CBSE Class 12 2025 Chemistry 56-4-2 Question Paper Set-2: Download Solutions with Answer Key

When alcohols are heated with concentrated sulphuric acid, the first step is the protonation of the -OH group by H\textsuperscript{+ from H\textsubscript{2SO\textsubscript{4. This converts the -OH group into a better leaving group (water), which facilitates the elimination process that eventually forms the alkene.
Quick Tip: Always remember that in acid-catalyzed dehydration of alcohols, protonation of the hydroxyl group is the initial step before elimination.

DDT (Dichlorodiphenyltrichloroethane) is a synthetic insecticide that was widely used in agriculture. However, it is not biodegradable and tends to persist in the environment, leading to bioaccumulation and ecological harm. Therefore, it is classified as a non-biodegradable insecticide.
Quick Tip: DDT is a classic example of a harmful persistent organic pollutant (POP). Always link DDT with non-biodegradability and environmental concerns.

In an alkaline medium, permanganate (MnO\textsubscript{4\textsuperscript{–) is a strong oxidizing agent and oxidizes iodide (I\textsuperscript{–) to periodate (IO\textsubscript{4\textsuperscript{–). This is a higher oxidation state of iodine (+7), which is typically formed under strongly oxidizing alkaline conditions.
Quick Tip: Remember that oxidation of iodide in alkaline medium by MnO\textsubscript{4}\textsuperscript{–} gives IO\textsubscript{4}\textsuperscript{–} (periodate), while in acidic medium it usually forms I\textsubscript{2}.

Aniline is first treated with NaNO\textsubscript{2 and HCl at 0–5°C which forms a diazonium salt (benzene diazonium chloride). This is a classic diazotization reaction. The compound Y in this case is the diazonium salt itself — C\textsubscript{6H\textsubscript{5–N\textsubscript{2\textsuperscript{+Cl\textsuperscript{–.
Quick Tip: Diazotization of aniline using NaNO\textsubscript{2} and HCl at low temperature yields diazonium salt — useful in Sandmeyer and coupling reactions.

2-Bromobutane has a chiral center — the second carbon is bonded to four different groups: H, Br, CH\textsubscript{3, and C\textsubscript{2H\textsubscript{5. This makes it optically active. The other molecules lack such a carbon with four distinct substituents.
Quick Tip: To check chirality, look for a carbon atom attached to four different groups — that makes it a chiral center.

The anomeric carbon is the carbon that was originally part of the carbonyl group (aldehyde or ketone) in the open-chain form of the sugar. When the sugar cyclizes into the Haworth form, this carbon becomes a new chiral center (bonded to OH and OR groups). In the given structure, carbon number 1 is the anomeric carbon.
Quick Tip: Always count from the carbon next to the oxygen in the ring. Carbon 1 is the anomeric carbon in aldoses like glucose.

Henry’s law states that the solubility of a gas in a liquid is inversely proportional to Henry’s constant. A higher \( K_H \) value indicates that the gas is less soluble in the liquid, and a lower \( K_H \) means higher solubility.
Quick Tip: Remember: High \( K_H \) → Low solubility, Low \( K_H \) → High solubility.

Most lanthanoids are not radioactive. Only some of the later actinoids exhibit strong radioactivity. Hence, statement (D) is incorrect. All other statements are correct, especially regarding atomic radii and lanthanoid contraction.
Quick Tip: Only actinoids are largely radioactive, not lanthanoids. Look for words like “all” or “always” in such tricky options.

As the redox reaction proceeds, Ag\textsuperscript{+ is consumed and Mg\textsuperscript{2+ is produced. Increase in [Mg\textsuperscript{2+] and decrease in [Ag\textsuperscript{+] both contribute to a decrease in E\textsubscript{cell, according to the Nernst equation. Hence, the voltage will decrease.
Quick Tip: Use the Nernst equation: more products (like Mg\textsuperscript{2+}) or fewer reactants (like Ag\textsuperscript{+}) cause a drop in voltage.

Mn\textsuperscript{2+ has the electron configuration [Ar] 3d\textsuperscript{5, which means 5 unpaired electrons. The magnetic moment is given by the formula \( \mu = \sqrt{n(n+2)} \). For n = 5, \( \mu = \sqrt{35} ≈ 5.92 \, \mu_B \), which is the highest among the given options.
Quick Tip: To find magnetic moment, count unpaired electrons and use the formula \( \mu = \sqrt{n(n+2)} \).

