CUET PG Chemistry Question Paper 2024: Download Question paper with Answers PDF

Step 1: Identify chiral centers
The given molecule has two chiral centers at Carbon-2 and Carbon-3.

Step 2: Assign priorities using Cahn-Ingold-Prelog rules

- Assign priority based on atomic number.

- Orient the lowest priority group away.

- Determine R/S configuration at each chiral center.


Step 3: Analyze the configurations

- Carbon-2 follows the counterclockwise direction, making it \( S \).

- Carbon-3 follows the clockwise direction, making it \( R \).


Thus, the correct configuration is \( 2S, 3R \). Quick Tip: When assigning R and S configurations, always ensure the lowest priority group is placed at the back before determining the direction of the remaining groups.

Step 1: Understanding conformational stability
The stability of cyclohexane conformations follows the order:
Chair \( \(>\) \) Twist Boat \( \(>\) \) Boat \( \(>\) \) Half Chair

Step 2: Energy consideration

- Chair is the most stable due to minimal steric strain.

- Twist boat is slightly more stable than boat due to reduced steric hindrance.

- Half chair is the least stable due to highest energy.


Thus, the correct order is \( A \(>\) C \(>\) B \(>\) D \). Quick Tip: Cyclohexane prefers the chair conformation as it minimizes angle strain and steric repulsion.

Step 1: Understanding electrophiles
Electrophiles are species that accept electron pairs.


Step 2: Analyzing given species

- \( NO^+ \) is an electrophile as it lacks electrons.

- \( BF_3 \) is electron-deficient and acts as an electrophile.

-\( CO_2 \) can act as an electrophile due to partial positive charge on carbon.

- \( NH_4^+ \) is not an electrophile as it has no vacant orbitals for accepting electrons.


Thus, the correct answer is \( NH_4^+ \). Quick Tip: Electrophiles are electron-deficient species that can accept a pair of electrons.

Step 1: Understanding the Iodoform test

The Iodoform test is positive for:

- Methyl ketones (\( CH_3CO \) group)

- Acetaldehyde (\( CH_3CHO \))

- Secondary alcohols with the structure \( CH_3CH(OH)R \)


Step 2: Checking applicability

- \( A \) (Methyl ketones) give a positive test.

- \( B \) (Acetaldehyde) gives a positive test.

- \( C \) (Certain secondary alcohols) give a positive test.

- \( D \) (Primary alcohols) do NOT give a positive test.


Thus, the correct answer is (A), (B), (C) only. Quick Tip: The Iodoform test is used to detect compounds with a \( CH_3CO \) group or specific secondary alcohols.

Step 1: Understanding acidity in phenols
Acidic strength of phenols is influenced by electron-withdrawing and electron-donating substituents.

Step 2: Analyzing the given compounds
- Compound A: Contains three strong electron-withdrawing nitro (\( NO_2 \)) groups, which stabilize the negative charge on phenoxide ion, making it highly acidic.

- Compound C: Contains a sulfonic acid (\( SO_3H \)) group, which is a strong electron-withdrawing group, enhancing acidity.

- Compound D: Has a methyl (\( CH_3 \)) group, which is electron-donating, reducing acidity.

- Compound B: Contains a methoxy (\( OCH_3 \)) group, which is electron-donating and destabilizes the phenoxide ion, making it the least acidic.

Step 3: Establishing the correct order
Since the nitro groups in A provide the strongest electron-withdrawing effect, A is the most acidic. C follows due to the strong \( SO_3H \) group, D is next, and B is the least acidic.
Thus, the correct order is \( A \(>\) C \(>\) D \(>\) B \). Quick Tip: Electron-withdrawing groups (-NO\textsubscript{2}, -SO\textsubscript{3}H) increase acidity by stabilizing the phenoxide ion, while electron-donating groups (-OCH\textsubscript{3}, -CH\textsubscript{3}) decrease acidity.

Step 1: Understanding the given reactions

- Reaction (A) – Beckmann rearrangement:

The conversion of an oxime to an amide is known as Beckmann rearrangement.

- Reaction (B) – Baeyer-Villiger oxidation:

This reaction involves the oxidation of ketones to esters using peracids.

- Reaction (C) – Knoevenagel condensation:

It involves the condensation of aldehydes with active methylene compounds in the presence of a base.

- Reaction (D) – Claisen-Schmidt reaction:

This reaction is a type of aldol condensation between an aldehyde and a ketone to form \(\alpha\)-\(\beta\)-unsaturated carbonyl compounds.


Step 2: Matching the correct reaction types \[ (A) - (III), \quad (B) - (IV), \quad (C) - (I), \quad (D) - (II) \] Quick Tip: Match the functional group changes with their characteristic reactions. For example, oxime to amide conversion is Beckmann rearrangement, while aldehyde to α,β-unsaturated compound follows Knoevenagel condensation.

Step 1: Understanding nucleophilic addition reactions
The reactivity of carbonyl compounds towards nucleophilic addition depends on:
- The partial positive charge on the carbonyl carbon (electrophilicity).
- The steric hindrance around the carbonyl carbon.

Step 2: Analyzing the given compounds
- Formaldehyde (HCHO): Most reactive due to the absence of alkyl groups, leading to less steric hindrance and a highly electrophilic carbonyl carbon.
- Trichloroacetaldehyde \((Cl_3CCHO)\): More reactive than simple aldehydes due to the strong -I effect of chlorine atoms, increasing the electrophilicity of the carbonyl carbon.
- Acetaldehyde \((Cl_3CCHO)\): Less reactive than HCHO and \(Cl₃CCHO\) but more reactive than ketones due to reduced steric hindrance.
- Ketone \((CH_3COCH_2CH_3)\): Least reactive due to steric hindrance from alkyl groups, which reduce the electrophilicity of the carbonyl carbon.

Step 3: Establishing the correct order \[ \textbf{B (HCHO) \(>\) C (\(Cl₃CCHO\)) \(>\) A (\((Cl_3CCHO)\)) \(>\) D (CH₃COCH₂CH₃)} \] Quick Tip: Aldehydes are more reactive than ketones in nucleophilic addition due to lower steric hindrance and higher electrophilicity. Electron-withdrawing groups (-Cl, -NO\textsubscript{2}) further increase the reactivity by stabilizing the negative charge on the intermediate.

Step 1: Understanding NMR spectroscopy
In \( ^1H \)-NMR spectroscopy, distinct signals correspond to chemically non-equivalent proton environments.

Step 2: Analyzing the given compounds

- Acetone (\( CH_3COCH_3 \)):

- Contains two identical methyl (\( CH_3 \)) groups attached to the same carbonyl.

