Step 1: Understanding the Concept:

Molecular solids are classified into three categories: non-polar, polar, and hydrogen-bonded molecular solids.

Polar molecular solids are formed by molecules that have a permanent dipole moment due to differences in electronegativity between bonded atoms.

These molecules are held together by relatively strong dipole-dipole interactions.


Step 2: Detailed Explanation:

1. Option A: \(SO_{2}\) is a polar molecule because of its bent shape and the electronegativity difference between Sulphur and Oxygen. \(NH_{3}\) is also a polar molecule with a pyramidal shape and undergoes hydrogen bonding (a strong type of dipole-dipole interaction). Therefore, both are polar molecular solids in their solid state.

2. Option B: \(SO_{2}\) is polar, but \(CO_{2}\) is a non-polar molecular solid because its linear structure results in a zero net dipole moment.

3. Option C: \(HCl\) is a polar molecular solid, but \(AlN\) (Aluminium Nitride) is a covalent network solid.

4. Option D: \(MgO\) is an ionic solid held by strong electrostatic forces, not a molecular solid.


Step 3: Final Answer:

Since both \(SO_{2}(s)\) and \(NH_{3}(s)\) consist of polar molecules held by dipole-dipole or hydrogen bonding forces, they are polar molecular solids.
Quick Tip: Remember that \(CO_{2}\), \(I_{2}\), and \(CH_{4}\) are common examples of non-polar molecular solids, while \(H_{2}O\) (ice) and \(NH_{3}\) are hydrogen-bonded polar molecular solids.

Step 1: Understanding the Concept:

According to Raoult's Law for dilute solutions, the relative lowering of vapour pressure is equal to the mole fraction of the solute.


Step 2: Key Formula or Approach:

For a very dilute solution, the formula can be simplified to:
\[ \frac{P^{o} - P_{s}}{P^{o}} = \frac{n_{solute}}{n_{solvent}} \]
Where:
\( P^{o} - P_{s} = \Delta P \) (Lowering of vapour pressure)
\( P^{o} = \) Vapour pressure of pure solvent
\( n_{solute} = \frac{w_{2}}{M_{2}} \) and \( n_{solvent} = \frac{w_{1}}{M_{1}} \)


Step 3: Detailed Explanation:

Given values:
\( \Delta P = 0.20 \) mm Hg
\( P^{o} = 54.2 \) mm Hg
\( w_{1} = 100 \) g (mass of water)
\( M_{1} = 18 \) g/mol (molar mass of water)
\( M_{2} = 180 \) g/mol (molar mass of glucose)
\( w_{2} = ? \) (mass of glucose)


Substituting the values into the formula:
\[ \frac{0.20}{54.2} = \frac{w_{2} / 180}{100 / 18} \] \[ \frac{0.20}{54.2} = \frac{w_{2}}{180} \times \frac{18}{100} \] \[ \frac{0.20}{54.2} = \frac{w_{2}}{10 \times 100} \] \[ \frac{0.20}{54.2} = \frac{w_{2}}{1000} \] \[ w_{2} = \frac{0.20 \times 1000}{54.2} \] \[ w_{2} = \frac{200}{54.2} \approx 3.69 g \]

Step 4: Final Answer:

The weight of glucose required is approximately 3.69 g.
Quick Tip: For very dilute solutions, using \( \frac{\Delta P}{P^{o}} = \frac{n}{N} \) is a safe approximation that saves calculation time compared to using the full mole fraction formula.

Step 1: Understanding the Concept:

Since the reaction is first order with respect to both A and B, the rate law is given by:
\[ Rate = k [A]^{1} [B]^{1} \]

Step 2: Key Formula or Approach:

Use the data from the first row to find the rate constant \(k\), then use it to find the unknown values \(x\) and \(y\).


Step 3: Detailed Explanation:

Calculation of k:

From Row 1: \( 0.10 = k(20)(0.5) \)
\( 0.10 = k(10) \)
\( k = \frac{0.10}{10} = 0.01 L mol^{-1} s^{-1} \)


Calculation of x:

From Row 2: \( 0.40 = k(x)(0.5) \)
\( 0.40 = 0.01 \times x \times 0.5 \)
\( 0.40 = 0.005x \)
\( x = \frac{0.40}{0.005} = \frac{400}{5} = 80 \)


Calculation of y:

From Row 3: \( 0.80 = k(40)(y) \)
\( 0.80 = 0.01 \times 40 \times y \)
\( 0.80 = 0.4y \)
\( y = \frac{0.80}{0.4} = 2 \)


Step 4: Final Answer:

The values are \( x = 80 \) and \( y = 2 \).
Quick Tip: When the concentration of one reactant is kept constant (like [B] in rows 1 and 2), the rate ratio equals the concentration ratio of the other reactant raised to its order.
Row 1 to 2: \( \frac{0.40}{0.10} = \frac{x}{20} \implies 4 = \frac{x}{20} \implies x = 80 \).

Step 1: Understanding the Concept:

Mass percentage is defined as the mass of the solute divided by the total mass of the solution, multiplied by 100.


Step 2: Key Formula or Approach:
\[ Mass % of X = \frac{Mass of X}{Mass of X + Mass of water} \times 100 \]

Step 3: Detailed Explanation:

Mass of solute (X) = 2 g.

Moles of water = 1 mole.

Molar mass of water (\(H_{2}O\)) = 18 g/mol.

Mass of water = \( 1 mole \times 18 g/mol = 18 g \).

Total mass of solution = \( 2 g (solute) + 18 g (solvent) = 20 g \).

