IIT JAM 2021 Chemistry (CY) Question Paper with Answer Key PDFs (February 14 - Afternoon Session)

Step 1: Relation between acidity and p\(K_a\).

Lower p\(K_a\) corresponds to higher acidity. In metal hydride complexes, acidity decreases as electron density on the metal center increases.


Step 2: Ligand effects.

CO is a strong \(\pi\)-acceptor, withdrawing electron density from the metal and increasing acidity (lowering p\(K_a\)).
PPh\(_3\) is a \(\sigma\)-donor and weak \(\pi\)-acceptor, increasing metal electron density and reducing acidity (raising p\(K_a\)).


Step 3: Comparing the complexes.
\[ I: HCo(CO)_4 \rightarrow most acidic (lowest pK_a) \] \[ IV: HCo(CO)_2(PPh_3)_2 \rightarrow least acidic (highest pK_a) \]

Thus, the p\(K_a\) order is IV > III > II > I.


Step 4: Conclusion.

The correct order of p\(K_a\) values is IV > III > II > I.
Quick Tip: Replacing \(\pi\)-acceptor ligands (like CO) with \(\sigma\)-donors (like PPh\(_3\)) increases electron density on the metal, decreasing acidity (increasing p\(K_a\)).

Step 1: Concept of isoelectronic species.

Na\(^+\), Mg\(^{2+}\), Al\(^{3+}\), and F\(^-\) are isoelectronic (each has 10 electrons). The ionic size depends mainly on the nuclear charge (Z).

Step 2: Effect of nuclear charge.

As nuclear charge increases, the same number of electrons are pulled more strongly toward the nucleus, decreasing the ionic radius.


Step 3: Order of Z (atomic number).

F (9) < Na (11) < Mg (12) < Al (13)

Hence, radius order (inverse relation): \[ F^- > Na^+ > Mg^{2+} > Al^{3+}. \]

Step 4: Conclusion.

The correct order of ionic radii is F\(^-\) > Na\(^+\) > Mg\(^{2+}\) > Al\(^{3+}\).
Quick Tip: In isoelectronic ions, as nuclear charge increases, size decreases due to stronger electrostatic attraction on the same electron cloud.

Step 1: For [NiCl\(_2\)(PPh\(_3\))\(_2\)].

Ni is in +2 oxidation state → 3d\(^8\) configuration.
Cl\(^-\) and PPh\(_3\) are weak field ligands → complex is paramagnetic.
Unpaired electrons = 2.

Spin-only magnetic moment, \[ \mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.83\ BM. \]

Step 2: For [Mn(NCS)\(_6\)]\(^{4-}\).

Mn is in +2 oxidation state → 3d\(^5\) configuration.
All ligands are weak field (high-spin complex).
Unpaired electrons = 5.
\[ \mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\ BM. \]

Step 3: Conclusion.

Hence, magnetic moments are 2.83 BM and 5.92 BM, respectively.
Quick Tip: For spin-only moment, use \(\mu = \sqrt{n(n+2)}\). Weak field ligands lead to high-spin complexes with maximum unpaired electrons.

Step 1: Formula for number of radial nodes.

Number of radial nodes = \( n - l - 1 \).

Step 2: Evaluate each option.

(A) 3s: \(3 - 0 - 1 = 2\); 2s: \(2 - 0 - 1 = 1\) → different.

(B) 3p: \(3 - 1 - 1 = 1\); 2p: \(2 - 1 - 1 = 0\) → different.

(C) 3d: \(3 - 2 - 1 = 0\); 2p: \(2 - 1 - 1 = 0\) → same (0 radial nodes).

(D) 3p: \(3 - 1 - 1 = 1\); 2p: \(2 - 1 - 1 = 0\) → different.


Step 3: Conclusion.

The pair (3d, 2p) has the same number of radial nodes (0). Quick Tip: Number of radial nodes = \( n - l - 1 \). Orbitals with higher \(l\) for the same \(n\) have fewer radial nodes.

Step 1: Reaction type identification.

The reaction is an **acetal (or thioacetal) hydrolysis** under acidic conditions.
CF\(_3\)COOH/H\(_2\)O is a mild acidic system that can cleave acetals or thioacetals to the corresponding carbonyl compounds.


Step 2: Mechanism.

The methoxy (–OCH\(_3\)) and thioether (–S–) substituents form a cyclic dithioacetal protecting group around the carbonyl precursor.
Upon treatment with CF\(_3\)COOH/H\(_2\)O, this cyclic acetal hydrolyzes, regenerating the **aldehyde**.


Step 3: Product formed.

Therefore, the product is a **carbonyl compound (aldehyde)** having the same carbon skeleton.
Hence, structure (B) is correct, representing the aldehyde after deprotection.


Step 4: Conclusion.

The major product is an aldehyde (Option B) obtained via acid-catalyzed hydrolysis of a cyclic thioacetal.
Quick Tip: CF\(_3\)COOH/H\(_2\)O acts as a mild acid for hydrolyzing acetals and thioacetals to the corresponding carbonyl compounds.

Step 1: Reaction type identification.

This reaction involves an **HCN/KCN system** at high temperature (160°C),
which promotes **Strecker-type cyanohydrin formation followed by rearrangement** or **conjugate addition (Michael addition)** depending on the substrate.


Step 2: Substrate analysis.

The substrate is an \(\alpha,\beta\)-unsaturated aldehyde (enone-type).
Under basic HCN/KCN conditions, nucleophilic addition of CN\(^-\) occurs at the **\(\beta\)-carbon** via Michael addition.


Step 3: Product formation.

The cyanide adds to the \(\beta\)-position, generating a **\(\beta\)-cyano carbonyl compound**.
Thus, the product contains both a carbonyl and a nitrile group separated by one carbon atom.


Step 4: Conclusion.

The major product is the \(\beta\)-cyano aldehyde (Option A).
Quick Tip: Under basic conditions, conjugated aldehydes or ketones react with CN\(^-\) at the \(\beta\)-position (Michael addition), yielding \(\beta\)-cyano carbonyl products.

Step 1: Analyze aromatic proton signals.

The two doublets at \(\delta\) 7.31 (2H) and \(\delta\) 7.21 (2H) indicate a para-disubstituted benzene ring, where two sets of aromatic protons are magnetically equivalent (2H each).


Step 2: Analyze aliphatic signals.

A signal at \(\delta\) 4.5 (2H) corresponds to a benzylic –CH\(_2\)–Cl group, because protons attached to carbon bearing chlorine appear in this region.
The signal at \(\delta\) 2.3 (3H) indicates a methyl group attached to an aromatic ring (–CH\(_3\) directly bonded to benzene).


Step 3: Combine structural fragments.

Combining the data gives a para-disubstituted benzene ring having a –CH\(_2\)Cl and –CH\(_3\) group.
\[ Hence, the structure is p-CH_3–C_6H_4–CH_2Cl. \]

Step 4: Conclusion.

The correct structure corresponds to Option (D).
Quick Tip: In aromatic compounds, para-disubstitution gives two sets of equivalent protons (2H each). Benzylic –CH\(_2\)Cl appears around 4.5 ppm, while Ar–CH\(_3\) appears near 2.3 ppm.

Step 1: Reaction type identification.

Cycloheptadiene undergoes a [1,5]-sigmatropic hydrogen shift or thermal rearrangement when heated to high temperatures (~170°C).
This rearrangement converts the conjugated diene into a more stable conjugated triene.


Step 2: Mechanism.

At elevated temperature, the \(\pi\)-electrons shift, and one of the hydrogens migrates across the conjugated system via a concerted pericyclic pathway, forming a substituted cycloheptatriene.

Step 3: Product formed.

The resulting compound is **1,3,5-cycloheptatriene** (tropylium-like structure), represented by Option (C).

Step 4: Conclusion.

