Step 1: Understanding the Concept:

The photoelectric effect describes the emission of electrons from a metal surface when light of a sufficient frequency shines on it.

The fundamental condition for this effect is that the energy of the incident photon (\(E = h\nu\)) must be equal to or greater than the work function (\(\Phi = h\nu_0\)) of the metal.


Step 2: Detailed Explanation:

Assertion A: One of the key observations of the photoelectric effect is that it is an instantaneous process.

There is no detectable time lag (usually less than \(10^{-9}\) s) between the incidence of radiation and the emission of electrons, provided \(\nu > \nu_0\).

Therefore, Assertion A is correct.

Reason R: In the context of the photoelectric effect, a photon does not transfer its energy to an electron unless its energy is sufficient to overcome the binding forces of the atom.

If the energy of the photon is less than the threshold energy (work function), no emission occurs because the energy transfer is not effective for ejection.

The statement says "photon of any energy", which is incorrect as the energy must be above the threshold for the characteristic photoelectric transfer to occur.

Therefore, Reason R is not correct.


Step 3: Final Answer:

Assertion A is a true statement, but Reason R is a false statement.
Quick Tip: The photoelectric effect proves the particle nature of light. Remember: Intensity affects the number of electrons (current), while frequency affects the kinetic energy of the electrons.

Step 1: Understanding the Concept:

Molecular shapes are determined by the VSEPR (Valence Shell Electron Pair Repulsion) theory, which considers both bonding and lone pairs of electrons around the central atom.


Step 2: Detailed Explanation:

A. \(H_3O^+\):

Central atom Oxygen has \(6\) valence electrons.

In \(H_3O^+\), Oxygen forms \(3\) sigma bonds and has \(1\) lone pair.

Steric Number = \(3 + 1 = 4\); Hybridization = \(sp^3\).

Geometry is Tetrahedral, but the shape (arrangement of atoms) is Pyramidal (III).


B. Acetylide anion (\(C_2^{2-}\)):

The structure is \([C \equiv C]^{2-}\).

Each carbon is \(sp\) hybridized.

The molecule is Linear (II).


C. \(NH_4^+\):

Central atom Nitrogen has \(5\) valence electrons.

In \(NH_4^+\), it forms \(4\) sigma bonds and has \(0\) lone pairs.

Steric Number = \(4 + 0 = 4\); Hybridization = \(sp^3\).

Since there are no lone pairs, both geometry and shape are Tetrahedral (I).


D. \(ClO_2^-\):

Central atom Chlorine has \(7\) valence electrons.

In \(ClO_2^-\), Chlorine forms \(2\) sigma bonds and has \(2\) lone pairs.

Steric Number = \(2 + 2 = 4\); Hybridization = \(sp^3\).

Geometry is Tetrahedral, but the shape is Bent (IV).


Step 3: Final Answer:

Matching the species with shapes: A-III, B-II, C-I, D-IV.
Quick Tip: A quick way to determine shape: for \(sp^3\) hybridization, 0 lone pairs = Tetrahedral, 1 lone pair = Pyramidal, 2 lone pairs = Bent.

Step 1: Understanding the Concept:

When \(AgNO_3\) is added to \(KI\), a precipitation reaction occurs to form silver iodide (\(AgI\)).
\[ AgNO_3 + KI \rightarrow AgI(s) + KNO_3 \]


Step 2: Key Formula or Approach:

Calculate the number of millimoles (mmol) of each reactant:
\[ mmol = Volume (mL) \times Molarity (M) \]


Step 3: Detailed Explanation:

Millimoles of \(AgNO_3 = 25 \times 1 = 25\) mmol.

Millimoles of \(KI = 25 \times 1.05 = 26.25\) mmol.

In the reaction \(Ag^+ + I^- \rightarrow AgI(s)\), one mole of \(Ag^+\) reacts with one mole of \(I^-\).

Here, \(Ag^+\) is the limiting reagent because \(25 < 26.25\).

After the reaction:

- \(Ag^+\) ions are almost completely precipitated. Only a trace remains due to the solubility product (\(K_{sp}\)).

- Excess \(I^-\) ions = \(26.25 - 25 = 1.25\) mmol.

- \(K^+\) and \(NO_3^-\) are spectator ions and remain in high concentrations.

In competitive exams, while \(Ag^+\) is the limiting ion, both ions forming the precipitate (\(Ag^+\) and \(I^-\)) are typically highlighted as the species in low concentration relative to the bulk ions (\(K^+\) and \(NO_3^-\)). In some contexts, this question refers to the equilibrium state where both are low, but \(Ag^+\) is significantly lower.


Step 4: Final Answer:

The ions in very small quantity are \(Ag^+\) and \(I^-\) (relative to the high concentration spectator ions).
Quick Tip: Always find the limiting reagent first. The concentration of the limiting ion in a precipitation reaction will be extremely low, governed by the \(K_{sp}\) of the product.

