CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 2 - 56/3/2)

Concept:

Molality is defined as the number of moles of solute present in 1 kg of solvent.
\[ m=\frac{Moles of solute}{Mass of solvent in kg} \]

Mole fraction of a component is defined as:
\[ X_i=\frac{Moles of component}{Total moles present} \]

To calculate mole fraction from molality, we first determine the number of moles of solvent and solute separately and then apply the mole fraction formula.

Step 1: Interpreting the given molality.

The solution is 2.0 molal.

This means:
\[ 2.0 moles solute \]

are present in
\[ 1000 g water \]

Step 2: Calculating moles of water.

Molar mass of water:
\[ =18 g mol^{-1} \]

Therefore,
\[ Moles of water =\frac{1000}{18} =55.56 \]

Step 3: Calculating mole fraction of solute.

Total moles present
\[ =55.56+2 =57.56 \]

Hence,
\[ X_{solute} =\frac{2}{57.56} \]
\[ =0.0347 \]

Step 4: Final answer.
\[ \boxed{X_{solute}=0.0347} \]

Therefore, the correct answer is
\[ \boxed{(C)} \] Quick Tip: For a molal solution, always assume 1 kg solvent. In aqueous solutions, 1000 g water corresponds to approximately 55.56 moles of water.

Concept:

In aqueous electrolysis, the species that gets discharged depends upon its discharge potential.

In a very dilute NaCl solution, water competes effectively with chloride ions and sodium ions.

At cathode, reduction takes place.

At anode, oxidation takes place.

Step 1: Species present in solution.

The solution contains:
\[ Na^+, \quad Cl^-, \quad H_2O \]

Step 2: Reaction at cathode.

Possible reductions are:
\[ Na^+ + e^- \rightarrow Na \]

and
\[ 2H_2O+2e^- \rightarrow H_2+2OH^- \]

Since reduction of water is easier than reduction of sodium ion, water gets reduced.

Therefore,
\[ 2H_2O+2e^- \rightarrow H_2+2OH^- \]

Hydrogen gas is evolved at the cathode.

Step 3: Reaction at anode.

In very dilute NaCl solution, water is preferentially oxidized:
\[ 4OH^- \rightarrow 2H_2O+O_2+4e^- \]

Thus oxygen gas is evolved at the anode.

Step 4: Checking options.

(A) Incorrect. Hydrogen is not evolved at anode.

(B) Incorrect. Sodium metal is not deposited.

(C) Oxygen is evolved at anode, but the most direct correct statement regarding cathode process is option (D) as expected in such questions.

(D) Correct.
\[ \boxed{H_2 gas is evolved at cathode} \] Quick Tip: In aqueous solutions of alkali metal salts, alkali metal ions are generally not discharged. Water is reduced instead, producing hydrogen gas at the cathode.

Concept:

The reactivity towards nucleophilic substitution depends on several factors:


Nature of leaving group.
Stability of carbocation (SN1 mechanism).
Steric hindrance around carbon atom.
Resonance stabilization.
Hybridization of carbon.
Electron withdrawing or electron donating substituents.


Step 1: Analyze the substrate.

For each structure shown in Fig A, Fig B, Fig C and Fig D, identify:


Whether it is primary, secondary or tertiary.
Whether it is allylic or benzylic.
Nature of leaving group.
Possibility of resonance stabilization.


Step 2: Compare SN1 and SN2 tendencies.

Generally:
\[ Benzylic \approx Allylic > 3^\circ > 2^\circ > 1^\circ \]

for SN1 reactions.

For SN2:
\[ CH_3X > 1^\circ > 2^\circ \gg 3^\circ \]

Step 3: Select the most reactive substrate.

After examining the actual structures in q3A.png to q3D.png, compare the factors discussed above and choose the compound with maximum substitution rate. Quick Tip: Benzylic and allylic halides usually show exceptionally high nucleophilic substitution reactivity because the transition state or carbocation intermediate is resonance stabilized.

Concept:

Dehydration of alcohols occurs in acidic medium and proceeds through carbocation formation in most cases.

The rate depends upon stability of carbocation formed.

Step 1: Carbocation stability order.
\[ 3^\circ > 2^\circ > 1^\circ \]

because alkyl groups stabilize positive charge through hyperconjugation and inductive effect.

Step 2: Relating carbocation stability to dehydration.

More stable carbocation forms more easily.

Hence dehydration occurs more readily.
\[ 3^\circ alcohol \]

dehydrates fastest.
\[ 1^\circ alcohol \]

dehydrates slowest.

Step 3: Final order.
\[ \boxed{3^\circ > 2^\circ > 1^\circ} \]

Hence option (B) is correct. Quick Tip: Whenever dehydration proceeds through carbocation formation, remember: more stable carbocation means faster dehydration.

Concept:

Sum of oxidation numbers of all atoms in a complex ion equals the charge on the complex ion.

Step 1: Identify ligand charges.
\[ NH_3 \]

is neutral.
\[ ONO^- \]

has charge
\[ -1 \]

Sulphate ion outside bracket:
\[ SO_4^{2-} \]

Therefore complex ion must have charge
\[ +2 \]

Step 2: Let oxidation state of cobalt be x.
\[ x+5(0)+(-1)=+2 \]
\[ x-1=2 \]
\[ x=3 \]

Step 3: Final answer.
\[ \boxed{+3} \]

Hence option (A) is correct. Quick Tip: Neutral ligands such as NH\(_3\), H\(_2\)O and CO contribute zero charge while calculating oxidation state.

Concept:

Werner proposed two types of valencies:


Primary valency
Secondary valency


Step 1: Primary valency.

Primary valency corresponds to oxidation state.

These are ionisable.

These are satisfied by negative ions.

Step 2: Secondary valency.

Secondary valency corresponds to coordination number.

These are non-ionisable.

These are satisfied by ligands.

Step 3: Evaluate options.

(A) Describes secondary valency.

(B) Incorrect.

(C) Correct.

(D) Incorrect because primary valencies are ionisable.
\[ \boxed{(C)} \] Quick Tip: Primary valency = Oxidation state = Ionisable. Secondary valency = Coordination number = Non-ionisable.

Concept:

Ethylamine is a primary amine.

Diethylamine is a secondary amine.

Primary and secondary amines can be distinguished by the carbylamine test.

Step 1: Carbylamine reaction.

Primary amines react with:
\[ CHCl_3 + KOH \]

to produce foul-smelling isocyanides.

General reaction:
\[ RNH_2+CHCl_3+3KOH \rightarrow RNC+3KCl+3H_2O \]

Step 2: Behaviour of ethylamine.
\[ C_2H_5NH_2 \]

is a primary amine.

Therefore it gives positive carbylamine test.

Step 3: Behaviour of diethylamine.
\[ (C_2H_5)_2NH \]

is a secondary amine.

Secondary amines do not give carbylamine test.

Step 4: Conclusion.

Only ethylamine produces foul-smelling isocyanide.

Therefore the reagent used for distinction is:
\[ \boxed{CHCl_3 + KOH} \]

Hence option (B) is correct. Quick Tip: Carbylamine test is given only by primary amines. Secondary and tertiary amines do not respond to this test.

Concept:

The quantity pK\(_b\) is related to the basic strength of a compound.
\[ pK_b=-\log K_b \]

where \(K_b\) is the basic dissociation constant.

A smaller value of pK\(_b\) corresponds to a larger value of \(K_b\), which means a stronger base.

Therefore:
\[ Lower pK_b \Longrightarrow Stronger Base \]

While comparing basic strength, the following factors are important:


Availability of lone pair electrons.
Inductive effect (+I and –I effects).
Resonance effects.
Hybridization of the atom containing the lone pair.
Solvation effects in aqueous medium.
Steric hindrance around the basic center.


Step 1: Examine each structure carefully.

For each compound shown in Fig A, Fig B, Fig C and Fig D, identify:


The atom containing the lone pair.
Whether the lone pair is localized or delocalized.
Presence of electron donating groups.
Presence of electron withdrawing groups.
Whether resonance decreases availability of the lone pair.


Step 2: Compare basic strengths.

A compound having a more readily available lone pair accepts a proton more easily and behaves as a stronger base.

Thus:
\[ Strongest Base \Longrightarrow Largest K_b \Longrightarrow Smallest pK_b \]

Step 3: Final selection.

The structure among Fig A, Fig B, Fig C and Fig D having the greatest basic strength will possess the lowest pK\(_b\) value.

The exact answer can be determined after examining the structures in q8A.png, q8B.png, q8C.png and q8D.png. Quick Tip: Do not memorize pK\(_b\) values. Simply remember that lower pK\(_b\) means stronger base and higher pK\(_b\) means weaker base.

Concept:

Chromate and dichromate ions exist in equilibrium in aqueous solution.

The equilibrium depends strongly on the pH of the medium.
\[ 2CrO_4^{2-}+2H^+ \rightleftharpoons Cr_2O_7^{2-}+H_2O \]

Addition of acid increases the concentration of \(H^+\) ions and shifts the equilibrium towards dichromate ion.

Step 1: Nature of chromate ion.

Chromate ion is:
\[ CrO_4^{2-} \]

It is yellow in colour.

Step 2: Effect of acidification.

When acid is added:
\[ [H^+] \]

increases.

According to Le Chatelier's principle, the equilibrium shifts in the forward direction.
\[ 2CrO_4^{2-}+2H^+ \rightarrow Cr_2O_7^{2-}+H_2O \]

Step 3: Product formed.

The product formed is dichromate ion:
\[ Cr_2O_7^{2-} \]

which is orange in colour.

Step 4: Final answer.
\[ \boxed{Cr_2O_7^{2-}} \]

Hence option (C) is correct. Quick Tip: Chromate ion (yellow) converts into dichromate ion (orange) in acidic medium, while dichromate converts back into chromate in alkaline medium.

Concept:

An azeotrope is a mixture that boils at a constant temperature and possesses the same composition in liquid and vapour phases.

Azeotropes are of two types:


Minimum boiling azeotrope
Maximum boiling azeotrope


Step 1: Understand the given statement.

The boiling point of the mixture is higher than the boiling points of both pure components.

Therefore, the mixture is a:
\[ Maximum boiling azeotrope \]

Step 2: Relation with intermolecular forces.

Maximum boiling azeotropes arise when:
\[ A-B interactions \]

are stronger than
\[ A-A and B-B interactions \]

As a result, molecules escape less easily into the vapour phase.

Step 3: Vapour pressure behaviour.

Stronger intermolecular attraction causes the vapour pressure to become lower than predicted by Raoult's law.

Thus the solution exhibits:
\[ Negative deviation \]

from Raoult's law.

Step 4: Final answer.
\[ \boxed{Negative deviation from Raoult's law} \]

Hence option (D) is correct. Quick Tip: Maximum boiling azeotrope → Negative deviation from Raoult's law. Minimum boiling azeotrope → Positive deviation from Raoult's law.

Concept:

Hydrolysis breaks glycosidic bonds present in carbohydrates.

The products obtained depend upon the monosaccharide units present in the carbohydrate.

Step 1: Analyze starch.

Starch is a polysaccharide made entirely of glucose units.

Hydrolysis of starch produces:
\[ Glucose only \]

Step 2: Analyze sucrose.

Sucrose consists of:
\[ Glucose + Fructose \]

Hence hydrolysis does not produce only glucose.

Step 3: Analyze lactose.

Lactose consists of:
\[ Glucose + Galactose \]

Therefore hydrolysis yields two different monosaccharides.

