Exercises — Atomic radius — covalent, metallic, van der Waals; trends across period and group

WHAT the definitions say. A radius is never a property of a lone atom — it is half of a measured distance between two nuclei. Which distance you halve depends on what holds the two atoms together (figure s01).

• (a) → covalent radius, $r_{cov}=\frac{d_{A-A}}{2}$ (the dash = a real bond). Here $d_{A-A}=2\times 66=132$ pm.
• (b) → van der Waals radius, $r_{vdW}=\frac{d_{A\cdots A}}{2}$ (the dots = no bond, just touching). Here $d_{A\cdots A}=2\times 152=304$ pm.

WHY (a) is smaller: a covalent bond shares electrons, so the two nuclei are dragged close → small $d$ → small $r$. Non-bonded atoms in (b) only touch cloud-to-cloud, held by weak Van der Waals Forces, so they sit far apart → large $d$ → large $r$. Answer: (a) covalent, $66$ pm; (b) van der Waals, $152$ pm.

$r_{cov}<r_{met}<r_{vdW}$ WHY this order: stronger nucleus-to-nucleus pull ⇒ closer nuclei ⇒ smaller half-distance (figure s02). Covalent bonding pulls hardest, a metallic "sea" shared among many neighbours pulls a little less per neighbour (see Metallic Bonding), and no bond at all (vdW) leaves atoms farthest apart. Mnemonic: "Cars Move Very slowly" → Covalent < Metallic < Van der Waals.

$r_{cov}=\frac{d_{A-A}}{2}=\frac{198}{2}=99\text{ pm}.$ WHY halve: by symmetry the two identical atoms share the bond distance equally, so each owns exactly half.

WHAT we do: for a bond between different atoms A and B, add their covalent radii — because each atom contributes its own half of the internuclear distance: $d_{A-B}\approx r_{cov}(A)+r_{cov}(B).$ $d_{\text{H–Cl}}\approx 37+99=136\text{ pm}.$ WHY it works: the covalent radius is defined precisely so that these halves stack up to give the internuclear separation. (Real value ≈ 127 pm; electronegativity difference shortens it — see Electronegativity.)

All four are Period 2 (same shell, $n=2$). Left→right across the period, $Z$ rises by one proton each step while the new electron enters the same shell and shields poorly, so $Z_{eff}=Z-S$ climbs and the cloud contracts. Positions left→right: Li (3) → C (6) → N (7) → F (9), so size shrinks in that direction. $\text{F}<\text{N}<\text{C}<\text{Li}.$ (Check against table: F 57 < N 71 < C 76 < Li 128 pm. ✓)

Two things change from Na to K:

• $n$ rises from 3 to 4 (a whole new outer shell opens). In $r\propto n^2/Z_{eff}$, $n^2$ jumps from $9$ to $16$ (see the shell-scaling in figure s03).
• $Z$ rises (11 → 19), but those extra protons are buried under extra full inner shells that shield strongly (see Shielding and Penetration), so $Z_{eff}$ barely moves for the outer electron.

The numerator ($n^2$) grows a lot while the denominator ($Z_{eff}$) grows little, so $r$ increases. The "more protons ⇒ smaller" instinct only holds within a period, where $n$ is fixed. Answer: K > Na because the new shell ($n^2$ effect) dominates the weakly-felt extra charge.

The catch: we are comparing different rulers. Fluorine forms covalent bonds, so $57$ pm is a covalent radius (nuclei pulled in by a bond). Neon is a noble gas — it forms no bonds, so its only measurable radius is the van der Waals radius, taken between non-bonded neighbours that merely touch cloud-to-cloud. Because $r_{vdW}$ is intrinsically the largest of the three radii ($r_{cov}<r_{met}<r_{vdW}$), Ne's number looks large. The period trend (shrinking) genuinely continues in $Z_{eff}$; it's the type of radius that flips, not the physics. Answer: Ne's value is a vdW radius, not comparable to F's covalent radius — different measurement, not a broken trend.

Using the metallic-radius formula ($d_{M\cdots M}$ = distance between two adjacent lattice nuclei; the dots = non-covalent lattice contact): $r_{met}=\frac{d_{M\cdots M}}{2}=\frac{372}{2}=186\text{ pm}.$ WHY larger than covalent would be: in a metal each atom shares its electrons with many neighbours via a delocalised electron sea, so the bonding pull per neighbour is looser than a single localised covalent bond. Looser pull ⇒ larger internuclear distance ⇒ larger radius. This is exactly the middle term of $r_{cov}<r_{met}<r_{vdW}$.

