Reagent classification — electrophiles, nucleophiles (hard - soft): Worked examples

The scenario matrix

Before solving anything, let us map the whole battlefield. Every HSAB question is one of these cells:

Formula The scenario matrix — every case class

#	Case class	What is being tested	Example that covers it
1	Classify one species hard vs soft	size + charge density reasoning	(a) — $F^{-}$ vs $I^{-}$
2	Pick the favoured pairing among ions	hard–hard / soft–soft matching	(b) — $L i F + C s I$ equilibrium
3	Ambident, soft electrophile	soft partner steers the soft donor atom	(c) — $C N^{-}$ + $C H_{3} I$
4	Ambident, hard electrophile	hard partner steers the hard donor atom	(d) — enolate + $R_{3} S i C l$
5	Neutral species (charge = 0) — the "sign is zero" trap	electrophile/nucleophile without net charge	(e) — $B F_{3}$ and $N H_{3}$
6	Borderline / degenerate case	when the answer is "it's in between"	(f) — $B r^{-}$ , $C u^{2 +}$
7	Two axes at once (basicity vs nucleophilicity)	limiting behaviour by solvent	(g) — $F^{-}$ vs $I^{-}$ in water
8	Real-world / word problem	applying HSAB to a real separation	(h) — why gold ore uses cyanide
9	Exam-style twist	condition change flips the product	(i) — $C N^{-}$ with $A g^{+}$ ( $S_{N} 1$ )

Cells 1–9 below are each labelled with the cell they hit. Together they touch every column of the matrix — no scenario is left unshown.

(a) — Cell 1 · Classify a single species

Worked example Is

$F^{-}$ or $I^{-}$ the softer base? Explain from first principles. Forecast: guess now — which one has the "fluffier", easier-to-squash electron cloud?

Step 1 — Compare the ionic radii. $F^{-}$ radius $\approx 133 pm$ ; $I^{-}$ radius $\approx 220 pm$ . Why this step? Polarisability (how much an electron cloud deforms when a nearby charge tugs it) grows with size, because outer electrons in a big ion are far from the nucleus and only weakly held. See the figure: the same negative charge is spread over a much larger balloon in $I^{-}$ .

Step 2 — Compare charge density. Both carry charge $- 1$ . Charge density $\propto \frac{1}{volume} \propto \frac{1}{r ^{3}}$ . $\frac{ρ ( F ^{-} )}{ρ ( I ^{-} )} = (\frac{r _{I}}{r _{F}})^{3} = (\frac{220}{133})^{3} \approx 4.5$ Why this step? High charge density = charge packed tight = hard. $F^{-}$ has roughly $4.5 \times$ the charge density of $I^{-}$ , so $F^{-}$ is much harder.

Step 3 — Conclude. $I^{-}$ is the soft base; $F^{-}$ is the hard base.

Verify: the group-17 trend $F^{-} < C l^{-} < B r^{-} < I^{-}$ (hard → soft) from the parent's cheat-grid agrees — going down a group means bigger, softer. ✔

(b) — Cell 2 · Pick the favoured pairing

Worked example Which side does

$L i F + C s I ⇌ L i I + C s F$ favour? Forecast: which two salts are the "comfortable couples"?

Step 1 — Classify all four ions. $L i^{+}$ = small, high charge density = hard acid. $C s^{+}$ = huge alkali cation = soft acid. $F^{-}$ = hard base. $I^{-}$ = soft base. Why this step? You can never pick a pairing until every partner has a hard/soft label.

Step 2 — Match like with like. Hard $L i^{+}$ wants hard $F^{-}$ → $L i F$ . Soft $C s^{+}$ wants soft $I^{-}$ → $C s I$ . Why this step? HSAB: hard–hard maximises the ionic term $\frac{q _{+} q _{-}}{4 π ε _{0} r}$ (tiny $r$ ), soft–soft maximises covalent overlap. Mixing (e.g. $L i I$ ) is a mismatch.

Step 3 — Read the direction. The stable pairs $L i F + C s I$ are the left side, so the equilibrium lies to the left.

Verify: sanity check with a rough lattice-energy argument — $L i F$ has by far the smallest interionic distance of the four salts, so it is the most stable single bond; keeping $L i$ with $F$ is favourable. ✔

(c) — Cell 3 · Ambident donor, SOFT electrophile

Worked example

$C N^{-}$ reacts with $C H_{3} I$ (via $S_{N} 2$ ). Which atom of $C N^{-}$ bonds — C or N? Forecast: the cyanide ion can attack through carbon or nitrogen. Guess which.

Step 1 — Identify the electrophilic centre. In $C H_{3} I$ , the electrophile is the carbon bearing the leaving group $I^{-}$ . A saturated $s p^{3}$ carbon undergoing $S_N2$ is a soft electrophilic centre (diffuse, polarisable). Why this step? You must classify the partner before you can choose the donor atom — an ambident nucleophile lets its partner decide.