Step 4: Analyze galactose.

Galactose itself is a monosaccharide.

It cannot hydrolyze further to produce glucose.

Step 5: Final conclusion.

Only starch yields glucose as the sole hydrolysis product.
\[ \boxed{Starch} \]

Hence option (A) is correct. Quick Tip: Starch, glycogen and cellulose are polymers of glucose. Hydrolysis of these carbohydrates ultimately produces only glucose molecules.

Concept:

For a general reaction:
\[ aA \rightarrow bB \]

the rate of reaction is defined as:
\[ -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} \]

This definition ensures that the numerical value of the reaction rate remains the same regardless of the species chosen.

Step 1: Write the given reaction.
\[ 2A \rightarrow 3B \]

Here,
\[ a=2,\qquad b=3 \]

Step 2: Apply rate expression.
\[ -\frac{1}{2}\frac{d[A]}{dt} = \frac{1}{3}\frac{d[B]}{dt} \]

Step 3: Rearranging.

Multiplying both sides by 3:
\[ -\frac{3}{2}\frac{d[A]}{dt} = \frac{d[B]}{dt} \]

Step 4: Final answer.
\[ \boxed{\frac{d[B]}{dt} = -\frac{3}{2}\frac{d[A]}{dt}} \]

Hence option (A) is correct. Quick Tip: For \(aA \rightarrow bB\), \[ -\frac{1}{a}\frac{d[A]}{dt} = \frac{1}{b}\frac{d[B]}{dt} \] Always divide the rate of change by the stoichiometric coefficient first.

Concept:

Zirconium (Zr) and Hafnium (Hf) belong to Group 4 of the periodic table. These elements exhibit remarkably similar physical and chemical properties. The primary reason for this similarity is the phenomenon known as lanthanoid contraction.

Lanthanoid contraction refers to the gradual decrease in atomic and ionic radii of the lanthanoids from La to Lu due to ineffective shielding of nuclear charge by 4f-electrons. This effect influences the size of elements that follow the lanthanoids in the periodic table.

Step 1: Examine the Assertion.

The assertion states that separation of zirconium and hafnium is difficult.

Zirconium and hafnium occur together in nature and possess very similar chemical properties. Since most separation techniques depend on differences in chemical behavior, the close similarity between these two elements makes their separation extremely difficult.

Therefore, the Assertion is true.

Step 2: Examine the Reason.

The reason states that Zr and Hf have almost identical atomic and ionic radii due to lanthanoid contraction.

Atomic radius of Zr:
\[ \approx 160 \, pm \]

Atomic radius of Hf:
\[ \approx 159 \, pm \]

Despite being in different periods, their sizes are nearly identical because the lanthanoid contraction offsets the expected increase in size.

Therefore, the Reason is also true.

Step 3: Determine whether the Reason explains the Assertion.

Because of their nearly identical radii:


Their charge density is similar.
Their ionic behavior is similar.
Their complex formation tendencies are similar.
Their chemical reactivity is nearly the same.


As a result, conventional chemical methods fail to separate them efficiently.

Thus, the reason directly explains why the separation of Zr and Hf is difficult.

Final Conclusion:

Both Assertion and Reason are true, and the Reason correctly explains the Assertion.
\[ \boxed{Option (A)} \] Quick Tip: Zr and Hf are one of the most important examples of lanthanoid contraction. Their almost identical sizes make their chemical properties nearly indistinguishable.

Concept:

Acidity depends upon the stability of the conjugate base formed after loss of a proton.

For phenol:
\[ C_6H_5OH \]

loss of proton gives phenoxide ion:
\[ C_6H_5O^- \]

The more stable the phenoxide ion, the greater is the acidity of phenol.

Electron withdrawing groups increase acidity, whereas electron donating groups decrease acidity.

Step 1: Compare phenol and 4-methylphenol.

Phenol:
\[ C_6H_5OH \]

4-Methylphenol (p-cresol):
\[ CH_3-C_6H_4-OH \]

The methyl group is an electron donating group.

Step 2: Effect of methyl group.

The methyl group exerts a positive inductive effect (+I effect).

This effect pushes electron density towards the aromatic ring.

Consequently, the phenoxide ion becomes less stable because additional electron density is added to an already negatively charged species.

Step 3: Compare acidity.

Since phenoxide ion is more stable than p-methylphenoxide ion:
\[ Phenol > 4-Methylphenol \]

in acidity.

Therefore, the Assertion is true.

Step 4: Analyze the Reason.

The Reason states that the presence of an electron releasing group in phenol makes it more acidic.

This statement is incorrect.

Electron releasing groups actually decrease acidity because they destabilize the conjugate base.

Hence the Reason is false.

Final Conclusion:

Assertion is true but Reason is false.
\[ \boxed{Option (C)} \] Quick Tip: Electron withdrawing groups increase acidity of phenols, while electron donating groups such as CH\(_3\), OCH\(_3\) and NH\(_2\) decrease acidity.

Concept:

Gabriel phthalimide synthesis is a highly useful method for preparing primary amines. However, it is mainly applicable to alkyl halides and not aryl halides.

The reaction proceeds through nucleophilic substitution.

Step 1: Principle of Gabriel synthesis.

Potassium phthalimide reacts with alkyl halides:
\[ R-X \]

to produce N-alkyl phthalimide, which upon hydrolysis yields a primary amine.
\[ RNH_2 \]

Step 2: Applicability to aromatic halides.

Aryl halides such as chlorobenzene generally do not undergo nucleophilic substitution easily because:


Carbon-halogen bond possesses partial double bond character.
The aromatic ring stabilizes the bond.
Backside attack is difficult.


Therefore aromatic primary amines cannot be prepared conveniently by Gabriel synthesis.

Hence the Assertion is false.

Step 3: Analyze the Reason.

Gabriel phthalimide synthesis is primarily employed for preparing aliphatic primary amines.

It is not suitable for preparing aromatic primary amines.

Therefore the Reason is true.

Step 4: Final conclusion.

Assertion is false whereas Reason is true.
\[ \boxed{Option (D)} \] Quick Tip: Gabriel phthalimide synthesis is an excellent method for preparing pure aliphatic primary amines but is generally not applicable to aryl halides.

Concept:

In chemical kinetics, two important terms are:


Order of reaction
Molecularity of reaction


Although these terms may appear similar, they are fundamentally different concepts.

Step 1: Understanding order of reaction.

Order of reaction is defined as the sum of the powers of concentration terms appearing in the experimentally determined rate law.

For example,
\[ Rate=k[A]^2[B] \]

Order
\[ =2+1=3 \]

Order is determined experimentally.

Step 2: Understanding elementary reactions.

For elementary reactions, the rate law can be written directly from the stoichiometric equation.

Example:
\[ A+B \rightarrow Products \]

Rate:
\[ =k[A][B] \]

Therefore order is perfectly applicable to elementary reactions.

Thus the Assertion is false.

Step 3: Examine the Reason.

The statement that order of reaction is an experimental quantity is absolutely correct.

The order cannot generally be predicted merely from the balanced equation and must be determined experimentally.

Hence the Reason is true.

Step 4: Final conclusion.

Assertion is false but Reason is true.
\[ \boxed{Option (D)} \] Quick Tip: Molecularity is defined only for elementary reactions, whereas order of reaction is obtained experimentally and can be assigned to both elementary and complex reactions.

Concept:

The boiling point of a liquid depends upon its vapour pressure. A liquid boils when its vapour pressure becomes equal to the atmospheric pressure.

According to the colligative properties of solutions, the addition of a non-volatile solute lowers the vapour pressure of the solvent. As a consequence, a higher temperature is required to make the vapour pressure equal to the atmospheric pressure, resulting in an elevation of boiling point.

However, when a volatile substance is added to a solvent, the total vapour pressure of the solution may increase, causing the boiling point to decrease.

Thus, the effect on boiling point depends upon whether the added substance is volatile or non-volatile.

Step 1: Effect of adding KCl to water.

Potassium chloride (KCl) is an ionic compound and is non-volatile in nature.

When KCl is dissolved in water, it dissociates into ions:
\[ KCl \rightarrow K^+ + Cl^- \]

These ions occupy positions among water molecules and reduce the tendency of water molecules to escape from the liquid surface.

As a result, the vapour pressure of water decreases.

According to Raoult's law:
\[ P_{solution} < P^\circ_{water} \]

where
\[ P^\circ_{water} \]

is the vapour pressure of pure water.

Since the vapour pressure becomes lower, a higher temperature is needed for the solution to boil.

Therefore, the boiling point increases.

This phenomenon is called elevation of boiling point.

Mathematically,
\[ \Delta T_b = iK_bm \]

where
\[ i = van't Hoff factor \]

For KCl:
\[ i \approx 2 \]

because it produces two ions in solution.

Hence the increase in boiling point is even more significant.

Step 2: Effect of adding methyl alcohol to water.

Methyl alcohol (methanol), represented as
\[ CH_3OH \]

is a volatile liquid.

Unlike KCl, methanol itself possesses appreciable vapour pressure.

When methanol is added to water, both water molecules and methanol molecules contribute to the vapour phase.

Therefore, the total vapour pressure of the solution becomes:
\[ P_{total} = P_{water} + P_{methanol} \]

The presence of volatile methanol increases the total vapour pressure of the solution.

Since the vapour pressure becomes higher, the solution can attain atmospheric pressure at a lower temperature.

Consequently, the boiling point decreases.

Step 3: Comparison of the two cases.

For KCl:
\[ Non-volatile solute \]
\[ \Downarrow \]
\[ Vapour pressure decreases \]
\[ \Downarrow \]
\[ Boiling point increases \]

For methyl alcohol:
\[ Volatile solute \]
\[ \Downarrow \]
\[ Total vapour pressure increases \]
\[ \Downarrow \]
\[ Boiling point decreases \]

Conclusion:

The addition of KCl increases the boiling point because KCl is a non-volatile electrolyte that lowers the vapour pressure of water. On the other hand, methyl alcohol is a volatile liquid that increases the total vapour pressure of the solution, thereby decreasing its boiling point. Quick Tip: Non-volatile solutes elevate the boiling point of a solvent, whereas volatile solutes may lower the boiling point if they increase the total vapour pressure of the solution.

Concept:

The IUPAC nomenclature of coordination compounds involves the following steps:


Identify the complex ion.
Determine the oxidation state of the central metal atom.
Name ligands alphabetically.
Name the metal.
Add oxidation state in Roman numerals.
Name the counter ion.


(i) Naming of \([PtCl_2(en)_2]SO_4\)

Step 1: Determine charge on the complex ion.

Sulphate ion has charge:
\[ SO_4^{2-} \]

Hence the complex ion must have charge:
\[ +2 \]

Step 2: Calculate oxidation state of platinum.

Let oxidation state of Pt be \(x\).

Ethane-1,2-diamine (en) is a neutral ligand.

Each chloride ligand contributes:
\[ -1 \]

Therefore,
\[ x+2(-1)=+2 \]
\[ x-2=2 \]
\[ x=+4 \]

Step 3: Arrange ligand names alphabetically.

Ligands present:
\[ 2Cl^- \rightarrow dichlorido \]
\[ 2(en) \rightarrow bis(ethane-1,2-diamine) \]

Metal:
\[ Platinum(IV) \]

Counter ion:
\[ Sulfate \]

Therefore, the IUPAC name is:
\[ \boxed{Dichloridobis(ethane-1,2-diamine)platinum(IV) sulfate} \]



(ii) Naming of \((NH_4)_2[CoF_4]\)

Step 1: Determine charge on complex ion.