Mass % of X = \( \frac{2}{20} \times 100 = \frac{1}{10} \times 100 = 10% \).


Step 4: Final Answer:

The mass percent of "X" is 10%.
Quick Tip: Always remember to calculate the total mass of the solution (solute + solvent) in the denominator, not just the solvent mass.

Step 1: Understanding the Concept:

This question pertains to the principles of extractive metallurgy, specifically the pre-treatment of ores before chemical reduction to the metal.


Step 2: Detailed Explanation:

Statement I: Sulphide ores are typically roasted in the presence of air to convert them into oxides (e.g., \(2ZnS + 3O_{2} \rightarrow 2ZnO + 2SO_{2}\)). This is a standard step in metallurgy. Thus, Statement I is correct.

Statement II: Thermodynamically, it is easier to reduce metal oxides using common reducing agents like Carbon (coke) or Carbon Monoxide compared to reducing metal sulphides. The reduction of sulphides with Carbon is energetically unfavorable (highly positive \(\Delta G^{\circ}\)) under standard conditions compared to oxides. Hence, Statement II is correct.


Step 3: Final Answer:

Since both statements accurately reflect metallurgical principles, option (A) is the correct choice.
Quick Tip: Oxides are easier to reduce because oxygen has a higher affinity for reducing agents like Carbon (forming \(CO_{2}\)) than sulphur does (forming \(CS_{2}\)). This can be verified via Ellingham diagrams.

Step 1: Understanding the Concept:

Stability of hydrides depends on the bond strength and the nature of the bond (ionic vs covalent). Alkali metal hydrides like \(LiH\) are very stable ionic solids. p-block hydrides like \(NH_{3}\) and \(HF\) are stable covalent molecules due to significant electronegativity differences.


Step 2: Detailed Explanation:

1. \(LiH\): It is an ionic hydride with high lattice energy and high thermal stability.

2. \(BeH_{2}\): Beryllium hydride is a covalent, electron-deficient polymeric hydride. Due to its polymeric nature and the small size of Beryllium, it has lower thermal stability compared to alkali metal hydrides and highly electronegative non-metal hydrides. It is often described as metastable.

3. \(NH_{3}\) and \(HF\): These are very stable due to strong covalent bonds and significant hydrogen bonding.


Step 3: Final Answer:

Among the given options, \(BeH_{2}\) is the least stable hydride.
Quick Tip: Thermal stability of metal hydrides generally decreases as the size of the cation increases within a group, but transition from ionic to covalent/polymeric nature (like in \(Be\)) often results in lower stability.

Step 1: Understanding the Concept:

White phosphorus (\(P_{4}\)) reacts with thionyl chloride (\(SOCl_{2}\)) to produce phosphorus trichloride. This is a common laboratory preparation for \(PCl_{3}\).


Step 2: Key Formula or Approach:

The balanced chemical equation for the reaction is:
\[ P_{4} + 8SOCl_{2} \rightarrow 4PCl_{3} + 4SO_{2} + 2S_{2}Cl_{2} \]

Step 3: Detailed Explanation:

Comparing the balanced equation with the question:

- moles of \(P_{4}\) = 1 (Matches)

- moles of \(SOCl_{2}\) = 8 (Matches)

- 4 moles of A \(\rightarrow\) A is \(PCl_{3}\).

- \(x\) moles of \(SO_{2} \rightarrow x = 4\).

- 2 moles of B \(\rightarrow\) B is \(S_{2}Cl_{2}\).


Step 4: Final Answer:

A = \(PCl_{3}\), B = \(S_{2}Cl_{2}\), and \(x = 4\).
Quick Tip: Distinguish this from the reaction with sulphuryl chloride (\(SO_{2}Cl_{2}\)), which produces phosphorus pentachloride (\(PCl_{5}\)): \(P_{4} + 10SO_{2}Cl_{2} \rightarrow 4PCl_{5} + 10SO_{2}\).

Step 1: Understanding the Concept:

Melting point in metals depends on the strength of metallic bonding. Metallic bond strength depends on the number of valence electrons and the size of the atom.


Step 2: Detailed Explanation:

For alkali metals (Group 1), all have one valence electron. As we move down the group from \(Li\) to \(Cs\), the atomic size increases.

The increase in size leads to weaker attraction between the nucleus and the sea of delocalized electrons, resulting in weaker metallic bonding.

Therefore, the melting point decreases down the group:
\( Li > Na > K > Rb > Cs \).


Step 3: Final Answer:

Among the given options, Cesium (\(Cs\)) has the largest atomic size and the weakest metallic bonding, thus it has the least melting point.
Quick Tip: Cesium (\(Cs\)) has a melting point of approximately \(28.5^{\circ}\)C, which is so low that it can melt in the palm of your hand on a warm day.

Step 1: Understanding the Concept:

The spin-only magnetic moment (\(\mu_{s}\)) is calculated using the formula:
\[ \mu_{s} = \sqrt{n(n+2)} B.M. \]
where \(n\) is the number of unpaired electrons.


Step 2: Key Formula or Approach:

1. Determine the oxidation state of Fe.

2. Determine the electronic configuration of \(Fe^{3+}\).

3. Determine if the ligand is strong-field or weak-field to find the number of unpaired electrons.


Step 3: Detailed Explanation:

In both complexes, the oxidation state of Fe is +3.
\(Fe (Z=26): [Ar] 3d^{6} 4s^{2}\)
\(Fe^{3+}: [Ar] 3d^{5}\)


For \([Fe(H_{2}O)_{6}]^{3+}\):

- \(H_{2}O\) is a weak-field ligand.