Step 2 — Match the donor atom's hardness. $C N^{-}$ : the C end is the soft end (diffuse carbanion lobe), the N end is harder (more electronegative, charge localised). Why this step? Soft electrophile ⇒ soft donor atom ⇒ carbon attacks. The curved arrow leaves the carbon lone pair of $C N^{-}$ and points to the $C H_{3}$ carbon.

Step 3 — Write the product. $C H_{3} - I +^{-} C \equiv N ⟶ C H_{3} - C \equiv N (acetonitrile, a nitrile)$

Verify: the C–C bond product (nitrile) is the observed major product of $S_{N} 2$ alkylation of cyanide. Soft–soft rule holds. ✔

(d) — Cell 4 · Ambident donor, HARD electrophile

Worked example An enolate ion reacts with

$R_{3} S i C l$ (a silyl chloride). O-product or C-product? Forecast: enolates alkylate at carbon with $R X$ — will silicon do the same?

Step 1 — Classify the enolate's two ends. Enolate is ambident: the O end is hard (charge localised on electronegative oxygen), the C end is soft (diffuse $π$ density). Why this step? Same logic as $C N^{-}$ — an ambident donor waits for its partner's character.

Step 2 — Classify the electrophile. Silicon in $R_{3} S i C l$ is small, high oxidation state, oxophilic — a hard electrophilic centre (unlike the soft $s p^{3}$ carbon of an alkyl halide). Why this step? Hard electrophile ⇒ pairs with the hard donor atom.

Step 3 — Predict. Hard Si + hard O ⇒ O-attack → a silyl enol ether ( $C = C - O - S i R_{3}$ ). Contrast: soft $R X$ ⇒ C-attack (C-alkylation).

Verify: silyl enol ethers are indeed the standard O-silylation product of enolates — consistent with hard–hard matching. The same enolate gives different atoms depending on the partner's hardness. ✔

(e) — Cell 5 · Neutral species (the "charge is zero" trap)

Worked example

$B F_{3}$ and $N H_{3}$ are both neutral. Which is the electrophile, which the nucleophile? Classify hard/soft. Forecast: if you think "no charge = neither reacts," you're about to be surprised.

Step 1 — Count electrons, not charge. Boron in $B F_{3}$ has only 6 valence electrons — an empty $p$ -orbital. Nitrogen in $N H_{3}$ has a lone pair. Why this step? The parent's key mistake: it's electron-pair availability, not net charge. Charge here is zero for both; that tells us nothing.

Step 2 — Assign roles. Empty orbital = accepts a pair = electrophile (Lewis acid) → $B F_{3}$ . Lone pair = donates a pair = nucleophile (Lewis base) → $N H_{3}$ . They form $H_{3} N \to B F_{3}$ . See Lewis Acids and Bases.

Step 3 — Hard/soft. $B F_{3}$ : small B, high effective positive charge = hard acid. $N H_{3}$ : small N donor, electronegative = hard base. Hard–hard ⇒ a strong, stable adduct.

Verify: the $H_{3} N - B F_{3}$ adduct is a real, isolable, strongly bonded solid — exactly what hard–hard predicts. Neutral species, real reaction. ✔

(f) — Cell 6 · Borderline / degenerate case

Worked example Classify

$B r^{-}$ and $C u^{2 +}$ . What if a species is neither clearly hard nor soft? Forecast: not every ion sits at an extreme — where do these two land?

Step 1 — Place $B r^{-}$ on the halide trend. $F^{-}$ (hard) $< C l^{-} < B r^{-} < I^{-}$ (soft). $B r^{-}$ sits in the middle → borderline base. Why this step? Softness is a continuum, not two boxes. Size grows smoothly down the group.

Step 2 — Place $C u^{2 +}$ . $C u^{+}$ (large, low charge) is soft, but $C u^{2 +}$ (higher charge, smaller) pulls toward hard → it is listed borderline in the parent's grid. Why this step? Raising the charge on the same element raises charge density = pushes it harder. Charge and size trade off.

Step 3 — How to answer a borderline in an exam. State it is borderline, then predict from conditions: borderline species react well with both hard and soft partners, so the other reactant and the solvent decide.

Verify: the parent's cheat-grid lists $B r^{-}$ and $C u^{2 +}$ explicitly in the borderline column. ✔

(g) — Cell 7 · Two independent axes (basicity vs nucleophilicity)

Worked example In water, which is the better

nucleophile: $F^{-}$ or $I^{-}$ ? Which is the stronger base? Forecast: they seem to be the same question — they are not.

