Fajan's rules — covalent character in ionic compounds: Worked examples

Before any symbol appears, here is the one number we lean on again and again.

Three symbols we will reuse — define them once, before any formula uses them:

$q$ = the cation's charge, in units of the elementary charge (so $+ 1$ , $+ 2$ , $+ 3$ — just the number).
$r_{cat}$ = the cation's ionic radius; $r_{an}$ = the anion's ionic radius. Both in picometres ( $1 pm = 1 0^{- 12} m$ ). Picture each as "how far the outer skin of that ion sits from its centre." When we write a plain $r$ next to an ion symbol (e.g. $r_{Li^{+}}$ ) it just means the radius of that specific ion — cation or anion — and equals $r_{cat}$ or $r_{an}$ depending on which ion it labels.
The sum $r_{cat} + r_{an}$ is the centre-to-centre distance between the two ions — the actual gap the field must cross, which is why the true field uses it, not the cation radius alone.

The scenario matrix

Every Fajan's question is really one of these cells. The examples below are labelled with the cell they hit.

Cell	The variable being tested	Covalent-character winner	Example
A — cation size	same $q$ , change $r_{cat}$	smaller cation	E1
B — cation charge	same $r_{cat}$ -ish, change $q$	higher charge	E2
C — anion size	same cation, change $r_{an}$	bigger anion	E3
D — anion charge	same cation, change anion charge	more-charged anion	E4
E — electron configuration (tie-breaker)	same $q$ , same $r$ , change core	pseudo-noble ( $d^{10}$ ) cation	E5
F — competing variables (charge vs size fight)	$q$ and $r$ both change	must compute/reason carefully	E6
G — degenerate / equal case	identical $ϕ$ , no difference	"no prediction from $ϕ$ alone"	E7
H — limiting / extreme case	huge $q$ , tiny $r$	fully covalent, molecular	E8
I — real-world word problem	property ⇄ covalent character	E9
J — exam twist (trap)	colour / conductivity misuse	E10

Radii we will reuse (pm). Read this caveat first: ionic radii are not fixed constants — they depend on the ion's coordination number (how many neighbours surround it) and the data set. The values below are Shannon effective ionic radii for octahedral, 6-coordinate environments (the standard textbook choice), rounded to whole pm. A different coordination (4-fold, 8-fold) shifts them by several pm, but the rankings we build below are robust to that shift, which is all Fajan's reasoning needs.

Ion	$r$ (pm, 6-coord)	Ion	$r$ (pm, 6-coord)
Li⁺	76	Na⁺	102
K⁺	138	Mg²⁺	72
Al³⁺	54	Ag⁺	115
Cu⁺	77	Be²⁺	45
Ca²⁺	100	F⁻	133
Cl⁻	181	Br⁻	196
I⁻	220	S²⁻	184

Cell A — cation size

Worked example E1 · Which is more covalent, LiCl or KCl?

Forecast: both cations are $+ 1$ ; guess which wins before reading on — the small one or the big one?

Step 1 — Compute $ϕ$ for each cation. $ϕ_{Li^{+}} = \frac{1}{76} \approx 0.01316 pm^{- 1}, ϕ_{K^{+}} = \frac{1}{138} \approx 0.00725 pm^{- 1}$ Why this step? $ϕ$ collapses "same charge, different size" into one comparable number. Larger $ϕ$ = harder pull.

Step 2 — Compare. $ϕ_{Li^{+}} > ϕ_{K^{+}}$ , so LiCl is more covalent. Why this step? The whole point of building $ϕ$ was to make the decision a single "which number is bigger?" — comparing is where the ranking actually gets made, so it must be stated, not assumed.

Step 3 — Cross-check with the true field. Using $E \propto q / (r_{cat} + r_{an})^{2}$ with $r_{Cl^{-}} = 181$ : $\frac{E _{Li}}{E _{K}} = (\frac{138 + 181}{76 + 181})^{2} = (\frac{319}{257})^{2} \approx 1.54$ Why this step? $ϕ$ is only a proxy; recomputing with the physically honest field guards against a case where the crude proxy exaggerates or reverses the verdict. Here it confirms it. Li⁺ produces about $1.5 \times$ the true polarising field.

