Step 1: Understanding the Concept:

This problem involves stoichiometry based on the reaction between Zinc and Sulfuric acid to produce Hydrogen gas.

We must first identify the limiting reagent by calculating the number of moles of each reactant available.


: Key Formula or Approach:

Chemical Equation: \[ Zn + H_2SO_4 \to ZnSO_4 + H_2 \uparrow \]

Mass of pure substance \(= Volume \times Density \times Purity / 100\).

Number of moles \(= Mass / Molar Mass\).

Volume of gas at STP \(= moles \times 22.4 L mol^{-1}\).


Step 2: Detailed Explanation:

1. Calculation of moles of Zn:

Mass of Zinc \(= 20 g\).

Molar mass of Zinc \(= 65 g mol^{-1}\).

Moles of Zinc \(= \frac{20}{65} \approx 0.3077 mol\).



2. Calculation of moles of pure H_2SO_4:

Volume of solution \(= 50 mL\).

Mass of solution \(= 50 mL \times 1.3 g mL^{-1} = 65 g\).

Mass of pure \(H_2SO_4 = 65 g \times \frac{50}{100} = 32.5 g\).

Molar mass of \(H_2SO_4 = (2 \times 1) + 32 + (4 \times 16) = 98 g mol^{-1}\).

Moles of \(H_2SO_4 = \frac{32.5}{98} \approx 0.3316 mol\).



3. Limiting Reagent Determination:

From the balanced equation, 1 mole of Zn reacts with 1 mole of \(H_2SO_4\).

Since we have 0.3077 moles of Zn and 0.3316 moles of \(H_2SO_4\), Zinc is the limiting reagent.



4. Volume of H_2 gas at STP:

Moles of \(H_2\) produced \(=\) Moles of Zn used \(= 0.3077 mol\).

Volume at STP \(= 0.3077 \times 22.4 L \approx 6.89248 L\).


Step 3: Final Answer:

The volume of hydrogen gas liberated at STP is 6.892 L.
Quick Tip: Always check for the limiting reagent in stoichiometric problems. If the volume of one reactant is provided with density and purity, first convert it to mass of pure substance before calculating moles. Standard molar volume at STP is taken as 22.4 L/mol in most JEE problems unless 22.7 L/mol is specified.

Step 1: Understanding the Concept:

This question tests fundamental principles of atomic structure, specifically the \((n+l)\) rule for orbital energy, orbital sizes, and the calculation of radial nodes.


Step 2: Detailed Explanation:

Statement A: This is the Aufbau principle's \((n+l)\) rule. If two orbitals have the same \((n+l)\) value, the one with the lower principal quantum number (\(n\)) is lower in energy (e.g., 3d has \(n+l=5\) and 4p has \(n+l=5\), but 3d is filled first). This is True.



Statement B: As the atomic number (\(Z\)) increases, the effective nuclear charge increases. This causes the orbital to be attracted more strongly toward the nucleus, which actually decreases the energy (makes it more negative). Thus, the statement is False.



Statement C: Orbital size is primarily determined by the principal quantum number \(n\). Higher \(n\) means the electron is likely further from the nucleus. Since \(n=3\) for \(3p_x\) and \(n=2\) for \(2p_x\), \(3p_x\) is larger. This is True.



Statement D: Radial nodes are calculated as \(n - l - 1\).

5f: \(5 - 3 - 1 = 1\).

6s: \(6 - 0 - 1 = 5\).

4d: \(4 - 2 - 1 = 1\).

5p: \(5 - 1 - 1 = 3\).

5d: \(5 - 2 - 1 = 2\).

Since the 5d orbital does have 2 radial nodes, the statement "none of the orbitals have 2 radial nodes" is False.


Step 3: Final Answer:

Statements A and C are correct. Therefore, the answer is Option (B).
Quick Tip: Remember the radial node formula: Radial nodes \(= n - l - 1\). Total nodes \(= n - 1\). Angular nodes \(= l\). These are high-yield formulas for competitive exams.

Step 1: Understanding the Concept:

Covalent radius is defined as half of the distance between the nuclei of two bonded atoms in a homonuclear molecule. For a heteronuclear molecule AB, the bond length is approximately the sum of the covalent radii of the two atoms.


Step 2: Detailed Explanation:

The covalent bond length (\(d_{AB}\)) in a molecule AB is the distance between the centers of the nuclei of atom A and atom B.

By definition, if \(r_A\) is the covalent radius of atom A and \(r_B\) is the covalent radius of atom B, then:
\[ Covalent Bond Length = r_A + r_B \]

In a simple diatomic molecule model, the "total length" of the molecule (from the outer boundary of the electron cloud of A to that of B) is effectively defined by the distance spanning both covalent radii when bonded. Thus, it is also taken as \((r_A + r_B)\).