- No pairing occurs. The \(d\)-electrons remain as 5 unpaired electrons (\(n=5\)).

- \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 B.M. \)


For \([Fe(CN)_{6}]^{3-}\):

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

- Pairing occurs in the \(t_{2g}\) orbitals.

- Configuration becomes \(t_{2g}^{5} e_{g}^{0}\). Number of unpaired electrons (\(n\)) = 1.

- \(\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.732 B.M. \)


Step 4: Final Answer:

The values are 5.92 B.M. and 1.732 B.M. respectively.
Quick Tip: A quick shortcut for magnetic moments: if \(n=1\), \(\mu \approx 1.7-1.8\); if \(n=2\), \(\mu \approx 2.8\); if \(n=3\), \(\mu \approx 3.8\); if \(n=4\), \(\mu \approx 4.9\); if \(n=5\), \(\mu \approx 5.9\). The first digit usually matches the number of unpaired electrons.

Step 1: Understanding the Concept:

The color of coordination compounds arises from d-d transitions.

The energy of light absorbed (\(E\)) is equal to the crystal field splitting energy (\(\Delta_{o}\)) for octahedral complexes.

The relationship between energy and wavelength (\(\lambda\)) is inverse:
\[ E = \Delta_{o} = \frac{hc}{\lambda} \]
Factors that increase \(\Delta_{o}\) will decrease the wavelength of light absorbed.


Step 2: Detailed Explanation:

Evaluating Assertion A:

1. Let's find the oxidation state of Cobalt in both complexes:

For \([CoCl(NH_{3})_{5}]^{2+}\): \(x + (-1) + 5(0) = +2 \implies x = +3 \).

For \([Co(NH_{3})_{5}(H_{2}O)]^{3+}\): \(x + 5(0) + 0 = +3 \implies x = +3 \).

Both have the same central metal ion, \(Co^{3+}\).

2. Now, compare the ligand strengths using the Spectrochemical Series:

Ligand strength order: \(Cl^{-} < H_{2}O < NH_{3} \).

3. In \([CoCl(NH_{3})_{5}]^{2+}\), one ligand is \(Cl^{-}\), which is a weaker field ligand than \(H_{2}O\) found in \([Co(NH_{3})_{5}(H_{2}O)]^{3+}\).

4. Therefore, the crystal field splitting energy (\(\Delta_{o}\)) for the aqua complex is greater than that for the chloro complex:
\[ \Delta_{o} ([Co(NH_{3})_{5}(H_{2}O)]^{3+}) > \Delta_{o} ([CoCl(NH_{3})_{5}]^{2+}) \]
5. Since \(\lambda \propto \frac{1}{\Delta_{o}}\), the aqua complex will absorb light at a lower wavelength (higher energy) than the chloro complex.

Assertion A states that the chloro complex absorbs at a lower wavelength, which is False.


Evaluating Reason R:

1. The magnitude of crystal field splitting (\(\Delta_{o}\)) indeed depends on several factors, including:

- The nature of the ligands.

- The oxidation state of the metal ion (Higher oxidation state usually leads to larger \(\Delta_{o}\)).

- The geometry of the complex.

- The position of the metal in the periodic table (3d vs 4d vs 5d).

2. Since \(\Delta_{o}\) depends on the oxidation state, and \(\lambda\) depends on \(\Delta_{o}\), the statement that absorbed wavelength depends on the oxidation state is True as a general chemical principle.


Step 3: Final Answer:

Assertion A is false because the chloro complex absorbs at a higher wavelength compared to the aqua complex due to its smaller crystal field splitting.

Reason R is a true statement in general coordination chemistry.

Thus, A is false but R is true.
Quick Tip: Remember: Strong Field Ligand \(\rightarrow\) Large Splitting (\(\Delta_{o}\)) \(\rightarrow\) High Energy Transition \(\rightarrow\) Short Wavelength (\(\lambda\)).
The spectrochemical series is crucial: \(I^{-} < Br^{-} < S^{2-} < SCN^{-} < Cl^{-} < F^{-} < OH^{-} < C_{2}O_{4}^{2-} < H_{2}O < NH_{3} < en < CN^{-} < CO\).

Step 1: Understanding the Concept:

Freons are a group of chlorofluorocarbons (CFCs) or hydrofluorocarbons (HFCs) used primarily as refrigerants and propellants.

By convention, standard Freons are saturated (alkane derivatives) and contain both chlorine and fluorine atoms.


Step 2: Detailed Explanation:

1. Option A: \(C_{2}Cl_{2}F_{4}\) is known as Freon-114 (dichlorotetrafluoroethane). It is a saturated CFC and fits the definition of a Freon.

2. Option B: \(C_{2}H_{2}F_{2}\) (difluoroethene or difluoroethane) is a hydrofluorocarbon (HFC). While sometimes called "Freons" in a broad industrial sense, classical Freons are CFCs.

3. Option C: \(C_{2}HF_{3}\) is a hydrochlorofluorocarbon (HCFC) or HFC derivative.

4. Option D: \(C_{2}F_{4}\) (tetrafluoroethene) is an alkene and is the monomer for Teflon, not a standard Freon refrigerant.


Step 3: Final Answer:

The most appropriate example of a classical Freon among the choices is \(C_{2}Cl_{2}F_{4}\).
Quick Tip: Freon nomenclature follows the "Rule of 90": add 90 to the Freon number. The resulting three digits represent (from left to right) the number of Carbon, Hydrogen, and Fluorine atoms. The remaining valencies of Carbon are filled by Chlorine.

