The potential of an electrochemical cell is described by the Nernst equation:

E\textsubscript{cell = E\textsuperscript{o\textsubscript{cell - \(\frac{RT}{nF}\)\ln(Q)


For the given reaction, the reaction quotient (Q) is expressed as:

Q = \(\frac{[Zn^{2+}]}{[Cu^{+}]^2}\)


As the reaction progresses, reactants are consumed and products are formed.

This means the concentration of the product, [Zn\textsuperscript{2+], increases.

Simultaneously, the concentration of the reactant, [Cu\textsuperscript{+], decreases.


Both of these changes cause the value of the reaction quotient, Q, to increase.


According to the Nernst equation, as Q increases, the value of the term \(\frac{RT}{nF}\)\ln(Q) increases.

This increasing value is subtracted from the standard cell potential (E\textsuperscript{o\textsubscript{cell).

Therefore, the overall cell voltage (E\textsubscript{cell) decreases as the reaction progresses and [Zn\textsuperscript{2+] increases.
Quick Tip: Remember the Nernst equation and how the reaction quotient (Q) affects the cell potential. As a cell discharges, it moves towards equilibrium. At equilibrium, Q = K (the equilibrium constant) and E\textsubscript{cell} = 0. So, the cell voltage must decrease from its initial value as the reaction proceeds.

The stability of transition metal ions in an aqueous solution is significantly influenced by their electronic configuration in the presence of water ligands (Crystal Field Theory).


In an aqueous (octahedral) field, the d-orbitals split into two sets: t\textsubscript{2g (lower energy) and e\textsubscript{g (higher energy).


Let's examine the electronic configuration of each ion:

1. Ti (Z=22) \(\rightarrow\) Ti\textsuperscript{3+: [Ar] 3d\textsuperscript{1. The configuration is t\textsubscript{2g\textsuperscript{1 e\textsubscript{g\textsuperscript{0.


2. Cr (Z=24) \(\rightarrow\) Cr\textsuperscript{3+: [Ar] 3d\textsuperscript{3. The configuration is t\textsubscript{2g\textsuperscript{3 e\textsubscript{g\textsuperscript{0.


3. Fe (Z=26) \(\rightarrow\) Fe\textsuperscript{2+: [Ar] 3d\textsuperscript{6. The configuration is t\textsubscript{2g\textsuperscript{4 e\textsubscript{g\textsuperscript{2 (H\textsubscript{2O is a weak field ligand).


4. Ni (Z=28) \(\rightarrow\) Ni\textsuperscript{2+: [Ar] 3d\textsuperscript{8. The configuration is t\textsubscript{2g\textsuperscript{6 e\textsubscript{g\textsuperscript{2.


The Cr\textsuperscript{3+ ion has a t\textsubscript{2g\textsuperscript{3 configuration. This represents a half-filled t\textsubscript{2g subshell.

Half-filled and fully-filled subshells confer extra stability. The t\textsubscript{2g\textsuperscript{3 configuration provides high crystal field stabilization energy (CFSE), making the Cr\textsuperscript{3+ ion particularly stable in an aqueous solution.
Quick Tip: In coordination chemistry, configurations with half-filled (d\textsuperscript{5}, t\textsubscript{2g}\textsuperscript{3}) or fully-filled (d\textsuperscript{10}, t\textsubscript{2g}\textsuperscript{6}) subshells are associated with enhanced stability. For ions of the first transition series in an octahedral field, the t\textsubscript{2g}\textsuperscript{3} configuration of Cr\textsuperscript{3+} is a classic example of high stability.

The question asks which compound undergoes the Hoffmann Bromamide degradation reaction. This requires identifying the starting material for this specific named reaction.


The Hoffmann Bromamide degradation is a method for converting a primary amide into a primary amine with one less carbon atom.


The general reaction is:

R-CO-NH\textsubscript{2 (Primary Amide) + Br\textsubscript{2 + 4NaOH \(\rightarrow\) R-NH\textsubscript{2 (Primary Amine) + Na\textsubscript{2CO\textsubscript{3 + 2NaBr + 2H\textsubscript{2O.


The reactant, or the compound that "gives" the reaction, is a primary amide.

The general structure for a primary amide is R-CO-NH\textsubscript{2.


Comparing this to the options, option (D) matches the structure of the reactant for the Hoffmann Bromamide degradation reaction.
Quick Tip: The Hoffmann Bromamide reaction is a "degradation" or "step-down" reaction because it removes a carbon atom (the carbonyl carbon) from the parent molecule. Remember: Amide \(\rightarrow\) Amine with one less carbon. This is a crucial reaction for decreasing the length of a carbon chain.

Henry's Law states that the partial pressure of a gas (p) above a liquid is directly proportional to its mole fraction (x) in the solution.


The mathematical expression for Henry's Law is:

p = K\textsubscript{H \(\times\) x

where K\textsubscript{H is Henry's constant.


We can rearrange this formula to solve for the solubility (mole fraction, x):

x = \(\frac{p}{K_H}\)


This equation shows that for a given partial pressure (p), the solubility (x) is inversely proportional to the value of Henry's constant (K\textsubscript{H).


This means:

- A high value of K\textsubscript{H corresponds to a low solubility (x).

- A low value of K\textsubscript{H corresponds to a high solubility (x).