Step 3: Final Answer:

Both the covalent bond length and the total length of the AB molecule are represented as \((r_A + r_B)\).
Quick Tip: For heteronuclear molecules, the Bond Length is simply \(r_A + r_B - 0.09 (\chi_A - \chi_B)\) according to the Schomaker-Stevenson rule, but in basic models, we simplify it to just the sum of radii.

Step 1: Understanding the Concept:

Gibbs Free Energy change for a reaction can be calculated using the relation \(\Delta_r G^\ominus = \Delta_r H^\ominus - T \Delta_r S^\ominus\). We first need to calculate the standard enthalpy and entropy change for the reaction using the provided formation data.


: Key Formula or Approach:
\[ \Delta_r H^\ominus = \sum \Delta_f H^\ominus (products) - \sum \Delta_f H^\ominus (reactants) \]
\[ \Delta_r S^\ominus = \sum S^\ominus (products) - \sum S^\ominus (reactants) \]
\[ \Delta_r G^\ominus = \Delta_r H^\ominus - T \Delta_r S^\ominus \]


Step 2: Detailed Explanation:

1. Calculation of \(\Delta_r H^\ominus\):

Reaction: \(X_2(g) + Y_2(g) \rightleftharpoons 2XY(g)\)
\(\Delta_r H^\ominus = [2 \times \Delta_f H^\ominus(XY)] - [\Delta_f H^\ominus(X_2) + \Delta_f H^\ominus(Y_2)]\)
\(\Delta_r H^\ominus = [2 \times 42] - [8 + 80] = 84 - 88 = -4 kJ mol^{-1}\).



2. Calculation of \(\Delta_r S^\ominus\):
\(\Delta_r S^\ominus = [2 \times S^\ominus(XY)] - [S^\ominus(X_2) + S^\ominus(Y_2)]\)
\(\Delta_r S^\ominus = [2 \times 200] - [140 + 250] = 400 - 390 = 10 J mol^{-1} K^{-1}\).



3. Calculation of \(\Delta_r G^\ominus\):

Convert \(\Delta_r S^\ominus\) to kJ units: \(10 J = 0.010 kJ\).
\(T = 600 K\).
\(\Delta_r G^\ominus = (-4 kJ mol^{-1}) - (600 K \times 0.010 kJ mol^{-1} K^{-1})\)
\(\Delta_r G^\ominus = -4 - 6 = -10 kJ mol^{-1}\).


Step 3: Final Answer:

The \(\Delta_r G^\ominus\) for the reaction at 600 K is \(-10\) kJ \(mol^{-1}\).
Quick Tip: Pay close attention to units! Enthalpy is usually given in kJ/mol while Entropy is in J/mol\(\cdot\)K. Forgetting to divide the entropy term by 1000 is the most common mistake in these problems.

Step 1: Understanding the Concept:

Molar heat capacity (\(C_m\)) depends on the degrees of freedom available to the substance in its current phase (solid, liquid, or gas) and its molecular complexity.


Step 2: Detailed Explanation:

1. Helium (g): Helium is a monatomic gas. Its molar heat capacity at constant pressure (\(C_{p,m}\)) is approximately \(\frac{5}{2}R \approx 20.8 J mol^{-1} K^{-1}\).

2. Copper (s): According to the Dulong-Petit Law, the molar heat capacity of most solid elements is approximately \(3R \approx 24.9 J mol^{-1} K^{-1}\).

3. Bromine (l): Bromine is a liquid at room temperature and is diatomic (\(Br_2\)). Liquids generally have much higher molar heat capacities than solids or gases because they possess many vibrational, translational, and rotational modes, along with strong intermolecular interactions. For Bromine (l), \(C_{p,m} \approx 75.7 J mol^{-1} K^{-1}\).



Comparing the values: \(75.7 > 24.9 > 20.8\).

Therefore, the order is Bromine(l) \(>\) Copper(s) \(>\) Helium(g).


Step 3: Final Answer:

The correct order of molar heat capacities is Bromine(l) \(>\) Copper(s) \(>\) Helium(g).
Quick Tip: General Rule: Liquid phase typically has the highest heat capacity due to complex interactions and degrees of freedom, followed by the solid phase (Dulong-Petit law), and finally the gas phase (based on kinetic theory).

Step 1: Understanding the Concept:
\(K_p\) is defined as the product of the partial pressures of the products divided by the partial pressure of the reactant at equilibrium. Partial pressure of a gas is the product of its mole fraction and the total pressure.


: Key Formula or Approach:
\[ P_i = \chi_i \times P_{total} \]
\[ K_p = \frac{P_B \cdot P_C}{P_A} \]


Step 2: Detailed Explanation:

Let's set up the equilibrium table:


\begin{tabular{lccc
& \(A(g)\) & \(\rightleftharpoons\) \(B(g)\) & \(+ \ C(g)\)

Initial moles: & \(a\) & 0 & 0

Change: & \(-x\) & \(+x\) & \(+x\)

At equilibrium: & \(a-x\) & \(x\) & \(x\)

\end{tabular

Total moles at equilibrium \(= (a - x) + x + x = a + x\).