Step 1: Understanding the Concept:

Ionization Enthalpy generally increases across a period from left to right due to increased nuclear charge and decreased atomic size. However, electronic configurations cause exceptions.


Step 2: Detailed Explanation:

Across the 2nd period: Li, Be, B, C, N, O, F, Ne.

General trend: \(Li < Be < B < C < N < O < F\).

Exceptions:

1. Be vs B: \(Be\) has configuration \(1s^2 2s^2\) (fully filled \(s\)-orbital). \(B\) has \(1s^2 2s^2 2p^1\). It is harder to remove an electron from a stable fully filled \(s\) orbital than from a \(2p\) orbital. Thus, \(IE(Be) > IE(B)\).

2. N vs O: \(N\) has configuration \(1s^2 2s^2 2p^3\) (stable half-filled \(p\)-orbital). \(O\) has \(1s^2 2s^2 2p^4\). Removing an electron from \(N\) requires more energy than removing a paired electron from \(O\). Thus, \(IE(N) > IE(O)\).

Combining these:
\(Li < B < Be < C < O < N < F\).


Step 3: Final Answer:

The correct order is \(Li < B < Be < C < O < N < F\).
Quick Tip: Remember the peaks in the ionization energy graph: Group 2 is higher than Group 13, and Group 15 is higher than Group 16.

Step 1: Understanding the Concept:

The extraction of copper from copper glance (\(Cu_2S\)) involve roasting and self-reduction in a Bessemer converter.


Step 2: Detailed Explanation:

1. Reaction (i) shows the partial roasting of \(Cu_2S\) to form \(Cu_2O\).

2. Reaction (ii) shows the self-reduction or auto-reduction process where the remaining \(Cu_2S\) reacts with the newly formed \(Cu_2O\) to produce metallic copper.

3. As the molten copper solidifies, sulfur dioxide (\(SO_2\)) gas escapes, forming bubbles or blisters on the surface of the metal.

4. This solidified copper with a blistered appearance is known as Blister copper. It is about \(98%\) pure.


Step 3: Final Answer:

The product is called Blister copper.
Quick Tip: Copper Matte is a mixture of \(Cu_2S\) and \(FeS\). Don't confuse it with Blister Copper, which is the final metallic product of the converter.

Step 1: Understanding the Concept:

This question tests preparation reactions of Hydrogen and Nitrogen gases.


Step 2: Detailed Explanation:

Statement I: The reaction of methane with steam at high temperatures in the presence of nickel is used for the industrial production of hydrogen. This is called steam reforming.
\[ CH_4(g) + H_2O(g) \xrightarrow[1270 K]{Ni} CO(g) + 3H_2(g) \]

So, Statement I is correct.


Statement II: In the laboratory preparation of dinitrogen (\(N_2\)), aqueous sodium nitrite (\(NaNO_2\)) reacts with ammonium chloride (\(NH_4Cl\)).
\[ NaNO_2(aq) + NH_4Cl(aq) \rightarrow NaCl(aq) + 2H_2O(l) + N_2(g) \]

So, Statement II is correct.


Step 3: Final Answer:

Both Statement I and Statement II are correct.
Quick Tip: Steam reforming produces "Syngas" (CO + \(H_2\)), while the reaction of \(NaNO_2\) and \(NH_4Cl\) is the standard laboratory method for making \(N_2\) gas.

Step 1: Understanding the Concept:

Match the s-block elements or their compounds with their specific industrial or biological applications.


Step 2: Detailed Explanation:

A. Potassium (K): Potassium ions are vital for biological processes, especially in the transmission of nerve signals through the Sodium-potassium pump (III).

B. Potassium Chloride (KCl): It is widely used as a potassium-rich Fertilizer (II) in agriculture.

C. Potassium Hydroxide (KOH): It is a much better Absorbent of \(CO_2\) (IV) than NaOH, especially in gas analysis.

D. Lithium (Li): Lithium isotopes are used in Thermonuclear reactions (I), such as those in hydrogen bombs.


Step 3: Final Answer:

The matching is A-III, B-II, C-IV, D-I.
Quick Tip: Alkali metals have very distinct applications. Potassium is always associated with fertilizers (KCl) and biological pumps.

Step 1: Understanding the Concept:
\(GaAlCl_4\) is an ionic compound. Gallium, due to the inert pair effect, can exist in the \(+1\) oxidation state. Aluminum typically exists in the \(+3\) oxidation state.


Step 2: Detailed Explanation:

1. The actual structure of \(GaAlCl_4\) is \([Ga]^+[AlCl_4]^-\).

2. In this salt, Gallium is in the \(+1\) oxidation state and Aluminum is in the \(+3\) oxidation state.

3. Electronegativity: Ga (\(\approx 1.81\)) is more electronegative than Al (\(\approx 1.61\)).

4. Ga exists as the cation (\(Ga^+\)), while Al is part of the complex anion (\(AlCl_4^-\)).

- Option (A) is wrong because Ga is \(+1\).

