Step 1: Understanding the Concept:

The molecular formula can be determined by finding the molar mass of the compound from its vapour density at STP and then using the mass percentage of chlorine to find the number of chlorine atoms per molecule.


Step 2: Key Formula or Approach:

1. Molar Mass (\(M\)) = Mass of 22.4 L (22400 mL) of vapour at STP.

2. Number of Chlorine atoms = \(\frac{Total mass of Chlorine in 1 mole}{Atomic mass of Chlorine}\).


Step 3: Detailed Explanation:

Given that 100 mL of vapour weighs 0.57 g at STP.

The mass of 22400 mL (1 mole) of the substance is:
\[ Molar Mass = \frac{0.57 g}{100 mL} \times 22400 mL/mol = 0.57 \times 224 = 127.68 g/mol \]

The mass of chlorine in one mole of the metal chloride is:
\[ Mass of Cl = 55% of 127.68 = \frac{55}{100} \times 127.68 = 70.224 g \]

Now, we find the number of chlorine atoms (\(n\)) in the molecule:
\[ n = \frac{Mass of Cl}{Atomic mass of Cl} = \frac{70.224}{35.5} \approx 1.978 \approx 2 \]

Since there are 2 chlorine atoms, the molecular formula is \(MCl_2\).


Step 4: Final Answer:

The molecular formula of the metal chloride is \(MCl_2\).
Quick Tip: At STP, remember that 1 mole of any gas occupies 22.4 L. Rapidly calculating the molar mass by multiplying the weight of 100 mL by 224 is a useful time-saving shortcut.

Step 1: Understanding the Concept:

According to Molecular Orbital Theory (MOT), species with the same number of total electrons (isoelectronic species) usually have the same bond order and magnetic properties.


Step 2: Key Formula or Approach:

Identify the total number of electrons in the acetylide ion (\(C_2^{2-}\)) and compare it with the given options.


Step 3: Detailed Explanation:

1. Acetylide Ion (\(C_2^{2-}\)):

Total electrons = \(2 \times 6 (from C) + 2 (negative charge) = 14 electrons\).

Configuration: \(\sigma 1s^2, \sigma^* 1s^2, \sigma 2s^2, \sigma^* 2s^2, \pi 2p_x^2 = \pi 2p_y^2, \sigma 2p_z^2\).

Bond Order = \(\frac{10 - 4}{2} = 3\).

Since all electrons are paired, it is diamagnetic.


2. Comparing with options:

(A) \(O_2^{+}\): \(16 - 1 = 15 e^-\) (Paramagnetic, Bond Order = 2.5).

(B) \(N_2^{+}\): \(14 - 1 = 13 e^-\) (Paramagnetic, Bond Order = 2.5).

(C) \(NO^{+}\): \(7 (N) + 8 (O) - 1 = 14 e^-\).

It is isoelectronic with \(C_2^{2-}\). It has a bond order of 3 and is diamagnetic.

(D) \(O_2^{-}\): \(16 + 1 = 17 e^-\) (Paramagnetic, Bond Order = 1.5).


Step 4: Final Answer:

The acetylide ion is isoelectronic with \(NO^{+}\), sharing the same bond order (3) and diamagnetic nature.
Quick Tip: Memorize the properties of 14-electron systems (\(N_2, CO, CN^-, NO^+\)): they all have a bond order of 3 and are diamagnetic.

Step 1: Understanding the Concept:

A lead storage battery consists of a lead anode and a lead dioxide (\(PbO_2\)) cathode during discharge. The charging process is the reverse of the discharging process.


Step 2: Detailed Explanation:

1. Electrode Composition: In a lead storage battery, the anode is lead (\(Pb\)) and the cathode is lead dioxide (\(PbO_2\)). Thus, statement C is false as it identifies \(PbO_2\) as the anode.

2. Electrolyte: The electrolyte is a \(\sim 38%\) solution of sulphuric acid. Thus, statement D is true.

3. Charging Process: During charging, the chemical reactions are reversed by an external power source.

- At the electrode where \(PbO_2\) is present (the positive electrode, which acts as the cathode during discharge):
\(PbSO_4(s) + 2H_2O \rightarrow PbO_2(s) + SO_4^{2-} + 4H^+ + 2e^-\).

In some contexts, especially in older textbook conventions or specific exam patterns, this plate is consistently referred to as the "cathode plate". Under this naming convention, \(PbSO_4\) on the "cathode" is converted to \(PbO_2\). This makes statement B true in that context.

- At the other electrode (negative electrode): \(PbSO_4(s) + 2e^- \rightarrow Pb(s) + SO_4^{2-}\).


Step 3: Final Answer:

Based on standard textbook problems, statements B and D are identified as correct.
Quick Tip: Remember: Discharge \(\rightarrow\) Lead is Anode, \(PbO_2\) is Cathode. Charging \(\rightarrow\) Reactions reverse. The concentration of \(H_2SO_4\) (38%) is a key factual detail often tested.

Step 1: Understanding the Concept:

The extent of physisorption of a gas on a solid adsorbent is directly proportional to its ease of liquefaction.


Step 2: Detailed Explanation:

The ease of liquefaction of a gas depends on its critical temperature (\(T_c\)).

A higher critical temperature implies stronger intermolecular forces between the gas molecules.

