Step 1: Understanding the Concept:

In the Bohr model of the hydrogen atom, the energy of an electron in a specific orbit is inversely proportional to the square of the principal quantum number (\(n\)).


Step 2: Key Formula or Approach:

The formula for the energy of an electron in the \(n^{th}\) orbit is:
\[ E_n = \frac{E_1}{n^2} \]

where \(E_1\) is the energy of the first Bohr orbit (\(n=1\)).


Step 3: Detailed Explanation:

Given that the energy in the first orbit is \(E_1 = -2.18 \times 10^{-18}\) J.

We need to find the energy in the third orbit (\(n=3\)).

Using the formula:
\[ E_3 = \frac{E_1}{(3)^2} \]
\[ E_3 = \frac{E_1}{9} \]

This shows that the energy in the third orbit is exactly one-ninth (\(\frac{1}{9}\)) of the energy in the first orbit.


Step 4: Final Answer:

The energy in the third Bohr orbit is \(\frac{1}{9}\) th of the energy in the first orbit.
Quick Tip: For any hydrogen-like species, always remember the proportionality \(E \propto \frac{Z^2}{n^2}\). If the orbit number \(n\) increases by a factor of \(k\), the energy decreases by a factor of \(k^2\).

Step 1: Understanding the Concept:

Molecular Orbital Theory (MOT) helps in determining the bond order and magnetic properties based on the arrangement of electrons in bonding and antibonding molecular orbitals.


Step 2: Key Formula or Approach:

Bond Order (B.O.) = \(\frac{1}{2} (N_b - N_a)\), where \(N_b\) is the number of bonding electrons and \(N_a\) is the number of antibonding electrons.

A species is paramagnetic if it has unpaired electrons and diamagnetic if all electrons are paired.


Step 3: Detailed Explanation:

1. Process \(NO \rightarrow NO^+\):

- \(NO\) has 15 electrons. Configuration: \((\sigma 1s)^2 (\sigma^* 1s)^2 (\sigma 2s)^2 (\sigma^* 2s)^2 (\sigma 2p_z)^2 (\pi 2p_x^2 = \pi 2p_y^2) (\pi^* 2p_x)^1\).

- B.O. = \(\frac{10 - 5}{2} = 2.5\). It has one unpaired electron (Paramagnetic).

- \(NO^+\) has 14 electrons. The \(\pi^*\) electron is removed.

- B.O. = \(\frac{10 - 4}{2} = 3\). All electrons are paired (Diamagnetic).

- Conclusion: Bond order increases and character changes from paramagnetic to diamagnetic.


2. Process \(O_2 \rightarrow O_2^+\):

- \(O_2\) (16e, B.O. = 2, Para) \(\rightarrow O_2^+\) (15e, B.O. = 2.5, Para). Magnetic character remains paramagnetic.


3. Process \(O_2 \rightarrow O_2^{2-}\):

- \(O_2\) (16e, B.O. = 2, Para) \(\rightarrow O_2^{2-}\) (18e, B.O. = 1, Dia). Bond order decreases.


4. Process \(N_2 \rightarrow N_2^+\):

- \(N_2\) (14e, B.O. = 3, Dia) \(\rightarrow N_2^+\) (13e, B.O. = 2.5, Para). Bond order decreases and becomes paramagnetic.


Step 4: Final Answer:

The process \(NO \rightarrow NO^+\) results in a bond order increase from 2.5 to 3 and a change from paramagnetic to diamagnetic.
Quick Tip: Species with 14 electrons (like \(N_2, CO, NO^+, CN^-\)) generally have a bond order of 3 and are diamagnetic. Removing an electron from an antibonding orbital (\(\pi^*\) or \(\sigma^*\)) always increases the bond order.

Step 1: Understanding the Concept:

Lyophobic sols are unstable and easily coagulated by small amounts of electrolytes. Lyophilic sols, on the other hand, are highly stable due to extensive solvation.


Step 2: Detailed Explanation:

When a lyophilic sol is added to a lyophobic sol, the particles of the lyophilic sol surround the particles of the lyophobic sol.

This forms a protective layer or "film" around the lyophobic particles.

This layer prevents the lyophobic particles from coming into direct contact with electrolyte ions, thereby protecting them from coagulation.

This phenomenon is known as the "protection of colloids," and the lyophilic sol is called a protective colloid.


Step 4: Final Answer:

A protective film of the lyophilic sol is formed over the lyophobic sol particles.
Quick Tip: The "Gold Number" is a measure used to quantify the protective power of a lyophilic colloid. A lower Gold Number indicates a more efficient protective colloid.

Step 1: Understanding the Concept:

This question tests the knowledge of periodic trends and chemical classifications of elements and their compounds.


Step 2: Detailed Explanation:

Statement A: The electron gain enthalpy of \(Cl\) (\(-349 kJ/mol\)) is actually more negative than that of \(F\) (\(-328 kJ/mol\)). This is because \(F\) has a very small size, leading to high inter-electronic repulsions in its compact \(2p\) subshell, which makes the addition of an electron less favorable compared to \(Cl\). So, Statement A is Incorrect.

Statement B: As we move down a group, the atomic size increases and the valence electrons are further from the nucleus, leading to a decrease in ionization enthalpy. This statement is Correct.

