Step 1: Understanding the Concept:

Significant figures are the digits in a measured value that are known with certainty plus one final digit that is somewhat uncertain. Rules for significant figures include:

1. All non-zero digits are significant.

2. Leading zeros (zeros to the left of the first non-zero digit) are not significant.

3. Zeros between non-zero digits are significant.

4. Trailing zeros after a decimal point are significant.


Step 2: Detailed Explanation:

Let us evaluate each given number:

A. 0.00253: The leading zeros are not significant. The digits 2, 5, and 3 are significant. Total = 3 significant figures.

B. 1.0003: The zeros are between non-zero digits (1 and 3), so they are significant. Total = 5 significant figures.

C. 15.0: The trailing zero after the decimal point is significant. Total = 3 significant figures.

D. 163: All non-zero digits are significant. Total = 3 significant figures.

Numbers A, C, and D all have 3 significant figures.


Step 3: Final Answer:

A, C and D have the same number of significant figures.
Quick Tip: Leading zeros serve only as placeholders to locate the decimal point; they are never significant. Trailing zeros are significant only if a decimal point is present in the number.

Step 1: Understanding the Concept:

The van der Waals constant 'a' represents the magnitude of intermolecular attractive forces between the molecules of a gas. A higher value of 'a' indicates stronger intermolecular forces. These forces depend on molecular size, polarizability, and polarity.


Step 2: Detailed Explanation:

- Ar (Argon): An inert noble gas with very small atoms and weak London dispersion forces. It has the lowest 'a' value among the given options.

- CH\(_{4}\) (Methane): A non-polar molecule larger than Ar, thus having slightly stronger London dispersion forces.

- H\(_{2}\)O (Water): A highly polar molecule that exhibits strong hydrogen bonding. This results in a significantly higher 'a' value compared to simple gases.

- C\(_{6}\)H\(_{6}\) (Benzene): Although non-polar, it is a much larger molecule with a large electron cloud, leading to very strong London dispersion forces. Its 'a' value is the highest in this group (typically around 18 L\(^{2}\cdot\)bar/mol\(^{2}\)).

Increasing order of 'a': Ar \(<\) CH\(_{4}\) \(<\) H\(_{2}\)O \(<\) C\(_{6}\)H\(_{6}\).


Step 3: Final Answer:

The correct increasing order is A, B, C, D.
Quick Tip: Van der Waals constant 'a' is generally proportional to the ease of liquefaction of a gas. Larger and more polar molecules are easier to liquefy and have higher 'a' values.

Step 1: Understanding the Concept:

A catalyst is a substance that increases the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy (\(E_{a}\)). A positive catalyst decreases the energy barrier, allowing more reactant molecules to cross the transition state per unit time.


Step 2: Detailed Explanation:

In a standard reaction coordinate diagram:

1. The y-axis represents Potential Energy and the x-axis represents the Reaction Coordinate.

2. The solid line usually represents the uncatalyzed reaction, which has a higher activation energy (\(E_{a1}\)).

3. The dashed (or alternate) line represents the catalyzed reaction, which has a lower activation energy (\(E_{a2}\)).

4. The starting energy level (reactants) and final energy level (products) remain identical for both paths, as the catalyst does not change the enthalpy of reaction (\(\Delta H\)).

Therefore, the correct diagram shows the "With catalyst" peak below the "Without catalyst" peak.


Step 3: Final Answer:

The diagram where the solid peak is higher than the dashed peak correctly represents the effect of a positive catalyst.
Quick Tip: A catalyst lowers the activation energy for both the forward and the reverse reactions to the same extent, thus it does not affect the equilibrium constant or the heat of reaction.

Step 1: Understanding the Concept:

Henry Moseley's work provided a physical basis for the periodic table by showing that atomic number, rather than atomic mass, is the fundamental property of an element.


Step 2: Key Formula or Approach:

Moseley's Law states that the square root of the frequency (\(\nu\)) of the characteristic X-rays emitted by an element is proportional to its atomic number (\(Z\)):
\[ \sqrt{\nu} = a(Z - b) \]

where \(a\) and \(b\) are constants characteristic of the particular spectral line series.


Step 3: Detailed Explanation:

According to the equation \(\sqrt{\nu} = aZ - ab\), if we plot the square root of frequency (\(\sqrt{\nu}\)) on the y-axis against the atomic number (\(Z\)) on the x-axis, we obtain a straight line. This observation allowed Moseley to predict undiscovered elements and correct positions in the periodic table.


Step 4: Final Answer:

The graph of \(\sqrt{\nu}\) vs \(Z\) is a straight line.
Quick Tip: Moseley's Law confirmed that \(Z\) represents the number of positive charges (protons) in the nucleus. Always remember the plot is \(\sqrt{\nu}\) versus \(Z\), not \(\nu\) versus \(Z\).

Step 1: Understanding the Concept:

The Hall-Heroult process is the major industrial process for smelting aluminium. It involves the electrolytic reduction of alumina (Al\(_{2}\)O\(_{3}\)) dissolved in molten cryolite.


Step 2: Detailed Explanation:

1. Al\(_{2}\)O\(_{3}\) is mixed with Na\(_{3}\)AlF\(_{6}\) (cryolite) and CaF\(_{2}\) (fluorspar) to lower the melting point and increase conductivity.

