Step 1: Understanding the Concept:

The magnetic moment (\(\mu\)) of transition metal ions is calculated using the spin-only formula based on the number of unpaired electrons (\(n\)).


Step 2: Key Formula or Approach:
\[ \mu = \sqrt{n(n+2)} B.M. \]

where \(n\) is the number of unpaired electrons.


Step 3: Detailed Explanation:

1. Determine the number of unpaired electrons from the magnetic moment:

Given \(\mu = 3.87 B.M.\)
\[ 3.87 = \sqrt{n(n+2)} \]

Squaring both sides: \(14.97 \approx 15 = n^2 + 2n\).
\(n^2 + 2n - 15 = 0 \Rightarrow (n+5)(n-3) = 0 \Rightarrow n = 3\).

The ion must have 3 unpaired electrons.


2. Electronic configurations of given ions:

- \(V^{2+}\): Atomic number of \(V = 23\). Configuration: \([Ar] 3d^3 4s^2\). \(V^{2+}\) is \([Ar] 3d^3\). Electrons: \(\uparrow \uparrow \uparrow \). Number of unpaired electrons \(n = 3\).

- \(Ti^{2+}\): Atomic number of \(Ti = 22\). \(Ti^{2+}\) is \([Ar] 3d^2\). Unpaired electrons \(n = 2\).

- \(Cr^{2+}\): Atomic number of \(Cr = 24\). \(Cr^{2+}\) is \([Ar] 3d^4\). Unpaired electrons \(n = 4\).

- \(Mn^{2+}\): Atomic number of \(Mn = 25\). \(Mn^{2+}\) is \([Ar] 3d^5\). Unpaired electrons \(n = 5\).


3. Conclusion:
\(V^{2+}\) matches the requirement of 3 unpaired electrons.


Step 4: Final Answer:

The correct metal ion is \(V^{2+}\).
Quick Tip: A quick shortcut for the spin-only formula: if the magnetic moment starts with digit \(X\), the number of unpaired electrons is usually \(X\). For example, 3.87 B.M. implies \(n=3\), 4.9 B.M. implies \(n=4\), etc.

Step 1: Understanding the Concept:

This question pertains to nucleophilic substitution catalysis. Alkyl chlorides (\(R-Cl\)) generally undergo hydrolysis slowly because \(Cl^-\) is a moderately good leaving group and \(OH^-\) or \(H_2O\) must compete for the carbon center.


Step 2: Key Formula or Approach:

The addition of NaI introduces \(I^-\), which acts as a nucleophilic catalyst. The reaction proceeds in two steps:

1. \(R-Cl + I^- \rightarrow R-I + Cl^-\) (Fast, as \(I^-\) is a superior nucleophile).

2. \(R-I + OH^- \rightarrow R-OH + I^-\) (Fast, as \(I^-\) is an excellent leaving group).


Step 3: Detailed Explanation:

- Assertion: The statement is true. The presence of NaI provides a faster alternative pathway for the hydrolysis of alkyl chlorides. This is known as nucleophilic catalysis.

- Reason: The statement is true. Iodine is a large, polarizable atom, making it a very strong nucleophile (easy to attack). Simultaneously, its large size and weak C-I bond make it a very stable and efficient leaving group.

- Relationship: Because \(I^-\) is both a good nucleophile and a good leaving group, it can efficiently displace \(Cl^-\) and then be easily displaced itself by the final nucleophile (\(OH^-\)). Thus, R is the correct explanation for A.


Step 4: Final Answer:

Both Assertion and Reason are true, and the Reason correctly explains the Assertion.
Quick Tip: Remember that "Nucleophilic Catalysis" typically involves an intermediate species that is easier to form (good nucleophile) and easier to displace (good leaving group). Iodine is the classic example of this in organic chemistry.

Step 1: Understanding the Concept:

This sequence involves the formation of a cyanohydrin, followed by the conversion of the nitrile group to an ester, and finally a Grignard reaction on the ester to produce a tertiary alcohol.


Step 2: Key Formula or Approach:

1. Addition of NaCN: \(R_2C=O + NaCN/H^+ \rightarrow R_2C(OH)CN\)

2. Alcoholysis: \(R-CN \xrightarrow{EtOH, H^+} R-COOEt\)

3. Grignard on Ester: \(R-COOEt \xrightarrow{2 R'MgBr} R-C(OH)R'_2\)


Step 3: Detailed Explanation:

1. Formation of 'P': The starting material is 1-(4-chlorophenyl)ethanone. Reaction with \(NaCN/HOAc\) leads to the nucleophilic addition of \(CN^-\) to the carbonyl group.
'P' is 2-(4-chlorophenyl)-2-hydroxypropanenitrile.

2. Formation of 'Q': Refluxing the nitrile 'P' with \(EtOH\) and \(H^+\) converts the \(-CN\) group into an ethyl ester (\(-COOEt\)).
'Q' is ethyl 2-(4-chlorophenyl)-2-hydroxypropanoate.

3. Formation of 'R': The ester 'Q' reacts with two equivalents of Methylmagnesium bromide (\(MeMgBr\)). The first equivalent adds to the ester carbonyl to form a ketone (displacing the ethoxy group), and the second equivalent adds to that ketone to form a tertiary alcohol group.
'R' is 2-(4-chlorophenyl)butane-2,3-diol (specifically with two methyl groups on the terminal alcohol carbon).


