Step 1: Understanding the Concept:

The common and most stable oxidation state for lanthanoids is +3. Species in oxidation states other than +3 tend to reach the +3 state by either losing or gaining electrons.


Step 2: Detailed Explanation:

- Oxidizing Agents: Substances that undergo reduction (gain electrons).

- Lanthanoids in +4 State: \(Ce^{4+}\) and \(Tb^{4+}\) are in a higher oxidation state than the stable +3. To attain stability, they gain one electron and act as strong oxidizing agents.

- \(Ce^{4+} + e^- \to Ce^{3+}\) (Stable configuration)

- \(Tb^{4+} + e^- \to Tb^{3+}\) (Half-filled \(f^7\) stability)

- Lanthanoids in +2 State: \(Sm^{2+}\) and \(Ce^{2+}\) are in a lower state than +3. They tend to lose electrons to become +3, acting as reducing agents.


Step 3: Final Answer:

Therefore, \(Ce^{4+}\) and \(Tb^{4+}\) are good oxidizing agents.
Quick Tip: Remember: \(Ce^{4+}\) is widely used in analytical chemistry as a volumetric oxidizing agent (cerimetry) because it is a very strong oxidant.

Step 1: Understanding the Concept:

Crystal Field Theory (CFT) is an electrostatic model that considers the metal-ligand bond to be purely ionic, treating ligands as point charges.


Step 2: Detailed Explanation:

- Successes of CFT: It successfully explains the magnetic properties (pairing vs. high spin), colors (d-d transitions), and thermodynamic stability (CFSE) of coordination complexes.

- Failures of CFT: Because it treats ligands as point charges, it predicts that anionic ligands should cause greater splitting than neutral ones. However, anionic ligands like \(OH^-\) and \(I^-\) are at the low end of the spectrochemical series, while neutral ligands like \(CO\) are strong field ligands. This discrepancy is due to the covalent nature of bonding (back-bonding), which CFT ignores.


Step 3: Final Answer:

The order of the spectrochemical series cannot be explained by CFT; it requires Ligand Field Theory.
Quick Tip: CFT's biggest drawback is its complete neglect of the covalent character in metal-ligand bonds. This is why it fails to justify why \(CO\) is a stronger ligand than \(Cl^-\).

Step 1: Understanding the Concept:

Acidified potassium dichromate is a strong oxidizing agent (orange color). When it reacts with a reducing agent, it is reduced to chromium(III) ions, which are green in color.


Step 2: Detailed Explanation:

Sulphur dioxide (\(SO_2\)) is a reducing agent. When passed through acidified \(K_2Cr_2O_7\) solution or paper, it reduces the \(Cr(VI)\) in dichromate to \(Cr(III)\).

The chemical equation is:
\[ K_2Cr_2O_7 (orange) + 3SO_2 + H_2SO_4 \to Cr_2(SO_4)_3 (green) + K_2SO_4 + H_2O \]

The change from \(+6\) (orange) to \(+3\) (green) oxidation state of Chromium confirms the presence of \(SO_2\).


Step 3: Final Answer:
The gas is Sulphur dioxide.
Quick Tip: This is a characteristic test for \(SO_2\) gas. While \(H_2S\) also turns it green, it would also produce a yellow precipitate of sulphur, whereas \(SO_2\) produces a clear green solution.

Step 1: Understanding the Concept:

The reactions describe two different methods of hydration of an alkene: Hydroboration-Oxidation and Oxymercuration-Demercuration.


Step 2: Detailed Explanation:

The starting material is 2-methylbut-2-ene.

- Reaction A (Hydroboration-Oxidation): Reagents \(BH_3, THF\) followed by \(H_2O_2/OH^-\). This proceeds via Anti-Markovnikov addition of water. The OH group attaches to the less substituted carbon of the double bond.

Product A: 3-methylbutan-2-ol.

- Reaction B (Oxymercuration-Demercuration): Reagents \(Hg(OAc)_2, H_2O\) followed by \(NaBH_4\). This proceeds via Markovnikov addition of water without any carbocation rearrangement. The OH group attaches to the more substituted carbon.

Product B: 2-methylbutan-2-ol.


Step 3: Final Answer:

The products are A: 3-methylbutan-2-ol and B: 2-methylbutan-2-ol.
Quick Tip: Hydroboration = Anti-Markovnikov (OH on less substituted C).
Oxymercuration = Markovnikov (OH on more substituted C).
This rule covers almost all alkene hydration questions in exams!

Step 1: Understanding the Concept:

This question tests periodic trends and chemical properties of Group 1 (alkali) metals, including electrode potentials, solubility, and thermal stability.


Step 2: Detailed Explanation:

- Statement A: Standard reduction potentials (\(E^\circ\)) are: \(Li (-3.05 V)\), \(Rb (-2.93 V)\), \(Na (-2.71 V)\). In terms of value: \(-2.71 > -2.93 > -3.05\). So \(Na > Rb > Li\) is correct.