Mole fractions at equilibrium:
\(\chi_A = \frac{a - x}{a + x}\)
\(\chi_B = \frac{x}{a + x}\)
\(\chi_C = \frac{x}{a + x}\)



Partial Pressures:
\(P_A = \frac{a - x}{a + x} p\)
\(P_B = \frac{x}{a + x} p\)
\(P_C = \frac{x}{a + x} p\)



Calculating \(K_p\):
\(K_p = \frac{\left( \frac{x}{a + x} p \right) \left( \frac{x}{a + x} p \right)}{\left( \frac{a - x}{a + x} p \right)} = \frac{x^2 p^2 / (a + x)^2}{(a - x)p / (a + x)}\)
\(K_p = \frac{x^2 p}{(a + x)(a - x)} = \frac{x^2}{a^2 - x^2} p\).


Step 3: Final Answer:

The expression for \(K_p\) is \(\frac{x^2}{a^2 - x^2} \times p\).
Quick Tip: Always simplify the total moles term first. If the number of moles increases (\(\Delta n_g > 0\)), \(K_p\) will be directly proportional to total pressure. If \(\Delta n_g < 0\), it is inversely proportional.

Step 1: Understanding the Concept:

We use the Nernst equation to find the cell potential. First, we identify the cathode and anode based on standard reduction potentials, then determine the overall cell reaction and standard EMF.


: Key Formula or Approach:

Standard Cell EMF: \(E^\ominus_{cell} = E^\ominus_{cathode} - E^\ominus_{anode}\).

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


Step 2: Detailed Explanation:

1. Identifying Electrodes:

Since \(E^\ominus_{Ag^+/Ag} (0.80 V) > E^\ominus_{Zn^{2+}/Zn} (-0.76 V)\), Silver acts as the cathode and Zinc acts as the anode.

Anode (Oxidation): \(Zn \to Zn^{2+} + 2e^-\)

Cathode (Reduction): \(2Ag^+ + 2e^- \to 2Ag\)

Overall reaction: \(Zn + 2Ag^+ \to Zn^{2+} + 2Ag\).

Here, \(n = 2\).



2. Calculating Standard EMF:
\(E^\ominus_{cell} = 0.80 V - (-0.76 V) = 1.56 V\).



3. Applying Nernst Equation:
\(E_{cell} = E^\ominus_{cell} - \frac{0.059}{2} \log \frac{[Zn^{2+}]}{[Ag^+]^2}\)
\(1.60 = 1.56 - \frac{0.059}{2} \log \frac{1}{x^2}\)
\(1.60 - 1.56 = -\frac{0.059}{2} \cdot (-2 \log x)\)
\(0.04 = 0.059 \log x\)
\(\log x = \frac{0.04}{0.059} = \frac{4}{5.9}\).


Step 3: Final Answer:

The value of \(\log [Ag^+]\) is \(\frac{4}{5.9}\).
Quick Tip: Watch the stoichiometry in the Nernst equation! If an ion has a coefficient (like 2 for \(Ag^+\)), its concentration must be squared in the reaction quotient \(Q\). Also, remember that \(\log (1/x^2) = -2 \log x\).

Step 1: Understanding the Concept:

Oxides are classified based on their reaction with water, acids, and bases into acidic (non-metals/high oxidation state metals), basic (low oxidation state metals), amphoteric (react with both), and neutral (react with neither).


Step 2: Detailed Explanation:

Analysis of Statement I:

1. \([Al_2O_3, Cr_2O_3]\): Both are amphoteric. (Same nature)

2. \([Cl_2O_7, Mn_2O_7]\): Both are strongly acidic (\(Mn\) in +7 is non-metallic in character). (Same nature)

3. \([Na_2O, V_2O_3]\): Both are basic (\(V\) in +3 oxidation state forms basic oxides). (Same nature)

4. \([CO, N_2O]\): Both are neutral oxides. (Same nature)

Total pairs \(= 4\). So, Statement I is True.



Analysis of Statement II:
\(Na_2O\) (alkali metal oxide) is strongly basic.
\(Al_2O_3\) is amphoteric.
\(CO\) is neutral.
\(Cl_2O_7\) (non-metal oxide) is strongly acidic.

Thus, \(Na_2O\) is the most basic and \(Cl_2O_7\) is the most acidic. So, Statement II is True.


Step 3: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: Transition metal oxides show a trend: low oxidation states are basic (\(V_2O_3\)), intermediate are amphoteric (\(V_2O_4\)), and high oxidation states are acidic (\(V_2O_5\), \(Mn_2O_7\)). Always check the oxidation state for d-block oxides.

Step 1: Understanding the Concept:

Statement I involves the reaction of amphoteric metals with strong bases. Statement II involves VSEPR theory and the prediction of molecular geometry.