Step 1: Understanding the Concept:

The geometry and properties of coordination complexes depend on the central metal ion's electronic configuration and its position in the periodic table (3d vs 4d/5d).


Step 2: Detailed Explanation:

1. Complex with \(Ni^{2+}\):
\(Ni^{2+}\) is a \(3d^{8}\) ion. With weak field ligands like \(Cl^{-}\) and \(Br^{-}\), it forms a tetrahedral complex (\(sp^{3}\) hybridization).

- Geometry: Tetrahedral.

- Geometrical Isomerism: Tetrahedral \(MA_{2}B_{2}\) complexes do not show GI.

- Magnetic Properties: It has two unpaired electrons, making it paramagnetic.


2. Complex with \(Pt^{2+}\):
\(Pt^{2+}\) is a \(5d^{8}\) ion. Metal ions from the 4d and 5d series have very high crystal field splitting (\(\Delta_{o}\)), causing all complexes to be square planar regardless of ligand strength (\(dsp^{2}\) hybridization).

- Geometry: Square Planar.

- Geometrical Isomerism: Square planar \(MA_{2}B_{2}\) complexes show cis and trans isomerism.

- Magnetic Properties: All electrons are paired, making it diamagnetic.


3. Optical Isomerism: Neither simple tetrahedral nor square planar \(MA_{2}B_{2}\) complexes show optical isomerism.


Step 3: Final Answer:

The properties that change are Geometry (A), Geometrical isomerism (B), and Magnetic properties (D).
Quick Tip: Always remember: All 4d and 5d metals with coordination number 4 form square planar complexes because their large size and high nuclear charge lead to very large \(\Delta_{o}\) values.

Step 1: Understanding the Concept:

This question tests knowledge of the characteristic colors of inorganic precipitates used in qualitative analysis.


Step 2: Detailed Explanation:

1. A. \(Mg(NH_{4})PO_{4}\): Magnesium ammonium phosphate is a characteristic white crystalline precipitate formed in the identification of Magnesium ions. (A \(\rightarrow\) II)

2. B. \(K_{3}[Co(NO_{2})_{6}]\): Potassium cobaltinitrite (Fischer's salt) is a yellow precipitate used for the detection of potassium ions. (B \(\rightarrow\) III)

3. C. \(MnO(OH)_{2}\): This represents hydrated Manganese(IV) oxide, which is brown in color (similar to \(MnO_{2}\)). (C \(\rightarrow\) I)

4. D. \(Fe_{4}[Fe(CN)_{6}]_{3}\): Ferric ferrocyanide is the chemical name for Prussian Blue, which has an intense deep blue color. (D \(\rightarrow\) IV)


Step 3: Final Answer:

Matching the items: A-II, B-III, C-I, D-IV. This corresponds to Option (A).
Quick Tip: Qualitative analysis precipitates like Prussian Blue (Blue), Fischer's Salt (Yellow), and \(Mg(NH_{4})PO_{4}\) (White) are high-yield facts for competitive exams.

Step 1: Understanding the Concept:

Reaction with alcoholic \(AgNO_{3}\) is a test for halide ions. A precipitate of \(AgBr\) forms if the C-Br bond can break to form a stable carbocation ( \(S_{N}1\) mechanism).


Step 2: Detailed Explanation:

1. Benzyl bromide: Forms the benzylic carbocation, which is resonance stabilized. It reacts readily.

2. Cinnamyl bromide (\(C_{6}H_{5}CH=CHCH_{2}Br\)): Forms an allylic/benzylic carbocation which is highly resonance stabilized. It reacts readily.

3. Cyclopropenyl bromide: Loss of \(Br^{-}\) results in the cyclopropenyl cation, which is aromatic (\(4n+2 \pi\) electrons where \(n=0\)) and extremely stable. It reacts very fast.

4. 1-bromocyclohexene derivative (Vinyl halide): Here, Bromine is attached to a \(sp^{2}\) hybridized carbon. The C-Br bond has partial double bond character due to resonance, making it much stronger. Additionally, the vinylic carbocation is highly unstable. Thus, it does not undergo nucleophilic substitution under normal conditions and does not form a precipitate.


Step 3: Final Answer:

The vinyl halide (Option D) will not produce a precipitate.
Quick Tip: Vinyl halides (\(C=C-X\)) and aryl halides (\(Ar-X\)) are inert toward nucleophilic substitution reactions because of the partial double bond character of the C-X bond and high energy of the resulting carbocations.

Step 1: Understanding the Concept:

This is a sequence of functional group transformations: Alcohol \(\rightarrow\) Alkyl Halide \(\rightarrow\) Grignard Reagent \(\rightarrow\) Deuterated Alkane.


Step 2: Detailed Explanation:

Reaction 1: \(CH_{3}CH_{2}CH(OH)CH_{3} \xrightarrow{NaI, H_{3}PO_{4}}\)
\(NaI/H_{3}PO_{4}\) generates \(HI\) in situ, which converts the alcohol to an alkyl iodide via \(S_{N}\) mechanism.

Product: \(CH_{3}CH_{2}CH(I)CH_{3}\) (2-iodobutane).


Reaction 2: \(CH_{3}CH_{2}CH(I)CH_{3} \xrightarrow{Mg, Dry ether}\)

Formation of Grignard reagent.

Product: \(CH_{3}CH_{2}CH(MgI)CH_{3}\) (sec-butyl magnesium iodide).


Reaction 3: \(CH_{3}CH_{2}CH(MgI)CH_{3} \xrightarrow{D_{2}O}\)

Grignard reagents are strong bases. They react with sources of acidic hydrogen/deuterium (like \(D_{2}O\)) to form alkanes.
\(R-MgI + D_{2}O \rightarrow R-D + Mg(OD)I\).