- Statement B: Solubility depends on lattice and hydration energies. Large cations with large anions (\(Cs^+\) and \(I^-\)) have low solubility due to size matching. \(CsI\) is poorly soluble. Incorrect.

- Statement C: \(Li_2CO_3\) is unstable because \(Li^+\) is small and polarizes the large \(CO_3^{2-}\) ion, causing it to decompose into \(Li_2O\) and \(CO_2\). Incorrect.

- Statement D: \textit{Dilute solutions are blue and paramagnetic. \textit{Concentrated solutions are bronze-colored and diamagnetic. Incorrect.

- Statement E: All alkali metal hydrides (\(LiH, NaH\), etc.) are ionic crystalline solids with high melting points. Correct.


Step 3: Final Answer:

Only statements A and E are correct.
Quick Tip: Lithium is the strongest reducing agent in aqueous solution due to its exceptionally high hydration energy, making its reduction potential the most negative.

Step 1: Understanding the Concept:

A substance acts as a reducing agent when it loses electrons (its oxidation state increases) and reduces another species. In \(H_2O_2\), oxygen is in the \(-1\) state; as a reducing agent, it is oxidized to \(O_2\) (\(0\) state).


Step 2: Detailed Explanation:

Let's analyze the oxidation states in reaction (A):

- In \(HOCl\), Chlorine is in \(+1\) state. In \(Cl^-\), it is in \(-1\) state. Chlorine is reduced.

- In \(H_2O_2\), Oxygen is in \(-1\) state. In \(O_2\), it is in \(0\) state. Oxygen is oxidized.

Since \(H_2O_2\) is oxidized while reducing \(HOCl\), it acts as a reducing agent.

In all other options (B, C, and D), \(H_2O_2\) oxidizes the other reactant (\(Mn^{2+} \to Mn^{4+}\), \(Fe^{2+} \to Fe^{3+}\), \(S^{2-} \to S^{6+}\)), thus acting as an oxidizing agent.


Step 3: Final Answer:

Reaction (A) shows \(H_2O_2\) acting as a reducing agent.
Quick Tip: Whenever you see \(O_2\) gas being evolved from \(H_2O_2\), it's a clear sign that \(H_2O_2\) is being oxidized and is thus acting as a reducing agent.

Step 1: Understanding the Concept:

This refers to the hydrometallurgical extraction process, specifically the cyanide process used for noble metals.


Step 2: Detailed Explanation:

Silver (Ag) is extracted from its ore (like Argentite, \(Ag_2S\)) using the MacArthur-Forrest cyanide process:

1. Oxidation/Leaching: The ore is treated with a dilute solution of \(NaCN\) in the presence of air (\(O_2\)). Silver is oxidized from its elemental or sulfide form to \(Ag^+\) in a complex.
\[ 4Ag + 8CN^- + O_2 + 2H_2O \to 4[Ag(CN)_2]^- + 4OH^- \]

2. Reduction/Precipitation: The complex is then treated with a more reactive metal like Zinc, which reduces \(Ag^+\) back to metallic silver.
\[ 2[Ag(CN)_2]^- + Zn \to [Zn(CN)_4]^{2-} + 2Ag \]


Step 3: Final Answer:

The metal is Silver (Ag).
Quick Tip: Noble metals like Silver and Gold are extracted via "Hydrometallurgy" using leaching. Remember that Zinc acts as the reducing agent in the final step.

Step 1: Understanding the Concept:

This question pertains to the physical properties and stability of primary aromatic amines (arylamines) like aniline.


Step 2: Detailed Explanation:

- Statement I: Pure aniline is a colorless oily liquid. Most other arylamines are also colorless when freshly prepared. This statement is correct.

- Statement II: On exposure to air and light, arylamines gradually turn brown or dark red. This is due to slow atmospheric oxidation (forming complex quinone-like structures), not reduction. Therefore, Statement II is incorrect.


Step 3: Final Answer:

Statement I is correct, but Statement II is incorrect.
Quick Tip: Aniline is very sensitive to oxidation. Even a small amount of impurity makes it look yellow or brown. Always store it in dark bottles away from air.

Step 1: Understanding the Concept:

Clemmensen and Wolff-Kishner are complementary methods for reducing aldehydes and ketones to alkanes. However, they use different reagents that may affect other sensitive functional groups in the molecule.


Step 2: Detailed Explanation:

- Statement I (Clemmensen Reduction): Uses \(Zn/Hg\) and concentrated \(HCl\). It successfully reduces the carbonyl group to a \(CH_2\) group. While amines protonate in acid, the primary reduction of the ketone still occurs. The statement depicts a valid transformation. Correct.

- Statement II (Wolff-Kishner Reduction): Uses \(NH_2NH_2\) and \(KOH\) (strong base). The ketone group should reduce to \(CH_2\). However, the molecule contains an alkyl chloride (\(Cl\) group). In the presence of a strong base like \(KOH\) at high temperatures, the alkyl halide will undergo dehydrohalogenation (elimination) to form an alkene. The product shown in the statement incorrectly retains the \(Cl\) atom. Therefore, Statement II is false.