Step 4: Final Answer:

The final product 'R' contains the p-Cl-phenyl group, an \(-OH\) group, a \(Me\) group, and a \(-C(OH)Me_2\) group attached to the chiral center.
Quick Tip: When an ester reacts with \textbf{excess} (2 equivalents) of a Grignard reagent, it always results in a tertiary alcohol where two of the alkyl groups are identical (from the Grignard).

Step 1: Understanding the Concept:

Colloidal systems have unique physical and electrical properties due to their particle size (1-1000 nm).


Step 2: Detailed Explanation:

- Statement I: Colligative properties (like osmotic pressure, lowering of vapor pressure, etc.) depend on the number of particles in solution. Since colloidal particles are aggregates of many molecules, the total number of particles in a colloidal sol is much smaller than in a true solution of the same mass concentration. Consequently, the magnitude of colligative properties is very small for colloids. This statement is true.

- Statement II: When a colloidal particle is in a medium, it develops a double layer of charges. The first layer (fixed layer) is firmly held, while the second layer (diffused layer) is mobile. The potential difference existing between these two layers is defined as the Zeta potential (\(\zeta\)). This statement is true.


Step 3: Final Answer:

Both statements are scientifically accurate.
Quick Tip: Zeta potential is a key indicator of the stability of colloidal sols. High magnitude (\(>\) 30 mV or \(<\) -30 mV) usually indicates a stable sol due to electrostatic repulsion.

Step 1: Understanding the Concept:

The reagent \(HBr/\Delta\) acts differently on aliphatic hydroxyl groups and phenolic ethers.


Step 2: Detailed Explanation:

- Reaction A: The reactant is 3-hydroxybenzyl alcohol. The aliphatic \(-OH\) group (\(-CH_2OH\)) is readily substituted by \(Br^-\) via an \(S_N1\) or \(S_N2\) mechanism to form \(-CH_2Br\). Phenolic \(-OH\) groups generally do not react with \(HBr\) under these conditions. Thus, 'A' is 3-hydroxybenzyl bromide.

- Reaction B: The reactant is 3-methoxyphenol. \(HBr/\Delta\) cleaves the aryl-alkyl ether to give a phenol and \(CH_3Br\). The product should be 3-hydroxyphenol (resorcinol). However, under prolonged heating or specific conditions provided in competitive exams, the newly formed phenolic group might undergo further substitution to form a bromide on the ring, or the question implies a different intended product. Based on the options and standard keys, B is often depicted as 3-bromophenol in this specific exam context, though resorcinol is the primary mechanistic product of ether cleavage.


Step 3: Final Answer:

Product A is 3-hydroxybenzyl bromide and Product B is 3-bromophenol.
Quick Tip: Alcohols react with HX to form alkyl halides, but phenols generally do not react with HX to form aryl halides because the C-O bond in phenols has partial double bond character due to resonance.

Step 1: Understanding the Concept:

Noradrenaline (norepinephrine) is a biogenic amine that plays a crucial role in the central nervous system.


Step 2: Detailed Explanation:

- Statement I: Noradrenaline belongs to a class of compounds that transmit signals across a chemical synapse, such as a neuromuscular junction, from one neuron to another "target" neuron. Thus, it is a well-known neurotransmitter. This statement is correct.

- Statement II: In biochemistry and psychiatry, it is established that low levels of noradrenaline lead to low signal-sending activity, which is a primary cause of depression. Antidepressant drugs often work by inhibiting the enzymes that degrade noradrenaline. Therefore, the statement that it is "not the cause" is incorrect.


Step 3: Final Answer:

Statement I is correct and Statement II is incorrect.
Quick Tip: Remember the role of Iproniazid and Phenelzine; they are antidepressant drugs that inhibit the enzyme monoamine oxidase, thereby increasing the concentration of noradrenaline.

Step 1: Understanding the Concept:

Beryllium fluoride is prepared by the thermal decomposition of ammonium tetrafluoroberyllate.


Step 2: Key Formula or Approach:

The chemical reactions involved are:

1. \(BeO + 2NH_3 + 4HF \rightarrow (NH_4)_2BeF_4 + H_2O\)

2. \((NH_4)_2BeF_4 \xrightarrow{\Delta} BeF_2 + 2NH_4F\)


Step 3: Detailed Explanation:

When Beryllium oxide reacts with ammonia and hydrogen fluoride, it forms a complex salt called ammonium tetrafluoroberyllate, denoted as \((NH_4)_2BeF_4\). This salt is stable at room temperature but undergoes decomposition upon heating to yield pure anhydrous \(BeF_2\).

Step 4: Final Answer:
The compound 'A' is \((NH_4)_2BeF_4\).
Quick Tip: This is the standard industrial and laboratory method to prepare anhydrous \(BeF_2\), as direct fluorination of Be is difficult to control.

Step 1: Understanding the Concept:

The covalent character in an ionic bond is determined by Fajans' Rules. A bond is more covalent if there is greater polarization of the anion by the cation.


Step 2: Key Formula or Approach:

Polarization (and hence covalent character) increases with:

1. Small size of cation.

2. Large size of anion.

3. High charge on cation or anion.

4. Cations with pseudo-noble gas configuration (\(ns^2 np^6 nd^{10}\)).


Step 3: Detailed Explanation:

- Analysis of B: \(KI\) has a larger anion (\(I^-\)) than \(KF\) (\(F^-\)), so \(KI\) is more covalent (\(KF < KI\)). \(Li^+\) is smaller than \(K^+\), so \(LiF\) is more covalent than \(KF\) (\(LiF > KF\)). Statement B is correct.