Hofmann bromamide degradation is a reaction in which a primary amide (like CH\textsubscript{3CONH\textsubscript{2) is treated with Br\textsubscript{2 and a strong base (NaOH), resulting in the formation of a primary amine with one carbon atom less — here forming CH\textsubscript{3NH\textsubscript{2. Thus, the correct reactant is CH\textsubscript{3CONH\textsubscript{2.
Quick Tip: In Hofmann degradation, always look for a primary amide that loses one carbon atom to form a primary amine.

The oxidation of water to oxygen involves the half-reaction: \[ 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \]
This indicates that 4 electrons are transferred per mole of O\textsubscript{2 formed. Since 1 Faraday (F) corresponds to 1 mole of electrons, 4 Faradays are required for the oxidation of 1 mole of H\textsubscript{2O to O\textsubscript{2.
Quick Tip: For oxidation reactions, balance the electrons lost. O\textsubscript{2} evolution from H\textsubscript{2}O requires 4e\textsuperscript{–}, i.e., 4F per mole.

Cuprous ion (Cu\textsuperscript{+) has an electronic configuration of [Ar]3d\textsuperscript{10, meaning all d-orbitals are fully filled, resulting in no unpaired electrons — hence it is diamagnetic. The reason is incorrect because the 3d orbitals are \textit{fully filled, not partially.
Quick Tip: Diamagnetism arises when all electrons are paired. Check the electronic configuration to confirm.

Acetanilide is \textit{less basic than aniline because the acetyl group withdraws electron density from nitrogen, reducing its ability to donate a lone pair. So, the assertion is incorrect. However, the reason is correct.
Quick Tip: Electron-withdrawing groups reduce basicity by delocalizing the nitrogen’s lone pair.

In aqueous NaCl electrolysis, water is reduced at the cathode to form H\textsubscript{2, and chloride ions are oxidized at the anode to form Cl\textsubscript{2. This occurs because chlorine has a higher oxidation potential than water, making it preferentially oxidized.
Quick Tip: Always compare standard electrode potentials to determine what gets oxidized or reduced.

Although the C–Cl bond is more polar, the boiling point depends more on molecular weight and van der Waals forces. n-Butyl bromide has a higher molecular mass and therefore exhibits stronger dispersion forces, giving it a higher boiling point.
Quick Tip: Polarity affects boiling point, but molecular mass and van der Waals forces often have greater impact.

Essential amino acids are those that cannot be synthesized by the human body and must be obtained from the diet. Examples include leucine, isoleucine, and lysine.

Amino acids are amphoteric in nature because they contain both acidic (–COOH) and basic (–NH\textsubscript{2) functional groups. This allows them to act both as acids and bases, depending on the pH of the medium.
Quick Tip: Remember: Essential amino acids must be taken through food, and amphoteric nature arises from the presence of both acidic and basic groups.

The concentration halves every hour (from 0.40 to 0.20 to 0.10 to 0.05), indicating a constant half-life. This is characteristic of a first-order reaction.

Rate law expression: \[ Rate = k[A] \]
Quick Tip: A constant half-life is a clear indicator of a first-order reaction.

Adding 1 mole of H\textsubscript{2 increases [H\textsubscript{2], which increases the rate of the reaction because the rate depends on the product of [H\textsubscript{2] and [I\textsubscript{2].

However, the rate constant \( k \) depends only on temperature and remains unchanged as temperature is constant.
Quick Tip: Rate changes with concentration; rate constant stays constant unless temperature changes.

The complex is \ce{K2[PtCl6]. The chloride ions are inside the coordination sphere and not free, so they don't react with AgNO\textsubscript{3.

IUPAC name: Potassium hexachloroplatinate(IV)
Quick Tip: If chloride is inside the coordination sphere, it won’t form a precipitate with AgNO\textsubscript{3}.

A fuel cell is a device that converts chemical energy of a fuel (like hydrogen) directly into electrical energy through redox reactions.

Advantages:
1. High efficiency in energy conversion.

2. Environmentally friendly — water is the only by-product (in hydrogen fuel cells).
Quick Tip: Fuel cells are clean and efficient energy sources, unlike conventional batteries.

This reaction is a reduction of carboxylic acids by aluminum metal. The expected product is the corresponding alkane. \[ 6R–COOH + 2Al \rightarrow 3R–CH_3 + 2Al(OH)_3 \]

(b) \chemfig{CH_3–CO–CH_2–CO–CH_3} + NaBH\textsubscript{4}

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Solution:

Sodium borohydride reduces only aldehydes and ketones (not esters or carboxylic acids). The carbonyl groups are reduced to alcohols: \[ Product: \chemfig{CH_3–CH(OH)–CH_2–CH(OH)–CH_3} \] Quick Tip: NaBH\textsubscript{4} selectively reduces aldehydes and ketones, not acids or esters.

