Step 1: Understanding the Question:

The question describes a chemical reaction (iodide with hydrogen peroxide) and asks to identify a specific indicator ('X') and the substance ('A') with which it forms a blue complex. This is a classic setup for an iodine clock reaction.


Step 2: Key Concepts:

1. The reaction between iodide ions (I\(^-\)) and hydrogen peroxide (H\(_2\)O\(_2\)) in an acidic solution produces iodine (I\(_2\)):

\[ 2I^- (aq) + H_2O_2 (aq) + 2H^+ (aq) \rightarrow I_2 (aq) + 2H_2O (l) \]
2. The standard chemical test for the presence of iodine (I\(_2\)) is the addition of a starch solution.

3. Starch forms a deep blue-black colored complex with iodine (specifically, with the triiodide ion, I\(_3^-\), which is in equilibrium with I\(_2\) and I\(^-\)).


Step 3: Detailed Explanation:

- In this experiment, as the reaction proceeds, iodine (I\(_2\)) is produced.

- To detect the formation of this iodine, an indicator is needed.

- The specific indicator that forms a blue complex with iodine is starch.

- Therefore, the indicator 'X' must be starch.

- The compound 'A' that is present in the solution and forms the complex with the indicator is the product, iodine (I\(_2\)).

- So, 'X' is starch and 'A' is iodine.


Step 4: Final Answer:

The indicator 'X' is Starch and compound 'A' is iodine. This corresponds to option (A). (Note: The candidate's chosen option (B) is incorrect as starch does not form a complex with H\(_2\)O\(_2\)).
Quick Tip: The starch-iodine test is a very common and specific chemical test. Remember that starch is the indicator FOR iodine, and it produces a characteristic dark blue/black color. This is a fundamental concept in redox titrations involving iodine (iodometry/iodimetry).

Step 1: Understanding the Question:

We need to match the terms related to colligative properties and electrokinetic phenomena (List I) with their correct definitions (List II).


Step 3: Detailed Explanation:

A. Osmosis: This is the spontaneous net movement of solvent molecules through a semi-permeable membrane into a region of higher solute concentration, in the direction that tends to equalize the solute concentrations on the two sides. This means solvent moves from the pure solvent side to the solution side, or from a dilute solution to a concentrated one. This matches description III.


B. Reverse Osmosis: This is a process where solvent molecules are forced to move through a semi-permeable membrane from a region of high solute concentration to a region of low solute concentration by applying an external pressure greater than the osmotic pressure. This means solvent moves from the solution side towards the pure solvent side. This matches description I.


D. Electrophoresis: This is the motion of dispersed particles (charged colloidal particles) relative to a fluid under the influence of a spatially uniform electric field. The charged particles move towards the oppositely charged electrode. This matches description II.


C. Electro-osmosis: This is the motion of liquid (the dispersion medium) through a porous material or membrane under the influence of an applied electric field. It occurs when the movement of the charged colloidal particles is prevented. The medium itself moves. This matches description IV.


Matching Summary:

- A \(\rightarrow\) III

- B \(\rightarrow\) I

- C \(\rightarrow\) IV

- D \(\rightarrow\) II


Step 4: Final Answer:

The correct matching is A-III, B-I, C-IV, D-II. This corresponds to option (D).
Quick Tip: Carefully distinguish between Osmosis and Reverse Osmosis (direction of solvent flow) and between Electrophoresis and Electro-osmosis (what moves in the electric field: the charged particles or the liquid medium). Creating a small comparison table can help solidify these concepts.

Step 1: Understanding the Question:

This is a factual question from environmental chemistry. We need to know the standard values for Dissolved Oxygen (DO) required for aquatic life and the Biochemical Oxygen Demand (BOD) for clean water.


Step 2: Key Concepts:

- Dissolved Oxygen (DO): The amount of oxygen dissolved in water. It is crucial for the survival of fish and other aquatic organisms. Low DO levels indicate pollution.

- Biochemical Oxygen Demand (BOD): The amount of dissolved oxygen needed by aerobic biological organisms to break down organic material present in a given water sample at certain temperature over a specific time period. A high BOD indicates a high level of organic pollution, as more oxygen is required to decompose the waste.


Step 3: Standard Values:

- For the survival and growth of fish, the concentration of dissolved oxygen (DO) should generally be above 6 ppm. Water with DO below 4-5 ppm is considered polluted and unsuitable for most fish. So, X = 6.

- For water to be considered clean or non-polluted, the BOD value should be low. A BOD value of less than 5 ppm is indicative of clean water. High values (e.g., > 10-15 ppm) indicate significant pollution. So, Y = 5.


Step 4: Final Answer:

The required values are X = 6 and Y = 5. This corresponds to option (A). (Note: The candidate's chosen option (B) represents conditions of polluted water, not suitable for fish growth).
Quick Tip: Remember these benchmark values for water quality: - \textbf{Clean Water}: DO \(>\) 6 ppm, BOD \(<\) 5 ppm. - \textbf{Polluted Water}: DO \(<\) 5 ppm, BOD \(>\) 5 ppm (often much higher). DO and BOD are inversely related; as organic pollution (BOD) increases, aerobic bacteria consume DO, causing its level to drop.

Step 1: Understanding the Reaction:

The starting material is a 4-substituted cyclohexa-2,5-dienone. The substituents at the C4 position are a methyl group and an epoxy-methyl group (an oxirane ring). The reagent is H\(_3\)O\(^+\), indicating an acid-catalyzed reaction. This is a classic setup for a Dienone-Phenol rearrangement, often accompanied by other rearrangements.


Step 2: Analyzing Possible Mechanisms:

The reaction in acidic medium is complex. Two main events are expected: the rearrangement of the dienone to a stable aromatic phenol and the opening of the strained epoxide ring.

Mechanism (Dienone-Phenol Rearrangement):

1. The carbonyl oxygen of the dienone gets protonated by H\(_3\)O\(^+\).

2. To achieve the stable aromatic phenol structure, one of the groups at the C4 position must migrate to an adjacent carbon (C3 or C5). This is the key step of the rearrangement.

3. Simultaneously or subsequently, the protonated carbonyl group becomes a hydroxyl group, and the ring aromatizes.

4. The epoxide ring will also be opened by the acid catalyst to form a diol.

The complexity lies in the sequence of these steps and the migratory aptitude of the groups.


Step 3: Evaluating the Outcome and Options:

This specific rearrangement is known to be very complex, often leading to multiple products. However, in the context of an exam question, we look for the most plausible major product based on carbocation stability and rearrangement principles.
A plausible pathway involves the rearrangement giving a substituted phenol, followed by the opening of the epoxide into a diol. Given the options, the reaction seems to involve the formation of a catechol (1,2-dihydroxybenzene) derivative and a significant rearrangement of the carbon skeleton, similar to a pinacol rearrangement.

Let's analyze the chosen answer, Option (C), which is 1,2-dihydroxy-4-(2-hydroxy-2-methylpropyl) benzene. The formation of this product from the starting material is not straightforward and likely involves multiple rearrangement steps that are beyond the typical scope. There appears to be an inconsistency in the number of atoms between the reactant and the product in option (C), specifically the presence of two methyl groups in the side chain.

Given the ambiguity and likely error in the question, a definitive mechanistic derivation is problematic. However, if we are to rationalize the given answer, one would have to assume a complex cascade of epoxide opening, dienone-phenol rearrangement, and a pinacol-type rearrangement of the resulting side chain.


Step 4: Final Answer:

Due to a likely error in the question's structure or options, it is not possible to derive the product in Option (C) through a standard, unambiguous mechanism. However, as it is the keyed answer, we select it.
Quick Tip: When faced with a very complex organic reaction in an exam that doesn't follow a clear-cut named reaction path, look for key transformations. Here, dienone \(\rightarrow\) phenol and epoxide \(\rightarrow\) diol are expected. If the options don't match, the question might be flawed. In such cases, try to eliminate options based on what is impossible (e.g., violation of atom conservation).

Step 1: Understanding the Question:

We need to identify which of the given ores is a sulphide ore. This requires knowledge of the chemical formulas of common ores.


Step 3: Detailed Explanation:

Let's analyze the chemical composition of each ore:

- (A) Malachite: It is a copper carbonate hydroxide mineral. Its formula is Cu\(_2\)CO\(_3\)(OH)\(_2\). This is a carbonate/hydroxide ore, not a sulphide.

