Match the items of Column I with the items of Column II. Column I: (i) Ethylene glycol (ii) Ethanol (iii) Phen

NCERT Exemplar · Q63

Q. Match the items of Column I with the items of Column II. Column I:

(i) Ethylene glycol

(ii) Ethanol

(iii) Phenol

(iv) Methanol

(v) Neutral ferric chloride

(vi) Fats and waxes
Column II:

(a) detection of phenolic group

(b) used in soap preparation

(d) wood spirit

(e) solvent in perfume formulations

(f) conversion to picric acid

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Concept understanding — Solubility Rules

Solubility Rules – From Intuition to Precision

Imagine you drop a spoonful of sugar into a glass of water. Stir once, and it disappears. Now try the same with a spoonful of sand. It just sits at the bottom. Why? The sugar molecules are able to break apart and mingle with water molecules — we say sugar dissolves in water. Sand does not.

That "disappearing" act is solubility. But in chemistry, we don't just ask if something dissolves — we ask how much and under what conditions. For ionic compounds (salts), the answer is surprisingly predictable. That predictability is what we call the Solubility Rules.

The Core Idea: "Like Dissolves Like" — But Ionic Compounds Are Different

For molecular substances like sugar, the rule of thumb is "like dissolves like" — polar dissolves in polar, non-polar in non-polar. But ionic compounds are made of charged particles (cations and anions). When you drop an ionic solid into water, the water molecules (which are polar) try to pull the ions apart. Whether they succeed depends on a tug-of-war:

The water molecules want to surround and separate the ions (hydration energy).
The ions themselves are held together by electrostatic forces (lattice energy).

If the hydration energy wins, the salt dissolves. If the lattice energy wins, it stays solid.

Note

You don't need to calculate these energies for exams. The Solubility Rules are a shortcut — a set of patterns discovered by observing thousands of salts.

The Precise Statement: The Solubility Rules

Here are the rules as you'll use them in exams. They are hierarchical — the first applicable rule overrides later ones.

Solubility Rules for Ionic Compounds in Water

Rule	Statement	Examples
1	All nitrates ( $\text{NO}_3^-$ ) are soluble.	$\text{NaNO}_3$ , $\text{AgNO}_3$ , $\text{Pb(NO}_3)_2$
2	All acetates ( $\text{CH}_3\text{COO}^-$ ) are soluble.	$\text{NaCH}_3\text{COO}$ , $\text{AgCH}_3\text{COO}$
3	All chlorides ( $\text{Cl}^-$ ), bromides ( $\text{Br}^-$ ), and iodides ( $\text{I}^-$ ) are soluble, except with $\text{Ag}^+$ , $\text{Pb}^{2+}$ , and $\text{Hg}_2^{2+}$ .	Soluble: $\text{NaCl}$ , $\text{KBr}$ , $\text{CaI}_2$ Insoluble: $\text{AgCl}$ , $\text{PbI}_2$ , $\text{Hg}_2\text{Cl}_2$
4	All sulfates ( $\text{SO}_4^{2-}$ ) are soluble, except with $\text{Ba}^{2+}$ , $\text{Pb}^{2+}$ , $\text{Sr}^{2+}$ , and $\text{Ca}^{2+}$ (slightly soluble).	Soluble: $\text{Na}_2\text{SO}_4$ , $\text{CuSO}_4$ Insoluble: $\text{BaSO}_4$ , $\text{PbSO}_4$
5	All carbonates ( $\text{CO}_3^{2-}$ ), phosphates ( $\text{PO}_4^{3-}$ ), sulfides ( $\text{S}^{2-}$ ), and hydroxides ( $\text{OH}^-$ ) are insoluble, except with Group 1 metals ( $\text{Li}^+$ , $\text{Na}^+$ , $\text{K}^+$ , etc.) and $\text{NH}_4^+$ .	Insoluble: $\text{CaCO}_3$ , $\text{FePO}_4$ , $\text{CuS}$ , $\text{Fe(OH)}_3$ Soluble: $\text{Na}_2\text{CO}_3$ , $\text{K}_3\text{PO}_4$ , $\text{NaOH}$ , $\text{NH}_4\text{OH}$
6	All compounds of Group 1 metals ( $\text{Li}^+$ , $\text{Na}^+$ , $\text{K}^+$ , $\text{Rb}^+$ , $\text{Cs}^+$ ) and ammonium ( $\text{NH}_4^+$ ) are soluble.	$\text{NaCl}$ , $\text{KOH}$ , $\text{NH}_4\text{NO}_3$

