CBSE Class 12 Chemistry Question Paper 2026 with Solutions (Set 1 - 56/3/1)

Concept:


Nucleophilic substitution reactions are easier when the carbon-halogen bond can break easily.

In alkyl halides and benzyl halides, the carbon-halogen bond is relatively easier to break.

But in aryl halides, the halogen is directly attached to the benzene ring.

In aryl halides, the \(C-Cl\) bond gets partial double bond character due to resonance.



Step 1: Identify the type of halide in each option.


Option (A), \(C_6H_5CH_2Cl\), is benzyl chloride.

It is highly reactive towards nucleophilic substitution because the benzyl carbocation is resonance stabilized.
\[ C_6H_5CH_2Cl \Rightarrow benzyl halide \]

Option (C), \(CH_3Cl\), is methyl chloride.

It can undergo \(S_N2\) reaction.

Option (D), chlorocyclohexane, is an alkyl halide.

Option (B), \(p\)-chlorotoluene, is an aryl halide.
\[ p-chlorotoluene \Rightarrow aryl chloride \]



Step 2: Explain why aryl chloride is least reactive.


In aryl chloride, chlorine is directly attached to the benzene ring.

The lone pair on chlorine participates in resonance with the benzene ring.

Because of this, the \(C-Cl\) bond develops partial double bond character.
\[ C-Cl \Rightarrow partial double bond character \]

This makes the bond shorter and stronger.

Therefore, breaking this bond becomes difficult.



Step 3: Final conclusion.


Since \(p\)-chlorotoluene is an aryl halide, it is least reactive towards nucleophilic substitution.

Hence: \[ \boxed{(B) p-Chlorotoluene} \]

which is (B) Fig B Quick Tip: Aryl halides are least reactive towards nucleophilic substitution because the \(C-X\) bond has partial double bond character due to resonance.

Concept:


Dehydration of alcohol means removal of water from alcohol to form alkene.

In acid-catalysed dehydration of alcohol, the reaction generally proceeds through three steps.

First, alcohol gets protonated.

Second, water leaves from the protonated alcohol.

Third, deprotonation gives alkene.

The slowest step is the removal of water, which forms a carbocation.



Step 1: Protonation of alcohol.


Alcohol first reacts with acid.
\[ ROH+H^+ \rightarrow ROH_2^+ \]

The product is protonated alcohol.

This makes water a good leaving group.



Step 2: Loss of water in slow step.


The protonated alcohol loses water.
\[ ROH_2^+ \rightarrow R^+ + H_2O \]

This step forms carbocation.

This is the slowest step because bond breaking occurs here.



Step 3: Formation of alkene.


The carbocation loses a proton to form alkene.
\[ R^+ \rightarrow Alkene \]

Therefore, the intermediate formed during the slowest step is carbocation.

Hence: \[ \boxed{(D) Carbocation} \] Quick Tip: In acid-catalysed dehydration of alcohol, the slowest step is loss of water and formation of carbocation.

Concept:


Oxidation number of central metal ion is calculated by using the charges of ligands and counter ions.

The ligand \(en\), ethylenediamine, is a neutral ligand.
\[ en = 0 \]

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



Step 1: Find total charge due to sulphate ions.


The compound is:
\[ [Co(en)_3]_2(SO_4)_3 \]

There are three sulphate ions.
\[ 3 \times (-2)=-6 \]

So the two complex cations must have total charge:
\[ +6 \]



Step 2: Find charge on one complex ion.


There are two complex ions.
\[ 2[Co(en)_3]^{x+}=+6 \]

So charge on one complex ion is:
\[ +3 \]



Step 3: Find oxidation state of Co.


Since \(en\) is neutral:
\[ x+3(0)=+3 \]
\[ x=+3 \]

Hence: \[ \boxed{(A) +3} \] Quick Tip: Ethylenediamine \((en)\) is neutral. Do not assign negative charge to it while calculating oxidation number.

Concept:


According to Werner's theory, a central metal atom or ion has two types of valencies:
\[ Primary valency \]

and
\[ Secondary valency \]

Primary valency corresponds to oxidation state.

Secondary valency corresponds to coordination number.



Step 1: Understand primary valency.


Primary valency is ionisable.

It is satisfied by negative ions.

For example:
\[ [Co(NH_3)_6]Cl_3 \]

Here, three chloride ions satisfy the primary valency of cobalt.



Step 2: Understand secondary valency.


Secondary valency is non-ionisable.

It is satisfied by ligands, which may be neutral molecules or negative ions.

Secondary valency is equal to coordination number.



Step 3: Choose correct statement.


Primary valencies are satisfied by negative ions.

Hence: \[ \boxed{(C) are satisfied by negative ions} \] Quick Tip: Werner theory: Primary valency = oxidation state = ionisable; Secondary valency = coordination number = non-ionisable.

Concept:


Hinsberg reagent is used to distinguish primary, secondary and tertiary amines.

Hinsberg reagent is:
\[ C_6H_5SO_2Cl \]

It is also called benzene sulphonyl chloride.



Step 1: Identify the types of amines.

\[ (C_2H_5)_2NH \]

is a secondary amine.
\[ (C_2H_5)_3N \]

is a tertiary amine.



Step 2: Apply Hinsberg test.


Secondary amine reacts with Hinsberg reagent to give sulphonamide.

This product is insoluble in alkali because it does not contain acidic hydrogen attached to nitrogen.

Tertiary amine does not react with Hinsberg reagent in the same way because it has no \(N-H\) bond.



Step 3: Choose the reagent.


Therefore, the reagent used is:
\[ C_6H_5SO_2Cl \]

Hence: \[ \boxed{(B) C_6H_5SO_2Cl} \] Quick Tip: Hinsberg reagent \((C_6H_5SO_2Cl)\) is used to distinguish primary, secondary and tertiary amines.

Concept:


Basic strength of amines depends on the availability of the lone pair of electrons on nitrogen.

If the lone pair is easily available for donation, the compound is more basic.

If the lone pair is involved in resonance, basicity decreases.



Step 1: Compare aniline-type compounds.


In aniline, the lone pair on nitrogen is delocalised into the benzene ring by resonance.
\[ C_6H_5NH_2 \]

Because the lone pair is less available for protonation, aniline is less basic than aliphatic amines.



Step 2: Effect of substituents on aniline.


In \(p\)-nitroaniline, the \(-NO_2\) group is strongly electron withdrawing.

It decreases electron density on nitrogen.

So it is very weakly basic.
\[ -NO_2 \Rightarrow -I,\;-R \]

In \(p\)-toluidine, the \(-CH_3\) group is electron releasing.

It increases basicity compared with aniline, but the nitrogen lone pair is still involved in resonance.



Step 3: Compare benzylamine.


Benzylamine is:
\[ C_6H_5CH_2NH_2 \]

Here, the \(-NH_2\) group is not directly attached to the benzene ring.

The lone pair on nitrogen is not delocalised into the ring.

Therefore, the lone pair is more available for protonation.

So benzylamine is the strongest base among the given options.



Hence: \[ \boxed{(B) Benzylamine} \] Quick Tip: Amines with nitrogen directly attached to benzene are less basic due to resonance. Benzylamine is more basic because \(-NH_2\) is not directly attached to the ring.

Concept:


Chromate and dichromate ions exist in equilibrium depending upon the pH of the medium.

In acidic medium, chromate ion changes to dichromate ion.

In alkaline medium, dichromate ion changes to chromate ion.



Step 1: Write dichromate-chromate equilibrium.


The equilibrium is:
\[ Cr_2O_7^{2-}+2OH^- \rightarrow 2CrO_4^{2-}+H_2O \]



Step 2: Understand colour and medium.


Dichromate ion is orange.
\[ Cr_2O_7^{2-} \Rightarrow orange \]

Chromate ion is yellow.
\[ CrO_4^{2-} \Rightarrow yellow \]

In alkaline medium, dichromate converts into chromate.



Step 3: Final answer.


Therefore:
\[ Cr_2O_7^{2-} \xrightarrow{alkaline medium} CrO_4^{2-} \]

Hence: \[ \boxed{(B) CrO_4^{2-}} \] Quick Tip: Dichromate changes to chromate in alkaline medium; chromate changes to dichromate in acidic medium.

Concept:


Azeotropes are constant boiling mixtures.

If an azeotrope has boiling point lower than both pure components, it is called a minimum boiling azeotrope.

Minimum boiling azeotropes show positive deviation from Raoult's law.



Step 1: Understand the given condition.


The question says that the boiling point of water-ethanol azeotropic mixture is less than that of pure water and pure ethanol.

So it is a minimum boiling azeotrope.
\[ T_b(mixture) < T_b(pure components) \]



Step 2: Relate boiling point with vapour pressure.


Lower boiling point means higher vapour pressure.

If the observed vapour pressure is greater than expected by Raoult's law, the solution shows positive deviation.
\[ P_{observed}>P_{Raoult} \]



Step 3: Final conclusion.


Since the azeotrope has lower boiling point, it shows positive deviation from Raoult's law.

Hence: \[ \boxed{(A) Positive deviation from Raoult's law} \] Quick Tip: Minimum boiling azeotrope shows positive deviation; maximum boiling azeotrope shows negative deviation.

Concept:


Molality means number of moles of solute present in \(1kg\) of solvent.

A \(2.0\) molal aqueous solution means:
\[ 2.0\;mol solute in 1kg water \]

Mole fraction of solute is:
\[ X_{solute}=\frac{n_{solute}}{n_{solute}+n_{solvent}} \]



Step 1: Find moles of solute.


Since the solution is \(2.0\) molal:
\[ n_{solute}=2 \]



Step 2: Find moles of water.


Mass of water:
\[ 1kg=1000g \]

Molar mass of water:
\[ 18g mol^{-1} \]
\[ n_{water}=\frac{1000}{18} \]
\[ n_{water}=55.56 \]



Step 3: Calculate mole fraction of solute.

\[ X_{solute}=\frac{2}{2+55.56} \]
\[ X_{solute}=\frac{2}{57.56} \]
\[ X_{solute}=0.0347 \]



Hence: \[ \boxed{(C) 0.0347} \] Quick Tip: For molality questions, always take \(1kg\) solvent as the base.