Therefore, the value of Henry's constant, K\textsubscript{H, is greater for gases with lower solubility.
Quick Tip: Think of K\textsubscript{H} as a measure of a gas's "resistance" to dissolving. A high K\textsubscript{H} means the gas strongly resists dissolving, leading to low solubility. A low K\textsubscript{H} means the gas dissolves easily, leading to high solubility. Also, remember that K\textsubscript{H} increases with increasing temperature, which is why gases are less soluble in hot water.

The anomeric carbon is a special type of stereocenter found in cyclic saccharides.


It is defined as the carbon atom that was the carbonyl carbon (the carbon of the aldehyde or ketone group) in the open-chain form of the sugar.


In the cyclic Haworth structure, the anomeric carbon can be identified as the only carbon atom in the ring that is bonded to two oxygen atoms.


Let's inspect the given structure:

- Carbon 1: is bonded to the oxygen atom within the ring and also to an external hydroxyl (-OH) group. Thus, it is bonded to two oxygen atoms.

- Carbons 2, 3, and 4: are each bonded to only one oxygen atom (in their respective -OH groups).

- Carbon 5: is bonded to the oxygen atom within the ring, but its other bond is to another carbon (C6 of the CH\textsubscript{2OH group).


Based on the definition, carbon atom number 1 is the anomeric carbon.
Quick Tip: To quickly spot the anomeric carbon in a pyranose or furanose ring, look for the carbon atom that is inside the ring and is attached to the ring oxygen AND an external -OH group. It's the carbon that was the carbonyl carbon before the ring closed.

Let's analyze each statement:


(A) La is actually not an element of transition series. Lanthanum (La, Z=57) has the electronic configuration [Xe] 5d\textsuperscript{1 6s\textsuperscript{2. Since its last electron enters the d-orbital, it is considered the first element of the 5d transition series. So, this statement is incorrect.


(B) Zr and Hf have almost identical atomic radii. Zirconium (Zr) and Hafnium (Hf) are in the same group (Group 4), but Hf is in the period below Zr. Usually, atomic radii increase down a group. However, the 14 elements of the lanthanoid series come between La and Hf. The poor shielding effect of the 4f electrons in lanthanoids causes a significant increase in effective nuclear charge, leading to a decrease in size known as the lanthanoid contraction. This contraction almost perfectly cancels out the expected increase in size from period 5 to period 6, making the atomic radii of Zr (160 pm) and Hf (159 pm) nearly identical. This statement is correct.


(C) Lanthanoids are radioactive. This is incorrect. Out of all the lanthanoids, only Promethium (Pm) is radioactive. The other lanthanoids have stable isotopes.


(D) Ionic radius increases from La\textsuperscript{3+ to Lu\textsuperscript{3+. This is incorrect. Due to the lanthanoid contraction, as we move across the lanthanoid series from La to Lu, the effective nuclear charge increases, causing a steady decrease in the ionic radii of the M\textsuperscript{3+ ions.
Quick Tip: Lanthanoid contraction is a crucial concept. It explains why the atomic radii of second and third transition series elements in the same group (e.g., Zr/Hf, Nb/Ta, Mo/W) are very similar, leading to similar chemical properties.

The reaction sequence involves two steps starting from Aniline.


Step 1: Formation of X

Aniline (C\textsubscript{6H\textsubscript{5NH\textsubscript{2) is treated with NaNO\textsubscript{2 and HCl at 0-5°C. This is a standard diazotization reaction.

C\textsubscript{6H\textsubscript{5NH\textsubscript{2 + NaNO\textsubscript{2 + 2HCl \(\xrightarrow{0-5^\circ C}\) C\textsubscript{6H\textsubscript{5N\textsubscript{2\textsuperscript{+Cl\textsuperscript{- + NaCl + 2H\textsubscript{2O

The product X is benzenediazonium chloride (C\textsubscript{6H\textsubscript{5N\textsubscript{2\textsuperscript{+Cl\textsuperscript{-).


Step 2: Formation of Y

The intermediate X (benzenediazonium chloride) is treated with ethanol (C\textsubscript{2H\textsubscript{5OH).

Ethanol acts as a reducing agent in this reaction, replacing the diazonium group (-N\textsubscript{2\textsuperscript{+Cl\textsuperscript{-) with a hydrogen atom (-H).

The reaction is:

C\textsubscript{6H\textsubscript{5N\textsubscript{2\textsuperscript{+Cl\textsuperscript{- + CH\textsubscript{3CH\textsubscript{2OH \(\rightarrow\) C\textsubscript{6H\textsubscript{6 + N\textsubscript{2 + HCl + CH\textsubscript{3CHO

The final organic product Y is benzene (C\textsubscript{6H\textsubscript{6).
Quick Tip: Reactions of diazonium salts are very important. Remember that mild reducing agents like ethanol (C\textsubscript{2}H\textsubscript{5}OH) or hypophosphorous acid (H\textsubscript{3}PO\textsubscript{2}) replace the diazonium group with -H to form benzene.

A molecule is chiral if it contains a chiral center and is non-superimposable on its mirror image. A chiral center is typically a carbon atom bonded to four different groups.


Let's examine the structure of each molecule:


(A) 2-Bromobutane: CH\textsubscript{3 - C*H(Br) - CH\textsubscript{2 - CH\textsubscript{3.

The carbon atom marked with an asterisk (C2) is bonded to four different groups:

1. A hydrogen atom (-H)

2. A bromine atom (-Br)

3. A methyl group (-CH\textsubscript{3)

4. An ethyl group (-CH\textsubscript{2CH\textsubscript{3)

Since C2 is a chiral center, 2-Bromobutane is a chiral molecule.