- (B) Calamine: It is an ore of zinc, with the formula ZnCO\(_3\). This is a carbonate ore. (Note: Historically, calamine could also refer to the silicate ore hemimorphite).

- (C) Sphalerite: It is the primary ore of zinc. Its chemical formula is (Zn,Fe)S, which is essentially zinc sulphide (ZnS). This is a sulphide ore.

- (D) Siderite: It is an ore of iron with the formula FeCO\(_3\). This is a carbonate ore.


Step 4: Final Answer:

Among the given options, only Sphalerite is a sulphide-based mineral. This corresponds to option (C).
Quick Tip: Memorizing the names and chemical formulas of important ores is crucial for the metallurgy chapter. Create a table classifying ores based on the anion (oxide, sulphide, carbonate, halide, sulphate) and the metal they contain. This will help in quickly answering such factual questions.

Step 1: Understanding the Question:

We need to evaluate two statements related to the ionization enthalpy trends in Group 13 elements (B, Al, Ga).


Step 3: Detailed Explanation:

Analyzing Statement I:

"The decrease in first ionization enthalpy from B to Al is much larger than that from Al to Ga."

Let's look at the actual values of the first ionization enthalpy (IE\(_1\)) in kJ/mol:

- Boron (B): 801

- Aluminum (Al): 577

- Gallium (Ga): 579

The decrease from B to Al is \(801 - 577 = 224\) kJ/mol. This is a significant decrease, expected due to the increase in atomic size and shielding.

The change from Al to Ga is an increase of \(579 - 577 = 2\) kJ/mol. Therefore, the "decrease" is -2 kJ/mol.

Comparing the magnitudes, the decrease from B to Al (224) is indeed much larger than the change from Al to Ga (-2). So, Statement I is correct.


Analyzing Statement II:

"The d orbitals in Ga are completely filled."

The electronic configuration of Gallium (Ga, Z=31) is [Ar] 3d\(^{10}\) 4s\(^2\) 4p\(^1\).

As seen from the configuration, the 3d subshell is completely filled with 10 electrons. So, Statement II is correct. In fact, the poor shielding effect of these filled d-orbitals is the reason why the ionization enthalpy of Ga is slightly higher than that of Al, contrary to the general trend.


Step 4: Final Answer:

Both Statement I and Statement II are factually correct. This corresponds to option (C). (Note: The candidate's chosen option (D) is incorrect because Statement II is correct).
Quick Tip: Group 13 trends are a notable exception to general periodic trends. Remember the anomaly: IE\(_1\) of Ga > IE\(_1\) of Al. This is due to the poor shielding by the 10 d-electrons in Gallium, which increases the effective nuclear charge experienced by the valence electrons.

Step 1: Understanding the Question:

We need to determine the color of a Co(II) ion solution in amyl alcohol. The color of transition metal complexes depends on the metal ion, its oxidation state, and the coordination environment (ligands and geometry).


Step 2: Key Concepts:

- Co(II) is a \(d^7\) ion.

- In aqueous solutions, Co(II) typically exists as the hexaaquacobalt(II) ion, \([Co(H_2O)_6]^{2+}\). This complex has an octahedral geometry and is pink in color.

- However, in the presence of other ligands or in certain solvents, Co(II) can form tetrahedral complexes.

- Tetrahedral complexes of Co(II), such as \([CoCl_4]^{2-}\), are characteristically an intense blue color. The color difference arises from the different splitting of d-orbitals in octahedral vs. tetrahedral fields (crystal field theory).

- Amyl alcohol is a relatively weak, bulky ligand. While it can coordinate to the Co(II) ion, the steric hindrance and solvent properties often favor the formation of a four-coordinate, tetrahedral species over a six-coordinate, octahedral one.


Step 3: Detailed Explanation:

When a cobalt(II) salt is dissolved in a non-aqueous solvent like an alcohol (e.g., amyl alcohol), the coordination environment changes from the aqueous one. The equilibrium often shifts towards the formation of a tetrahedral complex, \([Co(ROH)_4]^{2+}\) or similar species. Tetrahedral Co(II) complexes have a very intense absorption band in the orange/red part of the visible spectrum, which results in them appearing a deep blue color to our eyes. This is a well-known characteristic of Co(II) chemistry.


Step 4: Final Answer:

A solution of Co(II) in amyl alcohol is expected to be blue due to the formation of a tetrahedral complex. This corresponds to option (C).
Quick Tip: A useful rule of thumb for Co(II) complexes: octahedral is pink, tetrahedral is blue. This is often used in chemical indicators for water, where the anhydrous form (often tetrahedral) is blue and the hydrated form (octahedral) is pink.

Step 1: Understanding the Question:

We need to identify the correct thermodynamic relations from a given list.


Step 3: Detailed Explanation:

Relation (A): \(\Delta U = q + p\Delta V\)

The first law of thermodynamics is \(\Delta U = q + w\), where \(w\) is the work done on the system. For mechanical work against an external pressure, \(w = -p_{ext}\Delta V\). So, the equation should be \(\Delta U = q - p\Delta V\). The given relation has a positive sign, which would imply work done by the system. The sign convention can vary, but typically in chemistry, \(w = -p\Delta V\) is used. Thus, statement (A) is generally considered incorrect in the standard sign convention.


Relation (B): \(\Delta G = \Delta H - T\Delta S\)

This is the Gibbs-Helmholtz equation, which defines the change in Gibbs free energy (\(\Delta G\)) for a process occurring at constant temperature. This is a fundamental and correct thermodynamic relation.


Relation (C): \(\Delta S = \frac{q_{rev}}{T}\)

This is the thermodynamic definition of the change in entropy (\(\Delta S\)) for a reversible process occurring at a constant temperature T. This is a correct and fundamental relation.


Relation (D): \(\Delta H = \Delta U - \Delta nRT\)

The relationship between enthalpy change (\(\Delta H\)) and internal energy change (\(\Delta U\)) is given by \(\Delta H = \Delta U + \Delta(pV)\). For reactions involving ideal gases, this becomes \(\Delta H = \Delta U + (\Delta n_g)RT\), where \(\Delta n_g\) is the change in the number of moles of gas. The given relation has a negative sign. Thus, statement (D) is incorrect.


Step 4: Final Answer:

The correct relations are (B) and (C). This corresponds to option (C). (Note: The candidate's chosen option (D) is incorrect because relation (A) is incorrect).
Quick Tip: Be very careful with signs in thermodynamic equations. - First Law: \(\Delta U = q + w\); work done on the system is positive (\(w = -p_{ext}\Delta V\)). - Enthalpy-Internal Energy: \(\Delta H = \Delta U + (\Delta n_g)RT\). Remember H is 'bigger' than U for gas-producing reactions at constant pressure. - Gibbs Energy: \(\Delta G = \Delta H - T\Delta S\). These are some of the most fundamental equations in thermodynamics.

Step 1: Understanding the Question:

We need to determine the order of the spin-only magnetic moments for three coordination complexes. The magnetic moment depends on the number of unpaired electrons.


Step 2: Key Formula or Approach:

The spin-only magnetic moment (\(\mu\)) is calculated using the formula:
\[ \mu = \sqrt{n(n+2)} \, Bohr Magnetons (BM) \]
where \(n\) is the number of unpaired electrons. A larger \(n\) leads to a larger \(\mu\). We need to find \(n\) for each complex. This requires determining the metal's oxidation state, its d-electron count, and whether the complex is high-spin or low-spin based on the ligand field strength.


Step 3: Detailed Explanation:

1. \([FeF_6]^{3-}\):

- Oxidation state of Fe: Let it be x. \(x + 6(-1) = -3 \implies x = +3\). So, we have Fe\(^{3+}\).

- Electronic configuration of Fe (Z=26) is [Ar] 3d\(^6\) 4s\(^2\). Fe\(^{3+}\) is [Ar] 3d\(^5\).

- Ligand: F\(^-\) is a weak-field ligand. It will form a high-spin octahedral complex.

- For a d\(^5\) high-spin case, all 5 electrons are unpaired (\(t_{2g}^3 e_g^2\)). So, \(n=5\).

- \(\mu = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92\) BM.


2. \([CoF_6]^{3-}\):

- Oxidation state of Co: Let it be x. \(x + 6(-1) = -3 \implies x = +3\). So, we have Co\(^{3+}\).