Break it into two moves.

1. Within Period 3 (Na, Mg, Al), left→right: same shell $n=3$, $Z_{eff}$ rises each step, cloud contracts. So $\text{Al}<\text{Mg}<\text{Na}$.
2. K is Period 4, Group 1: it opens a new $n=4$ shell, so by the $n^2$ argument it is larger than any Period-3 atom here. Combine: $\boxed{\text{Al}<\text{Mg}<\text{Na}<\text{K}}$ Each "<" is earned: rightward shrink for the trio, extra-shell growth for K.

• X is above Y in the group, so X has fewer shells (smaller $n$) → smaller radius; Y is larger.
• Ionisation energy = energy to remove the outermost electron. That electron in Y sits in a higher, farther shell where the nucleus's pull ($\propto Z_{eff}/r^2$-type falloff) is weaker, so it is easier to remove → Y has the lower IE. Link: smaller radius ⇒ outer electron closer & more tightly held ⇒ higher IE. So X (smaller) has the larger IE; Y (larger) has the smaller IE — radius and IE move oppositely down a group.

Single-bond estimate: $d_{\text{C–O}}\approx r_{cov}(\text{C})+r_{cov}(\text{O})=76+66=142\text{ pm}.$ Why the double bond is shorter: a double bond shares two electron pairs between the same two nuclei — more shared electron density means a stronger net attraction pulling the nuclei together, so $d$ shrinks well below the single-bond sum. Covalent radii are tabulated for single bonds; multiple bonds always contract the distance. Bond order is continuous (edge case): bond order need not be a whole number. In benzene each C–C bond has bond order $\approx 1.5$ (a resonance blend of single and double), and its length ($\approx 139$ pm) sits between the single-bond ($\approx 154$ pm) and double-bond ($\approx 134$ pm) values. So the covalent radius you'd infer varies continuously with bond order: higher order → shorter bond → smaller effective radius (figure s04). Answer: estimate 142 pm; the real 116 pm is shorter because the extra shared pair pulls the nuclei closer.

Argon: a noble gas forms no covalent or metallic bonds, so its only measurable internuclear distance is between non-bonded, just-touching atoms → its radius is necessarily the van der Waals radius, the largest of the three. Chlorine: $99$ pm was obtained from bonded ${\text{Cl}}_2$ ($d=198$ pm halved) → a covalent radius. For comparison, Cl's own vdW radius is $\approx 180$ pm. Proof they differ: if we (wrongly) compared Ar's vdW $188$ pm to Cl's vdW $180$ pm, the two are close and the period trend (Ar slightly larger only because of its filled shell / no bond contraction) is respected. It is the covalent-vs-vdW mismatch that manufactures the illusion of a huge jump. Answer: Ar's $188$ pm is a vdW radius; Cl's $99$ pm is covalent — comparing them is comparing different rulers.

As $Z_{eff}\to 0$, the ratio $n^2/Z_{eff}\to \infty$, so the model predicts an unbounded (very large) radius — the outer electron feels almost no inward pull and drifts far out. What it captures: it explains why the most weakly held electrons (e.g. the lone valence electron of a heavy Group 1 metal like Cs, well shielded by many inner shells) sit far from the nucleus, giving very large atoms and very low ionisation energy. The limit is idealised (no atom has $Z_{eff}=0$), but the direction is real: less net pull ⇒ bigger atom.

Key insight (the twist): all four species have the same 10 electrons in the same shells (same $n$, same $S$). So $n$ and shielding are constants — the only variable left is the nuclear charge $Z$ pulling those 10 electrons in. See Ionic Radius. Nuclear charges: ${\text{O}}^{2-}$ (8) < ${\text{F}}^{-}$ (9) < $\text{Ne}$ (10) < ${\text{Na}}^{+}$ (11). More protons pulling the same electron cloud ⇒ tighter, smaller. So radius decreases as $Z$ rises: ${\text{Na}}^{+}<\text{Ne}<{\text{F}}^{-}<{\text{O}}^{2-}(\text{increasing size}).$ Why shell count is irrelevant here: normally down a group $n$ changes and dominates. In an isoelectronic set $n$ is frozen, so the usual $n^2$ effect is switched off and $Z_{eff}$ (essentially $Z$, since $S$ is fixed) alone decides size.