Atomic Structure: Question 10

Worked solution

Part (a): General trend across Period 2

First ionisation energy generally increases from lithium to neon. Moving across the period, one proton is added to the nucleus at each step, increasing the nuclear charge. At the same time, each new electron is added to the same outer shell ($n=2$), so the shielding provided by inner, complete shells stays roughly constant. With shielding essentially unchanged but nuclear charge increasing, the effective nuclear charge experienced by the outer electrons rises steadily across the period. This pulls the outer electrons closer and holds them more tightly, so progressively more energy is needed to remove the outermost electron. First ionisation energy increases overall from lithium to neon.

Part (b): The two exceptions

Beryllium to boron: Beryllium’s configuration is $1s^2\,2s^2$; its outermost electron is removed from a full $2s^2$ sub-shell. Boron’s configuration is $1s^2\,2s^2\,2p^1$; its outermost electron occupies the $2p$ sub-shell, which is at a slightly higher energy than $2s$, and is also shielded a little by the underlying, complete $2s^2$ electrons. Because boron’s outer electron is, on balance, easier to remove than a $2s$ electron, its first ionisation energy is lower than beryllium’s, despite boron having one more proton.

Nitrogen to oxygen: Nitrogen’s configuration is $1s^2\,2s^2\,2p^3$: all three $2p$ electrons occupy separate orbitals, unpaired, which is a comparatively stable, low-repulsion arrangement. Oxygen’s configuration is $1s^2\,2s^2\,2p^4$: the fourth $2p$ electron must pair up in an orbital that is already singly occupied. The two electrons sharing that orbital experience extra electron-electron repulsion, which makes one of them easier to remove than expected. This extra repulsion effect outweighs oxygen’s extra proton, so oxygen’s first ionisation energy is lower than nitrogen’s.

Part (c): Lithium versus sodium (down Group 1)

Lithium’s outermost electron occupies the $2s$ sub-shell ($n=2$), shielded only by the complete inner $1s^2$ shell. Sodium’s outermost electron occupies the $3s$ sub-shell ($n=3$), a completely new shell, further from the nucleus, and shielded by two complete inner shells ($1s^2\,2s^2\,2p^6$) rather than just one.

Although sodium has a much greater nuclear charge (11 protons) than lithium (3 protons), the large increase in distance from the nucleus and in shielding going down the group outweighs this increase in nuclear charge. Sodium’s outer electron is therefore held less strongly overall, and sodium’s first ionisation energy is lower than lithium’s. Consistent with the general trend that first ionisation energy decreases down a group.

Final answers

• (a) First IE generally increases across Period 2: constant shielding + rising nuclear charge $\Rightarrow$ rising effective nuclear charge.
• (b) Be$\to$B: outer electron moves to higher-energy, shielded $2p$. N$\to$O: extra repulsion from pairing a $2p$ electron.
• (c) Sodium’s outer electron is in a new, more distant, more heavily shielded shell ($n=3$ vs $n=2$), which outweighs its larger nuclear charge $\Rightarrow$ lower first IE than lithium.