Reagent classification — electrophiles, nucleophiles (hard - soft): Question bank

Before you start, a two-line reminder of the vocabulary so nothing below is a surprise:

Recall What the four core words mean

Electrophile = electron-pair acceptor (electron-poor: full $+$ , partial $δ^{+}$ , or empty orbital). Nucleophile = electron-pair donor (electron-rich: lone pair, $π$ bond, or $-$ charge). Hard = small, tight, non-polarisable electron cloud. Soft = large, loose, easily-distorted cloud. Polarisability = how easily an electron cloud gets pushed out of shape by a nearby charge.

True or false — justify

B F_{3}

carries no charge, so it cannot be an electrophile.

False. Electrophilicity is about an empty/incomplete octet, not net charge; boron in

B F_{3}

has only 6 electrons and an empty p-orbital, so it eagerly accepts a lone pair — a neutral yet strong electrophile.

N H_{3}

is neutral, so it is a poor nucleophile.

False. Nucleophilicity comes from an available electron pair, and

N H_{3}

has a lone pair on nitrogen. Neutral species with lone pairs (

H_{2} O

N H_{3}

R_{2} S

) are perfectly good donors.

Every electrophile is a Lewis acid and every nucleophile is a Lewis base.

True. An electrophile accepts an electron pair (the exact definition of a Lewis acid) and a nucleophile donates one (the definition of a Lewis base); the two vocabularies describe the same event from organic and inorganic viewpoints.

A soft nucleophile is always a weaker nucleophile than a hard one.

False. Softness measures polarisability, not strength. In protic solvents

I^{-}

(soft) out-competes

F^{-}

(hard) because its loose cloud reaches the transition state more easily; hardness and reactivity are independent axes.

Basicity and nucleophilicity are the same property under a different name.

False. Basicity is the thermodynamic tendency to grab

H^{+}

; nucleophilicity is the kinetic tendency to attack carbon.

F^{-}

is a strong base but a poor nucleophile in water, showing the two can point opposite ways.

Going down a group (F → Cl → Br → I) the ion gets softer.

True. Atomic size increases down a group, so the same charge spreads over a larger volume, lowering charge density and raising polarisability — the very definition of getting softer.

H^{+}

is a hard acid because it is small and highly charged.

True. A bare proton is the smallest possible acid with its full

+ 1

packed into essentially zero volume, giving maximum charge density and no polarisable cloud at all — the archetypal hard acid.

The curved arrow can validly be drawn starting at the electrophile.

False. The arrow tracks electron motion, and electrons flow from the electron-rich donor to the electron-poor acceptor, so it must start at the nucleophile and end at the electrophile (see Curved Arrow Notation).

Spot the error

C l^{-}

is soft, so it prefers soft electrophiles like

A g^{+}

Error.

C l^{-}

is borderline/hard-ish, not soft; softness in the halides really begins at

B r^{-}

and peaks at

I^{-}

. Ranking is

F^{-} < C l^{-} < B r^{-} < I^{-}

from hard to soft.

"Enolate always alkylates at oxygen because the negative charge sits on O."

Error. The O end is hard (localised charge) and the C end is soft (diffuse

π

). With a soft alkyl halide the soft–soft match wins, so C-alkylation dominates; only hard electrophiles (silyl, acyl,

H^{+}

) go to O.

A l C l_{3}

is a nucleophile because it donates

C l

to the substrate."

Error. In Friedel–Crafts

A l C l_{3}

accepts the lone pair of a chlorine (empty Al orbital, only 6 electrons), pulling

C l^{-}

off

R C l

to make

R^{+}

. Accepting electrons makes it a Lewis acid → electrophile.

"The HSAB rule says hard reacts with soft to balance out."

Error. It is the opposite: hard prefers hard, soft prefers soft. Like-with-like gives the most stable bond; hard–soft is the mismatched, weaker combination.

"Since

C N^{-}

is a soft nucleophile, it must attack every electrophile through nitrogen."

Error.

C N^{-}

is ambident: carbon is the softer, more polarisable end, so with a soft

S_{N} 2

carbon it bonds through C (giving the nitrile

R - C N

); N-attack (isocyanide) needs a harder, more ionic pathway (e.g.

A g^{+}

). See Ambident Nucleophiles.

"A carbocation is always a soft electrophile because carbon is soft."

Error. Hardness depends on charge density and stability of the specific cation. A small, localised, unstabilised carbocation is fairly hard; only large, resonance-stabilised

R_{3} C^{+}

(charge spread out, more polarisable) drifts toward soft — see Carbocation Stability and Inductive and Resonance Effects.