- Electronic configuration of Co (Z=27) is [Ar] 3d\(^7\) 4s\(^2\). Co\(^{3+}\) is [Ar] 3d\(^6\).

- Ligand: F\(^-\) is a weak-field ligand. It will form a high-spin octahedral complex.

- For a d\(^6\) high-spin case, there are 4 unpaired electrons (\(t_{2g}^4 e_g^2\)). So, \(n=4\).

- \(\mu = \sqrt{4(4+2)} = \sqrt{24} \approx 4.90\) BM.


3. \([Co(C_2O_4)_3]^{3-}\):

- Oxidation state of Co: Let it be x. Oxalate (C\(_2\)O\(_4\)\(^{2-}\)) has a -2 charge. \(x + 3(-2) = -3 \implies x = +3\). So, we have Co\(^{3+}\).

- Electronic configuration: Co\(^{3+}\) is [Ar] 3d\(^6\).

- Ligand: Oxalate is generally considered a strong-field ligand, causing electron pairing. It will form a low-spin octahedral complex.

- For a d\(^6\) low-spin case, all 6 electrons are paired up in the t\(_{2g}\) orbitals (\(t_{2g}^6 e_g^0\)). So, \(n=0\).

- \(\mu = \sqrt{0(0+2)} = 0\) BM.


Comparing the Magnetic Moments:
\(\mu([FeF_6]^{3-}) \approx 5.92\) BM
\(\mu([CoF_6]^{3-}) \approx 4.90\) BM
\(\mu([Co(C_2O_4)_3]^{3-}) = 0\) BM

The order is: \([FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}\).


Step 4: Final Answer:

The correct order of spin only magnetic moment is \([FeF_6]^{3-} > [CoF_6]^{3-} > [Co(C_2O_4)_3]^{3-}\). This corresponds to option (C).
Quick Tip: To solve these problems quickly, you just need to find the number of unpaired electrons, \(n\). The magnetic moment \(\mu\) increases with \(n\). You don't need to calculate the exact value of \(\mu\), just compare the values of \(n\). Here, \(n=5\), \(n=4\), and \(n=0\), so the order is clear. Remember the spectrochemical series to decide if a ligand is weak-field (high-spin) or strong-field (low-spin). Halides are typically weak, while ligands with C or N donors (like CN\(^-\), CO, en) and oxalate are typically strong.

Step 1: Understanding the Question:

We are given a multi-step organic synthesis starting from p-chlorobenzaldehyde. We need to identify the intermediate product A and the final product B.


Step 3: Detailed Explanation of the Reaction Sequence:


Step 1: Formation of A

Reactant: p-Chlorobenzaldehyde (\(p-Cl-C_6H_4-CHO\))

Reagent: NaCN. This is a source of the cyanide nucleophile, CN\(^-\).

The CN\(^-\) ion attacks the electrophilic carbonyl carbon of the aldehyde. Subsequent protonation (from the solvent, which is likely aqueous or alcoholic) of the resulting alkoxide ion yields a cyanohydrin.
\[ Product A = p-Cl-C_6H_4-CH(OH)CN \]
This compound is named p-chloromandelonitrile. This matches product A in option (A).


Step 2: Intermediate Reaction

Reactant: Product A (\(p-Cl-C_6H_4-CH(OH)CN\))

Reagent: EtOH, H\(_2\)O. This condition suggests the hydrolysis of the nitrile group (-CN) to a carboxylic acid group (-COOH). While strong acid or base is typically required, this is the most plausible transformation in this context.
\[ Intermediate = p-Cl-C_6H_4-CH(OH)COOH \]
This is p-chloromandelic acid.


Step 3: Formation of B

Reactant: p-chloromandelic acid

Reagent: MeMgBr (excess), followed by H\(_3\)O\(^+\) workup. MeMgBr is a Grignard reagent, a strong nucleophile and base.

1. Acid-Base Reactions: The Grignard reagent will first react with the two acidic protons of the hydroxyl (-OH) and carboxyl (-COOH) groups. Two equivalents of MeMgBr are consumed to form a dianion.

\(p-Cl-C_6H_4-CH(O^-)-COO^-\)

2. Nucleophilic Addition: A third equivalent of MeMgBr attacks the electrophilic carbon of the carboxylate group. This forms a tetrahedral intermediate.

3. Second Nucleophilic Addition: Upon collapse of this tetrahedral intermediate during the reaction (or workup), a ketone, \(p-Cl-C_6H_4-CH(OH)-CO-CH_3\), would be formed. Since MeMgBr is in excess, it will immediately attack this ketone.

4. A fourth equivalent of MeMgBr attacks the ketone's carbonyl carbon, forming another alkoxide.

5. Workup: The final step is adding H\(_3\)O\(^+\) (acidic workup), which protonates all the alkoxides to give hydroxyl groups.

The final result is the addition of two methyl groups to the original carboxyl carbon.
\[ Product B = p-Cl-C_6H_4-CH(OH)-C(OH)(CH_3)_2 \]
This product is a diol, specifically 2-(4-chlorophenyl)-3-methylbutane-2,3-diol. This matches product B in option (A).


Step 4: Final Answer:

Both the identified structures for A and B match those given in option (A).
Quick Tip: Remember the reactivity of Grignard reagents. They are strong bases and will react with any acidic protons first before acting as nucleophiles. When reacting with esters or carboxylic acids, they add twice to the carbonyl carbon (after the initial acid-base reaction for acids) to produce a tertiary alcohol.

Step 1: Understanding the Question:

We are dealing with the combustion of a hydrocarbon. From the stoichiometry of the reactants and products (oxygen and water), we need to determine the molecular formula of the hydrocarbon.


Step 2: Key Formula or Approach:

The general balanced equation for the combustion of a hydrocarbon C\(_x\)H\(_y\) is:
\[ C_xH_y + \left(x + \frac{y}{4}\right) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O \]
The term "equivalents" here refers to the molar ratio with respect to 1 mole of the hydrocarbon A.


Step 3: Detailed Explanation:

From the balanced equation, for 1 mole (or 1 equivalent) of C\(_x\)H\(_y\):

- Equivalents of oxygen required = \(x + \frac{y}{4}\)

- Equivalents of water produced = \(\frac{y}{2}\)


We are given:

- Equivalents of oxygen required = 9.5

- Equivalents of water produced = 3


Using the information about water:
\[ \frac{y}{2} = 3 \implies y = 6 \]
Now we know the number of hydrogen atoms in the hydrocarbon is 6.


Using the information about oxygen and the value of y we just found:
\[ x + \frac{y}{4} = 9.5 \] \[ x + \frac{6}{4} = 9.5 \] \[ x + 1.5 = 9.5 \] \[ x = 9.5 - 1.5 = 8 \]
Now we know the number of carbon atoms is 8.


The molecular formula of the hydrocarbon A is C\(_8\)H\(_6\).


Step 4: Final Answer:

The molecular formula is C\(_8\)H\(_6\). This corresponds to option (B). (Note: The candidate's chosen option (A) is incorrect).
Quick Tip: For combustion analysis problems, always start with the general balanced equation. The coefficients directly give the molar ratios (or equivalents) of reactants and products relative to the hydrocarbon. Solve for `y` first using the water produced, then solve for `x` using the oxygen consumed.

Step 1: Understanding the Question:

We are given the structure of a tetrapeptide and need to identify the sequence of amino acids using their standard one-letter codes. By convention, peptides are written from the N-terminus (free amino group) to the C-terminus (free carboxyl group).


Step 2: Identifying the Amino Acid Residues:

We need to identify the side chain (R-group) for each amino acid in the chain, starting from the left (N-terminus).


1. First Amino Acid (N-terminus): The side chain is -CH\(_2\)-C\(_6\)H\(_4\)-OH (a benzyl group with a hydroxyl group at the para position). This is the side chain for Tyrosine (Y).

2. Second Amino Acid: The side chain is -(CH\(_2\))\(_2\)-CONH\(_2\). This is the side chain for Glutamine (Q).

3. Third Amino Acid: The side chain is -CH\(_2\)-CH(CH\(_3\))\(_2\). This is the side chain for Leucine (L).

4. Fourth Amino Acid (C-terminus): The side chain is -CH\(_2\)-C\(_6\)H\(_5\) (a benzyl group). This is the side chain for Phenylalanine (F).


