Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a)

Exercises · 6.40

Q. Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

(a) OH –

(b) F–

(d) BCl3 .

Karnataka PUCTextbookSubjective· 3mImportance★★★★★est

44% · 68/155 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Brønsted Lowry Conjugate Pairs

From Intuition to Precision: Brønsted–Lowry Conjugate Pairs

Imagine you're at a party where people are passing around a single, very special coin. The coin represents a proton ( $\text{H}^+$ ). The game is simple: someone can give the coin to someone else, but only if the other person is willing to take it. You can't just throw the coin; you need a willing receiver.

In chemistry, acids and bases play exactly this game. An acid is the person who gives away the proton. A base is the person who accepts it. But here's the twist: the moment the acid gives away its proton, it transforms into something new — something that can now accept a proton back. That transformed form is called the conjugate base. Similarly, the base, after accepting the proton, becomes something that can donate it back — the conjugate acid.

This is the core of the Brønsted–Lowry theory: every acid-base reaction involves a pair of substances that are linked by the gain or loss of a single proton.

The Precise Statement

\text{Acid} \rightleftharpoons \text{Conjugate Base} + \text{H}^+

A conjugate acid-base pair consists of two species that differ by exactly one proton ( $\text{H}^+$ ). The acid has the proton; the conjugate base does not.

When an acid donates a proton, it becomes its conjugate base.

When a base accepts a proton, it becomes its conjugate acid.

Important

In any Brønsted–Lowry reaction, there are two conjugate pairs: one on the reactant side and one on the product side. They always appear together.

Seeing It in Action

Take the classic reaction between hydrochloric acid and water:

\text{HCl} + \text{H}_2\text{O} \rightarrow \text{Cl}^- + \text{H}_3\text{O}^+

Let's identify the pairs:

Pair 1: $\text{HCl}$ (acid) and $\text{Cl}^-$ (conjugate base). They differ by one $\text{H}^+$ .
Pair 2: $\text{H}_2\text{O}$ (base) and $\text{H}_3\text{O}^+$ (conjugate acid). They also differ by one $\text{H}^+$ .

Notice: water acted as a base here — it accepted the proton from HCl. But water can also act as an acid in other reactions. That's the beauty of the Brønsted–Lowry theory: a substance's role depends on the reaction, not on a fixed label.

Tip

To find the conjugate base of any acid, simply remove one $\text{H}^+$ and reduce the charge by +1.

To find the conjugate acid of any base, add one $\text{H}^+$ and increase the charge by +1.

A Quick Reference Table

Acid	Conjugate Base	Base	Conjugate Acid
$\text{HCl}$	$\text{Cl}^-$	$\text{NH}_3$	$\text{NH}_4^+$
$\text{H}_2\text{SO}_4$	$\text{HSO}_4^-$	$\text{H}_2\text{O}$	$\text{H}_3\text{O}^+$
$\text{NH}_4^+$	$\text{NH}_3$	$\text{OH}^-$	$\text{H}_2\text{O}$
$\text{H}_2\text{O}$	$\text{OH}^-$	$\text{CO}_3^{2-}$	$\text{HCO}_3^-$

Notice how water appears in both columns — it's amphoteric, meaning it can act as either an acid or a base depending on its partner.

Why This Matters …

Start your 14-day free trial to unlock the full solution →

Lewis acids accept an electron pair; Lewis bases donate an electron pair. OH⁻ and F⁻ are Lewis bases (they donate a lone pair), while H⁺ and BCl₃ are Lewis acids (they accept a lone pair).

The key to classifying species as Lewis acids or bases lies in understanding the electron pair — not protons, not charge alone. A Lewis acid is an electron-pair acceptor; a Lewis base is an electron-pair donor. This definition is broader than Brønsted-Lowry, covering species like BCl₃ that have no proton to donate or accept.

Let’s examine each species one by one.

1. OH⁻ (hydroxide ion)

OH⁻ has three lone pairs on oxygen and a negative charge — it is electron-rich. It readily donates a lone pair to form a coordinate bond. For example, when OH⁻ reacts with H⁺:

\text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}

Here, OH⁻ donates its lone pair to H⁺. So OH⁻ acts as a Lewis base.

Tip

Every Brønsted-Lowry base (proton acceptor) is also a Lewis base, because accepting a proton requires donating an electron pair to it.

2. F⁻ (fluoride ion)

F⁻ has four lone pairs and a full octet with a negative charge — again, electron-rich. It donates a lone pair to electron-deficient species. For instance, with BF₃:

\text{F}^- + \text{BF}_3 \rightarrow [\text{BF}_4]^-

F⁻ donates its lone pair to the empty p-orbital on boron. So F⁻ is a Lewis base.

