The correct decreasing order of basic strength of the following species is ____. $H_2O$, $NH_3$, $OH^-$, $NH_2^-$ (i) $NH_2^- > OH^- > NH_3 > H_2O$ (ii) $OH^- > NH_2^- > H_2O > NH_3$ (iii) $NH_3 > H_2O > NH_2^- > OH^-$ (iv) $H_2O > NH_3 > OH^- > NH_2^-$

The correct decreasing order of basic strength of the following species is __. H₂O, NH₃, OH^-, NH₂^- (i) NH₂^-

NCERT Exemplar · Q25

H_2O

(i)

NH_2^- > OH^- > NH_3 > H_2O

(ii)

OH^- > NH_2^- > H_2O > NH_3

(iii)

NH_3 > H_2O > NH_2^- > OH^-

(iv)

H_2O > NH_3 > OH^- > NH_2^-

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Concept understanding — Amine Basicity Order

Amine Basicity Order – From Intuition to Precision

You already know that a base is something that accepts a proton ( $H^+$ ). An amine does this through the lone pair on its nitrogen. The more readily that lone pair grabs a proton, the stronger the base. So the question becomes: what makes that lone pair more or less willing to accept a proton?

The answer depends on three competing effects: inductive effect (electron pushing/pulling from nearby groups), steric hindrance (bulky groups blocking the proton), and solvation (how water molecules stabilise the protonated amine). And critically, the order flips depending on whether you are in the gas phase or in water.

The Intuition – What Would You Expect?

Imagine a nitrogen with three hydrogens: ammonia ( $NH_3$ ). Now replace one hydrogen with a methyl group ( $CH_3$ ). Methyl is an electron-donating group — it pushes electron density toward the nitrogen. That makes the lone pair richer, so it should grab a proton more easily. So methylamine ( $CH_3NH_2$ ) should be a stronger base than ammonia.

Replace another hydrogen with a second methyl group. Now you have dimethylamine ( $(CH_3)_2NH$ ). Even more electron density on nitrogen — stronger base still. Replace the third hydrogen: trimethylamine ( $(CH_3)_3N$ ). Maximum electron donation — so you would expect it to be the strongest base of all.

That is the inductive effect prediction: more alkyl groups → stronger base. In the gas phase, this is exactly what happens. The order is:

Gas phase: $NH_3 < 1^\circ < 2^\circ < 3^\circ$

But in water, the observed order is different. Why?

The Complication – Solvation and Steric Hindrance

When an amine accepts a proton, it becomes a positively charged ammonium ion ( $RNH_3^+$ ). In water, that positive charge is stabilised by hydrogen bonding with water molecules. The more hydrogens on the nitrogen, the more hydrogen bonds it can form — and the more stable the protonated form becomes.

A primary amine ( $1^\circ$ ) has three N–H hydrogens after protonation. A secondary ( $2^\circ$ ) has two. A tertiary ( $3^\circ$ ) has only one. So solvation stabilises the protonated form in the order: $1^\circ > 2^\circ > 3^\circ$ .

At the same time, bulky alkyl groups physically block the approach of a proton to the nitrogen — this is steric hindrance. A tertiary amine has three bulky groups crowding the nitrogen, making it harder for $H^+$ to reach the lone pair.

So in water, two effects oppose the inductive effect: solvation (which favours more N–H bonds) and steric hindrance (which favours less crowding). The result is a compromise.

The Precise Statement – Aqueous Phase Order

In water, the typical order of basic strength for aliphatic amines is:

Aqueous phase: $NH_3 < 3^\circ < 1^\circ < 2^\circ$

That is, secondary amines are the strongest bases in water, followed by primary, then tertiary, then ammonia.

Important

The order in water is not simply "more alkyl groups = stronger base". The secondary amine wins because it has a good balance: enough electron donation from two alkyl groups, but still two N–H hydrogens for solvation, and not too much steric hindrance.

The Full Picture – A Table

Amine	Gas-phase order (inductive only)	Aqueous order (all effects)	Reason
$NH_3$	Weakest	Weakest	No alkyl donation; only 3 H-bonds
$1^\circ$ ( $RNH_2$ )	Second	Second	One alkyl group donates; 3 H-bonds after protonation
$2^\circ$ ( $R_2NH$ )	Third	Strongest	Two alkyl groups donate; 2 H-bonds; steric hindrance still low
$3^\circ$ ( $R_3N$ )	Strongest	Third	Three alkyl groups donate most, but only 1 H-bond and high steric hindrance

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The key idea is that basic strength depends on the availability of the lone pair for protonation. The conjugate base of a weaker acid is stronger. Here, $NH_2^-$ (conjugate base of $NH_3$ , a very weak acid) is the strongest base, followed by $OH^-$ , then $NH_3$ , and finally $H_2O$ as the weakest. The correct order is $NH_2^- > OH^- > NH_3 > H_2O$ , which is option (A).

Why basic strength follows this logic

Basic strength measures how readily a species donates a pair of electrons (usually a lone pair) to a proton. For neutral molecules like $H_2O$ and $NH_3$ , the lone pair is on oxygen and nitrogen respectively. For anions like $OH^-$ and $NH_2^-$ , the negative charge makes the lone pair even more available — so anions are generally stronger bases than their neutral parents.

The most reliable way to compare basic strength across different families is to look at the conjugate acid of each base. A stronger base has a weaker conjugate acid. So if we know the relative acid strengths of $H_2O$ , $NH_3$ , $H_2O$ (again, as the conjugate acid of $OH^-$ ), and $NH_3$ (as the conjugate acid of $NH_2^-$ ), we can rank the bases.