Step 3: Final Answer:

Statement I is true, and Statement II is false.
Quick Tip: Mnemonic: Clemmensen is Acidic (HCl), avoid with acid-sensitive groups. Wolff-Kishner is Basic (KOH), avoid with base-sensitive groups like halides (\(Cl, Br\)).

Step 1: Understanding the Concept:

The standard reduction potential (\(E^{\circ}\)) of a metal is a measure of its tendency to be reduced.

For metals, it is influenced by three main energetic factors: the enthalpy of atomization (\(\Delta H_{atom}\)), the ionization enthalpy (\(IE\)), and the hydration enthalpy (\(\Delta H_{hyd}\)).


Step 2: Key Formula or Approach:

The total enthalpy change for the process \(M(s) \rightarrow M^{2+}(aq) + 2e^{-}\) is:
\[ \Delta H_{total} = \Delta H_{atom} + IE_{1} + IE_{2} + \Delta H_{hyd} \]

A less negative \(E^{\circ}\) value indicates that the oxidation process is less favorable compared to other elements in the same group.


Step 3: Detailed Explanation:

1. Assertion analysis: Beryllium (\(Be\)) has a reduction potential of \(-1.97\,V\), while other alkaline earth metals like Magnesium (\(Mg\)) and Calcium (\(Ca\)) have values around \(-2.36\,V\) and \(-2.84\,V\) respectively.

Thus, \(Be\) indeed has a less negative value. Assertion A is correct.

2. Reason analysis: Due to its exceptionally small atomic and ionic size, Beryllium has a very high hydration energy.

However, this is offset by its very high enthalpy of atomization and exceptionally high ionization enthalpies (because the electrons are close to the nucleus).

The high energy required for atomization and ionization is not fully compensated by the hydration energy, leading to a less negative reduction potential overall.

Thus, Reason R is correct and provides the physical basis for the assertion.


Step 4: Final Answer:

Both statements are correct, and R explains A.
Quick Tip: For s-block elements, reduction potential is a balance of "costly" steps (sublimation, ionization) and "profitable" steps (hydration). For Be, the cost of ionization is too high.

Step 1: Understanding the Concept:

This question pertains to human physiology and environmental exposure levels discussed in environmental chemistry.


Step 2: Detailed Explanation:

1. On average, a human being consumes about \(1.5\) to \(2.0\) kg of food per day.

2. In contrast, an average adult breathes in approximately \(10,000\) to \(20,000\) liters of air per day.

3. Given that the density of air is approximately \(1.2\,kg/m^{3}\), the mass of air consumed is roughly \(12\) to \(24\) kg per day.

4. By comparing the masses: \( Mass of air / Mass of food \approx 20/1.5 \approx 13.3 \) to \(15\).

According to standard environmental chemistry facts (NCERT), a human consumes nearly \(15\) times more air than food by weight.


Step 3: Final Answer:

The mass of air consumed daily is significantly higher, approximately 15 times that of food.
Quick Tip: This fact explains why air pollution is considered more dangerous than food contamination; we take in a much larger "dose" of air pollutants daily.

Step 1: Understanding the Concept:

Deprotonation occurs most readily from the most acidic site. For active methylene compounds, acidity depends on the resonance stabilization of the resulting carbanion by the adjacent electron-withdrawing groups (EWG).


Step 2: Detailed Explanation:

The compounds are:

(a) Pentane-2,4-dione (1,3-diketone): The central \(CH_2\) is between two ketone groups.

(b) Dimethyl malonate (1,3-diester): The central \(CH_2\) is between two ester groups.

(c) Methyl acetoacetate (keto-ester): The central \(CH_2\) is between one ketone and one ester group.


1. Ketone groups (\(-COR\)) are more electron-withdrawing than ester groups (\(-COOR\)) because the alkoxy (\(-OR\)) group in an ester donates electrons into the carbonyl via resonance, reducing the carbonyl's ability to stabilize an external negative charge.

2. Therefore, the order of acidity is: 1,3-diketone \(>\) keto-ester \(>\) 1,3-diester.

3. Compound (a) has the most stable conjugate base and thus has the lowest \(pK_a\) (highest acidity).


Step 3: Final Answer:

Compound (a) is the most acidic and will deprotonate most readily in a basic medium.
Quick Tip: Acidity Order: \(-CHO > -COR > -COOR > -CONR_2\). More ketones surrounding a \(CH_2\) mean higher acidity.

Step 1: Understanding the Concept:

This question tests the knowledge of medicinal drug classifications from Chemistry in Everyday Life.