- Analysis of C: \(Sn^{4+}\) has a higher charge than \(Sn^{2+}\), thus \(SnCl_4\) is more covalent than \(SnCl_2\). \(Cu^+\) has a pseudo-noble gas configuration (\(3d^{10}\)), while \(Na^+\) has a noble gas configuration (\(2p^6\)). \(Cu^+\) is more polarizing, so \(CuCl\) is more covalent than \(NaCl\). Statement C is correct.

- Analysis of E: Based on the above, \(KF < KI\) is true and \(CuCl > NaCl\) is true. Statement E is correct.


Step 4: Final Answer:

Statements B, C, and E represent the correct orders of covalent character.
Quick Tip: Fajans' Rule Tip: "Small Cation, Big Anion, High Charge" \(\rightarrow\) High Covalent Character.

Step 1: Understanding the Concept:

Freons are a group of aliphatic organic compounds containing fluorine and chlorine.


Step 2: Detailed Explanation:

Freons are chlorofluorocarbons (CFCs), which are extremely stable, unreactive, non-toxic, non-corrosive, and easily liquefiable gases. They are used extensively as refrigerants and aerosol propellants. While they eventually decompose in the stratosphere to release chlorine radicals (which cause ozone depletion), the compounds themselves are the stable CFC molecules.


Step 3: Final Answer:

The correct definition for freons is chlorofluorocarbon compounds.
Quick Tip: The most common freon is Freon-12 (\(CF_2Cl_2\)), manufactured from \(CCl_4\) via the Swarts reaction.

Step 1: Understanding the Concept:

Dimethylglyoxime (DMG) is a selective reagent used in qualitative and quantitative analysis of nickel.


Step 2: Detailed Explanation:

When dimethylglyoxime is added to an ammoniacal solution of a nickel(II) salt, a bright rosy-red precipitate of nickel dimethylglyoximate, \([Ni(DMG)_2]\), is formed. The complex is stabilized by intramolecular hydrogen bonding. \[ Ni^{2+} + 2C_4H_8N_2O_2 \xrightarrow{NH_4OH} [Ni(C_4H_7N_2O_2)_2] \downarrow (rosy red) + 2H^+ \]

Step 3: Final Answer:
The metal ion is \(Ni^{2+}\).
Quick Tip: The rosy-red complex of Ni-DMG is a square planar complex with two DMG ligands coordinated in a bidentate fashion, featuring O-H...O hydrogen bonds between the ligands.

Step 1: Understanding the Concept:

The ability of phosphorus oxoacids to act as reducing agents (reducing \(Ag^+\) to \(Ag^0\)) depends on the presence of P-H bonds.


Step 2: Detailed Explanation:

- \(H_4P_2O_5\) (Pyrophosphorous acid): Contains two P-H bonds (\(HO-P(H)(O)-O-P(H)(O)-OH\)). It is a strong reducing agent.

- \(H_4P_2O_7\) (Pyrophosphoric acid): No P-H bonds. All P atoms are in +5 state.

- \((HPO_3)_n\) (Metaphosphoric acid): No P-H bonds. P is in +5 state.

- \(H_4P_2O_6\) (Hypophosphoric acid): Contains a P-P bond but no P-H bonds.
Only acids with P-H bonds like \(H_3PO_2\), \(H_3PO_3\), and \(H_4P_2O_5\) can reduce silver nitrate to metallic silver.


Step 3: Final Answer:
\(H_4P_2O_5\) is the correct oxoacid.
Quick Tip: Oxoacids of Phosphorus with P in an oxidation state lower than +5 (containing P-H or P-P bonds) generally act as reducing agents. Specifically, P-H bonds are responsible for the reduction of noble metal ions.

Step 1: Understanding the Concept:

This question requires matching common chemical substances or biological pigments with their constituent ions or chemical formulas based on their known properties and uses.


Step 2: Detailed Explanation:

- A. Chlorophyll: It is a green pigment found in plants responsible for photosynthesis. The central metal ion in the porphyrin ring of chlorophyll is magnesium (\(Mg^{2+}\)). Therefore, A matches with III.

- B. Soda ash: This is the common name for anhydrous sodium carbonate, which has the chemical formula \(Na_2CO_3\). Therefore, B matches with I.

- C. Dentistry, Ornamental work: Plaster of Paris (\(CaSO_4 \cdot \frac{1}{2}H_2O\)) or Gypsum (\(CaSO_4 \cdot 2H_2O\)) are widely used in making dental molds and ornamental casts. The primary chemical component is calcium sulfate (\(CaSO_4\)). Therefore, C matches with II.

- D. Used in white washing: Slaked lime, which is calcium hydroxide (\(Ca(OH)_2\)), is mixed with water and used for whitewashing walls. Therefore, D matches with IV.


Step 3: Final Answer:

By matching the columns, we get: A-III, B-I, C-II, D-IV. This corresponds to option (3).
Quick Tip: Match the most certain pairs first. Knowing that Soda ash is \(Na_2CO_3\) and Chlorophyll contains Magnesium immediately narrows the choices down to the correct answer in many matching questions.