Verify: $ϕ$ -ratio $= 138/76 \approx 1.82$ ; field-ratio $\approx 1.54$ . Both $> 1$ , both point to LiCl. ✅ (This is Rule 1 in action.)

Cell B — cation charge

Worked example E2 · Rank covalent character: NaCl, MgCl₂, AlCl₃.

Forecast: charge climbs $+ 1 \to + 2 \to + 3$ and the cation shrinks each step. Guess the covalent order first.

Step 1 — Tabulate $ϕ = q / r$ . $ϕ_{Na^{+}} = \frac{1}{102} \approx 0.00980, ϕ_{Mg^{2 +}} = \frac{2}{72} \approx 0.02778, ϕ_{Al^{3 +}} = \frac{3}{54} \approx 0.05556$ Why this step? Here charge and size cooperate — both push $ϕ$ up together — so a single number ranks them cleanly.

Step 2 — Order by $ϕ$ . $ϕ_{Al} > ϕ_{Mg} > ϕ_{Na}$ , so covalent character: AlCl₃ > MgCl₂ > NaCl. Why this step? Ranking is the deliverable the question asked for; turning three $ϕ$ values into an ordered list is the actual answer, and confirming the order is monotonic checks no two values accidentally swapped.

Verify: melting points fall as covalent character rises — NaCl $80 1^{\circ} C$ , MgCl₂ $71 4^{\circ} C$ , AlCl₃ sublimes $\sim 18 0^{\circ} C$ . Monotonic decrease matches. ✅

Cell C — anion size

Worked example E3 · Which is more covalent, AlF₃ or AlI₃?

Forecast: the cation (Al³⁺) is identical. Only the anion changes: F⁻ tiny, I⁻ huge. Which is squishier?

Step 1 — Note that $ϕ$ is useless here. $ϕ = q / r_{cat}$ is the same for both (same Al³⁺). Why this step? It flags that this cell tests polarisability of the anion, a property $ϕ$ deliberately ignores.

Step 2 — Compare anion size. $r_{I^{-}} = 220 pm ≫ r_{F^{-}} = 133 pm$ . Bigger anion → outer electrons farther from their nucleus → loosely held → more distortable. Why this step? Since $ϕ$ gave no separation, the only discriminating variable left is anion radius; measuring it is literally where the answer comes from in this cell.

Step 3 — Verdict. AlI₃ is far more covalent than AlF₃. Why this step? Translating "I⁻ more polarisable" into the compound-level statement "AlI₃ more covalent" is the step that answers the question as posed.

Verify: ratio of anion radii $220/133 \approx 1.65$ ; the iodide's cloud reaches much further and softer. Physically, AlI₃ is a low-melting covalent-ish solid; AlF₃ is a high-melting ionic lattice. ✅ (Rule 3.)

Cell D — anion charge

Worked example E4 · Which is more covalent, Na₂S or NaCl?

Forecast: same cation (Na⁺). S²⁻ and Cl⁻ are almost the same size ( $184$ vs $181$ pm). So what breaks the tie?

Step 1 — Compare anion charge. S²⁻ carries $2 -$ ; Cl⁻ carries $1 -$ . Why this step? With size nearly equal, only charge distinguishes them. An anion with two extra electrons over the same nuclear pull holds each electron more loosely → higher polarisability.

Step 2 — Verdict. Na₂S is more covalent than NaCl. Why this step? We isolated charge as the only live variable in Step 1; converting "more loosely held electrons" into "more covalent character" is the inference that actually settles the comparison.

Verify: radii nearly equal ( $∣184 - 181∣ = 3$ pm, negligible), so the difference must come from charge. Real trend: metal sulphides are darker and far less water-soluble than the chlorides — consistent with more covalent character. ✅ (This kills the common "only anion size matters" mistake from the parent note.)

Cell E — electron configuration (the tie-breaker)

Worked example E5 · Which is more covalent, CuCl or NaCl?

Forecast: Cu⁺ ( $77$ pm) and Na⁺ ( $102$ pm) are close in size and both $+ 1$ . $ϕ$ favours Cu⁺ only slightly — is that the whole story?

Step 1 — Compute $ϕ$ . $ϕ_{Cu^{+}} = 1/77 \approx 0.01299$ , $ϕ_{Na^{+}} = 1/102 \approx 0.00980$ . Why this step? To see how big the "size" advantage alone is — and it's modest.

