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

The distortion is polarisation. The cation's ability to cause it is its polarising power; the anion's tendency to suffer it is its polarisability. More polarisation ⇒ more covalent character — electron density piles up between the two nuclei, which is exactly what a shared (covalent) bond is. See Covalent bonding — electron sharing.

(a) Rule 1 — small cation (Li⁺ radius ≈ 76 pm < K⁺ ≈ 138 pm) ⇒ larger field. (b) Rule 2 — high cation charge ($+3$ vs $+1$). (c) Rule 3 — large anion, higher polarisability. (d) Rule 4 — pseudo-noble-gas ($3d^{10}$) core shields poorly ⇒ larger effective pull. See Effective nuclear charge and shielding.

What we do: take the ratio of fields; the constant $k$ and equal charge $q$ cancel. $\frac{E_{\text{Li}}}{E_{\text{K}}}=\frac{1/76^2}{1/138^2}={\left(\frac{138}{76}\right)}^2=1.816^2\approx 3.30$ Why squared: halving the radius quadruples the pull — the $r^2$ is the dominant effect, which is exactly why Rule 1 outranks intuition. Answer: ≈ 3.3× stronger.

${\varphi }_{\text{Na}}=\frac{1}{102}=0.00980,{\varphi }_{\text{Mg}}=\frac{2}{72}=0.02778,{\varphi }_{\text{Al}}=\frac{3}{53.5}=0.05607$ Ranking: ${\text{Al}}^{3+}>{\text{Mg}}^{2+}>{\text{Na}}^{+}$. Why: charge rises $1\to 2\to 3$ and radius shrinks — both push $\varphi$ up together, so the trend is unambiguous. Covalent character rises in the same order.

(a) Proxy: ${\varphi }_{\text{Li}}=1/76=0.01316$, ${\varphi }_{\text{Al}}=3/53.5=0.05607$. So ${\text{Al}}^{3+}\gg {\text{Li}}^{+}$. (b) Real field (drop the shared constant $k$; use $q/(r_c+r_a{)}^2$): $E_{\text{Li}}\propto \frac{1}{(76+181{)}^2}=\frac{1}{257^2}=1.514\times 10^{-5}$ $E_{\text{Al}}\propto \frac{3}{(53.5+181{)}^2}=\frac{3}{234.5^2}=5.455\times 10^{-5}$ Ratio $E_{\text{Al}}/E_{\text{Li}}\approx 3.60$. Both methods agree: Al³⁺ is the stronger polariser. Why they agree here: the large anion ($r_a=181$) dilutes the difference in cation radius, but Al's charge advantage ($\times 3$) survives easily. Look at the figure — even though the two cation circles differ little, the amber field arrows into Cl⁻ are visibly denser for Al³⁺.

Proxy: ${\varphi }_{\text{Li}}=1/76=0.01316$, ${\varphi }_{\text{Mg}}=2/72=0.02778$. Ratio $=2.11$. Real field: $E_{\text{Li}}\propto \frac{1}{(76+133{)}^2}=\frac{1}{209^2}=2.290\times 10^{-5},E_{\text{Mg}}\propto \frac{2}{(72+133{)}^2}=\frac{2}{205^2}=4.759\times 10^{-5}$ Ratio $E_{\text{Mg}}/E_{\text{Li}}=2.08$. Reading: proxy says Mg pulls $2.11\times$; real field says $2.08\times$ — nearly identical here because Li⁺ and Mg²⁺ have almost the same size, so only the charge separates them. The proxy neither exaggerates nor hides much in this special case — which is precisely why Li and Mg show similar covalent tendencies (the diagonal relationship).