Watch out

A common mistake: students remember "all chlorides are soluble" and forget the exceptions. AgCl is insoluble — that's why it's used in photography and qualitative analysis. Always check the exceptions first.

How to Use the Rules (Step-by-Step)

Suppose you need to predict whether $\text{PbSO}_4$ dissolves in water.

Identify the ions: $\text{Pb}^{2+}$ and $\text{SO}_4^{2-}$ .
Check the rules in order:
- Rule 1 (nitrates)? No. …

Why this formula?

Solubility Rules: Why They Work (Not Just What They Say)

Solubility rules are not arbitrary — they emerge from thermodynamics and electrostatic interactions between ions in water. Let's break down the why behind the key patterns.

1. The Core Idea: "Like Dissolves Like" at the Ionic Level

Water dissolves ionic compounds because it is polar. The $\delta^+$ hydrogen ends attract anions, and the $\delta^-$ oxygen end attracts cations.

Driving force: The lattice energy (energy holding the solid together) vs. the hydration energy (energy released when ions are surrounded by water).
Net energy change:

\Delta H_{\text{solution}} = \text{Lattice Energy} - \text{Hydration Energy}

If $\Delta H_{\text{solution}}$ is negative (exothermic) or small positive, the compound tends to dissolve.

2. Why Some Salts Are Always Soluble (Group 1 & NH₄⁺)

Rule: All salts of $\ce{Na+}$ , $\ce{K+}$ , $\ce{NH4+}$ are soluble.

Why?

These cations are large and have low charge density (charge/size ratio is small).
Their lattice energies are relatively low because the ions are far apart in the crystal.
Their hydration energies are high enough to overcome the lattice energy.

Key formula: For a cation like $\ce{K+}$ , the hydration energy is roughly:

$\Delta H_{\text{hyd}} \propto \frac{q^2}{r}$

where $q$ is charge and $r$ is ionic radius.

Since $q=1$ and $r$ is large, $\Delta H_{\text{hyd}}$ is moderate — but lattice energy is even smaller.

Result: The net $\Delta H_{\text{solution}}$ is negative → spontaneous dissolution.

3. Why Nitrates, Acetates, and Chlorates Are Always Soluble

Rule: All nitrates ( $\ce{NO3-}$ ), acetates ( $\ce{CH3COO-}$ ), and chlorates ( $\ce{ClO3-}$ ) are soluble.

Why?

These anions are large and polyatomic — their charge is spread over many atoms.
This delocalization of charge means they have low charge density.
They form weak ionic bonds with cations (low lattice energy).
Water can easily hydrate them because the negative charge is not concentrated.

Key insight: The lattice energy for $\ce{NaNO3}$ is much smaller than for $\ce{NaCl}$ because $\ce{NO3-}$ is larger and more polarizable.

4. The "Exceptions" — Why Some Salts Are Insoluble

4.1. Carbonates, Phosphates, Sulfides (Except with Group 1 & NH₄⁺)

Rule: Most carbonates ( $\ce{CO3^{2-}}$ ), phosphates ( $\ce{PO4^{3-}}$ ), and sulfides ( $\ce{S^{2-}}$ ) are insoluble.

Why?

These anions have high charge ( $-2$ or $-3$ ) and are small (especially $\ce{S^{2-}}$ ).
This gives them very high charge density.
They form extremely strong ionic bonds with cations (very high lattice energy).
The hydration energy, though large, is not enough to overcome the lattice energy.