Step 2 — Invoke Rule 4. Cu⁺ is $[Ar] 3 d^{10}$ — a pseudo-noble-gas core. Those $d^{10}$ electrons are diffuse and shield the nucleus poorly, so the anion feels a larger effective nuclear charge than the bare $+ 1$ suggests. Na⁺ is a true noble-gas core ( $[Ne]$ ) with tight, well-shielding electrons. Why this step? Step 1 showed $ϕ$ alone barely separates them, so we must reach for the variable $ϕ$ ignores — the electron core — otherwise we would badly under-predict CuCl's covalent character.

Step 3 — Verdict. CuCl is markedly more covalent — more than the small $ϕ$ gap alone would predict. Why this step? Naming the winner and flagging that it beats the $ϕ$ -based expectation is exactly the lesson of this cell: configuration can override a near-tie in the proxy.

Verify: $ϕ$ ratio $= 102/77 \approx 1.32$ (only 32% more) yet CuCl behaves far more covalently; the extra gap is the $d^{10}$ effect. Real: CuCl melts $\sim 43 0^{\circ} C$ vs NaCl $80 1^{\circ} C$ . ✅ (Rule 4.)

Cell F — competing variables (charge vs size fight)

Here $ϕ$ earns its keep: when charge and size push in opposite directions, you must actually compute.

Worked example E6 · Which is more covalent, LiCl or AlCl₃ — is it obvious?

Forecast: Li⁺ is tiny ( $76$ pm) but only $+ 1$ . Al³⁺ is even smaller ( $54$ pm) and $+ 3$ . Guess which wins.

Step 1 — Compute $ϕ$ . $ϕ_{Li^{+}} = 1/76 \approx 0.01316$ , $ϕ_{Al^{3 +}} = 3/54 \approx 0.05556$ . Why this step? Neither variable alone decides it — Li wins on nothing, Al wins on charge and size. Only the combined number is safe.

Step 2 — Compare. $ϕ_{Al} \approx 4.2 \times ϕ_{Li}$ , so AlCl₃ is more covalent by a wide margin. Why this step? When two effects could in principle oppose, forming the ratio is the only way to see whether one dominates or they nearly cancel — here it reveals a clear four-fold win, not a close call.

Step 3 — Interpret the picture. Notice Li⁺'s bar barely clears Na⁺'s (both $+ 1$ , only radius differs), while Al³⁺'s bar is in a different league — charge is the bigger lever here. Why this step? Seeing the magnitude of the gap (not just its sign) tells you AlCl₃ is decisively covalent, not marginally so, which matters when predicting real properties.

Verify: $3/54 = 0.05556$ and $1/76 = 0.01316$ ; ratio $= 0.05556/0.01316 \approx 4.22$ . Al wins. ✅ (Physically, AlCl₃ sublimes as covalent Al₂Cl₆ dimers; LiCl is a real, if slightly covalent, ionic solid melting $60 5^{\circ} C$ .)

Cell G — degenerate / equal case

Worked example E7 · Cation X⁺ has

$r = 100$ pm; cation Y²⁺ has $r = 200$ pm, both with the same anion and same noble-gas core. What does $ϕ$ predict? Forecast: one has double the charge, the other double the radius. Guess: tie, or does one win?

Step 1 — Compute $ϕ$ . $ϕ_{X} = 1/100 = 0.0100$ , $ϕ_{Y} = 2/200 = 0.0100$ . They are identical. Why this step? This is the degenerate cell — the proxy gives no separation, so you must say so honestly.

Step 2 — Interpret. $ϕ$ predicts equal polarising power, i.e. no prediction of a difference from $ϕ$ alone. If a real answer is needed, drop to the true field $E \propto q / (r_{cat} + r_{an})^{2}$ , which would separate them because the whole centre-to-centre distance $(r_{cat} + r_{an})$ is squared — so changing $r_{cat}$ moves that squared sum in a way $ϕ$ 's bare $1/ r_{cat}$ cannot capture. Why this step? Recognising that a proxy tie is not a physical tie is the whole teaching point of this cell — it tells you when to abandon $ϕ$ and escalate to the exact expression.