- Option (B) is wrong because Ga is an ion and is not coordinated like Al is.

- Option (D) is wrong as it is a salt, not a bridged dimer.


Step 3: Final Answer:

Option (C) is correct: Ga is more electronegative than Al and is present as the cationic part.
Quick Tip: Gallium can form \(Ga(I)\) and \(Ga(III)\) compounds. In salts like \(GaCl_2\) or \(GaAlCl_4\), it often acts as a \(Ga^+\) cation.

Step 1: Understanding the Concept:

Facial (\(fac\)) and Meridional (\(mer\)) isomers are a type of geometrical isomerism observed in octahedral complexes of the general formula \([MA_3B_3]\).


Step 2: Detailed Explanation:

1. In facial (fac) isomers, three identical ligands occupy the corners of one face of the octahedron.

2. In meridional (mer) isomers, three identical ligands occupy the corners of a meridian (a plane passing through the center).

3. Analyzing options:

- \([Co(NH_3)_3(NO_2)_3]\) is of type \([MA_3B_3]\), so it exists as \(fac\) and \(mer\) isomers.

- \([Co(en)_3]\) is \([M(AA)_3]\); it shows only optical isomerism.

- \([Co(en)_2Cl_2]\) is \([M(AA)_2B_2]\); it shows \(cis-trans\) isomerism.

- \([Pt(NH_3)_2Cl_2]\) is square planar \([MA_2B_2]\); it shows \(cis-trans\) isomerism.


Step 3: Final Answer:

The complex \([Co(NH_3)_3(NO_2)_3]\) can exist as a meridional isomer.
Quick Tip: Remember: \(fac-mer\) is specifically for octahedral \([MA_3B_3]\) complexes.

Step 1: Understanding the Concept:

An ambidentate ligand is a ligand that can coordinate to the central metal atom through two different atoms but only uses one at a time. Examples include \(SCN^-\) (bonding via S or N) and \(NO_2^-\) (bonding via N or O).


Step 2: Detailed Explanation:

Let's check the ligands in each set:

- Set (A): \(NCS^-\) is ambidentate.

- Set (B): \(NO_2^-\) is ambidentate.

- Set (C):

1. \(C_2O_4^{2-}\) (Oxalate) is a didentate ligand, bonding only through Oxygen atoms.

2. ethylene diammine (\(en\)) is a didentate ligand, bonding only through Nitrogen atoms.

3. \(H_2O\) is a monodentate ligand, bonding only through Oxygen.

None of these are ambidentate.

- Set (D): Both \(NO_2^-\) and \(NCS^-\) are ambidentate.


Step 3: Final Answer:

The set in option (C) does not have any ambidentate ligands.
Quick Tip: Ambidentate ligands like \(NO_2^-\) cause linkage isomerism in coordination compounds.

Step 1: Understanding the Concept:

Water quality is determined by various parameters including Biological Oxygen Demand (BOD), Dissolved Oxygen (DO), and the concentration of inorganic ions like Zinc (\(Zn^{2+}\)) and Nitrate (\(NO_3^-\)).


Step 2: Detailed Explanation:

Statement I:

Clean water typically has a BOD value of less than \(5\) ppm. A value of \(4\) ppm indicates relatively low organic pollution.

Healthy water for aquatic life usually requires a DO concentration above \(6\) ppm. A value of \(8\) ppm is considered excellent.

Thus, Statement I is correct.


Statement II:

According to international standards for drinking water:

1. The prescribed limit for Zinc (\(Zn\)) is \(5\) ppm (\(5\) mg/L).

2. The maximum limit for Nitrate (\(NO_3^-\)) is \(50\) ppm.

A concentration of \(5\) ppm for both is well within the safety limits for "good quality" or potable water.

Thus, Statement II is correct.


Step 3: Final Answer:

Since both statements are factually correct based on environmental chemistry standards, the most appropriate choice is (A).
Quick Tip: Remember the standard limits: BOD \(< 5\) ppm (Clean), Nitrate \(< 50\) ppm, Zinc \(< 5\) ppm, and Fluoride \(\approx 1\) ppm. High BOD indicates high organic pollution.

Step 1: Understanding the Concept:

In Thin Layer Chromatography (TLC), the distance traveled by a compound (\(R_f\) value) is inversely proportional to its adsorption on the stationary phase (silica gel).

Compounds that are least adsorbed travel the farthest (highest spot), while those that are strongly adsorbed stay near the base (lowest spot).


Step 2: Detailed Explanation:

In column chromatography, the elution order follows the same principle of adsorption:

1. The compound with the highest spot on the TLC plate (Spot A) is the least adsorbed. Consequently, it moves the fastest through the column and elutes first.

2. The compound with the middle spot (Spot C) has intermediate adsorption and elutes second.

3. The compound with the lowest spot (Spot B) is the most strongly adsorbed. It moves the slowest and elutes last.


Step 3: Final Answer:

Based on the visual evidence where A is highest, then C, then B, the elution order from the silica gel column will be A, C, B.
Quick Tip: Higher \(R_f\) in TLC \(=\) Faster elution in Column Chromatography. Always remember that the mobile phase carries the least polar (least adsorbed) components out first.