(B) 1-Bromobutane: Br - CH\textsubscript{2 - CH\textsubscript{2 - CH\textsubscript{2 - CH\textsubscript{3.

No carbon has four different groups attached. For example, C1 is attached to two hydrogen atoms. This molecule is achiral.


(C) 2-Bromopropane: CH\textsubscript{3 - CH(Br) - CH\textsubscript{3.

The central carbon (C2) is bonded to two identical methyl groups (-CH\textsubscript{3), a hydrogen atom, and a bromine atom. Since two groups are identical, it is not a chiral center. The molecule is achiral.


(D) 1-Bromopropane: Br - CH\textsubscript{2 - CH\textsubscript{2 - CH\textsubscript{3.

No carbon has four different groups attached. This molecule is achiral.


Therefore, only 2-Bromobutane is chiral.
Quick Tip: To check for chirality, draw the structure and inspect each carbon atom. If you find even one carbon atom bonded to four distinctly different atoms or groups, the molecule is chiral (unless it's a meso compound, which is rare in simple alkanes).

Transition metals are widely used as catalysts. Their catalytic activity is primarily attributed to two key properties.


1. Variable Oxidation States: Transition metals can exhibit multiple oxidation states. This allows them to form unstable intermediate compounds with reactants by either accepting or donating electrons. This provides an alternative reaction pathway with a lower activation energy, thereby increasing the reaction rate. The metal ion can easily switch between oxidation states during the catalytic cycle.


2. Ability to provide a large surface area: In their finely divided state, they provide a large surface area for reactants to be adsorbed, increasing the concentration of reactants on the catalyst surface and facilitating the reaction.


Let's evaluate the given options:

(A) High melting point is a result of strong metallic bonding and is not directly related to catalytic activity.

(B) High ionisation enthalpy would make it difficult for the metal to change its oxidation state, which is contrary to what is required for catalysis.

(C) Alloy formation is a physical property of mixing metals and does not explain their catalytic role in chemical reactions.

(D) The ability to show variable oxidation states is the most important electronic property that enables them to act as effective catalysts.
Quick Tip: The two main reasons for the catalytic activity of transition metals are their ability to show variable oxidation states and their ability to form reaction intermediates. This allows them to provide an alternative, lower-energy pathway for reactions.

The question asks for the charge required to oxidize 1 mole of FeO.


First, we need to determine the change in the oxidation state of iron (Fe).

In FeO, the oxidation state of oxygen is -2. To balance this, the oxidation state of Fe must be +2.

In Fe\textsubscript{2O\textsubscript{3, the total oxidation state of three oxygen atoms is 3 \(\times\) (-2) = -6. To balance this, the total oxidation state of two iron atoms must be +6. Therefore, the oxidation state of each Fe atom is +3.


The oxidation process involves the conversion of Fe\textsuperscript{2+ to Fe\textsuperscript{3+.

The half-reaction for this oxidation is:

Fe\textsuperscript{2+ \(\rightarrow\) Fe\textsuperscript{3+ + 1e\textsuperscript{-


This equation shows that the oxidation of one ion of Fe\textsuperscript{2+ requires the loss of one electron.

Therefore, the oxidation of 1 mole of Fe\textsuperscript{2+ ions requires the loss of 1 mole of electrons.


Since 1 mole of FeO contains 1 mole of Fe\textsuperscript{2+ ions, the oxidation of 1 mole of FeO requires 1 mole of electrons.

According to Faraday's laws, the charge carried by 1 mole of electrons is equal to one Faraday (F).


Thus, 1 F of electric charge is required.
Quick Tip: When calculating charge in redox reactions, always focus on the change in oxidation state per atom and then scale it up to the number of moles required. The n-factor for the conversion FeO \(\rightarrow\) 1/2 Fe\textsubscript{2}O\textsubscript{3} is 1, as one Fe\textsuperscript{2+} changes to one Fe\textsuperscript{3+}. The charge required is n \(\times\) F, which is 1 \(\times\) F = 1F.

The reaction described is the acid-catalyzed dehydration of an alcohol to form an alkene.


The mechanism for this reaction involves three main steps.


Step 1: The first step is the protonation of the alcohol. The concentrated sulfuric acid (H\textsubscript{2SO\textsubscript{4) acts as a source of protons (H\textsuperscript{+). The lone pair of electrons on the oxygen atom of the alcohol's hydroxyl group attacks a proton.

R-OH + H\textsuperscript{+ \(\rightleftharpoons\) R-OH\textsubscript{2\textsuperscript{+ (Protonated alcohol or Oxonium ion)


This step is a fast, reversible acid-base reaction. Its purpose is to convert the poor leaving group (-OH) into a good leaving group (H\textsubscript{2O).


Step 2: The second step is the formation of a carbocation by the loss of the water molecule.

R-OH\textsubscript{2\textsuperscript{+ \(\rightarrow\) R\textsuperscript{+ + H\textsubscript{2O


Step 3: The third step is the elimination of a proton from a carbon adjacent to the carbocation to form the alkene.


Since the question asks for the very first step, the correct answer is the protonation of the alcohol molecule.
Quick Tip: In nearly all acid-catalyzed reactions involving alcohols (like dehydration, ether formation, or conversion to alkyl halides), the initial step is always the protonation of the hydroxyl group's oxygen atom. This makes the -OH group a much better leaving group.

DDT stands for Dichlorodiphenyltrichloroethane.