Step 2: Detailed Explanation:

Analysis of Statement I:

Aluminium reacts with NaOH to form sodium tetrahydroxoaluminate(III) and hydrogen gas:
\[ 2Al + 2NaOH + 6H_2O \to 2Na[Al(OH)_4] + 3H_2 \uparrow \]

The predominant ion formed is the tetrahedral \([Al(OH)_4]^-\) ion, not the octahedral \([Al(OH)_6]^{3-}\). Thus, Statement I is False.



Analysis of Statement II:

1. \(ICl_4^-\): \(I\) has 7 valence electrons + 1 (charge) \(= 8\). 4 Bond pairs, 2 Lone pairs. Steric number \(= 6\) (\(sp^3d^2\)). Geometry: Square Planar.

2. \(ClO_3^-\): \(Cl\) has 7 valence electrons + 1 (charge) \(= 8\). 3 oxygen atoms use 6 electrons for bonds (assuming 1 double bond each). 1 Lone pair remains. Steric number \(= 4\). Geometry: Pyramidal.

3. \(IBr_2^-\): \(I\) has 7 valence electrons + 1 (charge) \(= 8\). 2 Bond pairs, 3 Lone pairs. Steric number \(= 5\) (\(sp^3d\)). Geometry: Linear.

Thus, Statement II is True.


Step 3: Final Answer:

Statement I is false but Statement II is true.
Quick Tip: For \(Al(OH)_x\), remember that in solution, it exists as \([Al(OH)_4]^-\). Also, for \(IBr_2^-\) and similar triatomic species with 3 lone pairs on the central atom (\(sp^3d\)), the lone pairs occupy equatorial positions, leaving the atoms in a linear arrangement.

Step 1: Understanding the Concept:

Statement I relates the strength of metallic bonding to electronic configuration. Statement II concerns the Crystal Field Splitting patterns for octahedral and tetrahedral complexes.


Step 2: Detailed Explanation:

Analysis of Statement I:

Transition metals have unpaired d-electrons which participate in interatomic metallic bonding. The greater the number of unpaired electrons, the stronger the metallic bond, and hence higher is the enthalpy of atomisation. This is True.



Analysis of Statement II:

1. \([Fe(H_2O)_6]^{3+}\) is an octahedral complex. In octahedral field, d-orbitals split into \(t_{2g} (d_{xy}, d_{xz}, d_{yz})\) and \(e_g (d_{x^2-y^2}, d_{z^2})\). The \(e_g\) set is higher in energy. The pattern given (\(d_{xy} = d_{xz} = d_{yz} < d_{x^2 - y^2} = d_{z^2}\)) is correct.

2. \([Ni(Cl)_4]^{2-}\) is a tetrahedral complex (due to weak field \(Cl^-\) ligand). In tetrahedral field, d-orbitals split into \(e (d_{x^2-y^2}, d_{z^2})\) and \(t_2 (d_{xy}, d_{xz}, d_{yz})\). The \(t_2\) set is higher in energy. The second pattern given (\(d_{x^2 - y^2} = d_{z^2} = d_{xy} < d_{xz} = d_{yz}\)) is incorrect because \(d_{xy}\) belongs to the \(t_2\) set, not \(e\).

Thus, Statement II is Incorrect.


Step 3: Final Answer:

Statement I is correct but Statement II is incorrect.
Quick Tip: Splitting Memory Hack:
Octahedral \(\to\) 2 orbitals up (\(d_{x^2-y^2}, d_{z^2}\)), 3 orbitals down.
Tetrahedral \(\to\) 3 orbitals up (\(d_{xy}, d_{xz}, d_{yz}\)), 2 orbitals down.
The order is exactly reversed!

Step 1: Understanding the Concept:

This question evaluates multiple concepts in coordination chemistry, including complex stability (chelate effect), hybridization (VBT/CFT), and types of isomerism/ligands.


Step 2: Detailed Explanation:

Statement A: Oxalate (\(C_2O_4^{2-}\)) is a chelating ligand, whereas \(OH^-\) and \(SCN^-\) are monodentate. Chelating complexes are significantly more stable than non-chelating ones due to the chelate effect. Thus, A is True.

Statement B: 'en' (ethylenediamine) is a bidentate chelating ligand, while \(NH_3\) is monodentate. According to the chelate effect, \([Cu(en)_2]^{2+}\) is more stable than \([Cu(NH_3)_4]^{2+}\). Thus, B is False.

Statement C: In \(K_4[Fe(CN)_6]\), iron is in the \(+2\) oxidation state (\(d^6\)). \(CN^-\) is a strong field ligand, causing electrons to pair up in the \(t_{2g}\) orbitals. This leaves two empty \(3d\) orbitals for hybridization. Thus, it undergoes \(d^2sp^3\) hybridization. Thus, C is True.

Statement D: \(NO_2^-\) is an ambidentate ligand (it can bind through N or O). Any complex containing an ambidentate ligand can exhibit linkage isomerism. Thus, D is True.