2. The electrolysis is carried out in a steel tank lined with carbon, which acts as the cathode.

3. Graphite rods are used as the anode. During the process, the oxygen evolved at the anode reacts with the graphite to form CO and CO\(_{2}\).

The overall reaction is: \(2Al_{2}O_{3} + 3C \rightarrow 4Al + 3CO_{2}\).

Here, carbon (in the form of graphite) effectively acts as the reducing agent that facilitates the conversion of Al\(^{3+}\) ions into Al metal.


Step 3: Final Answer:

Graphite is used as the reducing agent/electrode material in the Hall-Heroult process.
Quick Tip: Note that the graphite anodes are consumed during the reaction and must be replaced periodically. Na\(_{3}\)AlF\(_{6}\) and CaF\(_{2}\) are additives, not the reducing agent.

Step 1: Understanding the Concept:

Hydrogen peroxide (H\(_{2}\)O\(_{2}\)) is an unstable liquid that decomposes readily into water and oxygen. This decomposition is catalyzed by light, dust, rough surfaces, and traces of alkali.


Step 2: Detailed Explanation:

To prevent decomposition, H\(_{2}\)O\(_{2}\) is stored in dark-colored plastic or wax-lined glass bottles. Additionally, certain substances act as negative catalysts or "stabilizers" to inhibit the decomposition process. Examples of such stabilizers include urea, phosphoric acid, acetanilide, and sodium stannate.

- Alkali: Promotes decomposition.

- Urea: Acts as a stabilizer to reduce decomposition.

- Dust: Acts as a catalyst to increase decomposition.

- Glass containers: Traces of alkali in glass can catalyze the decomposition.


Step 3: Final Answer:

Urea can reduce the decomposition of H\(_{2}\)O\(_{2}\).
Quick Tip: H\(_{2}\)O\(_{2}\) is often described as "auto-oxidizing" and "auto-reducing." Keeping it away from light and metal ions is crucial for its stability.

Step 1: Understanding the Concept:

The abundance of elements in seawater depends on their geochemical behavior, biological uptake, and adsorption properties. Both sodium (\(Na\)) and potassium (\(K\)) are alkali metals.


Step 2: Detailed Explanation:

- Assertion A: In the oceans, the concentration of sodium ions (\(Na^{+}\)) is roughly 10.8 g/kg, while that of potassium ions (\(K^{+}\)) is roughly 0.39 g/kg. This makes sodium roughly 28–30 times more abundant than potassium. Thus, the assertion is True.

- Reason R: In the periodic table, potassium is below sodium in Group 1. As we move down a group, the atomic and ionic size increases. Therefore, potassium is indeed bigger in size than sodium. Thus, the reason is True.

- Relationship: The reason why sodium is more abundant in seawater is not its size. Instead, it is because potassium ions are more strongly adsorbed by clay minerals and are more effectively taken up by biological organisms and sediments compared to sodium ions. Therefore, R is not the correct explanation for A.


Step 3: Final Answer:

Both A and R are true but R is not the correct explanation of A.
Quick Tip: Sodium and potassium have similar solubilities, but their different interactions with clay and organisms in the crust lead to the drastic difference in oceanic abundance.

Step 1: Understanding the Concept:

Cement is a complex mixture of silicates and aluminates of calcium. Its properties, such as setting time and strength, depend heavily on the relative proportions of its constituent oxides.


Step 2: Detailed Explanation:

For high-quality Portland cement, the composition must be carefully balanced. Two critical ratios are generally monitored:

1. The ratio of silica (\(SiO_{2}\)) to alumina (\(Al_{2}O_{3}\)) should be between 2.5 and 4.0.

2. The ratio of lime (\(CaO\)) to the total of the acidic oxides (silicon dioxide \(SiO_{2}\), aluminium oxide \(Al_{2}O_{3}\), and iron oxide \(Fe_{2}O_{3}\)) is known as the Lime Saturation Factor. For good quality cement, this ratio should be as close as possible to 2.
\[ Ratio = \frac{% CaO}{% SiO_{2} + % Al_{2}O_{3} + % Fe_{2}O_{3}} \approx 2 \]


Step 3: Final Answer:

The ratio should be close to 2.
Quick Tip: If the lime content is too high, the cement becomes unsound and tends to crack. If it is too low, the cement is slow-setting and weak.

Step 1: Understanding the Concept:

The number of unpaired electrons in a coordination complex is determined by the oxidation state of the central metal, its \(d\)-electron configuration, and the strength of the ligands (Strong Field vs. Weak Field) according to Crystal Field Theory.


Step 2: Detailed Explanation:

A. [Cr(CN)\(_{6}\)]\(^{3-}\): Cr is in +3 state (\(3d^{3}\)). Regardless of ligand strength, for \(d^{1}, d^{2}, d^{3}\) in octahedral field, the configuration is \(t_{2g}^{3}e_{g}^{0}\). Unpaired electrons = 3. (A matches II)

B. [Fe(H\(_{2}\)O)\(_{6}\)]\(^{2+}\): Fe is in +2 state (\(3d^{6}\)). H\(_{2}\)O is a Weak Field Ligand (WFL). High spin configuration: \(t_{2g}^{4}e_{g}^{2}\). Unpaired electrons = 4. (B matches IV)

