CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 2 - 56/1/2)

Concept:


The basic or acidic nature of transition metal oxides depends mainly upon the oxidation state of the metal.

As the oxidation state increases:
\[ Basic character decreases \]

and
\[ Acidic character increases \]

Lower oxidation state oxides are generally ionic and basic, whereas higher oxidation state oxides are covalent and acidic.



Step 1: {\color{redFind oxidation state of Mn in each oxide.


For \(MnO\):
\[ Mn=+2 \]

For \(Mn_2O_3\):
\[ Mn=+3 \]

For \(MnO_2\):
\[ Mn=+4 \]

For \(Mn_2O_7\):
\[ Mn=+7 \]



Step 2: {\color{redCompare basic character.


Since \(MnO\) has the lowest oxidation state (\(+2\)), it possesses the maximum ionic character and therefore behaves as the most basic oxide.

Order:
\[ MnO > Mn_2O_3 > MnO_2 > Mn_2O_7 \]



Step 3: {\color{redChoose the correct option.


Therefore the most basic oxide is:
\[ \boxed{MnO} \]

Hence:
\[ \boxed{(C) MnO} \] Quick Tip: For transition metal oxides, lower oxidation state means more basic nature and higher oxidation state means more acidic nature.

Concept:


The order of a reaction determines how the concentration of reactants affects the rate of reaction.

For a first order reaction, the rate is directly proportional to the concentration of one reactant.
\[ Rate=k[A] \]

One of the most important characteristics of a first order reaction is that its half-life remains constant throughout the reaction.

The half-life expression is:
\[ t_{1/2}=\frac{0.693}{k} \]

where:


\(t_{1/2}\) = Half-life of the reaction
\(k\) = Rate constant




Step 1: {\color{redRecall the half-life formula for different orders.


For a zero order reaction:
\[ t_{1/2}=\frac{[A]_0}{2k} \]

which depends upon the initial concentration.

For a first order reaction:
\[ t_{1/2}=\frac{0.693}{k} \]

which is independent of concentration.

For a second order reaction:
\[ t_{1/2}=\frac{1}{k[A]_0} \]

which again depends upon concentration.



Step 2: {\color{redAnalyze the first order half-life equation.


Observe carefully that:
\[ t_{1/2}=\frac{0.693}{k} \]

contains only the rate constant \(k\).

There is no concentration term present in the equation.

Therefore, changing the concentration of reactant does not affect the half-life.



Step 3: {\color{redUnderstand the graphical implication.


Since the half-life remains the same irrespective of concentration,
\[ t_{1/2}=constant \]

Therefore, if we plot half-life on the y-axis and concentration on the x-axis, the graph will be a horizontal straight line parallel to the x-axis.



Step 4: {\color{redExamine the given graphs.


Among the four graphs shown:


Graph A shows half-life increasing with concentration.
Graph C shows half-life decreasing with concentration.
Graph D shows a curved relationship.
Graph B shows a constant horizontal value.


Only Graph B represents a constant half-life.



Step 5: {\color{redApply the property of first order reaction.


A constant half-life is the unique characteristic of a first order reaction.

Therefore, the graph corresponding to a first order reaction must be the horizontal straight line.



Hence:
\[ \boxed{Fig B} \]
\[ \boxed{(B)} \] Quick Tip: Remember the most important result: \[ t_{1/2}=\frac{0.693}{k} \] For a first order reaction, half-life is independent of concentration and always remains constant. Therefore, the \(t_{1/2}\) vs concentration graph is a horizontal straight line.

Concept:


Osmotic pressure is given by:
\[ \pi=iCRT \]

where
\[ i=Van't Hoff factor \]
\[ C=Molar concentration \]
\[ R=Gas constant \]
\[ T=Temperature \]

For equal concentration and temperature, osmotic pressure depends directly on the Van't Hoff factor.



Step 1: {\color{redDetermine dissociation of each solute.


For glucose:
\[ i=1 \]

For urea:
\[ i=1 \]

For KCl:
\[ KCl \rightarrow K^+ + Cl^- \]
\[ i=2 \]

For \(CaCl_2\):
\[ CaCl_2 \rightarrow Ca^{2+}+2Cl^- \]
\[ i=3 \]



Step 2: {\color{redCompare osmotic pressure.


Since osmotic pressure is proportional to \(i\),
\[ \pi_{CaCl_2} > \pi_{KCl} > \pi_{Glucose} = \pi_{Urea} \]



Step 3: {\color{redFinal answer.


Therefore:
\[ \boxed{0.1\,M\;CaCl_2} \]

Hence:
\[ \boxed{(B)} \] Quick Tip: More particles in solution means greater osmotic pressure. Always compare Van't Hoff factor.

Concept:


For a general reaction:
\[ aA+bB \rightarrow cC+dD \]

rate is written as:
\[ -\frac{1}{a}\frac{d[A]}{dt} = -\frac{1}{b}\frac{d[B]}{dt} = +\frac{1}{c}\frac{d[C]}{dt} \]



Step 1: {\color{redWrite the balanced equation.

\[ N_2+3H_2 \rightarrow 2NH_3 \]



Step 2: {\color{redUse stoichiometric coefficient.


The coefficient of ammonia is 2.

Therefore:
\[ Rate = +\frac{1}{2}\frac{\Delta[NH_3]}{\Delta t} \]

Positive sign is used because ammonia is being formed.



Step 3: {\color{redFinal answer.

\[ \boxed{+\frac{1}{2}\frac{\Delta[NH_3]}{\Delta t}} \]

Hence:
\[ \boxed{(D)} \] Quick Tip: Products always carry positive sign and reactants carry negative sign in rate expressions.

Concept:


One Faraday corresponds to one mole of electrons.

To determine Faradays required, balance the reduction half-reaction.



Step 1: {\color{redWrite reduction half reaction.

\[ Cr_2O_7^{2-}+14H^++6e^- \rightarrow 2Cr^{3+}+7H_2O \]



Step 2: {\color{redCount electrons.


The balanced equation shows:
\[ 6e^- \]

are required for reduction of one mole of dichromate ion.



Step 3: {\color{redConvert electrons into Faradays.

\[ 1F = 1\;mol\;e^- \]

Therefore:
\[ 6e^- = 6F \]

Hence:
\[ \boxed{(C) 6} \] Quick Tip: Always balance the redox half-reaction first. Number of electrons equals number of Faradays required.

Concept:


Methanal reacts with Grignard reagent to produce a primary alcohol.

General reaction:
\[ R-MgX + HCHO \rightarrow RCH_2OH \]



Step 1: {\color{redIdentify the required alcohol.


Cyclohexyl methanol:
\[ C_6H_{11}CH_2OH \]



Step 2: {\color{redFind the R-group.


Comparing with:
\[ RCH_2OH \]

we get:
\[ R=C_6H_{11} \]



Step 3: {\color{redWrite corresponding Grignard reagent.

\[ C_6H_{11}MgBr \]

which is cyclohexyl magnesium bromide.

Hence:
\[ \boxed{(C)} \] Quick Tip: Methanal + Grignard reagent always gives a primary alcohol containing one extra carbon.

Concept:


According to Werner's Coordination Theory:
\[ Secondary Valency = Coordination Number \]

Secondary valency represents the total number of donor atoms directly bonded to the central metal ion.



Step 1: {\color{redIdentify the central metal ion.


The given complex is:
\[ [Pt(en)_2Cl_2]^{2+} \]

The central metal ion is:
\[ Pt \]



Step 2: {\color{redDetermine the denticity of ethylenediamine.


Ethylenediamine (\(en\)) contains two nitrogen donor atoms.

Hence it is a bidentate ligand.
\[ 1\,en = 2 donor atoms \]



Step 3: {\color{redCalculate contribution of two \(en\) ligands.


Since two ethylenediamine molecules are present:
\[ 2\times2=4 \]

Thus, \(en\) contributes four coordination positions.



Step 4: {\color{redCount chloride ligands.


Each chloride ion is monodentate.
\[ 2Cl^- = 2 donor atoms \]



Step 5: {\color{redFind coordination number.


Total donor atoms attached to Pt are:
\[ 4+2=6 \]

Therefore,
\[ Secondary Valency = Coordination Number = 6 \]



Hence:
\[ \boxed{(A) 6} \] Quick Tip: Secondary valency = Coordination number. For polydentate ligands, count donor atoms, not just the number of ligand molecules.

Concept:


Unsymmetrical ethers react with hydrogen iodide \((HI)\) by cleavage of the \(C-O\) bond.

If one alkyl group is tertiary, cleavage occurs through the formation of a stable tertiary carbocation.



Step 1: {\color{redIdentify the ether.


The given ether is:
\[ (CH_3)_3C-OC_2H_5 \]

It contains:
\[ (CH_3)_3C- \]

tertiary butyl group and
\[ C_2H_5- \]

ethyl group.



Step 2: {\color{redUnderstand protonation by \(HI\).


The oxygen atom of ether gets protonated by \(HI\):
\[ (CH_3)_3C-OC_2H_5 + HI \rightarrow (CH_3)_3C-O^+H-C_2H_5 + I^- \]

This makes the \(C-O\) bond easier to break.



Step 3: {\color{redDecide where bond cleavage occurs.


Since the tertiary butyl group can form a stable tertiary carbocation,
\[ (CH_3)_3C^+ \]

the bond breaks on the tertiary side.



Step 4: {\color{redFormation of products.


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

The ethoxy part becomes ethanol:
\[ C_2H_5OH \]



Therefore, the products are:
\[ (CH_3)_3C-I \]

and
\[ C_2H_5OH \]



Hence:
\[ \boxed{(C) (CH_3)_3C-I and C_2H_5-OH} \] Quick Tip: In unsymmetrical ethers containing a tertiary alkyl group, cleavage with \(HI\) occurs at the tertiary carbon because a stable tertiary carbocation is formed.

Concept:


Aniline contains the \(-NH_2\) group.

Normally, \(-NH_2\) is an activating and ortho-para directing group.

But during direct nitration, the reaction medium is strongly acidic due to the presence of concentrated nitric acid and concentrated sulphuric acid.

In acidic medium, aniline gets protonated.



Step 1: {\color{redUnderstand the behaviour of aniline in acidic medium.


Aniline is basic because nitrogen has a lone pair of electrons.

In acidic medium, it accepts a proton:
\[ C_6H_5NH_2 + H^+ \rightarrow C_6H_5NH_3^+ \]

The group formed is:
\[ -NH_3^+ \]

This is called the anilinium ion.



Step 2: {\color{redEffect of anilinium ion on nitration.


The \(-NH_3^+\) group is strongly electron withdrawing.

It decreases electron density on the benzene ring.

Also, it becomes meta-directing.

Therefore, a considerable amount of meta-nitroaniline is formed.



Step 3: {\color{redWhy ortho product is also formed.


Not all aniline molecules are protonated.

Some free aniline molecules remain present.

Free \(-NH_2\) group is ortho-para directing.

So ortho and para products are also formed.



Step 4: {\color{redIdentify the major product distribution.


Due to protonation of aniline, meta derivative becomes the major product.