No, sodium methoxide and t-butyl bromide cannot be used effectively for the preparation of t-butyl methyl ether. This is because t-butyl bromide is a tertiary halide, which undergoes elimination rather than substitution with a strong base like sodium methoxide. It leads to the formation of alkenes via E2 mechanism.

Instead, methyl bromide (CH\textsubscript{3Br) and sodium t-butoxide should be used. Since CH\textsubscript{3Br is a primary halide, it undergoes SN2 reaction easily with t-butoxide ion, forming t-butyl methyl ether.
Quick Tip: Use primary alkyl halides with bulky alkoxides for ether synthesis to avoid elimination.

The IUPAC name of t-butyl methyl ether is: 2-methoxy-2-methylpropane.
Quick Tip: When naming ethers, choose the larger group as the parent chain and name the smaller as an alkoxy substituent.

Lower pK\textsubscript{b implies stronger base. Order of increasing pK\textsubscript{b (i.e., decreasing basicity):

(C\textsubscript{2H\textsubscript{5)\textsubscript{2NH \textless{ C\textsubscript{2H\textsubscript{5NH\textsubscript{2 \textless{ C\textsubscript{6H\textsubscript{5NHCH\textsubscript{3 \textless{ C\textsubscript{6H\textsubscript{5NH\textsubscript{2
Quick Tip: Aromatic amines are weaker bases due to resonance; aliphatic amines are stronger.

Cell notation: Al | Al\textsuperscript{3+ (0.001 M) || Ni\textsuperscript{2+ (0.001 M) | Ni

E\textsubscript{cell = E\textsuperscript{0\textsubscript{cathode – E\textsuperscript{0\textsubscript{anode
= (–0.25 V) – (–1.66 V) = 1.41 V
Quick Tip: E\textsubscript{cell} is always cathode minus anode. The more positive reduction potential is the cathode.

Allyl chloride undergoes faster SN1 reaction because the carbocation formed is resonance stabilized. n-Propyl chloride does not form such a stable carbocation, making its hydrolysis slower.


(ii) Isocyanides are formed when alkyl halides are treated with silver cyanide.


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Solution:

AgCN is covalent and favors the formation of isocyanides (R–NC) due to bond formation through nitrogen rather than carbon, unlike KCN which is ionic and gives nitriles (R–CN).


(iii) Methyl chloride reacts faster with OH\textsuperscript{–} ion in SN\textsubscript{2} reaction than ethyl chloride.


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Solution:

In SN2 reactions, less steric hindrance facilitates the reaction. Methyl chloride has no bulky groups attached to the carbon, making it more reactive than ethyl chloride, which has one additional methyl group.
Quick Tip: SN2 favors primary halides with minimal steric hindrance; resonance also stabilizes intermediates in SN1.

(i) CH\textsubscript{3CH = CH\textsubscript{2 + HBr (Peroxide) → CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2Br (‘A’)
→ (aq. KOH) → CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2OH (‘B’)


(ii) CH\textsubscript{3CH\textsubscript{2CH(Cl)CH\textsubscript{3 + alc. KOH → CH\textsubscript{3CH = CHCH\textsubscript{3 (‘A’)
→ HBr → CH\textsubscript{3CHBrCH\textsubscript{2CH\textsubscript{3 (‘B’)


(iii) CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2Br (‘A’) + Mg → CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2MgBr
→ H\textsubscript{2O/H\textsuperscript{+ → CH\textsubscript{3CH\textsubscript{2CH\textsubscript{3 (‘B’)
Quick Tip: In peroxide conditions, HBr adds via anti-Markovnikov rule. Grignard reagents form alkanes with protonation.

Mn\textsuperscript{3+ readily gains an electron to become the more stable Mn\textsuperscript{2+, hence acts as an oxidising agent.
Cr\textsuperscript{2+ loses an electron to form Cr\textsuperscript{3+, which is more stable due to half-filled t\textsubscript{2g orbitals (d\textsuperscript{3), thus acts as a reducing agent.


(b) Actinoid contraction is greater than lanthanoid contraction.


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Solution:

Actinoids have 5f orbitals that are more diffused and poorly shielded compared to 4f orbitals in lanthanoids. This leads to a greater effective nuclear charge and stronger contraction across the series.


(c) Transition metals form interstitial compounds with H, B, C, N.