Concept:


In electrolysis of aqueous sodium chloride solution, the discharge depends on concentration.

In concentrated NaCl solution, chloride ions are discharged at anode to give chlorine gas.

But in very dilute NaCl solution, water is oxidised at anode and oxygen gas is evolved.

At the cathode, water is reduced to hydrogen gas instead of sodium ion being reduced to sodium metal.



Step 1: Reaction at cathode.


At cathode, reduction takes place.

Sodium ion is not discharged easily because sodium has very high tendency to remain as \(Na^+\) in aqueous solution.

Instead, water is reduced.
\[ 2H_2O+2e^- \rightarrow H_2+2OH^- \]

So hydrogen gas is evolved at cathode.



Step 2: Reaction at anode.


At anode, oxidation takes place.

Since the solution is very dilute, water is oxidised more preferably than chloride ion.
\[ 2H_2O \rightarrow O_2+4H^+ +4e^- \]

So oxygen gas is evolved at anode.



Step 3: Important exam note.


Strictly, both of the following statements are correct:
\[ Oxygen gas is evolved at anode \]

and
\[ Hydrogen gas is evolved at cathode \]

Therefore, both (C) and (D) are chemically correct.

If the question expects one answer due to the phrase ``very dilute'', then the distinguishing answer is usually:
\[ \boxed{(C) O_2 gas is evolved at anode} \]

But do not ignore that cathodic hydrogen evolution is also correct.



Hence: \[ \boxed{(C) and (D) are chemically correct.} \] Quick Tip: In very dilute NaCl solution, water gives \(O_2\) at anode and \(H_2\) at cathode. In concentrated brine, \(Cl_2\) is evolved at anode.

Concept:


For a balanced chemical reaction, rate of reaction is written by dividing the rate of change of concentration by the stoichiometric coefficient.

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

the rate relation is:
\[ -\frac{1}{a}\frac{d[A]}{dt} = +\frac{1}{b}\frac{d[B]}{dt} \]

Reactant concentration decreases, so negative sign is used for reactant.

Product concentration increases, so positive sign is used for product.



Step 1: Write the given reaction.

\[ 3A \rightarrow 2B \]

Here, coefficient of \(A\) is \(3\), and coefficient of \(B\) is \(2\).



Step 2: Write the rate expression.

\[ -\frac{1}{3}\frac{d[A]}{dt} = +\frac{1}{2}\frac{d[B]}{dt} \]



Step 3: Find \(+\dfrac{d[B]}{dt}\).


Multiply both sides by \(2\):
\[ +\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt} \]



Hence: \[ \boxed{+\frac{d[B]}{dt} = -\frac{2}{3}\frac{d[A]}{dt}} \]
\[ \boxed{(B)} \] Quick Tip: For rate expression, always divide concentration change by stoichiometric coefficient.

Concept:


Disaccharides on hydrolysis give monosaccharide units.

Different disaccharides give different monosaccharides.
\[ Sucrose \rightarrow Glucose+Fructose \]
\[ Lactose \rightarrow Glucose+Galactose \]
\[ Maltose \rightarrow Glucose+Glucose \]



Step 1: Check sucrose.


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

So sucrose does not give only glucose.



Step 2: Check lactose.


Lactose gives glucose and galactose.
\[ Lactose+H_2O \rightarrow Glucose+Galactose \]

So lactose also does not give only glucose.



Step 3: Check maltose.


Maltose is made up of two glucose units.

On hydrolysis:
\[ Maltose+H_2O \rightarrow 2Glucose \]

Therefore, maltose gives only glucose.



Hence: \[ \boxed{(D) Maltose} \] Quick Tip: Maltose gives only glucose on hydrolysis. Sucrose gives glucose and fructose; lactose gives glucose and galactose.

Concept:


Zirconium \((Zr)\) and Hafnium \((Hf)\) are transition elements.

They have very similar chemical properties.

Their separation is difficult mainly because of lanthanoid contraction.

Due to lanthanoid contraction, the atomic and ionic radii of \(Zr\) and \(Hf\) become almost similar.



Step 1: Check the Assertion.


The assertion says that separation of \(Zr\) and \(Hf\) is difficult.

This is true.

They show very similar chemical behaviour.



Step 2: Check the Reason.


The reason says that \(Zr\) and \(Hf\) lie in the same group of the periodic table.

This is also true.

Both belong to group 4.



Step 3: Check whether Reason explains Assertion.


The main reason for difficult separation is not simply that they are in the same group.

The stronger and more precise reason is lanthanoid contraction, which makes their radii nearly equal.

Therefore, both Assertion and Reason are true, but Reason is not the correct explanation.



Hence: \[ \boxed{(B)} \] Quick Tip: Separation of \(Zr\) and \(Hf\) is difficult mainly due to lanthanoid contraction.

Concept:


Acidity of phenols depends on the stability of the phenoxide ion formed after loss of proton.

Electron-withdrawing groups increase acidity by stabilising phenoxide ion.

Electron-releasing groups decrease acidity by destabilising phenoxide ion.

The methyl group \((-CH_3)\) is electron releasing.



Step 1: Compare phenol and 4-methylphenol.


Phenol forms phenoxide ion:
\[ C_6H_5OH \rightarrow C_6H_5O^-+H^+ \]

4-methylphenol has a methyl group at para position.
\[ CH_3-group \Rightarrow electron releasing group \]

The methyl group increases electron density on the ring and destabilises the phenoxide ion.

So 4-methylphenol is less acidic than phenol.
\[ Phenol is more acidic than 4-methylphenol \]

Therefore, the Assertion is false.



Step 2: Check the Reason.


The reason says that electron releasing group makes phenol less acidic.

The principle is correct: electron releasing groups reduce acidity of phenol derivatives.

So Reason is considered true in concept.



Step 3: Final conclusion.


Assertion is false, but Reason is true.

Hence: \[ \boxed{(D)} \] Quick Tip: Electron-releasing groups decrease acidity of phenols; electron-withdrawing groups increase acidity.

Concept:


Gabriel phthalimide synthesis is used to prepare primary aliphatic amines.

It involves reaction of potassium phthalimide with alkyl halides followed by hydrolysis.

This reaction requires nucleophilic substitution.

Aryl halides do not undergo such substitution easily.



Step 1: Check the Assertion.


The assertion says that aromatic primary amines cannot be prepared by Gabriel phthalimide synthesis.

This is true.

Aryl halides like chlorobenzene do not undergo normal \(S_N2\) reaction with phthalimide ion.



Step 2: Check the Reason.


The reason says Gabriel phthalimide synthesis is used for preparation of primary aliphatic amines.

This is also true.
\[ R-X + Potassium phthalimide \rightarrow N-alkyl phthalimide \rightarrow RNH_2 \]



Step 3: Check whether Reason explains Assertion.


The exact reason aromatic amines cannot be prepared is that aryl halides do not undergo nucleophilic substitution easily because of partial double bond character in the \(C-X\) bond.

The Reason states a true fact, but it does not fully explain the Assertion.



Hence: \[ \boxed{(B)} \] Quick Tip: Gabriel phthalimide synthesis prepares primary aliphatic amines, not aromatic amines.

Concept:


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

For example:
\[ Rate=k[A]^m[B]^n \]

Then:
\[ Order=m+n \]

Order is not necessarily equal to stoichiometric coefficients.



Step 1: Check the Assertion.


Order of reaction can be defined for elementary reactions.

It can also be defined for complex reactions because every reaction can have an experimentally determined rate law.

Therefore, Assertion is true.



Step 2: Check the Reason.


Order of reaction is determined experimentally.

It cannot always be predicted from the balanced chemical equation.

Therefore, Reason is true.



Step 3: Check whether Reason explains Assertion.


Since order is based on experimentally obtained rate law, it can be applied to both elementary and complex reactions.

So Reason correctly explains Assertion.



Hence: \[ \boxed{(A)} \] Quick Tip: Order is experimental; molecularity is theoretical and applies only to elementary reactions.

Concept:



The \(S_N1\) reaction rate depends on the stability of the intermediate carbocation.
Tertiary carbocations are more stable than secondary, which are more stable than primary ones.
Stability order: \(3^{\circ} > 2^{\circ} > 1^{\circ}\).

Step 1: Identify the possible isomers of \(C_4H_9Br\)


The isomers include n-butyl bromide (1\(^{\circ}\)), isobutyl bromide (1\(^{\circ}\)), sec-butyl bromide (2\(^{\circ}\)), and tert-butyl bromide (3\(^{\circ}\)).


The reaction mechanism for \(S_N1\) involves the formation of a carbocation in the slow, rate-determining step.


Step 2: Evaluate carbocation stability


tert-butyl bromide (\((CH_3)_3CBr\)) loses a bromide ion to form the tert-butyl carbocation \((CH_3)_3C^{+}\).


This is a 3\(^{\circ}\) carbocation stabilized by 9 \(\alpha\)-hydrogens through hyperconjugation and three electron-releasing methyl groups through the +I effect.


Step 3: Conclusion on reactivity


Since tert-butyl bromide forms the most stable carbocation among all isomers, it is the most reactive towards \(S_N1\) displacement. Quick Tip: For \(S_N1\), reactivity is directly proportional to carbocation stability. Look for the carbon attached to the most alkyl groups.

Concept:



The reaction of alkyl halides with sodium in dry ether is the Wurtz reaction.
It couples two alkyl groups together to form a symmetrical alkane.
\(2R-X + 2Na \xrightarrow{ether} R-R + 2NaX\)


Step 1: Determine the structure of the product


The product is 2,5-dimethylhexane.


Its structural formula is \(CH_3-CH(CH_3)-CH_2-CH_2-CH(CH_3)-CH_3\).


Step 2: Identify the alkyl fragment \(R\)


By cutting the product symmetrically in the center, we find the repeating unit is the isobutyl group.


The group is \((CH_3)_2CH-CH_2-\).