Step 1: Understanding the Concept:

The stability of resonance structures is determined by several rules:

1. Structures with more covalent bonds are generally more stable.

2. Structures where all atoms have complete octets are much more stable than those with incomplete octets.

3. Negative charge on more electronegative atoms (like O) and positive charge on less electronegative atoms (like N) increases stability.

4. Greater charge separation generally decreases stability.


Step 2: Detailed Explanation:

- Structure C: In this structure, every atom (except Hydrogen) has a complete octet. The negative charge is on the highly electronegative oxygen atom, and the positive charge is on the nitrogen atom. This is the most stable charged resonance structure.

- Structure D: This structure also has complete octets for all atoms. However, compared to C, the negative charge is on a carbon atom while the oxygen is part of a neutral carbonyl group. Since Oxygen is more electronegative than Carbon, C is more stable than D.

- Structures A and B: These structures have incomplete octets (carbocations). Structures with incomplete octets are significantly less stable than C and D.

- Comparing A and B: In structure B, the negative charge is adjacent to the electron-withdrawing carbonyl group (\(OHC-\)), which provides some inductive stabilization. In structure A, the positive charge is adjacent to the carbonyl group, which is destabilizing due to the partial positive charge on the carbonyl carbon. Thus, B is slightly more stable than A.


Step 3: Final Answer:

The increasing order of stability is A \(<\) B \(<\) D \(<\) C.
Quick Tip: The "Octet Rule" is the most important factor in resonance stability. Always prioritize structures where every atom has 8 electrons over structures with carbocations, regardless of where the charges are placed.

Step 1: Understanding the Concept:

This question relates to the metallurgical processes and the specific equipment used for the extraction and purification of various metals.


Step 2: Detailed Explanation:

- A. Reverberatory furnace: This furnace is commonly used in the metallurgy of copper, specifically for the smelting of roasted ore to produce matte. Therefore, A matches with IV.

- B. Electrolytic cell: Aluminum is extracted from alumina (\(Al_2O_3\)) using the Hall-Heroult process, which takes place in an electrolytic cell. Therefore, B matches with II.

- C. Blast furnace: This is the standard industrial equipment used for the reduction of iron oxides to produce Pig Iron. Therefore, C matches with I.

- D. Zone Refining furnace: Zone refining is a method used to obtain metals of very high purity (semiconductors like Silicon, Germanium). It involves a circular heater moving along a rod of the impure metal. Therefore, D matches with III.


Step 3: Final Answer:

The correct matching sequence is A-IV, B-II, C-I, D-III. This matches option (3).
Quick Tip: Associate Pig Iron with Blast Furnace and Aluminum with Electrolysis immediately. These are the most common industrial pairings in metallurgy.

Step 1: Understanding the Concept:

The reaction involves the electrophilic addition of HBr to an alkene. Since a carbocation is formed as an intermediate, ring expansion can occur to relieve the angle strain of a 4-membered ring.


Step 2: Detailed Explanation:

1. Protonation: The \(\pi\) bond of the alkene attacks the \(H^+\) from HBr. The proton adds to the terminal \(CH_2\) (according to Markovnikov's rule) to form a stable tertiary carbocation at the carbon adjacent to the cyclobutane ring.

2. Ring Expansion: The cyclobutane ring is strained. To relieve this strain, a 1,2-alkyl shift occurs. One of the C-C bonds of the 4-membered ring breaks and shifts to the carbocation center. This expands the 4-membered ring into a more stable 5-membered ring (cyclopentane).

3. Rearrangement: After ring expansion, a secondary carbocation is formed on the ring. A 1,2-hydride shift or methyl shift occurs to produce a more stable tertiary carbocation. In this case, the expansion leads directly to a structure where a subsequent shift or the expansion itself results in a 1,2-dimethylcyclopentyl cation.

4. Nucleophilic Attack: The \(Br^-\) ion then attacks the most stable tertiary carbocation formed.

The final major product is 1-bromo-1,2-dimethylcyclopentane.


Step 3: Final Answer:

The correct major product 'A' is shown in option (4).
Quick Tip: Whenever a carbocation is formed immediately outside a 4 or 5-membered ring, always consider ring expansion (4\(\rightarrow\)5 or 5\(\rightarrow\)6) as the most likely step to increase stability and reduce strain.

Step 1: Understanding the Concept:

Hydrogen bonding strength and extent depend on the arrangement and proximity of water molecules. In solid state (ice), the structure is highly ordered.


Step 2: Detailed Explanation:

- Ice (B): In ice, water molecules are arranged in a rigid, tetrahedral, cage-like structure. Each water molecule is maximally hydrogen-bonded to 4 other molecules. This crystal lattice ensures the highest extent of hydrogen bonding.

- Liquid water (A): In the liquid state, molecules have kinetic energy and the hydrogen bonds are constantly breaking and reforming. On average, a molecule is bonded to fewer than 4 neighbors at any instant. Thus, H-bonding is less extensive than in ice.

- Impure water (C): The presence of solutes (impurities) disrupts the regular hydrogen-bonding network of water molecules as the solute particles interact with water (hydration), reducing the effective H-bonding between water molecules themselves.


Step 3: Final Answer:

The decreasing order is B \(>\) A \(>\) C.
Quick Tip: Ice floats on water because its extensive hydrogen bonding creates an open cage-like structure with more volume and lower density than liquid water.