Statement C: Electronegativity is not a fixed property of an element; it varies based on the oxidation state and hybridization of the atom, which are determined by the atoms it is bonded to. So, Statement C is technically Incorrect when treated as an inherent fixed constant in simple contexts, or often classified as a variable in advanced periodic property discussions. In this specific question's key, it is grouped with incorrect statements.

Statement D: \(Al_2O_3\) is an amphoteric oxide, but \(NO\) is a neutral oxide (along with \(CO\) and \(N_2O\)). Thus, the entire statement is Incorrect.


Step 4: Final Answer:

Statements A, C, and D are not correct.
Quick Tip: Always remember: Chlorine has the highest (most negative) electron gain enthalpy in the entire periodic table. Neutral oxides to memorize: \(H_2O, CO, NO, N_2O\).

Step 1: Understanding the Concept:

Refining of metals involves different physical or chemical techniques depending on the nature of the metal and the impurities.


Step 2: Detailed Explanation:

1. Titanium: It is refined by the van Arkel method, where it is converted into a volatile iodide and then decomposed. (Correct match)

2. Nickel: It is refined by the Mond process, involving the formation and decomposition of volatile nickel tetracarbonyl. (Correct match)

3. Zinc: Zinc is a volatile metal (low boiling point) and is typically refined by fractional distillation. Liquation is used for metals with low melting points like Tin (\(Sn\)) and Lead (\(Pb\)). So, Zinc - Liquation is a Mismatch.

4. Copper: High purity copper is obtained through electrolytic refining. (Correct match)


Step 4: Final Answer:

The mismatched pair is Zinc - Liquation.
Quick Tip: Liquation: Used for low melting point metals (\(Sn, Pb, Bi\)).
Distillation: Used for low boiling point (volatile) metals (\(Zn, Cd, Hg\)).
Vapor Phase Refining: Mond process (\(Ni\)) and van Arkel (\(Zr, Ti\)).

Step 1: Understanding the Concept:

Water softening methods like the Permutit (Zeolite) process and the Synthetic Resin method are used to remove hardness-causing \(Ca^{2+}\) and \(Mg^{2+}\) ions.


Step 2: Detailed Explanation:

Statement I: The Synthetic Resin method is actually far more efficient than the Permutit process. It can remove all dissolved mineral salts from water (producing deionized/demineralized water), whereas the Permutit process primarily exchanges cations and cannot produce completely demineralized water. Thus, Statement I is Incorrect.

Statement II: In the ion-exchange process using sodium-form resins (\(RNa\)), the calcium and magnesium ions in hard water are replaced by sodium ions. This results in the formation of soluble sodium salts in the water. Thus, Statement II is Correct.


Step 4: Final Answer:

Statement I is incorrect and Statement II is correct.
Quick Tip: Synthetic resin method is the most modern and efficient method because it allows for both cation and anion exchange, potentially removing all dissolved salts.

Step 1: Understanding the Concept:

Beryllium hydroxide (\(Be(OH)_2\)) is amphoteric, meaning it can react with both acids and bases. Strontium hydroxide (\(Sr(OH)_2\)) is a strong base.


Step 2: Detailed Explanation:

When \(Be(OH)_2\) reacts with an excess of base like \(Sr(OH)_2\), it acts as an acid to form a beryllate salt:
\[ Be(OH)_2 + Sr(OH)_2 \rightarrow Sr[Be(OH)_4] \]

In the salt \(Sr[Be(OH)_4]\):

1. The cation is \(Sr^{2+}\).

2. The anion is the complex \([Be(OH)_4]^{2-}\).

Therefore, Beryllium is part of the anionic complex, not the cationic part. This makes option (C) Incorrect.

In the \([Be(OH)_4]^{2-}\) ion, the Beryllium atom is \(sp^3\) hybridized and is tetrahedrally coordinated by four hydroxyl groups. Since an amphoteric substance reacts with a base, it follows an acid-base neutralization pathway.


Step 4: Final Answer:

Beryllium exists in the anionic part of the resulting salt, not the cationic part.
Quick Tip: Only Beryllium in Group 2 shows amphoteric behavior. Its small size and high charge density favor the formation of complex ions like \([Be(OH)_4]^{2-}\) and \([BeF_4]^{2-}\).

Step 1: Understanding the Concept:

The physical state and molecular geometry of interhalogen compounds depend on their intermolecular forces and electron pair arrangements (VSEPR theory).


Step 2: Key Formula or Approach:

Geometry is determined using the Steric Number (SN) = \(\frac{1}{2}\) [Valence electrons of central atom + Number of monovalent atoms].


Step 3: Detailed Explanation:

1. Physical State: Chlorine pentafluoride (\(ClF_5\)) has a boiling point of approximately \(-13.1^\circ C\). Therefore, at room temperature (\(25^\circ C\)), it exists as a colourless gas.

2. Geometry: For \(ClF_5\), Chlorine (the central atom) has 7 valence electrons.

- SN = \(\frac{1}{2} [7 + 5] = 6\).

- This corresponds to an octahedral electron pair geometry (\(sp^3d^2\) hybridization).

- Since there are 5 bond pairs and 1 lone pair, the molecular shape is Square Pyramidal.