"Because

S_{N} 2

needs a good nucleophile,

F^{-}

beats

I^{-}

in water."

Error. In a protic solvent, hard

F^{-}

is tightly solvated and shielded, while soft, polarisable

I^{-}

sheds its solvent shell easily and reaches carbon first — so

I^{-}

is the better

S_{N} 2

nucleophile there.

Why questions

Why do hard–hard pairs form strong bonds even though neither partner overlaps orbitals well?

Because their small size lets the ions approach at tiny separation

r

, and the electrostatic (ionic) energy

\propto q_{+} q_{-} / r

blows up as

r

shrinks — the bond is stabilised by charge attraction, not covalent overlap.

Why do soft–soft pairs bond well despite low charge density?

Their diffuse, polarisable clouds deform toward each other for excellent orbital overlap, and their matched (small) HOMO–LUMO energy gap makes the covalent stabilisation

\propto (overlap)^{2} /Δ E

large.

Why does a hard–soft combination give the weakest bond?

The small hard partner offers little polarisable cloud to overlap, while the diffuse soft partner cannot approach closely for a strong ionic term — so both stabilising channels are half-satisfied, leaving a mismatch.

Why does a small HOMO–LUMO gap favour covalent (soft–soft) bonding?

Second-order perturbation theory gives an interaction energy

\sim (matrix element)^{2} /Δ E

; a small energy gap

Δ E

(energies matched, as in soft–soft) makes that fraction large, so covalent stabilisation grows.

Why does changing

S_{N} 1

↔

S_{N} 2

conditions flip the product of an ambident nucleophile?

The two mechanisms present electrophilic centres of different hardness (a tight ion pair /

A g^{+}

-assisted centre is harder; a free

S_{N} 2

carbon is softer), and HSAB steers the nucleophile's soft or hard atom to whichever matches.

Why is net charge only a hint, not the law, for classifying reagents?

Reactivity is governed by electron-pair availability — a lone pair or empty orbital — which neutral species can also possess (

N H_{3}

donates,

B F_{3}

accepts), so charge correlates with but does not define the role.

Why does increasing atomic size increase polarisability (see Polarisability and Atomic Size)?

Larger atoms hold their outer electrons farther from the nucleus with weaker pull, so an approaching charge can push that loose cloud out of shape more easily — higher polarisability, hence softer.

Edge cases

H_{2} O

ever an electrophile rather than a nucleophile?

Yes — the oxygen lone pairs make it a nucleophile, but the

δ^{+}

hydrogens let it act as an electrophile/acid toward strong bases; the same molecule plays either role depending on the partner.

Where does

B r^{-}

sit on the hard–soft scale, and why is it awkward?

It is borderline, between hard

C l^{-}

and soft

I^{-}

, because its size and polarisability are intermediate — so it can partner reasonably with both hard and soft centres, giving less clear-cut predictions.

What happens with

C N^{-}

if the electrophile is genuinely borderline?

The C-vs-N selectivity becomes weak and product mixtures appear, because neither end enjoys a decisive HSAB match — ambident selectivity is sharpest only at clearly hard or clearly soft partners.

Can a species be a strong base yet a weak nucleophile at the same time?

Yes — a small, hard, tightly-solvated donor like

F^{-}

(or a bulky hindered base) grabs the small proton avidly (strong base) but reaches a crowded carbon poorly (weak nucleophile), because the two jobs test different things.

What is the "degenerate" case where HSAB gives no preference?

When both competing partners are the same hardness (e.g. two hard bases of equal charge density competing for a hard acid), HSAB predicts no selectivity and the outcome is decided by other factors — sterics, solvent, concentration.

Does a

π

bond count as a nucleophile even with no lone pair or charge?

Yes — the loosely-held

π

electrons of

C = C

are available to donate into an electron-poor centre, so alkenes act as (usually soft) nucleophiles despite being neutral and lone-pair-free.

Z n^{2 +}

hard or soft, and why is the answer "it depends"?

It is borderline: its

+ 2

charge pulls toward hard, but its moderately large ion size and filled-d character lend some polarisability, so its classification shifts with the ligand and conditions rather than being fixed.

Recall One-sentence summary to carry away

Charge is a hint but electron-pair availability is the law; hard/soft measures polarisability (not strength); like prefers like; and ambident reagents obey the partner's hardness, not a fixed habit.