Gases with higher \(T_c\) values are more easily liquefied and, consequently, are more strongly adsorbed on the surface of an adsorbent like charcoal.


Given critical temperatures:
\(T_c(A) = 5.3 K\)
\(T_c(B) = 33.2 K\)
\(T_c(C) = 126.0 K\)
\(T_c(D) = 154.3 K\)


Ordering by \(T_c\): \(D > C > B > A\).

Therefore, the extent of adsorption follows the same order: \(D > C > B > A\).


Step 3: Final Answer:

The correct order of adsorption is \(D > C > B > A\).
Quick Tip: Adsorption \(\propto\) Ease of liquefaction \(\propto\) Strength of van der Waals forces \(\propto\) Critical Temperature (\(T_c\)). Use this direct proportionality for all such gas adsorption problems.

Step 1: Understanding the Concept:

Density generally increases down a group in the periodic table because the increase in atomic mass usually outweighs the increase in atomic volume. However, alkali metals exhibit an anomaly.


Step 2: Detailed Explanation:

For Group 1 (Alkali Metals), the density values in \(g/cm^3\) are:

- Li: 0.53

- Na: 0.97

- K: 0.86

- Rb: 1.53

- Cs: 1.90



The anomaly occurs at Potassium (\(K\)). Potassium is less dense than Sodium (\(Na\)). This is because there is a sudden and large increase in atomic volume when moving from \(Na\) to \(K\) due to the presence of empty d-orbitals in the shell of \(K\), which leads to a less efficient packing compared to the mass increase.



The overall order of increasing density is: \(Li < K < Na < Rb < Cs\).

Comparing with the options provided, Option A represents this trend correctly for the four elements listed.


Step 3: Final Answer:

The correct order is \(K < Na < Rb < Cs\).
Quick Tip: "Sodium is heavier than Potassium" is a common trap. Always remember the \(Na/K\) density reversal in Group 1.

Step 1: Understanding the Concept:

An Ellingham diagram is a plot of \(\Delta G^\circ\) vs \(T\) for the formation of oxides. The slope of the line is given by \(-\Delta S^\circ\).


Step 2: Detailed Explanation:

1. Assertion A: In the Ellingham diagram, the plots are normally straight lines. A sharp change in the slope occurs at a particular temperature when there is a phase change (melting or boiling) of the metal or the oxide. For Magnesium, boiling occurs at \(\sim 1120^{\circ}C\). Thus, Assertion A is true.

2. Reason R: When a phase change occurs, such as from liquid to gas, the entropy of the system (\(\Delta S\)) increases significantly. Since the slope of the Ellingham curve is \(-\Delta S\), a sudden change in entropy results in a sudden change in the slope of the graph. Thus, Reason R is true.

3. Relationship: The change in entropy during the phase transition is precisely what causes the change in the slope of the line in the diagram. Therefore, Reason R correctly explains Assertion A.


Step 3: Final Answer:

Both Assertion A and Reason R are true, and Reason R is the correct explanation of Assertion A.
Quick Tip: In Ellingham diagrams, a slope becoming more positive indicates that the reactants are becoming more disordered (e.g., metal boiling), while a slope becoming more negative is rare for oxides.

Step 1: Understanding the Concept:

Hydrides are classified based on their bonding and the number of electrons available for bonding relative to the number of valence orbitals.


Step 2: Detailed Explanation:

1. Electron deficient hydrides: These have too few electrons for conventional Lewis structures. Group 13 hydrides like Diborane (\(B_2H_6\)) fall into this category. (A-III)

2. Electron rich hydrides: These have lone pairs of electrons on the central atom. Examples include Group 15, 16, and 17 hydrides like \(NH_3, H_2O, HF\). (B-II)

3. Electron precise hydrides: These have the exact number of electrons required to form normal covalent bonds. Group 14 hydrides like Methane (\(CH_4\)) are electron precise. (C-IV)

4. Saline (Ionic) hydrides: These are formed by highly electropositive s-block elements (except Li and Be). Magnesium hydride (\(MgH_2\)) is a saline/ionic hydride. (D-I)


Step 3: Final Answer:

The correct matching is A-III, B-II, C-IV, D-I.
Quick Tip: Group 13 = Deficient, Group 14 = Precise, Group 15-17 = Rich. s-block (metals) = Saline. Use this simple mapping for hydride questions.

Step 1: Understanding the Concept:

The reaction between a soluble metal salt and a soluble carbonate typically results in the precipitation of an insoluble metal carbonate.


Step 2: Detailed Explanation:

1. First Reaction:

When aqueous Calcium Chloride (\(CaCl_2\)) reacts with Sodium Carbonate (\(Na_2CO_3\)), a double displacement reaction occurs:
\[ CaCl_2 + Na_2CO_3 \rightarrow CaCO_3 \downarrow + 2NaCl \]

Here, \(X\) is Calcium Carbonate (\(CaCO_3\)) and \(Y\) is Sodium Chloride (\(NaCl\)).



2. Second Reaction:

To convert Calcium Carbonate (\(X\)) back into Calcium Chloride, it must react with an acid containing chloride ions. Dilute Hydrochloric Acid (\(HCl\)) is used for this purpose:
\[ CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2 \uparrow \]

Thus, \(Z\) is \(HCl\).