The approximate product distribution is:
\[ 51% meta \]
\[ 47% ortho \]
\[ 2% para \]



Hence:
\[ \boxed{(B) 51%-meta, 47%-ortho, 2%-para derivatives} \] Quick Tip: Direct nitration of aniline gives a large amount of meta product because aniline is protonated to anilinium ion \((C_6H_5NH_3^+)\), which is meta-directing.

Concept:


Enantiomers are non-superimposable mirror images of each other.

They possess identical physical properties except optical activity.



Step 1: {\color{redCompare physical properties.


Enantiomers have the same:
\[ Melting point \]
\[ Boiling point \]
\[ Density \]
\[ Refractive index \]



Step 2: {\color{redCompare chemical properties.


In achiral medium, they show identical chemical reactivity.



Step 3: {\color{redCompare optical rotation.


They rotate plane-polarized light in opposite directions.

If one rotates:
\[ +30^\circ \]

the other rotates:
\[ -30^\circ \]

Thus they do not have the same specific rotation.



Hence:
\[ \boxed{(B)} \] Quick Tip: Enantiomers have equal magnitude but opposite sign of optical rotation.

Concept:


Sugars are classified as reducing and non-reducing sugars on the basis of the presence or absence of a free anomeric carbon.

A reducing sugar has a free aldehydic group, ketonic group, or a free hemiacetal/hemi-ketal group which can reduce mild oxidising agents such as Tollen's reagent or Fehling's solution.

A non-reducing sugar does not have a free anomeric carbon because the anomeric carbons are involved in glycosidic linkage.



Step 1: {\color{redCheck glucose.


Glucose has a free anomeric carbon.

Therefore, glucose is a reducing sugar.
\[ Glucose \Rightarrow Reducing sugar \]



Step 2: {\color{redCheck lactose and maltose.


Lactose and maltose also have one free anomeric carbon.

Therefore, both are reducing sugars.
\[ Lactose \Rightarrow Reducing sugar \]
\[ Maltose \Rightarrow Reducing sugar \]



Step 3: {\color{redCheck sucrose.


In sucrose, the anomeric carbon of glucose and the anomeric carbon of fructose are both involved in glycosidic linkage.

So sucrose has no free anomeric carbon.

Therefore, sucrose cannot reduce Tollen's reagent or Fehling's solution.
\[ Sucrose \Rightarrow Non-reducing sugar \]



Hence: \[ \boxed{(B) Sucrose} \] Quick Tip: If a sugar has no free anomeric carbon, it is non-reducing. Sucrose is the common example.

Concept:


Benzene diazonium chloride undergoes coupling reaction with phenol in weakly basic medium.

This reaction is known as azo coupling.

In weakly basic medium, phenol forms phenoxide ion, which activates the benzene ring strongly toward electrophilic substitution.



Step 1: {\color{redUnderstand the reacting species.


Benzene diazonium chloride is:
\[ C_6H_5N_2^+Cl^- \]

Phenol is:
\[ C_6H_5OH \]

In weakly basic medium:
\[ C_6H_5OH + OH^- \rightarrow C_6H_5O^- + H_2O \]

Phenoxide ion is more activating than phenol.



Step 2: {\color{redIdentify the position of coupling.


The \(-OH\) group is an ortho-para directing group.

Due to less steric hindrance, coupling mainly occurs at the para position.

So benzene diazonium chloride couples with phenol at the para position.



Step 3: {\color{redWrite the product.


The product formed is:
\[ p-hydroxyazobenzene \]

It contains the azo linkage:
\[ -N=N- \]



Hence: \[ \boxed{(D) p-hydroxyazobenzene} \] Quick Tip: Diazonium salt plus phenol in weakly basic medium gives azo coupling product, mainly \(p\)-hydroxyazobenzene.

Concept:


In phenol, the \(-OH\) group is directly attached to the benzene ring.

Oxygen has lone pairs of electrons.

These lone pairs participate in resonance with the benzene ring and increase electron density at ortho and para positions.

Therefore, \(-OH\) is an activating and ortho-para directing group.



Step 1: {\color{redCheck the Assertion.


The assertion says that \(-OH\) group directs incoming group to ortho and para positions.

This is true.

In phenol:
\[ -OH \Rightarrow ortho-para directing group \]



Step 2: {\color{redCheck the Reason.


The reason says that \(-OH\) activates the aromatic ring towards electrophilic substitution.

This is also true.

The lone pair of oxygen is donated to the benzene ring by resonance.
\[ -OH \Rightarrow +R effect \]



Step 3: {\color{redCheck whether Reason explains Assertion.


Because \(-OH\) increases electron density at ortho and para positions, electrophiles prefer to attack these positions.

So the reason correctly explains why \(-OH\) directs incoming groups to ortho and para positions.



Hence: \[ \boxed{(A)} \] Quick Tip: Electron donating groups like \(-OH\), \(-NH_2\), and \(-OR\) generally activate benzene ring and direct electrophiles to ortho and para positions.

Concept:


Actinoids are \(5f\)-block elements.

Their electronic configurations are not always regular because the energies of \(5f\), \(6d\), and \(7s\) orbitals are very close to each other.

When orbital energies are comparable, electrons may enter different orbitals in a way that does not show a perfectly regular pattern.



Step 1: {\color{redCheck the Assertion.


The assertion says that actinoids show irregularities in electronic configurations.

This is true.

Actinoids often show irregular filling because of small energy differences among orbitals.



Step 2: {\color{redCheck the Reason.


The reason says that \(5f\), \(6d\), and \(7s\) orbitals are of comparable energies.

This is also true.
\[ 5f \approx 6d \approx 7s \]

in energy for actinoids.



Step 3: {\color{redCheck whether Reason explains Assertion.


Since these orbitals have comparable energies, electrons may occupy \(5f\), \(6d\), or \(7s\) orbitals depending on stability.

This causes irregular electronic configurations.

Therefore, Reason correctly explains Assertion.



Hence: \[ \boxed{(A)} \] Quick Tip: Actinoids show irregular configurations because \(5f\), \(6d\), and \(7s\) orbitals are very close in energy.

Concept:


Azeotropes are constant boiling mixtures.

They boil at a constant temperature and have the same composition in liquid phase and vapour phase.

Because of this property, their components cannot be separated by simple fractional distillation.



Step 1: {\color{redCheck the Assertion.


The assertion says that components of azeotropes are easily separated by fractional distillation.

This is false.

Azeotropes cannot be easily separated by fractional distillation because they behave like a single substance during boiling.



Step 2: {\color{redCheck the Reason.


The reason says that components of an azeotrope have the same composition in liquid and vapour phase.

This is true.
\[ Composition of liquid phase = Composition of vapour phase \]



Step 3: {\color{redFinal conclusion.


Since Assertion is false but Reason is true, the correct option is:
\[ \boxed{(D)} \] Quick Tip: Azeotropes cannot be separated by fractional distillation because liquid and vapour have the same composition.

Concept:


DNA is made up of two polynucleotide strands.

These two strands are held together by hydrogen bonds between nitrogenous bases.

The base pairing in DNA is specific.

Adenine always pairs with thymine, and guanine always pairs with cytosine.
\[ A=T \]
\[ G\equiv C \]

This fixed base pairing makes the two DNA strands complementary to each other.



Step 1: {\color{redCheck the Assertion.


The assertion says that the two strands of DNA are complementary to each other.

This is true.

If the base sequence of one strand is known, the base sequence of the other strand can be predicted.

For example, if one strand contains:
\[ A-T-G-C \]

then the complementary strand will contain:
\[ T-A-C-G \]



Step 2: {\color{redCheck the Reason.


The reason says that hydrogen bonds are formed between specific pairs of bases.

This is also true.

Adenine forms hydrogen bonds with thymine:
\[ A=T \]

Guanine forms hydrogen bonds with cytosine:
\[ G\equiv C \]



Step 3: {\color{redCheck whether Reason explains Assertion.


Because hydrogen bonding occurs only between specific base pairs, one strand determines the other strand.

This is why the two DNA strands are complementary.

So the Reason correctly explains the Assertion.



Hence: \[ \boxed{(A)} \] Quick Tip: In DNA, remember fixed base pairing: \(A\) pairs with \(T\), and \(G\) pairs with \(C\).

Concept:


Raoult's law states that the partial vapour pressure of each component of an ideal solution is directly proportional to its mole fraction.

For an ideal solution:
\[ P_A = P_A^\circ x_A \]
\[ P_B = P_B^\circ x_B \]

A solution shows positive deviation from Raoult's law when the intermolecular attraction between unlike molecules is weaker than that between like molecules.
\[ A-B < A-A and B-B \]

In such cases, molecules escape more easily into the vapour phase, so vapour pressure increases.



Step 1: {\color{redUnderstand interactions in pure ethanol.


Ethanol molecules are strongly associated due to intermolecular hydrogen bonding.
\[ C_2H_5OH \cdots HO-C_2H_5 \]

These hydrogen bonds hold ethanol molecules together strongly.



Step 2: {\color{redUnderstand what happens when acetone is added.


When acetone is mixed with ethanol, acetone molecules come between ethanol molecules.

This breaks or weakens the ethanol-ethanol hydrogen bonding.

The new ethanol-acetone interactions are weaker than the original ethanol-ethanol interactions.
\[ Ethanol-acetone interaction < Ethanol-ethanol interaction \]



Step 3: {\color{redDecide the deviation from Raoult's law.


Since intermolecular forces become weaker, molecules escape more easily into vapour phase.

Therefore, the observed vapour pressure is higher than the vapour pressure predicted by Raoult's law.
\[ P_{observed} > P_{Raoult} \]

Hence, the mixture shows positive deviation from Raoult's law.



Step 4: {\color{redEffect on boiling point.


Boiling occurs when vapour pressure becomes equal to atmospheric pressure.

If vapour pressure increases, the liquid reaches atmospheric pressure at a lower temperature.

Therefore, the boiling point decreases.



Hence: \[ \boxed{Positive deviation from Raoult's law and boiling point decreases.} \] Quick Tip: Positive deviation means weaker solute-solvent interaction, higher vapour pressure and lower boiling point.

Concept:


In coordination compounds, the cation is named first and the anion is named after it.

Ligands are named before the central metal ion.

The oxidation state of the metal is written in Roman numerals.

If the complex ion is anionic, the name of the metal generally ends with ``ate''.



Step 1: {\color{redName compound \((i)\).


The compound is:
\[ [Ag(NH_3)_2][Ag(CN)_2] \]

It contains a complex cation and a complex anion.

The cation is:
\[ [Ag(NH_3)_2]^+ \]

Here, \(NH_3\) is a neutral ligand and is named as ammine.

There are two ammine ligands, so the prefix used is:
\[ di \]

Thus:
\[ [Ag(NH_3)_2]^+ = diamminesilver(I) \]



Step 2: {\color{redName the anionic part.


The anion is:
\[ [Ag(CN)_2]^- \]

The ligand \(CN^-\) is named cyanido.

There are two cyanido ligands, so the prefix is:
\[ di \]

Since the complex is anionic, silver is named as argentate.