Step 2: Detailed Explanation:

1. Antifertility drugs: These are used for birth control. Norethindrone is a synthetic progesterone used in oral contraceptives. (A \(\rightarrow\) I)

2. Tranquilizers: These are used to treat stress and anxiety. Meprobamate is a mild tranquilizer. (B \(\rightarrow\) II)

3. Antihistamines: These are used to treat allergic reactions. Seldane (Terfenadine) is a common antihistamine. (C \(\rightarrow\) III)

4. Antibiotics: These are substances used to kill or inhibit bacteria. Ampicillin is a semi-synthetic antibiotic related to penicillin. (D \(\rightarrow\) IV)


Step 3: Final Answer:

The correct match is A-I, B-II, C-III, D-IV.
Quick Tip: Commonly asked drug classes: Equanil/Meprobamate (Tranquilizers), Terfenadine/Seldane (Antihistamines), Novestrol/Norethindrone (Antifertility).

Step 1: Understanding the Concept:

Benzoic acid (\(C_6H_5COOH\)) is a weak acid, and sodium hydroxide (\(NaOH\)) is a strong base. In a conductometric titration, we monitor the change in conductivity as the titrant is added.


Step 2: Detailed Explanation:

1. Initial Phase: Benzoic acid is weakly dissociated, so the starting conductance is low.

2. Addition of NaOH: As \(NaOH\) is added, poorly conducting acid is converted into highly dissociated salt (sodium benzoate). This leads to a steady increase in the number of ions and thus an increase in conductance.

3. Equivalence Point: At this point, all acid has been converted to salt.

4. Post-Equivalence: Excess \(NaOH\) adds highly mobile \(OH^{-}\) ions to the solution. This causes the conductance to rise much more steeply than before.

5. Graph Character: The resulting graph shows an initial slow linear rise followed by a sharp linear rise after the endpoint. This matches Graph 4.


Step 3: Final Answer:

Graph 4 is the correct representation for a Weak Acid vs. Strong Base conductometric titration.
Quick Tip: Strong Acid vs. Strong Base gives a V-shaped graph. Weak Acid vs. Strong Base starts low and keeps increasing, with a steeper slope after the equivalence point.

Step 1: Understanding the Concept:

Stability of benzene is attributed to its resonance energy. Resonance involves the delocalization of electrons across several nuclei.


Step 2: Detailed Explanation:

1. Assertion analysis: Benzene has a resonance energy of \(152\,kJ/mol\), meaning it is significantly more stable than a hypothetical molecule with three localized double bonds (cyclohexatriene). Assertion A is correct.

2. Reason analysis: In localized systems, electrons are held between two nuclei. In delocalized systems like benzene, the \(\pi\) electrons are shared by six carbon nuclei. This increased nuclear attraction lowers the potential energy of the electrons, enhancing stability. Reason R is correct.

3. Connection: The fundamental reason for the "resonance stabilization" mentioned in Assertion A is the increased attraction described in Reason R.


Step 3: Final Answer:

Both statements are correct and the Reason explains the Assertion.
Quick Tip: Delocalization = Lower Energy = Higher Stability. The energy of a system decreases as the volume occupied by its electrons increases.

Step 1: Understanding the Concept:

Protons determine the atomic number (\(Z\)). \(Z = 55\) corresponds to Cesium (Cs). A "unipositive state" refers to the \(Cs^{+}\) ion.


Step 2: Detailed Explanation:

1. Neutral Cs (Z=55) Configuration:
\(1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6} 3d^{10}, 4s^{2} 4p^{6} 4d^{10}, 5s^{2} 5p^{6}, 6s^{1}\).

2. Formation of \(Cs^{+}\): One electron is removed from the outermost shell (6s).

Configuration of \(Cs^{+}\): \(1s^{2}, 2s^{2} 2p^{6}, 3s^{2} 3p^{6} 3d^{10}, 4s^{2} 4p^{6} 4d^{10}, 5s^{2} 5p^{6}\).

3. Counting s-electrons:

- \(1s \rightarrow 2\)

- \(2s \rightarrow 2\)

- \(3s \rightarrow 2\)

- \(4s \rightarrow 2\)

- \(5s \rightarrow 2\)

Total s-electrons = \(2 + 2 + 2 + 2 + 2 = 10\).


Step 3: Final Answer:

The \(Cs^{+}\) ion contains 10 s-electrons.
Quick Tip: For any alkali metal \(M\) in period \(n\), the ion \(M^{+}\) will have \(2(n-1)\) s-electrons. Since Cs is in Period 6, it has \(2 \times (6-1) = 10\) s-electrons.

Step 1: Understanding the Concept:

We apply Valence Bond Theory (VBT) or Crystal Field Theory (CFT) to determine the electronic distribution in the complex.


Step 2: Detailed Explanation:

1. Oxidation State: \(Co\) in \([Co(NH_3)_6]^{3+}\) is \(Co^{3+}\).

2. Electronic Configuration: Neutral \(Co\) is \([Ar] 3d^{7} 4s^{2}\). \(Co^{3+}\) is \([Ar] 3d^{6}\).

3. Ligand strength: \(NH_3\) acts as a strong field ligand for \(Co^{3+}\), forcing the six electrons in the \(3d\) orbitals to pair up.