Step 3: Final Answer:

Comparing with the options: \(X = CaCO_3, Y = NaCl, Z = HCl\).
Quick Tip: Metal carbonates always react with dilute acids to produce the corresponding salt, water, and carbon dioxide gas. This "reversibility" to the original chloride salt is a standard chemical cycle.

Step 1: Understanding the Concept:

Boron is a unique non-metallic element in Group 13 with an exceptionally strong covalent network structure.


Step 2: Detailed Explanation:

1. Statement I: Boron exists in several allotropic forms, all consisting of \(B_{12}\) icosahedral units. These units are linked by very strong covalent bonds in a three-dimensional lattice. This makes Boron extremely hard (nearly as hard as diamond). While "lattice energy" is usually reserved for ionic compounds, in this context, it refers to the high cohesive/binding energy of the solid lattice. Thus, Statement I is considered correct.

2. Statement II: Due to its strong covalent network structure, a vast amount of energy is required to break the bonds to melt or boil it. Boron has the highest melting point (\(\sim 2450 K\)) and boiling point in Group 13. Other members like Gallium have unusually low melting points. Thus, Statement II is correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: Boron's properties are atypical for its group because of its small size and ability to form strong covalent networks. Remember the "icosahedral" structure as the reason for its extreme hardness.

Step 1: Understanding the Concept:

Covalent character in halides can be explained by Fajans' rules, and the stability of oxides depends on oxidation states and bonding nature.


Step 2: Detailed Explanation:

1. Statement I: According to Fajans' rules, a cation with a higher positive charge has higher polarizing power. In \(SbCl_5\), Antimony is in the +5 oxidation state, whereas in \(SbCl_3\), it is in +3. The higher charge density of \(Sb^{5+}\) polarizes the chloride ions more effectively, leading to greater covalent character. Thus, \(SbCl_5\) is more covalent. Statement I is correct.

2. Statement II: For halogens like Chlorine, the higher oxides (where the halogen is in a higher oxidation state, e.g., \(Cl_2O_7\)) are generally more stable than the lower oxides (e.g., \(Cl_2O\)). Higher oxidation states allow for better multiple bond character or resonance stabilization. For example, \(Cl_2O_7\) is much more stable than \(ClO_2\). Thus, Statement II is correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: Fajans' Rule: Higher Oxidation State \(\rightarrow\) Higher Polarizing Power \(\rightarrow\) More Covalent. This rule is a frequent visitor in inorganic chemistry questions.

Step 1: Understanding the Concept:

The electronic configuration and orbital spatial distribution of lanthanides (\(4f\) series) and actinides (\(5f\) series) differ significantly. This affects their chemical reactivity and bonding capabilities.


Step 2: Detailed Explanation:

1. Shielding and Penetration: In lanthanides, the \(4f\) orbitals are deeply "buried" within the atom and are effectively shielded by the \(5s\) and \(5p\) electrons. Consequently, \(4f\) electrons do not participate significantly in chemical bonding.

2. Actinides Characteristics: In contrast, the \(5f\) orbitals in actinides extend further from the nucleus. They are less effectively shielded by the \(6s\) and \(6p\) electrons compared to the \(4f\) electrons.

3. Bonding: Because the \(5f\) orbitals are more exposed (less buried), the electrons in these orbitals can interact more effectively with ligand orbitals, allowing them to participate in covalent bonding to a greater extent than \(4f\) electrons.

4. Conclusion: Both statements are true, and the fact that \(5f\) orbitals are less buried directly explains why they participate more in bonding.


Step 3: Final Answer:

Assertion A and Reason R are both correct, and Reason R provides the logical explanation for Assertion A.
Quick Tip: Remember that actinides show a wider range of oxidation states compared to lanthanides precisely because the \(5f, 6d,\) and \(7s\) electrons are closer in energy and more spatially available for bonding.

Step 1: Understanding the Concept:

Crystal Field Stabilization Energy (CFSE) for octahedral complexes is calculated based on the distribution of electrons in \(t_{2g}\) and \(e_g\) orbitals.


Step 2: Key Formula or Approach:
\[ CFSE = [n(t_{2g}) \times (-0.4) + n(e_g) \times (+0.6)] \Delta_o \]

Where \(n\) is the number of electrons in the respective orbitals.


Step 3: Detailed Explanation:

1. A. \([Cu(NH_3)_6]^{2+}\): \(Cu^{2+}\) is \(d^9\). Configuration: \(t_{2g}^6 e_g^3\).
\[ CFSE = [6(-0.4) + 3(0.6)]\Delta_o = [-2.4 + 1.8]\Delta_o = -0.6 \Delta_o (Matches I) \]

2. B. \([Ti(H_2O)_6]^{3+}\): \(Ti^{3+}\) is \(d^1\). Configuration: \(t_{2g}^1 e_g^0\).
\[ CFSE = [1(-0.4) + 0(0.6)]\Delta_o = -0.4 \Delta_o (Matches IV) \]

3. C. \([Fe(CN)_6]^{3-}\): \(Fe^{3+}\) is \(d^5\). \(CN^-\) is a strong field ligand, so it is low spin: \(t_{2g}^5 e_g^0\).
\[ CFSE = [5(-0.4) + 0(0.6)]\Delta_o = -2.0 \Delta_o (Matches II) \]

4. D. \([NiF_6]^{4-}\): \(Ni^{2+}\) is \(d^8\). \(F^-\) is a weak field ligand: \(t_{2g}^6 e_g^2\).
\[ CFSE = [6(-0.4) + 2(0.6)]\Delta_o = [-2.4 + 1.2]\Delta_o = -1.2 \Delta_o (Matches III) \]


Step 4: Final Answer:

The correct matching is A-I, B-IV, C-II, D-III.
Quick Tip: For \(d^8, d^9,\) and \(d^{10}\) systems, the configuration is independent of the ligand strength (strong or weak field) in octahedral fields. This simplifies calculations for \(Ni^{2+}\) and \(Cu^{2+}\).

