is true.

We would normally expect metals to form basic oxides, but we need to remember that aluminium oxide is amphoteric, which means it can react as an acid or base. Therefore statement 3 is true, but statement 2 is false.

Thus 1 and 3 only is the correct answer.

Halogens decrease in reactivity going down the group and this is explained in terms of the decreasing attraction between outer electrons and the nucleus as the atoms become larger. The outer electron is less easily gained as it is further from the nucleus and has increased ‘shielding’ from inner energy levels, so there is weaker electron-nucleus electrostatic attraction.

A more reactive halogen will displace a less reactive halogen from its compound. (A 'higher in the group' halogen will displace a lower halogen.) Therefore 1 and 2 won't react as chlorine is less reactive than flourine and bromine is less reactive than chlorine, but reaction 3 will react as written.

Thus 3 only is the correct answer.