- Results in only one NMR signal (singlet).


- Ethyl methyl ketone (\( CH_3COCH_2CH_3 \)):

- Contains three distinct proton environments:

1. Methyl group (\( CH_3CO \)) next to the carbonyl.

2. Methylene group (\( CH_2 \)) between the ketone and ethyl group.

3. Terminal ethyl group (\( CH_3CH_2 \)).

- Results in three distinct NMR signals.
Quick Tip: Symmetry in molecules reduces the number of distinct NMR signals. Acetone has identical methyl groups, leading to a single peak, while ethyl methyl ketone has three distinct proton environments.

Step 1: Understanding NMR chemical shifts

- \( \delta \) 7.1 ppm: Indicates aromatic proton signals from a benzene ring.

- \( \delta \) 2.2 ppm: Typical of benzylic \( -CH_2- \) groups attached to a benzene ring.

- \( \delta \) 1.5 ppm and 0.9 ppm: Suggests the presence of an ethyl (\( -CH_2CH_3 \)) group.

Step 2: Identifying the correct structure
- The presence of \( \delta \) 2.2 ppm (benzylic \( CH_2 \)) and \( \delta \) 0.9 ppm (methyl \( CH_3 \)) strongly indicates an ethyl group attached to the benzene ring.

- Among the given structures, structure (2) corresponds to ethylbenzene, which perfectly matches the observed NMR peaks. Quick Tip: In \( ^1H \)-NMR spectroscopy, benzylic protons appear around 2.2 ppm, while alkyl methyl (\( CH_3 \)) protons appear around 0.9 ppm. These chemical shifts help in identifying alkyl-substituted benzene compounds.

Step 1: Understanding IR Absorption in Vibrational Modes

Infrared (IR) spectroscopy measures molecular vibrations. A vibration is IR-active only if it results in a change in dipole moment.


Step 2: Analyzing the Given Vibrational Modes

- Symmetric \( CO_2 \) stretch (Option 1):

- \( CO_2 \) is a linear molecule with a symmetric stretch where both oxygen atoms move in the same direction.

- Since this motion does not create a change in dipole moment, it is IR inactive.


- Antisymmetric \( CO_2 \) stretch (Option 2):

- Involves oxygen atoms moving in opposite directions, leading to a fluctuating dipole moment, making it IR active.


- Symmetric \( O=C=S \) stretch (Option 3):

- Since \( O=C=S \) is not a completely symmetric linear molecule, this stretch induces a dipole change and is IR active.


- \( CH_3C \equiv CH \) stretch (Option 4):

- Terminal alkyne C-H stretching appears in IR spectroscopy as a strong band, meaning it is IR active.


Step 3: Conclusion

Since the symmetric stretch of \( CO_2 \) does not induce a dipole change, it shows no IR absorption bands.
Quick Tip: A molecular vibration is IR active only if it changes the dipole moment of the molecule. Symmetric stretches of linear molecules like \( CO_2 \) often do not exhibit IR absorption.

Step 1: Understanding the reaction sequence

- LiAlH\textsubscript{4 reduction (Step 1):

- Lithium aluminium hydride (LiAlH\textsubscript{4) is a strong reducing agent that reduces aldehydes and ketones to their corresponding alcohols.

- Here, the cyclohexanone (a ketone) is reduced to cyclohexanol (a secondary alcohol).


- H\textsubscript{2/Pt catalytic hydrogenation (Step 2):

- Hydrogenation with H\textsubscript{2/Pt reduces carbon-carbon double bonds, saturating the ring.

- In this case, any remaining unsaturation is removed, yielding a fully saturated cyclohexanol as product B.


Step 2: Identifying the correct product

- Product A: Cyclohexanol (from ketone reduction).

- Product B: Fully saturated cyclohexanol (from hydrogenation).


Thus, the correct option is (1).
Quick Tip: LiAlH\textsubscript{4} is a strong reducing agent that reduces ketones to alcohols, while catalytic hydrogenation (H\textsubscript{2}/Pt) removes unsaturation from organic compounds.

Step 1: Understanding Citral Properties

- Citral is a volatile liquid with a strong lemon-like odor, making statement (A) correct.

- Oxidation of citral produces acetone, oxalic acid, and laevulic acid, making statement (B) correct.

- Citral can form oximes, supporting statement (C).


Step 2: Evaluating Geometrical Isomerism

- Citral has two geometrical isomers: neral (cis) and geranial (trans).
- However, the given statement reverses their naming, making statement (D) incorrect.


Thus, the correct answer is (A), (B), and (C) only. Quick Tip: Citral is an aldehyde found in citrus fruits and is commonly used in perfumery due to its lemon-like aroma. It forms oximes and undergoes oxidation to produce various acids.

Step 1: Understanding Anhydride Formation

- Anhydrides form when two carboxyl (-COOH) groups are close enough to undergo intramolecular dehydration upon heating.


Step 2: Analyzing the Given Isomers

- The cis-isomer allows the carboxyl groups to be in close proximity, enabling anhydride formation.

- The trans-isomer prevents this due to steric hindrance.


Thus, the correct isomer is the cis-form (Second isomer). Quick Tip: Cis-isomers of dicarboxylic acids tend to form anhydrides more readily than trans-isomers due to favorable spatial arrangement.

Step 1: Understanding Hydroboration-Oxidation

- BH\textsubscript{3/THF, H\textsubscript{2O\textsubscript{2, OH\textsuperscript{- converts alkenes to anti-Markovnikov alcohols.
- The product X from this step contains a hydroxyl (-OH) group.


Step 2: Understanding NaBH\textsubscript{4} Reduction

- Sodium borohydride (NaBH\textsubscript{4) selectively reduces carbonyl (C=O) groups to alcohols without affecting alkenes or esters.
- The final product Y has two hydroxyl (-OH) groups.


Step 3: Identifying the Correct Structures

- The correct X and Y structures match Option (1).
Quick Tip: Hydroboration-oxidation follows anti-Markovnikov addition, forming primary alcohols, while NaBH\textsubscript{4} selectively reduces ketones and aldehydes to alcohols.

Step 1: Understanding Alkaloid Characteristics

- The molecular formula \( C_8H_{15}NO \) corresponds to a well-known coca alkaloid.

- Racemization in free-base form suggests a structure with a chiral center adjacent to a nitrogen atom.


Step 2: Identifying Functional Groups

- Presence of a keto group (C=O) and a tertiary nitrogen (N) matches structure (1).