Step 1: Understanding the Concept:

This question links common environmental pollutants with their primary ecological impacts.


Step 2: Detailed Explanation:

1. Nitrogen oxides (\(NO_x\)): These react with water and oxygen in the atmosphere to form nitric acid, which is a major component of Acid rain (A-IV).

2. Methane (\(CH_4\)): Methane is a potent greenhouse gas that traps heat in the atmosphere, leading to Global warming (B-III).

3. Carbon dioxide (\(CO_2\)): Normal rainwater has a pH of approximately 5.6 because \(CO_2\) dissolves in it to form weak carbonic acid (\(H_2CO_3\)) (C-II).

4. Phosphate fertilisers: When washed into water bodies, they provide excess nutrients, leading to excessive algal growth and oxygen depletion, known as Eutrophication (D-I).


Step 3: Final Answer:

The correct matches are: A-IV, B-III, C-II, D-I.
Quick Tip: Remember: Nitrogen/Sulphur oxides \(\rightarrow\) Acid Rain; \(CO_2/CH_4 \rightarrow\) Global Warming; Phosphates/Nitrates in water \(\rightarrow\) Eutrophication.

Step 1: Understanding the Concept:

Compound 'A' is squaric acid (3,4-dihydroxycyclobut-3-ene-1,2-dione). It is a strong dibasic acid that reacts with bases to form the squarate dianion.


Step 2: Detailed Explanation:

1. Reaction: Squaric acid (\(C_4H_2O_4\)) reacts with 2 equivalents of base (\(OH^-\)) to lose two protons, forming the squarate dianion (\(C_4O_4^{2-}\)).

2. Aromaticity (A): The squarate dianion is a planar, cyclic system with delocalized \(\pi\) electrons. The ring itself has 2 \(\pi\) electrons (from the double bond). According to Huckel's rule (\(4n+2\)), for \(n=0\), 2 \(\pi\) electrons make the system aromatic. Statement A is true.

3. Reaction Rate (B): Acid-base neutralizations are extremely fast ionic reactions. Statement B is false.

4. Tautomerism (C): Squaric acid exists in equilibrium with keto and enol forms (tautomerism). Statement C is true.

5. Bond Lengths (D): Due to complete resonance delocalization in the symmetric dianion 'B', all C-C bonds and all C-O bonds achieve equal length. Statement D is true.


Step 3: Final Answer:

Statements A, C, and D are correct.
Quick Tip: The squarate ion is a classic example of an "Oxocarbon" aromatic species. Its high stability is due to the perfect delocalization of negative charges over all four oxygen atoms.

Step 1: Understanding the Concept:

Ozonolysis followed by treatment with water (\(H_2O\)) alone (without a reducing agent like Zinc or Dimethyl sulfide) leads to oxidative workup.


Step 2: Detailed Explanation:

1. Substrate: 2-hexene has the structure \(CH_3-CH=CH-CH_2-CH_2-CH_3\).

2. Step 1 (Ozonolysis): The ozone molecule adds across the double bond to form an ozonide intermediate.

3. Step 2 (Hydrolysis): In the absence of a reducing agent like \(Zn\), the hydrolysis of the ozonide produces aldehydes and hydrogen peroxide (\(H_2O_2\)).

4. Oxidation: The \(H_2O_2\) produced is an oxidizing agent that further oxidizes the aldehydes to carboxylic acids.

- Acetaldehyde (\(CH_3CHO\)) \(\rightarrow\) Acetic acid (\(CH_3COOH\))

- Butanal (\(CH_3CH_2CH_2CHO\)) \(\rightarrow\) Butanoic acid (\(CH_3CH_2CH_2COOH\))


Step 3: Final Answer:

The final products are Butanoic acid and acetic acid.
Quick Tip: Always check the second step of ozonolysis. \((i) O_3 / (ii) Zn, H_2O\) gives Aldehydes/Ketones (Reductive). \((i) O_3 / (ii) H_2O\) or \(H_2O_2\) gives Carboxylic Acids (Oxidative).

Step 1: Understanding the Concept:

This is an intramolecular Grignard reaction where a halo-ketone reacts with Magnesium to form an organomagnesium intermediate which then attacks its own carbonyl group.


Step 2: Detailed Explanation:

1. Formation of Grignard: Magnesium reacts with the \(C-Br\) bond of 5-bromo-3-methylpentan-2-one to form a Grignard reagent at the terminal carbon: \(R-MgBr\).

2. Intramolecular Nucleophilic Attack: The carbanion (\(C^-\)) of the Grignard reagent attacks the electrophilic carbonyl carbon (\(C=O\)).

3. Ring Closure: Counting the carbons from the carbonyl carbon (\(C2\)) to the Grignard carbon (\(C5\)): \(C2-C3-C4-C5\). This results in a 4-membered ring.