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Solution:

Transition metals have vacant d-orbitals and lattice voids where small atoms like H, B, C, N can fit without disturbing the structure, forming interstitial compounds which are hard and have high melting points.
Quick Tip: Remember: Stability trends of oxidation states depend on electron configuration and shielding effect.

Given molar mass by formula = 40 g mol\textsuperscript{–1, observed = 25 g mol\textsuperscript{–1
Degree of dissociation (\( \alpha \)) is related as: \[ Observed molar mass = \frac{Theoretical molar mass}{1 + \alpha} \] \[ 25 = \frac{40}{1 + \alpha} \Rightarrow 1 + \alpha = \frac{40}{25} = 1.6 \Rightarrow \alpha = 0.6 \]
Percentage dissociation = \( \alpha \times 100 = 60% \)
Quick Tip: Lower observed molar mass implies dissociation; use van’t Hoff factor relations to find α.

Positive 2,4-DNP test and no reaction with Tollen’s or Fehling’s test indicate ‘A’ is a ketone.
C\textsubscript{4H\textsubscript{8O ketone → likely structure: Butan-2-one (CH\textsubscript{3COCH\textsubscript{2CH\textsubscript{3)
‘B’ is the sodium bisulfite addition product of the ketone.
On oxidation, ketones give carboxylic acids at the more oxidizable position, forming: CH\textsubscript{3COOH (ethanoic acid) and CO\textsubscript{2.
But since one carboxylic acid is obtained, structure of ‘C’ is likely CH\textsubscript{3COOH (ethanoic acid).

Structures:
‘A’ = Butan-2-one (CH\textsubscript{3COCH\textsubscript{2CH\textsubscript{3)

‘B’ = Bisulfite adduct of butan-2-one

‘C’ = Ethanoic acid (CH\textsubscript{3COOH)
Quick Tip: Use functional group tests like DNP and oxidation behavior to identify aldehydes vs ketones.

For a first-order reaction, the integrated rate law is:
\[ k = \frac{2.303}{t} \log \left( \frac{[R]_0}{[R]} \right) \]
Given:
\([R]_0 = 1.2 \times 10^{-2} \, mol L^{-1}\),
\([R] = 0.2 \times 10^{-2} \, mol L^{-1}\),
\(t = 60 \, min = 60 \times 60 = 3600 \, s\)

\[ k = \frac{2.303}{3600} \log \left( \frac{1.2 \times 10^{-2}}{0.2 \times 10^{-2}} \right) = \frac{2.303}{3600} \log(6) \] \[ k = \frac{2.303 \times 0.778}{3600} \approx \frac{1.792}{3600} \approx 4.98 \times 10^{-4} \, s^{-1} \] Quick Tip: Always use the first-order rate law equation with log when concentrations and time are given.

This is a reaction of semicarbazide with an aldehyde or ketone. The product is a semicarbazone.

Product: \chemfig{C(=N-NH-C(=O)-NH_2)-Ph


(II)
(CH\textsubscript{3)\textsubscript{2Cd + 2CH\textsubscript{3COCl \(\rightarrow\)

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Solution:

This is a cadmium-based nucleophilic acyl substitution yielding a ketone.

Product: CH\textsubscript{3COCH(CH\textsubscript{3)\textsubscript{2 (dimethyl ketone)


(III)
\chemfig{Ph-COCl \(\xrightarrow{H_2/Pd-BaSO_4}\) ?

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Solution:

This is Rosenmund reduction of acid chloride.

Product: \chemfig{Ph-CHO (Benzaldehyde)
Quick Tip: Cadmium reagents form ketones; Rosenmund reduction converts acid chlorides to aldehydes.

Add NaHCO\textsubscript{3 solution. Benzoic acid gives brisk effervescence (CO\textsubscript{2 gas), but ethyl benzoate does not react.


(II) Propanal and propanone

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Solution:

Use Tollen’s reagent. Propanal (an aldehyde) gives silver mirror test; propanone (a ketone) does not react.
Quick Tip: Use bicarbonate test for acids; Tollen’s test for aldehydes vs ketones.

Oxidation of ethylbenzene leads to benzoic acid.

Product: \chemfig{Ph-COOH (Benzoic acid)


(II) \chemfig{CH_2=CH-CH_2CH_2CH_3 \(\xrightarrow{O_3/H_2O}\) ?

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Solution:

Ozonolysis of alkenes splits the double bond into carbonyl compounds.

Product: Pentanal (CH\textsubscript{3(CH\textsubscript{2)\textsubscript{3CHO)


(III) \chemfig{Ph-CHO + [Ag(NH\textsubscript{3)\textsubscript{2]\textsuperscript{+ \(\rightarrow\) ?