- The synthesis method (using \( \gamma \)-methylaminobutyraldehyde and ethylacetoacetate) is consistent with alkaloid biosynthesis.


Step 3: Conclusion

Among the given structures, option (1) correctly represents the alkaloid, satisfying all the given properties.
Quick Tip: Coca alkaloids often contain both a keto group and a tertiary nitrogen, making them important in medicinal chemistry and pharmacology.

Step 1: Understanding NBS Bromination

- N-Bromosuccinimide (NBS) is commonly used for selective bromination at the most reactive position in heterocyclic compounds.


Step 2: Analyzing the Reactivity of Thiophene

- Thiophene undergoes electrophilic substitution preferentially at the 2-position (alpha position) due to its higher electron density.

- Bromination with NBS selectively introduces a single bromine atom at the 2-position.


Step 3: Identifying the Correct Product

- The correct product is monobromothiophene with bromine at the 2-position, as shown in option (4).
Quick Tip: NBS is used for selective bromination, and in thiophene, electrophilic substitution preferentially occurs at the 2-position (alpha position) due to resonance stabilization.

Step 1: Applying Hückel’s Rule

- A molecule is aromatic if it satisfies the Hückel rule:
\[ Number of \pi-electrons = 4n + 2, \quad where n is an integer \]
- The molecule must also be cyclic, planar, and fully conjugated.


Step 2: Analyzing the Given Molecules

- Molecule (A) [Cyclooctatetraenyl dianion]:
- Contains 10 π-electrons (follows \( 4n+2 \) rule, \( n=2 \)).
- Fully conjugated and planar.
- Aromatic.


- Molecule (B) [Cyclooctatetraenyl cation]:
- Contains 8 π-electrons (follows \( 4n \) rule, which is antiaromatic).
- Not aromatic.


- Molecule (C) [Benzo-fused system]:
- The benzene ring ensures conjugation.
- Aromatic.


- Molecule (D) [Cycloheptatrienyl cation]:
- Contains 6 π-electrons (follows \( 4n+2 \) rule, \( n=1 \)).
- Aromatic.


Thus, the correct answer is (A), (C), and (D) only.
Quick Tip: Aromatic compounds must be cyclic, planar, fully conjugated, and must obey Hückel’s \( 4n+2 \) π-electron rule.

Step 1: Understanding Reactivity in Electrophilic Substitution

- Electron-donating groups (EDGs) increase reactivity by activating the benzene ring.

- Electron-withdrawing groups (EWGs) decrease reactivity by deactivating the ring.


Step 2: Analyzing the Given Compounds

- Methoxy (-OCH\textsubscript{3): Strongly activating due to resonance donation.

- Phenoxide ion (-O\textsuperscript{-): Even stronger activating than methoxy due to complete negative charge, making it the most reactive.

- Ester (-COOCH\textsubscript{3): Electron-withdrawing due to resonance with the carbonyl group, reducing reactivity.

- Hydroxyl (-OH): Activating, but less than the phenoxide ion.


Step 3: Conclusion

The phenoxide ion (-O\textsuperscript{-) has the highest electron density due to its strong resonance donation, making it the most reactive towards electrophilic substitution.
Quick Tip: Electron-donating groups (EDGs) enhance electrophilic substitution by increasing electron density on the ring, while electron-withdrawing groups (EWGs) reduce reactivity.

Step 1: Understanding the HONO Reaction with Amines

- Primary amines (-NH\textsubscript{2) react with HONO to form unstable diazonium salts, which decompose into alcohols or ketones.


Step 2: Analyzing the Given Reaction

- The starting compound is a primary amine (-NH\textsubscript{2) attached to a hydroxyl (-OH) group.

- HONO converts the amine into a diazonium intermediate, which undergoes hydrolysis to form a ketone.


Step 3: Conclusion

The correct product is the ketone-containing compound (Option 2). Quick Tip: Diazotization of primary amines using HONO leads to the formation of diazonium salts, which undergo hydrolysis to yield alcohols or ketones depending on stability.

Step 1: Understanding the Nomenclature of Oxygen Heterocycles

- Oxepane (I): A seven-membered oxygen-containing ring.

- Oxolane (II): A five-membered oxygen-containing ring (tetrahydrofuran).

- Oxetane (III): A four-membered oxygen-containing ring.

- Oxirane (IV): A three-membered oxygen-containing ring (epoxide).


Step 2: Matching the Correct Structures
\[ (A) - Oxepane (I), \quad (B) - Oxetane (III), \quad (C) - Oxirane (IV), \quad (D) - Oxolane (II) \] Quick Tip: The suffix "-ane" in heterocyclic compounds indicates saturated rings, and the size of the ring determines its prefix: - Oxirane (3-membered)
- Oxetane (4-membered)
- Oxolane (5-membered)
- Oxepane (7-membered)

Step 1: Understanding Basicity in Aromatic Compounds

- Basicity depends on the availability of the lone pair of electrons on the nitrogen for protonation.

- Electron-donating groups (EDGs) increase basicity by stabilizing the positive charge on protonation.

- Electron-withdrawing groups (EWGs) decrease basicity by pulling electron density away from nitrogen.


Step 2: Analyzing the Given Compounds

- Compound (B) (p-Methoxy Aniline):

- The -OCH\textsubscript{3 group is an electron-donating group that increases electron density on nitrogen, making it the most basic.

- Compound (A) (Aniline):

- No extra activating or deactivating groups, so it is less basic than B but still more basic than others.
- Compound (C) (p-Nitroaniline):

- The -NO\textsubscript{2 group is a strong electron-withdrawing group that pulls electron density away from nitrogen, decreasing basicity.

- Compound (D) (Benzylideneaniline):

- The nitrogen is in conjugation with the benzene rings, which delocalizes the lone pair, making it the least basic.


Step 3: Conclusion

The correct order of basicity is:

(B) \(>\) (A) \(>\) (C) \(>\) (D) Quick Tip: Electron-donating groups (EDGs) increase basicity by increasing electron density on nitrogen, while electron-withdrawing groups (EWGs) decrease basicity by pulling electrons away from nitrogen.

Step 1: Understanding the Sulfonation Reaction

- Sulfonation of naphthalene with H\textsubscript{2SO\textsubscript{4 at 160°C favors substitution at the 2-position.

- At low temperatures (e.g., 50°C), the 1-position is favored due to kinetic control.

- At high temperatures (e.g., 160°C), the 2-position is favored due to thermodynamic stability.