4. Substitution: The carbonyl oxygen becomes an alcohol (\(OH\)) after the aqueous workup (\(H_2O\)). The methyl group at \(C2\) and the methyl group at \(C3\) remain.

5. Structure: The resulting product is 1,2-dimethylcyclobutanol.


Step 3: Final Answer:

The major product is 1,2-dimethylcyclobutanol.
Quick Tip: For intramolecular reactions, always count the number of atoms between the nucleophile and electrophile to determine the ring size correctly.

Step 1: Understanding the Concept:

Intramolecular Aldol condensation of diones follows the formation of the most stable ring (usually 5 or 6 membered).


Step 2: Detailed Explanation:

1. Octane-2,7-dione structure: \(CH_3-C(=O)-CH_2-CH_2-CH_2-CH_2-C(=O)-CH_3\).

2. Enolization: Base can abstract a proton from a methyl group (\(C1\) or \(C8\)) or a methylene group (\(C3\) or \(C6\)).

3. Ring Formation:

- If enolate from \(C1\) attacks \(C7\): A 7-membered ring forms (less stable).

- If enolate from \(C3\) attacks \(C7\): A 5-membered ring forms (more stable).

4. Mechanism: The enolate at \(C3\) attacks the carbonyl at \(C7\), followed by dehydration (\(-\Delta\)) to form an \(\alpha,\beta\)-unsaturated ketone.

5. Product: The resulting structure contains a 5-membered ring with a methyl group at \(C2\) and an acetyl group (\(-COCH_3\)) at \(C1\). This is 1-acetyl-2-methylcyclopentene.


Step 3: Final Answer:

The major product is 1-acetyl-2-methylcyclopentene.
Quick Tip: In intramolecular Aldol reactions, 5 and 6 membered rings are kinetically and thermodynamically favoured over 7 or 4 membered rings.

Step 1: Understanding the Concept:

This sequence involves transformation of an acid to an amide followed by reduction.


Step 2: Detailed Explanation:

1. Step (i): \(Ph-CH=CH-COOH\) reacts with \(SOCl_2\) to form the acid chloride: \(Ph-CH=CH-COCl\).

2. Step (ii): The acid chloride reacts with the primary amine \(R-NH_2\) to form a secondary amide: \(Ph-CH=CH-CONH-R\).

3. Step (iii): \(LiAlH_4\) is a strong reducing agent. It reduces the amide group (\(-CONH-\)) to an amine (\(-CH_2NH-\)).

4. Important Note: While \(LiAlH_4\) usually does not reduce isolated \(C=C\) bonds, it does reduce \(\alpha,\beta\)-unsaturated systems that are in conjugation with both a phenyl ring and a carbonyl group (cinnamic system).

5. Result: Both the \(C=C\) bond and the \(C=O\) group are reduced. The final product after workup is \(Ph-CH_2-CH_2-CH_2-NH-R\).


Step 3: Final Answer:

The product is 3-phenyl-N-alkylpropan-1-amine.
Quick Tip: Remember that \(LiAlH_4\) reduces the double bond in cinnamic acid derivatives (\(Ph-CH=CH-CO-\)) because the resonance makes the \(\beta\)-carbon electrophilic enough for hydride attack.

Step 1: Understanding the Concept:

Tertiary aromatic amines react with nitrous acid via electrophilic aromatic substitution rather than forming diazonium salts.


Step 2: Detailed Explanation:

1. Electrophile: \(NaNO_2 + HX\) generates the nitrosonium ion (\(NO^+\)) in situ. Statement C is correct.

2. Reaction: Due to the strong activating nature of the \(-N(Me)_2\) group, substitution occurs at the para position (or ortho if para is blocked).

3. Product: The reaction yields \(p\)-nitroso-\(N,N\)-dimethylaniline. This is a C-nitroso compound, not an N-nitroso compound. Statement A is correct.

4. Conditions: These nitrosation reactions are typically conducted at low temperatures (\(0-5^\circ C\)) to maintain stability. Statement B is correct.

5. Incorrect Statement: Statement D is false because the nitroso group attaches to the ring carbon, not the nitrogen atom. Tertiary amines cannot form stable N-nitroso compounds like secondary amines do.


Step 3: Final Answer:

The incorrect statement is D.
Quick Tip: Primary aromatic amines \(\rightarrow\) Diazonium salts; Secondary \(\rightarrow\) N-nitroso (Yellow oil); Tertiary \(\rightarrow\) p-Nitroso (Green/Blue compound). This is a standard test to distinguish amines.

Step 1: Understanding the Concept:

Polymers are classified based on their monomeric units, mode of polymerization, and physical properties.


Step 2: Detailed Explanation:

1. A. 2-chloro-1,3-butadiene (Chloroprene): Polymerizes to form Neoprene, which is a Synthetic Rubber (A-II).

2. B. Nylon 2-nylon 6: This is an alternating polyamide copolymer made from glycine and amino caproic acid. It is well known as a Biodegradable polymer (B-I).

3. C. Polyacrylonitrile (PAN): Formed by the peroxide-initiated addition polymerization of acrylonitrile. It is a classic example of an Addition Polymer (C-IV).

4. D. Dacron (Terylene): Produced by the condensation of ethylene glycol and terephthalic acid. It contains ester linkages and is thus a Polyester (D-III).