Statement E: \(NO_2^-\) and \(SCN^-\) are classic examples of ambidentate ligands because they have more than one potential donor atom. Thus, E is False.


Step 3: Final Answer:

Statements A, C, and D are correct.
Quick Tip: Chelation always increases stability. Always check for ambidentate ligands like \(NO_2^-, SCN^-, CN^-\) to identify linkage isomerism immediately.

Step 1: Understanding the Concept:

Purification techniques for organic compounds depend on physical properties like boiling point, volatility, and thermal stability.


Step 2: Detailed Explanation:

A. Simple distillation: Used for liquids that are stable at their boiling points and have a large difference in boiling points (typically \(>25\) K). (Matches II).

B. Fractional distillation: Used to separate liquids whose boiling points are close to each other, employing a fractionating column. (Matches IV).

C. Steam distillation: Specifically used for substances that are steam volatile and immiscible with water (e.g., aniline). (Matches I).

D. Distillation under reduced pressure (Vacuum Distillation): Used for liquids that have very high boiling points or those that decompose at or below their normal boiling points (e.g., glycerol). (Matches III).


Step 3: Final Answer:

The correct matching is A-II, B-IV, C-I, D-III.
Quick Tip: Vacuum distillation is the go-to for "thermally unstable" liquids. If the boiling points are "close," think "fractional."

Step 1: Understanding the Concept:

The stability of alkenes is primarily determined by the degree of substitution at the double bond. According to Saytzeff's rule and hyperconjugation, more substituted alkenes are generally more stable.


Step 2: Detailed Explanation:

Stability increases with the number of \(\alpha\)-hydrogens (hyperconjugative structures).

C. 2,3-Dimethylbut-2-ene: \((CH_3)_2C=C(CH_3)_2\). It is tetra-substituted and has 12 \(\alpha\)-H. (Most stable).

A. 2-Methylbut-2-ene: \((CH_3)_2C=CHCH_3\). It is tri-substituted and has 9 \(\alpha\)-H.

B. cis-But-2-ene: \(CH_3CH=CHCH_3\). It is di-substituted and has 6 \(\alpha\)-H.

D. Prop-1-ene: \(CH_3CH=CH_2\). It is mono-substituted and has 3 \(\alpha\)-H. (Least stable).

Order: C \(>\) A \(>\) B \(>\) D.


Step 3: Final Answer:

The stability order is C \(>\) A \(>\) B \(>\) D.
Quick Tip: Count the number of alkyl groups attached to the C=C bond. Stability: Tetra-substituted \(>\) Tri-substituted \(>\) Di-substituted \(>\) Mono-substituted.

Step 1: Understanding the Concept:

First, determine the molecular formula from the molar mass. Then, analyze the structure of the isomers to count the number of primary (\(1^\circ\)) carbon atoms.


Step 2: Detailed Explanation:

Molar mass of alkane \(C_nH_{2n+2}\) is \(14n + 2 = 72 \implies 14n = 70 \implies n = 5\). The formula is \(C_5H_{12}\).

Isomers of \(C_5H_{12}\):

1. n-pentane: \(CH_3-CH_2-CH_2-CH_2-CH_3\). Primary carbons = 2 (the ends).

2. 2-methylbutane (Isopentane): \(CH_3-CH(CH_3)-CH_2-CH_3\). The three \(CH_3\) groups at the ends are primary carbons. Total = 3.

3. 2,2-dimethylpropane (Neopentane): \(C(CH_3)_4\). All four methyl groups are primary carbons. Total = 4.

4. 1,1-dimethylcyclopropane has formula \(C_5H_{10}\), so its molar mass is 70, which is incorrect.


Step 3: Final Answer:

2-Methylbutane is the correct hydrocarbon with 3 primary carbons and molar mass 72.
Quick Tip: A primary carbon is attached to only one other carbon atom. In alkanes, every branch adds a new primary carbon (terminal methyl group).

Step 1: Understanding the Concept:

The sequence involves the complete hydrolysis of a nitrile to a carboxylic acid, followed by the Hell-Volhard-Zelinsky (HVZ) reaction.


Step 2: Detailed Explanation:

Steps (i) and (ii): Nitriles (\(R-CN\)) undergo complete hydrolysis in the presence of acid or base and heat to form carboxylic acids.
\(CH_3CH_2CN \xrightarrow{H_2O/OH^-/\Delta, H^+} CH_3CH_2COOH\) (Propanoic acid).

Steps (iii) and (iv): Carboxylic acids with \(\alpha\)-hydrogens react with \(Cl_2/Red P\) followed by hydrolysis to substitute the \(\alpha\)-hydrogen with a chlorine atom. This is the HVZ reaction.
\(CH_3CH_2COOH \xrightarrow{Cl_2/Red P, H_2O} CH_3CH(Cl)COOH\) (2-Chloropropanoic acid).


Step 3: Final Answer:

The major product P is 2-chloropropanoic acid.
Quick Tip: HVZ reaction is highly selective for the alpha-position of carboxylic acids. It will never substitute a hydrogen at the beta or gamma positions.