4. Distribution: The 6 electrons fill the three \(t_{2g}\) orbitals: \( (t_{2g})^{6} (e_g)^{0} \). All electrons are paired.

5. Hybridization: This leaves two \(3d\) orbitals empty for bonding. Hybridization involves two \(3d\), one \(4s\), and three \(4p\) orbitals \(\rightarrow d^{2}sp^{3}\).

6. Magnetic Behaviour: Since all electrons are paired, it is diamagnetic.


Step 3: Final Answer:

The complex is \(d^{2}sp^{3}\) hybridized and diamagnetic.
Quick Tip: Cobalt (III) complexes with strong field ligands like \(NH_3, CN^{-}\), and \(NO_2^{-}\) are almost always low-spin, inner-orbital, and diamagnetic.

Step 1: Understanding the Concept:

The half-life (\(t_{1/2}\)) of a reaction is related to the initial concentration (or pressure \(p\)) by the formula \(t_{1/2} \propto 1 / p^{n-1}\), where \(n\) is the order of the reaction.


Step 2: Key Formula or Approach:
\[ \frac{(t_{1/2})_1}{(t_{1/2})_2} = \left( \frac{p_2}{p_1} \right)^{n-1} \]


Step 3: Detailed Explanation:

Take any two points from the table:

Point 1: \(p_1 = 50, t_{1/2} = 4\)

Point 2: \(p_2 = 100, t_{1/2} = 2\)
\[ \frac{4}{2} = \left( \frac{100}{50} \right)^{n-1} \]
\[ 2 = (2)^{n-1} \]

By comparing exponents: \(n - 1 = 1 \Rightarrow n = 2\).

We can verify with other points (e.g., \(200/400\)): \(1/0.5 = 2 = (400/200)^{n-1} \Rightarrow 2 = 2^{n-1} \Rightarrow n=2\).


Step 4: Final Answer:

The order of the reaction is 2.
Quick Tip: If \(t_{1/2} \times p = constant\), the reaction is 2nd order. If \(t_{1/2} = constant\), it's 1st order. If \(t_{1/2} \propto p\), it's zero order.

Step 1: Understanding the Concept:

Molecular Orbital Theory (MOT) describes the electronic configuration and bond properties of diatomic species.


Step 2: Detailed Explanation:

1. \(N_2\) (14 \(e^{-}\)): Configuration: \(\sigma 1s^{2} \sigma^{*}1s^{2} \sigma 2s^{2} \sigma^{*}2s^{2} (\pi 2p_x^{2} = \pi 2p_y^{2}) \sigma 2p_z^{2}\). The HOMO is \(\sigma 2p_z\) with 2 electrons. Unpaired electrons = 0.

2. \(N_2^{+}\) (13 \(e^{-}\)): One electron is removed from the HOMO (\(\sigma 2p_z\)). Unpaired electrons = 1.

3. \(O_2\) (16 \(e^{-}\)): Configuration: \(\sigma 1s^{2} \sigma^{*}1s^{2} \sigma 2s^{2} \sigma^{*}2s^{2} \sigma 2p_z^{2} (\pi 2p_x^{2} = \pi 2p_y^{2}) (\pi^{*} 2p_x^{1} = \pi^{*} 2p_y^{1})\). The HOMO is \(\pi^{*}\) with 2 unpaired electrons (Hund's rule). Unpaired electrons = 2.

4. \(O_2^{+}\) (15 \(e^{-}\)): One electron is removed from the \(\pi^{*}\) orbital. Unpaired electrons = 1.


Step 3: Final Answer:

The number of unpaired electrons is 0, 1, 2, 1.
Quick Tip: Remember that \(O_2\) is one of the classic examples of a paramagnetic molecule despite having an even number of electrons because of the degenerate \(\pi^{*}\) molecular orbitals.

Step 1: Understanding the Concept:

The reaction is a diazo-coupling reaction between a benzenediazonium derivative and an aromatic amine to form an azo dye.


Step 2: Detailed Explanation:

1. The reactant on the left is diazotized sulphanilic acid.

2. It reacts with 1-naphthylamine. The coupling typically occurs at the para-position to the amine group.

3. The resulting product is an extended conjugated system containing the \(-N=N-\) (azo) group.

4. This specific product is a well-known azo dye. Azo dyes formed from naphthylamines are characterized by intense colours, typically in the red/orange spectrum. In the Griess test for nitrites, this exact reaction produces a distinctive red colour.


Step 3: Final Answer:

The product formed is a red azo dye.
Quick Tip: Benzene-based azo dyes with amines are often yellow/orange, but naphthalene-based azo dyes usually shift the absorption towards longer wavelengths (red).

Step 1: Understanding the Concept:

The number of \(\pi\)-bonds in an oxoacid of sulphur is determined by its molecular structure, specifically the number of \(S=O\) double bonds formed when sulphur is in its higher oxidation states.