Step 4: Final Answer:
\(ClF_5\) is a colourless gas with square pyramidal geometry.
Quick Tip: VSEPR shorthand: \(AX_5E_1\) type molecules (5 BP + 1 LP) always exhibit a Square Pyramidal shape. Common examples include \(ClF_5, BrF_5\), and \(IF_5\).

Step 1: Understanding the Concept:

Borazine (\(B_3N_3H_6\)) is often called "inorganic benzene" because its physical properties and structure are similar to benzene.


Step 2: Detailed Explanation:

- Cyclic: Borazine has a six-membered hexagonal ring structure. (Correct)

- Electronic delocalization: The lone pairs on Nitrogen are donated to the empty p-orbitals of Boron, leading to \(\pi\) delocalization across the ring. (Correct)

- Reaction with water: Unlike benzene, borazine is polar and chemically more reactive. It undergoes hydrolysis with water to produce boric acid and ammonia. (Correct)

- Banana bonds: Banana bonds (3-center-2-electron bonds) are characteristic of diborane (\(B_2H_6\)), not borazine. Borazine has standard 2-center-2-electron \(\sigma\) and \(\pi\) bonds. So, this is Incorrect.


Step 4: Final Answer:

Borazine does not contain banana bonds.
Quick Tip: Differentiate clearly: Diborane (\(B_2H_6\)) \(\rightarrow\) Banana bonds (3c-2e). Borazine (\(B_3N_3H_6\)) \(\rightarrow\) Cyclic, aromatic character, 2c-2e bonds.

Step 1: Understanding the Concept:

Third ionization energy (\(IE_3\)) involves the removal of an electron from the \(M^{2+}\) ion. If the \(M^{2+}\) ion has a stable half-filled (\(f^7\)) or fully-filled (\(f^{14}\)) configuration, \(IE_3\) will be exceptionally high.


Step 2: Detailed Explanation:

1. Europium (Eu): Configuration is \([Xe] 4f^7 6s^2\). \(Eu^{2+}\) is \([Xe] 4f^7\). Removing an electron from the stable half-filled \(f^7\) subshell requires a lot of energy. Thus, \(Eu\) has high \(IE_3\).

2. Ytterbium (Yb): Configuration is \([Xe] 4f^{14} 6s^2\). \(Yb^{2+}\) is \([Xe] 4f^{14}\). Removing an electron from the stable fully-filled \(f^{14}\) subshell requires a lot of energy. Thus, \(Yb\) has high \(IE_3\).

3. Gadolinium (Gd): Configuration is \([Xe] 4f^7 5d^1 6s^2\). \(Gd^{2+}\) is \([Xe] 4f^7 5d^1\). The third electron is easily removed from the \(5d\) orbital to reach the stable \(f^7\) state, so its \(IE_3\) is low.


Step 4: Final Answer:

Eu and Yb have high third ionization energies because their dipositive ions have stable \(f^7\) and \(f^{14}\) configurations.
Quick Tip: Always check for the stability of \(f^0, f^7\), and \(f^{14}\) when dealing with ionization energies or oxidation states of lanthanides and actinides.

Step 1: Understanding the Concept:

Coordination compounds find various applications in biology, medicine, and industrial chemistry.


Step 2: Detailed Explanation:

A. Chlorophyll - Co: Mismatch. Chlorophyll contains Magnesium (\(Mg\)). Cobalt (\(Co\)) is found in Vitamin \(B_{12}\) (Cyanocobalamin).

B. Water hardness - EDTA: Correct. EDTA is used to estimate \(Ca^{2+}\) and \(Mg^{2+}\) ions through complexometric titration.

C. Photography - \([Ag(CN)_2]^-\): Mismatch. In black and white photography, the "fixing" agent is hypo (sodium thiosulfate), which forms the complex \([Ag(S_2O_3)_2]^{3-}\). \([Ag(CN)_2]^-\) is used in the extraction of silver and gold.

D. Wilkinson catalyst: Correct. Its formula is \([(Ph_3P)_3RhCl]\) and it is used for the hydrogenation of alkenes.

E. Chelating ligand - D-Penicillamine: Correct. It is a chelating ligand used in medical treatment for copper poisoning.


Step 4: Final Answer:

The mismatched pairs are A and C.
Quick Tip: Important complexes to remember: Haemoglobin (\(Fe\)), Chlorophyll (\(Mg\)), Vitamin \(B_{12}\) (\(Co\)), Cis-platin (\(Pt\)).

Step 1: Understanding the Concept:

Ozone depletion in the stratosphere is a catalytic process involving free radicals that decompose \(O_3\) into \(O_2\).


Step 2: Detailed Explanation:

Chlorofluorocarbons (CFCs) are stable in the lower atmosphere but reach the stratosphere where they are broken down by high-energy UV radiation.

The reaction releases chlorine free radicals (\(Cl^\bullet\)):
\[ CF_2Cl_2 \xrightarrow{UV} Cl^\bullet + CF_2Cl^\bullet \]

The \(Cl^\bullet\) radical then reacts with ozone to form chlorine monoxide and oxygen:
\[ Cl^\bullet + O_3 \rightarrow ClO^\bullet + O_2 \]

The \(ClO^\bullet\) then reacts with atomic oxygen to regenerate the \(Cl^\bullet\) radical:
\[ ClO^\bullet + O \rightarrow Cl^\bullet + O_2 \]

Thus, a single chlorine radical can destroy thousands of ozone molecules.