Step 3: Final Answer:

The correct matching is A-II, B-I, C-IV, D-III.
Quick Tip: Dacron is the trade name for Terylene. Always link "Polyester" with Terylene/Dacron and "Polyamide" with Nylons in polymer questions.

Step 1: Understanding the Concept:

The speeds of gas molecules are defined by Maxwell-Boltzmann distribution. The root mean square speed (\(v_{rms}\)) and the most probable speed (\(v_{mp}\)) are given by specific formulas involving temperature (\(T\)), gas constant (\(R\)), and molar mass (\(M\)).


Step 2: Key Formula or Approach:

1. Root mean square speed: \( v_{rms} = \sqrt{\frac{3RT}{M}} \)

2. Most probable speed: \( v_{mp} = \sqrt{\frac{2RT}{M}} \)


Step 3: Detailed Explanation:

Given:

For gas X: \( T_X = 600 K \), \( M_X = 40 g/mol \)

For gas Y: \( T_Y = 90 K \), \( M_Y = ? \)



According to the problem: \[ v_{rms}(gas X at 600K) = v_{mp}(gas Y at 90K) \] \[ \sqrt{\frac{3 \times R \times 600}{40}} = \sqrt{\frac{2 \times R \times 90}{M_Y}} \]


Squaring both sides and cancelling \( R \): \[ \frac{3 \times 600}{40} = \frac{2 \times 90}{M_Y} \] \[ \frac{1800}{40} = \frac{180}{M_Y} \] \[ 45 = \frac{180}{M_Y} \] \[ M_Y = \frac{180}{45} = 4 g/mol \]

Step 4: Final Answer:

The molar mass of gas Y is 4.
Quick Tip: Speed ratios to remember: \( v_{mp} : v_{avg} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1 : 1.128 : 1.224 \). Identifying the correct multiplier (2 vs 3) inside the square root is crucial for these equality problems.

Step 1: Understanding the Concept:

Photoelectric effect occurs only if the energy of the incident photon (\( E \)) is greater than or equal to the work function (\( W_0 \)) of the metal (\( E \geq W_0 \)).


Step 2: Key Formula or Approach:
Energy of photon: \( E = \frac{hc}{\lambda} \)

Energy in electron-volts: \( E(eV) = \frac{hc}{\lambda \times e} \)


Step 3: Detailed Explanation:

Incident wavelength \( \lambda = 400 nm = 400 \times 10^{-9} m \).

Calculate energy of incident photon: \[ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} Joules \] \[ E = \frac{19.8 \times 10^{-26}}{400 \times 10^{-9}} = 0.0495 \times 10^{-17} = 4.95 \times 10^{-19} J \]


Convert energy to eV: \[ E(eV) = \frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 3.09 eV \]


Now, compare \( E = 3.09 eV \) with the given work functions:
1. Li: \( 2.42 eV < 3.09 eV \) (Show effect)

2. Na: \( 2.3 eV < 3.09 eV \) (Show effect)

3. K: \( 2.25 eV < 3.09 eV \) (Show effect)

4. Mg: \( 3.7 eV > 3.09 eV \) (No effect)

5. Cu: \( 4.8 eV > 3.09 eV \) (No effect)

6. Ag: \( 4.3 eV > 3.09 eV \) (No effect)



Total number of metals = 3 (Li, Na, K).


Step 4: Final Answer:

The number of metals that show the photoelectric effect is 3.
Quick Tip: Use the shortcut \( E(eV) \approx \frac{1240}{\lambda(nm)} \) for faster calculation in exams. Here, \( 1240 / 400 = 3.1 eV \), which gives the same result instantly.

Step 1: Understanding the Concept:

The change in entropy of the surroundings (\( \Delta S_{surr} \)) is calculated using the heat exchanged with the surroundings: \( \Delta S_{surr} = \frac{q_{surr}}{T} \). For an isothermal process of an ideal gas, \( \Delta U = 0 \), so \( q = -w \).


Step 2: Key Formula or Approach:

1. Work (Irreversible expansion against constant \( P_{ext} \)): \( w = -P_{ext}(V_2 - V_1) \)

2. \( q_{sys} = -w_{sys} \)

3. \( \Delta S_{surr} = -\frac{q_{sys}}{T_{surr}} \)


Step 3: Detailed Explanation:

Note: The question states "isothermal reversible expansion" but then specifies "against a constant pressure of 4 atm". In competitive exams, if a constant external pressure is given for an expansion, the work done is typically calculated using the irreversible formula, even if the phrasing is slightly contradictory.



1. Work done by the system: \[ w = -P_{ext} \Delta V = -4 atm \times (3.0 L - 2.0 L) = -4 L-atm \]
2. Convert work to Joules (using \( 1 L-atm \approx 101.325 J \)): \[ w = -4 \times 101.325 = -405.3 J \]
3. Since it is isothermal, \( q_{sys} = -w = 405.3 J \).
4. The heat lost by surroundings is \( q_{surr} = -q_{sys} = -405.3 J \).
5. Entropy change of surroundings: \[ \Delta S_{surr} = \frac{-405.3 J}{350 K} \approx -1.158 J K^{-1} \]
Rounding to the nearest integer, we get \(-1\).