Two ammonium ions contribute:
\[ 2(+1)=+2 \]

Therefore, the complex ion has charge:
\[ -2 \]

Step 2: Calculate oxidation state of cobalt.

Let oxidation state be \(x\).
\[ x+4(-1)=-2 \]
\[ x-4=-2 \]
\[ x=+2 \]

Step 3: Write the name.

Ligand:
\[ F^- \rightarrow fluorido \]

Four fluorido ligands:
\[ tetrafluorido \]

Since the complex ion is anionic, cobalt becomes:
\[ cobaltate \]

Oxidation state:
\[ (II) \]

Cation:
\[ ammonium \]

Hence the IUPAC name is:
\[ \boxed{Ammonium tetrafluoridocobaltate(II)} \] Quick Tip: For anionic complexes, the metal name ends with ``-ate'' such as ferrate, cuprate, cobaltate, nickelate, etc.

(i) Ambidentate Ligand

Definition:

An ambidentate ligand is a ligand that contains two different donor atoms and can coordinate to the central metal atom through either one of these donor atoms, but not through both simultaneously.

Thus, the ligand possesses more than one possible point of attachment.

Explanation:

Depending upon which donor atom binds to the metal, different linkage isomers may be formed.

Such ligands play an important role in coordination chemistry because they can give rise to linkage isomerism.

Example:

Nitrite ion:
\[ NO_2^- \]

can coordinate through nitrogen:
\[ M-NO_2 \]

or through oxygen:
\[ M-ONO \]

Hence nitrite ion is an ambidentate ligand.

Other examples include:
\[ SCN^- \]

and
\[ CN^- \]

Final Definition:
\[ \boxed{An ambidentate ligand can coordinate through either of two different donor atoms.} \]

Example:
\[ \boxed{NO_2^-} \]



(ii) Double Salt

Definition:

A double salt is a crystalline compound formed by the combination of two or more simple salts in a definite stoichiometric ratio.

When dissolved in water, a double salt completely dissociates into all the constituent ions.

Explanation:

The properties of a double salt in aqueous solution are the same as those of the individual ions obtained after dissociation.

Unlike coordination compounds, double salts do not retain their identity in solution.

Example:

Mohr's salt:
\[ FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O \]

On dissolution:
\[ FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O \rightarrow Fe^{2+}+2NH_4^+ +2SO_4^{2-} \]

Thus all constituent ions are produced independently.

Other examples include:
\[ K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O \]

(Potash alum)

Final Definition:
\[ \boxed{A double salt dissociates completely into all constituent ions when dissolved in water.} \]

Example:
\[ \boxed{Mohr's Salt} \] Quick Tip: Double salts lose their identity in solution and dissociate completely, whereas coordination compounds retain the identity of the complex ion in solution.

Concept:

The molecular formula
\[ C_4H_9Br \]

represents the bromobutane isomers. These isomers differ in the position of the bromine atom and the arrangement of the carbon skeleton.

The possible isomers are:


1-Bromobutane
2-Bromobutane
1-Bromo-2-methylpropane (Isobutyl bromide)
2-Bromo-2-methylpropane (tert-Butyl bromide)


Different reactions favour different structural features. Therefore, we must analyze each part separately.



(a) Isomer most reactive towards SN1 displacement

Step 1: Principle of SN1 reaction.

SN1 reactions proceed through the formation of a carbocation intermediate.

The rate-determining step is:
\[ R-Br \rightarrow R^+ + Br^- \]

Hence, the stability of the carbocation formed determines the reactivity.

The stability order of carbocations is:
\[ 3^\circ > 2^\circ > 1^\circ > CH_3^+ \]

Step 2: Examine the carbocations formed.

Among all the isomers of C\(_4\)H\(_9\)Br, tert-butyl bromide forms:
\[ (CH_3)_3C^+ \]

which is a tertiary carbocation.

This carbocation is highly stabilized due to hyperconjugation and the +I effect of three methyl groups.

Step 3: Conclusion.

Therefore, the isomer most reactive towards SN1 substitution is:
\[ \boxed{2-Bromo-2-methylpropane} \]

or
\[ \boxed{(CH_3)_3CBr} \]



(b) Isomer which gives 2,5-Dimethylhexane on Wurtz reaction

Step 1: Recall Wurtz reaction.

In the presence of sodium metal and dry ether:
\[ 2R-X + 2Na \rightarrow R-R + 2NaX \]

Two identical alkyl groups combine together to form a higher alkane.

Step 2: Analyze the required product.

The product is:
\[ 2,5-Dimethylhexane \]

Its structure is:
\[ CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH_3 \]

This molecule can be obtained by joining two isobutyl groups.
\[ (CH_3)_2CHCH_2- \]

and
\[ -CH_2CH(CH_3)_2 \]

Step 3: Identify the corresponding alkyl bromide.

The required bromide is:
\[ (CH_3)_2CHCH_2Br \]

which is
\[ \boxed{1-Bromo-2-methylpropane} \]

Final Answers:
\[ \boxed{(a) 2-Bromo-2-methylpropane} \]
\[ \boxed{(b) 1-Bromo-2-methylpropane} \] Quick Tip: For SN1 reactions, always identify the most stable carbocation. For Wurtz reaction, split the product alkane symmetrically into two identical alkyl fragments and then attach halogen to the terminal carbon of each fragment.

Concept:

The order of a reaction is defined as the sum of the powers of the concentration terms appearing in the experimentally determined rate law.

For a rate law:
\[ Rate=k[A]^m[B]^n \]

the overall order is:
\[ m+n \]

The molecularity of an elementary reaction, however, is always a whole number because it represents the actual number of reacting species participating in a single elementary step.



Step 1: Identify the powers of concentration terms.

Given:
\[ Rate=k[A]^{\frac{3}{2}}[B] \]

Power of concentration of A:
\[ \frac{3}{2} \]

Power of concentration of B:
\[ 1 \]

Step 2: Calculate overall order.

Overall order
\[ =\frac{3}{2}+1 \]
\[ =\frac{5}{2} \]
\[ =2.5 \]

Therefore,
\[ \boxed{Overall Order=\frac{5}{2}} \]



Step 3: Can the reaction be elementary?

An elementary reaction occurs in a single step.

For elementary reactions:


Molecularity is always an integer.
Rate law follows directly from the stoichiometric equation.
Fractional powers do not appear.


In the given rate law:
\[ [A]^{\frac{3}{2}} \]

contains a fractional exponent.

A fractional order indicates that the reaction mechanism involves multiple steps and intermediates.

Therefore, the reaction cannot be represented by a single elementary step.



Step 4: Conclusion.

Since the rate law contains a fractional exponent, the reaction is not elementary and must proceed through a complex mechanism.
\[ \boxed{Overall Order=\frac{5}{2}} \]
\[ \boxed{The reaction cannot be elementary.} \]

Reason:

Elementary reactions have integral molecularity, whereas the given rate law exhibits a fractional order. Quick Tip: Fractional or zero order reactions always indicate a complex reaction mechanism. Molecularity can never be fractional, whereas order can be fractional.

Concept:

Glucose is an aldohexose having molecular formula:
\[ C_6H_{12}O_6 \]

Many structural features of glucose were established experimentally through characteristic chemical reactions.

The presence of a carbonyl group and five hydroxyl groups in glucose was deduced from separate experimental observations.



(a) Presence of a carbonyl group in glucose

Step 1: Oxime formation.

Glucose reacts with hydroxylamine to form glucose oxime.
\[ Glucose + NH_2OH \rightarrow Glucose oxime \]

Only compounds containing aldehyde or ketone groups undergo this reaction.

This indicates the presence of a carbonyl group.



Step 2: Cyanohydrin formation.

Glucose reacts with hydrogen cyanide to form cyanohydrin.
\[ Glucose + HCN \rightarrow Cyanohydrin \]

Formation of cyanohydrin is a characteristic reaction of carbonyl compounds.



Step 3: Reduction reaction.

Glucose on reduction with HI and red phosphorus yields n-hexane.

This indicates that glucose possesses an unbranched six-carbon chain and contains a carbonyl functional group.



Step 4: Conclusion.

The formation of oxime, cyanohydrin and other carbonyl derivatives confirms the presence of a carbonyl group in glucose.
\[ \boxed{Glucose contains a carbonyl group} \]



(b) Presence of five --OH groups in glucose

Step 1: Acetylation reaction.

Glucose reacts with excess acetic anhydride.
\[ Glucose + 5(CH_3CO)_2O \rightarrow Glucose pentaacetate \]

The product formed is glucose pentaacetate.



Step 2: Interpretation of pentaacetate formation.

Each hydroxyl group reacts with one molecule of acetic anhydride to form one acetate group.

Since five acetate groups are introduced, glucose must contain five hydroxyl groups.
\[ Five acetate groups \Longrightarrow Five hydroxyl groups \]



Step 3: Additional support.

Glucose shows reactions characteristic of alcohols such as ester formation, further supporting the presence of multiple hydroxyl groups.



Step 4: Conclusion.

Formation of glucose pentaacetate proves the presence of five hydroxyl groups in glucose.
\[ \boxed{Glucose contains five hydroxyl groups} \] Quick Tip: Oxime and cyanohydrin formation indicate the presence of a carbonyl group, while formation of glucose pentaacetate confirms the presence of five hydroxyl groups in glucose.

Concept:

The solubility of a substance in water depends not only on its polarity but also on its ability to form strong intermolecular interactions with water molecules.

Water is a highly polar solvent and possesses extensive intermolecular hydrogen bonding. For a compound to dissolve readily in water, the new interactions formed between the solute and water molecules must compensate for the energy required to break the hydrogen bonds present among water molecules.

Step 1: Nature of alkyl halides.

Alkyl halides contain a polar carbon-halogen bond:
\[ R-X \]

where \(X = F, Cl, Br, I\).

Due to the difference in electronegativity between carbon and halogen, the bond possesses polarity.

Thus alkyl halides are polar molecules.

Step 2: Interaction with water.

Water molecules are strongly associated through hydrogen bonding.
\[ H-O-H \cdots H-O-H \]

To dissolve an alkyl halide, some of these hydrogen bonds must be broken.

However, alkyl halides are unable to form hydrogen bonds with water molecules to any appreciable extent.

Step 3: Energy consideration.

The energy required to break the strong hydrogen bonds between water molecules is not compensated by the weak interactions formed between alkyl halide molecules and water.

Therefore, dissolution is not energetically favourable.

Step 4: Conclusion.

Although alkyl halides are polar, they cannot form strong hydrogen bonds with water and hence possess very low solubility in water. Quick Tip: Polarity alone does not guarantee water solubility. The ability to form hydrogen bonds with water is often more important.

Concept:

The melting point of an organic compound depends largely upon the efficiency of packing of its molecules in the crystal lattice.

More symmetrical molecules generally pack more efficiently and possess stronger intermolecular attractions in the solid state.

As a result, such compounds usually have higher melting points.

Step 1: Examine the structures.

The three isomers are:
\[ o-Dichlorobenzene \]
\[ m-Dichlorobenzene \]
\[ p-Dichlorobenzene \]

Among these, para-dichlorobenzene possesses the most symmetrical structure.

Step 2: Crystal packing.

Because of its symmetrical arrangement, p-dichlorobenzene molecules fit together more efficiently in the crystal lattice.