Step 1: Understanding the Concept:

The rate of Aromatic Electrophilic Substitution (AES) depends on the electron density of the benzene ring.

Electron Donating Groups (EDGs) like \(-OH, -R, -OR\) activate the ring and increase the rate, while Electron Withdrawing Groups (EWGs) like \(-C=O\) deactivate the ring and decrease the rate.


Step 2: Detailed Explanation:

1. Compound c: This is a ketone derivative. The carbonyl group (\(C=O\)) is a strong deactivating group (\(-M\) and \(-I\) effect). Thus, it has the lowest rate.

2. Compound b: Phenol is activated by the \(-OH\) group via the \(+M\) effect.

3. Compound a: Here, the ring is activated by the \(-OH\) group AND the alkyl bridge of the cyclohexane ring (via inductive effect and hyperconjugation). This makes it more active than phenol.

4. Compound d: In this structure, the ring is activated by an oxygen atom that is part of a 5-membered cyclic ether. The lone pair of this oxygen is highly available for resonance due to planarity, providing a very high rate of substitution.


Step 3: Final Answer:

The increasing order of the rate is: \(c < a < b < d\). (Note: Depending on exact substituent positions, 'a' and 'b' are close, but the deactivating 'c' must be first and 'd' is highly activated).
Quick Tip: Rate of AES \(\propto\) Electron density in the ring. Look for \(+M\) groups to find the fastest and \(-M\) groups for the slowest reactions.

Step 1: Understanding the Concept:

Nucleophilic Aromatic Substitution (\(S_NAr\)) typically follows a bimolecular mechanism (\(ArS_N2\)).

The reaction involves two steps: addition of the nucleophile to form a Meisenheimer complex (rate-determining step), followed by the elimination of the leaving group.


Step 2: Detailed Explanation:

1. In Reaction (II), the presence of a strong electron-withdrawing group (\(-NO_2\)) at the para position stabilizes the carbanion intermediate, facilitating the reaction.

2. Even in Reaction (I), where a methoxy group is present, if the substitution occurs at the chlorine-bearing carbon, it must proceed through a similar bimolecular collision mechanism.

3. Since the rate-determining step involves both the aromatic substrate and the nucleophile, the kinetics for these types of aromatic substitutions is overall second order.


Step 3: Final Answer:

Both reactions follow a bimolecular pathway, hence both are of \(2^{nd}\) order.
Quick Tip: Aryl halides rarely undergo \(S_N1\). They mostly follow \(S_NAr\) (2nd order) or the Benzyne mechanism (2nd order).

Step 1: Understanding the Concept:

The reaction sequence involves two major transformations:

1. Oxidation of an alkyl side chain on a benzene ring using \(KMnO_4\).

2. Reduction of a carbonyl group using the Wolff-Kishner reduction (\(NH_2NH_2 / KOH\)).


Step 2: Detailed Explanation:

1. Reaction 1 (\(KMnO_4\)): Potassium permanganate is a strong oxidizing agent. It oxidizes any alkyl side chain containing at least one benzylic hydrogen to a carboxylic acid (\(-COOH\)) group. The internal ketone group in the chain remains relatively stable under these conditions or is part of the oxidized fragment. Thus, 'A' contains a carboxyl group.

2. Reaction 2 (Wolff-Kishner Reduction): The reagents hydrazine (\(NH_2NH_2\)) and \(KOH\) followed by heating specifically reduce a ketone or aldehyde carbonyl group (\(C=O\)) to a methylene group (\(CH_2\)).

3. Therefore, product 'B' will have the carboxylic acid group (from step 1) and an alkane chain where the ketone used to be.


Step 3: Final Answer:

'A' is the keto-acid and 'B' is the fully reduced fatty acid/alkanoic acid derivative. The correct match corresponds to option (C).
Quick Tip: Remember: \(KMnO_4\) turns the whole side chain into \(-COOH\) if a benzylic H is present. Wolff-Kishner reduces \(C=O\) to \(CH_2\) but leaves \(-COOH\) salts intact.

Step 1: Understanding the Concept:

Solubility of coordination complexes depends on their ionic nature and lattice energy. Many complexes used as qualitative tests are insoluble precipitates.


Step 2: Detailed Explanation:

1. Option (C): \(Fe_4[Fe(CN)_6]_3\) is Prussian Blue, a well-known insoluble precipitate used to detect iron or nitrogen.

2. Option (B): \(K_3[Co(NO_2)_6]\) (Potassium cobaltinitrite) is a yellow precipitate formed in the test for potassium ions.

3. Option (D): Ammonium phosphomolybdate (or arsenomolybdate) is a canary yellow precipitate used in the detection of phosphate/arsenate.