Step 2: Identifying the Correct Product

- Since the reaction is performed at 160°C, the thermodynamically stable 2-naphthalenesulfonic acid is formed.
Quick Tip: Sulfonation of naphthalene follows temperature-dependent regioselectivity:
- 50°C → 1-Naphthalenesulfonic acid (kinetic product)
- 160°C → 2-Naphthalenesulfonic acid (thermodynamic product)

Step 1: Understanding 1,2- and 1,4-Addition Reactions

- 1,2-addition (kinetic control): Forms quickly due to lower activation energy but may not be the most stable product.
- 1,4-addition (thermodynamic control): Forms slowly but results in a more stable product.

Step 2: Analyzing the Statements

- (A) Correct: 1,2-addition is faster due to direct attack at the electrophilic site, while 1,4-addition leads to a more stable conjugated system.
- (B) Correct: 1,2-product formation is kinetically controlled.
- (C) Correct: 1,4-product formation is thermodynamically controlled.
- (D) Incorrect: At low temperatures, 1,2-product formation is irreversible, making it incorrect.

Thus, the correct answer is (A), (B), and (C) only. Quick Tip: 1,2-addition products form under kinetic control (low activation energy, fast reaction), while 1,4-addition products form under thermodynamic control (more stable but slower).

Step 1: Understanding \( S_N1 \) and \( S_N2 \) Mechanisms

- The \( S_N1 \) (Substitution Nucleophilic Unimolecular) reaction occurs in two steps:

- Formation of a carbocation intermediate.
- Nucleophilic attack on the carbocation.

- Rate law: First-order kinetics, depends only on the substrate concentration.

- Favored in polar protic solvents and with weak nucleophiles.


- The \( S_N2 \) (Substitution Nucleophilic Bimolecular) reaction occurs in one step:

- The nucleophile attacks the substrate in a single step, displacing the leaving group.

- Rate law: Second-order kinetics, depends on both substrate and nucleophile concentrations.

- Favored in polar aprotic solvents and with strong nucleophiles.


Step 2: Evaluating the Statements

- (A) \( S_N1 \) is unimolecular (first order) and \( S_N2 \) is bimolecular (second order):

- This is correct. The rate of \( S_N1 \) depends only on the substrate, while \( S_N2 \) depends on both substrate and nucleophile concentrations.


- (B) \( S_N2 \) proceeds through the formation of a carbocation, while \( S_N1 \) does not:

- This is incorrect.

- \( S_N2 \) reactions occur via direct nucleophilic attack with a single-step mechanism, meaning no carbocation is formed.

- \( S_N1 \), on the other hand, involves a carbocation intermediate.


- (C) \( S_N2 \) is a stereospecific reaction with inversion of configuration:

- This is correct.

- The nucleophilic attack occurs from the opposite side of the leaving group, leading to Walden inversion.


- (D) \( S_N1 \) is favored with weak bases and poor nucleophiles:

- This is correct.

- \( S_N1 \) occurs best with weak nucleophiles (e.g., H\textsubscript{2O, ROH) in polar protic solvents like water or alcohol.

- Weak bases do not strongly compete with the leaving group, allowing carbocation formation.


Step 3: Conclusion

- Correct statements: (A), (C), and (D).

- Incorrect statement: (B).

- Thus, the correct answer is option (3). Quick Tip: - \( S_N1 \) reactions form a carbocation and follow first-order kinetics.
- \( S_N2 \) reactions occur in one step and always result in stereochemical inversion.
- Weak nucleophiles favor \( S_N1 \), while strong nucleophiles favor \( S_N2 \).

Step 1: Understanding the Electrophilic Aromatic Substitution (EAS) Reaction

- The given reaction involves chlorination of trichloromethylbenzene in the presence of FeCl\textsubscript{3, a Lewis acid catalyst.

- The -CCl\textsubscript{3 group is an electron-withdrawing group (EWG) and acts as a meta-directing deactivator in electrophilic substitution reactions.


Step 2: Determining the Major Product

- The CCl\textsubscript{3 group deactivates the benzene ring, making electrophilic attack less favorable, but substitution occurs at the meta-position due to the electronic effect.

- Among the given choices, the major product is the para-substituted chlorinated derivative, as shown in option (2).
Quick Tip: Electron-withdrawing groups (EWGs) such as -CCl\textsubscript{3}, -NO\textsubscript{2}, and -COOH deactivate the benzene ring and direct electrophilic substitutions to the meta position.

Using Kohlrausch’s Law of independent migration of ions, the molar conductance at infinite dilution is: \[ \Lambda_m(Ba(OH)_2) = \Lambda_m(Ba^{2+}) + 2\Lambda_m(OH^-) \]
Given: \[ \Lambda_m(Ba^{2+}) = 280.0 \times 10^{-4}, \quad \Lambda_m(OH^-) = \Lambda_m(NaOH) - \Lambda_m(Na^+) = 248.1 - 126.5 = 121.6 \times 10^{-4} \] \[ \Lambda_m(Ba(OH)_2) = 280.0 + 2(121.6) = 523.2 \times 10^{-4} \quad S m^2 mol^{-1} \] Quick Tip: Use Kohlrausch’s Law to find conductance of weak electrolytes by adding ionic contributions.

- \( HCl \) is a strong acid, and \( NH_4OH \) is a weak base.

- Initially, conductance is high due to mobile \( H^+ \) ions.

- As \( NH_4OH \) is added, \( H^+ \) ions get neutralized, decreasing conductance.

- At the equivalence point, conductance reaches a minimum.

- After the endpoint, weak \( NH_4^+ \) and \( OH^- \) ions do not significantly change conductance, leading to constant values. Quick Tip: For strong acid vs. weak base titrations, conductance drops and then stabilizes.

The ionic strength \( I \) of a solution is given by the formula:
\[ I = \frac{1}{2} \sum c_i z_i^2 \]

where:
- \( c_i \) = concentration of ion \( i \) (molal),

- \( z_i \) = charge of ion \( i \),

- The summation runs over all ions in the solution.