C. [Co(NH\(_{3}\))\(_{6}\)]\(^{3+}\): Co is in +3 state (\(3d^{6}\)). NH\(_{3}\) is a Strong Field Ligand (SFL) for Co\(^{3+}\). Low spin configuration: \(t_{2g}^{6}e_{g}^{0}\). Unpaired electrons = 0. (C matches I)

D. [Ni(NH\(_{3}\))\(_{6}\)]\(^{2+}\): Ni is in +2 state (\(3d^{8}\)). For \(d^{8}\) octahedral, configuration is always \(t_{2g}^{6}e_{g}^{2}\). Unpaired electrons = 2. (D matches III)


Step 3: Final Answer:

Matching: A-II, B-IV, C-I, D-III.
Quick Tip: For \(d^{3}\) and \(d^{8}\) octahedral complexes, the number of unpaired electrons is independent of the field strength of the ligands. Always identify the oxidation state of the metal first.

Step 1: Understanding the Concept:

Ozone depletion in the stratosphere occurs via a catalytic cycle involving halogen radicals, primarily chlorine radicals (\(\dot{C}l\)) derived from chlorofluorocarbons (CFCs).


Step 2: Detailed Explanation:

The standard mechanism for stratospheric ozone breakdown includes:

1. Initiation: UV radiation breaks down CFCs to produce chlorine radicals:
\(CF_{2}Cl_{2} \xrightarrow{uv} \dot{C}l + \dot{C}F_{2}Cl\) (Option A is part of the process).

2. Propagation Step 1: Chlorine radical reacts with ozone:
\(\dot{C}l + O_{3} \rightarrow Cl\dot{O} + O_{2}\) (Option B is part of the process).

3. Propagation Step 2: Chlorine monoxide radical reacts with atomic oxygen to regenerate the chlorine radical:
\(Cl\dot{O} + O \rightarrow \dot{C}l + O_{2}\) (Option C is part of the process).

Option D (\(2Cl\dot{O} \rightarrow ClO_{2} + \dot{Cl}\)) is not a part of the standard stratospheric catalytic cycle for ozone depletion. While other side reactions exist, this is not the fundamental breakdown pathway discussed in textbooks.


Step 3: Final Answer:

Reaction (D) is not part of the standard breakdown of ozone.
Quick Tip: The key to ozone depletion is the "catalytic" nature of \(\dot{Cl}\), where it is consumed in the first step and regenerated in the second, allowing one radical to destroy thousands of ozone molecules.

Step 1: Understanding the Concept:

The reaction is a halolactonization, specifically a bromolactonization. In the presence of \(NaHCO_{3}\), the carboxylic acid is converted into its carboxylate ion (\(RCOO^{-}\)). This nucleophilic oxygen then attacks an intermediate bromonium ion formed across the alkene double bond.


Step 2: Key Formula or Approach:

The formation of a lactone (cyclic ester) depends on the ring size stability. 5-membered rings (\(\gamma\)-lactones) are generally kinetically favored over 6-membered rings (\(\delta\)-lactones) due to the "5-exo-tet" cyclization pathway being more efficient according to Baldwin's rules.


Step 3: Detailed Explanation:

1. Deprotonation: \(NaHCO_{3}\) reacts with the carboxylic acid to form the carboxylate salt.

2. Bromonium Ion Formation: \(Br_{2}\) adds across the terminal double bond of hex-5-enoic acid to form a cyclic bromonium ion intermediate.

3. Intramolecular Cyclization: The carboxylate oxygen acts as an internal nucleophile. It attacks the more substituted carbon of the bromonium ion (C-5) to form a 5-membered ring.

4. Product Identification: The resulting product is 5-(bromomethyl)tetrahydrofuran-2-one. A 6-membered ring formation via attack at C-6 is less favorable under these conditions.


Step 4: Final Answer:

The major product is the 5-membered bromolactone.
Quick Tip: In halolactonization of \(\gamma,\delta\)-unsaturated acids, always prioritize the formation of the 5-membered ring (\(\gamma\)-lactone) as it is the kinetically favored product.

Step 1: Understanding the Concept:

IUPAC nomenclature requires identifying the principal functional group, selecting the longest carbon chain containing it, and numbering the chain to give substituents the lowest possible locants.


Step 2: Key Formula or Approach:

Priority of functional groups: \(-COOH > -CHO > C=O\). Since \(-COOH\) is present, it is the principal group, and the suffix is "-oic acid".


Step 3: Detailed Explanation:

1. Chain Selection and Numbering: The longest chain including the \(-COOH\) group and the ketone group has 6 carbons. Numbering starts at the carboxylic carbon (C-1).

C-1: \(-COOH\)

C-2: \(-CH(CH_{3})-\) (Methyl substituent at pos 2)

C-3: \(-CH_{2}-\)

C-4: \(-CH_{2}-\)

C-5: \(-C(=O)-\) (Oxo substituent at pos 5)

C-6: \(-CH_{3}\)

2. Naming: The base name for 6 carbons is hexanoic acid. Substituents are listed alphabetically: 2-Methyl and 5-Oxo.

3. Full Name: 2-Methyl-5-oxohexanoic acid.


Step 4: Final Answer:

The IUPAC name is 2-Methyl-5-oxohexanoic acid.
Quick Tip: Always remember that the carbon of the \(-COOH\) group is assigned locant 1. When a ketone is a secondary functional group, use the prefix "oxo".