4. Option (A): Basic ferric acetate \([Fe_3(OH)_2(OAc)_6]^{+}\) is a water-soluble complex that produces a deep red color in the solution when neutral \(FeCl_3\) is added to acetate ions. It only precipitates as a brown-red solid upon boiling.


Step 3: Final Answer:

The complex \([Fe_3(OH)_2(OAc)_6]Cl\) is the one that exists as a soluble deep red complex in aqueous medium.
Quick Tip: Most "colored tests" in qualitative analysis involve the formation of a precipitate. The acetate test with \(FeCl_3\) is unique because it first forms a \textbf{soluble red complex}.

Step 1: Understanding the Concept:

When aromatic primary diamines react with nitrous acid (\(HNO_2\)), diazotization occurs. If the amino groups are ortho to each other, internal cyclization is highly favored.


Step 2: Detailed Explanation:

1. The first amino group of o-phenylenediamine reacts with \(HNO_2\) at low temperature to form a monodiazonium ion.

2. Due to the proximity of the second \(-NH_2\) group at the ortho position, the lone pair on the nitrogen of the second amine attacks the diazonium group (\(-N_2^+\)).

3. This leads to the formation of a stable 5-membered nitrogen-containing heterocycle fused to the benzene ring.

4. The resulting product is Benzotriazole.


Step 3: Final Answer:

The major product 'X' is benzotriazole (represented by the structure in Option 36669413226).
Quick Tip: Orthodiamines \(+\) \(HNO_2\) \(\rightarrow\) Triazoles.
Metadiamines \(+\) \(HNO_2\) \(\rightarrow\) Bismarck Brown.
Paradiamines \(+\) \(HNO_2\) \(\rightarrow\) Bis-diazonium salts.

Step 1: Understanding the Concept:

Polymerization of ethene can lead to different types of polyethylene depending on the catalyst and conditions used.


Step 2: Detailed Explanation:

1. High Density Polythene (HDPE): This is produced by the coordination polymerization of ethene at low pressure (\(6\)-\(7\) atm) and moderate temperature (\(333\)-\(343\) K) in the presence of Ziegler-Natta catalyst.

2. The Ziegler-Natta catalyst is a mixture of triethylaluminium \([Al(C_2H_5)_3]\) and titanium tetrachloride \([TiCl_4]\).

3. HDPE consists of linear chains with very little branching. This allows the molecules to be closely packed, resulting in high density and high chemical robustness.


Step 3: Final Answer:

The description perfectly matches High density polythene.
Quick Tip: Ziegler-Natta catalyst \(=\) HDPE (Linear).
High pressure/Peroxide catalyst \(=\) LDPE (Branched).

Step 1: Understanding the Concept:

The compound is a tetrose (\(4\) carbons) with an aldehyde group (positive Schiff's test). Since it forms a triacetate, it has \(3\) hydroxyl (\(-OH\)) groups. This defines it as an aldotetrose.


Step 2: Detailed Explanation:

1. There are two aldotetroses: Erythrose and Threose. Both have \(2\) chiral centers.

2. Reduction with \(NaBH_4\): Erythrose reduces to erythritol (meso-compound, achiral). Threose reduces to threitol (chiral compound).

3. The problem states that product 'B' is a chiral compound. This confirms that 'X' must be Threose.

4. Since the question specifies the L-isomer, the starting material is L-Threose.


Step 3: Final Answer:

'X' is L-Threose (Option 36669413235 in the diagram).
Quick Tip: Erythro isomers have substituents on the same side and often lead to meso products upon symmetric reduction. Threo isomers have them on opposite sides and lead to chiral products.

Step 1: Understanding the Concept:

This question involves specific qualitative tests for acid radicals (anions).


Step 2: Detailed Explanation:

1. Brown Ring Test: Freshly prepared \(FeSO_4\) in the presence of concentrated \(H_2SO_4\) reacting with a salt solution to form a brown ring at the junction indicates the presence of the Nitrate ion (\(NO_3^-\)). The ring is due to the complex \([Fe(H_2O)_5(NO)]^{2+}\).

2. Neutral \(FeCl_3\) Test: When neutral ferric chloride is added to a solution containing Acetate ions (\(CH_3COO^-\)), a deep red color is produced due to the formation of a soluble complex, basic ferric acetate \([Fe_3(OH)_2(CH_3COO)_6]^{+}\).

3. Upon boiling this red solution, the complex decomposes to give a reddish-brown precipitate of basic ferric acetate.


Step 3: Final Answer:

The two ions identified are \(CH_3COO^-\) and \(NO_3^-\).
Quick Tip: Brown Ring \(=\) Nitrate. Deep Red with \(FeCl_3\) (disappearing on boiling) \(=\) Acetate. These are two of the most characteristic anion tests in inorganic chemistry.

Step 1: Understanding the Concept:

The mass percentage of a solution is calculated by dividing the total mass of the solute by the total mass of the solution and multiplying by 100.