Thus:
\[ [Ag(CN)_2]^- = dicyanidoargentate(I) \]

Therefore:
\[ [Ag(NH_3)_2][Ag(CN)_2] = diamminesilver(I) dicyanidoargentate(I) \]



Step 3: {\color{redName compound \((ii)\).


The compound is:
\[ K_3[Fe(C_2O_4)_3] \]

The cation is potassium.

The complex ion is:
\[ [Fe(C_2O_4)_3]^{3-} \]

Oxalate ion is:
\[ C_2O_4^{2-} \]

and it is named as oxalato.

There are three oxalato ligands, so the prefix used is:
\[ tris \]



Step 4: {\color{redFind oxidation state of iron.


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

Charge of three oxalate ligands:
\[ 3(-2)=-6 \]

Overall charge on complex ion:
\[ -3 \]

Therefore:
\[ x-6=-3 \]
\[ x=+3 \]

Since the complex ion is anionic, iron is named ferrate.

So:
\[ K_3[Fe(C_2O_4)_3] = potassium tris(oxalato)ferrate(III) \]



Hence: \[ \boxed{(i)\;Diamminesilver(I) dicyanidoargentate(I)} \]
\[ \boxed{(ii)\;Potassium tris(oxalato)ferrate(III)} \] Quick Tip: For anionic complexes, use metal names such as ferrate, argentate, cuprate and cobaltate.

Concept:


Ionisation isomers produce different ions in aqueous solution because different ions are present outside the coordination sphere.



Step 1: {\color{redTest for chloride ion.

\[ [Co(NH_3)_5SO_4]Cl \rightleftharpoons [Co(NH_3)_5SO_4]^+ + Cl^- \]

On adding \(AgNO_3\):
\[ Ag^+ + Cl^- \rightarrow AgCl\downarrow \]

A white precipitate of \(AgCl\) is obtained.



Step 2: {\color{redTest for sulphate ion.

\[ [Co(NH_3)_5Cl]SO_4 \rightleftharpoons [Co(NH_3)_5Cl]^{2+}+SO_4^{2-} \]

On adding \(BaCl_2\):
\[ Ba^{2+}+SO_4^{2-} \rightarrow BaSO_4\downarrow \]

A white precipitate of \(BaSO_4\) is obtained.



Hence:
\[ \boxed{ [Co(NH_3)_5SO_4]Cl gives AgCl\downarrow } \]
\[ \boxed{ [Co(NH_3)_5Cl]SO_4 gives BaSO_4\downarrow } \] Quick Tip: Ionisation isomers produce different ions in solution and therefore give different precipitation tests.

Concept:


The enhanced stability of complexes containing polydentate ligands compared to similar complexes containing monodentate ligands is called the chelate effect.



Step 1: {\color{redUnderstand chelation.


A polydentate ligand attaches to the metal ion through two or more donor atoms and forms ring structures called chelate rings.



Step 2: {\color{redState the effect.


Such chelate complexes are more stable than corresponding complexes formed by monodentate ligands.

This increased stability is known as the chelate effect.



Step 3: {\color{redGive an example.


Ethylenediamine (\(en\)) is a bidentate ligand.
\[ [Ni(en)_3]^{2+} \]

is a chelate complex.



Hence:
\[ \boxed{Chelate effect = Extra stability of complexes formed by polydentate ligands} \]

Example:
\[ \boxed{[Ni(en)_3]^{2+}} \] Quick Tip: Polydentate ligands form ring structures around the metal ion and increase the stability of the complex.

Concept:


Haloarenes are aryl halides in which the halogen atom is directly attached to an aromatic ring.

Example:
\[ C_6H_5Cl \]

Unlike haloalkanes, haloarenes do not easily undergo nucleophilic substitution.

This is because the carbon-halogen bond in haloarenes is stronger and the usual \(S_N1\) and \(S_N2\) mechanisms are not favourable.



Step 1: {\color{redReason 1: Resonance effect.


In haloarenes, the lone pair of electrons on halogen participates in resonance with the benzene ring.

Because of resonance, the \(C-X\) bond acquires partial double bond character.
\[ C-X \Rightarrow partial double bond character \]

A bond with partial double bond character is shorter and stronger than a normal single bond.

Therefore, it is difficult to break the \(C-X\) bond in haloarenes.



Step 2: {\color{redReason 2: \(sp^2\) hybridised carbon.


In haloarenes, the halogen is attached to an \(sp^2\) hybridised carbon of benzene.

An \(sp^2\) carbon has more \(s\)-character than an \(sp^3\) carbon.

More \(s\)-character makes the bond shorter and stronger.

So the aryl \(C-X\) bond is stronger than alkyl \(C-X\) bond.



Step 3: {\color{redReason 3: Phenyl cation is unstable.


For \(S_N1\) reaction, formation of carbocation is necessary.

But haloarenes would have to form phenyl cation:
\[ C_6H_5^+ \]

Phenyl cation is highly unstable.

Therefore, \(S_N1\) mechanism is not favoured.



Step 4: {\color{redReason 4: \(S_N2\) attack is difficult.


For \(S_N2\) reaction, backside attack of nucleophile is required.

But in haloarenes, the benzene ring blocks backside attack and the electron-rich ring repels nucleophiles.

Therefore, \(S_N2\) mechanism is also difficult.



Hence, haloarenes are less reactive because their \(C-X\) bond is stronger and the usual \(S_N1\) and \(S_N2\) reaction pathways are not favoured. Quick Tip: In haloarenes, resonance gives partial double bond character to \(C-X\), making the bond difficult to break.

Concept:


For an elementary reaction, the rate law can be written directly from the stoichiometric coefficients.

Given reaction:
\[ 2A+B\rightarrow 2C \]

Since the reaction takes place in one step, it is an elementary reaction.

Therefore, the rate law is:
\[ Rate=k[A]^2[B] \]

The total order of reaction is:
\[ 2+1=3 \]



Step 1: {\color{redWrite the initial rate expression.


Let the initial rate be \(r_1\).
\[ r_1=k[A]^2[B] \]



Step 2: {\color{redUnderstand the effect of decreasing volume.


Concentration is inversely proportional to volume.
\[ Concentration \propto \frac{1}{Volume} \]

If volume is decreased to one third:
\[ V_2=\frac{V_1}{3} \]

then concentration becomes three times.

So:
\[ [A]_2=3[A] \]
\[ [B]_2=3[B] \]



Step 3: {\color{redWrite the new rate expression.


New rate:
\[ r_2=k[3A]^2[3B] \]
\[ r_2=k(9[A]^2)(3[B]) \]
\[ r_2=27k[A]^2[B] \]

But:
\[ r_1=k[A]^2[B] \]

Therefore:
\[ r_2=27r_1 \]



Step 4: {\color{redCheck change in order of reaction.


The order of reaction depends on the powers of concentration terms in the rate law.
\[ Rate=k[A]^2[B] \]

Order:
\[ 2+1=3 \]

Changing volume changes concentration, but it does not change the rate law.

Therefore, the order remains the same.



Hence: \[ \boxed{Rate becomes 27 times the original rate.} \]
\[ \boxed{Order remains unchanged, i.e., 3.} \] Quick Tip: When volume is reduced to one third, concentration becomes three times. Substitute the new concentration into the rate law.

Concept:


Amino acids contain both amino \((-NH_2)\) and carboxyl \((-COOH)\) groups. Their acidic or basic character depends upon the additional functional group present in the side chain.



Step 1: {\color{redUnderstand acidic amino acids.


Acidic amino acids contain an extra carboxyl group \((-COOH)\) in their side chain.

Examples:
\[ Aspartic acid \]
\[ Glutamic acid \]



Step 2: {\color{redUnderstand basic amino acids.


Basic amino acids contain an extra amino group \((-NH_2)\) in their side chain.

Examples:
\text{Lysine,Histidine,Arginine








Hence: Acidic amino acids contain an extra \(-COOH\) group, whereas basic amino acids contain an extra \(-NH_2\) group. Quick Tip: Acidic amino acids have an extra carboxyl group, whereas basic amino acids have an extra amino group.

Concept:


Nucleosides and nucleotides are important components of nucleic acids such as DNA and RNA.



Step 1: {\color{redUnderstand nucleoside.


A nucleoside contains:
\[ Pentose Sugar+Nitrogenous Base \]

Example:
\[ Adenosine \]



Step 2: {\color{redUnderstand nucleotide.


A nucleotide contains:
\[ Pentose Sugar + Nitrogenous Base + Phosphate Group \]

Example:
\[ Adenylic Acid (AMP) \]







Hence:
\[ \boxed{Nucleoside = Sugar + Base} \]
\[ \boxed{Nucleotide = Sugar + Base + Phosphate Group} \] Quick Tip: Nucleotide = Nucleoside + Phosphate group.

Concept:


The mechanism of nucleophilic substitution depends on the structure of alkyl halide.

For \(S_N1\) reaction:
\[ Rate=k[RX] \]

The rate depends only on the concentration of alkyl halide.

For \(S_N2\) reaction:
\[ Rate=k[RX][OH^-] \]

The rate depends on both alkyl halide and nucleophile.

Also:
\[ S_N1 \Rightarrow racemisation \]
\[ S_N2 \Rightarrow inversion of configuration \]



Step 1: {\color{redIdentify compound \(X\).


The rate of reaction of \(X\) depends only on the concentration of \(X\).

Therefore, \(X\) follows \(S_N1\) mechanism.
\[ Rate=k[X] \]
\(S_N1\) mechanism is favoured by tertiary alkyl halides because they form stable tertiary carbocations.

For molecular formula \(C_4H_9Br\), the tertiary isomer is tert-butyl bromide:
\[ X=(CH_3)_3CBr \]



Step 2: {\color{redIdentify compound \(Y\).


The optically active isomer \(Y\) reacts with aqueous KOH and the rate depends on both \(Y\) and KOH.

Therefore, \(Y\) follows \(S_N2\) mechanism.
\[ Rate=k[Y][OH^-] \]

The optically active isomer of \(C_4H_9Br\) is 2-bromobutane:
\[ Y=CH_3CH(Br)CH_2CH_3 \]

The carbon attached to bromine is chiral because it is attached to four different groups:
\[ -H,\;-Br,\;-CH_3,\;-C_2H_5 \]



Step 3: {\color{redFind which compound undergoes racemisation.


Compound \(X\) undergoes \(S_N1\) reaction.

In \(S_N1\), carbocation intermediate is formed.

The carbocation is planar:
\[ sp^2 hybridised carbocation \]

So the nucleophile can attack from either side.

This gives a mixture of products with both configurations.

Therefore, racemisation takes place.
\[ X \Rightarrow racemisation \]



Step 4: {\color{redFind which compound gives inversion of configuration.


Compound \(Y\) undergoes \(S_N2\) reaction.

In \(S_N2\), nucleophile attacks from the backside.

Backside attack causes inversion of configuration.

This is known as Walden inversion.
\[ Y \Rightarrow inversion of configuration \]



Hence: \[ \boxed{X=(CH_3)_3CBr} \]
\[ \boxed{Y=CH_3CH(Br)CH_2CH_3} \]
\[ \boxed{X undergoes racemisation due to planar carbocation in S_N1.} \]
\[ \boxed{Y gives inversion due to backside attack in S_N2.} \] Quick Tip: \(S_N1\) gives racemisation through planar carbocation, while \(S_N2\) gives inversion due to backside attack.