Step 4: Final Answer:

The chlorine free radical (\(Cl^\bullet\)) is the main cause of ozone depletion via UV radiation.
Quick Tip: Radicals like \(NO^\bullet\) also contribute, but the question specifies "in the presence of UV radiations," which is the classic mechanism for CFC-derived chlorine radicals.

Step 1: Understanding the Concept:

Dipole moment (\(\mu\)) increases when a molecule can exist in a highly polar resonance form that is significantly stabilized by aromaticity.


Step 2: Detailed Explanation:

In Cyclopropenone, the \(C=O\) bond is strongly polarized. The resonance contributor has a single bond \(C-O^-\) and a positive charge on the ring carbon.

This creates a 2-pi electron system in the three-membered ring. According to Huckel's rule (\(4n+2\)), with \(n=0\), this cyclopropenyl cation is aromatic.

Because this polar resonance form is exceptionally stable due to aromaticity, the molecule exists with a very high degree of charge separation, resulting in a large dipole moment (\(\sim 4.4 D\)).


Step 4: Final Answer:

Cyclopropenone has the highest dipole moment due to its aromatic-stabilized dipolar resonance form.
Quick Tip: Look for cases where the separation of charge (\(+\) and \(-\)) makes a ring aromatic. Such molecules (like cyclopropenone, tropone, or azulene) always have high dipole moments.

Step 1: Understanding the Concept:

Anhydrous \(AlCl_3\) in the presence of \(HCl\) is a standard reagent used for the isomerization of alkanes.


Step 2: Detailed Explanation:

The starting material is n-hexane (\(CH_3(CH_2)_4CH_3\)).

When n-alkanes are heated with \(AlCl_3\) and \(HCl\), they undergo skeletal rearrangement to form branched-chain isomers.

n-Hexane isomerizes primarily to a mixture of 2-methylpentane and 3-methylpentane.

Option (D) represents 2-methylpentane (\(CH_3-CH(CH_3)-CH_2-CH_2-CH_3\)).


Step 4: Final Answer:

The major product 'X' is 2-methylpentane.
Quick Tip: Isomerization of alkanes increases the octane number of fuels, making them more efficient for combustion in engines.

Step 1: Understanding the Concept:

Substitution reactions depend on the nucleophile strength and the potential for carbocation rearrangement.


Step 2: Detailed Explanation:

The substrate is 2-methyl propyl bromide (isobutyl bromide), which is a primary alkyl halide.

1. Reaction with \(C_2H_5O^-\): Ethoxide is a strong nucleophile. Primary halides with strong nucleophiles prefer the \(S_N2\) mechanism. Because \(S_N2\) is a concerted one-step process, there is no rearrangement. The product is isobutyl ethyl ether (A).

2. Reaction with \(C_2H_5OH\): Ethanol is a weak nucleophile and acts as a solvent (solvolysis). This favors the \(S_N1\) mechanism.

- First, the \(1^\circ\) isobutyl carbocation \(((CH_3)_2CHCH_2^+)\) is formed.

- It immediately undergoes a 1,2-hydride shift to form the much more stable \(3^\circ\) tert-butyl carbocation \(((CH_3)_3C^+)\).

- Ethanol then attacks this \(3^\circ\) cation to form tert-butyl ethyl ether (B).


Step 4: Final Answer:

Product A is formed by \(S_N2\) (isobutyl ethyl ether) and Product B is formed by \(S_N1\) with rearrangement (tert-butyl ethyl ether).
Quick Tip: Strong nucleophiles promote \(S_N2\) (no rearrangement). Weak nucleophiles/solvents promote \(S_N1\) (rearrangement likely if a more stable cation can form).

Step 1: Understanding the Concept:

This reaction involves an acid-catalyzed dehydration of a tertiary alcohol situated at the junction of two four-membered rings.

Four-membered rings have significant angle strain, and their expansion to five-membered rings is highly thermodynamically favorable.


Step 2: Key Formula or Approach:

The mechanism follows these stages:

1. Protonation of the hydroxyl group by \(H^+\).

2. Loss of a water molecule to form a tertiary carbocation at the bridgehead position.

3. Ring expansion to relieve strain.


Step 3: Detailed Explanation:

Initially, we have a spiro or fused system of two cyclobutane rings (A and B) with an \(OH\) group.

Upon treatment with \(H^+\), the alcohol is protonated to form \(R-OH_2^+\).

Loss of \(H_2O\) creates a carbocation.

A bond from ring A migrates to the carbocation center, expanding ring A from a 4-membered ring to a 5-membered ring.

This generates a new carbocation on the adjacent carbon.

To further stabilize the system and reduce the strain in the remaining 4-membered ring B, a bond from ring B then migrates, expanding it into a 5-membered ring as well.

The final stable product consists of two fused five-membered rings (pentalene derivative).


Step 4: Final Answer:

Both four-membered rings undergo expansion to minimize angle strain, eventually resulting in two five-membered rings.
Quick Tip: Ring expansion is driven by the relief of angle strain. Moving from 3 \(\rightarrow\) 4 or 4 \(\rightarrow\) 5 membered rings is very common when a carbocation is formed adjacent to the ring. 5 and 6 membered rings are the most stable "sweet spots" in organic chemistry.