Step 2: Key Formula or Approach:

1. Find the mass of sugar in each solution: \( Mass of solute = Mass of solution \times Concentration \)

2. Total mass of sugar = \( w_1 + w_2 \)

3. Total mass of solution = \( m_1 + m_2 \)

4. Mass % = \( \left( \frac{Total mass of sugar}{Total mass of solution} \right) \times 100 \)


Step 3: Detailed Explanation:

Mass of sugar from the first solution: \[ w_1 = 200 \times \frac{25}{100} = 50 g \]
Mass of sugar from the second solution: \[ w_2 = 500 \times \frac{40}{100} = 200 g \]
Total mass of sugar: \[ W_{total} = 50 + 200 = 250 g \]
Total mass of the resulting solution: \[ M_{total} = 200 + 500 = 700 g \]
Mass percentage of the resulting solution: \[ Mass % = \frac{250}{700} \times 100 = \frac{250}{7} \approx 35.71% \]

Step 4: Final Answer:

The mass percentage is approximately 35.71. Rounding to the nearest integer gives 36.
Quick Tip: When mixing solutions, masses of solutes and solutions are additive. Do not average the percentages directly unless the masses of the solutions are equal.

Step 1: Understanding the Concept:

The density of a unit cell is related to the number of atoms (\(Z\)), molar mass (\(M\)), edge length (\(a\)), and Avogadro's number (\(N_A\)).


Step 2: Key Formula or Approach: \[ d = \frac{Z \cdot M}{a^3 \cdot N_A} \]
Rearranging for \(Z\): \[ Z = \frac{d \cdot a^3 \cdot N_A}{M} \]

Step 3: Detailed Explanation:
Given values:
\( d = 3.0 g/cm^3 \) (since 1 g/mL = 1 g/cm\(^3\))
\( M = 12 g/mol \)
\( a = 300 pm = 300 \times 10^{-10} cm = 3 \times 10^{-8} cm \)
\( N_A = 6.02 \times 10^{23} mol^{-1} \)


Substituting the values into the formula: \[ Z = \frac{3.0 \times (3 \times 10^{-8})^3 \times 6.02 \times 10^{23}}{12} \] \[ Z = \frac{3.0 \times 27 \times 10^{-24} \times 6.02 \times 10^{23}}{12} \] \[ Z = \frac{3.0 \times 27 \times 0.602}{12} \] \[ Z = \frac{48.762}{12} \approx 4.0635 \]

Step 4: Final Answer:
Rounding to the nearest integer, the number of atoms present in one unit cell is 4.
Quick Tip: Ensure all units are consistent. Convert pm to cm before calculating volume to match the density units of g/cm\(^3\).

Step 1: Understanding the Concept:

The thermite reaction is: \[ Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2Fe \]
The stoichiometric ratio of Fe\(_2\)O\(_3\) to Al is 1:2 by moles, which matches the problem statement.


Step 2: Key Formula or Approach:
1. \(\Delta H_{rxn} = \sum \Delta H_f^\circ(Products) - \sum \Delta H_f^\circ(Reactants)\)

2. Heat evolved per gram = \(\frac{|\Delta H_{rxn}|}{Total mass of stoichiometric reactants}\)


Step 3: Detailed Explanation:
Calculating \(\Delta H_{rxn}\): \[ \Delta H_{rxn} = [\Delta H_f^\circ(Al_2O_3) + 2\Delta H_f^\circ(Fe)] - [\Delta H_f^\circ(Fe_2O_3) + 2\Delta H_f^\circ(Al)] \]
Since elements in their standard state have \(\Delta H_f^\circ = 0\): \[ \Delta H_{rxn} = [-1700 + 0] - [-840 + 0] = -1700 + 840 = -860 kJ \]
The heat evolved for 1 mole of the reaction is 860 kJ.


Total mass of reactants for 1 mole of reaction:
Mass of 1 mole Fe\(_2\)O\(_3 = 2(56) + 3(16) = 112 + 48 = 160 g \)

Mass of 2 moles Al \( = 2(27) = 54 g \)

Total mass of mixture \( = 160 + 54 = 214 g \)


Heat evolved per gram: \[ q = \frac{860}{214} \approx 4.018 kJ/g \]

Step 4: Final Answer:
Rounding to the nearest integer, the heat evolved is 4 kJ/g.
Quick Tip: Always check if the given "ratio" is molar or by mass. In most thermodynamics problems like this, a ratio like 1:2 for these specific reactants implies the stoichiometric molar ratio.

Step 1: Understanding the Concept:

Isotonic solutions have the same osmotic pressure (\(\pi_1 = \pi_2\)). For dilute solutions, this means \(i_1 C_1 R T = i_2 C_2 R T \), which simplifies to \(i_1 C_1 = i_2 C_2\).


Step 2: Key Formula or Approach:
1. Van't Hoff factor for K\(_2\)SO\(_4\): \(i = 1 + (n-1)\alpha\).
2. For K\(_2\)SO\(_4 \rightarrow 2K^+ + SO_4^{2-}\), \(n = 3\).
3. For glucose (non-electrolyte), \(i = 1\).