Concept:


Phenol and anisole are activated aromatic compounds.

Phenol undergoes electrophilic substitution mainly at ortho and para positions because the \(-OH\) group is activating.

Anisole also undergoes electrophilic substitution mainly at ortho and para positions because the \(-OCH_3\) group is activating.



Step 1: {\color{redWrite Reimer-Tiemann reaction.


In Reimer-Tiemann reaction, phenol reacts with chloroform and aqueous sodium hydroxide.

The formyl group \(-CHO\) is introduced mainly at the ortho position.
\[ C_6H_5OH \xrightarrow[NaOH]{CHCl_3} o-HOC_6H_4CHO \]

The product is salicylaldehyde.
\[ o-HOC_6H_4CHO=Salicylaldehyde \]



Step 2: {\color{redWrite Kolbe's reaction.


In Kolbe reaction, sodium phenoxide reacts with carbon dioxide under pressure.

After acidification, salicylic acid is obtained.
\[ C_6H_5ONa + CO_2 \xrightarrow[pressure]{} o-HOC_6H_4COONa \]

On acidification:
\[ o-HOC_6H_4COONa + H^+ \rightarrow o-HOC_6H_4COOH \]

The final product is salicylic acid.
\[ o-HOC_6H_4COOH=Salicylic acid \]



Step 3: {\color{redWrite Friedel-Crafts acylation of anisole.


Anisole reacts with acetyl chloride in presence of anhydrous aluminium chloride.
\[ C_6H_5OCH_3 + CH_3COCl \xrightarrow{anhyd.\;AlCl_3} o- and p-methoxyacetophenone \]

The para product is major due to less steric hindrance.
\[ p-CH_3OC_6H_4COCH_3 \]



Hence: \[ \boxed{C_6H_5OH \xrightarrow[NaOH]{CHCl_3} o-HOC_6H_4CHO} \]
\[ \boxed{C_6H_5ONa + CO_2 \rightarrow o-HOC_6H_4COONa \xrightarrow{H^+} o-HOC_6H_4COOH} \]
\[ \boxed{C_6H_5OCH_3 + CH_3COCl \xrightarrow{AlCl_3} p-CH_3OC_6H_4COCH_3} \] Quick Tip: Reimer-Tiemann introduces \(-CHO\), Kolbe introduces \(-COOH\), and Friedel-Crafts acylation introduces \(-COCH_3\).

Concept:


The physical and chemical properties of carbonyl compounds and carboxylic acids are controlled by hydrogen bonding, resonance stabilisation and nucleophilicity.

Carboxylic acids show strong intermolecular association.

Aldehydes and ketones form resonance-stabilised enolate ions.

Ammonia derivatives act as nucleophiles only when the lone pair on nitrogen is available.



Step 1: {\color{redReason for higher boiling point of carboxylic acids.


Carboxylic acids contain the \(-COOH\) group.

Two molecules of carboxylic acid associate through intermolecular hydrogen bonding.

They form cyclic dimers.
\[ 2RCOOH \rightleftharpoons (RCOOH)_2 \]

Because of dimer formation, the effective molecular mass increases.

Also, strong hydrogen bonding requires more heat energy to break.

Therefore, carboxylic acids have higher boiling points than alcohols of comparable molecular masses.



Step 2: {\color{redReason for acidic nature of alpha hydrogens.


Aldehydes and ketones contain carbonyl group:
\[ >C=O \]

The hydrogen attached to the carbon adjacent to carbonyl carbon is called alpha hydrogen.

When alpha hydrogen is removed, an enolate ion is formed.
\[ RCOCH_2R' \rightarrow RCOCHR'^- + H^+ \]

This enolate ion is resonance stabilised.
\[ RCOCH^-R' \leftrightarrow RCO^- = CHR' \]

Due to resonance stabilisation of the conjugate base, alpha hydrogens are acidic.



Step 3: {\color{redReason why ammonia derivatives do not react in strongly acidic medium.


Ammonia and its derivatives have lone pair on nitrogen.

This lone pair is responsible for nucleophilic attack on carbonyl carbon.

In strongly acidic medium, ammonia or its derivative gets protonated.
\[ NH_3 + H^+ \rightarrow NH_4^+ \]

After protonation, the lone pair is no longer available for nucleophilic attack.

Therefore, nucleophilicity decreases.

Hence, nucleophilic addition to carbonyl group does not occur effectively in strongly acidic medium.



Hence: \[ \boxed{Carboxylic acids form hydrogen-bonded dimers and hence have higher boiling points.} \]
\[ \boxed{\alpha-hydrogens are acidic due to resonance stabilisation of enolate ion.} \]
\[ \boxed{In strongly acidic medium, NH_3 derivatives are protonated and lose nucleophilicity.} \] Quick Tip: Carboxylic acids form dimers, alpha hydrogens form resonance-stabilised enolates, and protonated amines are poor nucleophiles.

Concept:


Carbohydrates exist in open-chain and cyclic forms.

In cyclic forms, a new stereocentre is formed at the carbonyl carbon.

This carbon is called the anomeric carbon.

Carbohydrate molecules can join through glycosidic linkage to form disaccharides and polysaccharides.



Step 1: {\color{redDefine anomers.


Anomers are stereoisomers of cyclic sugars which differ only in configuration around the anomeric carbon.

For example, glucose exists as:
\[ \alpha-D-glucose \]

and
\[ \beta-D-glucose \]

These two forms differ only in the position of \(-OH\) group at the anomeric carbon.

Therefore, they are called anomers.



Step 2: {\color{redDefine invert sugar.


Sucrose on hydrolysis gives glucose and fructose.
\[ Sucrose+H_2O \rightarrow Glucose+Fructose \]

The equimolar mixture of glucose and fructose obtained after hydrolysis of sucrose is called invert sugar.

It is called invert sugar because the sign of optical rotation changes from dextrorotatory to laevorotatory after hydrolysis.



Step 3: {\color{redDefine glycosidic linkage.


Glycosidic linkage is the bond formed between two monosaccharide units by loss of water molecule.

It usually involves the anomeric carbon of one monosaccharide and the hydroxyl group of another monosaccharide.

The linkage is through oxygen.

For example, in maltose, two glucose units are joined by glycosidic linkage.
\[ C-O-C \]



Hence: \[ \boxed{Anomers differ only in configuration at anomeric carbon.} \]
\[ \boxed{Invert sugar is an equimolar mixture of glucose and fructose.} \]
\[ \boxed{Glycosidic linkage is an oxygen bridge joining two sugar units.} \] Quick Tip: Anomer means difference at anomeric carbon; invert sugar comes from sucrose hydrolysis; glycosidic linkage joins sugar units.

Concept:


For a first order reaction:
\[ k=\frac{2.303}{t}\log\frac{P_0}{P_t} \]

For gaseous decomposition:
\[ A(g)\rightarrow B(g)+C(g) \]

if initial pressure is \(P_0\) and total pressure at time \(t\) is \(P_{total}\), then pressure of reactant left is:
\[ P_A=2P_0-P_{total} \]



Step 1: {\color{redWrite the given data.


Initial pressure:
\[ P_0=0.30\;atm \]

Total pressure after \(30s\):
\[ P_{total}=0.50\;atm \]



Step 2: {\color{redFind pressure of reactant left.

\[ P_A=2P_0-P_{total} \]
\[ P_A=2(0.30)-0.50 \]
\[ P_A=0.60-0.50 \]
\[ P_A=0.10\;atm \]



Step 3: {\color{redApply first order rate constant formula.

\[ k=\frac{2.303}{t}\log\frac{P_0}{P_A} \]
\[ k=\frac{2.303}{30}\log\frac{0.30}{0.10} \]
\[ k=\frac{2.303}{30}\log 3 \]

Given:
\[ \log 3=0.48 \]
\[ k=\frac{2.303\times 0.48}{30} \]
\[ k=\frac{1.10544}{30} \]
\[ k=0.0368\;s^{-1} \]



Hence: \[ \boxed{k=3.68\times 10^{-2}\;s^{-1}} \] Quick Tip: For \(A(g)\rightarrow B(g)+C(g)\), pressure of reactant left is \(2P_0-P_{total}\).

Concept:


Electrochemical cells convert chemical energy into electrical energy. The relationship between equilibrium constant, electrode potential and Gibbs energy helps us understand the spontaneity of reactions.

Lead storage batteries are secondary batteries and their charging reaction is the reverse of the discharge reaction.



(i) Why is K_c related to E^\circ_{cell and not to E_{cell?




Step 1: {\color{redRelation between Gibbs energy and standard emf.


The standard Gibbs free energy change is related to standard cell potential by:
\[ \Delta G^\circ=-nFE^\circ_{cell} \]

where
\[ n=number of electrons transferred \]
\[ F=Faraday constant \]



Step 2: {\color{redRelation between Gibbs energy and equilibrium constant.


At equilibrium,
\[ \Delta G^\circ=-RT\ln K_c \]

Combining the two equations:
\[ nFE^\circ_{cell}=RT\ln K_c \]

Thus,
\[ E^\circ_{cell} = \frac{RT}{nF}\ln K_c \]



Step 3: {\color{redReason for using standard emf.

\(K_c\) is a constant defined under equilibrium and standard conditions.

Therefore it is related to the standard cell potential \(E^\circ_{cell}\).

The value of \(E_{cell}\) changes with concentration according to the Nernst equation and hence cannot be directly related to a fixed equilibrium constant.



Hence:
\(K_c\) is related to \(E^\circ_{cell}\) and not to \(E_{cell}\) because both \(K_c\) and \(E^\circ_{cell}\) correspond to standard equilibrium conditions.



(ii) Which metal will liberate hydrogen from dilute H_2SO_4?



Step 1: {\color{redRecall the criterion.


A metal can liberate hydrogen from dilute acids if its standard reduction potential is less than that of hydrogen.
\[ E^\circ(H^+/H_2)=0.00V \]



Step 2: {\color{redCompare the electrode potentials.


For metal A:
\[ E^\circ=-0.24V \]

For metal B:
\[ E^\circ=+0.80V \]

Since
\[ -0.24V < 0.00V \]

metal A is more reactive than hydrogen.

Therefore it can displace hydrogen from dilute sulphuric acid.



Hence:

Metal A will liberate hydrogen gas from dilute \(H_2SO_4\).



(iii) Cell reaction during charging of lead storage battery



Step 1: {\color{redRecall discharge reaction.


During discharge:
\[ Pb+PbO_2+2H_2SO_4 \rightarrow 2PbSO_4+2H_2O \]



Step 2: {\color{redWrite charging reaction.


Charging is exactly the reverse process.

Therefore:
\[ 2PbSO_4(s)+2H_2O(l) \rightarrow Pb(s)+PbO_2(s)+2H_2SO_4(aq) \]



Hence:
\[ 2PbSO_4(s)+2H_2O(l) \rightarrow Pb(s)+PbO_2(s)+2H_2SO_4(aq) \] Quick Tip: Remember: \[ nFE^\circ_{cell}=RT\ln K_c \] A metal with negative standard electrode potential can generally displace hydrogen from dilute acids. Charging of a lead storage battery is the reverse of its discharge reaction.