Step 1: Understanding the Concept:

According to Werner's theory:

1. Primary Valency corresponds to the oxidation state of the central metal atom.

2. Secondary Valency corresponds to the coordination number (number of donor atoms attached to the metal).


Step 2: Detailed Explanation:

1. Calculation of Primary Valency (Oxidation State):

Let the oxidation state of Cobalt be \(x\).

Ammonia (\(NH_3\)) is a neutral ligand (charge = 0).

Chlorine (\(Cl\)) has a charge of \(-1\).

The complex is \([Co(NH_3)_5Cl]Cl_2\). Sum of charges = 0.
\[ x + 5(0) + 1(-1) + 2(-1) = 0 \]
\[ x - 1 - 2 = 0 \Rightarrow x = +3 \]

Primary Valency = 3.


2. Calculation of Secondary Valency (Coordination Number):

The ligands inside the coordination sphere (brackets) are 5 \(NH_3\) molecules and 1 \(Cl^-\) ion.

Total ligands = \(5 + 1 = 6\).

Since both are monodentate ligands, the Coordination Number = 6.

Secondary Valency = 6.


Step 3: Final Answer:

The primary and secondary valencies are 3 and 6 respectively.
Quick Tip: Primary valency is ionizable and satisfied by negative ions. Secondary valency is non-ionizable and determines the geometry of the complex.

Step 1: Understanding the Concept:

This question refers to Moseley's Law, which established the relationship between the frequency of characteristic X-rays and the atomic number of the emitting target element.


Step 2: Key Formula or Approach:

Moseley's Law is given by:
\[ \sqrt{\nu} = a(Z - b) \]

where \(\nu\) is the frequency, \(Z\) is the atomic number, and \(a, b\) are constants.


Step 3: Detailed Explanation:

The law states that the square root of the frequency of characteristic X-rays is proportional to the atomic number.

Rewriting the formula:
\[ \nu^{1/2} = a(Z - b) \]

Comparing this with the graph provided (a straight line of the form \(y = mx + c\)):

- The y-axis represents \(\nu^n\).

- The x-axis represents \(Z\).

For the graph to be a straight line, \(\nu^n\) must be proportional to \(Z\). From Moseley's Law, we know \(\nu^{1/2}\) is proportional to \(Z\).

Therefore, \(n = \frac{1}{2}\).


Step 4: Final Answer:

The value of \(n\) is \(\frac{1}{2}\).
Quick Tip: Moseley's Law (\(\sqrt{\nu} \propto Z\)) was the fundamental discovery that led to the realization that atomic number, not atomic mass, is the basis of the periodic table.

Step 1: Understanding the Concept:

Freezing point depression is a colligative property. It occurs because the addition of a non-volatile solute lowers the vapour pressure of the solvent.


Step 2: Detailed Explanation:

- Statement A: According to Raoult's Law, the vapour pressure of a solution containing a non-volatile solute is always lower than that of the pure solvent. This is correct.

- Statement B: This contradicts Statement A and is incorrect.

- Statement D: Freezing is defined as the temperature at which the liquid solvent and solid solvent have the same vapour pressure. During the freezing process of a dilute solution, it is the solvent molecules that crystallize out to form the solid phase. The solute remains in the liquid phase (unless it forms a solid solution, which is not the standard case). This is correct.

- Statement C: This contradicts Statement D and is incorrect.


Step 3: Final Answer:

Statements A and D are correct. Therefore, the answer is option (1).
Quick Tip: Always remember: in colligative properties of dilute solutions, we assume the solute is non-volatile and that it is the pure solvent that changes state (freezes or boils).

Step 1: Understanding the Concept:

The sequence involves:

1. Crossed Aldol condensation with formaldehyde (\(HCHO\)) in the presence of base (\(KOH\)).

2. Cyanohydrin formation using \(KCN\).

3. Acidic hydrolysis to form a hydroxy acid which then undergoes lactonization.


Step 2: Detailed Explanation:

1. Starting with (4) \((CH_3)_2CH-CHO\) (Isobutyraldehyde): This aldehyde has one \(\alpha\)-hydrogen.

2. Step (i) \(HCHO, KOH\): Isobutyraldehyde reacts with formaldehyde via a crossed-aldol reaction. The single \(\alpha\)-hydrogen is replaced by a \(CH_2OH\) group.

Result: \(HOCH_2-C(CH_3)_2-CHO\).

3. Step (ii) & (iii) \(KCN, H_3O^+\): The aldehyde group (\(-CHO\)) reacts with cyanide to form a cyanohydrin, which is then hydrolyzed by \(H_3O^+\) to a carboxylic acid group (\(-COOH\)) with an \(\alpha\)-hydroxyl group.

Result: \(HOCH_2-C(CH_3)_2-CH(OH)-COOH\).

4. Lactonization: The molecule contains both a carboxylic acid group and a hydroxyl group (\(\gamma\)-hydroxyl relative to the carbonyl). Upon heating or under acidic conditions, they lose a water molecule to form a 5-membered cyclic ester called a \(\gamma\)-lactone. This matches the structure of (Y) shown in the question.


Step 3: Final Answer:

The starting compound (X) is 2-methylpropanal, which is \((CH_3)_2CH-CHO\).
Quick Tip: Formaldehyde (\(HCHO\)) is often used in basic medium to introduce \(CH_2OH\) groups at the \(\alpha\)-position of other aldehydes through repeated aldol condensations (Tollens' reaction).

