Periodic Trends - Physical and Chemical Properties

We are moving left to right across a period, so we consider what changes: proton number increases, but electrons are added to the same energy level (shell).

More protons means a stronger positive pull on the outer electrons. Because electrons stay in the same shell, there is no extra shielding added.

The stronger nuclear pull draws outer electrons closer to the nucleus, so atomic radius decreases. The correct answer states exactly this.

The choices “Atomic radius increases because more electrons are added to the outer shell” and “Atomic radius increases because shielding increases across a period” suggest atomic radius increases across a period. We know from the periodic trend that radius decreases left to right, so both are wrong.

The choice “Atomic radius decreases because more electron shells are added” says radius decreases because more shells are added — but more shells occur when going DOWN a group, not across a period. This explanation is incorrect for the given direction.

The choice “Atomic radius decreases because nuclear charge increases while electrons are added to the same shell” correctly states both the trend (decreases) and the reason (higher nuclear charge, same shell), making it the only fully correct answer.

Ionisation energy is the energy required to remove one electron from one mole of gaseous atoms, forming a positive ion.

The process must start with a gaseous atom $X(g)$ — not a solid or aqueous ion — because ionisation energy is defined for isolated gas-phase atoms.

Removing one electron from $X(g)$ gives $X^{+}(g)$ and a free electron $e^{-}$. This is exactly what the choice “$X(g)\to X^{+}(g)+e^{-}$” shows.

The choice “$X(s)\to X^{+}(aq)+e^{-}$” uses solid $X(s)$ and aqueous $X^{+}(aq)$ — ionisation energy refers to gaseous atoms, not solids or aqueous species.

The choice “$X(g)+e^{-}\to X^{-}(g)$” shows an electron being gained ($X+e^{-}\to X^{-}$), which is electron affinity, not ionisation energy.

The choice “$X^{+}(g)\to X^{2+}(g)+e^{-}$” shows removal of an electron from $X^{+}$, which is the second ionisation energy, not the first.

Only the choice “$X(g)\to X^{+}(g)+e^{-}$” correctly shows removal of one electron from a neutral gaseous atom to form a +1 ion, matching the definition of first ionisation energy.