Concept:


A mercury cell is a primary cell in which the chemical reaction is irreversible.

It provides a nearly constant voltage throughout its life and is widely used in watches, hearing aids and calculators.



Step 1: {\color{redIdentify the type of battery.


Mercury cell cannot be recharged after use.

Therefore, it is classified as a:
\[ \boxed{Primary Battery} \]



Step 2: {\color{redAdvantages over dry cell.


Mercury cell has several advantages:


It provides a constant cell potential throughout its life.
Internal resistance remains low.
Voltage does not decrease rapidly during use.
It has a longer shelf life.
It is more reliable for precision electronic devices.




Step 3: {\color{redHalf-cell reactions.


At the anode:
\[ Zn(Hg)+2OH^- \rightarrow ZnO+H_2O+2e^- \]

At the cathode:
\[ HgO+H_2O+2e^- \rightarrow Hg+2OH^- \]



Step 4: {\color{redOverall cell reaction.


Adding the two half reactions:
\[ Zn(Hg)+HgO \rightarrow ZnO+Hg \]



Hence:

Mercury cell is a primary battery and its overall reaction is:
\[ Zn(s)+HgO(s) \rightarrow ZnO(s)+Hg(l) \] Quick Tip: Mercury cell is a primary battery that delivers nearly constant voltage throughout its operation, making it superior to an ordinary dry cell.

Concept:


Elevation in boiling point is:
\[ \Delta T_b=iK_bm \]

For association or dimerisation:
\[ i=1-\frac{\alpha}{2} \]

where \(\alpha\) is degree of association.



Step 1: {\color{redCalculate moles of benzoic acid.

\[ Moles=\frac{0.61}{122} \]
\[ Moles=0.005 \]



Step 2: {\color{redCalculate mass of solvent in kg.

\[ 5g=0.005kg \]



Step 3: {\color{redCalculate molality.

\[ m=\frac{0.005}{0.005} \]
\[ m=1 \]



Step 4: {\color{redCalculate Van't Hoff factor.


Degree of dimerisation:
\[ \alpha=88%=0.88 \]

For dimerisation:
\[ i=1-\frac{\alpha}{2} \]
\[ i=1-\frac{0.88}{2} \]
\[ i=1-0.44 \]
\[ i=0.56 \]



Step 5: {\color{redCalculate elevation in boiling point.

\[ \Delta T_b=iK_bm \]
\[ \Delta T_b=0.56\times 2.3\times 1 \]
\[ \Delta T_b=1.288K \]



Step 6: {\color{redCalculate boiling point of solution.

\[ T_b=46.2+1.288 \]
\[ T_b=47.488^\circ C \]



Hence: \[ \boxed{T_b=47.49^\circ C} \] Quick Tip: For dimerisation, use \(i=1-\frac{\alpha}{2}\). Do not forget to convert solvent mass into kg.

Concept:


In octahedral complexes, five \(d\)-orbitals split into two sets:
\[ t_{2g}:d_{xy},d_{yz},d_{zx} \]
\[ e_g:d_{x^2-y^2},d_{z^2} \]

The \(e_g\) orbitals point directly toward ligands and experience more repulsion.



Step 1: {\color{redAnswer part (a).


In octahedral field, ligands approach along the axes.

The orbitals:
\[ d_{x^2-y^2} \]

and
\[ d_{z^2} \]

lie along the axes.

Therefore, they face maximum repulsion from ligands.

So their energies are raised.
\[ \boxed{e_g:d_{x^2-y^2},d_{z^2}} \]



Step 2: {\color{redFind oxidation state and configuration of Co.


For both complexes, cobalt is in \(+3\) oxidation state.
\[ Co: [Ar]3d^74s^2 \]
\[ Co^{3+}:3d^6 \]



Step 3: {\color{redConfiguration in \([CoF_6]^{3-}\).

\(F^-\) is a weak field ligand.

So it forms high-spin complex.

For \(d^6\) high-spin octahedral complex:
\[ t_{2g}^4e_g^2 \]
\[ \boxed{[CoF_6]^{3-}:t_{2g}^4e_g^2} \]



Step 4: {\color{redConfiguration in \([Co(NH_3)_6]^{3+}\).

\(NH_3\) is a stronger field ligand than \(F^-\).

So pairing occurs.

For \(d^6\) low-spin octahedral complex:
\[ t_{2g}^6e_g^0 \]
\[ \boxed{[Co(NH_3)_6]^{3+}:t_{2g}^6} \]



Step 5: {\color{redExplain magnetic behaviour of nickel complexes.


In \([NiCl_4]^{2-}\):
\[ Ni^{2+}=3d^8 \]
\(Cl^-\) is weak field ligand, so electrons remain unpaired.

Thus, it is paramagnetic.

In \([Ni(CO)_4]\):
\[ Ni^0=3d^84s^2 \]

CO is strong field ligand and causes pairing.

So all electrons become paired.

Thus, it is diamagnetic.



Hence: \[ \boxed{d_{x^2-y^2} and d_{z^2} have higher energy in octahedral field.} \]
\[ \boxed{[CoF_6]^{3-}:t_{2g}^4e_g^2,\quad [Co(NH_3)_6]^{3+}:t_{2g}^6} \]
\[ \boxed{[NiCl_4]^{2-} is paramagnetic, [Ni(CO)_4] is diamagnetic.} \] Quick Tip: In octahedral complexes, \(e_g\) orbitals face ligands directly, so their energy increases.

Concept:


The hybridization and magnetic behaviour of a coordination compound depend upon the oxidation state of the central metal ion, its electronic configuration and the strength of the ligand.
\(CN^-\) is a strong field ligand. It causes pairing of electrons in the \(3d\)-orbitals.



Step 1: {\color{redFind oxidation state of iron.


The given complex is:
\[ [Fe(CN)_6]^{3-} \]

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

Since each \(CN^-\) ligand has charge \(-1\):
\[ x+6(-1)=-3 \]
\[ x-6=-3 \]
\[ x=+3 \]

Therefore, the central metal ion is:
\[ Fe^{3+} \]



Step 2: {\color{redWrite electronic configuration of iron.


Atomic number of Fe is:
\[ 26 \]

Electronic configuration of neutral Fe is:
\[ Fe:[Ar]3d^64s^2 \]

For \(Fe^{3+}\), three electrons are removed.

First, two electrons are removed from \(4s\), then one electron from \(3d\).
\[ Fe^{3+}:[Ar]3d^5 \]



Step 3: {\color{redEffect of strong field ligand \(CN^-\).

\(CN^-\) is a strong field ligand.

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

For \(Fe^{3+}\):
\[ 3d^5 \]

After pairing:
\[ 3d:\uparrow\downarrow\;\uparrow\downarrow\;\uparrow\; \square\; \square \]

Thus, two inner \(3d\)-orbitals become vacant.



Step 4: {\color{redFind hybridization.


The complex has six ligands.

Therefore, coordination number is:
\[ 6 \]

Six hybrid orbitals are required.

The vacant orbitals used are:
\[ 2(3d)+1(4s)+3(4p) \]

Therefore, hybridization is:
\[ d^2sp^3 \]

Since inner \(3d\)-orbitals are used, it is an inner orbital complex.



Step 5: {\color{redFind magnetic behaviour.


After pairing, \(Fe^{3+}\) in \([Fe(CN)_6]^{3-}\) has one unpaired electron.

A complex with unpaired electron is paramagnetic.

Therefore:
\[ [Fe(CN)_6]^{3-} \]

is paramagnetic.



Hence:
\[ \boxed{Hybridization of [Fe(CN)_6]^{3-}=d^2sp^3} \]
\[ \boxed{Magnetic behaviour: Paramagnetic due to one unpaired electron} \] Quick Tip: \(CN^-\) is a strong field ligand. It causes pairing and forms inner orbital complexes with \(d^2sp^3\) hybridization in octahedral complexes.

Concept:


Basic strength of amines in aqueous solution depends mainly on three factors:
\[ +I effect \]
\[ solvation of conjugate acid \]
\[ steric hindrance \]

In gas phase, tertiary amines may appear more basic due to stronger \(+I\) effect, but in aqueous solution, solvation becomes very important.



Step 1: {\color{redCompare methylamine and trimethylamine.


Methylamine is:
\[ CH_3NH_2 \]

Trimethylamine is:
\[ (CH_3)_3N \]

Both contain nitrogen with a lone pair.

This lone pair accepts proton and shows basic character.



Step 2: {\color{redUnderstand solvation in aqueous medium.


In water, after accepting proton:
\[ CH_3NH_2+H^+ \rightarrow CH_3NH_3^+ \]
\[ (CH_3)_3N+H^+ \rightarrow (CH_3)_3NH^+ \]

The ion \(CH_3NH_3^+\) has more hydrogen atoms attached to nitrogen.

Therefore, it can form stronger hydrogen bonding with water.

So it is better solvated.

But \((CH_3)_3NH^+\) has bulky methyl groups around nitrogen.

These methyl groups create steric hindrance and reduce solvation.



Step 3: {\color{redConclusion for basicity.


Since \(CH_3NH_3^+\) is better stabilised by solvation, \(CH_3NH_2\) accepts proton more easily in aqueous solution.

Therefore:
\[ CH_3NH_2 is stronger base than (CH_3)_3N in aqueous solution. \]



Step 4: {\color{redFind compound A.


The reaction is:
\[ CH_3CONH_2 \xrightarrow{NaOBr} A \]

This is Hofmann bromamide degradation.

In this reaction, an amide is converted into a primary amine having one carbon atom less.
\[ RCONH_2 \xrightarrow{NaOBr} RNH_2 \]

Here:
\[ CH_3CONH_2 \rightarrow CH_3NH_2 \]

So:
\[ A=CH_3NH_2 \]



Step 5: {\color{redFind compound B.


Now \(A\) reacts with benzoyl chloride in presence of base:
\[ CH_3NH_2+C_6H_5COCl \xrightarrow{Base} C_6H_5CONHCH_3+HCl \]

Base removes \(HCl\) formed during the reaction.

The product is \(N\)-methylbenzamide.

So:
\[ B=C_6H_5CONHCH_3 \]



Step 6: {\color{redIdentify compound X using Hinsberg test.


Molecular formula of \(X\) is:
\[ C_3H_9N \]

Hinsberg reagent is:
\[ C_6H_5SO_2Cl \]

A secondary amine reacts with Hinsberg reagent to give a sulphonamide which is insoluble in alkali.

A suitable secondary amine with molecular formula \(C_3H_9N\) is:
\[ CH_3NHCH_2CH_3 \]

That is ethyl methyl amine.

Many school-level keys may also write a secondary amine representation as:
\[ (CH_3)_2NH \]

But strictly, \((CH_3)_2NH\) has formula \(C_2H_7N\), not \(C_3H_9N\).