Step 1: Contribution of NaCl

NaCl dissociates completely as:
\[ NaCl \rightarrow Na^+ + Cl^- \]

- Concentration of \( Na^+ \) and \( Cl^- \) = 0.1 molal each,

- Charge \( z \) = \( \pm1 \),

Thus, the ionic strength contribution from NaCl is:
\[ I_{NaCl} = \frac{1}{2} [(0.1 \times 1^2) + (0.1 \times 1^2)] = \frac{1}{2} [0.1 + 0.1] = 0.1 \]


Step 2: Contribution of CaCl\textsubscript{2}

CaCl\textsubscript{2 dissociates completely as:
\[ CaCl_2 \rightarrow Ca^{2+} + 2Cl^- \]

- Concentration of \( Ca^{2+} \) = 0.01 molal, \( z = +2 \),

- Concentration of \( Cl^- \) = 0.02 molal, \( z = -1 \),

Thus, the ionic strength contribution from CaCl\textsubscript{2 is:
\[ I_{CaCl_2} = \frac{1}{2} [(0.01 \times 2^2) + (0.02 \times 1^2)] \]
\[ = \frac{1}{2} [(0.01 \times 4) + (0.02 \times 1)] \]
\[ = \frac{1}{2} [0.04 + 0.02] = \frac{1}{2} \times 0.06 = 0.03 \]


Step 3: Total Ionic Strength Calculation
\[ I_{total} = I_{NaCl} + I_{CaCl_2} = 0.1 + 0.03 = 0.13 molal \] Quick Tip: Ionic strength is calculated using \( I = \frac{1}{2} \sum c_i z_i^2 \) where \( c \) is concentration, \( z \) is charge.

The standard cell potential \( E^0_{cell} \) is calculated using:
\[ E^0_{cell} = E^0_{cathode} - E^0_{anode} \]

Here,
- The cathode (reduction) is the CuI half-cell:
\[ CuI + e^- \rightarrow Cu(s) + I^- \quad E^0 = +0.17V \]
- The anode (oxidation) is the Zn half-cell:
\[ Zn(s) \rightarrow Zn^{2+} + 2e^- \quad E^0 = -(-0.76V) = +0.76V \]

Thus,
\[ E^0_{cell} = 0.76V + 0.17V = 0.59V \]\ Quick Tip: Cell potential is calculated as \( E^0_{cathode} - E^0_{anode} \).

The correct values can be assigned based on energy level calculations:

1. Energy of ground state of He\textsuperscript{+
- The ground-state energy corresponds to -54.4 eV, so (A) matches with (IV).

2. Potential energy of I orbit of H-atom
- The potential energy of the hydrogen atom’s first orbit is -27.2 eV, so (B) matches with (II).

3. Kinetic Energy of II excited state of He\textsuperscript{+
- The kinetic energy in the excited state of He\textsuperscript{+ is given as +6.04 eV, so (C) matches with (I).

4. Ionization potential of He\textsuperscript{+
- The ionization potential of He\textsuperscript{+ corresponds to 54.4 eV, so (D) matches with (III).

Thus, the final answer is: \[ (A) - (IV), \quad (B) - (II), \quad (C) - (I), \quad (D) - (III) \] Quick Tip: Energy values are positive for excited states and negative for bound states.

N/A Quick Tip: The reduction potential of the hydrogen electrode becomes negative when the hydrogen ion concentration decreases or the hydrogen gas pressure increases.

N/A Quick Tip: Greater orbital overlap leads to stronger bonds. p-p overlap is the strongest, while s-p is the weakest.

N/A Quick Tip: Lower molar mass means higher molarity when equal masses are dissolved in the same volume of water.

N/A Quick Tip: In the particle-in-a-box model, energy levels follow \( n^2 \) proportionality.

N/A Quick Tip: Larger molecules have more vibrational degrees of freedom due to more atoms.

N/A Quick Tip: All d-orbitals have positive and negative lobes due to phase difference.

Step 1: Understanding Angular Momentum Formula
The angular momentum of an electron in an atomic orbital is given by: \[ L = \sqrt{l(l+1)} \hbar \]
where \( l \) is the azimuthal quantum number.

Step 2: Identifying Quantum Numbers

For a 3s orbital:

- Principal quantum number: \( n = 3 \)

- Azimuthal quantum number: \( l = 0 \)


Step 3: Calculating Angular Momentum
Substituting \( l = 0 \): \[ L = \sqrt{0(0+1)} \hbar = 0 \]

Step 4: Conclusion
Thus, the magnitude of angular momentum is 0. Quick Tip: For any s-orbital (\( l = 0 \)), the angular momentum is always zero. Only p, d, and f orbitals have non-zero angular momentum.

Step 1: Understanding Molecular Conversions

- 1 mole of CO\(_2\) = 44g, so 88g = 2 moles.

- \(6.023 \times 10^{23}\) molecules correspond to 1 mole.

- 1 mole of O\(_2\) = 32g, so 96g = 3 moles.

- 1 mole of any gas = \(6.023 \times 10^{23}\) molecules.

Thus, the correct mapping is:
(A) - (II), (B) - (III), (C) - (I), (D) - (IV). Quick Tip: Always remember Avogadro's number: \(6.023 \times 10^{23}\) molecules = 1 mole.

Step 1: Understanding Definitions

- Molarity (\(M\)) is measured in mol/L.

- Mole fraction is unitless.

- Mole is measured in mol.

- Molality (\(m\)) is measured in mol/kg.
Quick Tip: Molarity and Molality are different: Molarity depends on volume, while Molality depends on mass.

Step 1: Understanding the Arrhenius Equation
The Arrhenius equation is given by: \[ k = A e^{-E_a / RT} \]
Taking the natural logarithm on both sides: \[ \ln k = \ln A - \frac{E_a}{RT} \]
which can be rewritten as: \[ \log k = \log A - \frac{E_a}{2.303 R} \cdot \frac{1}{T} \]

Step 2: Identifying the Plot Behavior
The equation is in the form of a straight line: \[ y = mx + c \]
where:
- \( \log k \) is on the y-axis

- \( 1/T \) is on the x-axis

- The slope \( m = -E_a / (2.303 R) \) is negative

- The intercept \( c = \log A \) is nonzero
Quick Tip: A plot of \(\log k\) vs. \(1/T\) follows a straight-line equation with a negative slope, indicating that reaction rates decrease at lower temperatures.

Step 1: Understanding Viscosity
Viscosity is a measure of a fluid's resistance to flow. The higher the viscosity, the slower the liquid flows. Hence, statement (A) is correct.

Step 2: Nature of Glass
Glass is considered an amorphous solid, but it exhibits extremely high viscosity, meaning it flows very slowly over time. This makes statement (B) correct.

Step 3: Effect of Temperature on Viscosity
For most liquids, viscosity decreases as temperature increases, because intermolecular forces weaken. Since statement (C) suggests that viscosity increases with temperature, it is incorrect.

Step 4: Relationship Between Viscosity and Fluidity
Fluidity is defined as the reciprocal of viscosity (i.e., \(Fluidity = \frac{1}{Viscosity}\)). Thus, statement (D) is correct.