Step 1: Understanding the Concept:

The starting material is a cyclic amide, specifically a \(\beta\)-lactam or \(\gamma\)-lactam derivative (depending on the ring size shown).

Base-catalyzed hydrolysis of lactams (cyclic amides) breaks the \(C-N\) bond to yield amino acids.


Step 2: Detailed Explanation:

The structure shown is a cyclic amide with a methyl group on nitrogen and a carbonyl group in the ring.

1. Reaction with NaOH, \(\Delta\): The hydroxide ion (\(OH^-\)) attacks the electrophilic carbonyl carbon.

2. The tetrahedral intermediate collapses, leading to the cleavage of the \(N-CO\) bond (the amide bond).

3. This opens the ring. For a 4-membered lactam, you get a 3-carbon chain between the nitrogen and the carboxylate. If it's a 5-membered lactam, you get a 4-carbon chain.

4. Acidification (\(H^+\)): The resulting carboxylate salt (\(-COO^-Na^+\)) is protonated to form a carboxylic acid (\(-COOH\)).

Looking at the carbon count in the starting ring (which appears to be 5-membered including the Nitrogen), the hydrolysis product is an \(\omega\)-amino acid.


Step 4: Final Answer:

The ring opening of the lactam yields the linear amino acid \(MeHN-(CH_2)_n-COOH\).
Quick Tip: Lactams behave just like acyclic amides but lead to bifunctional linear molecules (amino acids) upon hydrolysis. Heat (\(\Delta\)) is usually required to drive the reaction forward.

Step 1: Understanding the Concept:

This is a Pinacol-Pinacolone type rearrangement or a Demjanov-like rearrangement involving an alcohol and an amine in a complex fused/spiro system.


Step 3: Detailed Explanation:

The starting material has a tertiary alcohol and a primary amine group.

1. Reaction with \(HCl\): The amine is protonated to \(NH_3^+\). However, the alcohol is also a site for protonation. In these specific systems, protonation of \(OH\) followed by loss of water creates a carbocation at the carbon previously bearing the \(OH\).

2. Rearrangement: The carbocation is situated adjacent to a 5-membered spiro ring. To stabilize the cation and relieve ring strain, a bond migration occurs (ring expansion).

3. The 5-membered ring expands to a 6-membered ring.

4. Final Step with \(KOH\): The base helps in neutralizing any remaining salts or driving the elimination/deprotonation to yield the final stable hydrocarbon or amine scaffold.

Given the options, the most logical outcome for such a strained spiro system is the expansion to a more stable six-membered ring system.


Step 4: Final Answer:

The major product 'B' is the result of ring expansion of the 5-membered ring to a 6-membered ring.
Quick Tip: In complex polycyclic systems, always look for "ring expansion" to 6-membered rings if a carbocation is formed adjacent to a 4 or 5-membered ring. It is the most common thermodynamic driver in these problems.

Step 1: Understanding the Concept:

This question tests the knowledge of polymers, their monomers, and their structural classifications.


Step 2: Detailed Explanation:

1. Nylon 6: It is a polyamide produced by the ring-opening polymerization of its monomer, Caprolactam (III).

2. Vulcanized Rubber: The process of vulcanization involves heating natural rubber with sulfur, which introduces sulfur cross-links (II) between polymer chains to improve elasticity and strength.

3. cis-1, 4-polyisoprene: This is the chemical name for Natural Rubber (I). The trans-isomer is known as Gutta-percha.

4. Polychloroprene: It is a synthetic rubber commonly known as Neoprene (IV).


Step 4: Final Answer:

The correct matching is: a-III, b-II, c-I, d-IV.
Quick Tip: Monomers are the "building blocks." Caprolactam for Nylon 6 and Hexamethylene diamine + Adipic acid for Nylon 6,6 are extremely frequent exam topics.

Step 1: Understanding the Concept:

This reaction sequence describes the Kiliani-Fischer Synthesis used to extend the carbon chain of sugars, followed by oxidation to dicarboxylic acids (aldaric acids).


Step 2: Key Formula or Approach:

1. Addition of \(HCN\) to the aldehyde group creates a new chiral center at C2.

2. Hydrolysis of the nitrile (\(-CN\)) gives a carboxylic acid (\(-COOH\)).

3. Oxidation of the primary alcohol (\(-CH_2OH\)) by \(HNO_3\) also gives a carboxylic acid (\(-COOH\)).


Step 3: Detailed Explanation:

Starting with \(D\)-Glyceraldehyde, the configuration at C2 (now C3 in the product) is fixed (\(OH\) on the right).

Step (i) adds a new carbon and creates a new chiral center at the new C2. Two diastereomers are formed:

- Isomer 1: \(OH\) groups at C2 and C3 are on the same side (Erythro-type).

- Isomer 2: \(OH\) groups at C2 and C3 are on opposite sides (Threo-type).

After steps (ii) and (iii), both the top and bottom carbons become \(-COOH\), forming Tartaric Acid.

- The Erythro-type isomer becomes Meso-tartaric acid (the molecule has a plane of symmetry, making it optically inactive).

- The Threo-type isomer becomes \(L\)-(+)-tartaric acid (or \(D\)-tartaric acid depending on orientation), which is optically active.