Step 3: Write the isomer name


The corresponding alkyl halide is isobutyl bromide, also known as 1-bromo-2-methylpropane. Quick Tip: To find the starting material for a Wurtz reaction, split the product alkane into two equal identical halves. The halogen is attached to the carbon where the split occurred.

Concept:



The order of a reaction is the sum of the exponents of the concentration terms in the rate law.
Elementary reactions occur in a single step and their orders must be integers matching their molecularity.


Step 1: Calculate the overall order


In the expression \(Rate = k[A]^1[B]^{3/2}\):

Order with respect to A = 1

Order with respect to B = \(3/2\) or 1.5

Overall order = \(1 + 1.5 = 2.5\)


Step 2: Determine if the reaction is elementary


An elementary reaction cannot have a fractional order.


Molecularity and order are the same for elementary reactions and must be whole numbers.


Step 3: Final reasoning


Since the overall order is 2.5 (a fraction), this reaction cannot be an elementary reaction.


It must be a complex reaction occurring in multiple steps. Quick Tip: Elementary reactions always have integral orders. Fractional orders or orders greater than 3 always signify complex mechanisms.

Concept:



Boiling point elevation occurs with non-volatile solutes.
Boiling point depression (or lowering) can occur if a solute is more volatile than the solvent.


Step 1: Analyze the effect of \(KCl\)

\(KCl\) is a non-volatile electrolyte.


Its addition lowers the vapor pressure of the solvent (water) because solute particles occupy the surface area, hindering the evaporation of water molecules.


To bring the vapor pressure back to atmospheric pressure, a higher temperature is needed, causing an increase in boiling point (elevation).


Step 2: Analyze the effect of Methyl Alcohol


Methyl alcohol (\(CH_3OH\)) is a volatile liquid with a lower boiling point than water.


When added to water, it increases the total vapor pressure of the solution at a given temperature compared to pure water.


The solution's vapor pressure reaches atmospheric pressure at a lower temperature, resulting in a decrease in boiling point.


Step 3: Summary

\(KCl\) is non-volatile, causing vapor pressure lowering and boiling point elevation.


Methyl alcohol is more volatile than water, increasing solution vapor pressure and lowering the boiling point. Quick Tip: Non-volatile solutes (salts, sugars) always raise the boiling point. Volatile solutes with lower boiling points than the solvent decrease the overall boiling point of the mixture.

Concept:



Coordination compounds are named by listing ligands alphabetically followed by the metal and its oxidation state.
Prefix 'bis' is used for polydentate ligands like ethylenediamine.


Step 1: Find the oxidation state of the central metal


Let the oxidation state of Platinum be \(x\).

Chlorido ligand (\(Cl^-\)) has a charge of -1.

Ethylenediamine (\(en\)) is a neutral ligand (charge 0).

Sulfate (\(SO_4^{2-}\)) has a charge of -2.
\(x + 2(-1) + 2(0) + (-2) = 0 \implies x - 4 = 0 \implies x = +4\).


Step 2: Identify and alphabetize ligands


Ligands are: Chlorido and ethane-1,2-diamine (ethylenediamine).

'C' comes before 'e'.


Since 'ethane-1,2-diamine' contains a numerical prefix, we use 'bis' for two of them.


Step 3: Assemble the name


Name: Dichloridobis(ethane-1,2-diamine)platinum(IV) sulfate. Quick Tip: Always name the cation first, then the anion. If the complex is the cation, the metal retains its standard name.

Concept:



For anionic complexes, the metal name ends with the suffix '-ate'.
Cobalt becomes cobaltate in an anionic coordination sphere.

Step 1: Find the oxidation state of Cobalt


Let the oxidation state of Co be \(x\).

Ammonium (\(NH_4^+\)) has a charge of +1.

Fluorido (\(F^-\)) has a charge of -1.
\(2(+1) + [x + 4(-1)] = 0 \implies 2 + x - 4 = 0 \implies x = +2\).


Step 2: Name the cation and the complex anion


Cation: Ammonium.


Complex Anion: Tetrafluoridocobaltate(II).


Step 3: Assemble the final name


Name: Ammonium tetrafluoridocobaltate(II). Quick Tip: Neutralize the overall charge to find the metal's oxidation state. Don't forget the '-ate' suffix when the complex square bracket follows the positive ion.

Concept:



Ligands donate electron pairs to the metal.
Some ligands have multiple potential donor sites but only use one at a time.


Step 1: Provide the definition


An ambidentate ligand is a unidentate ligand that contains more than one donor atom and can coordinate to the central metal atom through either of these donor atoms.


Step 2: Explain with examples


Example 1: The nitrite ion (\(NO_2^-\)).


It can bond to the metal via Nitrogen (\(M-NO_2\), nitrito-N) or Oxygen (\(M-ONO\), nitrito-O).


Example 2: The thiocyanate ion (\(SCN^-\)).


It can bond via Sulfur (\(M-SCN\), thiocyanato-S) or Nitrogen (\(M-NCS\), isothiocyanato-N). Quick Tip: Ambidentate ligands give rise to linkage isomerism. Remember: one ligand, two possible donor atoms, but only one bond formed at a time.

Concept:



Addition compounds are formed by combining two stable salts.
They differ in their behavior in water.


Step 1: Provide the definition


A double salt is an addition compound that exists only in the solid state and completely dissociates into its constituent simple ions when dissolved in water.


They lose their identity in aqueous solution and show properties of all individual component ions.


Step 2: Provide an example


Example: Mohr's Salt (\(FeSO_4 \cdot (NH_4)_2SO_4 \cdot 6H_2O\)).


When dissolved in water, it breaks into \(Fe^{2+}\), \(NH_4^{+}\), and \(SO_4^{2-}\) ions and gives tests for all of them.


Another example is Potash Alum (\(K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O\)). Quick Tip: Double salts dissociate completely into ions. Complex salts (coordination compounds) retain the identity of the complex ion in solution.

Concept:



Chemical reactions are used to prove the skeletal structure of glucose.
Reducing agents can convert the oxygenated chain into a simple hydrocarbon.


Step 1: Describe the reaction with HI


Glucose is heated with concentrated hydroiodic acid (\(HI\)) and red phosphorus for a prolonged period.


Step 2: Analyze the reaction product


The reaction results in the formation of n-hexane.

\(C_6H_{12}O_6 + HI/P \xrightarrow{\Delta} CH_3-CH_2-CH_2-CH_2-CH_2-CH_3\)
Step 3: Draw the conclusion


The formation of n-hexane (a six-carbon unbranched chain) proves that all six carbon atoms in glucose are linked in a straight, continuous chain. Quick Tip: Reaction with \(HI\) is the standard test for the carbon skeleton. n-hexane formation implies a lack of any branching in the molecule.

Concept:



Acetylation reaction identifies the number of hydroxyl groups.
Geminal diols (two -OH on one carbon) are generally unstable.


Step 1: Describe the acetylation reaction


Glucose is reacted with acetic anhydride or acetyl chloride.


Step 2: Observe the stoichiometry


It forms glucose pentaacetate, indicating that five moles of acetic acid reacted with one mole of glucose.

\(C_6H_{12}O_6 + 5(CH_3CO)_2O \rightarrow Glucose pentaacetate + 5CH_3COOH\)


Step 3: Explain the distribution of groups


The existence of a stable pentaacetate indicates the presence of five hydroxyl (\(-OH\)) groups.


Since a single carbon atom with more than one hydroxyl group is unstable and undergoes dehydration, these five groups must be attached to five different carbon atoms. Quick Tip: Acetylation counts the number of replaceable hydrogens in hydroxyl groups. Stable glucose molecules follow the rule of one -OH per carbon atom.

Concept:

Radioactive decay follows first-order kinetics.
The rate constant (\(k\)) is related to the half-life (\(t_{1/2}\)) by the formula: \(k = \frac{0.693}{t_{1/2}}\).
The time (\(t\)) taken for a first-order reaction is given by: \(t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}\).


Step 1: Determine the rate constant \(k\)

Given \(t_{1/2} = 1.5 \times 10^{10}\) years.

We know that \(k = \frac{2.303 \log 2}{t_{1/2}}\).

Using \(\log 2 = 0.30\):
\[ k = \frac{2.303 \times 0.30}{1.5 \times 10^{10}} \]

Step 2: Set up the equation for time \(t\)

Initial activity \([A]_0 = 100\).

Final activity \([A] = 75\).

The formula for time is: \[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]
Substituting the value of \(k\): \[ t = \frac{2.303 \times (1.5 \times 10^{10})}{2.303 \times 0.30} \log \left( \frac{100}{75} \right) \] \[ t = \frac{1.5 \times 10^{10}}{0.30} \log \left( \frac{4}{3} \right) \]

Step 3: Calculate the final value
\[ \log \left( \frac{4}{3} \right) = \log 4 - \log 3 \]
Given \(\log 4 = 0.60\) and \(\log 3 = 0.48\):
\[ \log \left( \frac{4}{3} \right) = 0.60 - 0.48 = 0.12 \]
Now, calculate \(t\): \[ t = \frac{1.5 \times 10^{10}}{0.30} \times 0.12 \] \[ t = (5 \times 10^{10}) \times 0.12 = 0.60 \times 10^{10} \] \[ t = 6 \times 10^9 years \] Quick Tip: Radioactive decay is always first order. When using logs provided in the question, stick to them even if they are slightly rounded off (e.g., using 0.30 for 0.3010).

Concept:

Kohlrausch's Law: The molar conductivity of an electrolyte at infinite dilution is the sum of the individual contributions of its constituent ions.
Debye-Hückel-Onsager Equation: For strong electrolytes, \(\Lambda_m = \Lambda_m^\circ - A\sqrt{C}\), showing a linear relationship.


Step 1: Statement of the Law

Kohlrausch's law states that the limiting molar conductivity of an electrolyte (molar conductivity at infinite dilution) can be represented as the sum of the individual molar ionic conductivities of the anion and cation. \[ \Lambda_m^\circ = \nu_+ \lambda_+^\circ + \nu_- \lambda_-^\circ \]
where \(\nu_+\) and \(\nu_-\) are the number of cations and anions respectively.