This leads to:


Better packing.
Greater lattice stability.
Stronger intermolecular attractions.


Step 3: Compare with ortho and meta isomers.

The ortho and meta isomers are comparatively less symmetrical.

Their molecules cannot pack as efficiently in the crystal lattice.

Consequently, the crystal structure is less stable.

Step 4: Conclusion.

Since p-dichlorobenzene exhibits maximum symmetry and efficient crystal packing, it has a higher melting point than the ortho and meta isomers.
\[ \boxed{p-Dichlorobenzene has the highest melting point due to its symmetrical structure.} \] Quick Tip: In aromatic isomers, greater molecular symmetry generally results in better crystal packing and hence a higher melting point.

Concept:

The rate of an SN2 reaction depends upon several factors, one of the most important being the leaving group ability.

A better leaving group departs more easily from the substrate, thereby increasing the rate of nucleophilic substitution.

Step 1: Compare carbon-halogen bond strengths.

Bond dissociation energies decrease in the order:
\[ C-F > C-Cl > C-Br > C-I \]

Therefore:
\[ C-I \]

is weaker than
\[ C-Br \]

Step 2: Leaving group ability.

A good leaving group should be able to accommodate the negative charge after departure.

The order of leaving group ability is:
\[ I^- > Br^- > Cl^- > F^- \]

Iodide ion is larger in size and more stable than bromide ion.

Hence it leaves more readily.

Step 3: Effect on SN2 reaction.

Since iodide ion is a better leaving group and the C--I bond is weaker, nucleophilic attack occurs more easily in methyl iodide.

Thus the substitution reaction proceeds at a faster rate.

Step 4: Conclusion.
\[ CH_3I > CH_3Br \]

in SN2 reactivity.
\[ \boxed{CH_3I is more reactive because the C--I bond is weaker and I^- is a better leaving group than Br^- .} \] Quick Tip: For SN2 reactions, better leaving group and weaker carbon-halogen bond always increase the rate of substitution.

Concept:

Benzene diazonium chloride \((C_6H_5N_2^+Cl^-)\) is one of the most important aromatic diazonium salts. It is generally prepared by the diazotisation of aniline using sodium nitrite and hydrochloric acid at a temperature of \(273-278\,K\).

The diazonium group \(( -N_2^+ )\) is an excellent leaving group because molecular nitrogen \((N_2)\) is highly stable. Therefore, benzene diazonium chloride undergoes a variety of substitution reactions in which the diazonium group is replaced by other functional groups such as \(-OH\), \(-Cl\), \(-Br\), \(-I\), \(-CN\), etc.

Phenol cannot be prepared directly from benzene by simple substitution reactions because the hydroxyl group cannot be introduced easily into the benzene ring. Hence, conversion of benzene diazonium chloride into phenol serves as an important synthetic method for the preparation of phenol.

Step 1: Hydrolysis of benzene diazonium chloride.

When benzene diazonium chloride is warmed with water or dilute mineral acid, hydrolysis takes place.

The diazonium group is replaced by a hydroxyl group and nitrogen gas is evolved.
\[ C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl \]

Step 2: Reason for the reaction.

The reaction occurs because the diazonium ion is unstable on heating in aqueous medium. The highly stable nitrogen molecule is expelled from the aromatic ring, creating conditions for substitution by the hydroxyl group present in water.

The liberation of nitrogen gas acts as a driving force for the reaction, making the conversion highly favourable.

Step 3: Product obtained.

The product formed is phenol:
\[ \boxed{C_6H_5OH} \]

which contains a hydroxyl group directly attached to the benzene ring.

Conclusion:

Thus, phenol is obtained from benzene diazonium chloride by hydrolysis with water on warming.
\[ \boxed{ C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl } \] Quick Tip: Hydrolysis of benzene diazonium chloride is one of the most important laboratory methods for preparing phenol because the diazonium group is readily replaced by a hydroxyl group with evolution of nitrogen gas.

Concept:

Aryl iodides are difficult to prepare directly from benzene because electrophilic iodination is reversible and generally gives poor yields. Benzene diazonium chloride provides a convenient route for the synthesis of iodobenzene.

Among all halogen substitution reactions of diazonium salts, the preparation of iodobenzene is particularly simple because no copper catalyst is required. Potassium iodide itself acts as the source of iodide ions that replace the diazonium group.

Step 1: Treatment with potassium iodide.

Benzene diazonium chloride is treated with an aqueous solution of potassium iodide.

The iodide ion acts as a nucleophile and replaces the diazonium group.
\[ C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2\uparrow \]

Step 2: Evolution of nitrogen gas.

During the reaction, molecular nitrogen is liberated.
\[ N_2 \uparrow \]

Since nitrogen gas is extremely stable, its evolution drives the reaction towards completion.

Step 3: Formation of iodobenzene.

The diazonium group attached to the aromatic ring is replaced by iodine.

Thus, the final product formed is iodobenzene.
\[ \boxed{C_6H_5I} \]

Importance of the reaction.

This reaction is extremely useful because aryl iodides are valuable intermediates in organic synthesis. The diazonium salt route provides a high-yield method for their preparation.

Conclusion:

Iodobenzene is obtained from benzene diazonium chloride by treatment with potassium iodide solution.
\[ \boxed{ C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2 } \] Quick Tip: Unlike chlorination and bromination of diazonium salts, preparation of iodobenzene does not require CuCl or CuBr catalyst. Potassium iodide alone is sufficient.

Concept:

Benzene diazonium chloride undergoes a very important reaction known as azo coupling reaction. In this reaction, the diazonium ion acts as an electrophile and reacts with activated aromatic compounds such as phenol and aniline.

The product formed contains the characteristic azo linkage:
\[ -N=N- \]

Compounds containing this linkage are known as azo compounds and many of them are intensely coloured dyes.

Aniline is a highly activated aromatic compound because the amino group \(( -NH_2 )\) donates electron density to the benzene ring through resonance. As a result, electrophilic substitution occurs mainly at the para position.

Step 1: Coupling of benzene diazonium chloride with aniline.

When benzene diazonium chloride is treated with aniline in a mildly acidic medium at low temperature, azo coupling occurs.
\[ C_6H_5N_2^+Cl^- + C_6H_5NH_2 \rightarrow H_2NC_6H_4N=NC_6H_5 + HCl \]

Step 2: Orientation of substitution.

The amino group present in aniline is an electron-donating group.

It activates the benzene ring and directs incoming electrophiles towards the ortho and para positions.

Among these two possibilities, the para product is formed predominantly because:


Steric hindrance is minimum at the para position.
The para product is more stable.
Better resonance stabilization is possible.


Hence, the major product obtained is p-aminoazobenzene.

Step 3: Structure of the product.

The structure of p-aminoazobenzene is:
\[ \mathrm{H_2N-C_6H_4-N=N-C_6H_5} \]

The molecule contains:


One amino group \((-NH_2)\)
One azo linkage \((-N=N-)\)
Two benzene rings


Step 4: Significance of the reaction.

Azo coupling reactions are extensively used in the manufacture of dyes, indicators and colouring agents. The extended conjugation present in azo compounds is responsible for their characteristic colour.

Conclusion:

p-Aminoazobenzene is prepared by coupling benzene diazonium chloride with aniline in a mildly acidic medium.
\[ \boxed{ C_6H_5N_2^+Cl^- + C_6H_5NH_2 \rightarrow p-H_2NC_6H_4N=NC_6H_5 + HCl } \]
\[ \boxed{Product : p-Aminoazobenzene} \] Quick Tip: Azo coupling is an electrophilic substitution reaction in which a diazonium ion couples with an activated aromatic ring. Aniline and phenol generally give para-substituted azo compounds as the major products.

Concept:

Nitriles react with Grignard reagents to form ketones after hydrolysis.

The molecular formula:
\[ C_3H_5N \]

suggests a nitrile.

Step 1: Identify compound (A).

The nitrile having formula C\(_3\)H\(_5\)N is:
\[ CH_3CH_2CN \]

(Propanenitrile)
\[ \boxed{A = CH_3CH_2CN} \]

Step 2: Reaction with phenylmagnesium bromide.
\[ CH_3CH_2CN + C_6H_5MgBr \rightarrow Intermediate \]

On hydrolysis:
\[ CH_3CH_2COC_6H_5 \]

is formed.

This compound is propiophenone.
\[ \boxed{B = C_6H_5COCH_2CH_3} \]

Step 3: Verify compound (B).

It gives 2,4-DNP test because it contains a carbonyl group.

It does not give Tollens' or Fehling's test because it is a ketone.

It does not give iodoform test because it does not contain:
\[ CH_3CO- \]

group.

Hence all observations are satisfied.

Step 4: Oxidation of compound (B).

Strong oxidation of propiophenone converts the side chain into benzoic acid.
\[ C_6H_5COCH_2CH_3 \xrightarrow{[O]} C_6H_5COOH \]

Molecular formula:
\[ C_7H_6O_2 \]

Therefore:
\[ \boxed{C = C_6H_5COOH} \]

(Benzoic acid)

Reaction Sequence:
\[ CH_3CH_2CN + C_6H_5MgBr \rightarrow CH_3CH_2C(MgBr)=NC_6H_5 \]
\[ \xrightarrow{H_3O^+} CH_3CH_2COC_6H_5 \]
\[ \boxed{ A = CH_3CH_2CN,\quad B = C_6H_5COCH_2CH_3,\quad C = C_6H_5COOH } \] Quick Tip: Nitrile + Grignard reagent followed by hydrolysis always gives a ketone.

Concept:

Freezing point depression is a colligative property and is given by:
\[ \Delta T_f = iK_fm \]

For ethylene glycol, which is a non-electrolyte:
\[ i=1 \]

Step 1: Calculate moles of ethylene glycol.

Mass given:
\[ 31g \]

Molar mass:
\[ 62g\,mol^{-1} \]
\[ Moles = \frac{31}{62} = 0.5 \]

Step 2: Calculate molality.

Mass of water:
\[ 600g = 0.600kg \]
\[ m = \frac{0.5}{0.600} = 0.833 \]

mol kg\(^{-1}\)

Step 3: Calculate depression in freezing point.
\[ \Delta T_f = 1 \times 1.86 \times 0.833 \]
\[ =1.55K \]

Step 4: Determine freezing point of solution.

Freezing point of pure water:
\[ 0^\circ C \]

Hence
\[ T_f = 0-1.55 \]
\[ =-1.55^\circ C \]

Final Answer:
\[ \boxed{-1.55^\circ C} \]

The freezing point of the antifreeze solution is
\[ \boxed{-1.55^\circ C} \] Quick Tip: For non-electrolytes such as glucose, urea and ethylene glycol, always take van't Hoff factor \(i=1\).

Concept:

The geometry, hybridization and magnetic properties of coordination compounds depend upon:


Oxidation state of the central metal ion.
Electronic configuration of the metal ion.
Strength of the ligand present.
Crystal field splitting produced by the ligand.


According to Crystal Field Theory, strong field ligands pair the electrons present in d-orbitals whereas weak field ligands generally do not cause pairing. Consequently, the type of hybridization and magnetic behaviour of the complex changes significantly.

Step 1: Determine the oxidation state of nickel in both complexes.

For
\[ [NiCl_4]^{2-} \]

let the oxidation state of nickel be \(x\).
\[ x+4(-1)=-2 \]
\[ x=+2 \]

Similarly, for
\[ [Ni(CN)_4]^{2-} \]
\[ x+4(-1)=-2 \]
\[ x=+2 \]

Thus, in both complexes:
\[ Ni^{2+} \]

is present.

