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

Definition The three radii, defined before you use them

Every symbol below appears throughout this bank — pin them down first. In each case we measure $d$ , the internuclear distance (centre-to-centre separation of two nuclei), then split it.

Covalent radius $r_{co v} = \frac{d _{A - A}}{2}$ — half the distance between two atoms joined by a covalent bond (shared electron pair pulling the nuclei together).
Metallic radius $r_{m e t} = \frac{d _{M \dots M}}{2}$ — half the distance between two nearest nuclei in a metallic lattice (electrons delocalised into a shared "sea").
van der Waals radius $r_{v d W} = \frac{d _{A \dots A}}{2}$ — half the distance between two non-bonded atoms merely touching cloud-to-cloud (held by weak Van der Waals Forces).

Because tighter binding pulls nuclei closer, $r_{co v} < r_{m e t} < r_{v d W}$ . Look at Figure s01: three pairs of atoms at the same scale, showing how the same element yields three different half-distances.

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

Formula Where

$r \propto n^{2} / Z_{e f f}$ comes from (brief derivation) For a one-electron (hydrogen-like) atom the allowed orbit radius is exactly $r_{n} = \frac{n ^{2} a _{0}}{Z}, a_{0} = 52.9 pm (Bohr radius) .$ The $n^{2}$ appears because balancing the electron's angular momentum (quantised in units of $n$ ) against the Coulomb pull gives $r \propto n^{2}$ ; the $Z$ in the denominator is the nuclear charge doing the pulling — more charge, tighter orbit. A multi-electron atom is not one-electron, so we replace the bare $Z$ by the net charge the outer electron actually feels, $Z_{e f f} = Z - S$ : $r \propto \frac{n ^{2}}{Z _{e f f}}$ Now the two knobs are explicit: across a period $n$ is fixed but $Z_{e f f}$ rises → $r$ falls; down a group $n$ jumps by one → $n^{2}$ grows faster than $Z_{e f f}$ → $r$ rises. Figure s03 plots this model against real Period-2 radii so you see it is a genuine trend, not a slogan.

Recall A word on the numbers you'll see

Tabulated radii (e.g. Ne vdW $\approx 154$ pm, Cl covalent $99$ pm) vary by $\pm$ a few pm between sources — they are averages over many measured internuclear distances (X-ray crystallography, gas-phase spectroscopy), not exact constants. Use them for ranking and trends, not high-precision arithmetic; the ordering is robust even where an individual value is uncertain.

True or false — justify

An atom has one true, fixed radius we just measure imperfectly.

False — the radius is a distance between two atoms split in half, so the number you get depends on what kind of contact (bond, lattice, or mere touching) you measured; see Van der Waals Forces.

For the same element,

r_{v d W}

is always larger than

r_{co v}

True — no bond pulls the nuclei in, only weak attraction opposes full-cloud repulsion, so non-bonded atoms stay farther apart than bonded ones (

r_{co v} < r_{m e t} < r_{v d W}

Adding a proton always shrinks the atom.

False — a proton shrinks the atom only if no new shell opens; down a group you add a proton and a whole new shell, and the new shell's larger

n

wins, so the atom grows.

Across a period the outer electrons enter the same principal shell.

True — that is exactly why shielding barely rises while

Z

climbs, so Effective Nuclear Charge increases and the atom contracts.

Z_{e f f}

is the same as the atomic number

Z

False —

Z_{e f f} = Z - S

, where

S

is the screening by inner electrons; the outer electron feels the reduced net charge, not the full

Z

Metallic radius equals covalent radius for a metal that also bonds covalently.

False — in a metallic lattice each atom shares electrons with many neighbours, so the pull per neighbour is looser and

r_{m e t} > r_{co v}

Noble gases are the smallest atoms in their period.

False — they have no bonding partner, so their tabulated radius is a van der Waals radius (a bigger ruler), which makes them look larger than their covalent-radius neighbours.

Down a group,

Z_{e f f}

increases so much it should shrink the atom.

False — the added inner shells shield the extra protons well, so

Z_{e f f}

rises only modestly; the jump in

n

(and hence

n^{2}

r \propto n^{2} / Z_{e f f}

) dominates and the atom grows.

Same-shell electrons contribute the same to

S

as core electrons.

False — a same-shell partner contributes only about

0.35

S

while a core electron contributes near

1.00

; this gap is precisely why

Z_{e f f}

climbs across a period.

Spot the error

"F has covalent radius 57 pm and Ne has radius 38 pm, so Ne is smaller than F."

The error is mixing rulers: F's 57 pm is a covalent half-distance (measured to a bonded partner that drags nuclei in), while Ne bonds to nothing, so its real radius is a van der Waals value (

\approx 154

pm) measured to a merely-touching neighbour. A vdW ruler always reads larger than a covalent one, so the comparison is meaningless because the underlying physical contact is different, not just the label.

"Cl

_{2}

nuclei sit 198 pm apart, so the Cl atom is 198 pm across."

The 198 pm belongs to two atoms; by the symmetry of identical partners each owns exactly half, giving

r_{co v} = 198/2 = 99

pm. Reporting the full distance double-counts one atom's share.

"K has more protons than Na, so K must be smaller."

More protons only shrink an atom when the outer electron stays in the same shell. K adds electrons in a new

n = 4

shell that Na lacks, and K's 18 core electrons screen the extra 8 protons almost fully (

S

rises nearly in step with

Z

), so

Z_{e f f}

barely grows while

n^{2}

jumps — K ends up larger.