Step 2: Comparison using a curve

For strong electrolytes (like \(KCl\)), the curve of \(\Lambda_m\) versus \(\sqrt{C}\) is a straight line. It can be easily extrapolated to zero concentration to find the intercept \(\Lambda_m^\circ\).

For weak electrolytes (like \(CH_3COOH\)), as concentration decreases, the degree of dissociation (\(\alpha\)) increases. At very low concentrations, \(\alpha\) increases sharply, causing a sudden steep rise in \(\Lambda_m\).

Step 3: Explanation of the difficulty in extrapolation

As concentration approaches zero, the curve for weak electrolytes becomes nearly parallel to the y-axis (\(\Lambda_m\) axis).
Since the curve does not intersect the y-axis, we cannot find the value of \(\Lambda_m^\circ\) by simple graphical extrapolation. Quick Tip: To find \(\Lambda_m^\circ\) for weak electrolytes, we apply Kohlrausch's law using experimental \(\Lambda_m^\circ\) values of strong electrolytes. Strong electrolytes follow a linear \(\sqrt{c}\) plot, while weak electrolytes show a steep hyperbolic-like increase.

Concept:

Freezing point depression (\(\Delta T_f\)) is a colligative property.
Formula: \(\Delta T_f=K_f\times m\), where \(m\) is the molality.
Molality (\(m\)) \(=\dfrac{\text{Moles of solute}}{\text{Mass of solvent (kg)}}\).

Step 1: Calculate the moles of ethylene glycol

Mass of solute (\(w_2\)) = \(31\,\text{g}\)
Molar mass of solute (\(M_2\)) = \(62\,\text{g mol}^{-1}\)

Step 2: Calculate the molality of the solution

Mass of solvent (\(w_1\)) = \(600\,\text{g}=0.6\,\text{kg}\)

Step 3: Calculate the freezing point depression (\(\Delta T_f\))

\(K_f\) for water = \(1.86\,\text{K kg mol}^{-1}\)

Step 4: Determine the freezing point of the solution

Freezing point of pure water (\(T_f^\circ\)) = \(273.15\,\text{K}\) (or \(0^\circ\text{C}\))

Alternatively, in Celsius: \(0-1.55=-1.55^\circ\text{C}\).

Quick Tip: Always subtract \(\Delta T_f\) from the freezing point of the pure solvent. Also, ensure that the mass of the solvent is expressed in kilograms when calculating molality.

Concept:



Valence Bond Theory (VBT) determines hybridization based on ligand strength and the coordination number.
\(Ni^{2+}\) has the electronic configuration \([Ar]3d^8 4s^0\).


Step 1: Hybridization for \([NiCl_4]^{2-}\)

\(Ni\) is in \(+2\) oxidation state (\(3d^8\)).

\(Cl^-\) is a weak field ligand and cannot cause pairing of electrons.


The two unpaired electrons remain in \(3d\) orbitals.


To accommodate 4 pairs from ligands, one \(4s\) and three \(4p\) orbitals hybridize.


Hybridization: \(sp^3\) (Tetrahedral).


Step 2: Hybridization for \([Ni(CN)_4]^{2-}\)

\(Ni\) is in \(+2\) oxidation state (\(3d^8\)).

\(CN^-\) is a strong field ligand and causes the pairing of the two unpaired \(3d\) electrons.


This leaves one \(3d\) orbital vacant.


The central metal uses one \(3d\), one \(4s\), and two \(4p\) orbitals for hybridization.


Hybridization: \(dsp^2\) (Square Planar). Quick Tip: Strong field ligands (\(CN^-, CO\)) usually cause pairing, often leading to \(dsp^2\) or \(d^2sp^3\) hybridization. Weak field ligands (\(Cl^-, F^-\)) do not cause pairing.

Concept:



Inner orbital complexes use \((n-1)d\) orbitals for hybridization.
Outer orbital complexes use \(nd\) orbitals for hybridization.


Step 1: Analyze \([Ni(CN)_4]^{2-}\)


As determined in the hybridization step, it uses one \(3d\) orbital (an inner \(d\)-orbital) to form \(dsp^2\) hybrid orbitals.


Therefore, \([Ni(CN)_4]^{2-}\) is an inner orbital complex.


Step 2: Analyze \([NiCl_4]^{2-}\)


This complex utilizes \(4s\) and \(4p\) orbitals for \(sp^3\) hybridization. In many contexts of tetrahedral complexes, since no inner \(d\) orbitals are used, it is effectively the "outer" configuration compared to \(dsp^2\).


Note: For coordination number 4, the term "outer orbital" is less common than for CN=6, but here \([NiCl_4]^{2-}\) is considered the outer orbital complex in comparison to the square planar \(dsp^2\) version. Quick Tip: Inner orbital complexes involve the pairing of electrons to free up \((n-1)d\) orbitals. Outer orbital complexes do not involve \((n-1)d\) orbitals in the bonding hybridization.

Concept:



Paramagnetism arises from the presence of unpaired electrons.
Diamagnetism occurs when all electrons are paired.


Step 1: Determine magnetic behavior of \([NiCl_4]^{2-}\)


In the \(3d^8\) configuration of \(Ni^{2+}\), there are 2 unpaired electrons in the \(e_g\) orbitals (in tetrahedral field) because \(Cl^-\) is weak field.


Since unpaired electrons are present, it is paramagnetic.


Step 2: Determine magnetic behavior of \([Ni(CN)_4]^{2-}\)


In the presence of the strong ligand \(CN^-\), the electrons in the \(3d\) subshell are forced to pair up.


With zero unpaired electrons, it is diamagnetic. Quick Tip: Magnetic moment \(\mu = \sqrt{n(n+2)}\) BM, where \(n\) is the number of unpaired electrons. If \(n=0\), the complex is diamagnetic.

Concept:


Coordination compounds play vital roles in the metabolic and physiological processes of living organisms. They usually consist of a central metal ion bonded to organic molecules (ligands).


Chlorophyll: A coordination compound of Magnesium (\(Mg\)), essential for photosynthesis.
Hemoglobin: A coordination compound of Iron (\(Fe\)), responsible for oxygen transport in the blood.
Vitamin \(B_{12}\): A coordination compound of Cobalt (\(Co\)), essential for nerve tissue health and red blood cell production.



Step 1: Identify the first example


Chlorophyll is the green pigment in plants. It is a coordination compound where the central metal ion is Magnesium (\(Mg^{2+}\)). It captures light energy during photosynthesis.



Step 2: Identify the second example


Hemoglobin is the red pigment in blood. It is a coordination compound containing Iron (\(Fe^{2+}\)) in a heme group. It binds to oxygen in the lungs and releases it in the tissues. Quick Tip: Remember: Magnesium \(\rightarrow\) Chlorophyll; Iron \(\rightarrow\) Hemoglobin; Cobalt \(\rightarrow\) Vitamin \(B_{12}\). These are frequently asked examples of bio-inorganic coordination chemistry.

Concept:


The stability of a coordination complex is significantly enhanced when the central metal ion is bonded to a polydentate ligand (one that can bind through multiple donor atoms) to form a ring structure.


Chelating Ligand: A ligand that uses two or more donor atoms to bind to a single metal ion.
Chelate Effect: The enhanced stability of a chelated complex compared to a similar complex with unidentate ligands.



Step 1: Define the Chelate Effect


When a di- or polydentate ligand uses its two or more donor atoms to bind to a single metal ion, it forms a five- or six-membered ring. This process is called chelation.


Complexes containing such rings are much more stable than similar complexes containing only unidentate ligands. This stabilization is called the chelate effect.



Step 2: Provide a suitable example


Consider the complex formed by Nickel with ethylenediamine (\(en\)): \[ [Ni(H_2O)_6]^{2+} + 3en \rightarrow [Ni(en)_3]^{2+} + 6H_2O \]
The complex \([Ni(en)_3]^{2+}\) is significantly more stable than \([Ni(NH_3)_6]^{2+}\) because ethylenediamine is a bidentate ligand that forms three five-membered rings with the nickel ion. Quick Tip: Chelation increases the entropy of the system (\(\Delta S > 0\)) because more molecules/ions are released into the solution than are consumed, contributing to higher stability. Five- and six-membered rings are the most stable.

Concept:


The formation of a low spin or high spin complex depends on the magnitude of the crystal field splitting energy (\(\Delta\)) relative to the electron pairing energy (\(P\)).


If \(\Delta > P\), electrons pair up (Low Spin).
If \(\Delta < P\), electrons remain unpaired and occupy higher orbitals (High Spin).



Step 1: Analyze the magnitude of splitting in tetrahedral fields


In a tetrahedral geometry, the number of ligands is only four (compared to six in octahedral). Consequently, the interaction between ligands and metal \(d\)-orbitals is weaker.


The crystal field splitting for tetrahedral complexes (\(\Delta_t\)) is roughly only \(4/9\) of the splitting for octahedral complexes (\(\Delta_o\)). \[ \Delta_t \approx \frac{4}{9} \Delta_o \]



Step 2: Compare splitting energy with pairing energy


Because \(\Delta_t\) is relatively small, it is almost always less than the energy required to pair electrons (\(P\)). \[ \Delta_t < P \]


Since the energy required to jump to a higher energy orbital (\(t_2\)) is less than the energy required to pair up in a lower orbital (\(e\)), the electrons prefer to occupy the higher orbitals.



Step 3: Conclusion


Because the splitting energy rarely exceeds the pairing energy in tetrahedral environments, electrons do not pair up. This results in the formation of high spin complexes, making low spin tetrahedral complexes extremely rare. Quick Tip: Low spin = Pairing happens (Strong Field). High spin = No pairing (Weak Field). Tetrahedral splitting is so small that even "strong" ligands often fail to cause pairing.

Concept:

Directing influence depends on the net effect of Inductive (\(-I\)) and Resonance (\(+R\)) effects.
Deactivation depends primarily on the strength of the inductive effect.