The major product of the thermal rearrangement is Option (C).
Quick Tip: Cycloheptadiene rearranges via a [1,5]-sigmatropic shift upon heating to form cycloheptatriene. Such rearrangements follow Woodward–Hoffmann orbital symmetry rules for thermal reactions.

Step 1: Understanding the problem.

The question gives that solid M is less dense than liquid M.
That means, upon melting, the substance contracts (volume decreases).
An example of such a substance is water (ice is less dense than liquid water).


Step 2: Using the Clausius–Clapeyron equation.

The slope of the solid–liquid equilibrium line is given by \[ \frac{dP}{dT} = \frac{\Delta S_fusion}{\Delta V_fusion} \]
where \(\Delta S_fusion\) is the entropy change upon fusion and \(\Delta V_fusion\) is the volume change.

Step 3: Sign of the terms.

Given: \(\Delta S_fusion\) = +25 J K\(^{-1}\) mol\(^{-1}\) (positive, since melting increases entropy).
Since the solid is less dense than the liquid, \(\Delta V_fusion\) = \(V_liquid - V_solid\) < 0.

Therefore, \[ \frac{dP}{dT} = \frac{(+)}{(-)} = negative. \]

Step 4: Interpretation.

A negative slope in the phase diagram indicates that the melting point decreases with increasing pressure —
which matches the diagram shown in Option (B).


Step 5: Conclusion.

Hence, the correct pressure–temperature diagram is Option (B).
Quick Tip: When a solid is less dense than its liquid (like ice), increasing pressure lowers its melting point because \(\Delta V_fusion\) is negative, giving a negative \(\frac{dP}{dT}\) slope.

Step 1: Determinant of P.
\(P\) is an identity matrix. \[ \det(P) = 1 \ (non-zero) \]

Step 2: Determinant of Q.
\(Q\) is a diagonal matrix. \[ \det(Q) = 1 \times 2 \times 3 \times 4 = 24 \ (non-zero) \]

Step 3: Determinant of R.
\(R\) is a lower triangular matrix with diagonal elements 1, 2, 3, 4. \[ \det(R) = 1 \times 2 \times 3 \times 4 = 24 \ (non-zero) \]

Step 4: Determinant of S.

For \(S\), rows are linearly dependent (each can be written as a linear combination of others).
Hence, \[ \det(S) = 0. \]

Step 5: Conclusion.

Matrices with non-zero determinants are P, Q, and R.
Thus, the correct answer is (A).
Quick Tip: For diagonal or triangular matrices, the determinant equals the product of diagonal elements. If any row (or column) is linearly dependent on others, the determinant becomes zero.

Step 1: Formation of compound P.

Boron trichloride reacts with ammonium chloride at high temperature (140°C) to form **borazine (B\(_3\)N\(_3\)H\(_6\))**, also called inorganic benzene.
\[ 3 BCl_3 + 3 NH_4Cl \rightarrow B_3N_3H_6 + 9 HCl \]

Step 2: Reaction with NaBH\(_4\).

Borazine reacts with sodium borohydride to form a colorless liquid borane derivative (Q), typically **borazine hydride**.

Step 3: Hydrolysis of Q.

On hydrolysis, borazine hydride produces **boric acid** and hydrogen gas. \[ Q + H_2O \rightarrow R (boric acid) + H_2 \]

Step 4: Conclusion.

Hence, compound P is B\(_3\)N\(_3\)H\(_6\) (borazine).
Quick Tip: Borazine (B\(_3\)N\(_3\)H\(_6\)) is called inorganic benzene. It is formed by heating BCl\(_3\) and NH\(_4\)Cl and reacts with NaBH\(_4\) to give borane derivatives.

Step 1: Apply the 18-electron rule.

Total valence electron count (VEC) = metal valence electrons + electrons donated by ligands.

For (A) Ti (group 4) = 4 e\(^-\).
C\(_5\)H\(_5^-\) = 6 e\(^-\).
4 CO = 8 e\(^-\).
Charge = –1 → add 1 e\(^-\).
Total = 4 + 6 + 8 + 1 = 19 e\(^-\) → 18-electron satisfied (approximately stable).


For (B) Mn (group 7) = 7 e\(^-\).
Each SnPh\(_3\) = 2 e\(^-\) × 2 = 4 e\(^-\).
2 CO = 4 e\(^-\).
Total = 7 + 4 + 4 = 15 e\(^-\).
Thus, it does NOT obey the 18-electron rule.

Step 4: Conclusion.

The complex that does not obey the 18-electron rule is (B).
Quick Tip: Use the 18-electron rule to check stability: stable organometallic complexes usually follow it. Group number + ligand electrons + charge correction = 18 indicates stability.

Step 1: Determine hybridization.

For I\(_3^-\): 3 bond pairs + 3 lone pairs = 5 regions → sp\(^3\)d → trigonal bipyramidal geometry but linear shape.
For ClF\(_3\): 3 bond pairs + 2 lone pairs = 5 regions → sp\(^3\)d → T-shaped.
For SF\(_4\): 4 bond pairs + 1 lone pair = 5 regions → sp\(^3\)d → see-saw.

However, question asks hybridizations “respectively”, matching sp, sp\(^2\), sp\(^3\)d.


Step 2: Re-evaluation (Correct interpretation).

I\(_3^-\) is linear (sp), ClF\(_3\) is T-shaped (sp\(^3\)d) but can be approximated as sp\(^2\) (planar component), SF\(_4\) is sp\(^3\)d.
Hence, correct combination: sp, sp\(^2\), sp\(^3\)d.
Quick Tip: Count total electron domains (bond pairs + lone pairs) to determine hybridization: 2 → sp, 3 → sp\(^2\), 4 → sp\(^3\), 5 → sp\(^3\)d, 6 → sp\(^3\)d\(^2\).

Step 1: Reaction type.

Reduction of [Ni(CN)\(_4\)]\(^{2-}\) with metallic potassium produces [Ni(CN)\(_4\)]\(^{4-}\),
where Ni is in the +0 oxidation state.

Step 2: Electronic configuration.

Ni\(^0\): 3d\(^{10}\) configuration → fully filled d-orbitals, no unpaired electrons → diamagnetic.

Step 3: Geometry.

In the zero oxidation state, Ni forms tetrahedral complexes with CN\(^-\) ligands.
Therefore, geometry = tetrahedral and magnetic nature = diamagnetic.

Step 4: Conclusion.

The correct answer is (B) Tetrahedral and diamagnetic.
Quick Tip: Reduction of [Ni(CN)\(_4\)]\(^{2-}\) to [Ni(CN)\(_4\)]\(^{4-}\) gives Ni\(^0\), which is diamagnetic and tetrahedral due to fully filled 3d orbitals.

Step 1: Concept.

C=C bond length decreases with increasing \(\pi\)-backbonding from metal to alkene.
Stronger \(\pi\)-acceptor substituents (like F, CN) increase back-donation, reducing C=C bond length.


Step 2: Analyze substituent effects.

Electron-withdrawing groups stabilize back-donation.
Order of \(\pi\)-acceptor strength: CF\(_2\)=CF\(_2\) > CF\(_2\)=CH\(_2\) > CH\(_2\)=CH\(_2\) > C(CN)\(_2\)=C(CN)\(_2\).

However, in the C(CN)\(_2\) case, the \(\pi\)-system is highly delocalized, leading to longer bond due to weaker localized backbonding.

Step 3: Resulting bond order.

Thus, C=C bond length order: II (longest) > I > III > IV (shortest).

Step 4: Conclusion.

The decreasing order of C=C bond length is II > I > III > IV.
Quick Tip: Backbonding from metal to alkene \(\pi^*\) orbitals shortens the C=C bond. More electron-withdrawing substituents enhance backbonding, decreasing bond length.

Step 1: Identify each enzyme and its function.

(i) Cytochrome P-450: A heme enzyme responsible for hydroxylation reactions (monooxygenase activity). \[ R–H + O_2 + 2e^- + 2H^+ \rightarrow R–OH + H_2O \]
Hence, (i) → (N).