Step 2: Key Formula or Approach:

Draw the chemical structures for both acids:

1. Peroxodisulphuric acid (\(H_{2}S_{2}O_{8}\)), also known as Marshall's acid.

2. Pyrosulphuric acid (\(H_{2}S_{2}O_{7}\)), also known as Oleum.


Step 3: Detailed Explanation:

1. Peroxodisulphuric acid (\(H_{2}S_{2}O_{8}\)):

Its structure is \(HO-S(=O)_{2}-O-O-S(=O)_{2}-OH\).

Each sulphur atom is bonded to two oxygen atoms via double bonds (\(S=O\)).

Since there are two sulphur atoms, total \(\pi\)-bonds = \(2 + 2 = 4\).


2. Pyrosulphuric acid (\(H_{2}S_{2}O_{7}\)):

Its structure is \(HO-S(=O)_{2}-O-S(=O)_{2}-OH\).

Similarly, each sulphur atom forms two \(S=O\) double bonds.

Total \(\pi\)-bonds = \(2 + 2 = 4\).


3. Sum of \(\pi\)-bonds:
\[ Sum = 4 (in H_{2}S_{2}O_{8}) + 4 (in H_{2}S_{2}O_{7}) = 8 \]


Step 4: Final Answer:

The sum of \(\pi\)-bonds is 8.
Quick Tip: In oxoacids of sulphur where sulphur is in the +6 oxidation state (like \(H_{2}SO_{4}\), \(H_{2}S_{2}O_{7}\), \(H_{2}S_{2}O_{8}\)), each sulphur atom typically contributes 2 \(\pi\)-bonds from its two \(S=O\) groups.

Step 1: Understanding the Concept:

Concentration is a measure of the amount of solute present in a given amount of solvent or solution. Units of concentration relate the quantity of solute to the quantity of the solution or solvent.


Step 2: Detailed Explanation:

Let's evaluate each term in the list:

1. Mass percent: Expresses the mass of solute per 100g of solution. (Concentration unit)

2. Mole: A unit of the amount of substance (SI unit). It is a quantity, not a ratio or concentration. (Not a concentration unit)

3. Mole fraction: The ratio of moles of a component to the total moles in the mixture. (Concentration unit)

4. Molarity: Moles of solute per liter of solution. (Concentration unit)

5. ppm (parts per million): Expresses very dilute concentrations as parts of solute per million parts of solution. (Concentration unit)

6. Molality: Moles of solute per kilogram of solvent. (Concentration unit)


Counting the valid units: Mass percent, Mole fraction, Molarity, ppm, and Molality. Total = 5.


Step 3: Final Answer:

The number of concentration units is 5.
Quick Tip: Always distinguish between units of quantity (like mole, gram, liter) and units of concentration (which are always ratios of solute to solution/solvent).

Step 1: Understanding the Concept:

The number of monochloro derivatives depends on the number of chemically distinct sets of hydrogen atoms in the molecule and whether substitution at those positions creates chiral centers.


Step 2: Key Formula or Approach:

Draw the structure of 2,2,5,5-tetramethylhexane:
\[ CH_{3} - C(CH_{3})_{2} - CH_{2} - CH_{2} - C(CH_{3})_{2} - CH_{3} \]

Identify equivalent groups of hydrogens based on molecular symmetry.


Step 3: Detailed Explanation:

1. Symmetry Analysis: The molecule is highly symmetric. There is a center of symmetry in the middle of the \(C3-C4\) bond.

2. Set 1 (Methyl Hydrogens): All 18 methyl hydrogens (at \(C1, C6\) and the four branch methyls) are equivalent due to rotation and symmetry. Replacing any of these with Cl gives:
\(Cl-CH_{2}-C(CH_{3})_{2}-CH_{2}-CH_{2}-C(CH_{3})_{2}-CH_{3}\) (1-chloro-2,2,5,5-tetramethylhexane). This molecule is achiral. (1 isomer)

3. Set 2 (Methylene Hydrogens): The 4 hydrogens on \(C3\) and \(C4\) are equivalent. Replacing one H at \(C3\) with Cl gives:
\(CH_{3}-C(CH_{3})_{2}-CHCl-CH_{2}-C(CH_{3})_{2}-CH_{3}\).

4. Stereochemistry: The carbon atom where Cl is attached (\(C3\)) becomes a chiral center because it is bonded to four different groups: H, Cl, a tert-butyl group, and a neopentyl-like group. This results in a pair of enantiomers (R and S). (2 isomers)

5. Total Isomers: \(1 (achiral) + 2 (enantiomers) = 3\).


Step 4: Final Answer:

The maximum number of isomeric monochloro derivatives is 3.
Quick Tip: When a question asks for "isomers" without specifying "structural isomers," you must include stereoisomers (enantiomers and diastereomers) in your count.

Step 1: Understanding the Concept:

Black body radiation curves describe the intensity of radiation emitted by a black body at different wavelengths for various temperatures. Key laws involved are Wien's Displacement Law and Planck's Law.