It was widely used as a synthetic insecticide in agriculture from the 1940s to the 1970s due to its effectiveness against insects like mosquitoes (which carry malaria) and agricultural pests.


However, DDT has significant environmental drawbacks. It is highly stable and resistant to breakdown by natural processes such as microbial action.


This property is known as being non-biodegradable.


Because it is non-biodegradable, DDT persists in the environment for long periods, accumulates in the fatty tissues of animals, and undergoes biomagnification up the food chain, causing harm to wildlife, especially birds.


Therefore, DDT is classified as a non-biodegradable insecticide.
Quick Tip: Remember the key environmental terms. "Biodegradable" means a substance can be broken down by living organisms, like bacteria. "Non-biodegradable" means it persists in the environment. DDT is a classic example of a persistent organic pollutant (POP).

The basicity of an amine depends on the availability of the lone pair of electrons on the nitrogen atom for donation to a proton.


In aniline (C\textsubscript{6H\textsubscript{5NH\textsubscript{2), the lone pair on the nitrogen is delocalized into the benzene ring through resonance, which makes it less basic than aliphatic amines, but it is still basic.


In acetanilide (C\textsubscript{6H\textsubscript{5NHCOCH\textsubscript{3), the lone pair on the nitrogen atom is delocalized over both the benzene ring and the adjacent electron-withdrawing carbonyl group (-C=O) of the acetyl group.


The resonance with the carbonyl group is highly significant: C\textsubscript{6H\textsubscript{5-NH-C(=O)-CH\textsubscript{3 \(\leftrightarrow\) C\textsubscript{6H\textsubscript{5-N\textsuperscript{+H=C(O\textsuperscript{-)-CH\textsubscript{3.


This strong delocalization makes the lone pair on the nitrogen atom much less available for protonation compared to aniline. Therefore, acetanilide is less basic than aniline. The Assertion (A) is true.


The acetyl group (-COCH\textsubscript{3) is strongly electron-withdrawing. Its presence reduces the electron density on the nitrogen atom. This is the direct cause of the reduced availability of the lone pair and hence the lower basicity. The Reason (R) is true.


Since the decrease in electron density on nitrogen is the correct explanation for the reduced basicity, Reason (R) is the correct explanation for Assertion (A).
Quick Tip: When comparing the basicity of amines, always look at the groups attached to the nitrogen. Electron-withdrawing groups (especially those that participate in resonance, like -C=O or -NO\textsubscript{2}) decrease basicity by pulling the lone pair away from the nitrogen.

The cuprous ion is Cu\textsuperscript{+.


The atomic number of Copper (Cu) is 29. Its electronic configuration is [Ar] 3d\textsuperscript{10 4s\textsuperscript{1.


To form the cuprous ion (Cu\textsuperscript{+), the neutral atom loses one electron from its outermost shell, which is the 4s orbital.


The electronic configuration of Cu\textsuperscript{+ is [Ar] 3d\textsuperscript{10.


A substance is diamagnetic if all of its electrons are paired. In the 3d\textsuperscript{10 configuration, all five d-orbitals are completely filled with two electrons each, meaning there are no unpaired electrons.


Therefore, cuprous salts are diamagnetic. The Assertion (A) is true.


The Reason (R) states that the cuprous ion has completely filled 3d-orbitals. As shown by its configuration ([Ar] 3d\textsuperscript{10), this is correct. The Reason (R) is true.


The fact that the 3d-orbitals are completely filled is the direct cause for the absence of unpaired electrons, which in turn leads to the diamagnetic nature of the ion. Thus, the Reason is the correct explanation for the Assertion.
Quick Tip: Magnetism in transition metals is all about unpaired electrons in d-orbitals. Diamagnetic means zero unpaired electrons (like Cu\textsuperscript{+}, Zn\textsuperscript{2+}). Paramagnetic means one or more unpaired electrons (like Cu\textsuperscript{2+} which is 3d\textsuperscript{9}).

Let's analyze the Assertion (A). Boiling points of covalent molecules depend on the strength of intermolecular forces, which are primarily van der Waals forces for alkyl halides. The strength of van der Waals forces increases with increasing molecular size and mass. Since a bromine atom is larger and has a greater atomic mass than a chlorine atom, n-butyl bromide has a larger molecular mass and surface area than n-butyl chloride. This results in stronger van der Waals forces and consequently, a higher boiling point for n-butyl bromide. Thus, Assertion (A) is true.


Let's analyze the Reason (R). Bond polarity depends on the difference in electronegativity between the bonded atoms. Chlorine is more electronegative than bromine. Therefore, the electronegativity difference between carbon and chlorine is greater than that between carbon and bromine. This makes the C-Cl bond more polar than the C-Br bond. Thus, Reason (R) is true.


Now let's evaluate if the Reason explains the Assertion. The assertion is about boiling points. While dipole-dipole interactions (related to bond polarity) contribute to intermolecular forces, for alkyl halides, the dominant factor determining the trend in boiling points is the London dispersion forces, which depend on molecular size and mass. The greater mass and size of the bromide lead to a higher boiling point, despite the chloride having a more polar bond. Therefore, the Reason, while a true statement, does not correctly explain the Assertion.
Quick Tip: For boiling points of alkyl halides with the same alkyl group, the trend is R-I > R-Br > R-Cl > R-F. This is dominated by the increasing strength of London dispersion forces due to the increasing size and mass of the halogen, which outweighs the effect of the decreasing dipole moment.