Step 1: Understanding the Concept:

Statement I refers to the nucleophilic addition-elimination reaction of carbonyls with ammonia derivatives. Statement II refers to the instability of hemiacetals in acidic media.


Step 2: Detailed Explanation:

Statement I: Acetaldehyde (\(CH_3CHO\)) reacts with semicarbazide (\(NH_2NHCONH_2\)) to form a semicarbazone. The reaction is a condensation (loss of \(H_2O\)) and occurs optimally at pH 3.5. Thus, Statement I is True.

Statement II: The given molecule is a hemiacetal (formed from benzaldehyde and methanol). Hemiacetals are unstable and readily revert to the parent aldehyde and alcohol in the presence of dilute acid. Thus, Statement II is True.


Step 3: Final Answer:

Both statements are true.
Quick Tip: Carbonyl + Semicarbazide \(\to\) Semicarbazone. Hemiacetals are like "half-way" houses; they go back to aldehydes easily with a little acid and water.

Step 1: Understanding the Concept:

Statement I describes the Hoffmann bromamide degradation reaction. Statement II describes the unusual nitration behavior of aniline.


Step 2: Detailed Explanation:

Statement I: The Hoffmann bromamide reaction converts an amide (\(RCONH_2\)) to a primary amine (\(RNH_2\)) with one less carbon. Benzamide (\(C_6H_5CONH_2\)) reacts to form Aniline (\(C_6H_5NH_2\)), not benzylamine (\(C_6H_5CH_2NH_2\)). Thus, Statement I is False.

Statement II: In strongly acidic nitrating mixtures, aniline is protonated to the anilinium ion (\(-NH_3^+\)), which is a strongly deactivating meta-directing group. This leads to 47% m-nitroaniline, whereas o-nitroaniline is only 2%. Thus, Statement II is True.


Step 3: Final Answer:

Statement I is false but Statement II is true.
Quick Tip: Hoffmann bromamide: AMIDE \(\to\) AMINE (minus 1 carbon). Aniline nitration is a "trap" question; always remember the high yield of meta product due to the anilinium ion.

Step 1: Understanding the Concept:

Tertiary structure refers to the overall folding of the polypeptide chain. It is highly sensitive to the surrounding environment.


Step 2: Detailed Explanation:

Option A: True. Tertiary structure describes the 3D shape, which is often globular or fibrous.

Option B: True. These are the various intermolecular and intramolecular forces that hold the 3D shape together.

Option C: Incorrect. Proteins undergo denaturation when exposed to significant changes in pH or temperature. This process disrupts the tertiary and secondary structures.

Option D: True. Protein folding follows a hierarchical path from primary to secondary to tertiary.


Step 3: Final Answer:

Statement (C) is incorrect.
Quick Tip: Denaturation destroys secondary and tertiary structures but \textbf{never} the primary structure (the sequence of amino acids).

Step 1: Understanding the Concept:

Statement I deals with the cyclic structure of carbohydrates. Statement II compares the configurations of the lower carbons in aldohexoses and ketohexoses.


Step 2: Detailed Explanation:

Statement I: Anomers are cyclic diastereomers that differ in configuration only at the hemiacetal/hemiketal carbon (\(C_1\) for glucose). \(\alpha\) and \(\beta\) glucose are indeed anomers. Thus, Statement I is True.

Statement II: In the open chain structures of D-glucose and D-fructose, the chiral centers at \(C_3, C_4\), and \(C_5\) have identical configurations (OH on right, H on left for \(C_4/C_5\) and OH on left for \(C_3\) in Fischer projection). This is why they yield the same osazone. Thus, Statement II is True.


Step 3: Final Answer:

Both Statement I and Statement II are true.
Quick Tip: Glucose and Fructose share the same "bottom half" (\(C_3\) to \(C_6\)). This explains many shared chemical properties.

Step 1: Understanding the Concept:
\(SO_2\) is a strong reducing agent. It reacts with oxidizing agents, often resulting in a characteristic color change.


Step 2: Detailed Explanation:

Acidified Potassium Dichromate (\(K_2Cr_2O_7\)) is orange in color. When it reacts with sulfur dioxide (\(SO_2\)), the dichromate ion is reduced to the chromic ion (\(Cr^{3+}\)), which is green.

Reaction:
\[ Cr_2O_7^{2-} (orange) + 3SO_2 + 2H^+ \to 2Cr^{3+} (green) + 3SO_4^{2-} + H_2O \]


Step 3: Final Answer:

The compound X is \(K_2Cr_2O_7\).
Quick Tip: Orange to Green = Dichromate reduction. Pink to Colorless = Permanganate reduction. This is a standard test for reducing gases like \(SO_2\) and \(H_2S\).