Step 1: Understanding the Concept:

Nucleophilic Aromatic Substitution (\(S_{N}Ar\)) in haloarenes is facilitated by the presence of strong electron-withdrawing groups (EWGs) like \(-NO_{2}\) at ortho and para positions relative to the leaving group (halogen).


Step 2: Key Formula or Approach:

Reactivity \(\propto\) Number of electron-withdrawing groups at ortho/para positions. These groups stabilize the anionic Meisenheimer intermediate through resonance.


Step 3: Detailed Explanation:

1. Compound A (Chlorobenzene): Lacks EWGs; extremely slow reactivity.

2. Compound B (4-nitrochlorobenzene): One \(-NO_{2}\) group at the para position stabilizes the intermediate.

3. Compound C (2,4-dinitrochlorobenzene): Two \(-NO_{2}\) groups (ortho and para) provide significant stabilization.

4. Compound D (2,4,6-trinitrochlorobenzene): Three \(-NO_{2}\) groups (two ortho, one para) provide maximum stabilization. This compound reacts even with warm water.

5. Order: Reactivity increases with the number of nitro groups: D \(>\) C \(>\) B \(>\) A.


Step 4: Final Answer:

The correct order is D \(>\) C \(>\) B \(>\) A.
Quick Tip: Picryl chloride (D) is so reactive that it doesn't even require \(NaOH\); it hydrolyzes simply upon warming with water!

Step 1: Understanding the Concept:

Acidity of carboxylic acids is enhanced by the presence of electron-withdrawing groups (EWGs). EWGs stabilize the conjugate base (carboxylate ion) through the inductive (\(-I\)) effect.


Step 2: Key Formula or Approach:

Strength of \(-I\) effect: \(-F > -Cl > -Br\). Also, more EWGs result in greater acidity.


Step 3: Detailed Explanation:

1. Compound B (\(CF_{3}COOH\)): Three fluorine atoms provide the strongest \(-I\) effect. It is the most acidic.

2. Mono-halogenated acids: Compare electronegativity of halogens: \(F > Cl > Br\).

- \(FCH_{2}COOH\) (D) is more acidic than \(ClCH_{2}COOH\) (C).

- \(ClCH_{2}COOH\) (C) is more acidic than \(BrCH_{2}COOH\) (E).

3. Compound A (\(CH_{3}COOH\)): Methyl group has a \(+I\) effect, which destabilizes the conjugate base. It is the least acidic.

4. Descending Order: B \(>\) D \(>\) C \(>\) E \(>\) A.


Step 4: Final Answer:

The descending order of acidity is B \(>\) D \(>\) C \(>\) E \(>\) A.
Quick Tip: For acidity, look at the stability of the conjugate base. Any group that pulls electron density away from the \(O^{-}\) stabilizes the ion and increases acidity.

Step 1: Understanding the Concept:

Phenolphthalein is a common acid-base indicator synthesized by the condensation of phenol with phthalic anhydride in the presence of a dehydrating agent like concentrated sulfuric acid.


Step 2: Detailed Explanation:

1. Reactants: Phenol (also known as carbolic acid) reacts with phthalic anhydride.

2. Condition: Concentrated \(H_{2}SO_{4}\) acts as a catalyst and a dehydrating agent.

3. Product Formation: Two moles of phenol condense with one mole of phthalic anhydride to form phenolphthalein.

4. Indicator Property: Phenolphthalein ('Y') is colorless in acidic media and turns pink in basic media, making it a valuable indicator.

5. Conclusion: 'X' is carbolic acid (phenol) and 'Y' is phenolphthalein.


Step 3: Final Answer:

The compounds are carbolic acid and phenolphthalein.
Quick Tip: Phenol is often called carbolic acid. In phenolphthalein synthesis, the phenol molecules attach to the phthalic anhydride at their para positions.

Step 1: Understanding the Concept:

This is a sequence of aromatic substitution and functional group transformations starting from p-nitrotoluene.


Step 2: Detailed Explanation:

1. Step (i) \(Br_{2}, Fe\): Bromination occurs. The methyl group (\(+I, +M\)) is ortho/para directing, and the nitro group (\(-I, -M\)) is meta directing. Both groups direct the incoming Bromine to the position ortho to the methyl group. Product: 2-bromo-4-nitrotoluene.

2. Step (ii) \(H_{2}/Pd\): The nitro group is reduced to an amino group. Product: 2-bromo-4-methylaniline.

3. Step (iii) \(NaNO_{2}, HCl, 0^{\circ}C\): Diazotization of the primary aromatic amine. Product: 2-bromo-4-methylbenzenediazonium chloride.

4. Step (iv) \(H_{3}PO_{2}\): The diazonium group is replaced by a hydrogen atom (deamination). Product: 3-bromotoluene (m-bromotoluene).

5. Final Product: The structure corresponding to m-bromotoluene is shown in Option (A).


Step 3: Final Answer:

The final product is 3-bromotoluene (m-bromotoluene).
Quick Tip: \(H_{3}PO_{2}\) or \(CH_{3}CH_{2}OH\) are used to remove a diazonium group and replace it with H. This is a common strategy to synthesize meta-substituted products that are otherwise difficult to obtain.

Step 1: Understanding the Concept:

Drugs interact with receptors in two main ways: as agonists or as antagonists. Receptors are usually located on the cell surface.