"Radius scales as

n^{2} / Z_{e f f}

, and across a period

Z_{e f f}

falls, so atoms grow rightward."

Z_{e f f}

actually rises across a period, because each added proton is only weakly shielded (

\approx 0.35

) by its same-shell partner, so the numerator

n^{2}

is fixed while the denominator grows — the radius therefore shrinks; the reasoning has the sign of

Z_{e f f}

's change backwards.

"O is to the right of N, so O's covalent radius is smaller — but O is 66 and N is 71, so the trend is broken."

66 < 71, so O is smaller than N; the "break" is a misread of the inequality — the period trend holds exactly as expected because

Z_{e f f}

keeps rising rightward.

"Shielding is done by all other electrons, including ones in the same shell equally."

Screening depends on where the shielding electron sits: a same-shell electron is at the same radius, so it spends little time between the nucleus and the outer electron and contributes only

\approx 0.35

S

, whereas an inner electron sits directly in the line of pull and contributes

\approx 0.85

–

1.00

. Treating them as equal wrongly predicts no period contraction — see Shielding and Penetration.

Why questions

Why must one atom carry three different radii at all?

Because an atom has no sharp edge, we infer size from internuclear distance, and that distance changes with the type of contact — strong covalent bond, delocalised metallic lattice, or weak vdW touching.

Why does a covalent bond give the smallest of the three radii?

In a bond the two atoms' outer orbitals overlap, so a shared electron pair sits in the region between the nuclei; that build-up of negative charge simultaneously attracts both nuclei inward until repulsion balances it — the overlap physically shortens the internuclear distance below any non-overlapping (metallic or vdW) contact.

Why does poor same-shell shielding make periods contract?

A newly added same-shell electron contributes only

\approx 0.35

S

but its accompanying proton adds a full

+ 1

Z

, so

Z_{e f f} = Z - S

climbs by roughly

0.65

each step; the stronger net pull on the fixed-

n

shell contracts the whole cloud.

Why do we use half the distance rather than the full internuclear distance?

For two identical atoms the midpoint of the bond is a plane of symmetry, so the electron cloud is shared evenly and each atom's fair share of

d

is exactly

d /2

Why does a smaller radius usually mean higher ionisation energy?

A smaller atom holds its outer electron closer to the nucleus where the Coulomb attraction (

\propto 1/ r

) is stronger and

Z_{e f f}

is higher, so more energy is needed to pull that electron away; see Ionisation Energy.

Why does electronegativity broadly track the radius trend?

A small atom exposes a bonding electron to a high

Z_{e f f}

at short range, so it pulls shared electrons hard toward itself — high Electronegativity; big atoms with distant, well-shielded outer shells pull weakly.

Why can't we just say "more electrons = bigger atom"?

Number of electrons alone ignores whether they enter a new shell (

n ↑

, bigger) or the same shell (better

Z_{e f f}

, smaller); the shell they occupy, not the count, sets the size.

Edge cases

A hypothetical atom gains an electron in the same shell with no new proton — bigger or smaller?

Slightly bigger, because added same-shell electron–electron repulsion (and a tiny rise in shielding) with no extra nuclear charge loosens the cloud — this is the logic behind anion swelling in Ionic Radius.

Helium: does it have a covalent radius?

No — He forms no bonds, so it has only a van der Waals radius; there is no bonded He–He distance to halve.

Compare Ne (vdW radius) and Na (metallic radius) directly — is that fair?

Not really — they are measured with different rulers and different bonding contexts, so a raw number comparison is misleading; always match the radius type before ranking.

Two identical atoms in different molecules just touching — which radius does that internuclear distance give?

The van der Waals radius, since the atoms are non-bonded and held only by weak Van der Waals Forces, so half that distance is

r_{v d W}

Going from Ar to K, radius jumps up sharply — why the discontinuity?

K starts a brand-new

n = 4

shell while Ar's radius is a vdW value; the fresh, distant shell plus the ruler change together produce a large apparent increase.

Within a metallic lattice, if you counted a next-nearest neighbour instead of the nearest, would you get the metallic radius?

No — the metallic radius uses the nearest adjacent nuclei; using a farther pair overstates the distance and the radius, so nearest-neighbour contact is the defining case.

An atom just after a filled d-block subshell — why might its radius barely change from the one above?

Ten poorly-shielding d-electrons let

Z_{e f f}

rise more than usual (the d-block contraction), partly cancelling the new shell's expansion, so the radius stays nearly flat — a shielding edge case from Shielding and Penetration.

An atom just after the 4f (lanthanide) block — why is it almost the same size as the element directly above it?

The 14 f-electrons shield the growing nuclear charge very poorly, so

Z_{e f f}

swells sharply across the lanthanides — the lanthanide contraction — pulling the outer shell in enough to offset the extra

n

. This is why Hf (

n = 6

) is almost identical in size to Zr (

n = 5

) directly above it.

Recall One-line survival rules

Match the ruler before you rank (cov/met/vdW), let $Z_{e f f}$ handle periods, let the new shell ( $n$ ) handle groups, and remember the radius is a pair measurement, never a lone-atom property.

Connections

Effective Nuclear Charge — decides every period trap.
Shielding and Penetration — explains the "poor same-shell shielding" answers.
Ionic Radius — the anion-swell edge case.
Ionisation Energy — smaller radius, higher IE.
Electronegativity — tracks $Z_{e f f}$ and radius.
Van der Waals Forces — why the vdW ruler is the largest.