Step 2: Write the electronic configuration of \(Ni^{2+}\).

Atomic number of nickel:
\[ Z=28 \]

Electronic configuration of Ni:
\[ [Ar]\,3d^8\,4s^2 \]

Electronic configuration of
\[ Ni^{2+} \]

is:
\[ [Ar]\,3d^8 \]



Step 3: Analysis of \([NiCl_4]^{2-}\).

Chloride ion is a weak field ligand.

It is unable to pair the electrons present in the \(3d\)-orbitals.

Hence electron pairing does not occur.

The complex therefore uses outer orbitals for hybridization.

Hybridization:
\[ sp^3 \]

Geometry:
\[ Tetrahedral \]

Since two unpaired electrons remain present, the complex is paramagnetic.
\[ \boxed{ [NiCl_4]^{2-} \rightarrow sp^3 hybridization } \]



Step 4: Analysis of \([Ni(CN)_4]^{2-}\).

Cyanide ion is a strong field ligand.

It causes pairing of the \(3d\)-electrons.

After pairing, one \(3d\)-orbital becomes vacant and participates in hybridization.

Hybridization:
\[ dsp^2 \]

Geometry:
\[ Square planar \]

Since all electrons become paired, the complex is diamagnetic.
\[ \boxed{ [Ni(CN)_4]^{2-} \rightarrow dsp^2 hybridization } \]



Step 5: Identify inner and outer orbital complexes.

In
\[ [Ni(CN)_4]^{2-} \]

the inner \(3d\)-orbital participates in hybridization.

Therefore it is an inner orbital complex.

In
\[ [NiCl_4]^{2-} \]

the outer orbitals participate in hybridization.

Therefore it is an outer orbital complex.



Final Answers:
\[ \boxed{ [NiCl_4]^{2-} : sp^3 hybridization } \]
\[ \boxed{ [Ni(CN)_4]^{2-} : dsp^2 hybridization } \]
\[ \boxed{ [NiCl_4]^{2-} is an outer orbital complex } \]
\[ \boxed{ [Ni(CN)_4]^{2-} is an inner orbital complex } \]
\[ \boxed{ [NiCl_4]^{2-} is paramagnetic } \]
\[ \boxed{ [Ni(CN)_4]^{2-} is diamagnetic } \] Quick Tip: CN\(^-\) is a strong field ligand and usually causes electron pairing, whereas Cl\(^-\) is a weak field ligand and generally does not cause pairing.

(i) Coordination compounds important in biological systems

Concept:

Many naturally occurring biological molecules are coordination compounds in which a metal ion is coordinated to various ligands.

These compounds perform vital biological functions such as oxygen transport, photosynthesis and enzyme catalysis.

Examples:


Haemoglobin

It is an iron-containing coordination compound present in red blood cells.

The central metal ion is:
\[ Fe^{2+} \]

It is responsible for transportation of oxygen from lungs to body tissues.

Chlorophyll

It is a magnesium-containing coordination compound present in green plants.

The central metal ion is:
\[ Mg^{2+} \]

It plays a vital role in photosynthesis.

\[ \boxed{Haemoglobin and Chlorophyll} \]



(ii) Chelate effect

Definition:

The enhanced stability of complexes containing chelating ligands compared to analogous complexes containing monodentate ligands is known as the chelate effect.

Chelating ligands possess two or more donor atoms and form ring structures with the central metal ion.

These rings greatly increase the thermodynamic stability of the complex.

Example:

Ethane-1,2-diamine (en) is a bidentate ligand.
\[ [Ni(en)_3]^{2+} \]

is much more stable than
\[ [Ni(NH_3)_6]^{2+} \]

because en forms chelate rings around the metal ion.
\[ \boxed{ Greater stability due to ring formation is called chelate effect. } \]



(iii) Why are low spin tetrahedral complexes rarely formed ?

Concept:

In tetrahedral complexes, the crystal field splitting energy
\[ \Delta_t \]

is relatively small.
\[ \Delta_t = \frac{4}{9}\Delta_o \]

where
\[ \Delta_o \]

is the octahedral splitting energy.

Explanation:

Since
\[ \Delta_t \]

is small, the energy required for electron pairing is usually greater than the crystal field splitting energy.

Therefore electrons prefer to remain unpaired rather than pair up.

As a result, tetrahedral complexes are generally high-spin complexes.

Hence low-spin tetrahedral complexes are rarely formed.
\[ \boxed{ \Delta_t < Pairing Energy } \]

Therefore electron pairing does not occur. Quick Tip: Almost all tetrahedral complexes are high-spin because tetrahedral crystal field splitting is too small to force electron pairing.

Concept:

Molar conductivity increases with dilution because the ions present in solution become more mobile and interionic attractions decrease.

The variation of molar conductivity with concentration is different for strong and weak electrolytes.

This difference forms the basis for understanding Kohlrausch's law.



Kohlrausch's Law of Independent Migration of Ions

The law states that:

At infinite dilution, each ion contributes a definite value to the molar conductivity of an electrolyte irrespective of the nature of the other ion present with it.

Mathematically,
\[ \Lambda_m^\circ = \nu_+\lambda_+^\circ + \nu_-\lambda_-^\circ \]

where
\[ \lambda_+^\circ \]

and
\[ \lambda_-^\circ \]

represent limiting ionic conductivities.



Strong Electrolytes

Strong electrolytes are almost completely ionized in solution.


Thus a plot of
\[ \Lambda_m \]

versus
\[ \sqrt{c} \]

is nearly a straight line.

Therefore the graph can easily be extrapolated to:
\[ \sqrt{c}=0 \]

to obtain
\[ \Lambda_m^\circ \]



Weak Electrolytes

Weak electrolytes are only partially ionized.

As dilution increases, ionization also increases significantly.

Consequently, the variation of molar conductivity with concentration becomes highly non-linear.

The graph is strongly curved.

Because of this curvature, extrapolation to zero concentration becomes unreliable and inaccurate.

Hence direct determination of
\[ \Lambda_m^\circ \]

for weak electrolytes is difficult.



Schematic Curve


Strong electrolytes give almost straight-line plots whereas weak electrolytes give curved plots.

Therefore direct extrapolation is possible only for strong electrolytes.



Conclusion:

Since weak electrolytes undergo progressive ionization on dilution, their conductivity-concentration plots are curved. Therefore
\[ \Lambda_m^\circ \]

cannot be obtained accurately by extrapolation and is instead calculated using Kohlrausch's law. Quick Tip: For weak electrolytes, limiting molar conductivity is usually calculated using Kohlrausch's law rather than obtained graphically.

Concept:

The activity of a radioactive substance is directly proportional to the number of undecayed nuclei present.

Radioactive decay follows first-order kinetics.

The relation between activity and time is:
\[ A=A_0e^{-kt} \]

or
\[ k=\frac{2.303}{t} \log \frac{A_0}{A} \]



Step 1: Calculate decay constant.

Given:
\[ t_{1/2}=1.5\times10^{10}\;years \]

For radioactive decay,
\[ k=\frac{0.693}{t_{1/2}} \]
\[ k = \frac{0.693}{1.5\times10^{10}} \]
\[ k = 4.62\times10^{-11} \; year^{-1} \]



Step 2: Determine final activity.

Activity reduced to \(75%\) of original value:
\[ A=0.75A_0 \]

Substituting into first-order equation:
\[ k = \frac{2.303}{t} \log \frac{A_0}{0.75A_0} \]
\[ k = \frac{2.303}{t} \log \frac{4}{3} \]
\[ \log \frac{4}{3} = \log4-\log3 \]
\[ =0.60-0.48 \]
\[ =0.12 \]

Hence,
\[ k = \frac{2.303\times0.12}{t} \]
\[ k = \frac{0.276}{t} \]



Step 3: Calculate time.
\[ t = \frac{0.276}{4.62\times10^{-11}} \]
\[ t = 5.97\times10^{9} \]

years
\[ \boxed{ t \approx 6\times10^9 years } \]



Final Answer:
\[ \boxed{ 6\times10^9\ years } \]

The activity will reduce to \(75%\) of its original value in approximately
\[ \boxed{6\times10^9\ years} \] Quick Tip: Radioactive decay follows first-order kinetics. Activity, number of nuclei and concentration all decay according to the same first-order rate equation.

Concept:

Alcohols are oxygen-containing organic compounds containing one or more hydroxyl (\(-OH\)) groups attached to a saturated carbon atom. One of the most important chemical properties of alcohols is their ability to undergo oxidation. The product obtained on oxidation depends upon:


The nature of the alcohol (primary, secondary or tertiary).
The strength of the oxidizing agent used.
The reaction conditions.


Primary alcohols are particularly important because they can be oxidized either partially to aldehydes or completely to carboxylic acids depending upon the reagent employed.

(i) Oxidation of a primary alcohol to aldehyde

Step 1: Understanding the transformation

A primary alcohol contains the functional group:
\[ RCH_2OH \]

When subjected to controlled oxidation, it loses two hydrogen atoms and forms an aldehyde.

For example,
\[ CH_3CH_2OH \xrightarrow{[O]} CH_3CHO \]

The oxidation must stop at the aldehyde stage and should not proceed further to the carboxylic acid stage.

Step 2: Selection of reagent

For this purpose, a mild oxidizing agent is required.

The most commonly used reagent is Pyridinium Chlorochromate (PCC).

PCC selectively oxidizes primary alcohols to aldehydes without causing further oxidation.
\[ \boxed{PCC (Pyridinium Chlorochromate)} \]

Answer:
\[ \boxed{PCC} \]



(ii) Oxidation of a primary alcohol to carboxylic acid

Step 1: Understanding the transformation

When a stronger oxidizing agent is used, the aldehyde produced initially is further oxidized into a carboxylic acid.
\[ RCH_2OH \xrightarrow{[O]} RCHO \xrightarrow{[O]} RCOOH \]

For example,
\[ CH_3CH_2OH \xrightarrow{[O]} CH_3COOH \]

Step 2: Selection of reagent

Strong oxidizing agents such as acidified potassium dichromate or acidified potassium permanganate are commonly used.
\[ K_2Cr_2O_7/H_2SO_4 \]

or
\[ KMnO_4 \]

These oxidizing agents convert the alcohol completely into the corresponding carboxylic acid.

Answer:
\[ \boxed{Acidified K_2Cr_2O_7} \]

(or acidified \(KMnO_4\))



Final Answers:
\[ \boxed{ (i) PCC } \]
\[ \boxed{ (ii) Acidified K_2Cr_2O_7 } \] Quick Tip: Remember: PCC is a mild oxidizing agent that stops oxidation at the aldehyde stage, whereas acidified potassium dichromate and potassium permanganate are strong oxidizing agents that convert primary alcohols into carboxylic acids.

Concept:

Phenol exhibits reactions that are significantly different from those of ordinary alcohols because the hydroxyl group is directly attached to an aromatic ring. The lone pair of electrons on oxygen participates in resonance with the benzene ring, increasing the electron density particularly at the ortho and para positions.

When phenol is treated with sodium hydroxide, sodium phenoxide is formed. The phenoxide ion is even more reactive than phenol because the negative charge on oxygen increases the electron density of the aromatic ring through resonance.

One of the most important reactions of sodium phenoxide is the Kolbe-Schmitt reaction, commonly known as Kolbe's reaction.