Step 4: Final Answer:

The change in entropy of the surroundings is -1 J K\(^{-1}\).
Quick Tip: For surroundings, \( \Delta S \) is always calculated as \( q_{rev, surr}/T \). Since surroundings are massive, any process is "reversible" for them, hence \( \Delta S_{surr} = -q_{sys}/T \).

Step 1: Understanding the Concept:

The lowering of vapour pressure is a colligative property that depends on the total number of solute particles, which is determined by the Van't Hoff factor (\( i \)).


Step 2: Key Formula or Approach:

1. Van't Hoff factor: \( i = 1 + \alpha(n - 1) \)

2. Relative lowering of vapour pressure: \( \frac{P^0 - P_s}{P^0} = \chi_{solute} \approx \frac{i \times n_2}{n_1} \) (for dilute solutions).


Step 3: Detailed Explanation:

1. For \( MgCl_2 \rightarrow Mg^{2+} + 2Cl^- \), \( n = 3 \).
2. Degree of dissociation \( \alpha = 80% = 0.8 \).
3. \( i = 1 + 0.8(3 - 1) = 1 + 1.6 = 2.6 \).
4. Molality = 1.0 molal means 1 mole of solute in 1000g of water.
5. Moles of water (\( n_1 \)) = \( 1000 / 18 \approx 55.55 moles \).
6. Moles of solute (\( n_2 \)) = 1.0 mole.
7. Using the formula: \[ \frac{50 - P_s}{50} = \frac{i \times n_2}{n_1 + i \times n_2} \]
Since \( n_1 \gg i \times n_2 \), we use: \[ \frac{50 - P_s}{50} \approx \frac{2.6 \times 1}{55.55} = 0.0468 \] \[ 50 - P_s = 50 \times 0.0468 = 2.34 \] \[ P_s = 50 - 2.34 = 47.66 mm Hg \]
Rounding to the nearest integer, we get 48.


Step 4: Final Answer:

The vapour pressure of the solution is 48 mm Hg.
Quick Tip: For molal solutions (\( m \)), the mole fraction of solute can be approximated as \( \frac{m}{m + 55.5} \). Always multiply molality by \( i \) to account for dissociation/association.

Step 1: Understanding the Concept:

The pH of a strong acid solution is related to its molarity. Dilution decreases the concentration of \( [H^+] \) according to the law of conservation of moles (\( M_1V_1 = M_2V_2 \)).


Step 2: Detailed Explanation:

1. Initial solution: \( pH_1 = 1 \). \[ [H^+]_1 = 10^{-pH_1} = 10^{-1} = 0.1 M \]
Initial volume \( V_1 = 1 L = 1000 mL \).


2. Final solution: \( pH_2 = 2 \). \[ [H^+]_2 = 10^{-pH_2} = 10^{-2} = 0.01 M \]


3. Dilution equation: \[ M_1V_1 = M_2V_2 \] \[ 0.1 \times 1 = 0.01 \times V_2 \] \[ V_2 = \frac{0.1}{0.01} = 10 L \]


4. Volume of water added: \[ V_{added} = V_2 - V_1 = 10 L - 1 L = 9 L \] \[ V_{added} = 9 \times 1000 mL = 9000 mL \]

Step 3: Final Answer:

The volume of water needed is 9000 mL.
Quick Tip: Increasing the pH of a strong acid by 1 unit requires a 10-fold dilution. Increasing it by 2 units requires a 100-fold dilution. Here, a 10-fold dilution means the final volume is 10 times the initial, meaning 9 parts of water were added.

Step 1: Understanding the Concept:

The rate of a multi-step reaction is determined by the slowest step, called the Rate Determining Step (RDS). If the RDS involves an intermediate, we substitute its concentration from the preceding fast equilibrium step.


Step 2: Detailed Explanation:
1. From the mechanism, the second step is the RDS: \[ Rate = k[NOBr_2][NO] \]
2. However, \( NOBr_2 \) is an intermediate. From the first fast equilibrium step: \[ K_{eq} = \frac{[NOBr_2]}{[NO][Br_2]} \implies [NOBr_2] = K_{eq}[NO][Br_2] \]
3. Substitute the expression for \( [NOBr_2] \) into the rate law: \[ Rate = k(K_{eq}[NO][Br_2])[NO] \] \[ Rate = (k \cdot K_{eq})[NO]^2[Br_2] \]
4. The rate law is \( Rate = k'[NO]^2[Br_2]^1 \).
5. Overall order = sum of exponents = \( 2 + 1 = 3 \).

Step 3: Final Answer:

The overall order of the reaction is 3.
Quick Tip: When a mechanism has a fast equilibrium followed by a slow step, the overall order is simply the stoichiometric sum of reactants in all steps up to and including the slow step (minus any products formed in those steps).

Step 1: Understanding the Concept:

In Thin Layer Chromatography (TLC), the stationary phase (silica gel) is polar. More polar compounds interact more strongly with the stationary phase and travel shorter distances compared to less polar compounds.


Step 2: Key Formula or Approach:
\[ R_f = \frac{Distance travelled by the substance from baseline}{Distance travelled by the solvent from baseline} \]

Step 3: Detailed Explanation:

From the diagram:
- Solvent front distance = 8 cm.
- Compound A distance = 6 cm.
- Compound B distance = 4 cm.
- Compound C distance = 2 cm.


Since the stationary phase is polar and the solvent (hexane) is non-polar, the compound that travels the least distance is the most polar.
Therefore, Compound C is the most polar compound.