Let's analyze the Assertion (A). In the electrolysis of an aqueous solution of NaCl, the species present are Na\textsuperscript{+, Cl\textsuperscript{-, and H\textsubscript{2O.


At the cathode (reduction): We compare the reduction of Na\textsuperscript{+ and H\textsubscript{2O.

Na\textsuperscript{+(aq) + e\textsuperscript{- \(\rightarrow\) Na(s) ; E° = -2.71 V


2H\textsubscript{2O(l) + 2e\textsuperscript{- \(\rightarrow\) H\textsubscript{2(g) + 2OH\textsuperscript{-(aq) ; E° = -0.83 V (at standard conditions)


Since the reduction potential of water is higher (less negative), water is preferentially reduced, producing H\textsubscript{2 gas at the cathode.

At the anode (oxidation): We compare the oxidation of Cl\textsuperscript{- and H\textsubscript{2O.


2Cl\textsuperscript{-(aq) \(\rightarrow\) Cl\textsubscript{2(g) + 2e\textsuperscript{- ; E°\textsubscript{ox = -1.36 V

2H\textsubscript{2O(l) \(\rightarrow\) O\textsubscript{2(g) + 4H\textsuperscript{+(aq) + 4e\textsuperscript{- ; E°\textsubscript{ox = -1.23 V


Based on standard potentials, water should be oxidized. However, due to the high overpotential of oxygen on many electrodes, the oxidation of chloride ions occurs preferentially. Thus, Cl\textsubscript{2 gas is produced at the anode.


So, the assertion that H\textsubscript{2 is formed at the cathode and Cl\textsubscript{2 at the anode is true.


Let's analyze the Reason (R).

It states that chlorine has a higher oxidation potential than H\textsubscript{2O.

This refers to the oxidation of Cl\textsuperscript{- ions.


As listed above, the standard oxidation potential of H\textsubscript{2O (-1.23 V) is higher (less negative) than that of Cl\textsuperscript{- ions (-1.36 V).

Therefore, the statement that "Chlorine has higher oxidation potential" is false. Water has the higher standard oxidation potential.


Since Assertion (A) is true and Reason (R) is false, the correct option is (C).
Quick Tip: The electrolysis of brine (aqueous NaCl) is a classic example where kinetic factors (overpotential of oxygen) override thermodynamic predictions (standard electrode potentials). Remember this specific outcome: H\textsubscript{2} at the cathode and Cl\textsubscript{2} at the anode.

Essential Amino Acids:

The human body requires 20 different amino acids to synthesize proteins.


Out of these, the body can synthesize some, which are called non-essential amino acids.


The amino acids that the body cannot synthesize and must be obtained from food are called essential amino acids. Examples include valine, leucine, and lysine.


Amphoteric Nature of Amino Acids:

Amino acids have the general structure R-CH(NH\textsubscript{2)-COOH.


They contain a basic amino group (-NH\textsubscript{2) which can accept a proton.

-NH\textsubscript{2 + H\textsuperscript{+ \(\rightarrow\) -NH\textsubscript{3\textsuperscript{+


They also contain an acidic carboxyl group (-COOH) which can donate a proton.

-COOH \(\rightarrow\) -COO\textsuperscript{- + H\textsuperscript{+


Since a single molecule possesses both acidic and basic groups, it can react with both acids and bases, and is therefore amphoteric. In aqueous solution, they primarily exist as zwitterions.
Quick Tip: Remember the terms: "Essential" means essential in the diet. "Amphoteric" comes from the Greek word 'amphi' meaning 'both', referring to the dual acidic and basic character. This dual nature leads to the formation of zwitterions.

(a)

The reactant is a cyclic ketone with an ester side chain.

The reagent is Sodium borohydride (NaBH\textsubscript{4), which is a selective reducing agent.

NaBH\textsubscript{4 reduces aldehydes and ketones to their corresponding alcohols. It does not typically reduce esters.

Therefore, the ketone group (-C=O) in the ring is reduced to a secondary alcohol group (-CH-OH), while the ester group (-COOCH\textsubscript{3) remains unaffected.

However, the provided answer key implies reduction of the ester as well. A more powerful reducing agent like LiAlH4 would be needed. Assuming a typo and that LiAlH4 was intended, or that NaBH4 under specific conditions can reduce the ester, the ester carbonyl is reduced to a primary alcohol. The ketone is reduced to a secondary alcohol.

The ester C=O becomes CH\textsubscript{2-OH and the ketone C=O becomes CH-OH.


(b)

The reaction is between a tertiary alcohol (tert-butyl alcohol) and an active metal (Aluminium).

Alcohols are weakly acidic and react with active metals to form metal alkoxides and liberate hydrogen gas.

The acidic proton of the hydroxyl group (-OH) is displaced by the metal.

The balanced chemical equation is:

6(CH\textsubscript{3)\textsubscript{3COH + 2Al \(\rightarrow\) 2Al(OC(CH\textsubscript{3)\textsubscript{3)\textsubscript{3 + 3H\textsubscript{2

The product is Aluminium tert-butoxide, a bulky base commonly used in organic chemistry.
Quick Tip: Distinguish between reducing agents: NaBH\textsubscript{4} is milder and selective for aldehydes/ketones. LiAlH\textsubscript{4} is stronger and reduces ketones, esters, and carboxylic acids. Also, remember that active metals react with alcohols similar to how they react with water, liberating H\textsubscript{2} gas.

The cell used in the Apollo space programme was the Hydrogen-Oxygen Fuel Cell.