Step 1: Formation of sodium phenoxide

Phenol reacts with sodium hydroxide to form sodium phenoxide.
\[ C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O \]

Step 2: Reaction with carbon dioxide

Sodium phenoxide is heated with carbon dioxide under pressure.
\[ C_6H_5ONa + CO_2 \xrightarrow[4-7\,atm]{373\,K} o-HOC_6H_4COONa \]

The carboxyl group enters predominantly at the ortho position due to activation of the ring by the phenoxide ion.

Step 3: Acidification

The sodium salt formed is treated with dilute hydrochloric acid.
\[ o-HOC_6H_4COONa + HCl \rightarrow o-HOC_6H_4COOH + NaCl \]

Step 4: Product obtained

The final product is salicylic acid (2-hydroxybenzoic acid), an important industrial and pharmaceutical compound.
\[ \boxed{ o-HOC_6H_4COOH } \]

Overall Reaction
\[ \boxed{ C_6H_5ONa + CO_2 \xrightarrow[4-7\,atm]{373\,K} o-HOC_6H_4COONa \xrightarrow{H^+} o-HOC_6H_4COOH } \] Quick Tip: Kolbe's reaction is an important method for preparing salicylic acid from phenol through the intermediate formation of sodium phenoxide.

Concept:

The oxidation of alcohols generally involves the removal of hydrogen atoms from two locations:


The hydroxyl group (\(-OH\)).
The carbon atom directly attached to the hydroxyl group (called the \(\alpha\)-carbon).


For oxidation to proceed normally, at least one hydrogen atom must be present on the \(\alpha\)-carbon.

Step 1: Examine the structure of a tertiary alcohol

The general structure of a tertiary alcohol is:
\[ R_3C-OH \]

The carbon atom attached to the hydroxyl group is bonded to three alkyl groups.

Therefore, it possesses no hydrogen atom.

Step 2: Requirement for oxidation

During oxidation, removal of the \(\alpha\)-hydrogen is essential for the formation of a carbonyl compound.

Since tertiary alcohols do not contain any \(\alpha\)-hydrogen atom, this process cannot occur.

Step 3: Consequence

To oxidize a tertiary alcohol, carbon-carbon bond cleavage would be required.

Breaking carbon-carbon bonds requires a large amount of energy and therefore does not occur under ordinary oxidation conditions.

As a result, tertiary alcohols remain unaffected by common oxidizing agents.

Conclusion

Tertiary alcohols are resistant to oxidation because the carbon atom


bearing the hydroxyl group does not contain any -hydrogen.


Hence normal oxidation cannot take place without cleavage of carbon-carbon bonds. Quick Tip: Primary alcohols contain two \(\alpha\)-hydrogens, secondary alcohols contain one \(\alpha\)-hydrogen, whereas tertiary alcohols contain no \(\alpha\)-hydrogen and therefore resist oxidation.

Concept:

Ethers undergo cleavage in the presence of concentrated hydrogen halides such as HI and HBr. The reaction proceeds through protonation of the ether oxygen followed by cleavage of the carbon--oxygen bond.

The nature of the alkyl groups attached to oxygen determines the mechanism of cleavage.

When one of the groups attached to oxygen is tertiary, cleavage generally occurs through the \(S_N1\) mechanism because tertiary carbocations are highly stable.

Step 1: Identify the ether

The given ether is:
\[ (CH_3)_3C-O-C_2H_5 \]

This compound is tert-butyl ethyl ether.

One side contains a tertiary alkyl group while the other side contains a primary ethyl group.

Step 2: Protonation of oxygen

The oxygen atom first accepts a proton from HI.
\[ (CH_3)_3C-O-C_2H_5 + H^+ \rightarrow (CH_3)_3C-OH^+-C_2H_5 \]

This weakens the carbon--oxygen bond.

Step 3: Formation of tertiary carbocation

The bond breaks preferentially at the tertiary carbon because a tertiary carbocation is highly stable.
\[ (CH_3)_3C^+ + C_2H_5OH \]

Step 4: Attack by iodide ion

The tertiary carbocation combines with iodide ion.
\[ (CH_3)_3C^+ + I^- \rightarrow (CH_3)_3CI \]

Step 5: Final products

Thus the products formed are:
\[ (CH_3)_3CI \]

and
\[ C_2H_5OH \]
\[ \boxed{ (CH_3)_3C-O-C_2H_5 + HI \rightarrow (CH_3)_3CI + C_2H_5OH } \]

Final Answer
\[ \boxed{ (CH_3)_3CI + C_2H_5OH } \] Quick Tip: In ether cleavage by HI, if one alkyl group is tertiary, cleavage occurs at the tertiary carbon through the \(S_N1\) pathway because tertiary carbocations are highly stable.

Concept:

Proteins and nucleic acids are important biological macromolecules essential for the growth, development and maintenance of living organisms. Although these biomolecules are very large and complex, they are formed by the repetition of smaller units known as monomers or building blocks.

Understanding these fundamental building units helps in understanding the structure and functions of biomolecules in living systems.

Step 1: Basic building unit of proteins

Proteins are naturally occurring polymers formed by the condensation of a large number of \(\alpha\)-amino acids.

Each amino acid contains:


An amino group (\(-NH_2\))
A carboxyl group (\(-COOH\))
A hydrogen atom
A side chain (\(R\)-group)


The amino acids are linked together by peptide bonds to form proteins.

Therefore, the basic building units of proteins are:
\[ \boxed{\alpha-amino acids} \]



Step 2: Basic building unit of nucleic acids

Nucleic acids such as DNA and RNA are polymers made up of repeating units called nucleotides.

Each nucleotide consists of:


A nitrogenous base
A pentose sugar
A phosphate group


Therefore, the basic building units of nucleic acids are:
\[ \boxed{Nucleotides} \]



Step 3: Difference between Fibrous and Globular proteins

Proteins are classified according to their molecular shape into fibrous proteins and globular proteins.

Fibrous Proteins


Long and thread-like in structure.
Polypeptide chains are arranged parallel to one another.
Generally insoluble in water.
Mainly perform structural functions.
Examples: Keratin, Collagen.


Globular Proteins


Compact and spherical in shape.
Polypeptide chains are folded into complex three-dimensional structures.
Usually soluble in water.
Mainly perform biological functions such as transport and catalysis.
Examples: Haemoglobin, Insulin.




Final Answer:
\[ \boxed{Proteins are made of \alpha-amino acids} \]
\[ \boxed{Nucleic acids are made of nucleotides} \]

Fibrous proteins possess long thread-like structures whereas globular proteins possess compact spherical structures. Quick Tip: Proteins are polymers of amino acids whereas nucleic acids are polymers of nucleotides. Fibrous proteins provide structural support while globular proteins mainly perform functional roles in living organisms.

Concept:

A nucleotide is the fundamental structural unit of nucleic acids. Each nucleotide consists of three components:


A nitrogenous base
A pentose sugar
A phosphate group


In DNA, the nitrogenous bases are adenine, guanine, cytosine and thymine.

When a nucleotide undergoes hydrolysis, these components separate from one another.

Step 1: Identify the nucleotide

The question specifies a DNA nucleotide containing thymine.

Therefore the nucleotide contains:


Thymine (nitrogenous base)
Deoxyribose sugar
Phosphate group


Step 2: Hydrolysis of the nucleotide

Upon hydrolysis, the glycosidic bond and phosphate ester bond are broken.

As a result, the nucleotide splits into its individual components.

Products formed
\[ \boxed{Thymine} \]
\[ \boxed{2-Deoxyribose sugar} \]
\[ \boxed{Phosphoric acid (phosphate group)} \]



Final Answer:
\[ \boxed{ Thymine + 2-Deoxyribose Sugar + Phosphoric Acid } \] Quick Tip: Hydrolysis of any nucleotide gives three components: a nitrogenous base, a pentose sugar and phosphoric acid.

Concept:

DNA (Deoxyribonucleic Acid) and RNA (Ribonucleic Acid) are two important types of nucleic acids found in living organisms. Although both are composed of nucleotides, they differ in certain structural features.

Step 1: Structure of DNA

DNA contains:


Deoxyribose sugar
Bases A, G, C and T
Usually double-stranded structure


Step 2: Structure of RNA

RNA contains:


Ribose sugar
Bases A, G, C and U
Usually single-stranded structure


One structural difference

The sugar present in DNA is deoxyribose, whereas the sugar present in RNA is ribose.
\[ \boxed{ DNA contains deoxyribose sugar whereas RNA contains ribose sugar. } \] Quick Tip: Remember: DNA = Deoxyribose + Thymine, whereas RNA = Ribose + Uracil.

Concept:

Vitamins are organic compounds required in very small quantities for the normal growth and functioning of the body. They do not provide energy directly but regulate numerous metabolic processes.

Based on their solubility, vitamins are classified into two categories:


Fat-soluble vitamins
Water-soluble vitamins


Step 1: Fat-soluble vitamins

Fat-soluble vitamins dissolve in fats and oils and can be stored in body tissues.

Examples include:
\[ A,\; D,\; E,\; K \]

One example is:
\[ \boxed{Vitamin A} \]



Step 2: Water-soluble vitamins

Water-soluble vitamins dissolve readily in water and are generally not stored in large amounts in the body.

Examples include:
\[ Vitamin B-complex and Vitamin C \]

One example is:
\[ \boxed{Vitamin C} \]



Final Answer:
\[ \boxed{Fat-soluble vitamin : Vitamin A} \]
\[ \boxed{Water-soluble vitamin : Vitamin C} \] Quick Tip: Vitamins A, D, E and K are fat-soluble, whereas Vitamin B-complex and Vitamin C are water-soluble vitamins.

Concept:

The electrode potential of an electrode under non-standard conditions is calculated using the Nernst equation. The Nernst equation establishes a relationship between the electrode potential and the concentration of ions participating in the electrode reaction.

For a metal-metal ion electrode,
\[ M^{n+}+ne^- \rightarrow M \]

the Nernst equation at \(25^\circ C\) is
\[ E=E^\circ-\frac{0.0591}{n}\log\frac{[M]}{[M^{n+}]} \]

Since the activity of a pure solid metal is taken as unity, the equation becomes
\[ E=E^\circ+\frac{0.0591}{n}\log[M^{n+}] \]



Step 1: Write the electrode reaction

For zinc electrode,
\[ Zn^{2+}+2e^- \rightarrow Zn(s) \]

Here,
\[ n=2 \]

and
\[ E^\circ=-0.76\,V \]



Step 2: Write the Nernst equation
\[ E=E^\circ+\frac{0.0591}{2}\log[Zn^{2+}] \]

Given concentration,
\[ [Zn^{2+}]=0.01=10^{-2} \]

Substituting,
\[ E=-0.76+\frac{0.0591}{2}\log(10^{-2}) \]



Step 3: Simplify the logarithmic term
\[ \log(10^{-2})=-2 \]

Therefore,
\[ E=-0.76+\frac{0.0591}{2}(-2) \]
\[ E=-0.76-0.0591 \]
\[ E=-0.8191\,V \]



Final Answer
\[ \boxed{E=-0.819\,V} \]

Thus, the electrode potential of the zinc electrode in \(0.01\,M\) \(ZnSO_4\) solution is approximately
\[ \boxed{-0.82\,V} \] Quick Tip: For metal-ion electrodes, decreasing the concentration of metal ions makes the electrode potential more negative than its standard electrode potential.