Step 3: Determining the Sequence:

The sequence of the tetrapeptide from the N-terminus to the C-terminus is Tyrosine - Glutamine - Leucine - Phenylalanine.

Using the one-letter codes, this is represented as Y-Q-L-F.


Step 4: Final Answer:

The correct representation of the tetrapeptide is YQLF. This corresponds to option (A).
Quick Tip: To identify peptide sequences, first locate the peptide backbone (-N-C\(\alpha\)-C-). The groups attached to the alpha-carbons (C\(\alpha\)) are the side chains (R-groups) that define the amino acids. Always read the sequence from the N-terminus (free -NH\(_2\) or -NH\(_3^+\)) to the C-terminus (free -COOH or -COO\(^-\)). Memorizing the structures and codes of the 20 standard amino acids is essential.

Step 1: Understanding the Question:

We need to identify the product of the reaction between propanamide and a mixture of bromine and aqueous potassium hydroxide.


Step 2: Identifying the Named Reaction:

The reaction of a primary amide with bromine in an aqueous or ethanolic solution of sodium/potassium hydroxide is known as the Hofmann bromamide degradation reaction.


Step 3: Applying the Reaction Principle:

The key feature of the Hofmann bromamide degradation is that it converts a primary amide into a primary amine containing one carbon atom less than the original amide. The carbonyl carbon of the amide group is lost (as carbonate).

- The starting material is propanamide: CH\(_3\)CH\(_2\)CONH\(_2\). It contains 3 carbon atoms.

- The product will be a primary amine with \(3-1=2\) carbon atoms.

- The two-carbon primary amine is ethylamine: CH\(_3\)CH\(_2\)NH\(_2\).


The overall reaction is:
\[ CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O \]

Step 4: Final Answer:

The product of the reaction is Ethylamine. This corresponds to option (D).
Quick Tip: The Hofmann bromamide reaction is a "step-down" reaction, meaning it shortens the carbon chain by one. Just remove the C=O group from the amide to find the structure of the resulting amine. This is a quick way to identify the product in multiple-choice questions.

Step 1: Understanding the Question:

We need to match the chemical terms in List I with their correct definitions or related concepts in List II.


Step 3: Detailed Matching:

- A. van't Hoff factor, i: This factor accounts for the effect of solute dissociation or association on colligative properties. It is defined as the ratio of the observed colligative property to the calculated colligative property. Since colligative properties are inversely proportional to molar mass, \(i\) is also defined as the ratio of the normal (theoretical) molar mass to the abnormal (observed) molar mass. This matches with III.

- B. k\(_f\): This is the molal freezing point depression constant, also known as the cryoscopic constant. It is a property of the solvent. This matches with I.

- C. Solutions with same osmotic pressure: By definition, solutions that have the same osmotic pressure at a given temperature are called isotonic solutions. This matches with II.

- D. Azeotropes: These are liquid mixtures that have a constant boiling point and whose vapor has the same composition as the liquid. This means they are solutions with the same composition of vapour above it. This matches with IV.


Summary of Matches:

A \(\rightarrow\) III

B \(\rightarrow\) I

C \(\rightarrow\) II

D \(\rightarrow\) IV


Step 4: Final Answer:

The correct set of matches is A-III, B-I, C-II, D-IV. This corresponds to option (C). (Note: The candidate's chosen option (D) is incorrect).
Quick Tip: Create a quick reference sheet for the "Solutions" chapter with key definitions: van't Hoff factor, molal/molar constants (ebullioscopic, cryoscopic), isotonic/hypotonic/hypertonic solutions, azeotropes, and Henry's law. Matching questions are common and test direct knowledge of these definitions.

Step 1: Understanding the Question:

This is a knowledge-based question from the chapter "Chemistry in Everyday Life". We need to identify the therapeutic use of the drug Equanil.


Step 2: Identifying the Drug Class:

Equanil is the trade name for the drug Meprobamate. Meprobamate belongs to a class of drugs called tranquilizers.


Step 3: Function of Tranquilizers:

Tranquilizers are neurological drugs that are used to treat conditions such as stress, anxiety, and mental disorders. They act on the central nervous system to induce a sense of calm and well-being. Equanil, in particular, is used to control depression and hypertension (high blood pressure). It helps relieve anxiety and tension.


Step 4: Evaluating the Options:

- (A) Stomach ulcers and (B) Hyperacidity are treated with antacids (like ranitidine, cimetidine). So, these are incorrect.

- (C) While Equanil is used for anxiety and stress, option (D) is more specific and encompassing of its primary uses as listed in many textbooks.

- (D) Depression and hypertension are specific conditions for which Equanil is prescribed. This is the most accurate description of its use among the choices.


Step 4: Final Answer:

Equanil is used to treat depression and hypertension. This corresponds to option (D).
Quick Tip: For the "Chemistry in Everyday Life" chapter, create flashcards for important drugs with their class (e.g., antacid, antihistamine, tranquilizer, antibiotic) and their specific use. Questions are often direct recall of these facts. Equanil is a classic example of a tranquilizer.

Step 1: Understanding the Question:

We need to perform a dehydrohalogenation (E2 elimination) reaction on four different alkyl bromides and determine which one produces the highest number of unique isomeric alkenes (including constitutional and stereoisomers like E/Z or cis/trans).


Step 2: Analyzing Each Reactant:

Dehydrohalogenation involves removing H and Br from adjacent carbons to form a double bond. We need to check for all possible \(\beta\)-hydrogens.


- (A) 2-Bromopropane: CH\(_3\)-CH(Br)-CH\(_3\). The two \(\beta\)-carbons (C1 and C3) are equivalent. Removing H from either gives only one product: Propene (CH\(_3\)-CH=CH\(_2\)). Total products: 1.


- (B) 1-Bromo-2-methylbutane: CH\(_3\)-CH\(_2\)-CH(CH\(_3\))-CH\(_2\)Br. There is only one \(\beta\)-carbon (the CH group). Removing the H from this carbon gives only one constitutional isomer: 2-methylbut-1-ene (CH\(_3\)-CH\(_2\)-C(CH\(_3\))=CH\(_2\)). This alkene does not show geometric isomerism. Total products: 1. (Zaitsev elimination would give 2-methylbut-2-ene, but this requires rearrangement, which is excluded).


- (C) 2-Bromopentane: CH\(_3\)-CH\(_2\)-CH\(_2\)-CH(Br)-CH\(_3\). There are two different \(\beta\)-carbons: C1 and C3.

- Elimination of H from C1 gives: Pent-1-ene (CH\(_3\)-CH\(_2\)-CH\(_2\)-CH=CH\(_2\)). This does not have geometric isomers. (1 product)

- Elimination of H from C3 gives: Pent-2-ene (CH\(_3\)-CH\(_2\)-CH=CH-CH\(_3\)). This alkene has two different groups on each carbon of the double bond, so it can exist as geometric (cis/trans or E/Z) isomers. (2 products)

Total products: 1 + 2 = 3.


- (D) 2-Bromo-3,3-dimethylpentane: CH\(_3\)-CH\(_2\)-C(CH\(_3\))\(_2\)-CH(Br)-CH\(_3\). There are two \(\beta\)-carbons: C1 and C3.

- Elimination of H from C1 gives: 3,3-Dimethylpent-1-ene. (1 product)

- The C3 carbon has no hydrogen atoms, so elimination from this side is not possible.

Total products: 1.


Step 3: Comparing the Results:

- (A) gives 1 product.
- (B) gives 1 product.
- (C) gives 3 products.
- (D) gives 1 product.
The maximum number of isomeric alkenes is produced by 2-Bromopentane.


Step 4: Final Answer:

2-Bromopentane gives the maximum number of isomeric alkenes. This corresponds to option (C).
Quick Tip: To find the number of alkene products from elimination, identify all unique \(\beta\)-hydrogens. For each unique \(\beta\)-hydrogen, draw the resulting alkene. Then, check each alkene for the possibility of geometric isomerism (cis/trans or E/Z). An alkene C(R1,R2)=C(R3,R4) shows geometric isomerism only if R1 \(\neq\) R2 and R3 \(\neq\) R4.

Step 1: Understanding the Question:

We need to match the classification of polymers based on intermolecular forces (List I) with their corresponding examples (List II).


Step 3: Detailed Matching:

- A. Elastomeric polymer (Elastomer): These polymers have weak intermolecular forces, allowing them to be stretched. They possess elastic properties. Neoprene is a synthetic rubber and is a classic example of an elastomer. So, A matches with IV.