Step 4: Final Answer:

The reaction produces a mixture of one meso compound and one optically active compound.
Quick Tip: Kiliani-Fischer synthesis always produces a pair of diastereomers because the addition to the carbonyl group is not stereoselective. If the resulting dicarboxylic acid has a plane of symmetry, it's Meso.

Step 1: Understanding the Concept:

In an acidic medium, potassium permanganate (\(KMnO_4\)) acts as a strong oxidizing agent.

It reacts with ferrous ammonium sulphate (\(FeSO_4(NH_4)_2SO_4 \cdot 6H_2O\)), also known as Mohr's salt, in a redox reaction where \(Fe^{2+}\) is oxidized to \(Fe^{3+}\) and \(Mn^{7+}\) is reduced to \(Mn^{2+}\).


Step 2: Key Formula or Approach:

The complete balanced molecular equation for the reaction is:
\[ 2KMnO_4 + 8H_2SO_4 + 10[FeSO_4(NH_4)_2SO_4 \cdot 6H_2O] \rightarrow K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 10(NH_4)_2SO_4 + 68H_2O \]


Step 3: Detailed Explanation:

From the balanced chemical equation:

1. For every 2 moles of \(KMnO_4\), 8 moles of \(H_2SO_4\) are consumed, producing 8 moles of water from the reduction process (\(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\)).

2. Additionally, the Mohr's salt used is a hexahydrate, meaning each molecule contains 6 molecules of water of crystallization.

3. Since 10 molecules of Mohr's salt are required to react with 2 molecules of \(KMnO_4\), the total water of crystallization released is:
\[ 10 \times 6 = 60 molecules \]

4. Total water molecules produced in the final equation:
\[ 8 (from redox) + 60 (from Mohr's salt) = 68 molecules \]


Step 4: Final Answer:

The total number of water molecules produced for 2 molecules of \(KMnO_4\) is 68.
Quick Tip: In redox titrations involving hydrates, always account for the water of crystallization in the total count of water molecules in the final balanced equation.

Step 1: Understanding the Concept:

For a real gas, the ideal gas law is modified using the compressibility factor (\(Z\)), defined as \(Z = \frac{PV}{nRT}\).

Since the quantity (moles \(n\)) of the gas remains constant, we can relate the two states of the gas.


Step 2: Key Formula or Approach:

Using the equation for a fixed amount of gas:
\[ \frac{P_1 V_1}{Z_1 T_1} = \frac{P_2 V_2}{Z_2 T_2} \]


Step 3: Detailed Explanation:

Given data for State 1:
\(P_1 = 100 atm\), \(V_1 = 0.15 dm^3\), \(T_1 = 500 K\), \(Z_1 = 1.07\)

Given data for State 2:
\(P_2 = 300 atm\), \(T_2 = 300 K\), \(Z_2 = 1.4\), \(V_2 = ?\)

Substituting the values into the relation:
\[ \frac{100 \times 0.15}{1.07 \times 500} = \frac{300 \times V_2}{1.4 \times 300} \]
\[ \frac{15}{535} = \frac{V_2}{1.4} \]
\[ V_2 = \frac{15 \times 1.4}{535} = \frac{21}{535} \approx 0.039252 dm^3 \]

To express this in terms of \(10^{-4} dm^3\):
\[ V_2 = 392.52 \times 10^{-4} dm^3 \]


Step 4: Final Answer:

Rounding to the nearest integer, we get 393.
Quick Tip: Whenever conditions for a real gas change, remember that \(n = \frac{PV}{ZRT}\) is constant. This allows you to skip calculating the number of moles directly.

Step 1: Understanding the Concept:

The enthalpy of a reaction (\(\Delta H_r\)) can be calculated using the bond enthalpies of reactants and products.
\[ \Delta H_r = \sum Bond enthalpy (Reactants) - \sum Bond enthalpy (Products) \]


Step 2: Key Formula or Approach:

Reaction: \(A_2 + B_2 \rightarrow 2AB\)
\(\Delta H_r = 2 \times \Delta H_f^\circ = 2 \times (-200) = -400 kJ mol^{-1}\)


Step 3: Detailed Explanation:

Let the bond enthalpy of \(A_2\) be \(x\).

According to the given ratio 1:0.5:1:

Bond enthalpy of \(A_2\) = \(x\)

Bond enthalpy of \(B_2\) = \(0.5x\)

Bond enthalpy of \(AB\) = \(x\)

Applying the enthalpy formula:
\[ \Delta H_r = [BE(A_2) + BE(B_2)] - [2 \times BE(AB)] \]
\[ -400 = [x + 0.5x] - [2 \times x] \]
\[ -400 = 1.5x - 2x \]
\[ -400 = -0.5x \]
\[ x = \frac{400}{0.5} = 800 kJ mol^{-1} \]


Step 4: Final Answer:

The bond enthalpy of \(A_2\) is 800 kJ mol\(^{-1}\).
Quick Tip: Always double-check if the given \(\Delta H\) is for "formation" (per mole of product) or for the reaction as written. In this case, \(\Delta H_r = 2 \Delta H_f\) because 2 moles of \(AB\) are formed.

Step 1: Understanding the Concept:

Solutions having the same osmotic pressure at the same temperature are called isotonic solutions.