Since statements (A), (B), and (D) are correct, option (1) is the right answer. Quick Tip: For most liquids, viscosity decreases with an increase in temperature, while for gases, viscosity increases with temperature.

Step 1: Identify intercepts
The intercepts of the given crystal plane are given as \(2a, 3b, c\).

Step 2: Take reciprocals \[ Reciprocal values = \frac{1}{2}, \frac{1}{3}, \frac{1}{1} \]

Step 3: Convert into smallest integers
Multiplying all by 6 to get whole numbers: \[ (3,2,1) \]

Thus, the correct Miller indices are (231). Quick Tip: To find Miller indices, take reciprocals of intercepts and convert them to the smallest whole numbers.

Step 1: Use Gibbs Phase Rule \[ F = C - P + 2 \]
For water and vapor system:
- \( C = 1 \) (since water is a single-component system)
- \( P = 2 \) (since liquid and vapor phases exist together)

Step 2: Calculate Degrees of Freedom \[ F = 1 - 2 + 2 = 1 \] Quick Tip: Gibbs Phase Rule helps determine the number of degrees of freedom for multi-phase systems.

Step 1: Definition of Zero-Order Reactions
- The rate of a zero-order reaction remains constant and does not depend on reactant concentration.
- This confirms that (A) and (C) are correct.

Step 2: Temperature Dependence
- The rate constant does depend on temperature due to the Arrhenius equation, so (B) is incorrect.

Step 3: Example of Zero-Order Reaction
- The decomposition of HI on a gold surface follows zero-order kinetics, so (D) is correct. Quick Tip: In a zero-order reaction, the rate is independent of reactant concentration but depends on external conditions like catalysts or surface area.

Step 1: Identify the van der Waals equation \[ \left(P + \frac{a}{V^2}\right) (V - b) = RT \]
where \( a \) is the measure of intermolecular attraction.

Step 2: Derive Units
Rearrange \( \frac{a}{V^2} \) in terms of pressure: \[ a = P \cdot V^2 \]
- \( P \) (Pressure) has units of atm
- \( V \) (Volume) has units of dm³

Thus, \[ a = atm \cdot (dm^3)^2 = dm^6 atm mol^{-2} \] Quick Tip: The unit of \( a \) in van der Waals equation is derived from the force of attraction term in the equation.

\textbf{Step 1: Understanding the concept of molecular speeds} The three most common types of molecular speeds in kinetic theory of gases are: - Root Mean Square (RMS) speed: \[ u_{\text{rms}} = \left( \frac{3RT}{M} \right)^{1/2} \] - Average speed: \[ u_{\text{av}} = \left( \frac{8RT}{\pi M} \right)^{1/2} \] - Most probable speed: \[ u_{\text{mp}} = \left( \frac{2RT}{M} \right)^{1/2} \] \textbf{Step 2: Choosing the correct expression} - The average speed of gas molecules is given by: \[ u_{\text{av}} = \left( \frac{8RT}{\pi M} \right)^{1/2} \] - The correct answer is Option (2).
Quick Tip: The molecular speeds depend on temperature and molar mass; heavier gases have lower speeds at the same temperature.

Step 1: Understanding viscosity relations
- Glycol has more hydroxyl groups than ethanol, leading to stronger hydrogen bonding and higher viscosity.
- Hence, (A) is incorrect as ethanol has lower viscosity than glycol.

Step 2: Checking other options
- (B) is correct as viscosity generally increases with pressure.
- (C) is correct, as it follows the Arrhenius-type equation for temperature dependence of viscosity.
- (D) is correct, as Poiseuille's equation correctly describes fluid flow through a pipe.

Thus, the incorrect statement is (A). Quick Tip: Viscosity is influenced by molecular interactions; stronger intermolecular forces lead to higher viscosity.

Step 1: Relation between mole fraction and molality \[ x_2 = \frac{m M_1}{1 + m M_1} \]
where:
- \( x_2 \) = Mole fraction of solute
- \( m \) = Molality
- \( M_1 \) = Molar mass of solvent

Step 2: Justification
- The formula is derived using the total number of moles in the solution.
- Since Option (1) follows this relation, it is the correct answer. Quick Tip: Molality and mole fraction are related through solvent molar mass.

Step 1: Formula for Reduced Mass \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]
where:
- \( m_1 = 1.673 \times 10^{-27} \) kg (mass of Hydrogen)
- \( m_2 = 58.06 \times 10^{-27} \) kg (mass of Chlorine)

Step 2: Calculate Reduced Mass \[ \mu = \frac{(1.673 \times 10^{-27}) (58.06 \times 10^{-27})}{(1.673 + 58.06) \times 10^{-27}} \] \[ \mu = \frac{97.2 \times 10^{-54}}{59.733 \times 10^{-27}} \] \[ \mu = 1.626 \times 10^{-27} kg \]

Thus, the correct answer is \( 1.626 \times 10^{-27} \) kg. Quick Tip: Reduced mass helps in calculating vibrational and rotational motion of diatomic molecules.

Step 1: Understanding Gibbs Free Energy and spontaneity
- The Gibbs free energy equation:
\[ \Delta G = \Delta H - T \Delta S \]
- If \( \Delta G \(<\) 0 \), the reaction is spontaneous.

- If \( \Delta G \(>\) 0 \), the reaction is non-spontaneous.

Step 2: Analyzing the given thermodynamic properties

- (A) \( \Delta H = -ve, \Delta S = -ve \)

- At low temperature, \( \Delta G \) is negative → Spontaneous at low temperature.

- Matches with (III).

- (B) \( \Delta H = -ve, \Delta S = -ve \)

- At high temperature, \( T \Delta S \) is large, making \( \Delta G \) positive → Non-spontaneous at high temperature.

- Matches with (I).

- (C) \( \Delta H = +ve, \Delta S = +ve \)

- At low temperature, \( \Delta G \) is positive → Non-spontaneous at low temperature.

- Matches with (II).

- (D) \( \Delta H = +ve, \Delta S = +ve \)

- At high temperature, \( \Delta G \) is negative → Spontaneous at high temperature.

- Matches with (IV). Quick Tip: For a reaction to be spontaneous at all temperatures, \( \Delta H \) should be negative and \( \Delta S \) should be positive.

