Atomic radius — covalent, metallic, van der Waals; trends across period and group: Worked examples

This page is the practice ground for the parent topic. We will hit every kind of question the topic can throw at you: comparing across a period, down a group, mixing the two, handling the three different radii, degenerate/tricky cases (noble gases, transition-metal squeeze), a real-world word problem, and an exam-style twist. Guess first, then check.

Before we start, one reminder so no symbol sneaks in unexplained:

The scenario matrix

Every atomic-radius question is one of these cells. Each worked example below is tagged with the cell(s) it covers.

Cell	Case class	What decides the answer	Example
A	Pure period (same $n$ , vary $Z$ )	$Z_{e f f}$ only	Ex 1
B	Pure group (same group, vary $n$ )	$n$ dominates	Ex 2
C	Mixed — different period AND group	rank period first, then shift	Ex 3
D	Which radius? (cov vs met vs vdW)	nature of contact	Ex 4
E	Degenerate — noble gas / "wrong ruler"	vdW radius, not covalent	Ex 5
F	Anomaly / limiting — d-block squeeze, near-equal	small $Δ Z_{e f f}$ , filled d/f	Ex 6
G	Real-world word problem	convert measured $d$ to radius	Ex 7
H	Exam twist — combine with a sister trend	radius → IE / electronegativity	Ex 8

Example 1 — Cell A (pure period)

Worked example Arrange

C, N, O, F (all Period 2) by decreasing covalent radius. Forecast: which end is biggest — the left (C) or the right (F)? Jot your guess.

Steps

Note they share the shell $n = 2$ . Why this step? If $n$ is the same for all four, the $n^{2}$ term in $r \propto n^{2} / Z_{e f f}$ is frozen, so only $Z_{e f f}$ can move the size. We've killed one of the two knobs.
Track $Z$ left→right: C(6), N(7), O(8), F(9). Each step adds a proton, and the new electron lands in the same $n = 2$ shell where it shields poorly. Why this step? Same-shell electrons sit at roughly the same distance from the nucleus, so they barely screen one another; $S$ creeps up slowly while $Z$ jumps by a full 1 each time. Net: $Z_{e f f}$ rises rightward.
Rising $Z_{e f f}$ with fixed $n$ ⇒ tighter pull ⇒ smaller. So size falls C→N→O→F. Why this step? Direct read of $r \propto n^{2} / Z_{e f f}$ : denominator up, $r$ down.
Answer (decreasing): $C > N > O > F$ . Numerically (pm): 76 > 71 > 66 > 57.

Verify: Differences shrink but stay positive ( $76 - 71 = 5$ , $71 - 66 = 5$ , $66 - 57 = 9$ ) — a smooth contraction, no reversals. Units are picometres throughout. Matches the parent's Period-2 list. ✓

Example 2 — Cell B (pure group)

Worked example Arrange

Li, Na, K, Cs (Group 1) by increasing metallic radius. Forecast: does adding protons downward shrink them, or does something else win?

Steps

Each step down opens a new shell: $n = 2 (Li) \to 3 (Na) \to 4 (K) \to 6 (Cs)$ . Why this step? Down a group the outer electron moves to a higher principal shell — this is the $n$ knob turning, and it turns hard because $n$ is squared in $r \propto n^{2} / Z_{e f f}$ .
Yes, $Z$ also rises (3→11→19→55), but each new inner shell shields the outer electron well, so $Z_{e f f}$ on the valence electron stays modest. Why this step? We must check the other knob doesn't cancel the first — here it doesn't, because filled inner shells are excellent screens (see Shielding and Penetration).
$n^{2}$ growth beats the mild $Z_{e f f}$ rise ⇒ atoms get bigger downward. Why this step? Compare magnitudes: $n^{2}$ roughly quadruples over the range while $Z_{e f f}$ barely doubles.
Answer (increasing): $L i < N a < K < C s$ . In pm: 152 < 186 < 227 < 265.

Verify: Every gap is positive and growing early ( $186 - 152 = 34$ , $227 - 186 = 41$ ) then easing ( $265 - 248$ -ish region), consistent with $n$ jumping by 1 each period. Metallic radii used throughout (Group 1 are metals). ✓

Example 3 — Cell C (mixed period + group)

Worked example Order

Na, Mg, Al, K by increasing atomic radius. Forecast: Na, Mg, Al are neighbours in one row; K is dropped into a new row. Where does K land?

