Fajan's rules — covalent character in ionic compounds: Visual walkthrough

Step 1 — Draw the two ions before anything happens

WHAT. We place two objects side by side. On the left, a small hard ball carrying positive charge — this is the cation (a positive ion, "cat = paws-itive"). On the right, a big soft cloud of negative charge — this is the anion (a negative ion). The cloud is soft because its outer electrons are far from their own nucleus and only loosely held.

WHY. Before we can talk about distortion, we must see the undistorted starting shape. In an ideal ionic picture the anion cloud is a perfect sphere — the cation took one electron, walked away, and the two just sit there attracting each other as whole charges. Anything that dents that sphere is the new physics we are hunting.

PICTURE. Look at the figure: the blue ball (cation, $+$ ) is small and rigid; the orange fuzzy circle (anion, $-$ ) is large and drawn dashed to say "squishy". The gap between the two nuclei is labelled — that gap is where a covalent bond could later appear.

Figure — Fajan's rules — covalent character in ionic compounds

Step 2 — Turn on the cation's electric field

WHAT. Every charge fills the space around it with a pull-per-unit-charge called the electric field, written $E$ . For a single point charge it is:

$E = \frac{k q}{r ^{2}}$

$E$ — the strength of the pull felt at some point in space (bigger $E$ = harder yank).
$q$ — how much charge the cation carries ( $+ 1$ , $+ 2$ , $+ 3$ …).
$r$ — the distance from the cation to the point where we measure the pull.
$k$ — a fixed constant of nature ( $\approx 9 \times 1 0^{9}$ in SI units); it just sets the scale and never changes, so we can ignore it when we compare two ions.

WHY this tool and not another? We need one number that says "how hard does the cation tug on the anion's electrons, and how does that tug change if I move things?" The electric field is exactly that — force per unit charge — so it answers "how strong is the pull here?". We use $1/ r^{2}$ (not $1/ r$ ) because that is the honest law for a point charge: double the distance and the pull drops to a quarter, not a half. That squaring is the hero of Rule 1.

PICTURE. The figure draws blue field arrows shooting out from the cation. Notice they are long and dense close in and short and sparse far away — that shrinking is the $1/ r^{2}$ falloff drawn honestly. The anion's near edge sits inside the strong-arrow zone, so that edge gets pulled hardest.

Step 3 — Watch the cloud dent (Rule 1: small cation)

WHAT. Now let the field act. The cation's arrows grab the anion's near-side electrons and drag them inward, into the gap. The once-round orange cloud becomes egg-shaped, bulging toward the cation. We then ask: what happens if we swap the cation for a smaller one of the same charge?

WHY. Rule 1 says smaller cation → more covalent. Read it off the field law. At the anion's edge the distance from the cation is roughly the cation's own radius $r$ . Halve $r$ and $E = k q / r^{2}$ jumps by $2^{2} = 4$ . So a small cation drags the cloud far harder — the $r^{2}$ makes size the dominant knob.

Concretely, comparing $L i^{+}$ ( $r \approx 76$ pm) with $K^{+}$ ( $r \approx 138$ pm), both $+ 1$ :

$\frac{E _{Li}}{E _{K}} = \frac{k q /7 6 ^{2}}{k q /13 8 ^{2}} = (\frac{138}{76})^{2} \approx 3.3$

Each symbol: $q$ and $k$ cancel (same charge, same constant), leaving only the radius ratio squared — so Li⁺ pulls about $3.3 \times$ harder, and LiCl is more covalent than KCl.

PICTURE. Two panels side by side: small blue ball → deep egg-shaped dent (lots of orange in the gap); big blue ball → gentle dent (cloud still nearly round). The label "shared density" sits in the gap of the small-cation panel.

Step 4 — Crank the charge (Rule 2: high cation charge)

WHAT. Keep the cation's size fixed but change its charge from $+ 1$ to $+ 3$ . In $E = k q / r^{2}$ the charge $q$ sits on top, so triple $q$ ⇒ triple the field ⇒ triple the yank.

WHY. Rule 2 says higher cation charge → more covalent. Because $E \propto q$ is linear, this knob is gentler than the size knob (which was squared) — but still real. An $A l^{3 +}$ (small and $+ 3$ ) is a double win: small $r$ and big $q$ , so it dents any cloud hard. That is why $AlC l_{3}$ behaves covalently (it sublimes, dissolves in organic solvents) while $NaCl$ ( $+ 1$ ) stays a proper high-melting ionic solid.

PICTURE. Same cloud, two cations of equal size but charges $+ 1$ and $+ 3$ . The $+ 3$ panel has three times as many field arrows and a deeper dent. A little bar next to each shows dent-depth $\propto q$ — a straight-line growth, contrasting with Step 3's squared growth.

Step 5 — Soften the target (Rule 3: big / highly-charged anion)

WHAT. Now stop touching the cation and change the anion instead. A bigger anion holds its outer electrons far from its own nucleus, so those electrons are loosely held — the cloud is floppier. And a more negatively charged anion ( $2 -$ vs $1 -$ ) has extra electrons crammed in per unit of its own nuclear charge, again loosely held. Both make the cloud easier to dent — its polarisability is high.

WHY. The field from the cation might be identical, but the response depends on how squishy the target is. Same $A l^{3 +}$ : against tiny tight $F^{-}$ you get a shallow dent (AlF₃ ionic); against huge floppy $I^{-}$ you get a deep dent (AlI₃ largely covalent). Likewise $S^{2 -}$ (big and $2 -$ ) dents far more than $C l^{-}$ ( $1 -$ ), so sulphides run more covalent than chlorides.

