Atomic and ionic radii

Atomic radius is not a fixed physical boundary — it is defined operationally as half the distance between the centres of two adjacent touching atoms. Depending on how the atoms are held together, this is called the covalent radius (for a covalently bonded pair), the metallic radius (for atoms in a metal lattice), or the van der Waals radius (for atoms that merely touch without bonding, as in a solid noble gas). The single idea that explains almost every trend in this topic is effective nuclear charge, $Z_{eff}$: the net positive pull actually felt by an outer (valence) electron once the repulsion of the inner-shell electrons is accounted for.

where: $Z$ is the atomic number (total nuclear charge) and $\sigma$ is the shielding constant, the extent to which inner electrons screen the valence electron from the full nuclear pull. A larger $Z_{eff}$ pulls the valence shell in tighter, shrinking the atom; a smaller $Z_{eff}$ lets it relax outward.

Across a period (left to right), radius decreases. Each new element adds one proton to the nucleus and one electron to the same valence shell. Electrons sharing a shell shield each other only weakly, so $\sigma$ barely rises while $Z$ climbs steadily — $Z_{eff}$ increases across the whole period, and the electron cloud is pulled in tighter at every step.

Down a group (top to bottom), radius increases. Each new element adds an entirely new principal shell farther from the nucleus. $Z_{eff}$ does creep up slightly down a group too, but that effect is swamped by the much larger average distance of the new outermost shell — so the atom is bigger overall despite the higher nuclear charge.

Ionic radii follow the same logic. A cation (electron(s) removed) is always smaller than its parent atom: fewer electrons means less electron-electron repulsion and a higher $Z_{eff}$ per remaining electron, and often an entire shell disappears. An anion (electron(s) added) is always larger: more electrons means more repulsion and a lower $Z_{eff}$ per electron, so the cloud expands.

Isoelectronic species — different atoms/ions sharing the same electron count — give the cleanest test of this idea. Take $\text{N}^{3-}, \text{O}^{2-}, \text{F}^-, \text{Ne}, \text{Na}^+, \text{Mg}^{2+}$: all ten-electron species with an identical electron configuration. The only variable left is nuclear charge $Z$, and radius falls steadily as $Z$ rises, because more protons grip the same fixed electron cloud more tightly.

Question: Arrange Na, Mg, Al, Si (all Period 3) in order of decreasing atomic radius, justifying with $Z_{eff}$; then use the same reasoning to arrange the isoelectronic ten-electron series $\text{O}^{2-}$, $\text{F}^-$, $\text{Na}^+$, $\text{Mg}^{2+}$ in order of increasing radius.

Step 1. All four Period-3 elements add one proton and one electron to the same $n = 3$ valence shell as you move Na (11) to Mg (12) to Al (13) to Si (14) — no new shell appears anywhere in this row.

Step 2. Because same-shell shielding is weak, $Z_{eff}$ rises steadily across these four elements, so radius falls steadily too, giving decreasing atomic radius Na > Mg > Al > Si.

Step 3. For the isoelectronic series, first confirm the electron count: $\text{O}^{2-}$ has $8+2=10$ electrons, $\text{F}^-$ has $9+1=10$, $\text{Na}^+$ has $11-1=10$, $\text{Mg}^{2+}$ has $12-2=10$ — all four share the neon configuration $1s^2 2s^2 2p^6$, so $n$ and $\sigma$ are essentially identical across the set.

Step 4. With electron count and shielding fixed, only $Z$ varies: O ($Z=8$), F ($Z=9$), Na ($Z=11$), Mg ($Z=12$). Higher $Z$ means higher $Z_{eff}$ per electron, so radius falls as $Z$ rises — meaning radius rises as $Z$ falls.

Step 5. Ordering by increasing radius therefore runs from highest $Z$ to lowest $Z$: $\text{Mg}^{2+} < \text{Na}^+ < \text{F}^- < \text{O}^{2-}$.

Competitive stretch

A related question is to compare an atom directly with its own ion, e.g. Na versus $\text{Na}^+$. Removing the single $3s$ electron leaves $\text{Na}^+$ with the neon core alone — repulsion drops and an entire shell ($n=3$) disappears, so $\text{Na}^+$ is dramatically smaller than Na, consistent with the cation rule and its position at the small end of the series above.

• Assuming radius always increases with atomic number — true down a group, but the opposite happens across a period.
• Reversing the cation/anion rule — a cation is always smaller and an anion always larger than the neutral parent atom.
• Applying the isoelectronic shortcut (radius falls as $Z$ rises) to species with different electron counts — it only holds when the electron count is identical across the set.
• Blaming the across-period decrease on "more electrons repelling each other less" — the correct cause is rising $Z_{eff}$ from weak same-shell shielding, not an electron-electron effect.

Ask the tutor: if ranking an isoelectronic series or comparing an atom with its own ion is confusing you, ask the AI tutor to walk through the $Z_{eff}$ reasoning step by step with your specific example.