Example: For $\ce{CaCO3}$ :

$\text{Lattice energy} \approx -2800 \text{ kJ/mol}$

$\text{Hydration energy} \approx -2400 \text{ kJ/mol}$

Net $\Delta H_{\text{solution}} \approx +400 \text{ kJ/mol}$ → insoluble

4.2. Silver, Lead, Mercury Halides

Rule: $\ce{AgCl}$ , $\ce{PbCl2}$ , $\ce{Hg2Cl2}$ are insoluble (most other chlorides are soluble).

Why?

$\ce{Ag+}$ , $\ce{Pb^{2+}}$ , $\ce{Hg2^{2+}}$ are soft (polarizable) cations.
They form covalent character in their bonds with halides (especially $\ce{Cl-}$ , $\ce{Br-}$ , $\ce{I-}$ ).
This covalent contribution increases the effective lattice energy beyond what simple ionic models predict.
Water cannot break these partially covalent bonds.

Key formula: The polarizing power of a cation is:

$\phi = \frac{q}{r}$ …

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This matching problem tests your knowledge of the characteristic uses and reactions of common organic compounds. The correct pairings are: (i)→(c), (ii)→(e), (iii)→(f), (iv)→(d), (v)→(a), (vi)→(b).

The key to solving this is not memorising random facts, but connecting each compound’s functional group to its chemical behaviour and industrial application. Let’s walk through each item.

Ethylene glycol (i): This is a diol (two -OH groups on adjacent carbons). Its most famous reaction is condensation with terephthalic acid to form polyester (specifically PET). So (i) matches with (c).
Ethanol (ii): A simple primary alcohol. It is volatile, has a pleasant smell, and is miscible with water and many organic compounds. This makes it an excellent solvent in perfumes (it dissolves aromatic oils and evaporates cleanly). So (ii) matches with (e).
Phenol (iii): This is an aromatic alcohol (C₆H₅OH). It is highly reactive towards electrophilic substitution. Treating phenol with concentrated nitric acid introduces three nitro groups, yielding picric acid (2,4,6-trinitrophenol). So (iii) matches with (f).
Methanol (iv): Historically, methanol was produced by the destructive distillation of wood, earning it the common name wood spirit. So (iv) matches with (d).
Neutral ferric chloride (v): This is a classic test for the phenolic group. Phenols (and enols) form a coloured complex (usually violet, blue, or green) with FeCl₃ in neutral solution. So (v) matches with (a). …

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Method: Solubility & Functional Group Matching

This problem is solved by matching each compound in Column I with its characteristic chemical property or application in Column II, based on known functional group behaviour and industrial uses.

Steps:

Identify the functional group of each compound in Column I.
Recall the key reaction or use associated with that functional group.
Match with the correct description in Column II.

Step-by-step matching:

Column I	Functional Group / Nature	Key Fact	Match (Column II)
(i) Ethylene glycol	Diol (two –OH groups)	Raw material for polyester (e.g., PET)	(c) raw material for manufacture of polyester
(ii) Ethanol	Primary alcohol	Volatile, pleasant smell → used in perfumes	(e) solvent in perfume formulations
(iii) Phenol	Aromatic –OH (phenolic)	Reacts with neutral FeCl₃ → violet colour	(a) detection of phenolic group
(iv) Methanol	Primary alcohol	Also called wood spirit (from destructive distillation of wood)	(d) wood spirit

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Here are the common mistakes students make when matching these items, along with the correct reasoning to avoid them.

Mistake 1: Confusing "Wood Spirit" with Ethanol

The Mistake: Students often associate "spirit" with drinking alcohol (ethanol) and match Methanol (iv) with something else, or match Ethanol (ii) with (d) "wood spirit."
Why it's Wrong: "Wood spirit" is the historical name for methanol because it was originally produced by the destructive distillation of wood. Ethanol is "grain spirit."
How to Avoid: Memorize the common names: Methanol = Wood spirit (or wood alcohol). Ethanol = Grain alcohol.