Forecast: $\varphi$ rises ${\text{Na}}^{+}<{\text{Mg}}^{2+}<{\text{Al}}^{3+}$ (from L2.2: $0.0098<0.0278<0.0561$). More $\varphi$ ⇒ more covalent ⇒ weaker giant ionic lattice, more discrete molecules ⇒ lower melting point. Predicted MP order (high→low): $\text{NaCl}>{\text{MgCl}}_2>{\text{AlCl}}_3$. Verify (real data): NaCl 801 °C, MgCl₂ 714 °C, AlCl₃ sublimes ≈ 180 °C. ✅ Matches. See Ionic bonding — lattice energy.

Reason: ${\text{S}}^{2-}$ is both larger and carries a higher $2-$ charge, so it holds its outer electrons more loosely ⇒ greater polarisability (Rule 3) ⇒ more covalent character than ${\text{Cl}}^{-}$. Consequence: more covalent, less "ionic" solids dissolve less in water (weaker ion–dipole hydration payoff vs a covalent network). So sulphides are typically darker and less water-soluble than the matching chlorides. See Solubility and lattice/hydration energy and Electronegativity and bond polarity.

${\varphi }_{\text{Cu}}=1/95=0.01053$, ${\varphi }_{\text{Na}}=1/102=0.00980$. Ratio $=1.074$ — only ~7% apart. By Rules 1&2 alone they'd be near-twins. Rule 4 breaks the tie: ${\text{Cu}}^{+}$ has a $3d^{10}$ pseudo-noble-gas core. The $d$ electrons are diffuse and shield the nucleus poorly, so the anion feels an effective nuclear charge larger than $+1$ suggests. That extra real pull makes $\text{CuCl}$ much more covalent. See Effective nuclear charge and shielding.

Build it stepwise:

• $\text{NaF}$: $+1$ cation, small hard ${\text{F}}^{-}$ anion, noble-gas core → least covalent (weak on every lever).
• ${\text{AlF}}_3$: swap Na⁺→Al³⁺ (Rules 1&2: smaller, $+3$) — big jump up, but ${\text{F}}^{-}$ is a poor anion (small, tight).
• ${\text{AlI}}_3$: keep Al³⁺, swap F⁻→I⁻ (Rule 3: huge, squishy anion) — jumps above ${\text{AlF}}_3$.
• $\text{AgI}$: ${\text{Ag}}^{+}$ is only $+1$, but it has a $4d^{10}$ pseudo-noble-gas core (Rule 4) and is paired with the most polarisable anion ${\text{I}}^{-}$ (Rule 3). The combination of a poorly-shielding core with the softest anion pushes it highest. Order (most→least covalent): $\text{AgI}>{\text{AlI}}_3>{\text{AlF}}_3>\text{NaF}$. (Note: AlI₃ vs AgI can be argued either way in strict lists; the defensible core answer is that both sit far above the fluorides, with the $d^{10}$ + soft-iodide pairing of AgI making it exceptionally covalent.)

(a) No anion charge means no loosely-held excess electron cloud to distort in the ionic sense — Fajan's polarisation of an anion is undefined; you'd instead be describing ordinary covalent overlap from the start. (b) As separation $\to \infty$, $E\propto q/(r_c+r_a{)}^2\to 0$: the field vanishes, no distortion, the bond tends to ideal ionic (or no bond). This is the limiting sanity-check on Rule 1. (c) $\text{CsF}$: huge low-charge cation ($\varphi$ minimal) + small tight anion ⇒ minimal covalent character, the most nearly-ideal ionic compound. Fajan predicts CsF as the "purest" ionic salt — and indeed it is famously so.

Which single fact makes CuCl more covalent than NaCl despite equal charge and size? ::: The $3d^{10}$ pseudo-noble-gas core of Cu⁺ shields poorly, so the anion feels a larger effective nuclear charge (Rule 4). Why is $\varphi =q/r_{\text{cation}}$ only a rough proxy? ::: It uses $1/r$ not $1/r^2$ and ignores the anion's radius and charge entirely. In the limit $r_c+r_a\to \infty$, what happens to covalent character? ::: The polarising field $\to 0$, so the bond tends to ideal ionic.