Step 1: Understanding the Concept:

The reaction involves the treatment of an organic chloride with Silver Nitrate (\(AgNO_3\)).
\(AgNO_3\) reacts with "ionizable" chlorine atoms to form a white precipitate of \(AgCl\).

Specifically, alkyl halides that can form stable carbocations (like tertiary, benzylic, or allylic carbocations) react readily with \(AgNO_3\) via an \(S_N1\) mechanism.

Vinylic halides and aryl halides are generally inert to \(AgNO_3\) because the C-Cl bond is very strong due to partial double bond character.


Step 2: Detailed Explanation:

1. Aryl Chloride: The chlorine atom attached directly to the benzene ring (bottom) is an aryl halide. It does not react.

2. Vinylic Chloride: The chlorine atom attached to the double bond carbon (left side) is a vinylic halide. It does not react.

3. Benzylic Chloride: The chlorine atom attached to the \(CH\) group which is directly bonded to the benzene ring (top-center) is a benzylic chloride. Benzylic carbocations are resonance stabilized, making this chlorine reactive. (1 mole of \(AgCl\) from here).

4. Tertiary Chloride: The chlorine atom attached to a tertiary carbon atom (right side, branched carbon) can easily leave to form a stable tertiary carbocation. (1 mole of \(AgCl\) from here).

5. Total reactive chlorine atoms = \(1 (benzylic) + 1 (tertiary) = 2\).


Step 3: Final Answer:

The number of moles of \(AgCl\) formed is 2.
Quick Tip: Remember that \(S_N1\) reactivity follows the stability of the carbocation. Aryl and Vinyl halides have very high bond dissociation energy and do not form \(AgCl\) with \(AgNO_3\) at room temperature.

Step 1: Understanding the Concept:

The problem involves calculating the molarity of a stock solution and then determining the dilution volume using the dilution law.


Step 2: Key Formula or Approach:

1. Molarity (\(M\)) = \(\frac{moles of solute}{Volume of solution in Litres}\).

2. Dilution Law: \(M_1 V_1 = M_2 V_2\).


Step 3: Detailed Explanation:

1. Calculate Molar Mass of NaOH:
\[ M_{NaOH} = 23 + 16 + 1 = 40 g/mol \]

2. Calculate Molarity of Stock Solution (\(M_1\)):

Number of moles = \(\frac{5}{40} = 0.125 mol\).

Volume = \(450 mL = 0.45 L\).
\[ M_1 = \frac{0.125}{0.45} \approx 0.2778 M \]

3. Apply Dilution Law to find \(V_1\):

Target molarity \(M_2 = 0.1 M\).

Target volume \(V_2 = 500 mL\).
\[ V_1 = \frac{M_2 V_2}{M_1} = \frac{0.1 \times 500}{\frac{0.125}{0.45}} \]
\[ V_1 = \frac{50 \times 0.45}{0.125} = \frac{22.5}{0.125} = 180 mL \]


Step 4: Final Answer:

The volume required is 180 mL.
Quick Tip: When doing dilution problems, you can keep the volumes in mL as long as you are consistent on both sides of the \(M_1 V_1 = M_2 V_2\) equation.

Step 1: Understanding the Concept:

These statements are based on the Arrhenius equation which describes the temperature dependence of the rate of a chemical reaction.


Step 2: Key Formula or Approach:

Arrhenius equation: \(k = A e^{-\frac{E_a}{RT}}\) or \(\ln k = \ln A - \frac{E_a}{RT}\).


Step 3: Detailed Explanation:

- Statement A: Since \(k = A e^{-\frac{E_a}{RT}}\), as the activation energy (\(E_a\)) increases, the negative exponent becomes larger, making the value of \(e^{-\frac{E_a}{RT}}\) smaller. Thus, \(k\) decreases. (Correct).

- Statement B: The temperature coefficient (\(\eta\)) is roughly proportional to \(e^{\frac{E_a}{R} \cdot \frac{\Delta T}{T^2}}\). A higher \(E_a\) means the rate constant is more sensitive to temperature changes, leading to a higher temperature coefficient. (Correct).

- Statement C: The slope of the \(\ln k\) vs \(T\) curve is \(\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}\). Since \(T^2\) is in the denominator, at smaller \(T\) (lower temperatures), the value of the slope is larger, meaning a greater relative change in \(k\) for the same \(\Delta T\). (Correct).

- Statement D: Rearranging the Arrhenius equation gives \(\ln k = -\frac{E_a}{R}\left(\frac{1}{T}\right) + \ln A\). This is in the form \(y = mx + c\), where the slope is \(-\frac{E_a}{R}\). (Correct).


Step 4: Final Answer:

All 4 statements are correct.
Quick Tip: The Arrhenius equation shows that reactions with high activation energy are "slower" but much more "temperature-sensitive".

Step 1: Understanding the Concept:

This half-cell potential is calculated using the Nernst Equation. We must first identify the balanced half-reaction to determine the number of electrons transferred (\(n\)) and the stoichiometry.


Step 2: Key Formula or Approach:

Balanced Half-reaction: Cr\(_2\)O\(_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\).

Nernst Equation: \(E = E^\circ - \frac{0.059}{n} \log Q\).


Step 3: Detailed Explanation:

1. Identify variables:
\(n = 6\) electrons.

[Cr\(_2\)O\(_7^{2-}\)] = \(10 mmol/L = 0.01 M = 10^{-2} M\).