For non-electrolytes, isotonicity implies equal molar concentrations (\(C_1 = C_2\)).


Step 2: Key Formula or Approach:

Osmotic pressure \(\pi = CRT\).

Since \(\pi_A = \pi_{glucose}\) at the same \(T\), we have \(C_A = C_{glucose}\).


Step 3: Detailed Explanation:

1. Molarity of glucose solution = \(0.05 M\).

2. Molarity of non-electrolyte (A):
\[ C_A = \frac{Mass of A}{Molecular mass (M) \times Volume in L} \]

Given mass = \(12 g\), Volume = \(1000 mL = 1 L\).
\[ C_A = \frac{12}{M \times 1} \]

3. Equating the concentrations:
\[ \frac{12}{M} = 0.05 \]
\[ M = \frac{12}{0.05} = 240 g mol^{-1} \]


Step 4: Final Answer:

The molecular mass of the non-electrolyte A is 240 g.
Quick Tip: For isotonic solutions of non-electrolytes, the molar concentration must be equal. If one is an electrolyte, multiply its concentration by the van't Hoff factor (\(i\)).

Step 1: Understanding the Concept:

When two solutions are mixed, the concentrations of ions are diluted. The ionic product (\(Q\)) determines the state of saturation relative to the solubility product (\(K_{sp}\)).


Step 2: Key Formula or Approach:

1. Calculate the final concentration of each ion after mixing.

2. Calculate the ionic product \(Q = [Ba^{2+}][F^-]^2\).

3. Find the ratio \(\frac{Q}{K_{sp}}\).


Step 3: Detailed Explanation:

Total volume after mixing = \(25.0 + 25.0 = 50.0 mL\).

Final concentration of \(Ba^{2+}\) ions:
\[ [Ba^{2+}] = \frac{0.050 M \times 25.0 mL}{50.0 mL} = 0.025 M \]

Final concentration of \(F^-\) ions:
\[ [F^-] = \frac{0.020 M \times 25.0 mL}{50.0 mL} = 0.010 M \]

Calculating the ionic product (\(Q\)):
\[ Q = [Ba^{2+}][F^-]^2 = (0.025) \times (0.010)^2 \]
\[ Q = 0.025 \times 10^{-4} = 2.5 \times 10^{-6} \]

Calculating the ratio:
\[ Ratio = \frac{Q}{K_{sp}} = \frac{2.5 \times 10^{-6}}{0.5 \times 10^{-6}} = \frac{2.5}{0.5} = 5 \]


Step 4: Final Answer:

The ratio of the ionic product to \(K_{sp}\) is 5.
Quick Tip: Always account for the dilution effect when mixing solutions before calculating the ionic product. If \(Q > K_{sp}\), a precipitate will form.

Step 1: Understanding the Concept:

Faraday's first law of electrolysis states that the mass of a substance deposited is proportional to the quantity of electricity passed (\(w = Zit\)).


Step 2: Key Formula or Approach:

1. Mass deposited (\(w\)) = Volume \(\times\) Density = (Area \(\times\) thickness) \(\times\) Density.

2. \(w = \frac{E \cdot I \cdot t}{F}\), where \(E = \frac{Molar mass}{Valency factor}\).


Step 3: Detailed Explanation:

Area = \(100 cm^2\)

Thickness = \(0.001 mm = 10^{-4} cm\)

Volume of nickel = \(100 \times 10^{-4} = 0.01 cm^3\)

Mass of nickel deposited (\(w\)) = \(0.01 cm^3 \times 10 g/cm^3 = 0.1 g\)

For \(Ni(NO_3)_2\), Nickel exists as \(Ni^{2+}\), so \(n = 2\).

Equivalent weight of \(Ni = \frac{60}{2} = 30\).

Using Faraday's formula:
\[ 0.1 = \frac{30 \times 2 \times x}{96500} \]
\[ 60x = 0.1 \times 96500 = 9650 \]
\[ x = \frac{9650}{60} \approx 160.83 s \]


Step 4: Final Answer:

Rounding to the nearest integer, the value of \(x\) is 161.
Quick Tip: Ensure all physical quantities like thickness and area are in consistent units (cm and cm\(^2\)) before calculating volume and mass.

Step 1: Understanding the Concept:

For a first-order reaction, the time required to complete a certain percentage of the reaction depends on the half-life (\(t_{1/2}\) or \(t_{50}\)).


Step 2: Key Formula or Approach:

The amount of reactant remaining after \(n\) half-lives is given by:
\[ [A] = \frac{[A]_0}{2^n} \]


Step 3: Detailed Explanation:

1. For 50% completion, one half-life has passed.
\[ t_{50} = 1 \times t_{1/2} \]

2. For 87.5% completion, the amount remaining is:
\[ 100% - 87.5% = 12.5% \]

3. Expressing the remaining amount in terms of the initial amount:
\[ 12.5% = \frac{100%}{8} = \frac{100%}{2^3} \]

4. This means 3 half-lives have passed.
\[ t_{87.5} = 3 \times t_{1/2} \]

5. Comparing with the given relation \(t_{87.5} = x \times t_{50}\):
\[ 3 \times t_{1/2} = x \times t_{1/2} \Rightarrow x = 3 \]


Step 4: Final Answer:

The value of \(x\) is 3.
Quick Tip: Useful shortcuts for first-order reactions:
\(t_{75%} = 2 \times t_{1/2}\)
\(t_{87.5%} = 3 \times t_{1/2}\)
\(t_{93.75%} = 4 \times t_{1/2}\)

Step 1: Understanding the Concept:

Quantitative analysis of carbon and hydrogen is done by measuring the amount of \(CO_2\) and \(H_2O\) produced during combustion.