Steps

Handle the same-period trio first: Na, Mg, Al are Period 3, left→right, so $Z_{e f f} ↑$ with fixed $n = 3$ . Why this step? Same $n$ ⇒ Cell-A logic ⇒ size falls rightward: $A l < M g < N a$ .
Place K: it is Period 4, so its outer electron sits in $n = 4$ — a brand-new shell beyond Period 3. Why this step? An extra shell ( $n$ up by 1) dominates any leftover $Z_{e f f}$ difference (Cell-B logic), so K is bigger than all of Period 3's earlier members here.
Stitch together. Why this step? Chain the within-row order onto the cross-row jump.
Answer (increasing): $A l < M g < N a < K$ .

Verify: Covalent radii (pm): Al 118, Mg 141, Na 166, K 203 → 118 < 141 < 166 < 203, strictly increasing, matches our order exactly. ✓

Example 4 — Cell D (which radius?)

Worked example For

chlorine, its bonded neighbour sits $198$ pm away; a non-bonded Cl in the next molecule sits $\approx 360$ pm away. Find $r_{co v}$ and $r_{v d W}$ , and state which is which — and why they differ. Forecast: both come from "half the distance." Which half-distance is bigger, and does that make sense?

Steps

Both radii are half an internuclear distance — we just pick the right $d$ . Why this step? An atom has no hard edge, so "radius" always means half the centre-to-centre distance of a chosen contact.
Bonded pair (Cl–Cl in Cl $_{2}$ ): $r_{co v} = d /2 = 198/2 = 99$ pm. Why this step? A shared covalent bond drags the two nuclei in, giving the smallest contact distance.
Non-bonded pair (Cl of one molecule "touching" Cl of the next): $r_{v d W} = 360/2 = 180$ pm. Why this step? No bond pulls them together — only weak Van der Waals Forces against full-cloud repulsion — so they rest much further apart.
Answer: $r_{co v} = 99$ pm $<$ $r_{v d W} = 180$ pm.

Verify: $99 < 180$ obeys the ordering $r_{co v} < r_{m e t} < r_{v d W}$ . Sanity: bonded = squished = small; touching = loose = big. Units pm both sides. ✓

Example 5 — Cell E (degenerate: noble gas / wrong ruler)

Worked example

F (covalent radius 57 pm) is one column left of Ne. A student says "Ne is at the end of Period 2, so it must be even smaller than F — under 57 pm." Ne's listed radius is actually $\approx 154$ pm. Explain the trap. Forecast: is the student comparing like with like?

Steps

Identify F's radius type: F forms bonds, so 57 pm is a covalent radius (measured to a bonded neighbour). Why this step? You must name the ruler before comparing numbers.
Identify Ne's radius type: a noble gas forms no bonds, so its 154 pm is a van der Waals radius (measured to a non-bonded neighbour). Why this step? Cell-E is precisely the case where the standard "smallest at the right end" period rule silently switches rulers on you.
Compare correctly: $r_{co v}$ and $r_{v d W}$ are different kinds of distance; vdW is always larger because no bond pulls the nuclei in. Why this step? $r_{co v} < r_{v d W}$ for the same atom, so a right-end atom measured by vdW can easily out-size its neighbour measured by covalent.
Answer: The period trend (shrink rightward) is real only within one ruler. Ne's 154 pm being larger than F's 57 pm is not a violation — it's a ruler swap. The "decrease across a period" applies to same-type radii.

Verify: $154 > 57$ is expected once we accept vdW $>$ covalent. Had we compared Ne's hypothetical covalent radius, it would indeed be $<$ 57 pm — but no such bond exists to measure. Consistent, no contradiction. ✓

Example 6 — Cell F (anomaly / near-equal limiting case)

Worked example

Zr ( $r \approx 160$ pm, Period 5) and Hf ( $r \approx 159$ pm, Period 6) sit one below the other in Group 4. By the "down-a-group grows" rule Hf should be clearly bigger — yet it's essentially the same size. What limiting effect flattens the trend? Forecast: the group rule predicts growth; what almost-perfectly cancels it here?