Step 1: Understanding the Concept:

In an octahedral crystal field, the five d-orbitals split into two sets: the lower energy \(t_{2g}\) set (consisting of \(d_{xy}, d_{yz}, d_{zx}\)) and the higher energy \(e_g\) set (consisting of \(d_{x^2-y^2}, d_{z^2}\)).

The distribution of electrons depends on the crystal field splitting energy (\(\Delta_o\)) relative to the pairing energy (P).


: Key Formula or Approach:

For high spin (weak field), electrons fill all orbitals singly before pairing starts (Hund's rule).

For low spin (strong field), electrons fill the lower energy \(t_{2g}\) orbitals completely before moving to \(e_g\).


Step 2: Detailed Explanation:

We analyze each electronic configuration:

1. \(d^3\): Electrons fill \(t_{2g}\) orbitals singly.

Configuration: \(t_{2g}^3 e_g^0\). Unpaired electrons (\(n\)) = 3.

2. \(d^4\) (low spin): Electrons stay in \(t_{2g}\) and pair up.

Configuration: \(t_{2g}^4 e_g^0\). Two electrons are unpaired, one pair is formed (\(n\)) = 2.

3. \(d^5\) (high spin): Electrons fill all five orbitals singly.

Configuration: \(t_{2g}^3 e_g^2\). Unpaired electrons (\(n\)) = 5.

4. \(d^6\) (high spin): Five electrons are unpaired, the sixth pairs in \(t_{2g}\).

Configuration: \(t_{2g}^4 e_g^2\). Unpaired electrons (\(n\)) = 4.

5. \(d^7\) (low spin): \(t_{2g}\) is completely filled (6 electrons), the 7th is in \(e_g\).

Configuration: \(t_{2g}^6 e_g^1\). Unpaired electrons (\(n\)) = 1.



Total unpaired electrons = \(3 + 2 + 5 + 4 + 1 = 15\).


Step 3: Final Answer:

The total number of unpaired electrons is 15.
Quick Tip: For octahedral complexes, \(d^1, d^2, d^3\) and \(d^8, d^9, d^{10}\) always have the same number of unpaired electrons regardless of the field strength. Differences only arise in \(d^4\) through \(d^7\) configurations.

Step 1: Understanding the Concept:

Grignard reagents react with compounds containing active hydrogen (like water) to produce alkanes. The molar mass of the liberated gas identifies the alkyl group R. Subsequent reactions include the iodination of alkanes and the Wurtz reaction.


: Key Formula or Approach:

1. Reaction: \(RMgI + H_2O \to RH + Mg(OH)I\).

2. Molar mass \(M = \frac{Molar Volume at STP}{Specific Volume}\).

3. Wurtz Reaction: \(2R-X + 2Na \xrightarrow{ether} R-R + 2NaX\).


Step 2: Detailed Explanation:

1. Identification of the gas:

Molar volume at STP = 22.4 L/mol = 22.4 dm\(^3\)/mol.

Specific volume of the gas = 1.4 dm\(^3\)/g.

Molar mass of the gas (\(RH\)) = \(\frac{22.4 dm^3/mol}{1.4 dm^3/g} = 16 g/mol\).

The alkane with molar mass 16 is Methane (\(CH_4\)). Thus, the alkyl group \(R\) is Methyl (\(-CH_3\)).

2. Formation of Compound (X):

Methane reacts with \(I_2\) in the presence of \(HIO_3\) (which removes \(HI\) to prevent the reversible reaction):
\(CH_4 + I_2 \xrightarrow{HIO_3} CH_3I + HI\).

So, Compound (X) is Methyl iodide (\(CH_3I\)).

3. Formation of Compound (Y):

Methyl iodide undergoes the Wurtz reaction with Sodium in dry ether:
\(2CH_3I + 2Na \xrightarrow{ether} CH_3-CH_3 + 2NaI\).

So, Compound (Y) is Ethane (\(C_2H_6\)).

4. Molar mass of (Y):

Molar mass of \(C_2H_6 = 2(12) + 6(1) = 30 g/mol\).


Step 3: Final Answer:

The molar mass of compound (Y) is 30 g mol\(^{-1}\).
Quick Tip: Methane is the only alkane with a molar mass of 16. In Wurtz reactions, the number of carbon atoms in the resulting symmetrical alkane is always double the number of carbon atoms in the starting alkyl halide.

Step 1: Understanding the Concept:

Osmotic pressure (\(\pi\)) can be measured by the hydrostatic pressure exerted by the column of liquid at equilibrium, given by \(\pi = h\rho g\). This pressure is also related to the concentration of the solute via the Van't Hoff equation \(\pi = CRT\).


: Key Formula or Approach:

1. \(\pi = h\rho g\).

2. \(\pi = \frac{w}{MV} RT\).


Step 2: Detailed Explanation:

1. Calculate Osmotic Pressure (\(\pi\)):

Height \(h = 80.0 mm = 0.08 m\).

Density \(\rho = 1000 kg/m^3\).

Gravity \(g = 10 m/s^2\).
\(\pi = h\rho g = 0.08 \times 1000 \times 10 = 800 N/m^2 = 800 Pa\).