- B. Fibre Polymer: These polymers have strong intermolecular forces like hydrogen bonds or dipole-dipole interactions, which lead to close packing of chains and high tensile strength. They are used to make fibres. Polyester (like Dacron or Terylene) is a common fibre. So, B matches with III.

- C. Thermosetting Polymer: These polymers undergo extensive cross-linking when heated, leading to a rigid 3D network structure. Once set, they cannot be remelted. Urea formaldehyde resin is a thermosetting polymer. So, C matches with I.

- D. Thermoplastic Polymer: These polymers have intermediate intermolecular forces. They soften on heating and harden on cooling, and this process is reversible. Polystyrene is a widely used thermoplastic. So, D matches with II.


Summary of Matches:

A \(\rightarrow\) IV

B \(\rightarrow\) III

C \(\rightarrow\) I

D \(\rightarrow\) II


Step 4: Final Answer:

The correct set of matches is A-IV, B-III, C-I, D-II. This corresponds to option (D). (Note: The candidate's chosen option (A) seems to be based on an incorrect match).
Quick Tip: Remember the four main classes of polymers based on intermolecular forces: Elastomers (weakest forces, e.g., rubbers), Thermoplastics (intermediate forces, e.g., Polythene, PVC, Polystyrene), Fibres (strongest forces, e.g., Nylon, Polyester), and Thermosetting polymers (cross-linked, e.g., Bakelite, Urea-formaldehyde resin).

Step 1: Understanding the Question:

We need to evaluate the correctness of two independent statements, one about the catalytic uses of Nickel and the other about the types of hydrides formed by Silicon.


Step 3: Detailed Explanation:

Analyzing Statement I:

"Nickel is being used as the catalyst for producing syn gas and edible fats."

- Production of edible fats: Edible fats (like vanaspati ghee) are produced by the hydrogenation of vegetable oils. Finely divided Nickel (Raney Nickel) is the catalyst commonly used for this process. This part is correct.

- Production of syngas: Syngas (a mixture of CO and H\(_2\)) can be produced by the steam reforming of hydrocarbons like methane. This industrial process uses a Nickel catalyst at high temperatures. (\( CH_4 + H_2O \xrightarrow{Ni} CO + 3H_2 \)). This part is also correct.

Since both uses are correct, Statement I is correct.


Analyzing Statement II:

"Silicon forms both electron rich and electron deficient hydrides."

Hydrides are classified based on the number of valence electrons in the central atom compared to what is needed for normal covalent bonding.

- Electron deficient hydrides are formed by Group 13 elements (e.g., B\(_2\)H\(_6\)), which have fewer valence electrons than required for conventional bonding.

- Electron precise hydrides are formed by Group 14 elements (e.g., CH\(_4\), SiH\(_4\)). They have the exact number of electrons to form normal covalent bonds.

- Electron rich hydrides are formed by Group 15, 16, and 17 elements (e.g., NH\(_3\), H\(_2\)O, HF), which have lone pairs of electrons.

Silicon is in Group 14. Its hydride, silane (SiH\(_4\)), is an electron-precise hydride. It does not form electron-deficient or electron-rich hydrides. Therefore, Statement II is incorrect.


Step 4: Final Answer:

Statement I is correct but statement II is incorrect. This corresponds to option (A). (Note: The candidate's chosen option (B) is incorrect).
Quick Tip: Remember the classification of covalent hydrides based on their group in the periodic table: Group 13 forms electron-deficient, Group 14 forms electron-precise, and Groups 15-17 form electron-rich hydrides. This is a direct and reliable way to classify them.

Step 1: Understanding the Question:

We need to evaluate four statements about the oxidation states and properties of d-block elements and identify which of them are correct.


Step 3: Detailed Explanation:

- (i) Manganese exhibits +7 oxidation state in its oxide.
Manganese shows a wide range of oxidation states, with the highest being +7. This is famously exhibited in potassium permanganate (KMnO\(_4\)) and in its acidic oxide, dimanganese heptoxide (Mn\(_2\)O\(_7\)). So, this statement is correct.


- (ii) Ruthenium and Osmium exhibit +8 oxidation in their oxides.
Ruthenium (Ru) and Osmium (Os) are in the second and third transition series, respectively, below iron. They are known to exhibit the very high oxidation state of +8 in their oxides, RuO\(_4\) (ruthenium tetroxide) and OsO\(_4\) (osmium tetroxide). So, this statement is correct.


- (iii) Sc shows +4 oxidation state which is oxidizing in nature.
Scandium (Sc) has the electronic configuration [Ar] 3d\(^1\) 4s\(^2\). It loses all three of its valence electrons to form the Sc\(^{3+}\) ion, which has a stable noble gas configuration. Scandium exclusively shows the +3 oxidation state in its compounds. It does not show a +4 oxidation state. So, this statement is incorrect.


- (iv) Cr shows oxidising nature in +6 oxidation state.
Chromium in its +6 oxidation state, as found in compounds like potassium dichromate (K\(_2\)Cr\(_2\)O\(_7\)) and chromium trioxide (CrO\(_3\)), is a very strong oxidizing agent. For example, dichromate is widely used in redox titrations to oxidize Fe\(^{2+}\) to Fe\(^{3+}\) or I\(^-\) to I\(_2\). So, this statement is correct.


Step 4: Final Answer:

The correct statements are (i), (ii), and (iv). This corresponds to option (D). (Note: The candidate's chosen option (C) is incorrect because statement (iii) is false).
Quick Tip: Remember the trends in oxidation states for d-block elements. The maximum oxidation state generally increases up to the middle of the series (e.g., Mn shows +7). The highest oxidation states are typically found in oxides and fluorides. For heavier transition metals (like Ru, Os), higher oxidation states are more stable than for their lighter congeners.

Step 1: Understanding the Question:

We need to calculate the bond order for three different diatomic species using Molecular Orbital (MO) Theory.


Step 2: Key Formula or Approach:

The bond order (BO) is calculated using the formula:
\[ Bond Order = \frac{1}{2} (Number of bonding electrons - Number of antibonding electrons) \] \[ BO = \frac{1}{2} (N_b - N_a) \]
We need to determine the total number of electrons for each species and fill the MO energy level diagram.


Step 3: Detailed Explanation:

1. O\(_2^{2-}\) (Peroxide ion):

- Total electrons = 8(O) + 8(O) + 2(charge) = 18 electrons.

- MO configuration (for species with > 14 electrons):

\((\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\sigma_{2p_z})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\pi^*_{2p_x})^2 (\pi^*_{2p_y})^2\)

- Bonding electrons (\(N_b\)): 2 (from \(\sigma_{1s}\)) + 2 (from \(\sigma_{2s}\)) + 2 (from \(\sigma_{2p_z}\)) + 4 (from \(\pi_{2p}\)) = 10.

- Antibonding electrons (\(N_a\)): 2 (from \(\sigma^*_{1s}\)) + 2 (from \(\sigma^*_{2s}\)) + 4 (from \(\pi^*_{2p}\)) = 8.

- Bond Order = \(\frac{1}{2}(10 - 8) = \frac{2}{2} = 1\).


2. CO (Carbon Monoxide):

- Total electrons = 6(C) + 8(O) = 14 electrons.

- It is isoelectronic with N\(_2\).

- MO configuration (for species with \(\le\) 14 electrons):

\((\sigma_{1s})^2 (\sigma^*_{1s})^2 (\sigma_{2s})^2 (\sigma^*_{2s})^2 (\pi_{2p_x})^2 (\pi_{2p_y})^2 (\sigma_{2p_z})^2\)

- Bonding electrons (\(N_b\)): 10.

- Antibonding electrons (\(N_a\)): 4.

- Bond Order = \(\frac{1}{2}(10 - 4) = \frac{6}{2} = 3\).


3. NO\(^+\) (Nitrosonium ion):

- Total electrons = 7(N) + 8(O) - 1(charge) = 14 electrons.

- It is also isoelectronic with N\(_2\) and CO.

- The MO configuration and electron count are the same as for CO.

- Bond Order = \(\frac{1}{2}(10 - 4) = 3\).