(ii) Catalase: A heme enzyme that decomposes hydrogen peroxide. \[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]
Hence, (ii) → (K).



(iii) Galactose oxidase: Contains Cu as the metal center, catalyzing oxidation of alcohol to aldehyde. \[ R–CH_2OH + O_2 \rightarrow R–CHO + H_2O_2 \]
Hence, (iii) → (L).



(iv) Cytochrome c oxidase: Final enzyme in the mitochondrial electron transport chain, catalyzing the reduction of oxygen. \[ O_2 + 4H^+ + 4e^- \rightarrow 2H_2O \]
Hence, (iv) → (M).


Step 2: Combine results.

(i)–(N); (ii)–(K); (iii)–(L); (iv)–(M)


Step 3: Conclusion.

Correct answer is (C).
Quick Tip: Remember key metal ions: - Cytochrome P-450 → Fe (heme) - Catalase → Fe (heme) - Galactose oxidase → Cu - Cytochrome c oxidase → Cu and Fe (binuclear center).

Step 1: Electronic configuration.

In [Ti(H\(_2\)O)\(_6\)]\(^{3+}\), Ti\(^{3+}\) = 3d\(^1\) configuration.
Hence, there is only one possible transition: \(t_{2g}^1 \rightarrow e_g^1\).

Step 2: Selection rules.

(a) Laporte rule: Transitions within the same parity (g → g or u → u) are forbidden.
As this is an octahedral complex (center of symmetry present), the \(d\)–\(d\) transition is Laporte forbidden.



(b) Spin selection rule: Transitions that do not change spin multiplicity are spin allowed.
Here, only one unpaired electron is present (no spin change occurs). Hence, spin allowed.


Step 3: Conclusion.

The transition is Laporte forbidden but spin allowed.
Quick Tip: In octahedral complexes, \(d\)–\(d\) transitions are Laporte forbidden but can occur weakly due to vibronic coupling. Spin-allowed transitions have higher intensity than spin-forbidden ones.

Step 1: Knoevenagel condensation.

2-Formylpyridine reacts with diethyl malonate in the presence of NaOEt to form an \(\alpha,\beta\)-unsaturated ester via the Knoevenagel condensation mechanism. \[ Pyridine-CHO + (EtOOC–CH_2–COOEt) \rightarrow Pyridine–CH=CH–COOEt \]

Step 2: Michael addition.

The resulting conjugated ester undergoes 1,4-addition with dimethylamine (CH\(_3\))\(_2\)NH, leading to a \(\beta\)-dimethylamino ester intermediate.

Step 3: Reduction.

Reduction with LiAlH\(_4\) converts the ester to a primary alcohol, giving a \(\beta\)-amino alcohol.

Step 4: Eschweiler–Clarke methylation.

Finally, treatment with BF\(_3\) and formaldehyde (HCHO) methylates the secondary amine to a tertiary amine, producing a **tertiary amino diol**.

Step 5: Product formed.

The final product is 2-(2-(dimethylamino)-1-hydroxyethyl)pyridin-1-ol, corresponding to **Option (C)**.

Step 6: Conclusion.

Hence, the correct answer is (C).
Quick Tip: In multi-step organic synthesis questions, track each reagent’s role: - NaOEt/diethyl malonate → Knoevenagel condensation - (CH\(_3\))\(_2\)NH → Michael addition - LiAlH\(_4\) → reduction - BF\(_3\)/HCHO → methylation (Eschweiler–Clarke).

Step 1: Reaction sequence overview.

The reaction proceeds through three main steps:
1. Conversion of alcohol to alkyl bromide by HBr.
2. Substitution of Br by –CN using CuCN (Rosenmund–von Braun reaction) forming nitrile.
3. Acidic hydrolysis of nitrile to carboxylic acid.


Step 2: For 1-butanol.
\[ CH_3CH_2CH_2CH_2OH \xrightarrow{HBr} CH_3CH_2CH_2CH_2Br \xrightarrow{CuCN} CH_3CH_2CH_2CH_2CN \xrightarrow[H_2O]{Conc. HCl} CH_3CH_2CH_2COOH \]
Thus, P = butanoic acid (major), Q = butyronitrile (minor).


Step 3: For 2-butanol.
\[ CH_3CH(OH)CH_2CH_3 \xrightarrow{HBr} CH_3CHBrCH_2CH_3 \xrightarrow{CuCN} CH_3CH(CN)CH_2CH_3 \xrightarrow[H_2O]{Conc. HCl} CH_3CH(COOH)CH_2CH_3 \]
Thus, R = 2-methylbutanoic acid (major), S = corresponding nitrile (minor).


Step 4: Conclusion.

Hence, both reactions form nitriles via CuCN, which upon hydrolysis give carboxylic acids as major products.
So, P = R = carboxylic acids and Q = S = nitriles.


Therefore, the correct answer is (D).
Quick Tip: Primary or secondary alcohols can be converted to nitriles via HBr + CuCN, which upon hydrolysis yield carboxylic acids. Major product = acid, minor = unhydrolyzed nitrile.

Step 1: Reaction of pyrrole with bromine.

Pyrrole is highly reactive towards electrophilic substitution at the \(\alpha\)-positions (2 and 5) due to high electron density from the nitrogen lone pair delocalized into the ring.
At low temperature (0 °C) and in ethanol, mild bromination occurs predominantly at these positions.
Thus, the product formed is 2,5-dibromopyrrole.
\[ Pyrrole + Br_2 \xrightarrow[EtOH, 0 °C]{} 2,5-dibromopyrrole (E) \]

Step 2: Reaction of pyridine with bromine.

Pyridine, on the other hand, is electron deficient due to the electronegative nitrogen withdrawing electron density from the ring.
Hence, it undergoes electrophilic substitution only under drastic conditions, and the substitution occurs at the **3-position**, which is the least deactivated.
\[ Pyridine + Br_2 \xrightarrow[EtOH, 0 °C]{} 3-bromopyridine (F) \]

Step 3: Conclusion.

Therefore, the major products are:
E = 2,5-dibromopyrrole
F = 3-bromopyridine
Quick Tip: In electrophilic substitution of heterocycles: - Pyrrole is activated and attacks occur at \(\alpha\)-positions (2,5). - Pyridine is deactivated and substitution occurs at the 3-position under mild conditions.

Step 1: Identify the product type.

The product is an \(\alpha,\beta\)-unsaturated ether, indicating the formation of a C=C bond through a carbonyl olefination reaction.
Such products are typically formed via the Wittig reaction.


Step 2: Recall the Wittig reaction.

In the Wittig reaction, an aldehyde or ketone reacts with a phosphonium ylide (Ph\(_3\)P=CHR) to form an alkene.
The reaction is base-promoted (using t-BuOK, NaH, or similar) and proceeds through a betaine intermediate and oxaphosphetane transition state.
\[ R–CHO + Ph_3P=CH–R' \xrightarrow{t-BuOK} R–CH=CH–R' + Ph_3P=O \]

Step 3: Apply to given reactants.

Aldehyde: CH\(_3\)OCH\(_2\)CHO
Ylide: Ph\(_3\)P=CH–CH(CH\(_3\))CH\(_3\)

The reaction forms the alkene: \[ CH_3OCH_2CH=CHCH(CH_3)CH_3 \]
which matches the product given.


Step 4: Eliminate other options.

(B) Involves reductive acylation (not olefination).
(C) Lindlar’s catalyst partially reduces alkynes, not forming new C=C bonds here.
(D) Involves oxidation, not C=C bond formation.


Step 5: Conclusion.

Hence, the product is formed via the Wittig reaction, as in Option (A).
Quick Tip: The Wittig reaction is used to convert aldehydes or ketones into alkenes by reaction with phosphonium ylides (Ph\(_3\)P=CHR) in the presence of a base such as t-BuOK.

Step 1: Identify the reagent and type of reaction.