Step 3: Detailed Explanation:
Setting the isotonic condition: \[ i_{K_2SO_4} \cdot [C_{K_2SO_4}] = i_{glucose} \cdot [C_{glucose}] \] \[ i \cdot (0.004) = 1 \cdot (0.01) \] \[ i = \frac{0.01}{0.004} = 2.5 \]
Now calculate \(\alpha\): \[ i = 1 + (3-1)\alpha = 1 + 2\alpha \] \[ 2.5 = 1 + 2\alpha \] \[ 2\alpha = 1.5 \Rightarrow \alpha = 0.75 \]
Percentage dissociation \( = 0.75 \times 100 = 75% \).


Step 4: Final Answer:
The percentage dissociation of K\(_2\)SO\(_4\) is 75.
Quick Tip: For isotonic problems, always include the Van't Hoff factor (\(i\)) if the solute is an electrolyte. Remember that glucose, urea, and sucrose have \(i = 1\).

Step 1: Understanding the Concept:

The equilibrium constant \(K_c\) is the ratio of the products' concentrations to the reactants' concentrations at equilibrium.


Step 2: Key Formula or Approach: \[ K_c = \frac{[CO_2][H_2]}{[CO][H_2O]} \]
Since \(\Delta n_g = 0\), the volume term cancels out and we can use equilibrium moles directly.


Step 3: Detailed Explanation:
Reaction: \(CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)\)

Initial moles: 1 (CO), 1 (H\(_2\)O), 0 (CO\(_2\)), 0 (H\(_2\))

Moles reacted (40% of 1): \(x = 0.4\)

Equilibrium moles:
CO: \(1 - 0.4 = 0.6\)

H\(_2\)O: \(1 - 0.4 = 0.6\)

CO\(_2\): \(0.4\)

H\(_2\): \(0.4\)


Calculating \(K_c\): \[ K_c = \frac{0.4 \cdot 0.4}{0.6 \cdot 0.6} = \frac{0.16}{0.36} = \frac{4}{9} \approx 0.4444 \]
The value required is \(K_c \times 10^2 = 0.4444 \times 100 = 44.44\).


Step 4: Final Answer:
Rounding to the nearest integer, the answer is 44.
Quick Tip: When \(\Delta n_g = 0\), volume does not affect the value of \(K_c\). You can simplify your work by performing calculations using moles instead of concentrations.

Step 1: Understanding the Concept:

Standard electrode potentials cannot be added directly. We must use Gibbs free energy changes (\(\Delta G^\circ = -nFE^\circ\)), which are additive.


Step 2: Key Formula or Approach:
1. \(Pb^{2+} + 2e^- \rightarrow Pb \quad \Delta G_1^\circ = -2Fm \)
2. \(Pb^{4+} + 4e^- \rightarrow Pb \quad \Delta G_2^\circ = -4Fn \)
Target: \(Pb^{2+} \rightarrow Pb^{4+} + 2e^-\)


Step 3: Detailed Explanation:
We can obtain the target reaction by subtracting reaction (2) from reaction (1): \((Pb^{2+} + 2e^- \rightarrow Pb) - (Pb^{4+} + 4e^- \rightarrow Pb)\)
Result: \(Pb^{2+} - Pb^{4+} - 2e^- = 0 \Rightarrow Pb^{2+} \rightarrow Pb^{4+} + 2e^-\)


Therefore, \(\Delta G_3^\circ = \Delta G_1^\circ - \Delta G_2^\circ\).
Let \(E^\circ(Pb^{2+}/Pb^{4+}) = E_3^\circ\). \[ -2FE_3^\circ = -2Fm - (-4Fn) \] \[ -2FE_3^\circ = -2Fm + 4Fn \]
Divide by \(-2F\): \[ E_3^\circ = m - 2n \]
Comparing this to the given expression \(m - xn\), we find \(x = 2\).


Step 4: Final Answer:
The value of \(x\) is 2.
Quick Tip: Remember: \(\Delta G^\circ\) is an extensive property and is additive, whereas \(E^\circ\) is an intensive property and is not directly additive.

Step 1: Understanding the Concept:

The rate of reaction is defined in terms of the rate of disappearance of reactants or the rate of production of products, normalized by their stoichiometric coefficients.


Step 2: Key Formula or Approach: \[ Rate of Reaction = -\frac{1}{6} \frac{d[FeSO_4]}{dt} = \frac{1}{3} \frac{d[Fe_2(SO_4)_3]}{dt} \]
So, \(Rate of production of Fe_2(SO_4)_3 = \frac{1}{2} \times \left( -\frac{d[FeSO_4]}{dt} \right) \).


Step 3: Detailed Explanation:
Change in concentration of FeSO\(_4\): \[ \Delta [FeSO_4] = 8.8 - 10 = -1.2 M \]
Time interval \(t = 0.5 hour = 30 \times 60 = 1800 s \).