Step 4: Final Answer:

The bond orders for O\(_2^{2-}\), CO, and NO\(^+\) are 1, 3, and 3, respectively. This corresponds to option (C). (Note: The candidate's chosen option (A) is incorrect. The bond order of NO\(^+\) is 3, not 2).
Quick Tip: Memorize the bond orders for common diatomic species based on the total electron count. This pattern is a very powerful shortcut for 2nd-period diatomics:
10 e\(^-\) \(\rightarrow\) Bond Order = 1 (e.g., Be\(_2\))
11 e\(^-\) \(\rightarrow\) Bond Order = 1.5
12 e\(^-\) \(\rightarrow\) Bond Order = 2 (e.g., C\(_2\))
13 e\(^-\) \(\rightarrow\) Bond Order = 2.5
14 e\(^-\) \(\rightarrow\) Bond Order = 3 (e.g., N\(_2\), CO, NO\(^+\))
15 e\(^-\) \(\rightarrow\) Bond Order = 2.5 (e.g., NO)
16 e\(^-\) \(\rightarrow\) Bond Order = 2 (e.g., O\(_2\))
17 e\(^-\) \(\rightarrow\) Bond Order = 1.5 (e.g., O\(_2^-\))
18 e\(^-\) \(\rightarrow\) Bond Order = 1 (e.g., F\(_2\), O\(_2^{2-}\))

Step 1: Understanding the Question:

This is a stoichiometry problem involving two consecutive reactions. First, sodium reacts with water to produce NaOH. Second, this NaOH is neutralized by an HCl solution. We need to find the volume of HCl required.


Step 2: Writing the Balanced Equations:

1. Reaction of sodium with water: \(2Na(s) + 2H_2O(l) \rightarrow 2NaOH(aq) + H_2(g)\)

2. Neutralization reaction: \(NaOH(aq) + HCl(aq) \rightarrow NaCl(aq) + H_2O(l)\)


Step 3: Step-by-Step Calculation:

1. Calculate moles of Na:

Molar mass of Na = 23 g/mol.

Moles of Na = \(\frac{Mass}{Molar Mass} = \frac{0.69 g}{23 g/mol} = 0.03 mol\).


2. Calculate moles of NaOH produced:

From the stoichiometry of the first reaction (2Na \(\rightarrow\) 2NaOH), the molar ratio is 1:1.

Moles of NaOH = Moles of Na = 0.03 mol.


3. Calculate moles of HCl required:

From the stoichiometry of the second reaction (NaOH + HCl \(\rightarrow\) NaCl), the molar ratio is 1:1.

Moles of HCl required = Moles of NaOH = 0.03 mol.


4. Calculate the molarity of the HCl solution:

The solution contains 73 g of HCl per liter.

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol.

Molarity (M) = \(\frac{Mass per liter}{Molar Mass} = \frac{73 g/L}{36.5 g/mol} = 2 mol/L\).


5. Calculate the volume of HCl solution required:

Volume (L) = \(\frac{Moles}{Molarity} = \frac{0.03 mol}{2 mol/L} = 0.015 L\).


6. Convert the volume to mL:

Volume (mL) = \(0.015 L \times 1000 mL/L = 15 mL\).


Step 4: Final Answer:

The volume of HCl required is 15 mL. The nearest integer is 15. (Note: The candidate's given answer of 2 is incorrect and likely resulted from a misreading of the mass of sodium as 0.069g, which would give 1.5 mL, rounding to 2).
Quick Tip: In sequential reaction stoichiometry problems, the product of one reaction becomes the reactant for the next. The key is to carry forward the number of moles correctly from one step to the next using the balanced chemical equations.

Step 1: Understanding the Question:

This is a combustion analysis problem. We are given information about the combustion of a known amount (in moles) of an organic compound, and we need to determine its molar mass.


Step 2: Key Principle:

The Law of Conservation of Mass states that all the carbon atoms in the CO\(_2\) produced must have come from the original organic compound.


Step 3: Step-by-Step Calculation:

1. Calculate moles of CO\(_2\) produced:

Molar mass of CO\(_2\) = 12 + 2(16) = 44 g/mol.

Moles of CO\(_2\) = \(\frac{Mass}{Molar Mass} = \frac{4.4 g}{44 g/mol} = 0.1 mol\).


2. Calculate moles of Carbon atoms:

Each molecule of CO\(_2\) contains one atom of Carbon.

Therefore, moles of C atoms = Moles of CO\(_2\) = 0.1 mol.


3. Relate moles of C to moles of the compound:

We know that 0.01 mol of the organic compound produced 0.1 mol of C atoms.

Therefore, 1 mole of the organic compound must contain \(\frac{0.1 mol C}{0.01 mol compound} = 10\) moles of C atoms.

So, the molecular formula of the compound is of the form C\(_{10}\)H\(_y\)O\(_z\)...


4. Calculate the mass of Carbon in one mole of the compound:

Mass of C in 1 mole = (Number of C atoms) \(\times\) (Molar mass of C)

Mass of C = \(10 \times 12 g/mol = 120 g\).


5. Calculate the total molar mass of the compound:

We are given that the compound contains 60% carbon by mass.

Let M be the molar mass of the compound.

Mass of C in 1 mole = 60% of M.

\[ 120 g = 0.60 \times M \]
\[ M = \frac{120}{0.60} = 200 g/mol \]

Step 4: Final Answer:

The molar mass of the compound is 200 g/mol. The nearest integer is 200.
Quick Tip: In combustion analysis, first find the moles of products (CO\(_2\), H\(_2\)O). From this, find the moles of the constituent elements (C, H). Then, use the initial amount of the compound to find the number of atoms of each element per molecule (the empirical or molecular formula).

Step 1: Understanding the Question:

We are given the value and units of a rate constant (k) and asked to determine the order of the reaction.


Step 2: Key Formula or Approach:

The units of the rate constant for a reaction of order 'n' are given by the general formula:
\[ Units of k = (Concentration)^{1-n} (Time)^{-1} \]
Commonly, concentration is expressed in mol L\(^{-1}\) and time in s. So, the units are (mol L\(^{-1}\))\(^{1-n}\) s\(^{-1}\).


Step 3: Detailed Explanation:

The given units of the rate constant are L mol\(^{-1}\) s\(^{-1}\).

Let's match these units with the general formula.
\[ L mol^{-1} s^{-1} = (mol L^{-1})^{1-n} s^{-1} \]
First, let's rewrite the given units to match the (mol L\(^{-1}\)) format:
\[ L mol^{-1} = (mol L^{-1})^{-1} \]
So, the given units are \((mol L^{-1})^{-1} s^{-1}\).

Now, we can equate the exponents of the concentration term:
\[ -1 = 1 - n \]
Solving for n:
\[ n = 1 - (-1) = 1 + 1 = 2 \]
Therefore, the reaction is of the second order.


Step 4: Final Answer:

The order of the reaction is 2.
Quick Tip: You can memorize the units of the rate constant for common reaction orders:
- Zero order: mol L\(^{-1}\) s\(^{-1}\)
- First order: s\(^{-1}\)
- Second order: L mol\(^{-1}\) s\(^{-1}\)
- Third order: L\(^2\) mol\(^{-2}\) s\(^{-1}\)
Recognizing these patterns allows you to determine the reaction order instantly.

Step 1: Understanding the Question:

We need to identify the products formed upon the thermal decomposition of lithium nitrate (LiNO\(_3\)) and count how many of those products are present in the given list.


Step 2: Thermal Decomposition of Alkali Metal Nitrates:

- Most alkali metal nitrates (like NaNO\(_3\), KNO\(_3\)) decompose upon heating to yield the corresponding metal nitrite and oxygen gas. Example: \(2NaNO_3 \rightarrow 2NaNO_2 + O_2\).

- Lithium nitrate is an exception. Due to the small size and high polarizing power of the Li\(^+\) ion, it has a diagonal relationship with magnesium (Mg). Therefore, lithium nitrate decomposes in a manner similar to Group 2 metal nitrates.


Step 3: Writing the Reaction and Identifying Products:

The thermal decomposition of lithium nitrate yields lithium oxide, nitrogen dioxide, and oxygen.

The balanced chemical equation is:
\[ 4LiNO_3(s) \xrightarrow{\Delta} 2Li_2O(s) + 4NO_2(g) + O_2(g) \]
The products formed are:

1. Lithium oxide (Li\(_2\)O)

2. Nitrogen dioxide (NO\(_2\))

3. Oxygen (O\(_2\))


Step 4: Comparing with the Given List:

The given list is: {Li\(_2\)O, N\(_2\), O\(_2\), LiNO\(_2\), NO\(_2\).