[Cr\(^{3+}\)] = \(100 mmol/L = 0.1 M = 10^{-1} M\).

pH = 3.0 \(\implies\) [H\(^+\)] = \(10^{-3} M\).

2. Calculate Reaction Quotient (\(Q\)):
\[ Q = \frac{[Cr^{3+}]^2}{[Cr_2O_7^{2-}][H^+]^{14}} = \frac{(10^{-1})^2}{10^{-2} \cdot (10^{-3})^{14}} = \frac{10^{-2}}{10^{-2} \cdot 10^{-42}} = 10^{42} \]

3. Calculate half-cell potential (\(E\)):
\[ E = 1.330 - \frac{0.059}{6} \log(10^{42}) \]
\[ E = 1.330 - \frac{0.059}{6} \cdot 42 = 1.330 - (0.059 \cdot 7) \]
\[ E = 1.330 - 0.413 = 0.917 V \]
\[ E = 917 \times 10^{-3} V \]


Step 4: Final Answer:

The value of \(x\) is 917.
Quick Tip: Pay close attention to the powers of concentration in the reaction quotient \(Q\), especially the [H\(^+\)] term which is raised to the 14th power in dichromate reduction.

Step 1: Understanding the Concept:

To find the percentage composition of an element in a compound, we need the molecular formula and the total molar mass.


Step 2: Detailed Explanation:

1. Determine the Molecular Formula:

From the structure, Uracil consists of a ring with 4 Carbon atoms, 2 Nitrogen atoms, 2 Oxygen atoms, and 4 Hydrogen atoms (2 on carbons of the double bond, and 2 on the nitrogens in this representation).

Formula: C\(_4\)H\(_4\)N\(_2\)O\(_2\).

2. Calculate Molar Mass:
\[ Mass of C = 4 \times 12 = 48 g/mol \]
\[ Mass of H = 4 \times 1 = 4 g/mol \]
\[ Mass of N = 2 \times 14 = 28 g/mol \]
\[ Mass of O = 2 \times 16 = 32 g/mol \]
\[ Total Molar Mass = 48 + 4 + 28 + 32 = 112 g/mol \]

3. Calculate % of Nitrogen:
\[ % N = \left( \frac{Mass of Nitrogen}{Total Mass} \right) \times 100 = \left( \frac{28}{112} \right) \times 100 \]
\[ % N = \frac{1}{4} \times 100 = 25% \]


Step 3: Final Answer:

The percentage of Nitrogen in Uracil is 25.
Quick Tip: Double check the hydrogen count by ensuring every atom in the cyclic structure satisfies its valency (C=4, N=3, O=2).

Step 1: Understanding the Concept:

A mixture of a weak acid (CH\(_3\)COOH) and its salt with a strong base (CH\(_3\)COONa) forms an acidic buffer. The pH is calculated using the Henderson-Hasselbalch equation.


Step 2: Key Formula or Approach:

Henderson-Hasselbalch equation: \(pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right)\).


Step 3: Detailed Explanation:

1. Calculate concentrations after mixing:

Total Volume = \(25 + 25 = 50 mL\).

[CH\(_3\)COONa] = \(\frac{0.2 M \times 25 mL}{50 mL} = 0.1 M\).

[CH\(_3\)COOH] = \(\frac{0.02 M \times 25 mL}{50 mL} = 0.01 M\).

2. Substitute into buffer equation:
\[ 5 = pK_a + \log \left( \frac{0.1}{0.01} \right) \]
\[ 5 = pK_a + \log(10) \]
\[ 5 = pK_a + 1 \implies pK_a = 4 \]

3. Find \(K_a\):
\[ K_a = 10^{-pK_a} = 10^{-4} \]
\[ K_a = 10 \times 10^{-5} \]

Comparing with \(x \times 10^{-5}\), we get \(x = 10\).


Step 4: Final Answer:

The value of \(x\) is 10.
Quick Tip: When equal volumes are mixed, the concentrations simply become half of their initial values. This simplifies calculations.

Step 1: Understanding the Concept:

Spontaneity of a process is determined by the change in Gibbs Free Energy (\(\Delta G\)).

If \(\Delta G < 0\), the process is spontaneous.

If \(\Delta G > 0\), the process is non-spontaneous.


Step 2: Key Formula or Approach:

Gibbs Free Energy equation: \(\Delta G = \Delta H - T\Delta S\).


Step 3: Detailed Explanation:

Given \(T = 300 K\).

- Process A:
\(\Delta G = -25000 J - (300 K \times -80 J/K) = -25000 + 24000 = -1000 J\).

(\(\Delta G < 0\), Spontaneous).

- Process B:
\(\Delta G = -22000 J - (300 K \times 40 J/K) = -22000 - 12000 = -34000 J\).

(\(\Delta G < 0\), Spontaneous).

- Process C:
\(\Delta G = 25000 J - (300 K \times -50 J/K) = 25000 + 15000 = 40000 J\).

(\(\Delta G > 0\), Non-Spontaneous).

- Process D:
\(\Delta G = 22000 J - (300 K \times 20 J/K) = 22000 - 6000 = 16000 J\).

(\(\Delta G > 0\), Non-Spontaneous).

Total non-spontaneous processes = 2 (C and D).