Rate of disappearance of FeSO\(_4\): \[ -\frac{d[FeSO_4]}{dt} = \frac{1.2}{1800} M/s \]
Rate of production of Fe\(_2\)(SO\(_4\))\(_3\): \[ Rate = \frac{1}{2} \times \frac{1.2}{1800} = \frac{0.6}{1800} = \frac{1}{3000} M/s \] \[ Rate \approx 3.333 \times 10^{-4} mol L^{-1} s^{-1} \]
Expressing in terms of \(10^{-6}\): \[ 333.33 \times 10^{-6} mol L^{-1} s^{-1} \]

Step 4: Final Answer:
Rounding to the nearest integer, the answer is 333.
Quick Tip: Always check the stoichiometry carefully. Here, 6 moles of FeSO\(_4\) produce 3 moles of Fe\(_2\)(SO\(_4\))\(_3\), so the rate of product formation is exactly half the rate of reactant disappearance.

Step 1: Understanding the Concept:

The spin-only magnetic moment is given by \(\mu = \sqrt{n(n+2)}\) BM, where \(n\) is the number of unpaired electrons.


Step 2: Detailed Explanation:
For both complexes, the central metal ion is \(Cr^{3+}\).
Electronic configuration of Cr is [Ar] \(3d^5 4s^1\).
Cr\(^{3+}\) has configuration [Ar] \(3d^3\).


1. \([Cr(CN)_6]^{3-}\):
CN\(^-\) is a strong field ligand.
For a \(d^3\) configuration in an octahedral field, the electrons fill the \(t_{2g}\) orbitals first: \(t_{2g}^3 e_g^0\).
Number of unpaired electrons \(n = 3\). \(\mu_1 = \sqrt{3(3+2)} = \sqrt{15}\) BM.


2. \([Cr(H_2O)_6]^{3+}\):
H\(_2\)O is a weak field ligand.
For a \(d^3\) configuration, the filling is the same: \(t_{2g}^3 e_g^0\).
Number of unpaired electrons \(n = 3\). \(\mu_2 = \sqrt{3(3+2)} = \sqrt{15}\) BM.


The ratio is \(\frac{\sqrt{15}}{\sqrt{15}} = 1\).


Step 3: Final Answer:
The ratio is 1.
Quick Tip: For \(d^1, d^2, and d^3\) octahedral complexes, the number of unpaired electrons is always the same regardless of whether the ligand is strong field or weak field.

Step 1: Understanding the Concept:

Acid-catalyzed dehydration or substitution of an alcohol begins with protonation of the \(-OH\) group followed by loss of water to form a carbocation. Hyperconjugation depends on the number of \(\alpha\)-hydrogens adjacent to the carbocation center.


Step 2: Detailed Explanation:
1. Protonation of 1-methyl-1-hydroxydecalin leads to the formation of a tertiary carbocation at the 1st position (where the methyl and hydroxyl groups were).
2. The carbocation is located at a tertiary carbon fused in the bicyclic system.
3. Adjacent (\(\alpha\)) carbons are:
- The methyl group (C atoms attached to the cation): \(3 \alpha-H\) atoms.
- The C-2 methylene group in the ring: \(2 \alpha-H\) atoms.
- The C-9 bridgehead carbon: \(1 \alpha-H\) atom.
4. Total number of \(\alpha\)-hydrogens \( = 3 + 2 + 1 = 6\).
5. Since each \(\alpha\)-hydrogen contributes one hyperconjugative structure, there are 6 structures.


Step 3: Final Answer:
The number of hyperconjugation structures is 6.
Quick Tip: To find hyperconjugative structures, identify the carbocation and count every hydrogen on the carbons directly bonded to the positively charged carbon.

Step 1: Understanding the Concept:
Grignard reagents (\(RMgX\)) react with acidic hydrogens (like those in alcohols) in an acid-base reaction and also add to carbonyl groups (aldehydes/ketones).


Step 2: Detailed Explanation:
1. The substrate (y moles) contains one aldehyde (\(-CHO\)) group and two alcohol (\(-OH\)) groups.
2. Reaction with -OH groups (Acid-Base): Each mole of hydroxyl group consumes 1 mole of MeMgBr to form methane. Total = 2 moles.
3. Reaction with -CHO group (Addition): 1 mole of aldehyde consumes 1 mole of MeMgBr to form a secondary alcohol (after workup). Total = 1 mole.
4. Total moles of MeMgBr (\(x\)) required for complete reaction of 1 mole of substrate (\(y\)) is \(2 + 1 = 3\).
5. Ratio \(x/y = 3/1 = 3\).


Step 3: Final Answer:
The ratio \(x/y\) is 3.
Quick Tip: Grignard reagents are strong bases as well as nucleophiles. They will always react with acidic protons (from \(OH, NH, SH\)) before attacking a carbonyl group.