Step 2: Detailed Explanation:

1. Statement A: Antagonists are drugs that bind to the receptor site. This is True. By binding, they occupy the site and prevent the natural chemical messenger from binding.

2. Statement B: Most receptors are on the surface, and drugs act without being transferred inside the cell. Antagonists primarily act by blocking the surface receptor. This is False.

3. Statement C: Because they block the binding of natural messengers, they inhibit the natural communication process. This is True. This is useful when the natural messenger is overactive.

4. Statement D: Drugs that mimic the natural messenger are called Agonists, not antagonists. This is False.

5. Conclusion: Statements A and C are true.


Step 3: Final Answer:

The true statements are A and C.
Quick Tip: Antagonist = Block (like a wrong key stuck in a lock).
Agonist = Mimic (like a master key that opens the lock).

Step 1: Understanding the Concept:

Each of the 20 standard natural amino acids is represented by a specific three-letter code and a unique one-letter code.


Step 2: Detailed Explanation:

1. Glutamic acid: Its one-letter code is 'E'. (A matches III)

2. Glutamine: Its one-letter code is 'Q'. (B matches I)

3. Tyrosine: Its one-letter code is 'Y'. (C matches IV)

4. Tryptophan: Its one-letter code is 'W'. (D matches II)

5. Matching: A-III, B-I, C-IV, D-II.


Step 3: Final Answer:

The correct matching is A-III, B-I, C-IV, D-II.
Quick Tip: Try to remember codes for similar-sounding names: Aspartic acid (D) vs Glutamic acid (E); Asparagine (N) vs Glutamine (Q). Tryptophan uses 'W' because it contains a "double" ring.

Step 1: Understanding the Concept:

Methyl orange is a common pH indicator. The color change in indicators is explained by the change in molecular structure (benzenoid vs. quinonoid) upon gain or loss of a proton.


Step 2: Detailed Explanation:

1. Statement I: Methyl orange is typically described as a weak base indicator (it is a sulfonic acid salt, but its indicator action involves the basic amino group). Some texts might loosely call it a weak acid, but in standard JEE/NCERT context, it is treated as a weak base. Thus, Statement I is generally considered incorrect.

2. Statement II: According to the Ostwald/Quinonoid theory, the quinonoid form (found in acidic medium for methyl orange, appearing red) is more deeply colored due to higher conjugation and charge distribution compared to the benzenoid form (found in basic medium, appearing yellow). Thus, Statement II is incorrect as it claims the benzenoid form is more intense.

3. Conclusion: Both statements are incorrect.


Step 3: Final Answer:

Both Statement I and Statement II are incorrect.
Quick Tip: Indicators like Phenolphthalein are weak acids, while Methyl Orange is a weak base. Quinonoid structures always produce deeper colors because of the nature of the extended chromophore.

Step 1: Understanding the Concept:

Titration indicators work by changing color in response to a specific change in the chemical environment at the equivalence point.


Step 2: Detailed Explanation:

1. Statement I: In redox titrations, indicators (like diphenylamine or ferroin) respond to changes in the electrode potential (oxidation-reduction potential) of the solution, not the pH. Thus, Statement I is incorrect.

2. Statement II: In acid-base titrations, indicators (like phenolphthalein or methyl orange) respond to changes in the pH (hydrogen ion concentration) of the solution, not the oxidation potential. Thus, Statement II is incorrect.

3. Conclusion: Both statements are incorrect because they swap the fundamental mechanisms of the two types of indicators.


Step 3: Final Answer:

Both Statement I and Statement II are incorrect.
Quick Tip: Redox \(\rightarrow\) Potential change.
Acid-Base \(\rightarrow\) pH change.
Remember this simple mapping to avoid confusion in statement-based questions.

Step 1: Understanding the Concept:

Radial nodes are regions in an atom where the probability of finding an electron is zero. The number of radial nodes in an atomic orbital depends on the principal quantum number (\(n\)) and the azimuthal quantum number (\(l\)).


Step 2: Key Formula or Approach:

The formula to calculate the number of radial nodes for any orbital is:
\[ Number of Radial Nodes = n - l - 1 \]

where:
\(n\) = Principal quantum number
\(l\) = Azimuthal quantum number (s=0, p=1, d=2, f=3)


Step 3: Detailed Explanation:

Let us calculate the number of radial nodes for each given orbital:


7s: \(n = 7, l = 0\). Nodes \(= 7 - 0 - 1 = 6\).

7p: \(n = 7, l = 1\). Nodes \(= 7 - 1 - 1 = 5\). (Matches)

6s: \(n = 6, l = 0\). Nodes \(= 6 - 0 - 1 = 5\). (Matches)

8p: \(n = 8, l = 1\). Nodes \(= 8 - 1 - 1 = 6\).

8d: \(n = 8, l = 2\). Nodes \(= 8 - 2 - 1 = 5\). (Matches)


The orbitals having 5 radial nodes are 7p, 6s, and 8d.

The total number of such orbitals is 3.


Step 4: Final Answer:

The number of atomic orbitals having 5 radial nodes is 3.
Quick Tip: Remember the total number of nodes is always \(n - 1\), where angular nodes \(= l\) and radial nodes \(= n - l - 1\).

Step 1: Understanding the Concept:

The number of lone pairs on a central atom can be determined using the valence shell electron pair repulsion (VSEPR) theory. Xenon (Xe) is a noble gas with 8 valence electrons.