So the correct compound matching \(C_3H_9N\) is:
\[ CH3NHCH2CH3 \]



Hence:
\[ A=CH_3NH_2 \]
\[ B=C_6H_5CONHCH_3 \]
\[ X=CH_3NHCH_2CH_3 \] Quick Tip: In aqueous solution, solvation can dominate over \(+I\) effect. Hinsberg reagent identifies amines: secondary amines give products insoluble in alkali.

Concept:


Aromatic primary amines can be converted into diazonium salts.

Diazonium salts are highly useful intermediates because the diazonium group can be replaced by many groups such as \(Cl\), \(Br\), \(CN\), \(OH\), and \(H\).



Step 1: {\color{redConvert aniline into benzene diazonium chloride.


Aniline is treated with sodium nitrite and hydrochloric acid at \(0-5^\circ C\).
\[ C_6H_5NH_2 \xrightarrow[0-5^\circ C]{NaNO_2/HCl} C_6H_5N_2^+Cl^- \]

This reaction is called diazotisation.



Step 2: {\color{redConvert diazonium salt into benzonitrile.


Benzene diazonium chloride is treated with cuprous cyanide.
\[ C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN+N_2 \]

The product is benzonitrile.



Thus:
\[ C_6H_5NH_2 \xrightarrow[0-5^\circ C]{NaNO_2/HCl} C_6H_5N_2^+Cl^- \xrightarrow{CuCN} C_6H_5CN \]



Step 3: {\color{redReason for acetylation of aniline before nitration.


Aniline contains the \(-NH_2\) group.

The \(-NH_2\) group is strongly activating and ortho-para directing.

So direct nitration of aniline gives a mixture of products and may lead to oxidation or tarry products.

Also, nitration medium is strongly acidic.

In acidic medium, aniline gets protonated:
\[ C_6H_5NH_2+H^+ \rightarrow C_6H_5NH_3^+ \]

The \(-NH_3^+\) group is meta-directing.

So direct nitration gives a mixture of ortho, meta and para products.



Step 4: {\color{redHow acetylation helps.


Aniline is first acetylated to acetanilide:
\[ C_6H_5NH_2+(CH_3CO)_2O \rightarrow C_6H_5NHCOCH_3+CH_3COOH \]

The \(-NHCOCH_3\) group is less activating than \(-NH_2\).

It controls the reaction and mainly gives para product during nitration.

After nitration, hydrolysis can regenerate the \(-NH_2\) group.



Hence:

Aniline is acetylated before nitration to protect the \(-NH_2\) group, reduce its excessive activation, and avoid unwanted side products. Quick Tip: To convert aniline into benzonitrile, use diazotisation followed by \(CuCN\). Before nitration, aniline is acetylated to control reactivity and orientation.

Concept:


For an electrochemical cell:
\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

The Nernst equation at \(298K\) is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q \]

Also:
\[ \Delta G=-nFE_{cell} \]



Step 1: {\color{redIdentify anode and cathode.


Magnesium has lower reduction potential:
\[ E^\circ_{Mg^{2+}/Mg}=-2.37V \]

So magnesium acts as anode.

Silver has higher reduction potential:
\[ E^\circ_{Ag^+/Ag}=+0.80V \]

So silver acts as cathode.



Step 2: {\color{redWrite the cell reaction.


Oxidation at anode:
\[ Mg(s)\rightarrow Mg^{2+}(aq)+2e^- \]

Reduction at cathode:
\[ Ag^+(aq)+e^-\rightarrow Ag(s) \]

Multiply silver reaction by 2:
\[ 2Ag^+(aq)+2e^-\rightarrow 2Ag(s) \]

Overall reaction:
\[ Mg(s)+2Ag^+(aq)\rightarrow Mg^{2+}(aq)+2Ag(s) \]

Here:
\[ n=2 \]



Step 3: {\color{redCalculate standard cell potential.

\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
\[ E^\circ_{cell} = 0.80-(-2.37) \]
\[ E^\circ_{cell}=3.17V \]



Step 4: {\color{redWrite reaction quotient.


For the reaction:
\[ Mg(s)+2Ag^+(aq)\rightarrow Mg^{2+}(aq)+2Ag(s) \]

Solids are not included in \(Q\).
\[ Q=\frac{[Mg^{2+}]}{[Ag^+]^2} \]

Substitute:
\[ Q=\frac{0.01}{(0.001)^2} \]
\[ Q=\frac{10^{-2}}{(10^{-3})^2} \]
\[ Q=\frac{10^{-2}}{10^{-6}} \]
\[ Q=10^4 \]



Step 5: {\color{redApply Nernst equation.

\[ E_{cell} = 3.17-\frac{0.0591}{2}\log(10^4) \]
\[ \log(10^4)=4 \]
\[ E_{cell} = 3.17-\frac{0.0591}{2}\times4 \]
\[ E_{cell} = 3.17-0.1182 \]
\[ E_{cell}=3.0518V \]

Approximately:
\[ E_{cell}=3.05V \]



Step 6: {\color{redCalculate \(\Delta G\).

\[ \Delta G=-nFE_{cell} \]
\[ \Delta G=-2\times96500\times3.0518 \]
\[ \Delta G=-5,89,000\;Jmol^{-1} \]
\[ \Delta G=-5.89\times10^5\;Jmol^{-1} \]



Hence:
\[ E_{cell}=3.05V \]
\[ \Delta G=-5.89\times10^5\;Jmol^{-1} \] Quick Tip: In Nernst equation, never include solids in \(Q\). For spontaneous cell reaction, \(\Delta G\) is negative.

Concept:


The relation between standard Gibbs free energy and standard emf is:
\[ \Delta G^\circ=-nFE^\circ_{cell} \]

The Nernst equation at \(298K\) is:
\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n}\log Q \]



Step 1: {\color{redFind number of electrons transferred.


The reaction is:
\[ 2AgCl(s)+H_2(g)\rightarrow 2Ag(s)+2H^++2Cl^- \]

Hydrogen changes from oxidation state \(0\) in \(H_2\) to \(+1\) in \(H^+\).
\[ H_2\rightarrow2H^++2e^- \]

So:
\[ n=2 \]



Step 2: {\color{redCalculate standard emf.

\[ \Delta G^\circ=-nFE^\circ_{cell} \]
\[ E^\circ_{cell} = -\frac{\Delta G^\circ}{nF} \]
\[ E^\circ_{cell} = -\frac{-43500}{2\times96500} \]
\[ E^\circ_{cell} = \frac{43500}{193000} \]
\[ E^\circ_{cell}=0.225V \]



Step 3: {\color{redWrite reaction quotient.


For the reaction:
\[ 2AgCl(s)+H_2(g)\rightarrow 2Ag(s)+2H^++2Cl^- \]

Solids \(AgCl\) and \(Ag\) are not included in \(Q\).
\[ Q=\frac{[H^+]^2[Cl^-]^2}{P_{H_2}} \]

Substitute values:
\[ Q=\frac{(0.1)^2(0.2)^2}{0.4} \]
\[ Q=\frac{(0.01)(0.04)}{0.4} \]
\[ Q=\frac{0.0004}{0.4} \]
\[ Q=0.001 \]
\[ Q=10^{-3} \]



Step 4: {\color{redApply Nernst equation.

\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{2}\log Q \]
\[ E_{cell} = 0.225-\frac{0.0591}{2}\log(10^{-3}) \]
\[ \log(10^{-3})=-3 \]
\[ E_{cell} = 0.225-\frac{0.0591}{2}(-3) \]
\[ E_{cell} = 0.225+0.08865 \]
\[ E_{cell}=0.31365V \]



Hence:
\[ E_{cell}\approx0.314V \] Quick Tip: First calculate \(E^\circ_{cell}\) from \(\Delta G^\circ=-nFE^\circ_{cell}\), then apply the Nernst equation carefully.

\textcolor{red{Step 1: Concept
Distinct chemical tests are used to identify the specific nature of carbonyl compounds (aldehydes vs. ketones, and methyl ketones).

\textcolor{red{Step 2: Meaning
A negative Tollen's test indicates the compound is a ketone. A positive iodoform test confirms the presence of a methyl ketone group (\(-COCH_3\)). An Aldol condensation requires \(\alpha\)-hydrogens. A Cannizzaro reaction requires the complete absence of \(\alpha\)-hydrogens.

\textcolor{red{Step 3: Analysis

(I) A five-carbon ketone that is a methyl ketone is pentan-2-one.

(II) A five-carbon ketone that is not a methyl ketone but possesses \(\alpha\)-hydrogen atoms is pentan-3-one.

(III) A five-carbon aldehyde possessing no \(\alpha\)-hydrogens must have a highly branched adjacent carbon. This compound is 2,2-dimethylpropanal.

A methyl ketone is a ketone containing the functional group:
\[ -COCH_3. \]

Among the ketones having the molecular formula
\[ C_5H_{10}O, \]
pentan-2-one contains the
\[ -COCH_3 \]
group and therefore qualifies as a methyl ketone:
\[ CH_3COCH_2CH_2CH_3. \]

Pentan-3-one has the structure:
\[ CH_3CH_2COCH_2CH_3. \]

Since the carbonyl carbon is bonded to two ethyl groups instead of a methyl group, it is not a methyl ketone.

However, both carbon atoms adjacent to the carbonyl group contain hydrogen atoms, so pentan-3-one possesses \(\alpha\)-hydrogens.

A five-carbon aldehyde without any \(\alpha\)-hydrogen must have no hydrogen attached to the carbon adjacent to the aldehyde group.

This is possible only when the \(\alpha\)-carbon is fully substituted by carbon atoms.

The required compound is:
\[ (CH_3)_3CCHO, \]
whose IUPAC name is
\[ \boxed{2,2-dimethylpropanal}. \]

Since the \(\alpha\)-carbon carries no hydrogen atom, this aldehyde cannot undergo reactions that require \(\alpha\)-hydrogen atoms, such as aldol condensation.


\textcolor{red{Step 4: Conclusion
The structural formulas correspond exactly to the kinetic and qualitative test constraints provided.



\textcolor{red{Final Answer:

(I) Pentan-2-one: \(CH_3-CO-CH_2-CH_2-CH_3\)

(II) Pentan-3-one: \(CH_3-CH_2-CO-CH_2-CH_3\)

(III) 2,2-Dimethylpropanal: \(CH_3-C(CH_3)_2-CHO\) Quick Tip: Positive Iodoform = Methyl Ketone. Cannizzaro = Zero \(\alpha\)-hydrogens.

\textcolor{red{Step 1: Concept
Vigorous oxidation of alkyl or acyl side chains attached to an aromatic ring.

\textcolor{red{Step 2: Meaning
Any side chain containing at least one benzylic hydrogen is completely oxidized to a carboxyl group (\(-COOH\)) upon treatment with strong oxidizing agents.

\textcolor{red{Step 3: Analysis

Both acetophenone (which contains a \(-COCH_3\) group) and ethylbenzene (which contains a \(-CH_2CH_3\) group) possess benzylic hydrogen atoms. When refluxed with alkaline potassium permanganate (\(KMnO_4/KOH\)), the entire side chain is oxidized to form potassium benzoate.

Alkaline potassium permanganate is a powerful oxidizing agent.

Any alkyl side chain attached to a benzene ring can be completely oxidized provided it contains at least one benzylic hydrogen atom.