Step 3 — True field with, say, $r_{an} = 150$ pm. $\frac{E _{Y}}{E _{X}} = \frac{2/ ( 200 + 150 ) ^{2}}{1/ ( 100 + 150 ) ^{2}} = \frac{2/122500}{1/62500} = 2 \times \frac{62500}{122500} \approx 1.02$ Why this step? Actually evaluating the escalated formula shows the "hidden" difference $ϕ$ concealed — and quantifies it as tiny, vindicating $ϕ$ 's near-tie verdict rather than overturning it.

Verify: $ϕ_{X} = ϕ_{Y} = 0.01$ exactly. Field ratio $\approx 1.02$ , essentially degenerate. ✅ (Lesson: $ϕ$ hides the anion radius, so exact ties in $ϕ$ need the full field to resolve.)

Cell H — limiting / extreme case

Worked example E8 · Take charge to the extreme: is "BeCl₂" still ionic, versus a normal group-2 chloride like CaCl₂?

Forecast: as $q$ grows and $r_{cat}$ shrinks, $ϕ \to$ large. At the limit, is the compound still "ionic"?

Step 1 — Push $ϕ$ up. Using the master table: Be²⁺ ( $r_{cat} = 45$ pm, $q = + 2$ ) gives $ϕ_{Be^{2 +}} = 2/45 \approx 0.0444$ ; Ca²⁺ ( $r_{cat} = 100$ pm) gives $ϕ_{Ca^{2 +}} = 2/100 = 0.0200$ . Why this step? Be²⁺ is the smallest realistic $+ 2$ cation, so it gives the highest realistic group-2 $ϕ$ — the closest we get to the covalent extreme within real ions.

Step 2 — Limiting behaviour. As $ϕ \to \infty$ (imagine a point-like, highly charged cation), polarisation is so complete that electron density fully floods the gap — the "ionic" description breaks down entirely and the bond is essentially covalent / molecular. Why this step? Examining the mathematical limit tells you what the end of the covalent-character axis looks like, so you can place any real compound on the sliding scale between ionic and covalent.

Step 3 — Verdict. Be²⁺'s $ϕ$ is more than double Ca²⁺'s, so BeCl₂ sits near the covalent extreme for a group-2 chloride; it forms covalent chain/molecular structures, unlike ionic CaCl₂ or BaCl₂. Why this step? Landing the abstract limit back onto a concrete, memorable pair (BeCl₂ vs CaCl₂) is what makes the extreme case usable in an exam.

Verify: $ϕ_{Be} = 2/45 \approx 0.0444 > ϕ_{Ca} = 2/100 = 0.0200$ ; higher $ϕ$ ⇒ more covalent. Real: BeCl₂ is a low-melting ( $40 5^{\circ} C$ ), covalent, polymeric solid that sublimes; CaCl₂ is ionic (MP $77 2^{\circ} C$ ). ✅

Cell H′ — beyond $d^{10}$ : where Fajan's rules strain (advanced edge case)

Common mistake "Rule 4 only ever means the

$d^{10}$ pseudo-noble-gas core; every other cation is 'normal'." Why it feels right: the parent note only listed $d^{10}$ examples (Cu⁺, Ag⁺, Zn²⁺, ...), so it's tempting to think that is the only non-noble-gas core worth flagging. The fix: Rule 4 is really "poorly-shielding cores polarise more," and $d^{10}$ is just the commonest culprit. Two important extensions:

Inert-pair / $n s^{2}$ cores (e.g. Tl⁺, Pb²⁺, Sn²⁺, Bi³⁺): these have an outer $n s^{2}$ pair over a $(n - 1) d^{10}$ shell that shields poorly, so they polarise even harder than a plain $d^{10}$ ion — PbI₂ and Bi₂S₃ are strongly covalent/coloured, more than $ϕ$ or a $d^{10}$ -only view predicts.
$f$ -block ions (lanthanide/actinide, e.g. La³⁺, Ce³⁺, U⁴⁺): the $4 f$ / $5 f$ electrons are buried deep and shield the nucleus very poorly, yet they are also contracted and core-like, so they do not participate in bonding the way $d$ electrons do. The net result is that simple Fajan reasoning (and the crude $ϕ$ ) becomes unreliable for $f$ -block ions: effective nuclear charge arguments still apply, but the geometry of $f$ orbitals and relativistic effects mean you should treat any Fajan prediction for $f$ -block compounds as a weak hint, not a verdict.