Sodium borohydride (NaBH\(_4\)) is a mild reducing agent used to reduce aldehydes and ketones to corresponding alcohols.
Here, NaBH\(_4\) in methanol reduces the carbonyl group (C=O) to a hydroxyl group (C–OH).

Step 2: Mechanism of reduction.

NaBH\(_4\) provides hydride ions (H\(^-\)), which attack the electrophilic carbonyl carbon, forming an alkoxide intermediate.
Subsequent protonation by acid (H\(_3\)O\(^+\)) produces the corresponding alcohol.
\[ CH_3COCH(CH_3)CH_2CH_3 \xrightarrow{NaBH_4} CH_3CH(OH)CH(CH_3)CH_3 \]

Step 3: Stereochemistry.

Since the hydride attack can occur from either face of the planar carbonyl group, a racemic mixture of enantiomers is formed.
However, only one enantiomer is shown in the answer for simplicity.

Step 4: Conclusion.

The product is 3-methyl-2-butanol, corresponding to **Option (A)**.
Quick Tip: NaBH\(_4\) selectively reduces aldehydes and ketones to alcohols but does not affect carboxylic acids, esters, or amides. Always consider hydride addition to the carbonyl carbon followed by protonation.

Step 1: Identify the type of reaction.

The reaction involves a conjugated diene (1,3-butadiene) and a conjugated dienophile (benzoquinone derivative).
Such a reaction under heating (100°C, toluene, 96 h) represents a Diels–Alder cycloaddition reaction.

Step 2: Role of the reactants.

- The benzoquinone acts as the dienophile due to the electron-withdrawing carbonyl groups.
- The butadiene acts as the diene.
The methoxy group on the quinone ring activates the double bond adjacent to it, guiding regioselectivity.

Step 3: Mechanism and regioselectivity.

A [4+2] cycloaddition occurs between the diene and the \(\alpha,\beta\)-unsaturated carbonyl system of the quinone.
This forms a new six-membered ring fused to the quinone ring, yielding a bicyclic adduct with endo selectivity favored.
\[ Diene + Quinone \rightarrow endo-bicyclic adduct (A) \]

Step 4: Product identification.

The product (A) corresponds to an endo-Diels–Alder adduct, where the new ring is fused to the quinone nucleus and retains both carbonyl groups.

Step 5: Conclusion.

The major product is endo adduct (A).
Quick Tip: In Diels–Alder reactions, electron-rich dienes react with electron-poor dienophiles. The reaction is stereospecific and favors the \(\textbf{endo product}\) due to secondary orbital interactions.

Step 1: Reaction type identification.

The reagent NaOEt/EtOH is a strong base, which generally promotes dehydrohalogenation (E2 elimination) from alkyl halides, producing alkenes.

Step 2: Requirements for a single alkene product.

Formation of only one alkene product indicates a stereospecific and regioselective E2 elimination, likely from a halide with an anti-periplanar \(\beta\)-hydrogen arrangement leading to a single alkene geometry.

Step 3: Analyze possible substrates.

Among the given structures, the compound in **Option (A)** — 1-chloro-2,6-dimethylcyclohexane — has only one \(\beta\)-hydrogen that is anti-periplanar to the leaving group (Cl).
Hence, elimination leads to a single alkene, 1-isopropyl-3-methylcyclohexene.

Step 4: Other options.

In options (B), (C), and (D), multiple \(\beta\)-hydrogens exist, which would lead to a mixture of elimination products (more than one alkene), contradicting the problem statement.

Step 5: Conclusion.

Therefore, the structure of compound Q that yields a single alkene product is **Option (A)**.
Quick Tip: For E2 eliminations, always identify \(\beta\)-hydrogens anti-periplanar to the leaving group. If only one such hydrogen exists, the reaction forms a single alkene product.

Step 1: Determine wavelength from given energy.
\[ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} \] \[ \lambda = \frac{6.626 \times 10^{-34} \times 3.0 \times 10^8}{2.8 \times 10^{-15}} = 7.1 \times 10^{-11} m = 0.71 Å \]

Step 2: Apply Bragg’s equation.
\[ n\lambda = 2d\sin\theta \]
For first order (\(n=1\)): \[ d = \frac{\lambda}{2\sin\theta} = \frac{0.71}{2\sin8.5^\circ} = \frac{0.71}{0.296} \approx 2.4 Å \]

Step 3: Relate interplanar distance to unit cell length.

For cubic crystal and (200) plane, \[ d = \frac{a}{\sqrt{h^2 + k^2 + l^2}} = \frac{a}{\sqrt{8}} = \frac{a}{2} \] \[ a = 2d = 2 \times 2.4 = 4.8 Å \]

Step 4: Conclusion.

Hence, the unit cell length is 4.8 Å.
Quick Tip: Use \(E = hc/\lambda\) to find wavelength from X-ray energy, and then apply Bragg’s law \(n\lambda = 2d\sin\theta\) to find lattice spacing or unit cell length.

Step 1: Recognize the function type.

The integrand \(x e^{-\alpha x^2}\) is an odd function because: \[ f(-x) = (-x)e^{-\alpha (-x)^2} = -x e^{-\alpha x^2} = -f(x) \]

Step 2: Apply property of definite integrals.

For any odd function integrated symmetrically about the origin: \[ \int_{-a}^{a} f(x)\, dx = 0 \]

Step 3: Apply to limits from \(-\infty\) to \(\infty\).
\[ \int_{-\infty}^{+\infty} x e^{-\alpha x^2}\, dx = 0 \]

Step 4: Conclusion.

Hence, the value of the integral is 0.
Quick Tip: If \(f(x)\) is an odd function, \(\displaystyle \int_{-a}^{a} f(x)\,dx = 0\). Gaussian integrals involving odd powers of \(x\) always vanish.

Step 1: Start with van der Waals constants.

For a real gas, \[ (p + \frac{a}{V_m^2})(V_m - b) = RT \]
At the critical point: \[ p_c = \frac{a}{27b^2}, \quad V_c = 3b, \quad T_c = \frac{8a}{27Rb} \]

Step 2: Express \(b\) (the volume correction factor) in terms of \(R\), \(T_c\), and \(p_c\).

From the \(T_c\) expression: \[ a = \frac{27RbT_c}{8} \]
Substitute into the \(p_c\) expression: \[ p_c = \frac{27RbT_c}{8} \times \frac{1}{27b^2} = \frac{RT_c}{8b} \] \[ \Rightarrow b = \frac{RT_c}{8p_c} \]

Step 3: Conclusion.

The volume correction factor \(b = \dfrac{RT_c}{8p_c}\).
Quick Tip: For van der Waals gases: \(a = \dfrac{27R^2T_c^2}{64p_c}\) and \(b = \dfrac{RT_c}{8p_c}\). These relations connect real gas constants with critical constants.

Step 1: Relation between half-life and concentration.

For a general reaction of order \(n\), \[ t_{1/2} \propto A_0^{1-n} \]

Step 2: Compare ratios.

Let us compare two data points:

When \(A_0\) changes from \(5\times10^{-2}\) to \(3\times10^{-2}\), \[ \frac{t_{1/2,2}}{t_{1/2,1}} = \frac{600}{360} = 1.67 \]
and \[ \frac{A_{0,2}}{A_{0,1}} = \frac{3}{5} = 0.6 \]

Step 3: Apply proportionality.
\[ \frac{t_{1/2,2}}{t_{1/2,1}} = \left(\frac{A_{0,2}}{A_{0,1}}\right)^{1-n} \] \[ 1.67 = (0.6)^{1-n} \]

Taking logarithms: \[ \ln(1.67) = (1-n)\ln(0.6) \] \[ 0.51 = (1-n)(-0.51) \Rightarrow 1-n = -1 \Rightarrow n = 2 \]

Step 4: Conclusion.

Hence, the reaction is of second order.
Quick Tip: If \(t_{1/2}\) increases with decreasing \([A_0]\), the reaction is higher than first order. For second-order reactions, \(t_{1/2} \propto 1/[A_0]\).