Acetophenone contains the side chain:
\[ -COCH_3, \]
in which the methyl group provides benzylic hydrogen atoms.

Ethylbenzene contains the side chain:
\[ -CH_2CH_3, \]
where the benzylic carbon also possesses hydrogen atoms.

During oxidation, the entire side chain is removed irrespective of its length.

The carbon directly attached to the benzene ring is ultimately converted into a carboxylate group.

In alkaline medium, the product formed is:
\[ \boxed{C_6H_5COOK} \]
(potassium benzoate).

On subsequent acidification, potassium benzoate is converted into benzoic acid:
\[ C_6H_5COOK+HCl \rightarrow C_6H_5COOH+KCl. \]

Thus, both acetophenone and ethylbenzene yield the same oxidation product because both possess benzylic hydrogen atoms.


\textcolor{red{Step 4: Conclusion
Subsequent acidification of this intermediate yields benzoic acid. The identical reagent works for both starting materials.



\textcolor{red{Final Answer:

(I) Acetophenone \(\xrightarrow{KMnO_4, KOH, \Delta} Potassium benzoate \xrightarrow{H_3O^+} Benzoic acid\)

(II) Ethylbenzene \(\xrightarrow{KMnO_4, KOH, \Delta} Potassium benzoate \xrightarrow{H_3O^+} Benzoic acid\) Quick Tip: Alkaline \(KMnO_4\) is the universal "eraser" for aromatic side chains, reducing them all down to a benzoic acid group as long as one benzylic H exists.

\textcolor{red{Step 1: Concept
Nucleophilic addition-elimination reaction between a carbonyl compound and an ammonia derivative.

\textcolor{red{Step 2: Meaning
Benzaldehyde reacts with 2,4-dinitrophenylhydrazine (2,4-DNP) to eliminate a molecule of water and form a colored hydrazone derivative.

\textcolor{red{Step 3: Analysis

The reaction involves the carbonyl oxygen (\(=O\)) of benzaldehyde (\(C_6H_5CHO\)) and the two hydrogen atoms of hydrazine (\(NH_2NH_2\)). Removal of one molecule of water links the carbonyl carbon directly to nitrogen through a double bond, producing a hydrazone.

Benzaldehyde reacts with hydrazine in a condensation reaction.

The nucleophilic nitrogen atom of hydrazine attacks the electrophilic carbonyl carbon of benzaldehyde.

An unstable addition intermediate is first formed.

This intermediate subsequently eliminates one molecule of water:
\[ C_6H_5CHO+NH_2NH_2 \rightarrow C_6H_5CH=NNH_2+H_2O. \]

The product formed contains the characteristic functional group:
\[ -CH=N-NH_2, \]
which is known as a hydrazone.

The carbonyl oxygen atom and two hydrogen atoms from hydrazine together constitute the water molecule removed during the condensation.

Hydrazone formation is widely used for the identification and characterization of aldehydes and ketones.


\textcolor{red{Step 4: Conclusion
This yields the specific imine structure known as a hydrazone.



\textcolor{red{Final Answer: \(C_6H_5-CH=N-NH-C_6H_3(NO_2)_2\) Quick Tip: To draw hydrazones, simply remove the \(=O\) from the carbonyl and \(H_2\) from the hydrazine, then connect the remaining fragments with a double bond (\(C=N\)).

\textcolor{red{Step 1: Concept
Reactivity of carbonyl compounds towards nucleophilic addition reactions.

\textcolor{red{Step 2: Meaning
Reactivity is primarily governed by steric hindrance (crowding around the carbonyl carbon) and the inductive effect (electron donation by alkyl groups).

\textcolor{red{Step 3: Analysis

Aldehydes are generally more reactive than ketones because they possess only one alkyl group, resulting in less steric hindrance and a smaller positive inductive (\(+I\)) effect. Consequently, the carbonyl carbon is more electrophilic. Acetaldehyde is therefore more reactive than ketones. Among the ketones, di-tert-butyl ketone contains extremely bulky substituents that severely hinder the approach of the nucleophile compared with acetone.

Nucleophilic addition reactions occur at the electrophilic carbonyl carbon atom.

The reactivity of the carbonyl group depends mainly on:

The electrophilic character of the carbonyl carbon.
Steric hindrance around the carbonyl group.


Aldehydes possess only one alkyl group, whereas ketones possess two alkyl groups.

Therefore, aldehydes experience less steric hindrance and a weaker electron-donating (\(+I\)) effect than ketones.

As a result, the carbonyl carbon in aldehydes carries a larger partial positive charge and is attacked more readily by nucleophiles.

Among the aldehydes, acetaldehyde is highly reactive because it contains only one small methyl group.

Acetone contains two methyl groups, which reduce its reactivity slightly through both steric effects and the \(+I\) effect.

Di-\textit{tert-butyl ketone possesses two bulky \textit{tert-butyl groups surrounding the carbonyl carbon.

These bulky groups strongly hinder the approach of nucleophiles such as
\[ CN^-, \]
making nucleophilic addition extremely difficult.

Hence, the order of reactivity is:
\[ \boxed{Acetaldehyde>Acetone>Di-\textit{tert-butyl ketone.} \]


\textcolor{red{Step 4: Conclusion
Therefore, di-tert-butyl ketone is the least reactive.



\textcolor{red{Final Answer: Di-tert. butyl ketone \(<\) Acetone \(<\) Acetaldehyde Quick Tip: Less steric crowding and less \(+I\) effect = Higher reactivity towards nucleophiles. Aldehydes \textgreater\ Ketones.

\textcolor{red{Step 1: Concept
Identification of the carboxylic acid functional group using mild bases.

\textcolor{red{Step 2: Meaning
Carboxylic acids are sufficiently acidic to decompose sodium bicarbonate, whereas esters (like ethyl benzoate) are neutral and do not react.

\textcolor{red{Step 3: Analysis

When an aqueous solution of sodium bicarbonate (\(NaHCO_3\)) is added to benzoic acid, an acid--base reaction occurs, rapidly liberating carbon dioxide gas in the form of brisk effervescence. Ethyl benzoate, lacking an acidic proton, shows no such reaction.

Benzoic acid contains the acidic carboxyl group:
\[ -COOH. \]

Sodium bicarbonate is a weak base that reacts readily with carboxylic acids.

The reaction is:
\[ C_6H_5COOH+NaHCO_3 \rightarrow C_6H_5COONa+CO_2+H_2O. \]

Carbon dioxide gas is evolved rapidly, producing brisk effervescence.

This test is commonly used to distinguish carboxylic acids from many other organic compounds.

Ethyl benzoate is an ester and does not contain a replaceable acidic hydrogen atom.

Therefore, it does not react with sodium bicarbonate and no effervescence is observed.

Hence, the bicarbonate test provides a simple method to distinguish benzoic acid from ethyl benzoate.


\textcolor{red{Step 4: Conclusion
The visual confirmation of gas bubbles makes this an effective distinguishing test.



\textcolor{red{Final Answer: Sodium bicarbonate (\(NaHCO_3\)) test. Add aqueous \(NaHCO_3\) to both compounds. Benzoic acid will produce a brisk effervescence of \(CO_2\) gas. Ethyl benzoate will not react. Quick Tip: The \(NaHCO_3\) test is the standard qualitative chemical test for the \(-COOH\) group.

\textcolor{red{Step 1: Concept
Controlled partial reduction of a nitrile group (\(-CN\)) to an aldehyde group (\(-CHO\)).

\textcolor{red{Step 2: Meaning
A strong reducing agent would reduce the nitrile all the way to an amine. A specialized, milder reagent is required to stop the reduction at the intermediate imine stage.

\textcolor{red{Step 3: Analysis

Two standard chemical methods convert nitriles into aldehydes. The Stephen reaction utilizes stannous chloride and hydrochloric acid (\(SnCl_2/HCl\)) to form an imine hydrochloride, which is subsequently hydrolysed with water. Alternatively, DIBAL-H (diisobutylaluminium hydride) selectively reduces the nitrile to an imine, followed by aqueous hydrolysis.

Nitriles can be selectively converted into aldehydes without complete reduction to primary amines.

One important method is the Stephen reduction.

In this reaction, the nitrile is treated with stannous chloride and concentrated hydrochloric acid:
\[ RCN \xrightarrow{SnCl_2/HCl} RCH=NH\cdot HCl. \]

The intermediate imine hydrochloride is then hydrolysed with water:
\[ RCH=NH\cdot HCl+H_2O \rightarrow RCHO+NH_4Cl. \]

Another useful method employs
\[ \boxed{DIBAL-H} \]
(diisobutylaluminium hydride).

DIBAL-H selectively reduces the nitrile to an imine intermediate under controlled low-temperature conditions.

Subsequent aqueous hydrolysis converts the imine into the corresponding aldehyde:
\[ RCN \xrightarrow{DIBAL-H} RCH=NH \xrightarrow{H_2O} RCHO. \]

Both methods stop the reduction at the aldehyde stage and prevent complete reduction to the corresponding primary amine.


\textcolor{red{Step 4: Conclusion
Either reagent accurately satisfies the conversion requirement.



\textcolor{red{Final Answer: DIBAL-H (Diisobutylaluminium hydride) followed by \(H_2O\). (Alternatively, \(SnCl_2\) and \(HCl\) followed by \(H_3O^+\), known as Stephen reduction). Quick Tip: DIBAL-H is an excellent reagent for selectively stopping the reduction of nitriles and esters exactly at the aldehyde stage.

\textcolor{red{Step 1: Concept
Oxidation of secondary alcohols.

\textcolor{red{Step 2: Meaning
Chromic anhydride (\(CrO_3\)) in an acidic medium (Jones reagent) acts as a strong oxidizing agent.

\textcolor{red{Step 3: Analysis

The provided compound is cyclohexanol, a saturated six-membered cyclic alcohol containing a secondary hydroxyl (\(-OH\)) group. When a secondary alcohol is oxidized with chromium trioxide (\(CrO_3\)), it loses two hydrogen atoms (one from the hydroxyl group and one from the \(\alpha\)-carbon) to form the corresponding ketone.

Cyclohexanol is a cyclic alcohol having the molecular formula:
\[ C_6H_{11}OH. \]

The carbon atom bearing the hydroxyl group is attached to two other carbon atoms.

Therefore, cyclohexanol is classified as a
\[ \boxed{secondary alcohol}. \]

Chromium trioxide (\(CrO_3\)) is a powerful oxidizing agent commonly used for the oxidation of alcohols.

During oxidation, one hydrogen atom is removed from the hydroxyl group and another hydrogen atom is removed from the carbon bearing the hydroxyl group.

The removal of these two hydrogen atoms forms a carbonyl group:
\[ -CHOH- \longrightarrow -CO-. \]

Thus, cyclohexanol is oxidized to cyclohexanone:
\[ \boxed{Cyclohexanol \xrightarrow{CrO_3} Cyclohexanone.} \]

Unlike primary alcohols, secondary alcohols are normally oxidized only up to ketones because further oxidation would require breaking a carbon--carbon bond.


\textcolor{red{Step 4: Conclusion
The saturated ring remains fully intact, and the secondary \(-OH\) transforms into a carbonyl \(=O\), yielding cyclohexanone.