Step 1: Explain the Inductive effect

Chlorine is highly electronegative and exerts a strong \(-I\) (inductive) effect. This withdraws electrons from the benzene ring, making the ring less reactive (deactivated) compared to benzene.

Step 2: Explain the Resonance effect

Chlorine has lone pairs of electrons. Through the \(+R\) (resonance) effect, it can donate these lone pairs into the ring system. This increases the electron density specifically at the ortho and para positions.

Step 3: Conclusion

Although the \(-I\) effect is stronger than the \(+R\) effect (causing overall deactivation), the resonance effect determines the orientation. Because the ortho and para positions are relatively more electron-rich than the meta position, the electrophile is directed to these positions. Quick Tip: Chlorine is a rare case: it is deactivating but ortho-para directing. For most groups, if they are deactivating, they are meta-directing.

Concept:



The reactivity of the \(C-OH\) bond in phenols is significantly different from alcohols due to resonance.


Step 1: Analyze the resonance in Phenol


In phenol, the lone pair of electrons on the oxygen atom is in conjugation with the \(\pi\)-electrons of the benzene ring. This resonance results in a partial double bond character in the \(C-O\) bond.


Step 2: Evaluate bond strength


A double bond is shorter and much stronger than a single bond. Breaking the \(C-O\) bond in phenol requires significantly more energy than in aliphatic alcohols.


Step 3: Conclusion


Since electrophilic/nucleophilic cleavage of this strong bond is difficult, phenol does not easily undergo substitution of the \(-OH\) group by a halide ion when treated with \(HX\), \(PCl_3\), or \(PCl_5\). Quick Tip: Resonance makes the aryl-oxygen bond much stronger than an alkyl-oxygen bond. This is why phenols don't undergo simple nucleophilic substitution like alcohols do.

Concept:



Chloroform (\(CHCl_3\)) is susceptible to oxidation by atmospheric oxygen.


Step 1: Describe the chemical reaction


Chloroform, when exposed to air and sunlight, undergoes slow oxidation to form a highly poisonous gas called carbonyl chloride (also known as phosgene).

\[ 2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl \]



Step 2: Explain the storage conditions


Dark-coloured bottles are used to cut off the supply of light, which acts as a catalyst for this oxidation reaction. Closing the bottles tightly prevents the entry of air (oxygen).


Step 3: Conclusion


To prevent the formation of toxic phosgene and ensure safety, chloroform must be stored in dark, airtight containers. Quick Tip: Phosgene (\(COCl_2\)) is a chemical warfare agent; its prevention is critical. Sometimes a small amount of ethanol is added to chloroform to convert any formed phosgene into harmless diethyl carbonate.

Concept:



Benzenediazonium chloride (\(C_6H_5N_2^+Cl^-\)) is a versatile intermediate in organic synthesis.
The replacement of the diazonium group by a halide ion using copper(I) salts is known as the Sandmeyer reaction.
Alternatively, using copper powder and \(HCl\) is known as the Gattermann reaction.


Step 1: Describe the Sandmeyer Reaction


When a freshly prepared aqueous solution of benzenediazonium chloride is treated with cuprous chloride (\(Cu_2Cl_2\)) dissolved in hydrochloric acid (\(HCl\)), the diazonium group is replaced by a chlorine atom.


Step 2: Write the chemical equation

\[ C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2 / HCl} C_6H_5Cl + N_2 \uparrow \]
In this reaction, nitrogen gas is evolved, and chlorobenzene is formed as the primary organic product. Quick Tip: Sandmeyer reaction generally gives better yields than the Gattermann reaction. Always ensure the diazonium salt is kept cold (0-5\(^\circ\)C) before the reaction as it is unstable at room temperature.

Concept:



The conversion of a diazonium salt to an arene is a reduction process.
Mild reducing agents like hypophosphorous acid (\(H_3PO_2\)) or ethanol (\(CH_3CH_2OH\)) are used.


Step 1: Describe the reduction using Hypophosphorous acid


Benzenediazonium chloride reacts with hypophosphorous acid in the presence of water to produce benzene. In this process, \(H_3PO_2\) is oxidized to phosphorous acid (\(H_3PO_3\)).


Step 2: Write the chemical equation

\[ C_6H_5N_2Cl + H_3PO_2 + H_2O \rightarrow C_6H_6 + N_2 + H_3PO_3 + HCl \]

Step 3: Alternative method using Ethanol


Alternatively, ethanol can act as a reducing agent, where it is oxidized to ethanal (acetaldehyde). \[ C_6H_5N_2Cl + CH_3CH_2OH \rightarrow C_6H_6 + N_2 + CH_3CHO + HCl \] Quick Tip: To "deaminate" an aromatic ring (remove the \(-NH_2\) group via diazonium), reduction to benzene is the key step. \(H_3PO_2\) is often preferred in labs for higher efficiency in this reduction.

Concept:



The introduction of a cyano group (\(-CN\)) into the benzene ring is an application of the Sandmeyer reaction.
Cyanide ions (\(CN^-\)) act as the nucleophile in the presence of a copper(I) catalyst.


Step 1: Describe the reaction with Cuprous Cyanide


When an aqueous solution of benzenediazonium chloride is treated with a mixture of cuprous cyanide (\(CuCN\)) and potassium cyanide (\(KCN\)), the diazonium group is substituted by the nitrile group.


Step 2: Write the chemical equation

\[ C_6H_5N_2Cl \xrightarrow{CuCN / KCN} C_6H_5CN + N_2 \uparrow \]
The product formed is Benzonitrile (also known as Phenyl cyanide). Quick Tip: Benzonitrile is a very useful intermediate as it can be hydrolyzed to benzoic acid or reduced to benzylamine. The use of \(KCN\) provides the nucleophile while \(CuCN\) acts as the catalyst.

Concept:

Compounds with formula \(C_nH_{2n-1}N\) are typically nitriles (alkane nitriles).
Reaction of nitriles with Grignard reagents followed by hydrolysis yields ketones.
Chemical tests: 2,4-DNP (carbonyl group detection), Iodoform (methyl ketone detection), Tollens/Fehling (aldehyde detection), \(Br_2/H_2O\) (unsaturation detection).


Step 1: Identify Compound (A)

Molecular formula is \(C_3H_5N\). Based on the properties, it is Ethyl cyanide (Propionitrile). \[ Structure of (A): CH_3CH_2C \equiv N \]

Step 2: Identify Compound (B)

Reaction of (A) with phenylmagnesium bromide (\(C_6H_5MgBr\)): \[ CH_3CH_2CN + C_6H_5MgBr \rightarrow CH_3CH_2C(=NMgBr)C_6H_5 \xrightarrow{H_2O/H^+} CH_3CH_2-CO-C_6H_5 \]
Compound (B) is Propiophenone (1-Phenylpropan-1-one).

2,4-DNP test: Positive (it's a ketone).
Iodoform test: Negative (it is not a methyl ketone; it is an ethyl ketone).
Tollens/Fehling: Negative (it is a ketone, not an aldehyde).
Bromine water: Negative (no \(C=C\) or \(C \equiv C\) in the aliphatic chain).


Step 3: Identify Compound (C)

Drastic oxidation of propiophenone (\(C_6H_5COCH_2CH_3\)) with \(CrO_3\) involves the cleavage of the alkyl chain. The aromatic part remains intact to form benzoic acid.
Formula \(C_7H_6O_2\) matches Benzoic acid (\(C_6H_5COOH\)).

Step 4: Write the chemical equations

Reaction of Ethyl cyanide with Grignard reagent: \[ CH_3CH_2-C \equiv N + C_6H_5MgBr \xrightarrow{dry ether} CH_3CH_2-C(C_6H_5)=NMgBr \]
Acid hydrolysis of the intermediate: \[ CH_3CH_2-C(C_6H_5)=NMgBr + 2H_2O \xrightarrow{H^+} CH_3CH_2-CO-C_6H_5 + NH_3 + Mg(OH)Br \]

Step 5: Summary identification

(A) = Ethyl cyanide (\(CH_3CH_2CN\))

(B) = Propiophenone (\(C_6H_5COCH_2CH_3\))

(C) = Benzoic acid (\(C_6H_5COOH\)) Quick Tip: Nitriles + Grignard \(\rightarrow\) Ketone. If the iodoform test is negative for a ketone, it means the carbonyl group is not adjacent to a methyl group (no \(CH_3-CO-\)). Drastic oxidation of any alkyl benzene derivative or phenyl ketone usually yields benzoic acid.

Concept:

Biological macromolecules are polymers of smaller monomeric units.
Protein classification depends on the molecular shape and folding of polypeptide chains.


Step 1: Identify basic building units


Proteins: The basic building units are \(\alpha\)-amino acids. They are linked by peptide bonds.
Nucleic acids: The basic building units are nucleotides. Each nucleotide consists of a nitrogenous base, a pentose sugar, and a phosphate group.


Step 2: Differentiate between Fibrous and Globular proteins


Fibrous Proteins:

Polypeptide chains run parallel and are held together by hydrogen and disulfide bonds.
They have a fiber-like structure.
They are generally insoluble in water.
Examples: Keratin (hair, wool), Myosin (muscles).

Globular Proteins:

Polypeptide chains coil around to give a spherical or ball-like shape.
They are generally soluble in water.
They usually perform catalytic or regulatory functions.
Examples: Insulin, Albumin. Quick Tip: Fibrous = Structural, Insoluble, Linear. Globular = Functional (Enzymes/Hormones), Soluble, Spherical. DNA/RNA building blocks are 'nucleotides', while the base+sugar unit alone is a 'nucleoside'.

Concept:



Complete hydrolysis of a nucleotide breaks all the covalent bonds between its three components.


Step 1: Identify the components of a DNA nucleotide


A nucleotide in DNA consists of:



A nitrogenous base (given as Thymine).
A pentose sugar (which is \(\beta\)-D-2-deoxyribose in DNA).
A phosphate group.