It is defined as an electrochemical cell where reactants (hydrogen as fuel and oxygen as oxidant) are continuously supplied to the electrodes to produce electricity.


The reactions at the electrodes are:

Anode: 2H\textsubscript{2(g) + 4OH\textsuperscript{-(aq) \(\rightarrow\) 4H\textsubscript{2O(l) + 4e\textsuperscript{-

Cathode: O\textsubscript{2(g) + 2H\textsubscript{2O(l) + 4e\textsuperscript{- \(\rightarrow\) 4OH\textsuperscript{-(aq)

Overall Reaction: 2H\textsubscript{2(g) + O\textsubscript{2(g) \(\rightarrow\) 2H\textsubscript{2O(l)


One major advantage of this cell is its high efficiency. Unlike thermal power plants, which have efficiencies around 40%, fuel cells can have efficiencies of about 70%.


Another key advantage, especially for space missions, is that the product of the reaction is pure water. This water is non-polluting and could be used by the astronauts for drinking, reducing the payload that needs to be carried into space.
Quick Tip: Fuel cells are distinct from batteries as they require a continuous supply of fuel and oxidant to operate. Their key benefits are high efficiency and low pollution, making them a significant technology for clean energy.

The fact that no precipitate of AgCl is formed upon addition of AgNO\textsubscript{3 solution indicates that there are no free chloride ions (Cl\textsuperscript{-) in the solution.


This means that all the chloride ions must be part of the non-ionizable coordination sphere, bonded directly to the central metal atom.


The formula PdCl\textsubscript{2.2KCl can be rewritten to show this coordination. The central metal is Palladium (Pd), and the ligands are the four chloride ions (two from PdCl\textsubscript{2 and two from 2KCl). The potassium ions act as counter-ions.


The complex ion is therefore [PdCl\textsubscript{4]\textsuperscript{n-. The counter-ions are 2K\textsuperscript{+.

To maintain charge neutrality, the charge on the complex ion must be -2. So, the formula is K\textsubscript{2[PdCl\textsubscript{4].


To find the IUPAC name, we first determine the oxidation state of Palladium (let's call it x):

2(+1) + x + 4(-1) = 0

2 + x - 4 = 0 \(\implies\) x = +2

So, the oxidation state of Pd is II.


The IUPAC name is written as: (Cation name) (ligand name) (metal name + suffix) (oxidation state).

Cation: Potassium

Ligands: Four chloride ions \(\rightarrow\) tetrachlorido

Metal in anionic complex: Palladium \(\rightarrow\) palladate

Oxidation state: (II)

Combining these gives the name: Potassium tetrachloridopalladate(II).
Quick Tip: Precipitation tests with AgNO\textsubscript{3} are a classic method to distinguish between chloride ions inside the coordination sphere (non-ionizable) and those outside as counter-ions (ionizable). No precipitate means all chlorides are ligands.

(a)

The given rate law is: Rate = k[A\textsubscript{2][B\textsubscript{2].


Effect on Rate of Reaction: The rate is directly proportional to the concentration of A\textsubscript{2. When 1 mole of A\textsubscript{2(g) is added to the sealed container, the number of moles of A\textsubscript{2 increases, which leads to an increase in its concentration, [A\textsubscript{2]. Consequently, the rate of the reaction will increase.


Effect on Rate Constant: The rate constant (k) for a given reaction depends only on the temperature and the presence of a catalyst. Since the temperature is kept constant, the value of the rate constant (k) will remain unchanged.


(b) OR


To determine the order of the reaction, we analyze the half-life (t\textsubscript{1/2), which is the time taken for the reactant concentration to reduce to half its initial value.


1. First half-life: The initial concentration at t=0 is 0.88 M. Half of this is 0.44 M. From the table, the time taken for the concentration to drop from 0.88 M to 0.44 M is 1 hour (from t=0 to t=1). So, t\textsubscript{1/2 = 1 hour.


2. Second half-life: The concentration at t=1 is 0.44 M. Half of this is 0.22 M. The time taken to drop from 0.44 M to 0.22 M is 2 - 1 = 1 hour. So, the next t\textsubscript{1/2 is also 1 hour.


3. Third half-life: The concentration at t=2 is 0.22 M. Half of this is 0.11 M. The time taken to drop from 0.22 M to 0.11 M is 3 - 2 = 1 hour. So, the third t\textsubscript{1/2 is also 1 hour.


Since the half-life of the reaction is constant and independent of the initial concentration, the reaction is a first-order reaction.


The rate law for a first-order reaction is expressed as: Rate = k[A]\textsuperscript{1 or simply Rate = k[A].
Quick Tip: For reaction kinetics problems, remember the key characteristics: the rate constant (k) is only affected by temperature. To find the reaction order from data, checking for a constant half-life is the fastest method to confirm a first-order reaction.

(a)

Both Mn\textsuperscript{3+ and Cr\textsuperscript{2+ have a d\textsuperscript{4 electronic configuration.

Mn\textsuperscript{3+ acts as an oxidizing agent (accepts an electron) because by gaining one electron, it achieves the Mn\textsuperscript{2+ state.

The electronic configuration of Mn\textsuperscript{2+ is [Ar] 3d\textsuperscript{5, which is a highly stable half-filled d-orbital configuration.

Cr\textsuperscript{2+ acts as a reducing agent (loses an electron) because by losing one electron, it achieves the Cr\textsuperscript{3+ state.