Step 1 — Separate the two questions. Basicity = tendency to grab $H^{+}$ (thermodynamic, toward a proton). Nucleophilicity = kinetic tendency to attack carbon. Why this step? These are independent axes. A species can be strong on one, weak on the other.

Step 2 — Basicity. $F^{-}$ holds its electrons tightly and bonds hard $H^{+}$ strongly → $F^{-}$ is the stronger base. Why this step? Hard–hard ( $F^{-}$ + $H^{+}$ ) is a strong ionic bond.

Step 3 — Nucleophilicity in protic water. Water hydrogen-bonds tightly around small hard $F^{-}$ , "caging" it. Big soft $I^{-}$ is poorly solvated and its polarisable cloud reaches the transition state early → $I^{-}$ is the better nucleophile in water. Why this step? This is the limiting solvent case: in protic solvent, solvation flips the order. (In aprotic solvent, the caging is removed and $F^{-}$ recovers.) See Polarisability and Atomic Size.

Verify: the parent states exactly this — " $I^{-}$ is a good nucleophile but a weak base; $F^{-}$ a strong base but poor nucleophile (in water)." Two axes confirmed. ✔

(h) — Cell 8 · Real-world word problem

Worked example Gold is extracted from crushed ore by stirring with dilute

$N a C N$ solution, dissolving gold as $[A u (C N)_{2}]^{-}$ . Explain the choice of cyanide using HSAB. Forecast: why cyanide, of all the cheap anions ( $C l^{-}$ , $O H^{-}$ )?

Step 1 — Classify the metal. $A u^{+}$ is a large, low-charge, highly polarisable cation → a soft acid (like $A g^{+}$ , $C u^{+}$ in the parent's grid). Why this step? To pick a good ligand we first fix the metal's hardness.

Step 2 — Pick a matching donor. $C N^{-}$ donates through its soft carbon end → soft base. Soft $A u^{+}$ + soft $C N^{-}$ ⇒ a very stable covalent complex. Why this step? Hard bases like $F^{-}$ / $O H^{-}$ would bind $A u^{+}$ weakly (mismatch) and fail to pull gold into solution.

Step 3 — Conclude. The strong soft–soft $A u$ – $C$ bonding makes $[A u (C N)_{2}]^{-}$ so stable that gold dissolves selectively — the basis of industrial cyanidation.

Verify: the linear $[A u (C N)_{2}]^{-}$ ion is the actual species in gold cyanidation, and $A u^{+}$ / $A g^{+}$ are classic soft acids — soft–soft prediction matches industrial reality. ✔

(i) — Cell 9 · Exam twist (change the condition, flip the product)

Worked example

$C N^{-}$ gave a nitrile ( $R$ –CN) with $C H_{3} I$ in example (c). Now $A g^{+}$ is added. What changes, and why? Forecast: same cyanide, same alkyl halide — can the product really switch atoms?

Step 1 — What $A g^{+}$ does. Soft $A g^{+}$ grabs the leaving $I^{-}$ (soft–soft), pulling it off early and generating a carbocation $R^{+}$ — an $S_N1$-like pathway. See Carbocation Stability. Why this step? Removing the halide changes the mechanism and therefore the electrophile's character.

Step 2 — Reclassify the electrophile. A free carbocation $R^{+}$ is a harder, more ionic electrophilic centre than the diffuse $S_{N} 2$ carbon. Why this step? Harder electrophile now prefers the harder donor atom of $C N^{-}$ = nitrogen.

Step 3 — New product. N-attack ⇒ isocyanide $R - N \equiv C$ (an isonitrile), not the nitrile. $C H_{3} - I A g^{+}, C N^{-} soft \to S_{N} 1 C H_{3} - N \equiv C$

Verify: classic result — $S_{N} 2$ conditions give nitriles (C-attack), $A g^{+}$ / $S_{N} 1$ conditions give isocyanides (N-attack). One condition change flips the atom, exactly as HSAB predicts. ✔

Recall Self-test — cover the answers

A neutral species with an empty orbital is a(n) …? ::: electrophile (Lewis acid), regardless of zero charge. Which end of $C N^{-}$ attacks a soft $S_{N} 2$ carbon? ::: the carbon end (soft–soft) → nitrile. Which end attacks under $A g^{+}$ / $S_{N} 1$ ? ::: the nitrogen end (harder electrophile) → isocyanide. Is $F^{-}$ a better base or better nucleophile in water? ::: stronger base, but poorer nucleophile than $I^{-}$ . Ratio of charge densities $ρ (F^{-}) / ρ (I^{-})$ (radii 133, 220 pm)? ::: about $4.5$ . Which side does $L i F + C s I ⇌ L i I + C s F$ favour? ::: the left (hard $L i F$ + soft $C s I$ ).

Related: Ambident Nucleophiles · Lewis Acids and Bases · Inductive and Resonance Effects · Polarisability and Atomic Size · Hinglish version