Calculate \( R_f \) for C: \[ R_f(C) = \frac{2 cm}{8 cm} = 0.25 \]
To match the required format: \[ 0.25 = 25 \times 10^{-2} \]

Step 4: Final Answer:

The value is 25.
Quick Tip: Remember: Polar stays low, non-polar goes high on standard silica TLC plates. High \( R_f \) = Low polarity.

Step 1: Understanding the Concept:

This complex synthesis pathway involves Strecker synthesis, hydrolysis, nitration, acetylation, esterification, reduction, and Sandmeyer-type replacement to create a thyroxine-like molecule.


Step 2: Detailed Explanation:

1. Starting material: \( p-ethoxybenzaldehyde derivative \).
2. Step (i) \& (ii): Strecker synthesis followed by hydrolysis converts the aldehyde group (\( -CHO \)) into an \( \alpha-amino acid \) group (\( -CH(NH_2)COOH \)).
3. Step (iii): Nitration with 2 equivalents of \( HNO_3 \) adds two nitro groups (\( -NO_2 \)) to the benzene ring ortho to the ethoxy group.
4. Step (iv) \& (v): Acetylation of the amine and esterification of the carboxylic acid with ethanol results in \( -CH(NHCOCH_3)COOC_2H_5 \).
5. Step (vi): Hydrogenation reduces \( -NO_2 \) groups to \( -NH_2 \).
6. Step (vii) \& (viii): Diazotization and reaction with \( NaI \) replaces the \( -NH_2 \) groups with Iodine (\( I \)).


Counting Carbons in 'D':
- Ethoxy group (\( C_2H_5O- \)): 2 carbons.
- Benzene ring: 6 carbons.
- Side chain propanoic core: 3 carbons.
- Acetyl group (\( -COCH_3 \)) on Nitrogen: 2 carbons.
- Ethyl group (\( -C_2H_5 \)) from esterification: 2 carbons.
Total carbons \( x = 2 + 6 + 3 + 2 + 2 = 15 \).

Formula check: \( C_{15}H_{19}NO_4I_2 \).

Step 3: Final Answer:

The value of \( x \) is 15.
Quick Tip: This problem describes the synthesis of an artificial amino acid. Tracking the carbon counts for each functional group added (acetyl, ethyl ester) is the safest way to find \( x \).

Step 1: Understanding the Concept:

A peptide consists of amino acid residues linked by peptide bonds (\( -CO-NH- \)). Hybridization of carbon atoms is \( sp^2 \) if they are part of a double bond (like in carbonyl \( C=O \)) or a benzene ring.


Step 2: Detailed Explanation:

The peptide sequence is Ala - Gly - Phe - Ile.
1. Ala (Alanine): Provides 1 carbonyl carbon (\( sp^2 \)).
2. Gly (Glycine): Provides 1 carbonyl carbon (\( sp^2 \)).
3. Phe (Phenylalanine):
- 1 carbonyl carbon (\( sp^2 \)).
- 6 carbons in the phenyl ring (all are \( sp^2 \)).
- Total for Phe = 7.
4. Ile (Isoleucine): It is the C-terminal amino acid, so its carboxyl group (\( -COOH \)) remains as a carbonyl. It provides 1 carbonyl carbon (\( sp^2 \)).


Total \( sp^2 \) carbons = 1 (from Ala) + 1 (from Gly) + 7 (from Phe) + 1 (from Ile) = 10.


Step 3: Final Answer:

The number of \( sp^2 \) hybridized carbons is 10.
Quick Tip: Every amino acid in a chain contributes exactly 1 \( sp^2 \) carbon (the carbonyl). Then simply add the \( sp^2 \) carbons from side chains (like Benzene in Phe, Tyrosine, or Tryptophan).

Step 1: Understanding the Concept:

Tollen's test involves the reduction of \( [Ag(NH_3)_2]^+ \) by an aldehyde. For every mole of aldehyde oxidized, ammonia is released from the complex.


Step 2: Key Formula or Approach:

The balanced equation for Tollen's test: \[ RCHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow RCOO^- + 2Ag + 4NH_3 + 2H_2O \]
1 mole of aldehyde produces 4 moles of \( NH_3 \).


Step 3: Detailed Explanation:

1. Molar mass of Cyclohexanecarbaldehyde (\( C_7H_{12}O \)):
\( M = (7 \times 12) + (12 \times 1) + 16 = 84 + 12 + 16 = 112 g/mol \).
2. Moles of aldehyde:
\( Mass = 131.8 kg = 131800 g \).
\( Moles = \frac{131800}{112} \approx 1176.78 mol \).
3. Moles of \( NH_3 \):
From stoichiometry, \( 1 mole aldehyde \rightarrow 4 moles NH_3 \).
\( Moles of NH_3 = 4 \times 1176.78 \approx 4707.12 mol \).
4. Mass of \( NH_3 \):
Molar mass of \( NH_3 = 17 g/mol \).
\( Mass = 4707.12 \times 17 \approx 80021 g \approx 80 kg \).


Step 4: Final Answer:

The mass of \( NH_3 \) produced is 80 kg.
Quick Tip: In Tollen's test, the ratio of Aldehyde to Ammonia is 1:4 because each silver ion is coordinated to 2 ammonia ligands, and 2 silver ions are reduced per aldehyde group.