Product: \(CH_{3}CH_{2}CH(D)CH_{3}\).


Step 3: Final Answer:

The final product is 2-deuteriobutane.
Quick Tip: Grignard reagents (\(RMgX\)) are excellent tools for introducing isotopes like Deuterium or Tritium into a specific position on a carbon chain.

Step 1: Understanding the Concept:
\(Zn(Hg)/HCl\) is the reagent for Clemmensen reduction, which reduces aldehydes/ketones to alkanes. However, the presence of an alcohol and acidic conditions can lead to further reactions.


Step 2: Detailed Explanation:

1. Reduction of Carbonyl: The aldehyde group (\(-CHO\)) is reduced to a methyl group (\(-CH_{3}\)).

2. Acid-catalyzed Dehydration: The reactant is a \(\beta\)-hydroxy aldehyde. Under the acidic (\(HCl\)) and heating (\(\Delta\)) conditions of the Clemmensen reduction, benzylic alcohols undergo rapid dehydration to form a more stable, conjugated alkene system.

3. Reaction Path:
\(C_{6}H_{5}-CH(OH)-CH(CH_{3})-CH_{2}-CHO \xrightarrow{reduction} C_{6}H_{5}-CH(OH)-CH(CH_{3})-CH_{2}-CH_{3} \xrightarrow[(-H_{2}O)]{H^{+}/\Delta} C_{6}H_{5}-CH=C(CH_{3})-CH_{2}-CH_{3} \).

4. The formation of the alkene (B) is preferred because the double bond is in conjugation with the benzene ring, providing significant resonance stabilization.


Step 3: Final Answer:

The major product is the conjugated alkene, corresponding to B.
Quick Tip: In Clemmensen reduction, acid-sensitive groups like alcohols often undergo dehydration. If you want to preserve the alcohol, Wolff-Kishner reduction (basic) might be better, but \(\beta\)-hydroxy carbonyls are sensitive to both!

Step 1: Understanding the Concept:

Wolff-Kishner reduction involves heating a carbonyl compound with hydrazine (\(NH_{2}NH_{2}\)) and a strong base like \(KOH\).


Step 2: Detailed Explanation:

1. Reason R: It is a fundamental fact that Wolff-Kishner reduction converts a carbonyl group (\(C=O\)) into a methylene group (\(CH_{2}\)). This statement is True.

2. Assertion A: Chloro-ketones contain a halogen atom. Halogens (especially \(Cl, Br, I\)) are very sensitive to strong bases. Under the basic conditions of the Wolff-Kishner reaction (\(KOH\)), the chloro-ketone will undergo dehydrohalogenation (elimination) to form an alkene or potentially nucleophilic substitution. Therefore, Wolff-Kishner is generally not suitable for reducing halogenated ketones if the halogen is to be preserved. Clemmensen reduction (acidic) is typically preferred for such molecules. Thus, Assertion A is False.


Step 3: Final Answer:

Assertion A is false, and Reason R is true.
Quick Tip: Remember: Wolff-Kishner (Basic) is used for acid-sensitive compounds. Clemmensen (Acidic) is used for base-sensitive compounds (like halo-ketones).

Step 1: Understanding the Concept:

The sequence involves diazotization followed by substitution of the diazonium group with a thiol group.


Step 2: Detailed Explanation:

1. Step 1 (Diazotization): The aromatic amine (\(-NH_{2}\)) reacts with \(NaNO_{2}/HCl\) at low temperature to form the benzenediazonium salt (Compound A).

2. Step 2 (Substitution): Reaction of diazonium salts with hydrosulphide ions (\(SH^{-}\) from \(NH_{4}SH\)) replaces the \(-N_{2}^{+}\) group with a thiol (\(-SH\)) group.

3. The other substituents on the ring (isopropyl and hydroxy groups) remain unaffected by these specific reagents.


Step 3: Final Answer:

The resulting product has an \(-SH\) group at the position where the \(-NH_{2}\) group was originally located. This corresponds to the structure in Option (B).
Quick Tip: Diazonium salts are versatile intermediates. Replacing \(-N_{2}^{+}\) with \(-SH\) is a standard way to synthesize thiophenols.

Step 1: Understanding the Concept:

This question combines the Hell-Volhard-Zelinsky (HVZ) reaction with basic stereochemistry principles.


Step 2: Detailed Explanation:

1. Evaluating Assertion A: Glycine is \(H_{2}N-CH_{2}-COOH\). Reaction with \(Cl_{2}/Red\ P\) (HVZ reaction conditions) results in \(\alpha\)-halogenation of the carboxylic acid.

The product is \(\alpha\)-chloro glycine: \(H_{2}N-CH(Cl)-COOH\).

The \(\alpha\)-carbon is now bonded to four different groups: \(-H\), \(-Cl\), \(-NH_{2}\), and \(-COOH\). Therefore, it is a chiral carbon atom. Assertion A is True.

2. Evaluating Reason R: A molecule with two or more chiral carbons is not always optically active. If the molecule has an internal plane of symmetry (like meso compounds), it is optically inactive despite having chiral centers. Reason R is False.


Step 3: Final Answer:

Assertion A is true, but Reason R is false.
Quick Tip: Meso-tartaric acid is the classic example used to disprove Reason R. It has two chiral centers but is achiral overall due to symmetry.

Step 1: Understanding the Concept:

This question deals with the industrial preparation and classification of common polymers.