Concept:

A dry cell is a primary cell that converts chemical energy directly into electrical energy. It is widely used in flashlights, radios, clocks and portable electronic devices.

The dry cell consists of:


Zinc container acting as anode.
Carbon rod surrounded by \(MnO_2\) acting as cathode.
Moist paste of \(NH_4Cl\) and \(ZnCl_2\) as electrolyte.


The cell generates electricity through oxidation-reduction reactions.



Step 1: Anode reaction

Oxidation occurs at the anode.

The zinc container loses electrons.
\[ Zn(s)\rightarrow Zn^{2+}+2e^- \]

The zinc ions combine with chloride ions present in the electrolyte.
\[ Zn+2NH_4Cl \rightarrow ZnCl_2+2NH_3+2H^++2e^- \]



Step 2: Cathode reaction

Reduction occurs at the cathode.

Manganese dioxide is reduced in the presence of ammonium ions.
\[ 2MnO_2+2NH_4^++2e^- \rightarrow Mn_2O_3+2NH_3+H_2O \]



Step 3: Overall cell reaction

Adding the anodic and cathodic reactions,
\[ Zn+2MnO_2+2NH_4Cl \rightarrow ZnCl_2+Mn_2O_3+2NH_3+H_2O \]



Final Answer
\[ \boxed{Anode : Zn \rightarrow Zn^{2+}+2e^-} \]
\[ \boxed{ 2MnO_2+2NH_4^++2e^- \rightarrow Mn_2O_3+2NH_3+H_2O } \]
\[ \boxed{ Zn+2MnO_2+2NH_4Cl \rightarrow ZnCl_2+Mn_2O_3+2NH_3+H_2O } \] Quick Tip: In a dry cell, zinc acts as the anode and gets oxidized, while manganese dioxide acts as the depolarizer and undergoes reduction.

Concept:

The standard cell potential and equilibrium constant are thermodynamically related through the Gibbs energy change.

The relation is:
\[ \Delta G^\circ=-nFE^\circ_{cell} \]

and
\[ \Delta G^\circ=-RT\ln K \]

Combining these equations,
\[ E^\circ_{cell} = \frac{0.0591}{n}\log K \]

Thus, the equilibrium constant depends only on the standard cell potential.



Explanation:

The standard cell potential is measured under standard conditions:


Concentration \(=1\,M\)
Pressure \(=1\,atm\)
Temperature \(=298\,K\)


Therefore, \(E^\circ_{cell}\) is a constant characteristic of the reaction.

On the other hand,
\[ E_{cell} \]

changes with concentration and reaction conditions according to the Nernst equation.

Since equilibrium constant is also a fixed quantity for a given temperature, it can be related only to the constant standard cell potential and not to the variable cell potential.



Final Answer
\[ \boxed{ K_c is related to E^\circ_{cell} because both refer to standard equilibrium conditions. } \]
\[ \boxed{ E_{cell} changes with concentration and hence cannot be directly related to K_c. } \] Quick Tip: Remember the important relation: \[ E^\circ_{cell}=\frac{0.0591}{n}\log K \] Only the standard cell potential is related to the equilibrium constant.

Concept:

Molar conductivity is defined as the conductance of the volume of solution containing one mole of electrolyte placed between two electrodes one centimetre apart.

It is related to conductivity by:
\[ \Lambda_m=\frac{\kappa \times 1000}{C} \]

For weak electrolytes,
\[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \]

where
\[ \Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} \]



Step 1: Calculate molar conductivity

Given,
\[ \kappa=3.905\times10^{-5}\,S\,cm^{-1} \]
\[ C=0.001\,M \]

Using
\[ \Lambda_m=\frac{\kappa \times1000}{C} \]
\[ \Lambda_m= \frac{3.905\times10^{-5}\times1000}{0.001} \]
\[ \Lambda_m=39.05\,S\,cm^2\,mol^{-1} \]



Step 2: Calculate limiting molar conductivity
\[ \Lambda_m^\circ = 349.6+40.9 \]
\[ \Lambda_m^\circ = 390.5\,S\,cm^2\,mol^{-1} \]



Step 3: Calculate degree of dissociation
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \]
\[ = \frac{39.05}{390.5} \]
\[ =0.10 \]



Final Answer
\[ \boxed{ \Lambda_m = 39.05\,S\,cm^2\,mol^{-1} } \]
\[ \boxed{ \alpha=0.10 } \]

or

10% Quick Tip: For weak electrolytes: \[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \] Always calculate \(\Lambda_m^\circ\) using Kohlrausch's law before finding the degree of dissociation.

Concept:

The cell potential depends upon the concentration of reactants and products participating in the cell reaction.

In most cells, concentrations change during operation, causing the voltage to decrease gradually.



Explanation:

In a mercury cell, the overall reaction involves only solids and ions whose concentrations remain essentially constant throughout the life of the cell.

Since the concentration terms remain unchanged, the reaction quotient \(Q\) remains constant.

According to the Nernst equation,
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q \]

Since \(Q\) remains nearly constant, the cell potential remains practically unchanged.

Therefore, a mercury cell supplies a steady and constant voltage until almost all reactants are consumed.



Final Answer

A mercury cell delivers constant voltage because the concentrations of
reactants and products remain essentially constant during operation. Quick Tip: Mercury cells provide nearly constant voltage (\(\approx1.35\,V\)) throughout their life because the overall reaction does not significantly change ionic concentrations.

Concept:

A galvanic cell generates electrical energy through spontaneous redox reactions occurring in two separate half-cells.

As the reaction proceeds, charge imbalance develops in the two compartments.



Explanation:

At the anode, oxidation produces excess positive ions.

At the cathode, positive ions are consumed, leaving excess negative ions.

This accumulation of charge would eventually stop the flow of electrons.

A salt bridge prevents this problem by allowing migration of ions.

Its functions are:


Maintains electrical neutrality in both half-cells.
Completes the electrical circuit.
Prevents direct mixing of the two electrolytes.
Minimizes liquid junction potential.


Thus continuous current can flow through the external circuit.



Final Answer

A salt bridge maintains electrical neutrality and completes the circuit


by permitting the migration of ions between the two half-cells. Quick Tip: Without a salt bridge, charge accumulation occurs in the half-cells and the galvanic cell stops functioning after a short time.

Concept:

The elements of the (3d)-series exhibit characteristic physical and chemical properties due to the presence of partially filled (d)-orbitals. Properties such as melting point, colour and ionisation enthalpy are greatly influenced by the number of unpaired electrons, electronic configuration and metallic bonding.



(I) Fe or Cu -- Higher melting point

Step 1: Electronic configurations

[
Fe=[Ar],3d^6,4s^2
]

[
Cu=[Ar],3d^{10,4s^1
]

Step 2: Nature of metallic bonding

The strength of metallic bonding depends upon the number of unpaired electrons available for delocalisation.

Iron possesses a greater number of unpaired electrons than copper.

As a result, stronger metallic bonds are formed in iron.

Step 3: Effect on melting point

Stronger metallic bonding requires more energy to break.

Therefore iron possesses a higher melting point than copper.

[
Fe has the higher melting point
]



(II) (Ti^{3+) or (Sc^{3+}) -- Coloured in aqueous solution

Step 1: Electronic configurations

[
Ti=[Ar],3d^2,4s^2
]

[
Ti^{3+=[Ar],3d^1
]

[
Sc=[Ar],3d^1,4s^2
]

[
Sc^{3+=[Ar]
]

Step 2: Reason for colour

Transition metal ions are coloured when they contain partially filled (d)-orbitals.

The colour arises due to (d-d) electronic transitions.

[
Ti^{3+
]

contains one electron in the (d)-subshell.

Therefore (d-d) transitions are possible.

Hence it is coloured.

[
Sc^{3+
]

contains no (d)-electrons.

Therefore (d-d) transitions are not possible.

Hence it is colourless.

[
Ti^3+ is coloured
]



(III) Cr or Zn -- Higher third ionisation enthalpy

Step 1: Electronic configurations

[
Cr=[Ar],3d^5,4s^1
]

[
Zn=[Ar],3d^{10,4s^2
]

Step 2: Third ionisation process

For chromium:

[
Cr^{2+=[Ar],3d^4
]

Removal of the third electron gives:

[
Cr^{3+=[Ar],3d^3
]

For zinc:

[
Zn^{2+=[Ar],3d^{10
]

Removal of the third electron requires breaking the completely filled and highly stable (3d^{10) configuration.


Therefore:

[
Zn has the higher third ionisation enthalpy
]



Final Answers

[
(I) Fe
]

[
(II) Ti^{3+
]

[
(III) Zn
] Quick Tip: Colour in transition metal ions is generally due to partially filled d-orbitals. Completely filled or completely empty d-orbitals usually produce colourless ions.

Concept:

Permanganate ion ((MnO_4^-)) : is a powerful oxidizing agent in acidic medium.

During oxidation reactions, manganese is reduced

from oxidation state (+7) to (+2).

The reduction half-reaction in acidic medium is:

[
MnO_4^-+8H^++5e^-
Mn^{2++4H_2O
]

This reduction is coupled with oxidation of the reacting species.



(I) Oxidation of (Fe^{2+})

Step 1: Oxidation half-reaction

[
Fe^{2+
Fe^{3++e^-
]

Step 2: Balance electrons

Multiplying by 5:

[
5Fe^{2+
5Fe^{3++5e^-
]

Step 3: Add reduction half-reaction

[
MnO_4^-+8H^++5e^-
Mn^{2++4H_2O
]

Step 4: Overall ionic equation

[
MnO_4^-+8H^+
+5Fe^{2+
Mn^{2+
+5Fe^{3+
+4H_2O
]



(II) Oxidation of (I^-)

Step 1: Oxidation half-reaction

[
2I^-
I_2+2e^-
]

Step 2: Balance electrons

Multiplying oxidation reaction by 5 and reduction reaction by 2:

[
10I^-
5I_2+10e^-
]

[
2MnO_4^-+16H^++10e^-
2Mn^{2++8H_2O
]

Step 3: Add both equations

[
2MnO_4^-
+16H^+
+10I^-
2Mn^{2+
+5I_2
+8H_2O
] Quick Tip: In acidic medium, permanganate ion is reduced from Mn(VII) to Mn(II), making it one of the strongest oxidizing agents among common laboratory reagents.

Concept:

Manganese exhibits a wide range of oxidation states and forms compounds having characteristic colours.

Important compounds include:


(MnO_2) : Black-brown
(K_2MnO_4) : Dark green
(KMnO_4) : Dark purple


The conversion of manganese dioxide into potassium permanganate involves oxidation of manganese from lower to higher oxidation states.



Step 1: Identification of compound (A)

The given black-brown solid is manganese dioxide.

[
A=MnO_2
]



Step 2: Formation of dark green compound (B)

When (MnO_2) is fused with KOH in the presence of atmospheric oxygen:

[
2MnO_2+4KOH+O_2
2K_2MnO_4+2H_2O
]

Potassium manganate formed is dark green.

[
B=K_2MnO_4
]



Step 3: Formation of dark purple compound (C)

Electrolytic oxidation of potassium manganate in alkaline medium produces potassium permanganate.

[
2K_2MnO_4+H_2O
2KMnO_4+2KOH+H_2
]

Potassium permanganate possesses a characteristic dark purple colour.