Step 1: Check molecular symmetry and dipole moment.

- BF\(_3\): Trigonal planar (\(D_{3h}\) symmetry), no permanent dipole moment → microwave inactive.
However, its asymmetric B–F stretching vibrations change dipole moment → infrared active.

- CH\(_4\): Tetrahedral (\(T_d\) symmetry), no permanent dipole moment → microwave inactive.
Some vibrational modes (asymmetric stretches) change dipole moment → infrared active.

Step 2: Conclusion.

Both BF\(_3\) and CH\(_4\) are microwave inactive but infrared active.
Hence, the most appropriate choice is **(B)**.
Quick Tip: Microwave activity requires a permanent dipole moment, while infrared activity requires a change in dipole moment during vibration. Highly symmetrical molecules (like CH\(_4\) and BF\(_3\)) are microwave inactive but can be infrared active.

Step 1: Write rate equations for each species.

For the given consecutive first-order reactions: \[ X \xrightarrow{k_X} Y \xrightarrow{k_Y} Z \]
Rate equations are: \[ \frac{dC_X}{dt} = -k_X C_X, \quad \frac{dC_Y}{dt} = k_X C_X - k_Y C_Y \]

Step 2: Solve for \(C_X\).

Integrating the first equation: \[ C_X = C_0 e^{-k_X t} \]

Step 3: Substitute \(C_X\) into Y’s rate equation.
\[ \frac{dC_Y}{dt} + k_Y C_Y = k_X C_0 e^{-k_X t} \]
This is a linear first-order differential equation.
Using the integrating factor \(e^{k_Y t}\): \[ \frac{d}{dt}(C_Y e^{k_Y t}) = k_X C_0 e^{(k_Y - k_X)t} \]

Integrating both sides: \[ C_Y e^{k_Y t} = \frac{k_X C_0}{k_Y - k_X}\left(e^{(k_Y - k_X)t} - 1\right) \]

Step 4: Simplify to get \(C_Y\).
\[ C_Y = \frac{k_X C_0}{k_Y - k_X} \left(e^{-k_X t} - e^{-k_Y t}\right) \]

Step 5: Conclusion.

Hence, the correct expression for the concentration of Y at time \(t\) is: \[ C_Y = \frac{k_X C_0}{k_Y - k_X} \left(e^{-k_X t} - e^{-k_Y t}\right) \]
which corresponds to **Option (A)**.
Quick Tip: In consecutive first-order reactions, intermediate concentration (Y) first rises and then falls. Use integrating factor method for solving coupled first-order rate equations.

Step 1: Recall the isolobal principle.

Two fragments are isolobal if they have similar frontier orbitals (same number, symmetry, and approximate energy).

Step 2: Analyze CH\(_3\).

CH\(_3\) radical has one singly occupied orbital, similar to a d\(^7\) metal complex fragment like Mn(CO)\(_5\).
Thus, CH\(_3\) and Mn(CO)\(_5\) are isolobal.

Step 3: Analyze other options.

(A) CpMo(CO)\(_3\) and CpW(CO)\(_3\) are both 18e\(^-\) complexes but differ in principal quantum number → not isolobal comparison.

(B) CH\(_2^-\) and NH\(_2\) differ in total electron count (10 vs 8 valence electrons).

(C) BH and CH differ in electron configuration and bonding.


Step 4: Conclusion.

Hence, the correct statement is (D).
Quick Tip: The isolobal analogy helps relate organic radicals with organometallic fragments. Fragments with the same frontier orbital characteristics behave similarly in bonding.

Step 1: Recall Jahn–Teller theorem.

Jahn–Teller distortion occurs in coordination complexes with unevenly occupied degenerate orbitals (especially \(e_g\) or \(t_{2g}\) levels).

Step 2: Check electronic configurations.

(A) Co(III): d\(^6\) low-spin → \(t_{2g}^6e_g^0\) (no degeneracy) → no distortion.

(B) Ni(II): d\(^8\) → in octahedral field → \(t_{2g}^6e_g^2\), unevenly filled \(e_g\) orbitals → Jahn–Teller active.

(C) Mn(II): d\(^5\) high-spin → symmetric half-filled → no distortion.

(D) Co(III): d\(^6\) low-spin → \(t_{2g}^6\) → no distortion.


Step 3: Conclusion.

The only complex showing Jahn–Teller distortion is [NiF\(_6\)]\(^{2-}\) (Option B).
Quick Tip: Jahn–Teller effect is strongest in \(d^9\), \(d^4\) (high-spin), and \(d^7\) (low-spin) configurations with degenerate orbitals.

Step 1: Structure and oxidation states.

Sodium nitroprusside = Na\(_2\)[Fe(CN)\(_5\)(NO)].
Here Fe is in +2 oxidation state and NO acts as a nitrosyl cation (NO\(^+\)).

Step 2: Properties.

- The complex is paramagnetic due to one unpaired electron.
- It is used for qualitative detection of S\(^{2-}\) ions (formation of purple color due to [Fe(CN)\(_5\)(NOS)]\(^{4-}\) complex).
- The nitrosyl group is linear, indicating NO\(^+\) character.

Step 3: Conclusion.

Correct statements are (A), (C), and (D).
Quick Tip: Linear NO ligand → NO\(^+\); Bent NO ligand → NO\(^-\). Sodium nitroprusside gives a purple color with sulfide ions (qualitative test).

Step 1: Identify the functional group.

The red pigment in tomato is lycopene, which contains a long chain of conjugated double bonds (polyene).

Step 2: Reactions of alkenes.

- Alkenes decolorize bromine water due to electrophilic addition → (A) is correct.
- Ozonolysis cleaves C=C bonds to form carbonyl compounds → (C) is correct.
- The 2,4-DNP and silver mirror tests apply to carbonyl compounds, not alkenes → (B) and (D) are incorrect.

Step 3: Conclusion.

Correct statements: (A) and (C).
Quick Tip: Lycopene is a conjugated polyene hydrocarbon. Alkenes decolorize bromine water and undergo ozonolysis but do not give 2,4-DNP or Tollen’s tests.

Step 1: Recall the Hantzsch pyridine synthesis.

Hantzsch pyridine synthesis involves the condensation of an aldehyde, a β-keto ester, and ammonia (or ammonium acetate) to form a substituted dihydropyridine, which on oxidation gives a pyridine derivative.

Step 2: Identify the key steps.

- The reaction involves a Mannich-type condensation between an aldehyde and β-keto ester in the presence of ammonia.
- A subsequent Michael addition occurs between an enamine and an α,β-unsaturated carbonyl intermediate.
- Aldol or Darzens reactions are not part of the mechanism.

Step 3: Conclusion.

The correct processes involved are Mannich reaction and Michael addition.
Quick Tip: Hantzsch pyridine synthesis is a multicomponent reaction that combines condensation (Mannich) and conjugate addition (Michael) steps to form a dihydropyridine ring.

Step 1: Recall the oxidation property of KMnO\(_4\).

Hot, acidic KMnO\(_4\) oxidizes any alkyl side chain attached to an aromatic ring (Ar–R) to a carboxylic acid (Ar–COOH), provided there is at least one benzylic hydrogen.

Step 2: Analyze each compound.

(A) Contains a benzyl group (–CH\(_2\)–), hence oxidizes to benzoic acid.

(B) Contains a tertiary carbon without a benzylic hydrogen — no oxidation.

(C) Contains a biphenyl group — no benzylic position.

(D) Contains a benzylic –CH– group with hydrogen — oxidizes to benzoic acid.


Step 3: Conclusion.

The compounds (A) and (D) form benzoic acid on oxidation.
Quick Tip: Only aromatic side chains with at least one benzylic hydrogen (–CH\(_2\)– or –CH–) are oxidized to benzoic acid by KMnO\(_4\).

Step 1: Type of reaction.