Step 2: Detailed Explanation:

1. Statement I: The combination of \(AlEt_{3}\) and \(TiCl_{4}\) is known as the Ziegler-Natta catalyst. Polymerization of ethene using this catalyst at low temperatures and pressures results in High-Density Polyethylene (HDPE), not Low-Density Polyethylene (LDP). LDP requires extremely high pressures (1000-2000 atm) and a free-radical initiator. Thus, Statement I is False.

2. Statement II: Nylon 6 is manufactured by heating caprolactam with water at 533-543 K. The water acts as an initiator, opening the ring to form an amino acid which then undergoes polymerization. While it has characteristics of ring-opening polymerization, Nylon 6 is structurally a polyamide and its formation is classified as a step-growth process in many contexts. Statement II is True.


Step 3: Final Answer:

Statement I is false and Statement II is true.
Quick Tip: HDPE is produced at low pressures using Z-N catalysts and is linear. LDPE is produced at high pressures using oxygen/peroxides and is highly branched.

Step 1: Understanding the Concept:

The reaction between magnesium metal and hydrochloric acid is a redox reaction that produces hydrogen gas.

Stoichiometry allows us to relate the mass of the limiting reactant (magnesium) to the volume of the gas produced at standard temperature and pressure (STP).


Step 2: Key Formula or Approach:

1. Write the balanced chemical equation.

2. Calculate the number of moles of Magnesium.

3. Use the mole ratio to find the moles of Hydrogen gas.

4. Calculate the volume at STP: \( V = n \times 22.4 L \).


Step 3: Detailed Explanation:

The balanced chemical equation for the reaction is:
\[ Mg(s) + 2HCl(aq) \rightarrow MgCl_{2}(aq) + H_{2}(g) \]
From the equation, 1 mole of \(Mg\) produces 1 mole of \(H_{2}\).


Given:

Mass of \(Mg = 2.4 g \)

Molar mass of \(Mg = 24 g mol^{-1} \)

Moles of \(Mg = \frac{Mass}{Molar mass} = \frac{2.4}{24} = 0.1 mol \)


According to stoichiometry:

Moles of \(H_{2}\) produced = Moles of \(Mg = 0.1 mol \)


Volume of \(H_{2}\) at STP:
\[ V = 0.1 \times 22.4 L = 2.24 L \]

To express this in the form \( \dots \times 10^{-2} L \):
\[ 2.24 = 224 \times 10^{-2} L \]

Step 4: Final Answer:

The value to be filled in the blank is 224.
Quick Tip: At STP (old convention), 1 mole of any ideal gas occupies 22.4 L. Always ensure your chemical equation is balanced before performing stoichiometric calculations.

Step 1: Understanding the Concept:

The number of lone pairs on a central atom can be determined using VSEPR theory by calculating the total valence electrons and subtracting those used in bonding.


Step 2: Key Formula or Approach:

Number of Lone Pairs (LP) = \( \frac{V + M - C + A}{2} - (Number of Bonded Atoms) \)

Where \(V\) = valence electrons of central atom, \(M\) = monovalent atoms, \(C\) = cation charge, \(A\) = anion charge.


Step 3: Detailed Explanation:

1. \(ClO_{3}^{-}\): Central atom is \(Cl\) (7 valence \(e^{-}\)).

- Structure: Chlorine forms one single bond (to \(O^{-}\)) and two double bonds (to \(O\)). Total 5 electrons used for bonding.

- Lone Pairs = \( \frac{7 - 5 + 1}{2} = 1 LP \). (Or simply: 8 valence electrons, 6 used for bonds, 2 left = 1 LP).


2. \(XeF_{4}\): Central atom is \(Xe\) (8 valence \(e^{-}\)).

- monovalent atoms \(M = 4\).

- Electron pairs = \( \frac{8+4}{2} = 6 \).

- Lone Pairs = \( 6 - 4 = 2 LPs \).


3. \(SF_{4}\): Central atom is \(S\) (6 valence \(e^{-}\)).

- monovalent atoms \(M = 4\).

- Electron pairs = \( \frac{6+4}{2} = 5 \).

- Lone Pairs = \( 5 - 4 = 1 LP \).


4. \(I_{3}^{-}\): Central atom is \(I\) (7 valence \(e^{-}\)).

- monovalent atoms \(M = 2\) (other two Iodine atoms), charge \(A = 1\).

- Electron pairs = \( \frac{7+2+1}{2} = 5 \).

- Lone Pairs = \( 5 - 2 = 3 LPs \).


Step 4: Final Answer:

Comparing the lone pairs: \(ClO_{3}^{-}(1)\), \(XeF_{4}(2)\), \(SF_{4}(1)\), \(I_{3}^{-}(3)\). The maximum number is 3.
Quick Tip: For \(I_{3}^{-}\), remember it consists of an \(I_{2}\) molecule with an \(I^{-}\) ion. The central Iodine atom expands its octet and adopts a linear geometry with 3 lone pairs in the equatorial positions of a trigonal bipyramidal arrangement.

Step 1: Understanding the Concept:

This question tests the understanding of atomic structure, specifically the properties of wave functions (\(\psi\)) and probability density (\(\psi^{2}\)) for different orbitals.


Step 2: Detailed Explanation:

A. True: For all s-orbitals, the probability density (\(\psi^{2}\)) is maximum at the nucleus (\(r=0\)).

B. False: For a 2s orbital, the probability density starts at a maximum at the nucleus, decreases to zero (at the radial node), then increases to a smaller secondary maximum before finally decaying to zero at infinity. It does not "first increase".