\textcolor{red{Final Answer: 'X' is Cyclohexanone. (A six-membered saturated carbon ring with a double-bonded oxygen attached to one of the carbons). Quick Tip: \(CrO_3\) efficiently oxidizes \(1^\circ\) alcohols to carboxylic acids, and \(2^\circ\) alcohols exclusively to ketones.

\textcolor{red{Step 1: Concept
Standard electrode potentials (\(E^\circ\)) of 3d transition metals.

\textcolor{red{Step 2: Meaning
The overall \(E^\circ\) value is determined by the net sum of enthalpy of sublimation, ionization enthalpy, and hydration enthalpy.

\textcolor{red{Step 3: Analysis

(I) The irregular trend across the first transition series arises from the irregular variation in the sum of the first and second ionization enthalpies together with the sublimation enthalpies.

(II) For copper, the high energy required for atomization and ionization is not sufficiently compensated by its hydration enthalpy, resulting in a positive standard reduction potential.

(III) Manganese attains the exceptionally stable half-filled \(3d^5\) configuration in the \(+2\) oxidation state, making the formation of \(Mn^{2+}\) highly favourable.

The standard electrode potentials of transition elements do not show a regular increasing or decreasing trend across the series.

This irregularity arises because several energy factors contribute simultaneously, including:

Sublimation enthalpy.
First and second ionization enthalpies.
Hydration enthalpy of the metal ion.


The combined variation of these quantities produces the observed irregular pattern of standard electrode potentials.

In the case of copper, considerable energy is required to convert the metal atom into
\[ Cu^{2+}, \]
owing to its relatively high sublimation and ionization enthalpies.

Although hydration releases energy, it is insufficient to compensate for these large energy requirements.

Consequently, copper possesses a positive standard reduction potential:
\[ E^\circ(Cu^{2+}/Cu)=+0.34\;V. \]

Manganese, on the other hand, forms the particularly stable
\[ Mn^{2+}(3d^5) \]
ion.

The half-filled \(3d^5\) configuration possesses extra stability due to exchange energy and symmetrical electron distribution.

Therefore, formation of
\[ Mn^{2+} \]
is highly favourable, giving manganese a comparatively large tendency to undergo oxidation.


\textcolor{red{Step 4: Conclusion
The delicate balance of these specific thermodynamic energy terms dictates the observed standard potentials.



\textcolor{red{Final Answer:

(I) Due to irregular variations in ionization enthalpies and sublimation enthalpies across the series.

(II) Its high enthalpies of atomization and ionization are not fully compensated by its hydration enthalpy.

(III) Due to the extra thermodynamic stability of the exactly half-filled \(d^5\) configuration of the \(Mn^{2+}\) ion. Quick Tip: Positive \(E^\circ\) for Cu means it cannot displace \(H_2\) gas from acids. Half-filled \(d^5\) always grants extra stability.

\textcolor{red{Step 1: Concept
Redox behavior of the permanganate ion (\(MnO_4^-\)) is strictly dependent on the pH of the medium.

\textcolor{red{Step 2: Meaning
It oxidizes iodide (\(I^-\)) to different products based entirely on whether the environment is acidic or alkaline.

\textcolor{red{Step 3: Analysis

In an acidic medium, iodide ions (\(I^-\)) are oxidized to iodine (\(I_2\)), while permanganate ions (\(MnO_4^-\)) are reduced to \(Mn^{2+}\). In a neutral or faintly alkaline medium, iodide ions are oxidized further to iodate (\(IO_3^-\)), while permanganate ions are reduced to manganese dioxide (\(MnO_2\)).

Potassium permanganate is a powerful oxidizing agent whose reduction products depend upon the reaction medium.

In acidic solution, permanganate ions are reduced according to:
\[ MnO_4^-+8H^++5e^- \rightarrow Mn^{2+}+4H_2O. \]

Simultaneously, iodide ions are oxidized:
\[ 2I^- \rightarrow I_2+2e^-. \]

Hence, iodine is liberated in acidic medium.

In neutral or faintly alkaline solution, the reduction half-reaction becomes:
\[ MnO_4^-+2H_2O+3e^- \rightarrow MnO_2+4OH^-. \]

Under these conditions, iodide ions undergo stronger oxidation and are converted into iodate ions:
\[ I^- \rightarrow IO_3^-. \]

Thus, the oxidation products of iodide and the reduction products of permanganate depend strongly upon the pH of the reaction medium.


\textcolor{red{Step 4: Conclusion
Balancing the specific electron transfers in both media yields the respective balanced ionic equations.



\textcolor{red{Final Answer:

Acidic solution: \(2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O\)

Alkaline solution: \(2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-\) Quick Tip: Acidic medium yields \(I_2\) (iodine). Alkaline/neutral medium yields \(IO_3^-\) (iodate).

\textcolor{red{Step 1: Concept
Distinct physical and chemical properties of f-block elements based on electronic configuration.

\textcolor{red{Step 2: Meaning
Stability of empty (\(f^0\)), half-filled (\(f^7\)), or fully filled (\(f^{14}\)) subshells dictates the unusual oxidation states.

\textcolor{red{Step 3: Analysis

Cerium (Ce) loses four electrons to attain the highly stable noble-gas configuration with an empty \(4f\) shell (\(4f^0\)), while europium (Eu) loses two electrons to achieve the particularly stable half-filled \(4f^7\) configuration.

Lanthanoids generally exhibit the oxidation state:
\[ +3. \]

However, certain lanthanoids display exceptional oxidation states because of the extra stability associated with specific \(4f\) electron configurations.

Cerium readily loses four electrons to form:
\[ Ce^{4+}. \]

This produces the stable electronic configuration:
\[ 4f^0, \]
corresponding to an empty \(4f\) subshell.

Europium preferentially forms:
\[ Eu^{2+}, \]
by losing only two electrons.

The resulting electronic configuration is:
\[ 4f^7, \]
which represents a completely half-filled \(4f\) subshell.

Half-filled and completely empty subshells possess extra stability due to symmetrical electron distribution and exchange energy.

Therefore, cerium commonly exhibits the
\[ +4 \]
oxidation state, whereas europium commonly exhibits the
\[ +2 \]
oxidation state.



\textcolor{red{Step 4: Conclusion
These specific electronic configurations provide direct reasoning for these observed oxidation states.



\textcolor{red{Final Answer:

(I) Cerium (Ce)

(II) Europium (Eu) Quick Tip: Cerium likes +4 to empty its f-shell (\(f^0\)). Europium likes +2 to half-fill its f-shell (\(f^7\)).

\textcolor{red{Step 1: Concept
Catalytic properties of d-block elements.

\textcolor{red{Step 2: Meaning
Catalysts lower the activation energy of a reaction by providing an alternative pathway.

\textcolor{red{Step 3: Analysis

Transition metals act as excellent catalysts because they possess variable oxidation states and provide large surface areas for adsorption of reactant molecules.

Transition metals exhibit several oxidation states because both the \((n-1)d\) and \(ns\) electrons can participate in bonding.

This enables them to form unstable intermediate oxidation states during chemical reactions.

These intermediate species provide an alternative reaction pathway with a lower activation energy.

Consequently, the reaction proceeds at a much faster rate.

In addition, transition metals possess large surface areas, particularly when finely divided.

Reactant molecules become adsorbed on the metal surface.

Adsorption weakens existing chemical bonds and brings reactant molecules into close proximity, increasing the frequency of effective collisions.

Since the activation energy decreases while the catalyst itself remains chemically unchanged, transition metals are highly effective catalysts in numerous industrial processes.


\textcolor{red{Step 4: Conclusion
These features facilitate the breaking and forming of bonds during chemical reactions.



\textcolor{red{Final Answer: Because they exhibit multiple variable oxidation states to form unstable intermediates and provide a large solid surface area for the adsorption of reactant molecules. Quick Tip: Variable valency (for intermediates) + Large surface area (for adsorption) = Good Catalyst.

\textcolor{red{Step 1: Concept
Melting points and metallic bonding in transition metals.

\textcolor{red{Step 2: Meaning
Stronger metallic bonds require more thermal energy to break, resulting in a higher melting point.

\textcolor{red{Step 3: Analysis

Chromium (\(Cr\)) has the electronic configuration \(3d^5\,4s^1\), containing six unpaired electrons that contribute to strong metallic bonding. Manganese (\(Mn\)) has the configuration \(3d^5\,4s^2\); its \(3d\) electrons are more localized and participate less effectively in metallic bonding.

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

Chromium has the electronic configuration:
\[ 3d^5\,4s^1. \]

All six valence electrons participate effectively in metallic bonding.

The presence of many bonding electrons results in exceptionally strong metallic bonds.

Consequently, chromium possesses a very high melting point and high enthalpy of atomization.

Manganese has the electronic configuration:
\[ 3d^5\,4s^2. \]

The half-filled \(3d^5\) subshell is particularly stable and its electrons are relatively less available for metallic bonding.

Therefore, manganese forms comparatively weaker metallic bonds than chromium.

This explains why chromium has a higher melting point and greater hardness than manganese.



\textcolor{red{Step 4: Conclusion
The stronger metallic bonding in Cr directly leads to its higher melting point compared to Mn.



\textcolor{red{Final Answer: Cr has more unpaired electrons (\(3d^5 4s^1\)) participating in strong interatomic metallic bonding compared to Mn (\(3d^5 4s^2\)), resulting in a higher melting point. Quick Tip: Strong metallic bonds require a high number of unpaired electrons. Cr has the maximum (6) in the 3d series.

\textcolor{red{Step 1: Concept
Thermodynamic stability of the permanganate ion in acidic media.

\textcolor{red{Step 2: Meaning
\(KMnO_4\) is a strong oxidizing agent that can slowly react with the solvent (water) if left standing.

\textcolor{red{Step 3: Analysis

Acidified potassium permanganate is thermodynamically unstable and slowly oxidizes water to oxygen gas on standing. During this process, permanganate ions are reduced to manganese dioxide while water is oxidized to oxygen.

Potassium permanganate is a very powerful oxidizing agent.

Even in the absence of another reducing agent, acidified potassium permanganate slowly decomposes on standing.

In this process, water molecules undergo oxidation:
\[ 2H_2O \rightarrow O_2+4H^++4e^-. \]

Simultaneously, permanganate ions undergo reduction to manganese dioxide:
\[ MnO_4^- \rightarrow MnO_2. \]

As a result, oxygen gas is gradually evolved.

The formation of brown manganese dioxide causes the purple colour of the permanganate solution to fade with time.

Therefore, acidified potassium permanganate solutions are not stable for prolonged storage and are generally prepared fresh before use in analytical and laboratory work.



\textcolor{red{Step 4: Conclusion
This slow decomposition is an example of a redox reaction.



\textcolor{red{Final Answer: It slowly decomposes to release oxygen gas. Equation: \(4MnO_4^- + 4H^+ \rightarrow 4MnO_2 + 2H_2O + 3O_2\). This is a redox (decomposition) reaction. Quick Tip: Acidic \(KMnO_4\) slowly oxidizes water to \(O_2\) over time.