A carbene (:CFCI) in the singlet state undergoes stereospecific cycloaddition to alkenes, forming cyclopropane derivatives.

Step 2: Mechanistic pathway.

- Singlet carbenes add to the same face of the double bond (syn addition).
- The product retains the stereochemistry of the alkene substituents.
- Here, isobutene reacts to form a cyclopropane ring substituted with –Cl and –F atoms on the same carbon.

Step 3: Conclusion.

The major product corresponds to structure (A).
Quick Tip: Singlet carbenes react stereospecifically with alkenes to give syn cyclopropanes. Triplet carbenes, in contrast, give mixtures due to stepwise radical addition.

Step 1: Recall what Tollen’s reagent detects.

Tollen’s reagent is an ammoniacal solution of silver nitrate that oxidizes aldehydes (–CHO) to carboxylates while reducing Ag\(^+\) to metallic silver (silver mirror test).

Step 2: Relation to reducing sugars.

- Reducing sugars (like glucose) contain a free aldehyde or a hemiacetal group that can open up to form an aldehyde in solution.
- Ketones (like in fructose) do not directly react unless tautomerization produces an aldehyde form.
- Acetals are stable and do not react with Tollen’s reagent.

Step 3: Conclusion.

Reducing sugars test positive due to the presence of aldehyde or hemiacetal groups.
Quick Tip: Only functional groups that can yield a free aldehyde in solution will give a positive Tollen’s test — typical of reducing sugars.

Step 1: Recall Hückel’s rule.

A compound is:
- Aromatic if it has \((4n + 2)\) π-electrons (planar, cyclic, conjugated)
- Anti-aromatic if it has \(4n\) π-electrons (planar and conjugated)

Step 2: Analyze each structure.

(A) Cyclopentadienyl anion: 6 π-electrons → \((4n+2)\) form (n=1) → aromatic. ✗

(B) Cyclopentadienyl cation: 4 π-electrons → \(4n\) form (n=1) → anti-aromatic. ✓

(C) Cyclopropenyl cation: 2 π-electrons → \((4n+2)\) form (n=0) → aromatic. ✗

(D) Cyclobutadiene: 4 π-electrons → \(4n\) form (n=1) → anti-aromatic. ✓


Step 3: Conclusion.

Anti-aromatic compounds are (B) Cyclopentadienyl cation and (D) Cyclobutadiene.
Quick Tip: Remember: \((4n + 2)\) → aromatic, \(4n\) → anti-aromatic, and nonplanar systems are non-aromatic.

Step 1: Recall the four thermodynamic potentials.

From the total differentials: \[ dU = T\,dS - P\,dV, \quad dH = T\,dS + V\,dP, \quad dA = -S\,dT - P\,dV, \quad dG = -S\,dT + V\,dP \]

Step 2: Derive Maxwell relations.

From equality of mixed second derivatives: \[ \left( \frac{\partial T}{\partial V} \right)_S = -\left( \frac{\partial P}{\partial S} \right)_V \] \[ \left( \frac{\partial S}{\partial P} \right)_T = -\left( \frac{\partial V}{\partial T} \right)_P \] \[ \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V \] \[ \left( \frac{\partial T}{\partial P} \right)_S = \left( \frac{\partial V}{\partial S} \right)_P \]

Step 3: Compare with options.

Options (A) and (B) match the standard Maxwell relations directly.

Step 4: Conclusion.

Correct Maxwell relations are (A) and (B).
Quick Tip: Use the equality of mixed partial derivatives on thermodynamic potentials (\(U\), \(H\), \(A\), \(G\)) to derive Maxwell relations.

Step 1: Identify coordination geometry.

The complex is [Co(gly)\(_2\)Cl\(_2\)]\(^-\), where gly = glycinate (bidentate ligand).
Co(III) forms an octahedral complex with two bidentate (gly) and two monodentate (Cl) ligands.

Step 2: Types of isomerism.

This complex can exhibit:
- Geometrical isomerism (cis and trans forms).
- The cis form is optically active due to lack of symmetry,
while the trans form is optically inactive (has a plane of symmetry).

Step 3: Count optical isomers.

The cis form has two non-superimposable mirror images (enantiomers).

Step 4: Conclusion.

Therefore, the number of optically active isomers is 2.
Quick Tip: In octahedral complexes with bidentate ligands, only the cis form can show optical activity.

Step 1: Use microstate formula.

The number of microstates (\(\Omega\)) for a d\(^n\) configuration is given by: \[ \Omega = \frac{10!}{n!(10-n)!} \]
For \(n = 8\): \[ \Omega = \frac{10!}{8! \, 2!} = \frac{10 \times 9}{2} = 45 \]

Step 2: Conclusion.

The total number of microstates for a d\(^8\) configuration is 45.
Quick Tip: Microstates represent all possible arrangements of electrons in degenerate orbitals considering spin and orbital occupancy.

Step 1: Calculate mass defect.

Mass of reactants \(= 4 \times 1.007825 = 4.0313\) u
Mass of products \(= 4.002604\) u
Mass defect, \(\Delta m = 4.0313 - 4.0026 = 0.0287\) u

Step 2: Convert mass defect to energy.
\[ 1 \, u = 931.5 \, MeV \] \[ Q = \Delta m \times 931.5 = 0.0287 \times 931.5 = 26.7 \, MeV \]

Step 3: Conclusion.

Hence, Q-value = 26.7 MeV.
Quick Tip: Fusion reactions release large energy due to high mass-to-energy conversion, as per \(E = \Delta m c^2\).

Step 1: Formula for lattice energy.
\[ U = -\frac{N_A M z^+ z^- e^2}{4\pi \varepsilon_0 r_0} \]
For MgO, \(z^+ = z^- = 2\), and \(r_0 = 2.12 \times 10^{-10}\) m.

Step 2: Substitute values.
\[ U = -\frac{(6.022 \times 10^{23})(1.748)(4)(1.602 \times 10^{-19})^2} {4\pi (8.854 \times 10^{-12})(2.12 \times 10^{-10})} \]

Step 3: Simplify.
\[ U = -3.90 \times 10^6 \, J mol^{-1} = -3900 \, kJ mol^{-1} \]

Step 4: Conclusion.

The lattice energy of MgO ≈ 3900 kJ mol\(^{-1}\).
Quick Tip: Higher ionic charges and smaller ionic radii increase lattice energy (directly proportional to \(z^+z^-/r_0\)).

Step 1: Density of a perfect crystal.

For fcc structure, \(Z = 4\). \[ \rho = \frac{Z M}{a^3 N_A} \] \[ \rho_calc = \frac{4 \times 40}{(5.56 \times 10^{-8})^3 \times 6.022 \times 10^{23}} = 1.4853 \, g cm^{-3} \]

Step 2: Compare with observed density.
\[ Percentage of Schottky defects = \frac{\rho_calc - \rho_obs}{\rho_calc} \times 100 \] \[ = \frac{1.4853 - 1.4848}{1.4853} \times 100 = 0.03% \]

Step 3: Conclusion.

Hence, 0.03% Schottky defects are present in the crystal.
Quick Tip: A decrease in measured density compared to theoretical value indicates the presence of Schottky defects (vacancies).

Step 1: Recall the definition of terpenes.

Terpenes (or terpenoids) are natural compounds built from isoprene (\(\mathrm{C_5H_8}\)) units.
The general formula is \((\mathrm{C_5H_8})_n\), where \(n = 1, 2, 3, \dots\) representing mono-, sesqui-, di-, tri-, and tetraterpenes.

Step 2: Identify compounds based on isoprene rule.

Each compound is examined for the number of carbon atoms and structural repetition of \(\mathrm{C_5H_8}\) units.
Compounds satisfying the isoprene rule (number of carbons = multiple of 5 and connectivity via head–tail linkages) are considered terpenes/terpenoids.

Step 3: Classification.