Step 2: List the hydrolysis products


Upon complete hydrolysis, the nucleotide yields:



Thymine (the nitrogenous base).
\(\beta\)-D-2-deoxyribose (the sugar molecule).
Phosphoric acid (\(H_3PO_4\)). Quick Tip: DNA contains deoxyribose sugar (lacks oxygen at C-2), whereas RNA contains ribose sugar. Thymine is unique to DNA; Uracil is unique to RNA.

Concept:



DNA (Deoxyribonucleic acid) and RNA (Ribonucleic acid) differ in their sugar composition, nitrogenous bases, and physical structure.


Step 1: Provide structural differences



Sugar difference: In DNA, the pentose sugar is \(\beta\)-D-2-deoxyribose, whereas in RNA, the sugar is \(\beta\)-D-ribose.


Base difference: DNA contains the base Thymine, while in RNA, Thymine is replaced by Uracil.


Chain structure: DNA typically exists as a


double-stranded helix, while RNA is generally single-stranded. Quick Tip: The "Deoxy" in DNA refers to the absence of an -OH group at the 2' position of the ribose sugar. DNA uses A, G, C, T; RNA uses A, G, C, U.

Concept:



Vitamins are classified based on their solubility in water or fats/oils.


Step 1: Fat-soluble vitamins


These are soluble in fats and oils but insoluble in water. They are stored in liver and adipose tissues.


Examples: Vitamin A, D, E, or K. (Any one can be cited).


Step 2: Water-soluble vitamins


These are soluble in water and must be supplied regularly in the diet because they are readily excreted in urine.


Examples: Vitamin C or the Vitamin B-complex group. (Any one can be cited). Quick Tip: Mnemonic for fat-soluble vitamins: KEDA. Vitamin \(B_{12}\) is an exception as it is water-soluble but can be stored in the body for longer periods.

Concept:

A primary alcohol can be oxidized either to an aldehyde or to a carboxylic acid depending on the oxidizing agent used.
To obtain an aldehyde, a mild oxidizing agent is required so that further oxidation is prevented.


Step 1: Identify the reagent

The reagent commonly used for the oxidation of a primary alcohol to an aldehyde is Pyridinium Chlorochromate (PCC) in dichloromethane (\(CH_2Cl_2\)).


PCC is a mild oxidizing agent that converts primary alcohols into aldehydes without further oxidation to carboxylic acids.
\[ \mathrm{RCH_2OH \xrightarrow[PCC]{CH_2Cl_2} RCHO} \]



Quick Tip: PCC is the most commonly used reagent for converting a primary alcohol into an aldehyde because it prevents over-oxidation to a carboxylic acid.

Concept:

Strong oxidizing agents oxidize primary alcohols completely to carboxylic acids.
The aldehyde formed initially undergoes further oxidation to give the corresponding carboxylic acid.


Step 1: Identify the reagent

The oxidation of a primary alcohol to a carboxylic acid is carried out using strong oxidizing agents such as:


Acidified potassium dichromate \(\left(K_2Cr_2O_7/H_2SO_4\right)\), or
Acidified potassium permanganate \(\left(KMnO_4/H_2SO_4\right)\).


These oxidizing agents convert the primary alcohol first into an aldehyde and then further oxidize it into the corresponding carboxylic acid.
\[ \mathrm{RCH_2OH \xrightarrow[K_2Cr_2O_7/H_2SO_4]{Oxidation} RCOOH} \]



Quick Tip: Strong oxidizing agents such as acidified \(K_2Cr_2O_7\) and acidified \(KMnO_4\) convert primary alcohols completely into carboxylic acids.

Concept:

Phenols react with sodium hydroxide to form sodium phenoxide, which is more reactive than phenol towards electrophilic substitution.
It reacts with a weak electrophile like carbon dioxide (\(CO_2\)).


Step 1: Formation of Sodium Phenoxide

Phenol is treated with aqueous \(NaOH\). \[ C_6H_5OH + NaOH \rightarrow C_6H_5ONa + H_2O \]

Step 2: Carboxylation

Sodium phenoxide is heated with carbon dioxide (\(CO_2\)) at 4-7 atm and around 400 K. The electrophile \(CO_2\) attacks the ortho position. \[ C_6H_5ONa + CO_2 \rightarrow o-Hydroxybenzoic acid sodium salt \]

Step 3: Acidification

Acidification of the salt yields Salicylic acid (2-Hydroxybenzoic acid). \[ \mathrm{o\!-\!HOC_6H_4COONa + HCl \longrightarrow o\!-\!HOC_6H_4COOH + NaCl} \]

Overall Reaction: \[ Phenol \xrightarrow{(i) NaOH, (ii) CO_2, (iii) H^+} Salicylic acid \] Quick Tip: Kolbe's reaction is used specifically to introduce a carboxyl group into the ortho position of phenol. Salicylic acid is the starting material for making Aspirin (acetylsalicylic acid).

Concept:



Oxidation of alcohols involves the removal of hydrogen from the carbon bearing the hydroxyl group (\(\alpha\)-hydrogen).


Step 1: Analyze the structure of tertiary alcohols


In a tertiary (\(3^\circ\)) alcohol, the carbon atom attached to the \(-OH\) group is also attached to three other carbon atoms.
Example: \( (CH_3)_3C-OH \).


Step 2: Absence of \(\alpha\)-hydrogen


There is no hydrogen atom directly bonded to the \(\alpha\)-carbon (the carbon carrying the \(-OH\) group).
In primary and secondary alcohols, the first step of oxidation involves breaking the \(\alpha\)-C-H bond.


Step 3: Conclusion


Since there is no \(\alpha\)-H bond to break, tertiary alcohols do not undergo oxidation under normal (neutral or alkaline) conditions. Under drastic conditions (acidic strong oxidants), they may undergo dehydration followed by cleavage of \(C-C\) bonds to form a mixture of ketones and carboxylic acids with fewer carbon atoms. Quick Tip: Oxidation requires \(\alpha\)-hydrogens. \(1^\circ\) has 2, \(2^\circ\) has 1, \(3^\circ\) has 0. \(3^\circ\) alcohols prefer dehydration to alkenes when treated with acidic oxidizing agents.

Concept:

Cleavage of ethers with \(HI\) depends on the nature of the alkyl groups.
If one of the alkyl groups is tertiary, the reaction proceeds via the \(S_N1\) mechanism.


Step 1: Protonation of the ether

The oxygen atom is protonated by the acid to form an oxonium ion. \[ (CH_3)_3C - O^+(H) - C_2H_5 \]

Step 2: Formation of carbocation

Because the tert-butyl group can form a highly stable tertiary carbocation, the bond between the \(3^\circ\) carbon and the oxygen breaks first (\(S_N1\) pathway). \[ (CH_3)_3C - O^+(H) - C_2H_5 \rightarrow (CH_3)_3C^+ + C_2H_5OH \]

Step 3: Nucleophilic attack

The iodide ion (\(I^-\)) then attacks the stable carbocation. \[ (CH_3)_3C^+ + I^- \rightarrow (CH_3)_3C-I \]

Final Products:
The products are tert-Butyl iodide and Ethanol. \[ (CH_3)_3C-O-C_2H_5 + HI \rightarrow (CH_3)_3C-I + C_2H_5OH \] Quick Tip: Standard \(S_N2\) (small group gets iodine) applies for \(1^\circ\) and \(2^\circ\) alkyl groups. If a \(3^\circ\) group is present, \(S_N1\) occurs and the \(3^\circ\) group ALWAYS gets the iodine.

Concept:



The melting point of transition elements depends upon the strength of metallic bonding.
Stronger metallic bonding results from a greater number of unpaired d-electrons available for overlap.


Step 1: Compare the electronic configurations

Iron (\(Fe\)) has the electronic configuration \[ [Ar]\,3d^6\,4s^2 \]
and possesses more unpaired d-electrons than copper.

Copper (\(Cu\)) has the configuration \[ [Ar]\,3d^{10}\,4s^1 \]
where the \(3d\)-subshell is completely filled.

Step 2: Reason for higher melting point

Since iron has more unpaired d-electrons, stronger metallic bonding exists between its atoms. Therefore, more energy is required to separate them.

Hence, Fe has a higher melting point than Cu.



Quick Tip: More unpaired \(d\)-electrons \(\Rightarrow\) stronger metallic bonding \(\Rightarrow\) higher melting point.

Concept:



Colour in transition metal ions arises due to \(d\)-\(d\) electronic transitions.
Such transitions are possible only when partially filled \(d\)-orbitals are present.


Step 1: Compare electronic configurations

\[ Ti^{3+}=[Ar]\,3d^1 \]
contains one electron in the \(d\)-orbital.

\[ Sc^{3+}=[Ar]\,3d^0 \]
has no \(d\)-electrons.

Step 2: Reason for colour


The single \(d\)-electron in \(Ti^{3+}\) undergoes \(d\)-\(d\) transition by absorbing visible light, making the ion coloured.


Since \(Sc^{3+}\) has no \(d\)-electrons, no \(d\)-\(d\) transition is possible and it remains colourless.


Hence, \(Ti^{3+}\) is coloured in aqueous solution.



Quick Tip: Ions with \(d^0\) or \(d^{10}\) configuration are generally colourless because \(d\)-\(d\) transitions are not possible.

Concept:



Ionisation enthalpy depends upon the stability of the electronic configuration after electron removal.
Completely filled subshells possess exceptional stability.


Step 1: Compare the electronic configurations


After losing two electrons,
\[ Cr^{2+}=[Ar]\,3d^4 \]

whereas
\[ Zn^{2+}=[Ar]\,3d^{10}. \]

Step 2: Reason for higher ionisation enthalpy


Removing the third electron from \(Zn^{2+}\) disturbs the completely filled and highly stable \(3d^{10}\) configuration.


In contrast, removal of the third electron from \(Cr^{2+}\) does not involve breaking such a stable configuration.


Therefore, much more energy is required to remove the third electron from zinc.


Hence, Zn has the higher third ionisation enthalpy.



Quick Tip: Completely filled (\(d^{10}\)) and half-filled (\(d^5\)) subshells are especially stable and require more energy for electron removal.