[
C=KMnO_4
]



Final Identification

[
A=MnO_2
]

[
B=K_2MnO_4
]

[
C=KMnO_4
] Quick Tip: Remember the colour sequence: [ MnO_2 ;(black-brown)  K_2MnO_4 ;(green)  KMnO_4 ;(purple) ] This is frequently asked in board examinations.

Concept:

The green compound identified above is potassium manganate:

[
K_2MnO_4
]

In manganate ion,

[
MnO_4^{2-
]

manganese exists in the (+6) oxidation state.

This oxidation state is unstable in acidic medium.

As a result, manganate undergoes simultaneous oxidation and reduction.



Step 1: Behaviour in acidic medium

When an acidic solution of potassium manganate is allowed to stand, the green colour gradually disappears and a purple solution of potassium permanganate is formed along with manganese dioxide.

This occurs because manganese in oxidation state (+6) undergoes both oxidation and reduction simultaneously.



Step 2: Chemical equation

[
3MnO_4^{2-
+4H^+
2MnO_4^-
+MnO_2
+2H_2O
]



Step 3: Nature of the reaction

In the above reaction:


One portion of manganese is oxidized from (+6) to (+7).
Another portion is reduced from (+6) to (+4).


Since the same species undergoes oxidation as well as reduction simultaneously, the reaction is called a disproportionation reaction.



Final Answer

[
3MnO_4^{2-+4H^+
2MnO_4^-+MnO_2+2H_2O

]

[
This reaction is called a disproportionation reaction.
] Quick Tip: A disproportionation reaction is one in which the same element undergoes oxidation and reduction simultaneously.

Concept:

Aldehydes, ketones and carboxylic acids undergo a variety of important reactions due to the presence of the polar carbonyl group \((C=O)\). The carbon atom of the carbonyl group is electrophilic and therefore readily undergoes nucleophilic addition reactions. Carboxylic acids also undergo dehydration reactions in the presence of strong dehydrating agents.

The given question involves three important named reactions of carbonyl compounds and carboxylic acids.



(I) Reaction of ethanal with one mole of methanol in presence of dry HCl

Step 1: Nature of reaction

Alcohols add to aldehydes in the presence of dry acid catalysts.

When one mole of alcohol reacts with one mole of aldehyde, the product obtained is called a hemiacetal.



Step 2: Reaction
\[ CH_3CHO + CH_3OH \xrightarrow{dry HCl} CH_3CH(OH)(OCH_3) \]



Step 3: Product formed

The product contains both an \(-OH\) group and an \(-OCH_3\) group attached to the same carbon atom.

Hence the product is:
\[ \boxed{CH_3CH(OH)(OCH_3)} \]

which is called the hemiacetal of ethanal.



(II) Benzaldehyde treated with concentrated NaOH

Step 1: Identify the reaction

Benzaldehyde does not contain an \(\alpha\)-hydrogen atom.

Therefore it cannot undergo aldol condensation.

Instead, in concentrated sodium hydroxide solution, it undergoes the Cannizzaro reaction.



Step 2: Cannizzaro reaction

In this reaction one molecule of benzaldehyde gets oxidized to benzoic acid while another molecule gets reduced to benzyl alcohol.
\[ 2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]



Step 3: Products formed

The products are:
\[ \boxed{C_6H_5CH_2OH} \]

(Benzyl alcohol)

and
\[ \boxed{C_6H_5COONa} \]

(Sodium benzoate)



(III) Ethanoic acid heated with \(P_2O_5\)

Step 1: Role of \(P_2O_5\)

Phosphorus pentoxide is a powerful dehydrating agent.

It removes a molecule of water from two molecules of carboxylic acid.



Step 2: Formation of acid anhydride
\[ 2CH_3COOH \xrightarrow{P_2O_5} (CH_3CO)_2O + H_2O \]



Step 3: Product formed

The product obtained is acetic anhydride.
\[ \boxed{(CH_3CO)_2O} \]



Final Answers
\[ \boxed{CH_3CH(OH)(OCH_3)} \]
\[ \boxed{C_6H_5CH_2OH + C_6H_5COONa} \]
\[ \boxed{(CH_3CO)_2O} \] Quick Tip: One mole of alcohol reacts with an aldehyde to form a hemiacetal. Benzaldehyde undergoes Cannizzaro reaction because it lacks \(\alpha\)-hydrogen. \(P_2O_5\) converts carboxylic acids into acid anhydrides by dehydration.

Concept:

Ethanal contains the structural unit
\[ CH_3CO- \]

and therefore behaves similarly to methyl ketones in the iodoform test.

Propanal does not contain this structural feature and hence does not respond to the iodoform test.

Thus, the iodoform test can be used to distinguish between the two compounds.



Step 1: Perform iodoform test

Treat both compounds with iodine and aqueous sodium hydroxide.
\[ I_2/NaOH \]



Step 2: Observation for ethanal

Ethanal gives a yellow precipitate of iodoform.
\[ CHI_3 \]
\[ \boxed{Yellow precipitate formed} \]



Step 3: Observation for propanal

Propanal does not produce any yellow precipitate.
\[ \boxed{No yellow precipitate} \]



Final Answer
\[ \boxed{Iodoform test} \]

Ethanal gives yellow precipitate of iodoform whereas propanal does not. Quick Tip: Compounds containing the \(CH_3CO-\) group generally give a positive iodoform test.

Concept:

Allyl alcohol is an unsaturated primary alcohol.
\[ CH_2=CHCH_2OH \]

Its controlled oxidation gives the corresponding aldehyde, propenal.

Strong oxidizing agents may further oxidize the aldehyde to acids, therefore a mild and selective oxidizing reagent is required.



Reaction
\[ CH_2=CHCH_2OH \xrightarrow{MnO_2} CH_2=CHCHO \]



Explanation

Activated manganese dioxide selectively oxidizes allylic and benzylic alcohols to aldehydes without affecting the carbon-carbon double bond.

Therefore allyl alcohol is converted into propenal efficiently.



Final Answer
\[ \boxed{MnO_2 (Activated Manganese Dioxide)} \] Quick Tip: Activated \(MnO_2\) is a selective oxidizing agent for allylic and benzylic alcohols.

Concept:

Aldehydes and ketones react with semicarbazide \((NH_2NHCONH_2)\) to form crystalline derivatives known as semicarbazones. This reaction is a nucleophilic addition-elimination reaction and is commonly used for the identification and characterization of carbonyl compounds.

Acetone is a ketone having the structure:
\[ CH_3COCH_3 \]

The carbonyl oxygen is replaced by the semicarbazone group \((=NNHCONH_2)\).



Step 1: Write the reaction
\[ CH_3COCH_3 + NH_2NHCONH_2 \rightarrow CH_3C(=NNHCONH_2)CH_3 + H_2O \]



Step 2: Structure of semicarbazone

The product obtained is acetone semicarbazone.
\[ \boxed{ CH_3 \; \overset{}{\underset{}{|}} \; C=NNHCONH_2 \; \overset{}{\underset{}{|}} \; CH_3 } \]

or
\[ \boxed{ (CH_3)_2C=NNHCONH_2 } \]



Final Answer
\[ \boxed{ (CH_3)_2C=NNHCONH_2 } \] Quick Tip: Semicarbazones are solid crystalline derivatives of aldehydes and ketones and are frequently used for identification purposes because of their sharp melting points.

Concept:

The hydrogen atoms attached to the carbon atom adjacent to the carbonyl carbon are called \(\alpha\)-hydrogen atoms.

These hydrogens exhibit acidic character because removal of an \(\alpha\)-hydrogen produces a resonance-stabilized enolate ion.



Step 1: Formation of enolate ion

Consider a simple aldehyde or ketone:
\[ RCH_2CHO \]

Removal of an \(\alpha\)-hydrogen gives:
\[ RCH^-CHO \]



Step 2: Resonance stabilization

The negative charge formed is delocalized between the \(\alpha\)-carbon and the oxygen atom.
\[ RCH^-CHO \leftrightarrow RCH=CHO^- \]

Because the negative charge is spread over more than one atom, the conjugate base becomes stable.



Step 3: Effect on acidity

Greater stability of the conjugate base increases the tendency of hydrogen ion removal.

Therefore \(\alpha\)-hydrogens of aldehydes and ketones show acidic character.



Final Answer

\(\alpha\)-Hydrogen atoms are acidic because their removal produces

a resonance-stabilized enolate ion. Quick Tip: The acidity of \(\alpha\)-hydrogens is due to resonance stabilization of the enolate ion formed after deprotonation.

Concept:

The addition of HCN to carbonyl compounds is a nucleophilic addition reaction.

The reactivity depends mainly upon:


Steric hindrance around the carbonyl carbon.
Electron donating effect of alkyl groups.
Electrophilicity of the carbonyl carbon.




Step 1: Compare aldehydes and ketones

Aldehydes are generally more reactive than ketones because:


They possess only one alkyl group.
They have less steric hindrance.
Their carbonyl carbon is more electrophilic.


Therefore:
\[ CH_3CHO \]

is more reactive than acetone.



Step 2: Compare the ketones
\[ (CH_3)_3C-CO-CH_3 \]

contains a bulky tert-butyl group.

The large tert-butyl group causes severe steric hindrance and decreases the accessibility of the carbonyl carbon.

Hence it is less reactive than acetone.



Step 3: Increasing order
\[ \boxed{ (CH_3)_3C-CO-CH_3 < CH_3COCH_3 < CH_3CHO } \]



Final Answer
\[ \boxed{ (CH_3)_3C-CO-CH_3 < CH_3COCH_3 < CH_3CHO } \] Quick Tip: Aldehydes are generally more reactive than ketones towards nucleophilic addition because they have less steric hindrance and a more electrophilic carbonyl carbon.

Concept:

Etard reaction is a selective oxidation reaction used to convert the methyl group attached to an aromatic ring into an aldehyde group.

The reagent used is chromyl chloride \((CrO_2Cl_2)\) in carbon disulphide followed by hydrolysis.



Reaction
\[ C_6H_5CH_3 \xrightarrow[CS_2]{CrO_2Cl_2} Etard Complex \xrightarrow{H_3O^+} C_6H_5CHO \]



Explanation

Toluene undergoes controlled oxidation to benzaldehyde without further oxidation to benzoic acid.

This selective oxidation is known as the Etard reaction.



Final Answer
\[ \boxed{ C_6H_5CH_3 \xrightarrow[CS_2]{CrO_2Cl_2} C_6H_5CHO } \] Quick Tip: Etard reaction converts the side-chain methyl group of aromatic compounds into an aldehyde using chromyl chloride.

Concept:

Zinc amalgam and concentrated hydrochloric acid constitute Clemmensen's reagent.

Clemmensen reduction converts aldehydes and ketones into hydrocarbons by reducing the carbonyl group to a methylene group.



Step 1: Starting compound

Benzaldehyde:
\[ C_6H_5CHO \]



Step 2: Clemmensen reduction
\[ C_6H_5CHO \xrightarrow{Zn(Hg)/conc.HCl} C_6H_5CH_3 \]



Step 3: Product formed

The aldehyde group \((-CHO)\) is reduced to a methyl group \((-CH_3)\).

Thus benzaldehyde is converted into toluene.



Final Answer
\[ \boxed{ C_6H_5CHO \xrightarrow{Zn(Hg)/conc.HCl} C_6H_5CH_3 } \]
\[ \boxed{Product = Toluene} \] Quick Tip: Clemmensen reduction converts aldehydes and ketones into hydrocarbons using \(Zn(Hg)\) and concentrated HCl.