Let’s list each base and its conjugate acid:

Base	Conjugate Acid	Acid strength (relative)
$NH_2^-$	$NH_3$	Very weak ( $pK_a \approx 38$ )
$OH^-$	$H_2O$	Weak ( $pK_a \approx 15.7$ )
$NH_3$	$NH_4^+$	Moderate ( $pK_a \approx 9.25$ )
$H_2O$	$H_3O^+$	Strong ( $pK_a \approx -1.74$ )

The weaker the conjugate acid, the stronger the base. So $NH_3$ (conjugate acid $pK_a \approx 38$ ) is an extremely weak acid — meaning $NH_2^-$ is an extremely strong base. $H_2O$ (conjugate acid $pK_a \approx 15.7$ ) is a weak acid, so $OH^-$ is a moderately strong base. $NH_4^+$ ( $pK_a \approx 9.25$ ) is a stronger acid than water, so $NH_3$ is a weaker base than $OH^-$ . And $H_3O^+$ ( $pK_a \approx -1.74$ ) is a strong acid, so $H_2O$ is a very weak base.

Watch out

A common mistake is to compare the electronegativity of the atom bearing the lone pair and conclude that the more electronegative atom holds the lone pair tighter, making it a weaker base. That works within a period for neutral molecules (e.g., $NH_3$ is more basic than $H_2O$ ), but it fails when comparing anions to neutral molecules. The negative charge on $OH^-$ and $NH_2^-$ overpowers the electronegativity effect — they are both stronger bases than their neutral parents.

Step-by-step reasoning

Identify the conjugate acid of each base.
- $NH_2^-$ accepts a proton to become $NH_3$ .
- $OH^-$ accepts a proton to become $H_2O$ .
- $NH_3$ accepts a proton to become $NH_4^+$ .
- $H_2O$ accepts a proton to become $H_3O^+$ .
Recall the relative acid strengths of these conjugate acids.

The $pK_a$ values (in water) are: …

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✗ Mistake 1: Confusing basicity with nucleophilicity or pH

What students do:

They think “stronger base” means “more OH⁻ produced” or “higher pH.” Here, $OH^-$ is already a base, but $NH_2^-$ is even stronger.

Why it’s wrong:

Basicity is about proton affinity — how strongly a species accepts a proton ( $H^+$ ). It’s not about how many OH⁻ ions are released.

How to avoid:

Always ask: “How easily does this species grab an H⁺?” The more negative the charge and the more available the lone pair, the stronger the base.

✗ Mistake 2: Ignoring charge vs. neutral species

What students do:

They compare $NH_3$ and $NH_2^-$ as if charge doesn’t matter.

Why it’s wrong:

A negatively charged species ( $NH_2^-$ , $OH^-$ ) is always a stronger base than its neutral conjugate acid ( $NH_3$ , $H_2O$ ). Charge greatly increases electron density and proton affinity.

How to avoid:

Separate into two groups first:

Anions: $NH_2^-$ , $OH^-$
Neutral: $NH_3$ , $H_2O$

Anions > neutrals in basic strength.

✗ Mistake 3: Forgetting the conjugate acid strength rule

What students do:

They try to memorize orders without reasoning.

Why it’s wrong:

The key relationship is:

Weaker conjugate acid → Stronger conjugate base

So compare the acids:

$NH_3$ (conjugate acid of $NH_2^-$ ) is a very weak acid ( $pK_a \approx 38$ )
$H_2O$ (conjugate acid of $OH^-$ ) is a stronger acid ( $pK_a \approx 15.7$ )

Thus:

$NH_2^-$ (from weaker acid) > $OH^-$ (from stronger acid)

How to avoid:

Always convert to conjugate acids and compare their $pK_a$ values. Higher $pK_a$ of conjugate acid = stronger base.

✗ Mistake 4: Mixing up periodic trends for neutral molecules

What students do:

They think $NH_3$ is less basic than $H_2O$ because N is more electronegative.

Why it’s wrong:

For neutral molecules, lone pair availability matters more than electronegativity.

In $NH_3$ , N has a more concentrated negative charge and a more available lone pair (less held by the nucleus than O’s lone pair in $H_2O$ ).
So $NH_3$ is a stronger base than $H_2O$ .

How to avoid: …

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Q.The correct decreasing order of basic strength of the following species is ____.
$H_2O$ , $NH_3$ , $OH^-$ , $NH_2^-$

Concept understanding — Amine Basicity Order

Amine Basicity Order – From Intuition to Precision

The Intuition – What Would You Expect?

The Complication – Solvation and Steric Hindrance

The Precise Statement – Aqueous Phase Order

The Full Picture – A Table

Why basic strength follows this logic

Step-by-step reasoning

Method: Conjugate Base Stability & Electronegativity–Polarizability Analysis

Steps

✗ Mistake 1: Confusing basicity with nucleophilicity or pH

✗ Mistake 2: Ignoring charge vs. neutral species

✗ Mistake 3: Forgetting the conjugate acid strength rule

✗ Mistake 4: Mixing up periodic trends for neutral molecules

Q.The correct decreasing order of basic strength of the following species is ____. H2OH_2OH2​O, NH3NH_3NH3​, OH−OH^-OH−, NH2−NH_2^-NH2−​

Why basic strength follows this logic

Step-by-step reasoning

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Q.The correct decreasing order of basic strength of the following species is ____.
$H_2O$ , $NH_3$ , $OH^-$ , $NH_2^-$

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