PICTURE. One cation, three anions in a row: small-tight (F⁻, barely dented), large (I⁻, deeply dented), and large-plus-2⁻ (S²⁻, dented deepest). Springs drawn on each cloud — stiff spring for F⁻, loose spring for S²⁻ — show "polarisability = softness".

Step 6 — The sneaky core (Rule 4: pseudo-noble-gas cation)

WHAT. Take two cations that are the same size and same charge — $C u^{+}$ ( $\approx 95$ pm) and $N a^{+}$ ( $\approx 102$ pm), both $+ 1$ — yet $CuCl$ is far more covalent than $NaCl$ . Rules 1–3 predict a tie. Something else is deciding.

WHY. The difference is the cation's inner shell. $C u^{+}$ has a $3 d^{10}$ ( $(n - 1) d^{10} n s^{0}$ ) pseudo-noble-gas core; $N a^{+}$ has a neat noble-gas core. The $d$ electrons are diffuse and non-spherical, so they shield the nucleus poorly — they don't cleanly cover the positive nucleus. The result is that the anion feels a bigger effective nuclear charge than the $+ 1$ label suggests. Bigger effective pull ⇒ deeper dent ⇒ more covalent. Configuration is the tie-breaker.

PICTURE. Two same-sized cations. Na⁺'s core drawn as a smooth grey shell fully hiding its nucleus (good shielding, weak leak). Cu⁺'s core drawn as a lumpy $d$ -shell with gaps through which red "leaked field" lines escape toward the anion — so the anion feels more pull despite equal nominal charge.

Step 7 — Compress it into one number, honestly (ionic potential)

WHAT. The physically correct distorting field uses the whole cation-to-anion separation, not just the cation's own radius:

$E \propto \frac{q _{cation}}{( r _{cation} + r _{anion} ) ^{2}}$

$q_{cation}$ — cation charge (Rule 2 lives here, on top).
$r_{cation} + r_{anion}$ — the full centre-to-centre gap the field must cross (Rules 1 and 5 live here, on the bottom).
squared — the honest point-charge falloff.

Chemists then boil it down to a quick ranking proxy, the ionic potential:

$ϕ = \frac{q _{cation}}{r _{cation}}$

$ϕ$ (phi) — one number: bigger $ϕ$ ⇒ generally more covalent, all else equal.
$q_{cation}$ on top — captures "high charge helps".
$r_{cation}$ on bottom — captures "small cation helps".

WHY only a proxy? $ϕ$ throws away three things we drew: it drops $r_{anion}$ (ignores anion size), it uses $1/ r$ not $1/ r^{2}$ (softens the size effect), and it says nothing about anion charge (Rule 3) or the $d$ -shell leak (Rule 4). So $ϕ$ ranks fast but lies at the edges — never treat it as exact.

PICTURE. A number line of $ϕ$ with $N a^{+}, M g^{2 +}, A l^{3 +}$ climbing left-to-right, and below it the matching melting points falling — the covalent trend made visible as $ϕ$ rises.

Recall Forecast-then-verify: melting points of NaCl, MgCl₂, AlCl₃

Forecast: charge $+ 1 \to + 2 \to + 3$ , cation shrinks Na⁺>Mg²⁺>Al³⁺, so $ϕ$ rises ⇒ covalent character rises ⇒ MP should fall. Verify: NaCl $80 1^{\circ}$ C, MgCl₂ $71 4^{\circ}$ C, AlCl₃ sublimes $\sim 18 0^{\circ}$ C. ✅

The one-picture summary

Everything on one canvas. A single cation-anion pair with four knobs labelled: shrink the cation (↑field, squared), raise its charge (↑field, linear), swell/charge the anion (↑softness), and swap in a $d$ -shell core (↑leaked pull). Turn any knob toward "more" and the orange cloud dents deeper into the gap — that growing shared bulge is covalent character.

Recall Feynman retelling of the whole walkthrough

Two ions sit side by side: a small hard positive ball and a big soft ball of electrons. The positive ball fills space with a pull that is fierce up close and fades fast with distance (that's the $1/ r^{2}$ — Step 2). That pull drags the soft ball's near edge into the gap, denting it egg-shaped; the dent in the gap is a covalent bond sneaking in (Step 3). Four ways to make the dent bigger: use a smaller ball (distance shrinks, pull grows fast because it's squared — Rule 1), a more charged ball (pull grows, but only in step — Rule 2), a bigger or more negative soft ball (floppier, easier to dent — Rule 3), or a positive ball with a leaky $d$ -shell core that lets extra pull escape even at the same size and charge (Rule 4). Chemists shorthand the first two into one number $ϕ = q / r$ , handy but rough because it forgets the anion. Deep dent = covalent = lower melting point, more soluble in oil than water. That's the whole story in one picture.

See also: Coulomb's law and electric fields, Electronegativity and bond polarity, Effective nuclear charge and shielding, Ionic bonding — lattice energy, Covalent bonding — electron sharing, Solubility and lattice/hydration energy, Charge-transfer transitions and colour.

Recall Quick self-test

Why does the cation being small matter more than it being highly charged? ::: Field size effect is squared ( $1/ r^{2}$ ) while charge effect is linear ( $\propto q$ ), so shrinking the radius changes the field faster. What does $ϕ = q / r$ deliberately ignore? ::: The anion's radius, the true $1/ r^{2}$ (it uses $1/ r$ ), anion charge/polarisability, and cation d-shell configuration. Cu⁺ and Na⁺ are the same size and charge — why is CuCl more covalent? ::: Cu⁺'s $3 d^{10}$ core shields poorly, so the anion feels a larger effective nuclear charge → stronger pull.