Steps

Apply the naive group rule: Hf ( $n = 6$ ) below Zr ( $n = 5$ ) should be noticeably larger via $n^{2}$ . Why this step? Establish the expectation so the anomaly is visible.
Between them lies the lanthanide series (14 f-block elements filling $4 f$ ). Why this step? Those 14 extra protons are added while the $4 f$ electrons shield poorly — f-orbitals are diffuse and screen badly (see Shielding and Penetration).
Poor $4 f$ shielding ⇒ $Z_{e f f}$ on Hf's valence electrons is much larger than the simple group trend suggests ⇒ extra contraction that almost exactly offsets the $n = 5 \to 6$ growth. This is the lanthanide contraction. Why this step? It's the limiting case where the two knobs ( $n ↑$ vs $Z_{e f f} ↑$ ) nearly cancel, giving $Δ r \approx 0$ .
Answer: $r (Zr) \approx r (Hf)$ ; the expected downward growth is wiped out by the lanthanide contraction. Difference $\approx 1$ pm.

Verify: $∣160 - 159∣ = 1$ pm — negligible versus the $\sim 20$ – $40$ pm jumps seen in Group 1 (Ex 2). The near-cancellation is the hallmark of this degenerate cell. ✓

Example 7 — Cell G (real-world word problem, geometric)

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

Steps

Model each Cu atom as a sphere; nearest neighbours touch, so the line between the two nuclei passes through the point where their surfaces meet. Why this step? "Touching spheres" means the centre-to-centre distance equals the sum of the two radii — geometry pins the contact to the midpoint for identical atoms.

Identical atoms ⇒ equal radii ⇒ each owns exactly half of $d$ : $r_{m e t} = d /2 = 256/2 = 128$ pm. Why this step? Splitting the internuclear distance in half is the definition of metallic radius (half the distance between two adjacent lattice nuclei).

Compare with covalent: $r_{m e t} = 128 > r_{co v} = 132$ ? Check ordering. Why this step? We must confirm $r_{co v} \leq r_{m e t}$ ; here the tabulated covalent value (132) is close, showing that in a metal each atom bonds weakly to many neighbours, so the per-neighbour pull is gentle — metallic radius is not much smaller than covalent, and can be comparable.

Answer: $r_{m e t} (Cu) = 128$ pm.

Verify: $2 \times 128 = 256 = d$ ✓ (reconstructing the measured spacing). Units: pm halved gives pm. The value sits in the sensible transition-metal range (~120–140 pm). ✓

Example 8 — Cell H (exam twist: chain to a sister trend)

Worked example Without a table: which has the higher

first ionisation energy, Na or Cl? Use only radius reasoning. Forecast: guess which atom holds its outer electron tighter — the big one or the small one?

Steps

Both are Period 3; going Na→Cl is left→right, so $Z_{e f f} ↑$ with fixed $n = 3$ ⇒ Cl is smaller than Na (Cell-A logic). Why this step? Radius is our lever; a period trend fixes the ordering.
Link radius to grip: a smaller radius means the outer electron sits closer to the nucleus and feels a larger $Z_{e f f}$ , so it is held more tightly. Why this step? Ionisation energy = energy to remove the outer electron; tighter grip ⇒ more energy needed (see Ionisation Energy and Electronegativity, which track the same $Z_{e f f}$ ).
Therefore Cl (smaller, tighter) has the higher first ionisation energy. Why this step? We convert the size comparison into the energy comparison via the shared driver $Z_{e f f}$ .
Answer: $IE_{1} (Cl) > IE_{1} (Na)$ .

Verify: Actual values — Na $\approx 496$ kJ/mol, Cl $\approx 1251$ kJ/mol ⇒ $1251 > 496$ ✓. Smaller radius ⇒ higher IE, as predicted. Units kJ/mol both sides. ✓

Recall Quick self-test across the matrix

Which cell is each? "Rank F, Cl, Br by size." ::: Cell B (pure group) — size increases F<Cl<Br as $n$ grows. Which cell? "Compare Ar's radius to Cl's." ::: Cell E — wrong-ruler trap; Ar uses vdW, Cl covalent. Which cell? "Why is Hf ≈ Zr in size?" ::: Cell F — lanthanide contraction cancels the group growth. One-line reason smaller atom ⇒ higher IE. ::: Outer electron sits closer, feels larger $Z_{e f f}$ , harder to pull off.

Connections

Effective Nuclear Charge — the single lever behind Cells A, C, F, H.
Shielding and Penetration — why f-electrons screen poorly (Cell F).
Ionic Radius — the natural next step once you remove/add electrons.
Ionisation Energy — the sister trend chained in Cell H.
Electronegativity — also tracks $Z_{e f f}$ and radius.
Metallic Bonding — geometry behind the metallic radius (Cell G).
Van der Waals Forces — why vdW radius is largest (Cells D, E).