Step 2: Detailed Explanation:

Analyze each statement:

A. Incorrect: Looking at the graph, \(T_{1}\) has the highest peak and the peak wavelength is shifted most to the left (shortest \(\lambda\)). According to Wien's law, higher temperature means shorter peak wavelength and higher intensity. Thus, \(T_{1} > T_{2} > T_{3} > T_{4}\).

B. Incorrect: While Planck's model used "oscillators" (resonators), it is a theoretical construct for the walls of the cavity. Saying the black body "consists of particles performing SHM" is an oversimplification and not a defining characteristic of the spectrum.

C. Correct: Wien's Displacement Law (\(\lambda_{m}T = b\)) states that as temperature increases, the peak wavelength \(\lambda_{m}\) decreases.

D. Incorrect: Wien's law in terms of frequency is \(\frac{\nu_{m}}{T} = constant\). The statement says the ratio is not constant.

E. Correct: The classical Rayleigh-Jeans law failed to explain the spectrum (ultraviolet catastrophe). Planck successfully explained it by assuming energy is quantized (\(E = nh\nu\)).


Correct statements are C and E. Total = 2.


Step 3: Final Answer:

The number of correct statements is 2.
Quick Tip: Remember the two main trends in black body graphs as T increases: 1. The total area under the curve increases (\(E \propto T^{4}\)). 2. The peak shifts toward the left (shorter \(\lambda\), higher \(\nu\)).

Step 1: Understanding the Concept:

Work done in a thermodynamic cycle is the area enclosed by the \(P-V\) loop. Work for individual processes is calculated using \(W = -\int P dV\).


Step 2: Detailed Explanation:

From the graph, identify the cycle \(1 \rightarrow 2 \rightarrow 3 \rightarrow 1\):

- Process \(1 \rightarrow 2\): Isobaric expansion at \(P = 1.0\) bar.
\(W_{12} = -P \Delta V = -1.0 \times (40 - 20) = -20\) bar\(\cdot\)L.

- Process \(2 \rightarrow 3\): Isochoric cooling at \(V = 40\) L.
\(W_{23} = 0\) (as \(\Delta V = 0\)).

- Process \(3 \rightarrow 1\): Curved path. Check if it's isothermal:

At point 1: \(P \times V = 1.0 \times 20 = 20\).

At point 3: \(P \times V = 0.5 \times 40 = 20\).

Since \(PV\) is constant, it is an isothermal compression.
\(W_{31} = -nRT \ln\left(\frac{V_{final}}{V_{initial}}\right) = -P_{1}V_{1} \ln\left(\frac{V_{1}}{V_{3}}\right) = -20 \ln\left(\frac{20}{40}\right) = 20 \ln 2\).


Step 3: Calculation:

1. \(\ln 2 = 2.3 \times \log 2 = 2.3 \times 0.3 = 0.69\).

2. \(W_{31} = 20 \times 0.69 = 13.8\) bar\(\cdot\)L.

3. \(Net Work W = W_{12} + W_{23} + W_{31} = -20 + 0 + 13.8 = -6.2\) bar\(\cdot\)L.

4. \(Magnitude = 6.2\) bar\(\cdot\)L.

5. Convert to Joules (\(1 bar\cdotL = 100 J\)):
\(W = 6.2 \times 100 = 620\) J.


Step 4: Final Answer:

The magnitude of work done is 620 J.
Quick Tip: For an ideal gas, if \(P_{1}V_{1} = P_{2}V_{2}\), the process is isothermal. Use the formula \(W = 2.303 nRT \log(V_{1}/V_{2})\) carefully, keeping track of signs for expansion (-) and compression (+).

Step 1: Understanding the Concept:

A tripeptide is formed by the linkage of three amino acids. If we are "mixing" two amino acids, each of the three positions in the tripeptide chain can be occupied by either of the two available amino acids.


Step 2: Detailed Explanation:

1. Let the two amino acids be \(A\) (Valine) and \(B\) (Proline).

2. A tripeptide has 3 positions: \(Pos 1 - Pos 2 - Pos 3\).

3. Number of choices for Position 1 = 2 (either Val or Pro).

4. Number of choices for Position 2 = 2.

5. Number of choices for Position 3 = 2.

6. Total combinations = \(2 \times 2 \times 2 = 2^{3} = 8\).


The possible tripeptides are: Val-Val-Val, Val-Val-Pro, Val-Pro-Val, Pro-Val-Val, Val-Pro-Pro, Pro-Val-Pro, Pro-Pro-Val, Pro-Pro-Pro.


Step 3: Final Answer:

The total number of tripeptides is 8.
Quick Tip: If \(n\) different amino acids are available to form a peptide of length \(k\), the total number of possible peptides is \(n^{k}\). Here, \(n=2\) and \(k=3\).

Step 1: Understanding the Concept:

For an ideal solution, the total vapour pressure (\(P_{T}\)) is given by Raoult's Law: \(P_{T} = P_{A}^{\circ}X_{A} + P_{B}^{\circ}X_{B}\), where \(P^{\circ}\) are pure component pressures and \(X\) are mole fractions.