3. H⁺ (proton)

H⁺ has no electrons at all — just a bare nucleus. It desperately needs an electron pair to become stable (as in H₂O or NH₄⁺). When it accepts a lone pair from a base:

\text{H}^+ + :\text{NH}_3 \rightarrow \text{NH}_4^+

H⁺ accepts the electron pair. So H⁺ is a Lewis acid. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2023Set D-21 markMCQ
$\mathrm{SbH}_3 > \mathrm{AsH}_3 > \mathrm{PH}_3 > \mathrm{NH}_3$
›Reveal solutionSolution
Basicity = availability of the lone pair; the lone pair becomes more diffuse (less available) as the central atom's size increases down the group.

Step 1 — What makes a hydride basic

Each group-15 hydride $\mathrm{EH_3}$ (E = N, P, As, Sb, Bi) is pyramidal and carries one lone pair on the central atom. A Lewis base donates that lone pair, so:
$\text{basicity} \propto \text{electron density of the lone pair (its "availability")}$
Step 2 — The trend down the group

Going $\mathrm{N \to P \to As \to Sb}$ :

The atomic size increases sharply.

The lone pair therefore sits in a progressively larger, more diffuse orbital, so its charge density is spread over a bigger volume.

A diffuse lone pair overlaps poorly with the acceptor's empty orbital, so it is donated less readily.

Additionally, the E–H bond angle falls from $107^\circ$ in $\mathrm{NH_3}$ towards $\sim 90^\circ$ in the heavier hydrides, meaning the lone pair acquires more s-character and is held closer to the nucleus — again reducing its donor ability.

Step 3 — Order

Nitrogen, being the smallest, has the most concentrated and most available lone pair, so $\mathrm{NH_3}$ is by far the strongest base of the set:
$\mathrm{NH_3} > \mathrm{PH_3} > \mathrm{AsH_3} > \mathrm{SbH_3}\ (> \mathrm{BiH_3})$
Step 4 — Reject the rest …
Start your 14-day free trial to see the full solution →

KCET 2020Set A-11 markMCQ

\text{CH}_3\text{O}^-

›Reveal solution

The strongest base is the anion whose negative charge is least stabilised — and the electron-donating $\text{CH}_3$ group makes methoxide's oxygen the most electron-rich of the four.

Step 1 — The governing principle

A base is strong when it holds its lone pair loosely and donates it readily. Equivalently:

\text{More stable anion} \;\Longrightarrow\; \text{weaker base}

And since $\text{acid} \rightleftharpoons \text{H}^+ + \text{conjugate base}$ :

\text{Stronger conjugate acid} \;\Longrightarrow\; \text{weaker conjugate base}

Three factors stabilise (and hence weaken) an anion: resonance delocalisation, electronegativity/size of the charge-bearing atom, and electron-withdrawing (−I) groups. Conversely, an electron-donating (+I) group intensifies the charge and strengthens the base.

Step 2 — Rank the four anions by their conjugate acids

Base	Conjugate acid	Acid strength
$Cl^-$	$\mathrm{HCl}$	Very strong acid ( $\mathrm{p}K_a \approx -7$ )
$\mathrm{CH_3COO^-}$	$\mathrm{CH_3COOH}$	Weak acid ( $\mathrm{p}K_a \approx 4.8$ )
$\mathrm{OH^-}$	$\mathrm{H_2O}$	Extremely weak acid
$\mathrm{CH_3O^-}$	$\mathrm{CH_3OH}$	Extremely weak acid

The weaker the acid, the stronger its conjugate base — so immediately $Cl^- \;<\; \mathrm{CH_3COO^-} \;\ll\; \{\mathrm{OH^-},\ \mathrm{CH_3O^-}\}$ .

Step 3 — Eliminate (B) and (C) properly

(C) $Cl^-$ — conjugate base of the strong acid HCl. Chloride is large, and the charge is spread over a big, polarisable ion; it is an exceedingly weak base (which is why $\mathrm{NaCl}$ solution is neutral). ✗
(B) $\mathrm{CH_3COO^-}$ — the negative charge is delocalised by resonance over two equivalent oxygen atoms:

\mathrm{CH_3\text{–}C(=O)\text{–}O^-} \;\longleftrightarrow\; \mathrm{CH_3\text{–}C(\text{–}O^-)=O}

Resonance stabilisation makes it a weak base. ✗

Step 4 — Decide between $\mathrm{OH^-}$ and $\mathrm{CH_3O^-}$ …

Start your 14-day free trial to see the full solution →

←Previous question · EXERCISES6.39 — The species: H2O, HCO3 –, HSO4 – and NH3 can act both as Brönsted acid…→Next question · EXERCISES6.41 — The concentration of hydrogen ion in a sample of soft drink is 3.8 × 1…

Q.Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base:

1. OH⁻ (hydroxide ion)

2. F⁻ (fluoride ion)

3. H⁺ (proton)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days