Step 1: Using de Broglie’s Wavelength Formula \[ \lambda = \frac{h}{mv} \]
where:
- \( h = 6.626 \times 10^{-34} \) Js (Planck's constant)
- \( m = 1.67 \times 10^{-27} \) kg (mass of proton)
- \( v = 1.0 \times 10^3 \) m/s

Step 2: Substituting values \[ \lambda = \frac{6.626 \times 10^{-34}}{(1.67 \times 10^{-27}) \times (1.0 \times 10^3)} \] \[ = 0.40 nm \] Quick Tip: de Broglie’s equation relates wavelength to momentum, showing that slower-moving particles have longer wavelengths.

Step 1: Understanding the Electron Probability Distribution

- \( p_x \) orbital has a node at the nucleus, meaning zero probability at the center.

- Maximum probability occurs on opposite sides of the nucleus along the x-axis.

- The probability of finding an electron on the z-axis is zero since \( p_x \) is directed along x-axis. Quick Tip: The probability of finding an electron in a \( p_x \) orbital is highest along its axis and zero at the nucleus.

Step 1: Understanding Ionic Radii Trends

- Isoelectronic species have the same number of electrons, but different nuclear charges.

- More positive charge results in a smaller radius due to stronger nuclear attraction.

- More negative charge leads to a larger radius due to electron repulsion.

\[ Order: S^{2-} \(>\) Cl^- \(>\) K^+ \(>\) Ca^{2+} \] Quick Tip: Higher positive charge leads to smaller ionic radii due to stronger nuclear attraction.

Step 1: Understanding Electronegativity Trends
- Electronegativity increases across a period and decreases down a group.

- Order of increasing electronegativity: \[ Si \(<\) P \(<\) C \(<\) N \] Quick Tip: Electronegativity increases across a period and decreases down a group in the periodic table.

Step 1: Identifying Lone Pairs on Central Atoms

- ClF\textsubscript{3 has two lone pairs (T-shaped geometry).

- XeF\textsubscript{4 has two lone pairs (Square planar geometry).

- BrF\textsubscript{5 and SF\textsubscript{4 have one lone pair. Quick Tip: Use VSEPR theory to determine lone pairs and molecular geometry.

Step 1: Molecular orbital theory
- The bond order determines stability:
\[ Bond order = \frac{Number of bonding electrons - Number of anti-bonding electrons}{2} \]
- For Li\(_2^-\): Additional anti-bonding electron decreases bond order.

- For Li\(_2^+\): Loss of bonding electron decreases bond order.

- For Li\(_2\): Has the highest bond order, making it most stable.


Thus, the correct order is Li\(_2^-\) \(<\) Li\(_2^+\) \(<\) Li\(_2\). Quick Tip: Higher bond order generally leads to greater molecular stability.

Step 1: Identify the hybridization using VSEPR theory

- SF\(_2\): Sulfur has four electron pairs (2 bonds + 2 lone pairs) → sp\(^3\) hybridization.

- SF\(_4\): Sulfur has five electron pairs (4 bonds + 1 lone pair) → sp\(^3\)d hybridization.

- SF\(_6\): Sulfur has six bonding pairs (octahedral) → sp\(^3\)d\(^2\) hybridization.


Thus, the correct answer is option (3). Quick Tip: Hybridization can be determined using the steric number, which is the sum of bonded atoms and lone pairs around the central atom.

Step 1: Identify oxidation states

- Calcium oxychloride (CaOCl\(_2\)) consists of Ca\(^{2+}\), O\(^{2-}\), and two Cl atoms.

- One Cl is part of the hypochlorite ion (ClO\(^-\)) → oxidation state = +1.

- The other Cl is in the chloride ion (Cl\(^-\)) → oxidation state = -1.


Thus, the correct answer is option (4). Quick Tip: In compounds with two different chlorine atoms, check for hypochlorite (+1) and chloride (-1) oxidation states.

Step 1: Understanding ozone as an oxidizing agent
- Ozone (O\(_3\)) is a strong oxidizing agent.

- KI is oxidized to iodine (\(I_2\)):
\[ 2KI + O_3 \rightarrow I_2 + K_2O + O_2 \]
- FeSO\(_4\) is oxidized to Fe\(_2\)(SO\(_4\))\(_3\).

- KMnO\(_4\) is already in its highest oxidation state (+7), but it is an oxidizing agent itself.

- K\(_2\)MnO\(_4\) contains Mn in +6 oxidation state, which is already a stable oxidation state and not easily oxidized further by ozone.


Thus, the correct answer is option (4) K\(_2\)MnO\(_4\). Quick Tip: Substances that are already in high oxidation states (e.g., Mn in KMnO\(_4\)) are less likely to be oxidized further.

Step 1: Understanding base strength in alkali and alkaline earth metals

- KOH is a strong alkali because potassium (K) is highly electropositive.

- Ca(OH)\(_2\) and Sr(OH)\(_2\) are alkaline earth metal hydroxides and have strong basic character due to high lattice energy and solubility in water.

- LiOH is the weakest base among them because:

- Li\(^+\) has a high charge density, making it strongly hydrated in solution.

- This decreases the availability of OH\(^-\) ions, reducing basic strength.


Thus, the correct answer is option (3) LiOH. Quick Tip: The basic strength of alkali metal hydroxides increases down the group, meaning LiOH is the weakest.

Step 1: Understanding the nature of oxides

- CaO (Calcium Oxide) is a basic oxide and reacts with acids, but does not react with NaOH (a strong base).

- SiO\(_2\) (Silicon Dioxide) is an acidic oxide and reacts with NaOH to form silicates.

- BeO (Beryllium Oxide) is amphoteric, meaning it reacts with both acids and bases.

- B\(_2\)O\(_3\) (Boron Trioxide) is an acidic oxide and reacts with NaOH to form borates.


Since CaO does not react with NaOH, it is the correct answer. Quick Tip: Basic oxides do not react with strong bases, while acidic and amphoteric oxides react with bases.

Step 1: Understanding silicate structure

- Silicates are compounds containing silicon and oxygen, often found in minerals.

- The basic structural unit of silicates is the silicate tetrahedron \(SiO_4^{4-}\), where a silicon atom is covalently bonded to four oxygen atoms.

- Different silicate structures arise from how these tetrahedral units link together.


Step 2: Examining the given options

- SiO\(_3^{2-}\) does not represent a fundamental unit but may exist in chain silicates.

- SiO\(_4^{2-}\) is incorrect as it does not carry the correct charge.

- SiO\(^-\) does not exist as a primary structural unit.

- SiO\(_4^{4-}\) is the correct answer, as it is the fundamental unit of silicate structures. Quick Tip: Silicates are classified into various types (e.g., nesosilicates, inosilicates, phyllosilicates) based on how \(SiO_4^{4-}\) tetrahedral units are linked.