Take-away: Fajan's four rules are a first-order guide built around $s$ -, $p$ - and $d$ -cores. For inert-pair and $f$ -block cations, the direction of the trend (poor shielding → more covalent) still holds, but the magnitude and reliability degrade — cross-check with real data.

Cell I — real-world word problem

Worked example E9 · A lab bottle of "AgI" is a bright yellow, water-insoluble powder; "AgF" is white and dissolves freely in water. Explain using Fajan's rules.

Forecast: same cation (Ag⁺). Guess which halide is more covalent, and connect it to solubility.

Step 1 — Identify the changing variable. Only the anion changes: F⁻ ( $133$ pm) vs I⁻ ( $220$ pm). This is a Cell-C (anion size) problem wearing a word-problem coat. Why this step? Word problems hide the physics; naming the live variable first turns a paragraph of description into a clean single-variable comparison you already know how to do.

Step 2 — Rank polarisability. I⁻ is much bigger and softer → far more polarisable → AgI is much more covalent than AgF. (Ag⁺ is also $d^{10}$ , so both start from a high polarising power — Rule 4 — which amplifies the effect.) Why this step? Establishing the covalent-character order is the bridge between "what the ions are" and "how the compound behaves" — without it the property claim in Step 3 would be unjustified.

Step 3 — Link to solubility. More covalent character ⇒ weaker attraction to polar water molecules and a large lattice-vs-hydration mismatch ⇒ low water solubility (see Solubility and lattice/hydration energy). Ionic AgF dissolves readily; covalent-ish AgI does not. Why this step? The question asked us to explain the observation, so we must complete the chain all the way to the measurable property (solubility), not stop at "more covalent."

Verify: anion-radius ratio $220/133 \approx 1.65$ → AgI clearly more covalent → less water-soluble. Solubility data: AgF very soluble ( $\sim 1800$ g/L), AgI essentially insoluble ( $\sim 3 \times 1 0^{- 6}$ g/L). Monotonic with predicted covalent character. ✅

Cell J — exam twist (the colour trap)

Worked example E10 · TRAP: "AgI is yellow

because Fajan polarisation lowers the electronic transition energy into the visible range." True or false — and what's the correct statement? Forecast: it sounds right (more distortion → lower energy gap → visible absorption). Guess before deciding.

Step 1 — Spot the overreach. Fajan's rules tell you AgI is more covalent than AgF, and colour correlates with that. But correlation is not the mechanism. Why this step? Exams plant a true-sounding causal claim; you must separate the trend (Fajan) from the mechanism (physics of light absorption).

Step 2 — State the real mechanism. The yellow colour comes from a charge-transfer / band-structure transition in the solid — an electron jumping from a mostly-iodide level to a mostly-silver level (see Charge-transfer transitions and colour). Polarisation facilitates strong orbital mixing, but the colour is not "Fajan lowering a single transition energy." Why this step? Naming the correct physical cause is what lets you replace the wrong claim with a right one, rather than just declaring it false.

Step 3 — Verdict and corrected statement. The statement is False as a causal claim. The correct version is: "AgI's yellow colour is associated with its high covalent character, but the colour itself arises from a charge-transfer (band-structure) transition in the solid, not directly from Fajan polarisation lowering an excitation energy. Use colour as a hint of covalent character, never as Fajan 'proof'." Why this step? Producing the full corrected sentence — not a dangling half-idea — is exactly what an exam answer must deliver: reject the trap and supply the defensible replacement.

Verify: Consistency check — AgF (ionic, low covalent character) is white/colourless, AgCl white, AgBr pale-cream, AgI yellow. The colour deepens with covalent character (the correlation is real), yet the explanation is charge-transfer. Both halves check out. ✅

Recall Which matrix cell is each example? (self-test)

E1 — cation size ::: Cell A E2 — cation charge ::: Cell B E3 — anion size ::: Cell C E4 — anion charge ::: Cell D E5 — configuration tie-breaker ::: Cell E E6 — competing variables ::: Cell F E7 — degenerate/equal ::: Cell G E8 — limiting extreme ::: Cell H E9 — word problem ::: Cell I E10 — colour trap ::: Cell J Cores beyond d10 (inert-pair, f-block) ::: Cell H′ — Fajan strains, treat as weak hint