Concept:

In acidic medium, permanganate ion acts as a strong oxidizing agent and is reduced to \(Mn^{2+}\).


Step 1: Write the half-reactions


Reduction: \[ MnO_4^-+8H^++5e^- \rightarrow Mn^{2+}+4H_2O \]

Oxidation: \[ Fe^{2+}\rightarrow Fe^{3+}+e^- \]

Step 2: Balance electrons and combine

Multiplying the oxidation half-reaction by 5 and adding,
\[ \boxed{ MnO_4^-+5Fe^{2+}+8H^+ \rightarrow Mn^{2+}+5Fe^{3+}+4H_2O } \]



Quick Tip: In acidic medium, \(MnO_4^-\) is reduced from oxidation state \(+7\) to \(+2\).

Concept:

Permanganate ion oxidizes iodide ions to iodine in acidic medium.


Step 1: Write the half-reactions


Reduction: \[ MnO_4^-+8H^++5e^- \rightarrow Mn^{2+}+4H_2O \]

Oxidation: \[ 2I^- \rightarrow I_2+2e^- \]

Step 2: Balance electrons and combine

Multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5 to balance 10 electrons.

The balanced ionic equation is
\[ \boxed{ 2MnO_4^-+10I^-+16H^+ \rightarrow 2Mn^{2+}+5I_2+8H_2O } \]



Quick Tip: Remember: \(I^-\) is oxidized to \(I_2\), while \(MnO_4^-\) is reduced to \(Mn^{2+}\) in acidic medium.

Concept:

Pyrolusite ore (\(MnO_2\)) is the starting material for the preparation of potassium permanganate.
Oxidation of \(MnO_2\) yields manganate (\(MnO_4^{2-}\)), which is then further oxidized to permanganate (\(MnO_4^-\)).


Step 1: Identify the compounds

(A) is Manganese dioxide (\(MnO_2\)) (Black-brown solid).
(B) is Potassium manganate (\(K_2MnO_4\)) (Dark green compound).
(C) is Potassium permanganate (\(KMnO_4\)) (Dark purple compound).

Step 2: Write the reaction for the formation of (B)

When \(MnO_2\) is fused with \(KOH\) in the presence of atmospheric oxygen: \[ 2MnO_2 + 4KOH + O_2 \rightarrow 2K_2MnO_4 + 2H_2O \]

Step 3: Write the reaction for the formation of (C)

Potassium manganate undergoes electrolytic oxidation in alkaline solution to form potassium permanganate: \[ At Anode: MnO_4^{2-} \rightarrow MnO_4^- + e^- \] \[ Overall: 2K_2MnO_4 + H_2O + [O] \xrightarrow{electrolysis} 2KMnO_4 + 2KOH \] Quick Tip: Green = Manganate (\(MnO_4^{2-}\), oxidation state +6). Purple = Permanganate (\(MnO_4^-\), oxidation state +7). Black-brown solid in this context is almost always the pyrolusite ore, \(MnO_2\).

Concept:



Manganate ions are only stable in strongly alkaline solutions.
In acidic or neutral media, they undergo a specific type of redox reaction where the same element is both oxidized and reduced.


Step 1: Describe the observation


When an acidic solution of the green potassium manganate (B) is allowed to stand, the green colour disappears and the solution turns purple with the formation of a brown precipitate of \(MnO_2\).


Step 2: Write the chemical equation


The manganate ion (\(MnO_4^{2-}\)) reacts with hydrogen ions as follows: \[ 3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O \]


Step 3: Name the reaction type


In this reaction, Manganese in the +6 state is reduced to +4 (in \(MnO_2\)) and simultaneously oxidized to +7 (in \(MnO_4^-\)).
This type of reaction is called a Disproportionation reaction. Quick Tip: Disproportionation: Same species is oxidized and reduced. Manganate is stable only in alkaline medium; it "disproportionates" in acidic or neutral medium.

Concept:

Aldehydes react with one mole of alcohol in the presence of dry \(HCl\) to form hemiacetals.
If excess alcohol is used, acetals are formed.


Step 1: Identify the reaction

Ethanal reacts with one mole of methanol in the presence of dry hydrochloric acid. Since only one mole of alcohol is present, the reaction stops at the hemiacetal stage.

Step 2: Write the product

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

Hence, the product formed is 1-Methoxyethanol (Hemiacetal).



Quick Tip: One mole of alcohol forms a hemiacetal, whereas excess alcohol forms an acetal.

Concept:



Aldehydes that do not possess an \(\alpha\)-hydrogen undergo the Cannizzaro reaction in concentrated alkali.
In this reaction, one molecule is oxidized while the other is reduced.


Step 1: Identify the reaction


Benzaldehyde does not contain an \(\alpha\)-hydrogen atom. Therefore, it undergoes the Cannizzaro reaction with concentrated sodium hydroxide.


Step 2: Write the products

\[ 2C_6H_5CHO + NaOH \rightarrow C_6H_5CH_2OH + C_6H_5COONa \]


Hence, the products formed are Benzyl alcohol and Sodium benzoate.



Quick Tip: Only aldehydes without \(\alpha\)-hydrogen (such as benzaldehyde and formaldehyde) undergo the Cannizzaro reaction.

Concept:

Phosphorus pentoxide (\(P_2O_5\)) is a strong dehydrating agent.
It removes water from two molecules of carboxylic acid to produce the corresponding acid anhydride.


Step 1: Identify the reaction


When ethanoic acid is heated with \(P_2O_5\), two molecules of the acid lose one molecule of water.


Step 2: Write the product

\[ 2CH_3COOH \xrightarrow{P_2O_5,\Delta} (CH_3CO)_2O + H_2O \]


Hence, the product formed is Ethanoic anhydride (Acetic anhydride).



Quick Tip: \(P_2O_5\) is a powerful dehydrating agent that converts carboxylic acids into acid anhydrides.

Concept:



Both are aldehydes, so they give positive Tollens' and Fehling's tests.
They can be distinguished based on the presence of a methyl keto group (\(CH_3CO-\)) or its equivalent.


Step 1: Identify the reagent


The Iodoform test is used to distinguish them. Ethanal (\(CH_3CHO\)) contains the \(CH_3CO-\) group bonded to hydrogen, making it the only aldehyde that gives a positive iodoform test. Propanal (\(CH_3CH_2CHO\)) does not have this group.


Step 2: Describe the test for Ethanal


When ethanal is heated with iodine (\(I_2\)) and sodium hydroxide (\(NaOH\)), a yellow precipitate of iodoform (\(CHI_3\)) is formed. \[ CH_3CHO + 3I_2 + 4NaOH \rightarrow CHI_3 \downarrow (yellow) + HCOONa + 3NaI + 3H_2O \]


Step 3: Describe the test for Propanal


Propanal does not give a yellow precipitate when treated with iodine and \(NaOH\). \[ CH_3CH_2CHO + I_2 + NaOH \rightarrow No yellow precipitate \]
Quick Tip: The Iodoform test is specific for methyl ketones (\(R-CO-CH_3\)) and secondary alcohols with a methyl group (\(R-CH(OH)-CH_3\)). Ethanal is the only aldehyde that gives a positive Iodoform test.

Concept:



Allyl alcohol is \(CH_2=CH-CH_2OH\) (a primary unsaturated alcohol).
Propenal (Acrolein) is \(CH_2=CH-CHO\).
The conversion requires an oxidizing agent that oxidizes the alcohol to an aldehyde without affecting the carbon-carbon double bond.


Step 1: Identify the specific reagent


A mild and selective oxidizing agent is required. The best reagent for this transformation is Pyridinium chlorochromate (PCC) in dichloromethane (\(CH_2Cl_2\)).


Step 2: Explain the choice


PCC selectively oxidizes primary alcohols to aldehydes and secondary alcohols to ketones. Unlike stronger oxidizing agents (like acidic \(KMnO_4\)), it does not over-oxidize the aldehyde to a carboxylic acid and does not react with the \(C=C\) double bond.


Step 3: Mention alternatives


An alternative reagent is Manganese dioxide (\(MnO_2\)), which is highly selective for the oxidation of allylic and benzylic alcohols. Quick Tip: PCC is the "gentle" oxidant for primary alcohols. \(MnO_2\) is specifically useful for allylic (\(C=C-C-OH\)) and benzylic (\(Ar-C-OH\)) systems.

Concept:



Carbonyl compounds react with semicarbazide (\(NH_2NHCONH_2\)) to form semicarbazones.
This is a nucleophilic addition-elimination reaction involving the loss of a water molecule.


Step 1: Identify the reactants


Acetone: \(CH_3-CO-CH_3\)

Semicarbazide: \(NH_2-NH-CO-NH_2\)


Step 2: Describe the reaction


The lone pair on the \(NH_2\) group attached to the \(NH\) group (the hydrazine end) attacks the carbonyl carbon of acetone. Note that the \(NH_2\) group of the amide part (\(-CONH_2\)) is less nucleophilic due to resonance with the carbonyl group and does not participate.



Step 3: Draw the structure


The resulting semicarbazone has the structure: \[ (CH_3)_2C = N - NH - CO - NH_2 \] Quick Tip: In semicarbazide, only one of the two \(-NH_2\) groups is nucleophilic; the other is deactivated by resonance with the adjacent carbonyl group. Water is eliminated from the \(>C=O\) of the ketone and the \(-NH_2\) of the reagent.

Concept:

Acidity of a hydrogen atom depends on the stability of the conjugate base formed after its removal.


Step 1: Inductive Effect

The carbonyl group (\(>C=O\)) is strongly electron-withdrawing due to its \(-I\) effect. It withdraws electron density from the adjacent \(\alpha\)-carbon, which in turn weakens the \(\alpha\)-C-H bond, making it easier for a base to remove the hydrogen as a proton.