Step 2: Key Formula or Approach:

1. \(% C = \frac{12}{44} \times \frac{Mass of CO_2}{Mass of compound} \times 100\)

2. \(% H = \frac{2}{18} \times \frac{Mass of H_2O}{Mass of compound} \times 100\)


Step 3: Detailed Explanation:

Let the mass of the organic compound be \(w\).

Given \(% C = 24\).
\[ 24 = \frac{12}{44} \times \frac{0.220}{w} \times 100 \]
\[ 24 = \frac{12}{44} \times \frac{22}{w} = \frac{6}{w} \]
\[ w = \frac{6}{24} = 0.25 g \]

Now, calculate the % of Hydrogen:
\[ % H = \frac{2}{18} \times \frac{0.126}{0.25} \times 100 \]
\[ % H = \frac{1}{9} \times \frac{0.126}{0.25} \times 100 = \frac{0.014}{0.25} \times 100 = 5.6% \]

To express in \(\times 10^{-1}\):
\[ 5.6 = 56 \times 10^{-1} \]


Step 4: Final Answer:

The value is 56.
Quick Tip: Use the provided percentage of one element to find the total mass of the compound first, then use that mass to find the percentage of the other element.

Step 1: Understanding the Concept:

Dehydration of alcohols in acidic medium proceeds through a carbocation intermediate. Secondary carbocations can rearrange to more stable tertiary carbocations via shifts.


Step 3: Detailed Explanation:

The starting alcohol is 2,2,4-trimethylpentan-3-ol.

Protonation and loss of water leads to the secondary carbocation: \(Me_3C-\overset{+}{C}H-CHMe_2\).

Rearrangement to tertiary carbocations occurs in two main ways:

1. Methyl Shift: Creates the 2,3,4-trimethylpentan-2-yl cation (\(Me_2\overset{+}{C}-CH(Me)-CHMe_2\)). This cation yields 3 alkenes:

- 2,3,4-trimethylpent-1-ene (chiral, so 2 enantiomers).

- 2,3,4-trimethylpent-2-ene (no E/Z isomers due to identical groups on one end).

Total from this path = 3 products.

2. Hydride Shift: Creates the 2,2,4-trimethylpentan-2-yl cation (\(Me_3C-CH_2-\overset{+}{C}Me_2\)). This cation yields 2 alkenes:

- 2,4,4-trimethylpent-1-ene.

- 2,4,4-trimethylpent-2-ene.

Total from this path = 2 products.

Summing the constitutional and stereoisomers derived strictly from tertiary rearranged cations:

Total products = \(3 + 2 = 5\).


Step 4: Final Answer:

The total number of possible products formed by tertiary carbocations is 5.
Quick Tip: When counting "possible products," check for both constitutional isomers and stereoisomers (E/Z and enantiomers) unless the question specifies otherwise.

Step 1: Understanding the Concept:

The concentration of a base can be found by titrating it against a standard acid. Once the base concentration is known, it can be used to determine the concentration of an unknown acid solution.


Step 2: Key Formula or Approach:

At the equivalence point:
\[ M_{acid} \cdot V_{acid} \cdot n_{acid} = M_{base} \cdot V_{base} \cdot n_{base} \]


Step 3: Detailed Explanation:

Part 1: Standardizing the Base \(Ca(OH)_2\)

Volume of \(Ca(OH)_2\) used for \(HCl\) = \(35.5 - 25.5 = 10 mL\).

Applying the law of equivalents for \(HCl\) (\(n=1\)) and \(Ca(OH)_2\) (\(n=2\)):
\[ M_{HCl} \cdot V_{HCl} \cdot 1 = M_{Ca(OH)_2} \cdot V_{Ca(OH)_2} \cdot 2 \]
\[ 0.5 \times 20 \times 1 = M_{Ca(OH)_2} \times 10 \times 2 \]
\[ 10 = 20 \cdot M_{Ca(OH)_2} \Rightarrow M_{Ca(OH)_2} = 0.5 M \]


Part 2: Finding concentration of \(H_2SO_4\)

Reaction with 10 mL \(H_2SO_4\) (\(n=2\)) used 20 mL \(Ca(OH)_2\) (\(n=2\)):
\[ M_{H_2SO_4} \cdot V_{H_2SO_4} \cdot 2 = M_{Ca(OH)_2} \cdot V_{Ca(OH)_2} \cdot 2 \]
\[ M_{H_2SO_4} \times 10 = 0.5 \times 20 \]
\[ 10 \cdot M_{H_2SO_4} = 10 \Rightarrow M_{H_2SO_4} = 1 M \]


Step 4: Final Answer:

The concentration of \(H_2SO_4\) is 1 M.
Quick Tip: In acid-base titrations, identify the \(n\)-factor (basicity of acid or acidity of base) correctly. For \(H_2SO_4\) and \(Ca(OH)_2\), the \(n\)-factor is 2.