Step 2: Key Formula or Approach:

We have two unknowns (\(P_{A}^{\circ}\) and \(P_{B}^{\circ}\)) and two equations based on the two given conditions.


Step 3: Detailed Explanation:

1. Case 1: \(X_{A} = 0.7, X_{B} = 0.3, P_{T} = 350\).
\[ 0.7 P_{A}^{\circ} + 0.3 P_{B}^{\circ} = 350 \quad ---(1) \]

2. Case 2: \(X_{A} = 0.2, X_{B} = 0.8, P_{T} = 410\).
\[ 0.2 P_{A}^{\circ} + 0.8 P_{B}^{\circ} = 410 \quad ---(2) \]

3. Solving the equations:

Multiply equation (1) by 8 and equation (2) by 3 to eliminate \(P_{B}^{\circ}\):
\[ 5.6 P_{A}^{\circ} + 2.4 P_{B}^{\circ} = 2800 \]
\[ 0.6 P_{A}^{\circ} + 2.4 P_{B}^{\circ} = 1230 \]

Subtracting the two:
\[ 5.0 P_{A}^{\circ} = 1570 \Rightarrow P_{A}^{\circ} = \frac{1570}{5} = 314 mm Hg \]


Step 4: Final Answer:

The vapour pressure of pure A is 314 mm Hg.
Quick Tip: In Raoult's Law problems with two conditions, always set up a system of linear equations. If you need \(P_{A}^{\circ}\), eliminate \(P_{B}^{\circ}\) by matching its coefficients.

Step 1: Understanding the Concept:

Calcium lactate is a salt of a weak acid (lactic acid) and a strong base (calcium hydroxide). The pH of such a salt solution is calculated using the salt hydrolysis formula.


Step 2: Key Formula or Approach:

1. Salt concentration \(C = [Lactate^{-}]\). Since calcium lactate is \((CH_{3}CH(OH)COO)_{2}Ca\), one mole of salt gives 2 moles of lactate ions.

2. \(pH = 7 + \frac{1}{2}(pK_{a} + \log C)\).


Step 3: Detailed Explanation:

1. Calculate Ion Concentration:
\(Molarity of Calcium Lactate = 0.005 M\).
\([Lactate^{-}] = 2 \times 0.005 = 0.01 M = 10^{-2} M\).

2. Apply pH Formula:
\(pH = 7 + \frac{1}{2}(5 + \log 10^{-2})\)
\(pH = 7 + \frac{1}{2}(5 - 2) = 7 + \frac{3}{2} = 7 + 1.5 = 8.5\).

3. Convert to required format:
\(8.5 = 85 \times 10^{-1}\).


Step 4: Final Answer:

The value is 85.
Quick Tip: Don't forget the stoichiometry! For salts of divalent metals like \(Ca^{2+}\), the concentration of the anion (the species that hydrolyzes) is twice the molarity of the salt.

Step 1: Understanding the Concept:

Physisorption (physical adsorption) involves weak van der Waals forces between the adsorbate and adsorbent, unlike chemisorption which involves chemical bonding.


Step 2: Detailed Explanation:

Evaluate each statement:

A. Incorrect: Physisorption is not specific; any gas can be adsorbed to some extent on any solid. Chemisorption is highly specific.

B. Incorrect: Enthalpy is low (20-40 kJ/mol) because of weak forces.

C. Correct: Since adsorption is exothermic, according to Le Chatelier's principle, it decreases with increasing temperature.

D. Incorrect: It usually forms multimolecular layers. Unimolecular layers are characteristic of chemisorption.

E. Correct: Physisorption occurs almost instantaneously and does not require activation energy.


Characteristics are C and E. Total = 2.


Step 3: Final Answer:

The number of characteristic statements is 2.
Quick Tip: Think of Physisorption as "surface condensation." It's non-specific, low energy, and multilayered—just like a liquid film forming on a surface.

Step 1: Understanding the Concept:

The Andrews isotherm for \(CO_{2}\) shows the transition from gas to liquid. The horizontal portion of the curve represents the region where phase transition (liquefaction) occurs at constant pressure.


Step 2: Detailed Explanation:

Based on standard Andrews isotherm nomenclature:

A. Correct: From point (a) to point (b), the \(CO_{2}\) is being compressed as a gas. Liquefaction starts exactly at point (b).

B. Incorrect: Liquid appears for the first time at point (b), not (c). Point (c) is where the entire gas has been converted to liquid.

C. Correct: The horizontal line between (b) and (c) represents the equilibrium where both liquid and gaseous phases coexist.

D. Incorrect: As volume decreases from (b) to (c), more gas is converted to liquid, so the amount of liquid increases.


Correct statements are A and C. Total = 2.


Step 3: Final Answer:

The number of correct statements is 2.
Quick Tip: In a P-V isotherm, a horizontal line always indicates a phase change. During this change, pressure remains constant while the ratio of the two phases changes as volume is altered.