Let's check which of our products are on this list:

- Li\(_2\)O is on the list.

- NO\(_2\) is on the list.

- O\(_2\) is on the list.

- N\(_2\) and LiNO\(_2\) are not produced.


The number of compounds from the list that are produced is 3.

*(Note: The candidate's answer was 2. This might be based on the assumption that it decomposes to nitrite and oxygen, which would give LiNO\(_2\) and O\(_2\) (2 compounds). However, the complete decomposition to the oxide is the standard reaction taught for LiNO\(_3\).)*


Step 5: Final Answer:

The reaction produces 3 compounds from the given list.
Quick Tip: Remember the anomalous behavior of lithium in Group 1. Its properties (like the decomposition of its nitrate and carbonate) often resemble those of magnesium (Group 2) due to the diagonal relationship. This is a common point tested in exams.

Step 1: Understanding the Question:

We need to find the total number of voids (both tetrahedral and octahedral) in a given amount (in moles) of a substance that crystallizes in a hexagonal close-packed (HCP) structure.


Step 2: Key Concepts of Close-Packed Structures:

In any close-packed structure (HCP or CCP/FCC), for a lattice containing N atoms:

- The number of octahedral voids = N

- The number of tetrahedral voids = 2N

- The total number of voids = (Octahedral voids) + (Tetrahedral voids) = N + 2N = 3N.


Step 3: Step-by-Step Calculation:

1. Calculate the number of atoms (N) in 0.02 mol:

Number of atoms (N) = Moles \(\times\) Avogadro's number (N\(_A\))

\[ N = 0.02 mol \times (6.02 \times 10^{23} atoms/mol) \]
\[ N = 0.1204 \times 10^{23} atoms = 1.204 \times 10^{22} atoms \]

2. Calculate the total number of voids:

Total voids = 3N

\[ Total voids = 3 \times (1.204 \times 10^{22}) = 3.612 \times 10^{22} \]

3. Express the answer in the required format:

The question asks for the answer in the form of `_____ \(\times 10^{21}\)`.

\[ 3.612 \times 10^{22} = 36.12 \times 10^{21} \]

4. Round to the nearest integer:

The nearest integer to 36.12 is 36.


Step 4: Final Answer:

The total number of voids is 36 \(\times 10^{21}\).
Quick Tip: For any close-packed structure (HCP, CCP, FCC), the relationship between the number of atoms (N) and voids is fixed: N octahedral voids and 2N tetrahedral voids. So, the total number of voids is always 3N. This is a crucial fact for solid-state chemistry problems.

Step 1: Understanding the Question:

We need to classify a given list of oxides as acidic, basic, neutral, or amphoteric, and then count the number of acidic oxides.


Step 2: General Rules for Oxide Classification:

- Acidic Oxides: Generally formed by non-metals. They react with water to form acids or with bases to form salts. Examples: SO\(_2\), CO\(_2\), N\(_2\)O\(_5\).

- Basic Oxides: Generally formed by metals (especially alkali and alkaline earth metals). They react with water to form bases or with acids to form salts. Examples: Na\(_2\)O, CaO.

- Neutral Oxides: Non-metal oxides that show neither acidic nor basic properties. The common examples are CO, NO, and N\(_2\)O.

- Amphoteric Oxides: Oxides that can react with both acids and bases. Examples: Al\(_2\)O\(_3\), ZnO, SnO, PbO.


Step 3: Classifying the Given Oxides:

- N\(_2\)O\(_3\): Dinitrogen trioxide. An oxide of a non-metal. It reacts with water to form nitrous acid (HNO\(_2\)). It is acidic.

- NO\(_2\): Nitrogen dioxide. An oxide of a non-metal. It is a mixed anhydride, reacting with water to form both nitric acid (HNO\(_3\)) and nitrous acid (HNO\(_2\)). It is acidic.

- N\(_2\)O: Dinitrogen monoxide (nitrous oxide). It is a well-known neutral oxide.

- Cl\(_2\)O\(_7\): Dichlorine heptoxide. An oxide of a non-metal in a high oxidation state. It reacts with water to form perchloric acid (HClO\(_4\)). It is strongly acidic.

- SO\(_2\): Sulfur dioxide. An oxide of a non-metal. It reacts with water to form sulfurous acid (H\(_2\)SO\(_3\)). It is acidic.

- CO: Carbon monoxide. It is a well-known neutral oxide.

- CaO: Calcium oxide. An oxide of an alkaline earth metal. It is a basic oxide.

- Na\(_2\)O: Sodium oxide. An oxide of an alkali metal. It is a strongly basic oxide.

- NO: Nitrogen monoxide (nitric oxide). It is a well-known neutral oxide.


Step 4: Counting the Acidic Oxides:

The acidic oxides from the list are N\(_2\)O\(_3\), NO\(_2\), Cl\(_2\)O\(_7\), and SO\(_2\).

The total count is 4.

*(Note: The candidate's given answer of 3 is incorrect. All four listed are definitively acidic oxides according to standard chemical literature.)*


Step 5: Final Answer:

The total number of acidic oxides is 4.
Quick Tip: To quickly classify oxides, remember the key categories. Memorize the three common neutral oxides: CO, NO, N\(_2\)O. Oxides of alkali/alkaline earth metals are strongly basic. Other metal oxides can be basic or amphoteric. Non-metal oxides are almost always acidic (except for the neutral ones).

Step 1: Understanding the Question:

We are given three equilibrium reactions with their equilibrium constants (K\(_1\), K\(_2\), K\(_3\)). We need to find the equilibrium constant (K\(_target\)) for a target reaction by algebraically manipulating the given reactions.


Step 2: Manipulating the Given Reactions:

Let's label the given reactions:

(1) N\(_2\)(g) + 3H\(_2\)(g) \(\rightleftharpoons\) 2NH\(_3\) (g), K\(_1\) = 4 \(\times\) 10\(^5\)

(2) N\(_2\)(g) + O\(_2\)(g) \(\rightleftharpoons\) 2NO (g), K\(_2\) = 1.6 \(\times\) 10\(^{12}\)

(3) H\(_2\)(g) + \(\frac{1}{2}\)O\(_2\)(g) \(\rightleftharpoons\) H\(_2\)O (g), K\(_3\) = 1.0 \(\times\) 10\(^{-13}\)

Target Reaction: 2NH\(_3\) (g) + \(\frac{5}{2}\)O\(_2\) (g) \(\rightleftharpoons\) 2NO (g) + 3H\(_2\)O (g)


To obtain the target reaction, we perform the following steps:

- Reverse Reaction (1): We need 2NH\(_3\) on the reactant side. Reversing (1) gives:

(1') 2NH\(_3\)(g) \(\rightleftharpoons\) N\(_2\)(g) + 3H\(_2\)(g). The new constant is K\(_1'\) = 1/K\(_1\).

- Use Reaction (2) as is: We need 2NO on the product side. Reaction (2) already has this.

(2') N\(_2\)(g) + O\(_2\)(g) \(\rightleftharpoons\) 2NO (g). The constant is K\(_2'\) = K\(_2\).

- Multiply Reaction (3) by 3: We need 3H\(_2\)O on the product side. Multiplying (3) by 3 gives:

(3') 3H\(_2\)(g) + \(\frac{3}{2}\)O\(_2\)(g) \(\rightleftharpoons\) 3H\(_2\)O (g). The new constant is K\(_3'\) = (K\(_3\))\(^3\).