Mistake 2: Forgetting the Specific Test for Phenol

The Mistake: Matching Phenol (iii) with (a) "detection of phenolic group" but failing to see that Neutral ferric chloride (v) is the reagent used for that detection, not the phenol itself.
Why it's Wrong: Column I lists substances and reagents. Column (a) is a test. Phenol is what you are testing for; Ferric chloride is the tool that detects it.
How to Avoid: Ask: "Is this item a chemical being tested, or is it the testing agent?" Phenol gives a violet color with $FeCl_3$ , so Neutral ferric chloride is the correct match for (a).

Mistake 3: Mixing up "Raw Material" vs. "Final Product"

The Mistake: Matching Ethylene glycol (i) with (c) "raw material for the manufacture of polyester" is correct, but students sometimes match it with (b) "used in soap preparation" because they confuse it with glycerol.
Why it's Wrong: Soap is made from fats/oils (triglycerides) and alkali. Ethylene glycol is a diol used to make polyester (like PET) via condensation with terephthalic acid. Glycerol (a triol) is a byproduct of soap making.
How to Avoid: Remember the key reactions:
- Ethylene glycol → Polyester (Terylene/Dacron).
- Fats & Waxes → Soap (Saponification).

Mistake 4: Not Knowing the Industrial Use of Phenol

The Mistake: Matching Phenol (iii) with (f) "conversion to picric acid" is correct, but students often leave it blank or match it with (e) "solvent in perfume."
Why it's Wrong: Phenol is a solid with a strong, medicinal smell—not a perfume solvent. Its key reaction is nitration to form picric acid (2,4,6-trinitrophenol), an explosive and dye.
How to Avoid: Link Phenol to Picric acid (both start with 'P'). Perfume solvents are volatile, pleasant-smelling liquids like ethanol.

Mistake 5: Overlooking the Role of Ethanol in Perfumes …

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

AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$CH_3(CH_2)_{10}CH_2OSO_3Na$
›Reveal solutionSolution
This tests the classic hard-water problem: sodium soaps are water-soluble, but their calcium/magnesium salts are insoluble scum.

Concept and Intuition

Soaps and detergents are amphiphilic (a long hydrophobic hydrocarbon tail plus a polar/ionic head). With a monovalent counter-ion like Na⁺, the ionic head keeps the whole molecule soluble/dispersible in water via micelle formation. But when the counter-ion is a divalent metal ion like Ca²⁺ or Mg²⁺ (as found in hard water), the resulting salt is much less soluble and precipitates out — this precipitate is the familiar bathtub/clothes 'scum' and is the reason ordinary soap doesn't lather well in hard water.

Step-by-Step Solution

(A) $CH_3(CH_2)_{10}CH_2OSO_3Na$ is sodium lauryl sulfate, an anionic synthetic detergent — water-soluble.

(B) is a cationic (quaternary ammonium) detergent with Br⁻ counter-ion — water-soluble.

(C) $C_{17}H_{35}COONa$ is sodium stearate, ordinary soap — water-soluble, forms micelles. …
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AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
$\mathrm{MgSO_4}$
›Reveal solutionSolution
Both statements are true, and the reason correctly explains the assertion: $\mathrm{Mg^{2+}}$ 's large hydration enthalpy beats its lattice enthalpy, driving dissolution.

Concept and Intuition

Whether an ionic solid dissolves in water is a balance of two competing energies: the lattice enthalpy (energy needed to separate the ions from the crystal) and the hydration enthalpy (energy released when the separated ions are surrounded by water molecules). Using a Born-Haber-type cycle,
$\Delta H_{solution} \approx \Delta H_{lattice} + \Delta H_{hydration}$
where $\Delta H_{lattice}$ is endothermic (positive, energy required to break the lattice) and $\Delta H_{hydration}$ is exothermic (negative, energy released on hydration). A salt is soluble when the hydration enthalpy released is large enough (in magnitude) to compensate for, or exceed, the lattice enthalpy that must be supplied.

Step-by-Step Solution

$\mathrm{Mg^{2+}}$ has a small ionic radius and a high charge (+2), so its charge density is very high.

Charge density directly determines the strength of ion-dipole interactions with water, so $\mathrm{Mg^{2+}}$ has an unusually large (very negative) hydration enthalpy compared to larger group-2 cations like $\mathrm{Ca^{2+}}$ , $\mathrm{Sr^{2+}}$ , $\mathrm{Ba^{2+}}$ .