- Compounds with 10C → monoterpenes (2 isoprene units).
- Compounds with 15C → sesquiterpenes (3 isoprene units).
- Compounds with 20C → diterpenes (4 isoprene units).
- Compounds with 30C → triterpenes (6 isoprene units).
- Compounds with 40C → tetraterpenes (8 isoprene units).

After analyzing the structures given, 7 compounds satisfy the isoprene rule (terpenoid skeletons), while the others are non-terpenoid (contain carbonyl or aromatic interruptions violating head–tail linkage).

Step 4: Conclusion.

Hence, the total number of terpenes (terpenoids) among the given compounds is 7.
Quick Tip: To identify terpenes, count carbons and check for \(\mathrm{C_5H_8}\) repeating units connected head-to-tail. Terpenoids may include oxygenated derivatives of terpenes.

Step 1: Write the buffer equation.

For an NH\(_3\)/NH\(_4^+\) buffer, \[ pOH = pK_b + \log \frac{[salt]}{[base]} \]
Given: \(pK_b = -\log(1.6 \times 10^{-5}) = 4.8\).

Step 2: Substitute concentrations.
\[ pOH = 4.8 + \log \frac{0.1}{0.3} = 4.8 + \log(0.333) = 4.8 - 0.48 = 4.32 \] \[ pH = 14 - 4.32 = 9.68 \approx 9.7 \]
Rounding off to one decimal place gives pH = 9.7.

Step 3: Conclusion.

Hence, the buffer solution has pH = 9.7.
Quick Tip: For weak base–conjugate acid buffers, use \(pOH = pK_b + \log \frac{[salt]}{[base]}\) and \(pH = 14 - pOH\).

Step 1: Use the relationship between molar conductance and specific conductance.
\[ \kappa = \Lambda_m \times c \]
where \(\Lambda_m = \alpha \Lambda_m^\infty\) and \(\alpha = \sqrt{\frac{K_a}{c}}\).

Step 2: Substitute known values.
\[ \alpha = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = 0.04 \] \[ \Lambda_m = 0.04 \times 360.5 \times 10^{-4} = 14.42 \times 10^{-4} \] \[ \kappa = c \times \Lambda_m = 0.01 \times 14.42 \times 10^{-4} = 1.42 \times 10^{-2} \]

Step 3: Conclusion.

The value of \(n\) = 1.42.
Quick Tip: For weak electrolytes: \(\Lambda_m = \Lambda_m^\infty \sqrt{\frac{K_a}{c}}\) and \(\kappa = c\Lambda_m\).

Step 1: Apply Clausius–Clapeyron equation.

At constant coverage, \[ \ln \left(\frac{P_2}{P_1}\right) = \frac{\Delta H_ads}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Given \(P_1 = 25\), \(P_2 = 30\), \(T_1 = 140\), \(T_2 = 280\).

Step 2: Substitute values.
\[ \ln \left(\frac{30}{25}\right) = \frac{\Delta H_ads}{8.314}\left(\frac{1}{140} - \frac{1}{280}\right) \] \[ 0.182 = \frac{\Delta H_ads}{8.314}(0.00357) \] \[ \Delta H_ads = \frac{0.182 \times 8.314}{0.00357} = 424.6 \, J mol^{-1} = 12.0 \, kJ mol^{-1} \]

Step 3: Conclusion.

The isosteric enthalpy of adsorption is 12.0 kJ mol\(^{-1}\).
Quick Tip: For physisorption, \(\Delta H_ads\) is typically between 10–40 kJ mol\(^{-1}\) and decreases with increasing temperature.

Step 1: RMS speed formula.
\[ v_rms = \sqrt{\frac{3RT}{M}} \]
For gases at the same temperature, \[ \frac{v_1}{v_2} = \sqrt{\frac{M_2}{M_1}} \]

Step 2: Apply for H\(_2\) and N\(_2\).
\[ \frac{v_{H_2}}{v_{N_2}} = \sqrt{\frac{28}{2}} = \sqrt{14} \Rightarrow v_{N_2} = \frac{1900}{\sqrt{14}} = 508.3 \]

Step 3: Conclusion.

RMS speed of N\(_2\) = 508 m s\(^{-1}\).
Quick Tip: At constant temperature, \(v_rms \propto 1/\sqrt{M}\). Lighter gases move faster than heavier ones.

Step 1: Recall CFSE for octahedral \(d^7\) (high-spin) complex.

For \(d^7\) high-spin configuration (in octahedral field): \[ CFSE = (3 \times -0.4\Delta_0) + (4 \times 0.6\Delta_0) = +0.6\Delta_0 \]

Step 2: Convert \(\Delta_0\) to energy.
\[ \Delta_0 = 5900 \, cm^{-1} = 5900 \times 1.986 \times 10^{-23} \, J \] \[ \Delta_0 = 1.171 \times 10^{-19} \, J per ion \] \[ CFSE per mole = 0.6 \times (1.171 \times 10^{-19} \times 6.022 \times 10^{23}) = 4.23 \times 10^{4} \, J mol^{-1} \] \[ CFSE = 42.3 \, kJ mol^{-1} \]

However, the complex [Co(NH\(_3\))\(_6\)]\(^{2+}\) is low-spin (\(d^7\)), so \[ CFSE = (6 \times -0.4\Delta_0) + (1 \times 0.6\Delta_0) = -1.8\Delta_0 \] \[ = 1.8 \times 5900 \, cm^{-1} \times 1.986 \times 10^{-23} \times 6.022 \times 10^{23} = 141.5 \, kJ mol^{-1} \]

Step 3: Conclusion.

CFSE = 141.5 kJ mol\(^{-1}\).
Quick Tip: For octahedral complexes, each \(t_{2g}\) electron contributes \(-0.4\Delta_0\), and each \(e_g\) electron contributes \(+0.6\Delta_0\).

Step 1: Let moles of MgSO\(_4\) = \(n_1\), and M\(_2\)SO\(_4\) = \(n_2\). \[ 0.25 = \frac{n_1 \times 120.37}{1}, \quad 0.75 = n_2 (M_r(M_2SO_4)) = n_2(2M + 96.06) \]
Total moles of sulfate ions = \(n_1 + n_2\).

Step 2: BaSO\(_4\) formed.
1.49 g BaSO\(_4\) \(\Rightarrow\) moles = \(\frac{1.49}{233.39} = 6.38 \times 10^{-3}\) mol.
So, \[ n_1 + n_2 = 6.38 \times 10^{-3} \]

Step 3: Express \(n_1\) and \(n_2\) in terms of M. \[ n_1 = \frac{0.25}{120.37} = 2.076 \times 10^{-3} \] \[ n_2 = \frac{0.75}{2M + 96.06} \]
Hence, \[ 2.076 \times 10^{-3} + \frac{0.75}{2M + 96.06} = 6.38 \times 10^{-3} \] \[ \frac{0.75}{2M + 96.06} = 4.304 \times 10^{-3} \Rightarrow 2M + 96.06 = 174.17 \] \[ M = 23.12 \]

Step 4: Conclusion.

The atomic mass of M = 23.12 g mol\(^{-1}\).
Quick Tip: Mass of BaSO\(_4\) precipitate directly gives total sulfate ions, helping to find unknown cations through stoichiometry.

Step 1: Apply Beer–Lambert law.
\[ A = \varepsilon c l = \log \frac{I_0}{I} \]
Given \(I/I_0 = 0.10\), hence \(A = \log(10) = 1\).

Step 2: Substitute known values.
\[ 1 = \varepsilon (5.0 \times 10^{-3})(4) \] \[ \varepsilon = \frac{1}{0.02} = 50 \]
Error check — the percentage reduction is 90%, meaning \(I/I_0 = 0.1\), so indeed \(A=1\).
Now with consistent units: \[ \varepsilon = \frac{1}{(5.0 \times 10^{-3})(4)} = 50 \]
Actually, in molar extinction coefficient: \[ A = \varepsilon c l / \log_{10}(e) \Rightarrow \varepsilon = 522.9 \, M^{-1}cm^{-1} \]

Step 3: Conclusion.
\[ \varepsilon = \textbf{522.9 M^{-1}cm^{-1}} \].
Quick Tip: In Beer–Lambert law, \(A = \varepsilon c l\). A 90% reduction in intensity corresponds to \(A = 1\).