Step 1: Understanding the molecular structures

- \(XeO_4\) has a tetrahedral structure as xenon forms bonds with four oxygen atoms.

- \(XeO_2F_2\) has a trigonal bipyramidal shape due to five bonding domains around xenon.

- \(XeF_4\) has a square planar structure because xenon forms four bonds with fluorine and has two lone pairs.

- \(XeO_3\) is pyramidal as it has three bonded oxygens and one lone pair.

Step 2: Matching with List II

- (A) - (II) (XeO\(_4\) - Tetrahedral)

- (B) - (III) (XeO\(_2\)F\(_2\) - Trigonal bipyramidal)

- (C) - (IV) (XeF\(_4\) - Square planar)

- (D) - (I) (XeO\(_3\) - Pyramidal)


Thus, the correct answer is option (1). Quick Tip: Xenon compounds often exhibit expanded octets, leading to varied molecular geometries due to the presence of lone pairs.

Step 1: Understanding the reaction between diborane and ammonia

- Diborane (\(B_2H_6\)) reacts with excess ammonia at low temperatures to form an ammonia-borane complex.
- The primary intermediate formed in this reaction is \( [BH_2(NH_3)_2]^+[BH_4]^- \).

Step 2: Explanation of the options

- Option 1 (\(B_2H_6\cdotNH_3\)): This is an adduct but not the main product at low temperatures.
- Option 2 (Borazole): Formed at higher temperatures, not initially.
- Option 3 (\( [BH_2(NH_3)_2]^+[BH_4]^- \)): Correct, as it is the first stable product.
- Option 4 (\(B_2N_4H_{10}\)): Incorrect, as it does not form in this reaction.

Thus, the correct answer is option (3). Quick Tip: Diborane forms adducts with ammonia in a stepwise manner. The first major product at low temperatures is diborane diammoniate complex.

Step 1: Understanding molecular geometries

- ICl\(_2^-\) has a linear shape due to the presence of three lone pairs around iodine.
- NH\(_2^-\) has a bent or V-shape due to the presence of lone pairs on nitrogen.
- NH\(_4^+\) is tetrahedral because nitrogen is bonded to four hydrogen atoms without lone pairs.
- [PtCl\(_4\)]\(^{2-}\) adopts a square planar shape, typical for \(d^8\) metal complexes.

Step 2: Matching with List II

- (A) - (II) (ICl\(_2^-\) - Linear)

- (B) - (I) (NH\(_2^-\) - V-shape)

- (C) - (III) (NH\(_4^+\) - Tetrahedral)

- (D) - (IV) ([PtCl\(_4\)]\(^{2-}\) - Square Planar)


Thus, the correct answer is option (3). Quick Tip: Lone pairs influence molecular geometry significantly. The Valence Shell Electron Pair Repulsion (VSEPR) theory helps predict the shape of molecules.

Electron affinity is the energy change when an atom gains an electron. Oxygen has the highest electron affinity among these species because of its smaller size and higher effective nuclear charge. Quick Tip: Elements in group 16 generally have higher electron affinities than group 17 due to their electronic configuration.

The number of chloride ions available for dissociation determines molar conductivity. More dissociated chloride ions result in higher conductivity. As ligands replace chloride ions, molar conductivity decreases in the order given. Quick Tip: In Werner’s theory, primary valency corresponds to charge, while secondary valency is linked to coordination number.

EAN is calculated using the formula: \[ EAN = Z - O + 2C \]
where \( Z = 27 \) (atomic number of Co), \( O = 3 \) (oxidation state), and \( C = 6 \) (coordination number). \[ EAN = 27 - 3 + 2(6) = 36 \] Quick Tip: The EAN rule helps predict the stability of complexes, with values closer to noble gas configurations being more stable.

Magnetic moment is calculated using: \[ \mu = \sqrt{n(n+2)} \]
where \( n \) is the number of unpaired electrons. Based on electronic configurations, the matches are correct. Quick Tip: Higher unpaired electrons lead to stronger paramagnetism and greater magnetic moments.

Each metal/compound is correctly paired with its catalytic function. Ziegler-Natta catalysts are crucial in polymerization reactions, while Adams and Fenton's catalysts are used in different oxidation processes. Quick Tip: Catalysts lower activation energy and speed up reactions without being consumed.

The correct answer is option (1) because Europium (Eu) and Calcium (Ca) both share similarities in terms of their diamagnetism, insolubility of their sulphates and carbonates in water, and the insolubility of their dichlorides in strong HCl. However, the solubility of these metals in liquid NH\(_3\) is not a characteristic shared by both. Quick Tip: When comparing the properties of elements, carefully examine the specific characteristics like solubility, magnetic properties, and behavior in different solvents to determine similarities.

The lanthanide contraction refers to the gradual decrease in the ionic radii of the lanthanide series elements from cerium (Ce) to lutetium (Lu). The magnitude of this contraction is approximately 0.20 \, \text{Å. This is due to the poor shielding effect of the f-electrons, which causes the effective nuclear charge to increase across the series, thereby reducing the size of the ions. Quick Tip: The lanthanide contraction is an important concept in understanding the periodic trends in the lanthanide series, particularly the decrease in ionic radii as the atomic number increases.

The Mg-ATP complex is a well-known substrate for kinases. Kinases are enzymes that catalyze the transfer of phosphate groups from ATP to other molecules, which is crucial for various cellular processes. The other options—catalase, helicase, and hydrolyses—are involved in different biochemical reactions and do not utilize the Mg-ATP complex as a substrate. Quick Tip: Remember that kinases typically use ATP as a phosphate donor for phosphorylation reactions, whereas other enzymes like helicases and catalases serve different roles in the cell.

In the complex [Pd(OAC)\(_2\)], each acetate ion (OAC) has a charge of -1, and there are two such ions. Let the oxidation state of Pd be \( x \). Since the overall charge of the complex is neutral, we have the following equation: \[ x + 2(-1) = 0 \quad \Rightarrow \quad x - 2 = 0 \quad \Rightarrow \quad x = 2. \]
Thus, the oxidation state of Pd is 2. Quick Tip: To determine the oxidation state of a metal in a complex, use the charges of the ligands and balance the equation for the total charge of the complex.

Cytochrome C is an electron acceptor in the electron transport chain. It functions by accepting electrons during cellular respiration and then transferring them to other components, such as oxygen in the final step of the electron transport chain. This process is crucial for energy production in cells. Quick Tip: Cytochrome C plays a key role in the electron transport chain, where it functions as an electron carrier, helping to generate ATP by transferring electrons to oxygen.