Step 2: Key Formula or Approach:

Number of lone pair electrons \(= [Valence electrons of Xe - electrons shared in bonds - charge]\)

Number of lone pairs \(= \frac{Lone pair electrons}{2}\)


Step 3: Detailed Explanation:

Let's analyze each species:


XeF\(_{5}^{+}\): Xe has 8e\(^{-}\). Minus 1e\(^{-}\) for positive charge \(= 7\)e\(^{-}\). 5e\(^{-}\) used for 5 Xe-F bonds. Remaining \(= 2\)e\(^{-}\) (1 Lone Pair).

XeO\(_{3}\): Xe has 8e\(^{-}\). 3 Oxygen atoms form double bonds using 6e\(^{-}\). Remaining \(= 2\)e\(^{-}\) (1 Lone Pair).

XeO\(_{2}\)F\(_{2}\): Xe has 8e\(^{-}\). 2 Oxygen atoms (4e\(^{-}\)) + 2 Fluorine atoms (2e\(^{-}\)) used 6e\(^{-}\). Remaining \(= 2\)e\(^{-}\) (1 Lone Pair).

XeF\(_{5}^{-}\): Xe has 8e\(^{-}\). Plus 1e\(^{-}\) for negative charge \(= 9\)e\(^{-}\). 5e\(^{-}\) used for 5 bonds. Remaining \(= 4\)e\(^{-}\) (2 Lone Pairs).

XeO\(_{3}\)F\(_{2}\): Xe has 8e\(^{-}\). 3 Oxygen atoms (6e\(^{-}\)) + 2 Fluorine atoms (2e\(^{-}\)) used 8e\(^{-}\). Remaining \(= 0\)e\(^{-}\) (0 Lone Pairs).

XeOF\(_{4}\): Xe has 8e\(^{-}\). 1 Oxygen atom (2e\(^{-}\)) + 4 Fluorine atoms (4e\(^{-}\)) used 6e\(^{-}\). Remaining \(= 2\)e\(^{-}\) (1 Lone Pair).

XeF\(_{4}\): Xe has 8e\(^{-}\). 4 Fluorine atoms used 4e\(^{-}\). Remaining \(= 4\)e\(^{-}\) (2 Lone Pairs).


The species with a single lone pair are XeF\(_{5}^{+}\), XeO\(_{3}\), XeO\(_{2}\)F\(_{2}\), and XeOF\(_{4}\). Total count is 4.


Step 4: Final Answer:

The number of species carrying a single lone pair on Xenon is 4.
Quick Tip: Oxygen is divalent and forms double bonds in Xenon oxyfluorides, consuming 2 valence electrons per Oxygen atom.

Step 1: Understanding the Concept:

A bomb calorimeter measures heat at constant volume, which corresponds to the change in internal energy (\(\Delta U\)). The heat of combustion at constant pressure (\(\Delta H\)) is related to \(\Delta U\) by the work done due to gas expansion. For equilibrium, the Gibbs free energy change \(\Delta G\) must be zero.


Step 2: Key Formula or Approach:

1. \(\Delta H = \Delta U + \Delta n_g RT\)

2. At equilibrium, \(\Delta G = \Delta H - T\Delta S = 0 \implies T\Delta S = \Delta H\)

where \(\Delta n_g\) is the change in moles of gaseous products and reactants.


Step 3: Detailed Explanation:

The combustion reaction is:
\[ C_{2}H_{4}(g) + 3O_{2}(g) \rightarrow 2CO_{2}(g) + 2H_{2}O(l) \]

Calculate \(\Delta n_g\):
\[ \Delta n_g = moles of gaseous products - moles of gaseous reactants \] \[ \Delta n_g = 2 - (1 + 3) = -2 \]

Given \(\Delta U = -1406\) kJ mol\(^{-1}\) (heat produced is negative), \(T = 300\) K, \(R = 8.3 \times 10^{-3}\) kJ K\(^{-1}\) mol\(^{-1}\).

Calculate \(\Delta H\):
\[ \Delta H = -1406 + (-2 \times 8.3 \times 10^{-3} \times 300) \] \[ \Delta H = -1406 - 4.98 = -1410.98 kJ mol^{-1} \]

For equilibrium, \(\Delta G = 0 \implies T\Delta S = \Delta H\).
\[ T\Delta S = -1410.98 kJ \]

The question asks for the minimum value as \((-\) \underline{\hspace{1cm kJ.

Rounding to the nearest integer, the value is 1411.


Step 4: Final Answer:

The minimum value of \(T\Delta S\) is \(-\)1411 kJ.
Quick Tip: Remember that bomb calorimeter \(\rightarrow\) \(\Delta U\) and open vessel/calorimeter \(\rightarrow\) \(\Delta H\). Always check the physical states in the reaction for \(\Delta n_g\).

Step 1: Understanding the Concept:

The boiling point elevation constant (\(K_b\)), also known as the ebullioscopic constant, is a property of the solvent that depends on its boiling point and enthalpy of vaporization.


Step 2: Key Formula or Approach:

The formula for \(K_b\) is:
\[ K_b = \frac{R \cdot M \cdot T_b^2}{1000 \cdot \Delta H_{vap}} \]

where \(R\) is the gas constant, \(M\) is the molecular weight of the solvent, \(T_b\) is the boiling point, and \(\Delta H_{vap}\) is the enthalpy of vaporization.