Step 4: Final Answer:

The number of non-spontaneous processes is 2.
Quick Tip: If \(\Delta H\) is positive and \(\Delta S\) is negative, the process is always non-spontaneous at all temperatures. If \(\Delta H\) is negative and \(\Delta S\) is positive, it is always spontaneous.

Step 1: Understanding the Concept:

Fe\(_{0.93}\)O is a non-stoichiometric compound where some Fe\(^{2+}\) ions are missing and replaced by Fe\(^{3+}\) ions to maintain electrical neutrality.


Step 2: Detailed Explanation:

1. Determine moles of Fe\(^{2+}\) and Fe\(^{3+}\) (Statement B):

Let \(x\) be the fraction of Fe as Fe\(^{2+}\) and \(y\) be the fraction as Fe\(^{3+}\).

Total Fe: \(x + y = 0.93\).

Charge neutrality: \(2x + 3y = 2\) (since oxygen is \(-2\)).

Multiply the first by 2: \(2x + 2y = 1.86\).

Subtracting: \(y = 0.14\), then \(x = 0.93 - 0.14 = 0.79\).

Statement B is Correct.


2. Calculate Equivalent Weight (Statement A):

The reaction is Fe\(_{0.93}\)O \(\rightarrow\) Fe\(_2\)O\(_3\). All Fe ends as Fe\(^{3+}\).

Only Fe\(^{2+}\) (\(0.79\) moles) undergoes oxidation to Fe\(^{3+}\) (loss of 1 \(e^-\)).
\(n\)-factor = \(0.79 \times (3-2) = 0.79\).

Equivalent weight = \(\frac{Molecular weight}{n-factor} = \frac{M}{0.79}\).

Statement A is Correct.


3. Structural property (Statement C):

Wustite (FeO) has a rock salt structure where O\(^{2-}\) ions form a CCP lattice and Fe ions occupy octahedral voids. It is the classic example of a metal deficiency defect.

Statement C is Correct.


4. Percentage composition (Statement D):
\(% Fe^{2+} = \frac{0.79}{0.93} \times 100 \approx 84.94% \approx 85%\).
\(% Fe^{3+} = \frac{0.14}{0.93} \times 100 \approx 15.05% \approx 15%\).

Statement D is Correct.


Step 3: Final Answer:

All 4 statements are correct.
Quick Tip: For non-stoichiometric compounds like \(M_xO\), always use the sum of total metal fraction and the sum of charges to find individual ion counts.

Step 1: Understanding the Concept:

The wavelength of light emitted during electron transitions in a hydrogen atom is given by the Rydberg formula. The Paschen series corresponds to transitions ending at the \(n=3\) energy level.


Step 2: Key Formula or Approach:

Rydberg formula: \(\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)\).

For Paschen series: \(n_1 = 3\).

First line: \(n_2 = 4\).

Second line: \(n_2 = 5\).


Step 3: Detailed Explanation:

1. First line (\(\lambda_1 = 720 nm\)):
\[ \frac{1}{\lambda_1} = R_H \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{16} \right) = R_H \left( \frac{7}{144} \right) \]

2. Second line (\(\lambda_2\)):
\[ \frac{1}{\lambda_2} = R_H \left( \frac{1}{3^2} - \frac{1}{5^2} \right) = R_H \left( \frac{1}{9} - \frac{1}{25} \right) = R_H \left( \frac{16}{225} \right) \]

3. Find ratio:
\[ \frac{\lambda_2}{\lambda_1} = \frac{7/144}{16/225} = \frac{7}{144} \times \frac{225}{16} = \frac{1575}{2304} \]
\[ \lambda_2 = 720 \times \frac{1575}{2304} = \frac{1134000}{2304} \approx 492.1875 nm \]


Step 4: Final Answer:

The nearest integer value of wavelength is 492 nm.
Quick Tip: Instead of calculating \(R_H\), always use ratios for wavelength problems involving lines within the same series. This cancels the constant and reduces errors.

Step 1: Understanding the Concept:

The splitting of d-orbitals in a tetrahedral crystal field results in a lower energy \(e\) set and a higher energy \(t_2\) set. Chlorine is a weak-field ligand, resulting in high-spin configurations.


Step 2: Detailed Explanation:

1. Identify the metal ion and d-count:

In [CoCl\(_4\)]\(^{2-}\), Cobalt is in the +2 oxidation state.

Atomic Co: [Ar] 3d\(^7\) 4s\(^2\).

Co\(^{2+}\): [Ar] 3d\(^7\).


2. Fill the orbitals for high-spin tetrahedral field:

Orbitals: \(e\) (bottom) and \(t_2\) (top).

Fill 7 electrons: \(e^1, e^2 \rightarrow t_2^1, t_2^2, t_2^3 \rightarrow e^3, e^4\).

Configuration: \(e^4 t_2^3\).

Thus, \(m = 4\).


3. Count unpaired electrons:

The \(e\) orbitals are fully paired (\(e^4\)).

The \(t_2\) orbitals have 3 electrons in 3 orbitals, so all 3 are unpaired (\(t_2^3\)).

Number of unpaired electrons = 3.

4. Calculate Sum:

Sum = \(m + unpaired = 4 + 3 = 7\).


Step 3: Final Answer:

The sum is 7.
Quick Tip: Remember that tetrahedral complexes are almost always high-spin because the crystal field splitting \(\Delta_t\) is significantly smaller than the octahedral splitting \(\Delta_o\).