Step 2: Resonance stabilization

When a base removes an \(\alpha\)-hydrogen, a carbanion (enolate ion) is formed. This negative charge is delocalized over the \(\alpha\)-carbon and the oxygen atom through resonance. \[ [ >C^--C=O \longleftrightarrow >C=C-O^- ] \]

Step 3: Conclusion

The combination of the electron-withdrawing effect of the carbonyl group and the substantial resonance stabilization of the resulting enolate ion makes the \(\alpha\)-hydrogen atoms significantly acidic. Quick Tip: Acidity is all about "Base Stability". If the conjugate base (enolate) is stable via resonance, the parent hydrogen is acidic. The oxygen atom, being more electronegative, stabilizes the negative charge better in the enolate form.

Concept:



The reaction of \(HCN\) with carbonyl compounds is a Nucleophilic Addition reaction.
Reactivity decreases as the steric hindrance around the carbonyl carbon increases.
Reactivity decreases as the magnitude of the positive charge on the carbonyl carbon decreases (Inductive effect).


Step 1: Analyze Ethanal (\(CH_3CHO\))


Ethanal has only one methyl group causing minor steric hindrance and a small \(+I\) effect. Aldehydes are generally more reactive than ketones because they have less steric hindrance and a more positive carbonyl carbon.


Step 2: Analyze Acetone (\(CH_3COCH_3\))


Acetone has two methyl groups. This increases steric hindrance for the incoming nucleophile (\(CN^-\)) and provides a larger \(+I\) effect, which reduces the electrophilicity of the carbonyl carbon.


Step 3: Analyze Pinacolone (\((CH_3)_3C-CO-CH_3\))


This compound (3,3-dimethyl-2-butanone) has a bulky tert-butyl group on one side. The massive steric hindrance significantly blocks the approach of the nucleophile, making it the least reactive.


Final Order (Increasing Reactivity):

\[ (CH_3)_3C-CO-CH_3 < CH_3COCH_3 < CH_3CHO \] Quick Tip: Nucleophilic addition reactivity: Formaldehyde > Aldehydes > Ketones. Bulky groups act as "shields" for the carbonyl carbon, slowing down the reaction.

Concept:



The Etard reaction is a method to selectively oxidize a methyl group on an aromatic ring to an aldehyde group.


Step 1: Reaction description


Toluene is reacted with chromyl chloride (\(CrO_2Cl_2\)) in a solvent like carbon disulfide (\(CS_2\)).


Step 2: Formation of intermediate


Chromyl chloride forms a brown-coloured chromium complex with the methyl group.

\[ C_6H_5CH_3 + 2CrO_2Cl_2 \rightarrow C_6H_5CH(OCr(OH)Cl_2)_2 \]


Step 3: Hydrolysis


Mild acid hydrolysis of this chromium complex yields


Benzaldehyde.

\[ C_6H_5CH(OCr(OH)Cl_2)_2 \xrightarrow{H_3O^+} C_6H_5CHO \] Quick Tip: Etard reaction stops at the aldehyde stage because the intermediate chromium complex is resistant to further oxidation. This is a standard industrial method for producing benzaldehyde.

Concept:



The reagent \(Zn(Hg)\) in concentrated \(HCl\) is used for the Clemmensen reduction.
This reaction reduces the carbonyl group (\(>C=O\)) of aldehydes or ketones to a methylene group (\(>CH_2\)).


Step 1: Identify the reactant structure


Cyclohexanecarbaldehyde consists of a cyclohexyl ring attached to an aldehyde group (\(-CHO\)).

\[ Structure: C_6H_{11}-CHO \]

Step 2: Apply the reduction


The aldehyde group (\(-CHO\)) is converted into a methyl group (\(-CH_3\)). \[ C_6H_{11}-CHO \xrightarrow{Zn(Hg)/conc. HCl} C_6H_{11}-CH_3 \]



Step 3: Identify the product


The product is Methylcyclohexane. Quick Tip: Clemmensen reduction is ideal for acid-stable molecules. It turns "CHO" into "CH3" and "C=O" into "CH2".

Concept:
 

Concept:

The electrode potential (\(E\)) of a half-cell at any concentration and temperature is given by the Nernst equation.

For the reduction half-reaction: \(M^{n+}(aq)+ne^- \rightarrow M(s)\), the equation is:

At \(25^\circ\mathrm{C}\) (\(298\,\mathrm{K}\)), the value of \(\dfrac{2.303RT}{F}\) is \(0.0591\,\mathrm{V}\).

Concept:

A dry cell (Leclanché cell) is a primary cell consisting of a zinc container (anode) and a carbon (graphite) rod (cathode) surrounded by powdered \(MnO_2\) and carbon.
The electrolyte is a moist paste of \(NH_4Cl\) and \(ZnCl_2\).


Step 1: Write the Anode reaction

The zinc container acts as the anode where oxidation takes place. \[ Anode: Zn(s) \rightarrow Zn^{2+}(aq) + 2e^- \]

Step 2: Write the Cathode reaction

At the cathode, manganese is reduced from the +4 oxidation state to +3. \[ Cathode: MnO_2(s) + NH_4^+(aq) + e^- \rightarrow MnO(OH)(s) + NH_3(g) \]

Step 3: Write the Overall reaction

To get the overall reaction, multiply the cathode reaction by 2 to balance electrons and add it to the anode reaction: \[ Zn(s) + 2MnO_2(s) + 2NH_4^+(aq) \rightarrow Zn^{2+}(aq) + 2MnO(OH)(s) + 2NH_3(g) \]
Note: The \(NH_3\) gas produced reacts with \(Zn^{2+}\) to form a complex ion \([Zn(NH_3)_4]^{2+}\), which prevents pressure build-up inside the cell. Quick Tip: Dry cells are not rechargeable because the chemical reactions are irreversible. The voltage of a dry cell is approximately 1.5 V.

Concept:

\(E_{cell}\) is the potential of the cell at a specific concentration and temperature.
\(E^\circ_{cell}\) is the standard potential (constant for a given cell).


Step 1: Analyze the cell at equilibrium

When a chemical reaction in a cell reaches equilibrium, the net flow of electrons stops. At this stage, the potential of the cell (\(E_{cell}\)) becomes zero. \[ E_{cell} = 0 at equilibrium \]

Step 2: Apply the Nernst Equation at equilibrium

The Nernst equation is: \(E_{cell} = E^\circ_{cell} - \frac{2.303RT}{nF} \log Q_c\).
At equilibrium, \(Q_c = K_c\) and \(E_{cell} = 0\). \[ 0 = E^\circ_{cell} - \frac{2.303RT}{nF} \log K_c \implies E^\circ_{cell} = \frac{2.303RT}{nF} \log K_c \]

Step 3: Conclusion
\(E_{cell}\) is a variable that changes with concentration and eventually becomes zero at equilibrium. On the other hand, \(E^\circ_{cell}\) is a constant for a given reaction. Therefore, the equilibrium constant \(K_c\), which is also a constant at a given temperature, is mathematically related to the constant \(E^\circ_{cell}\) through the thermodynamic state of the system at equilibrium. Quick Tip: At equilibrium, a battery is "dead," so \(E_{cell} = 0\). The relationship \(\Delta G^\circ = -nFE^\circ_{cell} = -RT \ln K_c\) perfectly links standard potential to the equilibrium constant.

Concept:

Molar conductivity (\(\Lambda_m\)) is calculated as: \(\Lambda_m = \frac{\kappa \times 1000}{M}\).
Limiting molar conductivity (\(\Lambda_m^\circ\)) is the sum of ionic conductivities (Kohlrausch's Law).
Degree of dissociation (\(\alpha\)) = \(\frac{\Lambda_m}{\Lambda_m^\circ}\).


Step 1: Calculate Molar Conductivity (\(\Lambda_m\))

Given: \(\kappa = 3.905 \times 10^{-5} S cm^{-1}\) and \(M = 0.001 M = 10^{-3} M\). \[ \Lambda_m = \frac{3.905 \times 10^{-5} \times 1000}{0.001} \] \[ \Lambda_m = \frac{3.905 \times 10^{-2}}{10^{-3}} = 3.905 \times 10 = 39.05 S cm^2 mol^{-1} \]

Step 2: Calculate Limiting Molar Conductivity (\(\Lambda_m^\circ\))

Using Kohlrausch's law: \[ \Lambda_m^\circ(CH_3COOH) = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} \] \[ \Lambda_m^\circ = 349.6 + 40.9 = 390.5 S cm^2 mol^{-1} \]

Step 3: Calculate Degree of Dissociation (\(\alpha\))
\[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \] \[ \alpha = \frac{39.05}{390.5} = 0.1 \] Quick Tip: The unit of \(\Lambda_m\) depends on the units of \(\kappa\). If \(\kappa\) is in S cm\(^{-1}\), use the factor 1000 in the numerator. \(\alpha\) is a dimensionless ratio and should always be \(\leq 1\).

Concept:



The emf of an electrochemical cell depends on the concentration of ions participating in the cell reaction.


Step 1: Reason


In a mercury cell, the overall cell reaction is:


\[ Zn(Hg) + HgO(s) \rightarrow ZnO(s) + Hg(l) \]

During this reaction, no ions are produced or consumed in the electrolyte. Therefore, the concentration of the electrolyte remains unchanged throughout the operation of the cell.


Since the ionic concentration remains constant, the cell potential also remains constant. Hence, a mercury cell delivers a nearly constant voltage of about 1.35 V until all the reactants are completely consumed.




Quick Tip: Mercury cells provide a constant voltage because the electrolyte concentration does not change during the cell reaction.

Concept:



A galvanic cell can function continuously only if electrical neutrality is maintained in both half-cells.


Step 1: Reason


A salt bridge is necessary in a galvanic cell because it:



Maintains electrical neutrality by allowing the migration of ions into the two half-cells, thereby preventing the accumulation of excess positive or negative charges.


Completes the electrical circuit by providing a path for the flow of ions between the two half-cells, allowing the continuous flow of electrons through the external circuit.


Without a salt bridge, charge would quickly accumulate in the half-cells, causing the flow of electrons to stop and the cell to cease functioning.



Quick Tip: A salt bridge usually contains an inert electrolyte such as \(KCl\) or \(KNO_3\) in agar-agar gel. It completes the circuit and maintains charge balance.