C. False: Boundary surface diagrams are typically drawn to enclose the region where the probability of finding the electron is high (usually 90% or 95%). It can never be 100% because the probability density only reaches zero at infinity.

D. True: The number of angular nodes is equal to the azimuthal quantum number (\(l\)). For p-orbitals, \(l=1\) (1 angular node). For d-orbitals, \(l=2\) (2 angular nodes).

E. True: For p, d, and f orbitals, the wave function contains a term dependent on \(r\) that vanishes at the nucleus. Therefore, the probability density for these orbitals is zero at the nucleus.


Step 3: Final Answer:

Statements A, D, and E are correct. The number of correct statements is 3.
Quick Tip: Radial nodes = \(n - l - 1\). Angular nodes = \(l\). Total nodes = \(n - 1\). Only s-orbitals have non-zero probability at the nucleus.

Step 1: Understanding the Concept:

- Intensive properties are those which do not depend on the quantity or size of matter present in the system (e.g., density, temperature).

- Extensive properties are those which depend on the quantity or size of matter present in the system (e.g., mass, volume).


Step 2: Detailed Explanation:

1. Volume: Depends on the amount of substance. (Extensive)

2. Molar heat capacity: The heat capacity per mole. Since it is a ratio of two extensive properties (Heat capacity / Moles), it is intensive. (Intensive)

3. Molarity: Concentration (moles/volume). It is a ratio of two extensive properties. (Intensive)

4. \(E^{\theta}\) cell: Standard electrode potential is independent of the amount of substance or the size of the cell. (Intensive)

5. Gibbs free energy change (\(\Delta G\)): Depends on the number of moles of reactants (\(\Delta G = -nFE_{cell}\)). (Extensive)

6. Molar mass: Mass per mole of substance. It is constant for a given substance. (Intensive)

7. Mole: Represents the quantity of substance. (Extensive)


Step 3: Final Answer:

The intensive properties are: Molar heat capacity, Molarity, \(E^{\theta}\) cell, and Molar mass. Total count = 4.
Quick Tip: A useful rule: The ratio of any two extensive properties is always an intensive property (e.g., Mass/Volume = Density).

Step 1: Understanding the Concept:

For a reversible reaction, the equilibrium constant (\(K_{c}\)) is the ratio of the product of the concentrations of products to the product of the concentrations of reactants, each raised to the power of their stoichiometric coefficients.


Step 2: Key Formula or Approach:
\[ K_{c} = \frac{[HI]^{2}}{[H_{2}][I_{2}]} \]
Use an ICE (Initial, Change, Equilibrium) table to find equilibrium concentrations.


Step 3: Detailed Explanation:

Reaction: \(H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)\)


\begin{tabular{|l|c|c|c|
\hline
& \(H_{2}\) & \(I_{2}\) & \(HI\)
\hline
Initial moles & 4.5 & 4.5 & 0
\hline
Change & \(-x\) & \(-x\) & \(+2x\)
\hline
Equilibrium moles & \(4.5 - x\) & \(4.5 - x\) & \(2x\)
\hline
\end{tabular


Given: Moles of \(HI\) at equilibrium = 3.

So, \(2x = 3 \implies x = 1.5 \).


Equilibrium moles:

Moles of \(H_{2} = 4.5 - 1.5 = 3 \).

Moles of \(I_{2} = 4.5 - 1.5 = 3 \).

Moles of \(HI = 3 \).


Volume of vessel (\(V\)) = 10 L.

Equilibrium concentrations:
\( [H_{2}] = 3/10 = 0.3 M \).
\( [I_{2}] = 3/10 = 0.3 M \).
\( [HI] = 3/10 = 0.3 M \).

\[ K_{c} = \frac{(0.3)^{2}}{(0.3)(0.3)} = \frac{0.09}{0.09} = 1 \]

Step 4: Final Answer:

The equilibrium constant \(K_{c}\) is 1.
Quick Tip: For reactions where \(\Delta n_{g} = 0\) (sum of coefficients of products = sum of coefficients of reactants), the volume term cancels out in the \(K_{c}\) expression. You can directly use the number of moles.

Step 1: Understanding the Concept:

This question covers the fundamental principles of electrochemistry, including thermodynamics of cells, standard reduction potentials, and Faraday's laws of electrolysis.


Step 2: Detailed Explanation:

A. True: Cell potential (\(E_{cell}\)) does not depend on the amount of material; doubling the size of the cell does not change the voltage.

B. True: A negative standard reduction potential (\(E^{\theta} < 0\)) indicates that the species is more easily oxidized than hydrogen. Thus, the reduced form of the couple is a better reducing agent than \(H_{2}\).

C. True: According to Faraday's second law, the amount of substance produced is related to the moles of electrons (\(n\)) in the balanced half-reaction (\(Q = nF\)).

D. True: This is a direct statement of Faraday's First Law of Electrolysis (\(m \propto Q\)).


Step 3: Final Answer:

All four statements (A, B, C, D) are correct. The number is 4.
Quick Tip: Remember that while \(E_{cell}\) is intensive, Gibbs free energy (\(\Delta G\)) is extensive. The relation is \(\Delta G = -nFE_{cell}\).

Step 1: Understanding the Concept:

Heterogeneous catalysis involves a solid catalyst and gaseous or liquid reactants. The modern adsorption theory describes the mechanism through several steps.


Step 2: Detailed Explanation:

A. False: In reality, it is the reactants that diffuse to the surface of the catalyst, not the catalyst diffusing over the reactants.