Step 3: Detailed Explanation:

Since molecular weights (\(M\)) and \(R\) are constant for both solvents, we can write:
\[ K_b \propto \frac{T_b^2}{\Delta H_{vap}} \]

Given:
\(\frac{T_b(X)}{T_b(Y)} = \frac{2}{1}\) and \(\frac{\Delta H_{vap}(X)}{\Delta H_{vap}(Y)} = \frac{1}{2}\)

Now, find the ratio \(\frac{K_b(X)}{K_b(Y)}\):
\[ \frac{K_b(X)}{K_b(Y)} = \frac{(T_b(X))^2}{(T_b(Y))^2} \times \frac{\Delta H_{vap}(Y)}{\Delta H_{vap}(X)} \] \[ \frac{K_b(X)}{K_b(Y)} = \left( \frac{2}{1} \right)^2 \times \left( \frac{2}{1} \right) \] \[ \frac{K_b(X)}{K_b(Y)} = 4 \times 2 = 8 \]

Given \(K_b(X) = m \cdot K_b(Y)\), comparing with the ratio, we find \(m = 8\).


Step 4: Final Answer:

The value of \(m\) is 8.
Quick Tip: For comparative problems involving constants, identify the variables that cancel out to simplify the ratio equation immediately.

Step 1: Understanding the Concept:

Solubility is decreased in the presence of a common ion due to the common ion effect. BaSO\(_{4}\) and K\(_{2}\)SO\(_{4}\) both provide the sulphate ion (\(SO_4^{2-}\)).


Step 2: Key Formula or Approach:

1. \(K_{sp} = [Ba^{2+}][SO_4^{2-}]\)

2. Solubility (\(g L^{-1}\)) = Molar Solubility (\(mol L^{-1}\)) \(\times\) Molar Mass (\(g mol^{-1}\))


Step 3: Detailed Explanation:

K\(_{2}\)SO\(_{4}\) is a strong electrolyte:
\[ K_{2}SO_{4} \rightarrow 2K^{+} + SO_4^{2-} \]
Concentration of \(SO_4^{2-}\) from K\(_{2}\)SO\(_{4}\) is 0.1 M.

Let the molar solubility of BaSO\(_{4}\) be \(s\).
\[ BaSO_{4} \rightleftharpoons Ba^{2+} + SO_4^{2-} \]
Total \([SO_4^{2-}] = s + 0.1 \approx 0.1\) M (as \(s\) is very small).
\[ K_{sp} = [s][0.1] = 1 \times 10^{-10} \] \[ s = 1 \times 10^{-9} mol L^{-1} \]

Calculate solubility in \(g L^{-1}\):
\[ Solubility = (1 \times 10^{-9} mol L^{-1}) \times (233 g mol^{-1}) \] \[ Solubility = 233 \times 10^{-9} g L^{-1} \]

The value in the blank is 233.


Step 4: Final Answer:

The solubility of BaSO\(_{4}\) is \(233 \times 10^{-9}\) g L\(^{-1}\).
Quick Tip: In the presence of a relatively concentrated common ion, you can usually neglect the concentration of that ion coming from the sparsely soluble salt during addition.

Step 1: Understanding the Concept:

This question involves concepts from electrochemistry and thermodynamics, specifically the relations between cell potential, Gibbs energy, entropy, and experimental conditions.


Step 2: Detailed Explanation:


Statement A: For a reversible electrochemical cell at constant T and P, the electrical work (\(W_{elec}\)) done by the system is equal to the decrease in Gibbs free energy. Mathematically, \(\Delta G = -nFE_{cell} = W_{max, non-PV}\). This statement is considered correct in a magnitude context.

Statement B: \(E_{cell}^{0}\) is the standard cell potential. By definition, standard states refer to fixed conditions (1 bar pressure for gases, 1 M concentration). Therefore, standard potentials are independent of pressure. This statement is incorrect.

Statement C: From \(\Delta G^{0} = -nFE_{cell}^{0}\), differentiating with respect to T:

\[ \frac{d(\Delta G^{0})}{dT} = -nF \frac{dE_{cell}^{0}}{dT} \]
Since \(\left(\frac{\partial \Delta G}{\partial T}\right)_P = -\Delta S\), we have \(-\Delta S^{0} = -nF \frac{dE_{cell}^{0}}{dT} \implies \frac{dE_{cell}^{0}}{dT} = \frac{\Delta S^{0}}{nF}\). This statement is correct.

Statement D: An electrochemical cell operates reversibly only when an infinitesimally small current is drawn, which occurs when the cell emf is infinitesimally larger than an opposing potential. If exactly balanced, no current flows, representing the reversible equilibrium point. This statement is correct.


Only statement B is incorrect. The count of incorrect statements is 1.


Step 3: Final Answer:

The number of incorrect statements is 1.
Quick Tip: "Standard" (\(^{0}\)) values are always defined at fixed conditions (like 1 bar or 1 M), so they do not vary with changes in experimental pressure or concentration.

Step 1: Understanding the Concept:

Coagulating value (or flocculation value) is the minimum concentration of an electrolyte required to cause the coagulation of a sol. Coagulating power is the reciprocal of the coagulating value.