Step 3: Combining the Reactions and Constants:

Now, add the manipulated reactions (1'), (2'), and (3'):

(2NH\(_3\)) + (N\(_2\) + O\(_2\)) + (3H\(_2\) + \(\frac{3}{2}\)O\(_2\)) \(\rightleftharpoons\) (N\(_2\) + 3H\(_2\)) + (2NO) + (3H\(_2\)O)

Cancel the species that appear on both sides (N\(_2\) and 3H\(_2\)):

2NH\(_3\) + O\(_2\) + \(\frac{3}{2}\)O\(_2\) \(\rightleftharpoons\) 2NO + 3H\(_2\)O

Combining the O\(_2\) terms (1 + 3/2 = 5/2):

2NH\(_3\)(g) + \(\frac{5}{2}\)O\(_2\)(g) \(\rightleftharpoons\) 2NO (g) + 3H\(_2\)O (g)

This matches the target reaction. The equilibrium constant for the combined reaction is the product of the constants of the individual manipulated reactions:
\[ K_{target} = K_1' \times K_2' \times K_3' = \left(\frac{1}{K_1}\right) \times (K_2) \times (K_3)^3 \]

Step 4: Calculation:
\[ K_{target} = \left(\frac{1}{4 \times 10^5}\right) \times (1.6 \times 10^{12}) \times (1.0 \times 10^{-13})^3 \] \[ K_{target} = (0.25 \times 10^{-5}) \times (1.6 \times 10^{12}) \times (1.0 \times 10^{-39}) \] \[ K_{target} = (0.25 \times 1.6) \times 10^{-5 + 12 - 39} \] \[ K_{target} = 0.4 \times 10^{-32} = 4 \times 10^{-33} \]
The question asks for the answer in the form `_____ \(\times\) 10\(^{-33}\)`. The value is 4.


Step 5: Final Answer:

The value of the equilibrium constant is 4 \(\times\) 10\(^{-33}\).
Quick Tip: When combining equilibria (Hess's Law for K): - If you reverse a reaction, the new K is 1/K\(_old\). - If you add reactions, the new K is the product of the old K's. - If you multiply a reaction by a factor 'n', the new K is (K\(_old\))\(^n\). Carefully apply these rules to each step of the manipulation.

Step 1: Understanding the Question:

We need to determine the denticity of the ligand that complexes with Cu\(^{2+}\) ions in Fehling's reagent. Denticity refers to the number of donor atoms in a single ligand that bind to the central metal ion.


Step 2: Composition of Fehling's Reagent:

Fehling's reagent is prepared by mixing two solutions:

- Fehling's A: An aqueous solution of copper(II) sulfate (CuSO\(_4\)).

- Fehling's B: An alkaline solution of sodium potassium tartrate (NaKC\(_4\)H\(_4\)O\(_6\), also known as Rochelle salt).

When mixed, the Cu\(^{2+}\) ions from Fehling's A form a deep blue complex with the tartrate ions from Fehling's B. This complex prevents the precipitation of copper(II) hydroxide in the alkaline medium.


Step 3: Identifying the Ligand and its Structure:

The ligand is the tartrate ion, C\(_4\)H\(_4\)O\(_6^{2-}\). Its structure is:
\[ [-OOC-CH(OH)-CH(OH)-COO-]^{2-} \]
In the alkaline solution of Fehling's B, the hydroxyl groups are deprotonated to form alkoxides. The complexing species is
\([ OOC-CH(O)-CH(O)-COO ]^{4-}\).

The tartrate ion acts as a chelating ligand. It binds to the central Cu\(^{2+}\) ion through two donor atoms. In the complex, it typically acts as a bidentate ligand, coordinating through two of its oxygen atoms (e.g., the oxygen atoms from the two deprotonated hydroxyl groups).


Step 4: Final Answer:

Since the tartrate ligand binds to the metal center through two donor atoms, its denticity is 2.
Quick Tip: Remember the composition of common organic test reagents. Fehling's reagent and Benedict's reagent both use Cu\(^{2+}\) complexed with a ligand (tartrate in Fehling's, citrate in Benedict's) to keep it soluble in an alkaline solution for testing reducing sugars. The ligand in both cases is bidentate.

Step 1: Understanding the Question:

We are given the equilibrium constant (K) for a redox reaction and the standard reduction potential of one half-cell (Zn\(^{2+}\)/Zn). We need to find the standard reduction potential of the other half-cell (Sn\(^{2+}\)/Sn).


Step 2: Key Formulas:

1. The relationship between the standard cell potential (E\(^\circ\)\(_{cell}\)) and the equilibrium constant (K) is given by the Nernst equation at equilibrium:

\[ E^\circ_{cell} = \frac{2.303RT}{nF} \log K \]
2. The standard cell potential is the difference between the standard reduction potentials of the cathode and the anode:

\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Step 3: Step-by-Step Calculation:

1. Calculate E\(^\circ\)\(_{cell}\):

For the reaction Zn(s) + Sn\(^{2+}\)(aq) \(\rightarrow\) Zn\(^{2+}\)(aq) + Sn(s), two electrons are transferred (n=2).

Given K = 1 \(\times\) 10\(^{20}\) and \(\frac{2.303RT}{F} = 0.059\) V.

\[ E^\circ_{cell} = \frac{0.059}{n} \log K = \frac{0.059}{2} \log(10^{20}) \]
\[ E^\circ_{cell} = \frac{0.059}{2} \times 20 = 0.059 \times 10 = 0.59 V \]

2. Identify Anode and Cathode:

In the given reaction, Zinc (Zn) is being oxidized (Zn \(\rightarrow\) Zn\(^{2+}\)), so it is the anode.

Tin ion (Sn\(^{2+}\)) is being reduced (Sn\(^{2+}\) \(\rightarrow\) Sn), so it is the cathode.


3. Calculate E\(^\circ\)\(_{Sn^{2+}/Sn}\):

\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]
\[ E^\circ_{cell} = E^\circ_{Sn^{2+}/Sn} - E^\circ_{Zn^{2+}/Zn} \]
Substitute the known values:

\[ 0.59 V = E^\circ_{Sn^{2+}/Sn} - (-0.76 V) \]
\[ 0.59 = E^\circ_{Sn^{2+}/Sn} + 0.76 \]
\[ E^\circ_{Sn^{2+}/Sn} = 0.59 - 0.76 = -0.17 V \]

4. Express the Answer in the Required Format:

The question asks for the magnitude of the standard electrode potential, which is |-0.17 V| = 0.17 V.

We need to express this in the format `______ \(\times\) 10\(^{-2}\) V`.

\[ 0.17 V = 17 \times 10^{-2} V \]
The nearest integer is 17.


Step 4: Final Answer:

The value is 17.
Quick Tip: The link between thermodynamics and electrochemistry is \( \Delta G^\circ = -nFE^\circ_{cell} \) and \( \Delta G^\circ = -RT\ln K \). Combining these gives the crucial relation \( E^\circ_{cell} = (RT/nF)\ln K \). Always identify the anode (oxidation) and cathode (reduction) correctly from the overall cell reaction to apply \( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \).

Step 1: Understanding the Question:

We are asked to calculate the radius of the third orbit of a Helium ion (He\(^+\)) using the Bohr model, given a specific value for the radius of the first orbit of a Hydrogen atom.


Step 2: Key Formula from Bohr's Model:

The radius of the n\(^{th}\) orbit in a hydrogen-like species (with atomic number Z) is given by the formula:
\[ r_n = a_0 \frac{n^2}{Z} \]
where:

- \(r_n\) is the radius of the n\(^{th}\) orbit.

- \(a_0\) is the radius of the first Bohr orbit of the hydrogen atom (often called the Bohr radius).

- n is the principal quantum number (the orbit number).

- Z is the atomic number of the element.


Step 3: Step-by-Step Calculation:

1. Identify the given values:

- Radius of the first Bohr orbit of hydrogen, \(a_0 = 0.6\) \AA.

- We need to find the radius for the third orbit, so \(n = 3\).

- The species is the Helium ion (He\(^+\)), so its atomic number is \(Z = 2\).


2. Apply the formula:

\[ r_3(He^+) = a_0 \frac{n^2}{Z} = (0.6 \AA) \frac{3^2}{2} \]
\[ r_3(He^+) = 0.6 \times \frac{9}{2} = 0.6 \times 4.5 \]
\[ r_3(He^+) = 2.7 \AA \]

3. Convert the result to picometers (pm):

The conversion factor is 1 \AA = 100 pm.

\[ r_3(He^+) = 2.7 \AA \times 100 \frac{pm}{\AA} = 270 pm \]

Step 4: Final Answer:

The radius is 270 pm. The nearest integer is 270. (Note: The candidate's given answer of 2 is physically incorrect and may be a data entry error).
Quick Tip: For Bohr model calculations, remember the dependencies of key quantities on n and Z: - Radius: \(r_n \propto \frac{n^2}{Z}\) - Energy: \(E_n \propto -\frac{Z^2}{n^2}\) - Velocity: \(v_n \propto \frac{Z}{n}\) These proportionalities are very useful for solving ratio-based problems quickly.