B. True: Adsorption of reactant molecules on the catalyst surface is the second major step.

C. True: Reactants react on the surface to form an intermediate complex which then yields the products.

D. True: The modern theory successfully combines the ideas of surface adsorption (Old Theory) with chemical intermediate formation (Intermediate Theory).

E. True: By considering the availability of active sites on the surface, this theory explains how promoters increase efficiency (by increasing sites) and poisons decrease it (by blocking sites).


Step 3: Final Answer:

Statements B, C, D, and E are correct. The number is 4.
Quick Tip: The 5 steps of heterogeneous catalysis are: 1. Diffusion of reactants to surface \(\rightarrow\) 2. Adsorption \(\rightarrow\) 3. Chemical reaction/Intermediate formation \(\rightarrow\) 4. Desorption of products \(\rightarrow\) 5. Diffusion of products away.

Step 1: Understanding the Concept:

Hydration of alkaline earth metal salts typically decreases down the group as the size of the cation increases and its charge density decreases.


Step 2: Detailed Explanation:

1. Magnesium Nitrate: Due to the small size and high charge density of the \(Mg^{2+}\) ion, it has a strong tendency to be hydrated. It typically crystallizes as a hexahydrate: \(Mg(NO_{3})_{2} \cdot 6H_{2}O\). Therefore, \(X = 6\).

2. Barium Nitrate: As we go down the group to Barium (\(Ba^{2+}\)), the size of the cation increases significantly. Barium nitrate does not form stable crystalline hydrates and usually exists in its anhydrous form: \(Ba(NO_{3})_{2}\). Therefore, \(Y = 0\).


Calculation:

Sum = \(X + Y = 6 + 0 = 6\).


Step 3: Final Answer:

The sum of X and Y is 6.
Quick Tip: For Group 2 nitrates: \(Be\) and \(Mg\) form hexahydrates (\(6H_{2}O\)), \(Ca\) forms a tetrahydrate (\(4H_{2}O\)), while \(Sr\) and \(Ba\) form anhydrous salts. This is a common trend in hydration energy.

Step 1: Understanding the Concept:

Benedict's solution contains cupric ions (\(Cu^{2+}\)) in a basic medium.

Sugars that have a free aldehyde or ketone group (or can form one through tautomerization) are called reducing sugars.

Reducing sugars reduce \(Cu^{2+}\) to \(Cu^{+}\), which precipitates as cuprous oxide (\(Cu_{2}O\)), an orange-red solid.


Step 2: Detailed Explanation:

Let's analyze each compound:

1. Glucose: A monosaccharide aldose. It is a reducing sugar and gives a positive test.

2. Maltose: A disaccharide with a free hemiacetal group. It is a reducing sugar and gives a positive test.

3. Sucrose: A disaccharide where the glycosidic bond involves the anomeric carbons of both glucose and fructose. It has no free hemiacetal group. It is a non-reducing sugar and will not produce a precipitate.

4. Ribose: A monosaccharide aldose. It is a reducing sugar and gives a positive test.

5. 2-deoxyribose: A monosaccharide aldose. It is a reducing sugar and gives a positive test.

6. Amylose: A polysaccharide (starch component). Although it has one reducing end, the number of reducing units is negligible compared to the overall mass, making it effectively a non-reducing sugar. It will not produce a detectable precipitate.

7. Lactose: A disaccharide with a free hemiacetal group. It is a reducing sugar and gives a positive test.


Step 3: Final Answer:

The compounds that will not produce a precipitate are Sucrose and Amylose.

Total count = 2.
Quick Tip: All monosaccharides (aldoses and ketoses) are reducing sugars. Among disaccharides, sucrose is the most famous non-reducing sugar, while maltose and lactose are reducing. Polysaccharides like starch, glycogen, and cellulose are non-reducing.

Step 1: Understanding the Concept:

The reaction involves the electrophilic addition of \(H^{+}\) to the double bond to form a carbocation, followed by the attack of \(Cl^{-}\). Isomeric products include both structural isomers and stereoisomers.


Step 2: Detailed Explanation:

1. Carbocation Formation:

Protonation of \(CH_{2}=CH-CH(Cl)-CH_{3}\) at \(C_{1}\) yields a secondary carbocation:
\[ CH_{3}-CH^{+}-CH(Cl)-CH_{3} \quad (Secondary Carbocation I) \]
2. Rearrangement:

Carbocation I can undergo a 1,2-hydride shift from \(C_{3}\) to \(C_{2}\) to form a more stable carbocation:
\[ CH_{3}-CH_{2}-C^{+}(Cl)-CH_{3} \quad (Secondary Carbocation II) \]
Carbocation II is highly stabilized by the resonance (+M effect) of the chlorine atom's lone pairs.

3. Formation of Products:

- Attack of \(Cl^{-}\) on Carbocation I yields 2,3-dichlorobutane:
\(CH_{3}-CHCl-CHCl-CH_{3}\). This molecule has two identical chiral centers.

Possible stereoisomers: \((2R, 3R)\), \((2S, 3S)\), and the meso form \((2R, 3S)\). (Total = 3 isomers)

- Attack of \(Cl^{-}\) on Carbocation II yields 2,2-dichlorobutane:
\(CH_{3}-CH_{2}-CCl_{2}-CH_{3}\). This molecule is achiral. (Total = 1 isomer)


Step 3: Final Answer:

The total number of possible isomeric products is \(3 + 1 = 4\).
Quick Tip: When dealing with "isomeric products", always count structural isomers first and then check each for stereoisomerism (enantiomers and meso forms). Carbocation stability order is enhanced significantly by nearby heteroatoms with lone pairs due to resonance (+M).