Convert to kPa: \(\pi = 0.8 kPa\).

2. Calculate Molar Mass (M):

Mass of solute \(w = 20 g\).

Volume \(V = 1 L = 1 dm^3\).

Temperature \(T = 300 K\).
\(R = 8.3 kPa dm^3 K^{-1} mol^{-1}\).

Using \(\pi = \frac{w}{MV} RT\):
\(0.8 = \frac{20}{M \cdot 1} \cdot 8.3 \cdot 300\).
\(M = \frac{20 \cdot 8.3 \cdot 300}{0.8} = \frac{49800}{0.8} = 62250 g/mol\).

3. Convert to kg/mol:
\(M = 62.25 kg/mol\).


Step 3: Final Answer:

The molar mass of hemoglobin is 62 kg mol\(^{-1}\) to the nearest integer.
Quick Tip: When using \(R = 8.3 kPa dm^3 K^{-1} mol^{-1}\), ensure your pressure is in kPa and volume is in dm\(^3\) (Liters). The hydrostatic pressure formula \(\pi = h\rho g\) gives the answer in Pascals (N/m\(^2\)).

Step 1: Understanding the Concept:

Molar conductivity (\(\Lambda_m\)) is related to electrolytic conductivity (\(\kappa\)) and molarity (\(M\)). We first calculate \(\kappa\) using conductance and the cell constant, then find the molarity, and finally relate molarity to the mass percentage.


: Key Formula or Approach:

1. \(\kappa = G \cdot \frac{l}{a}\) (Conductance \(\times\) Cell Constant).

2. \(\Lambda_m = \frac{\kappa \times 1000}{M}\).

3. \(M = \frac{10 \times % \times d}{Mw}\).


Step 2: Detailed Explanation:

1. Calculate Conductivity (\(\kappa\)):
\(G = 1.9 \times 10^{-3} S\).

Cell Constant (\(G^*\)) = 1.3 cm\(^{-1}\).
\(\kappa = 1.9 \times 10^{-3} \times 1.3 = 2.47 \times 10^{-3} S cm^{-1}\).

2. Calculate Molarity (M):
\(\Lambda_m = 123.5 S cm^2 mol^{-1}\).
\(123.5 = \frac{2.47 \times 10^{-3} \times 1000}{M} = \frac{2.47}{M}\).
\(M = \frac{2.47}{123.5} = 0.02 mol/L\).

3. Calculate Mass Percentage (x):

Density \(d = 1.0 g/mL\).

Molar mass \(Mw = 75 g/mol\).

Using \(M = \frac{10 \cdot x \cdot d}{Mw}\):
\(0.02 = \frac{10 \cdot x \cdot 1.0}{75}\).
\(x = \frac{0.02 \times 75}{10} = \frac{1.5}{10} = 0.15\).

The question asks for the value of \(x\) as \(\dots \times 10^{-2}\).
\(0.15 = 15 \times 10^{-2}\).


Step 3: Final Answer:

The value of \(x\) is 15.
Quick Tip: The shortcut formula \(M = \frac{10 \times mass % \times density}{Molar mass}\) is a life-saver for concentration conversions. Always check the units of \(\kappa\) (usually S/cm) and \(\Lambda_m\) (S cm\(^2\)/mol).

Step 1: Understanding the Concept:

The graph of \(t_{1/2}\) versus \([A]_o\) is a straight line passing through the origin. This implies that \(t_{1/2} \propto [A]_o\), which is the defining characteristic of a zero-order reaction.


: Key Formula or Approach:

For a zero-order reaction: \(t_{1/2} = \frac{[A]_o}{2k}\).

Since the graph is a straight line through the origin, the slope is constant:

Slope = \(\frac{t_{1/2, 1}}{[A]_{o, 1}} = \frac{t_{1/2, 2}}{[A]_{o, 2}}\).


Step 2: Detailed Explanation:

We have two points from the graph:

Point 1: \([A]_{o, 1} = 4 \times 10^{-3} mol/L, t_{1/2, 1} = 240 s\).

Point 2: \([A]_{o, 2} = 1.5 \times 10^{-3} mol/L, t_{1/2, 2} = x s\).

Using the constant slope property:
\(\frac{x}{1.5 \times 10^{-3}} = \frac{240}{4 \times 10^{-3}}\).
\(x = \frac{240 \times 1.5 \times 10^{-3}}{4 \times 10^{-3}}\).
\(x = \frac{240 \times 1.5}{4} = 60 \times 1.5 = 90 s\).


Step 3: Final Answer:

The value of \(x\) is 90.
Quick Tip: Remember the \(t_{1/2}\) trends:
Zero order: \(t_{1/2} \propto [A]_o\) (Line through origin).
First order: \(t_{1/2} = constant\) (Horizontal line).
Second order: \(t_{1/2} \propto 1/[A]_o\) (Hyperbolic curve).