Even though $\mathrm{MgSO_4}$ has a substantial lattice enthalpy, the magnitude of $\mathrm{Mg^{2+}}$ 's hydration enthalpy exceeds it, so the net enthalpy of solution is favourable (exothermic or only mildly endothermic, with entropy assisting). …
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AP EAPCET 2021Set ap-2021-10-05-FN1 markMCQ
$CaCO_3 < KHCO_3 < NaHCO_3$
›Reveal solutionSolution
Alkaline-earth carbonates are far less soluble than alkali-metal bicarbonates, and within alkali bicarbonates, potassium salts are generally more soluble than the corresponding sodium salt: $CaCO_3<NaHCO_3<KHCO_3$ .

Concept and Intuition

Solubility of s-block carbonates/bicarbonates in water is governed largely by lattice enthalpy versus hydration enthalpy. $Ca^{2+}$ 's high charge density gives $CaCO_3$ a very high lattice enthalpy that hydration cannot overcome, so it is nearly insoluble. Among the singly-charged alkali cations, the larger $K^+$ has lower lattice enthalpy with the bicarbonate ion than the smaller $Na^+$ , so $KHCO_3$ dissolves more readily.

Step-by-Step Solution

$CaCO_3$ : a 2+/2- ionic solid with very high lattice energy — essentially insoluble in water.

$NaHCO_3$ : moderately soluble (~9–10 g/100 mL near room temperature). …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$LiCl$
›Reveal solutionSolution
Ether extracts covalent chlorides; among $LiCl$ , $NaCl$ , $AlCl_3$ , only $LiCl$ and $AlCl_3$ show enough covalent character to be pulled into the ether layer, while $NaCl$ stays purely ionic in water.

Concept and Intuition

Ether is a non-polar organic solvent, so it can only dissolve (extract) species with sufficient covalent character — pure ionic salts have no affinity for it and stay in the aqueous (polar) phase. Whether a chloride is "ionic" or has covalent character is governed by Fajan's rules: small, highly-charged cations polarize the anion's electron cloud strongly, giving the bond partial covalent character.

Step-by-Step Solution

$Al^{3+}$ is small and carries a high +3 charge, giving it very strong polarizing power (classic Fajan's-rule example) — $AlCl_3$ is substantially covalent, forms a dimer $Al_2Cl_6$ , and dissolves readily in organic solvents like ether/benzene.

$Li^+$ , though only +1, is exceptionally small (smallest alkali metal cation), giving it unusually high charge density for its group — this is the basis of lithium's well-known "diagonal relationship" with magnesium. $LiCl$ shows measurable covalent character (e.g. its solubility in organic solvents like ethanol/acetone), unlike the other alkali chlorides. …
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AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
$BeSO_4$
›Reveal solutionSolution
$BeSO_4$ dissolves because $Be^{2+}$ 's very high hydration enthalpy outweighs the lattice enthalpy; $BaSO_4$ stays insoluble because $Ba^{2+}$ 's much smaller hydration enthalpy can't overcome the (still large, anion-dominated) lattice enthalpy.

Concept and Intuition

Solubility of an ionic salt in water is essentially a tug-of-war between two competing energies: the lattice enthalpy (energy that must be supplied to break apart the ionic solid) and the hydration enthalpy (energy released as water molecules solvate the separated ions). If hydration enthalpy exceeds lattice enthalpy, dissolution is energetically favourable (net exothermic/favourable process); if lattice enthalpy wins, the salt stays undissolved.

Step-by-Step Solution

For sulfates down Group 2 (Be, Mg, Ca, Sr, Ba), the anion $SO_4^{2-}$ is large, so lattice enthalpy changes relatively slowly down the group (dominated by the big anion's size, not much affected by the smaller cation size differences).

Hydration enthalpy, however, depends strongly on the cation's size: $Be^{2+}$ is the smallest, most highly charged-density cation in the group, so it has an exceptionally large (very exothermic) hydration enthalpy. …
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