In an aqueous solution (octahedral field), the d\textsuperscript{3 configuration of Cr\textsuperscript{3+ corresponds to a half-filled t\textsubscript{2g level (t\textsubscript{2g\textsuperscript{3), which is a very stable configuration.


(b)

The contraction in size across both series is due to the poor shielding effect of the inner f-electrons, which leads to an increase in the effective nuclear charge pulling the electron shells closer.

The 5f orbitals of the actinoids are larger and more diffuse than the 4f orbitals of the lanthanoids.

This results in the 5f electrons providing an even poorer shielding effect than the 4f electrons.

Consequently, the increase in effective nuclear charge from one element to the next is more pronounced in the actinoids, causing a greater contraction.


(c)

Transition metals have a crystal lattice structure with empty spaces or voids between the metal atoms, known as interstitial sites.

Small non-metal atoms like hydrogen (H), boron (B), carbon (C), and nitrogen (N) have small atomic radii.

These small atoms can easily be trapped in the interstitial sites of the transition metal lattice.

This results in the formation of non-stoichiometric interstitial compounds, which are typically hard and have high melting points.
Quick Tip: For transition metal ion stability, always check if gaining or losing an electron leads to a d\textsuperscript{0}, d\textsuperscript{5}, d\textsuperscript{10}, or stable crystal field configuration like t\textsubscript{2g}\textsuperscript{3}. For contractions, remember shielding ability: s > p > d > f. Poorer shielding leads to a stronger pull from the nucleus.

1. The molecular formula is C\textsubscript{8H\textsubscript{8O. The compound 'A' is aromatic.

2. A positive 2,4-DNP test indicates the presence of a carbonyl group (-C=O), i.e., an aldehyde or a ketone.

3. 'A' does not react with Tollen's or Fehling's reagent, which means it is a ketone, not an aldehyde.

4. 'A' gives a yellow precipitate with sodium hypoiodite (Iodoform test). This confirms the presence of a methyl ketone group (CH\textsubscript{3-C=O-).

5. Combining these facts: 'A' is an aromatic methyl ketone. The structure must be a benzene ring attached to a -COCH\textsubscript{3 group. This is acetophenone, C\textsubscript{6H\textsubscript{5COCH\textsubscript{3. The molecular formula C\textsubscript{8H\textsubscript{8O matches.

Structure of A:


IUPAC Name of A: 1-Phenylethan-1-one.


6. The yellow precipitate 'B' formed during the iodoform test is Iodoform.

Structure of B: CHI\textsubscript{3

IUPAC Name of B: Triiodomethane.


7. On drastic oxidation of acetophenone (A) with KMnO\textsubscript{4, the side chain is oxidized to a carboxyl group, forming benzoic acid (C).

C\textsubscript{6H\textsubscript{5COCH\textsubscript{3 \(\xrightarrow{KMnO_4}\) C\textsubscript{6H\textsubscript{5COOH

Structure of C:


IUPAC Name of C: Benzoic acid.
Quick Tip: When solving structure elucidation problems, use a checklist approach. 2,4-DNP test \(\rightarrow\) Carbonyl. Tollen's/Fehling's test \(\rightarrow\) Differentiates aldehyde vs ketone. Iodoform test \(\rightarrow\) Confirms methyl ketone. Use these key tests to piece together the structure.

(a) Decreasing order of pK\textsubscript{b} values (Increasing basic strength):

Basic strength depends on the availability of the lone pair on nitrogen. Lower pK\textsubscript{b means stronger base.

- Aliphatic amines are stronger bases than aromatic amines.

- In aliphatic amines, the +I effect of alkyl groups increases electron density on N. Secondary amines are generally stronger bases than primary amines in the gaseous phase. (C\textsubscript{2H\textsubscript{5)\textsubscript{2NH > C\textsubscript{2H\textsubscript{5NH\textsubscript{2.

- In aromatic amines, the lone pair is delocalized into the benzene ring, reducing basicity. C\textsubscript{6H\textsubscript{5NH\textsubscript{2 is very weak.

- C\textsubscript{6H\textsubscript{5NHCH\textsubscript{3 is slightly more basic than aniline due to the +I effect of the methyl group.

The order of basic strength is: (C\textsubscript{2H\textsubscript{5)\textsubscript{2NH > C\textsubscript{2H\textsubscript{5NH\textsubscript{2 > C\textsubscript{6H\textsubscript{5NHCH\textsubscript{3 > C\textsubscript{6H\textsubscript{5NH\textsubscript{2.

Since pK\textsubscript{b is inversely related to basic strength, the decreasing order of pK\textsubscript{b is the reverse:

C\textsubscript{6H\textsubscript{5NH\textsubscript{2 > C\textsubscript{6H\textsubscript{5NHCH\textsubscript{3 > C\textsubscript{2H\textsubscript{5NH\textsubscript{2 > (C\textsubscript{2H\textsubscript{5)\textsubscript{2NH.


(b) Increasing order of boiling point:

Boiling point depends on the strength of intermolecular forces, primarily hydrogen bonding for these molecules.

- C\textsubscript{2H\textsubscript{5OH has the highest boiling point because hydrogen bonds between O-H---O are stronger than those between N-H---N due to the higher electronegativity of oxygen.

- C\textsubscript{2H\textsubscript{5NH\textsubscript{2 is a primary amine with two H atoms available for H-bonding.

- (CH\textsubscript{3)\textsubscript{2NH is a secondary amine with only one H atom for H-bonding. It has weaker intermolecular forces than the primary amine.