Step 2: Key Formula or Approach:
\[ Coagulating Power \propto \frac{1}{Coagulating Value} \] \[ \frac{Power (AlCl_3)}{Power (NaCl)} = \frac{Value (NaCl)}{Value (AlCl_3)} \]


Step 3: Detailed Explanation:

Given:

Coagulating value of AlCl\(_{3}\) \(= 0.09\)

Coagulating value of NaCl \(= 50.04\)

Ratio of powers (\(x\)):
\[ x = \frac{50.04}{0.09} \] \[ x = \frac{5004}{9} \] \[ x = 556 \]


Step 4: Final Answer:

The value of \(x\) is 556.
Quick Tip: Hardy-Schulze rule states that the greater the valence of the flocculating ion, the greater is its power to cause precipitation. Here, Al\(^{3+}\) has a much higher power than Na\(^{+}\), which is reflected in the much lower coagulating value for AlCl\(_{3}\).

Step 1: Understanding the Concept:

To find the number of sigma (\(\sigma\)) and pi (\(\pi\)) bonds, we must determine the chemical structure of pyrophosphoric acid. Pyrophosphoric acid is formed by the condensation of two molecules of orthophosphoric acid (H\(_{3}\)PO\(_{4}\)) with the loss of one water molecule.


Step 2: Detailed Explanation:

The formula of pyrophosphoric acid is H\(_{4}\)P\(_{2}\)O\(_{7}\).

Structure:
\[ (HO)_2P(=O)-O-P(=O)(OH)_2 \]

Let's count the bonds:


P-O single bonds: Each Phosphorus is bonded to 2 -OH groups and 1 bridging -O- atom. Total \(P-O\) single bonds \(= (2 \times 2) + 2 (bridge) = 6\).

P=O double bonds: Each Phosphorus has one double bond to Oxygen. Total \(= 2\).

O-H single bonds: There are 4 -OH groups. Total \(O-H\) bonds \(= 4\).


Now, count \(\sigma\) and \(\pi\):
\(\sigma\) bonds \(= 6 (P-O) + 2 (from P=O) + 4 (O-H) = 12\).
\(\pi\) bonds \(= 2 (from P=O)\).

Ratio \(\sigma/\pi = 12 / 2 = 6\).


Step 3: Final Answer:

The ratio of sigma to pi bonds in pyrophosphoric acid is 6.
Quick Tip: Drawing the structure correctly is key. Remember "pyro" means heat; pyrophosphoric acid is \(2 \times H_3PO_4 - H_2O\).

Step 1: Understanding the Concept:

The oxidation state of an atom in a compound is the formal charge it would carry if all bonds were ionic. The sum of oxidation states in a neutral molecule is zero, and in a polyatomic ion, it equals the charge on the ion.


Step 2: Detailed Explanation:


Fe(CO)\(_{5}\): CO (carbonyl) is a neutral ligand. Let oxidation state of Fe be \(x\).

\(x + 5(0) = 0 \implies x = 0\).

VO\(^{2+}\): Let oxidation state of V be \(y\). Oxygen usually has an oxidation state of \(-2\).

\(y + (-2) = +2 \implies y = +4\).

WO\(_{3}\): Let oxidation state of W be \(z\).

\(z + 3(-2) = 0 \implies z = +6\).


Sum of oxidation states \(= 0 + 4 + 6 = 10\).


Step 3: Final Answer:

The sum of oxidation states of the metals is 10.
Quick Tip: Metals in metal carbonyls always have an oxidation state of zero. In oxo-cations like \(MO^{n+}\), the oxygen contributes \(-2\) to the total charge equation.

Step 1: Understanding the Concept:

The magnetic moment (\(\mu\)) of a transition metal complex is related to the number of unpaired electrons (\(n\)) in the central metal atom. The charge on the complex (\(x\)) can be determined once the oxidation state of the metal is known.


Step 2: Key Formula or Approach:

1. \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons (BM)

2. \(Charge x = Oxidation state of Mn + Charge on ligands\)


Step 3: Detailed Explanation:

Given \(\mu = 6.06\) BM.

Using the spin-only formula:
\[ \sqrt{n(n+2)} \approx 6.06 \] \[ n(n+2) = (6.06)^2 \approx 36.7 \]
For \(n = 5\), \(\mu = \sqrt{5(5+2)} = \sqrt{35} = 5.92\) BM.

For \(n = 6\), \(\mu = \sqrt{6(6+2)} = \sqrt{48} = 6.93\) BM.

The value 6.06 BM corresponds to 5 unpaired electrons.

Manganese (Mn) has an atomic number of 25. Its valence shell configuration is \(3d^5 4s^2\).

To have 5 unpaired electrons, Mn must be in the \(+2\) oxidation state (\(3d^5\) configuration, high spin since NCS\(^-\) is usually a weak field ligand).

Oxidation state of Mn \(= +2\).

The ligand NCS\(^-\) has a charge of \(-1\).

For the complex \([Mn(NCS)_6]^x\):
\[ x = (+2) + 6(-1) = +2 - 6 = -4 \]

The numerical value of \(x\) is 4.


Step 4: Final Answer:

The numerical value of \(x\) is 4.
Quick Tip: The number of unpaired electrons is usually the integer part of the magnetic moment value. For example, if \(\mu = 5.9\) BM, \(n = 5\). Here 6.06 BM is slightly higher than 5.9 but closest to 5 electrons.