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Solution:

This is Tollen’s reagent test, oxidizing aldehydes to acids.

Product: Benzoic acid (Ph–COOH)
Quick Tip: KMnO\textsubscript{4} oxidizes side chains; ozonolysis splits alkenes; Tollen’s oxidizes aldehydes.

Step 1: Convert benzaldehyde to benzyl alcohol via reduction (e.g., NaBH\textsubscript{4)

Step 2: Friedel-Crafts acylation of benzene using benzoyl chloride (from benzoic acid or benzaldehyde) in the presence of AlCl\textsubscript{3.

Final product: Benzophenone (diphenyl ketone)


(II) Benzaldehyde to 3-phenyl propanol

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Solution:

Use Clemmensen or Wolff-Kishner reduction to get ethylbenzene.

Then, perform hydroboration-oxidation or Grignard reaction with formaldehyde followed by reduction to obtain 3-phenylpropanol.
Quick Tip: Use multistep reductions and Grignard additions for carbon chain extension and alcohol synthesis.

Ni has atomic number 28. So its configuration is [Ar] 3d\textsuperscript{8 4s\textsuperscript{2.

In [Ni(CO)\textsubscript{4], Ni is in 0 oxidation state. CO is a strong field ligand, causes pairing of electrons. All electrons are paired. Hence, it is diamagnetic.

In [NiCl\textsubscript{4]\textsuperscript{2−, Ni is in +2 oxidation state, i.e., 3d\textsuperscript{8. Cl\textsuperscript{− is a weak field ligand, does not cause pairing. Thus, unpaired electrons remain. Hence, it is paramagnetic.


(II) CO is a stronger complexing agent than NH\textsubscript{3}.

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Solution:

CO has better \(\pi\)-acceptor properties than NH\textsubscript{3. It engages in back bonding with metal orbitals, leading to stronger bonding.

NH\textsubscript{3 only donates lone pair through \(\sigma\) bonding, hence less effective.


(III) The trans isomer of complex [Co(en)\textsubscript{2}Cl\textsubscript{2}]\textsuperscript{+} is optically inactive.

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Solution:

The trans form has a plane of symmetry due to identical ligands opposite each other.

Due to symmetry, it cannot show optical activity (no chirality). Only cis form is optically active.
Quick Tip: Strong field ligands cause pairing of electrons and tend to form low-spin complexes. Optical activity requires chirality (no symmetry).

Fe\textsuperscript{3+ \(\Rightarrow\) [Ar] 3d\textsuperscript{5.

Strong field ligand causes pairing in t\textsubscript{2g orbitals.

So configuration: t\textsubscript{2g\textsuperscript{5 e\textsubscript{g\textsuperscript{0

Number of unpaired electrons = 1


(II) Weak field ligand

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Solution:

Weak field ligand does not cause pairing.

So configuration: t\textsubscript{2g\textsuperscript{3 e\textsubscript{g\textsuperscript{2

Number of unpaired electrons = 5
Quick Tip: Strong field ligands result in fewer unpaired electrons (low spin), while weak field ligands maintain high spin state.

This shows ionisation isomerism.

Isomer: [Co(CN)\textsubscript{6][Cr(NH\textsubscript{3)\textsubscript{6]


(II) [Co(en)\textsubscript{3}]\textsuperscript{3+}

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Solution:

This shows optical isomerism due to chirality from en ligands.

Isomers: d- and l- forms (non-superimposable mirror images).


(III) [Co(NH\textsubscript{3})\textsubscript{3}(NO\textsubscript{2})\textsubscript{3}]

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Solution:

This shows linkage isomerism. NO\textsubscript{2 ligand can bind through N or O.

Isomers: Nitro (NO\textsubscript{2-N) and Nitrito (NO\textsubscript{2-O) complexes.
Quick Tip: Isomerism types depend on ligand identity, binding atoms, and spatial arrangements.

- Strong field ligands (e.g., CN\textsuperscript{−, CO) cause large crystal field splitting (\(\Delta\)). They promote electron pairing in t\textsubscript{2g orbitals \(\Rightarrow\) Low spin complexes.

- Weak field ligands (e.g., F\textsuperscript{−, Cl\textsuperscript{−) cause small splitting. Electrons remain unpaired \(\Rightarrow\) High spin complexes.

- Spin state is determined by the relative magnitude of crystal field splitting (\(\Delta\)) vs pairing energy.
Quick Tip: Stronger ligands reduce unpaired electrons and form low spin complexes; weak ligands lead to high spin configurations.