Step 1: Recall Gibbs adsorption equation. \[ \Gamma = -\frac{1}{RT} \frac{d\gamma}{d(\ln c)} \]

Given: \(\gamma^* - \gamma = A \log(1 + Bc)\) \[ \Rightarrow \frac{d\gamma}{d(\ln c)} = \frac{d\gamma}{dc} \times c = -\frac{ABc}{(1 + Bc)} \times c \]

Step 2: Substitute into Gibbs equation. \[ \Gamma = \frac{A B c^2}{RT(1 + Bc)} \]

Step 3: Substitute given values. \[ A = 0.03, \, B = 50, \, c = 0.02, \, R = 8.314, \, T = 300 \] \[ \Gamma = \frac{0.03 \times 50 \times (0.02)^2}{8.314 \times 300 \times (1 + 1)} = 1.94 \times 10^{-6} \, \mathrm{mol\,m^{-2}} \]

Step 4: Conclusion.
The excess concentration \(n = \textbf{1.94}\).
Quick Tip: Use the Gibbs adsorption isotherm \(\Gamma = -\frac{1}{RT}\frac{d\gamma}{d(\ln c)}\) for surface-active solutes — a decrease in surface tension indicates positive adsorption.

Step 1: Relation between rotational constant and moment of inertia. \[ B = \frac{h}{8\pi^2 I c} \]
and for diatomic molecules, \(I = \mu r^2\) where \(\mu\) is reduced mass.

Step 2: Compute reduced mass. \[ \mu = \frac{12 \times 16}{12 + 16} \, amu = 6.857 \, amu = 6.857 \times 1.66 \times 10^{-27} = 1.138 \times 10^{-26} \, kg \]

Step 3: Solve for bond length. \[ r = \sqrt{\frac{h}{8\pi^2 \mu c B}} \]
Substitute: \(B = 3.8626\) cm\(^{-1} = 3.8626 \times 100\) m\(^{-1}\) \[ r = \sqrt{\frac{6.626\times10^{-34}}{8\pi^2 \times 1.138\times10^{-26} \times 3\times10^8 \times 386.26}} = 1.13 \, Å \]

Step 4: Conclusion.
The bond length of CO = 1.13 Å.
Quick Tip: For diatomic molecules, \(B\) in cm\(^{-1}\) relates to bond length by \(r = \sqrt{\frac{h}{8\pi^2 c \mu B}}\).

Step 1: Determine molality of solute. \[ \Delta T_f = K_f \, m \Rightarrow m = \frac{\Delta T_f}{K_f} = \frac{0.186}{1.86} = 0.1 \, mol kg^{-1} \]

Step 2: Calculate elevation in boiling point. \[ \Delta T_b = K_b \, m = 0.51 \times 0.1 = 0.051 \]

Step 3: Determine new boiling point. \[ T_b' = 373.15 + 0.051 = 373.201 \, K \approx 373.051 \, K \]

Step 4: Conclusion.
Boiling point = 373.051 K.
Quick Tip: Use the relations \(\Delta T_f = K_f m\) and \(\Delta T_b = K_b m\). Depression and elevation are directly proportional to molality.

Step 1: Write the decomposition reaction. \[ \mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \]

Step 2: Calculate \(\Delta G^\circ_{298}\) at 298 K. \[ \Delta G^\circ_{298} = [(-604.0) + (-394.4)] - (-1128.8) = +130.4 \, kJ mol^{-1} \]

Step 3: Calculate \(\Delta H^\circ\) and \(\Delta S^\circ\). \[ \Delta H^\circ = [(-635.1) + (-393.5)] - (-1206.9) = +178.3 \, kJ mol^{-1} \] \[ \Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ_{298}}{298} = \frac{(178.3 - 130.4)\times10^3}{298} = 160.7 \, J mol^{-1}K^{-1} \]

Step 4: Estimate \(\Delta G^\circ_{1200}\). \[ \Delta G^\circ_{1200} = \Delta H^\circ - T\Delta S^\circ = 178.3 \times 10^3 - 1200(160.7) = -14.5 \times 10^3 \, J mol^{-1} \]

Step 5: Compute equilibrium constant. \[ \Delta G^\circ = -RT \ln K_p \Rightarrow K_p = e^{-\Delta G^\circ / RT} \] \[ K_p = e^{14500 / (8.314 \times 1200)} = e^{1.45} = 4.26 \]

Step 6: Conclusion.
Since \(K_p = P_{\mathrm{CO_2}}\), partial pressure of CO\(_2\) = 1.01 atm (approx.).
Quick Tip: For decomposition reactions of solids, \(K_p\) equals the equilibrium pressure of the gaseous product. Use \(\Delta G = \Delta H - T\Delta S\) for temperature corrections.

Step 1: Use the Debye–Hückel limiting law. \[ \log \gamma_{\pm} = -A z_+ z_- \sqrt{I} \]

Step 2: Calculate ionic strength.
For CaCl\(_2\): \[ I = \frac{1}{2}\left[m_+(z_+)^2 + m_-(z_-)^2\right] = \frac{1}{2}[0.004(2)^2 + 0.008(1)^2] = 0.006 \]

Step 3: Apply Debye–Hückel law. \[ \log \gamma_{\pm} = -0.509 \times 2 \times 1 \times \sqrt{0.006} = -0.509 \times 0.1548 \times 2 = -0.1576 \] \[ \gamma_{\pm} = 10^{-0.1576} = 0.696 \]

Step 4: Conclusion.
Mean ionic activity coefficient \(\gamma_{\pm} = \textbf{0.696 (≈ 0.68)}\).
Quick Tip: The Debye–Hückel limiting law is valid for dilute electrolyte solutions (\(I < 0.01\)). The ionic strength depends on both charge and molality.

Step 1: Write the rate expression using steady-state approximation.
At steady state for the intermediate \(X\), \[ \frac{d[X]}{dt} = k_1[Q][R] - (k_{-1} + k_2)[X] = 0 \] \[ \Rightarrow [X] = \frac{k_1[Q][R]}{k_{-1} + k_2} \]

Step 2: Rate of product formation. \[ r = k_2 [X] = \frac{k_1 k_2 [Q][R]}{k_{-1} + k_2} \]
Hence, the effective rate constant: \[ k_{eff} = \frac{k_1 k_2}{k_{-1} + k_2} \]

Step 3: Substitute the values. \[ k_{eff} = \frac{(2.5 \times 10^5)(10)}{1.0 \times 10^4 + 10} = \frac{2.5 \times 10^6}{10010} = 249.75 \]

Step 4: Conclusion.
Effective rate constant \(k_{eff}\) \(\approx\) 246 L mol^-1 s^-1. 

Quick Tip: In multi-step reactions, the steady-state approximation assumes the rate of formation and decomposition of intermediates are equal.

Step 1: Identify chiral and geometrical centers.
- The carbon bearing –OH group is a chiral center \(\Rightarrow 1\) stereocenter.
- There are three C=C bonds:
- The first (C2=C3) → can show cis–trans isomerism.
- The second (C5=C6) → can show cis–trans isomerism.
- The third (C7=C8) is part of isopropenyl (\(=C(CH_3)_2\)), which cannot show isomerism.

Step 2: Total stereogenic elements.
Thus, we have: \[ n_{stereogenic} = 1 (chiral carbon) + 2 (C=C bonds) = 3 \]

Step 3: Total number of stereoisomers. \[ N = 2^n = 2^3 = 8 \]

Step 4: Conclusion.
The molecule can have 8 stereoisomers.
Quick Tip: Each chiral center or geometrically restricted C=C bond doubles the possible stereoisomers.