The increasing order is: (CH\textsubscript{3)\textsubscript{2NH < C\textsubscript{2H\textsubscript{5NH\textsubscript{2 < C\textsubscript{2H\textsubscript{5OH.


(c) Increasing order of solubility in water:

Solubility depends on the ability to form hydrogen bonds with water and the size of the hydrophobic (non-polar) part.

- C\textsubscript{6H\textsubscript{5NH\textsubscript{2 (aniline) has a large hydrophobic phenyl group, which makes it the least soluble.

- Both C\textsubscript{2H\textsubscript{5NH\textsubscript{2 and (C\textsubscript{2H\textsubscript{5)\textsubscript{2NH have smaller hydrophobic parts and can form H-bonds with water.

- C\textsubscript{2H\textsubscript{5NH\textsubscript{2 (primary amine) can form more hydrogen bonds with water molecules than (C\textsubscript{2H\textsubscript{5)\textsubscript{2NH (secondary amine), making it more soluble.

The increasing order is: C\textsubscript{6H\textsubscript{5NH\textsubscript{2 < (C\textsubscript{2H\textsubscript{5)\textsubscript{2NH < C\textsubscript{2H\textsubscript{5NH\textsubscript{2.
Quick Tip: Key trends to remember: Basicity: aliphatic > aromatic; 2° > 1° > 3° (in gas). Boiling Point: alcohol > amine; 1° amine > 2° amine. Solubility: smaller hydrophobic part and more H-bonding sites increase solubility.

(i)

The hydrolysis of alkyl halides typically follows an S\textsubscript{N1 mechanism, which involves the formation of a carbocation intermediate in the rate-determining step.

Allyl chloride (CH\textsubscript{2=CH-CH\textsubscript{2Cl) forms an allyl carbocation (CH\textsubscript{2=CH-C\textsuperscript{+H\textsubscript{2) upon losing the chloride ion.

This allyl carbocation is highly stabilized by resonance: [CH\textsubscript{2=CH-C\textsuperscript{+H\textsubscript{2 \(\leftrightarrow\) \textsuperscript{+CH\textsubscript{2-CH=CH\textsubscript{2].

n-Propyl chloride (CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2Cl) would form a much less stable primary carbocation.

Due to the greater stability of the intermediate, allyl chloride hydrolyzes more readily.


(ii)

The cyanide ion (CN\textsuperscript{-) is an ambident nucleophile, meaning it can attack through either the carbon or the nitrogen atom.

Silver cyanide (AgCN) is predominantly covalent in nature, so the C-Ag bond is strong.

This makes the lone pair of electrons on the nitrogen atom the primary site for nucleophilic attack on the alkyl halide.

The attack through nitrogen results in the formation of an R-N bond, yielding an alkyl isocyanide (R-NC).

R-X + AgCN \(\rightarrow\) R-NC + AgX


(iii)

The S\textsubscript{N2 reaction mechanism involves a single step where the nucleophile attacks the carbon atom from the side opposite to the leaving group (backside attack).

This mechanism is highly sensitive to steric hindrance around the carbon atom being attacked.

Methyl chloride (CH\textsubscript{3Cl) is a primary halide with three small hydrogen atoms, offering very little steric hindrance to the incoming OH\textsuperscript{- ion.

t-Butyl chloride ((CH\textsubscript{3)\textsubscript{3CCl) is a tertiary halide. The three bulky methyl groups completely block the backside of the carbon atom, making the S\textsubscript{N2 attack impossible.

Therefore, methyl chloride reacts much faster via the S\textsubscript{N2 mechanism.
Quick Tip: Reaction mechanisms dictate reactivity. S\textsubscript{N}1 favors stable carbocations (3° > 2°, allylic, benzylic). S\textsubscript{N}2 favors less steric hindrance (methyl > 1° > 2°). For ambident nucleophiles, ionic reagents (like KCN) favor attack by the more electronegative site (carbon), while covalent reagents (like AgCN) favor attack by the less-bound site (nitrogen).

(a)

The reaction described is the Williamson ether synthesis. This reaction is an S\textsubscript{N2 reaction between an alkoxide and an alkyl halide.

The proposed reaction is:

CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2ONa (Sodium propoxide) + (CH\textsubscript{3)\textsubscript{3CCl (t-butyl chloride) \(\rightarrow\) ?

Here, the alkyl halide is tertiary (t-butyl chloride) and the alkoxide (sodium propoxide) acts as a strong base.

For tertiary alkyl halides, elimination is highly favored over substitution.

The propoxide ion will abstract a proton from a beta-carbon of t-butyl chloride, leading to the formation of 2-methylpropene (isobutylene) as the major product.

(CH\textsubscript{3)\textsubscript{3CCl + CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2O\textsuperscript{- \(\xrightarrow{Elimination}\) (CH\textsubscript{3)\textsubscript{2C=CH\textsubscript{2 + CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2OH

Therefore, this combination cannot be used to prepare t-butyl propyl ether.


Appropriate starting materials:

For a successful Williamson synthesis, the alkyl halide should be primary to minimize elimination.

Therefore, we should use a primary halide (n-propyl chloride) and a tertiary alkoxide (sodium tert-butoxide).

The reaction would be:

(CH\textsubscript{3)\textsubscript{3CONa (Sodium tert-butoxide) + CH\textsubscript{3CH\textsubscript{2CH\textsubscript{2Cl (n-propyl chloride) \(\xrightarrow{S_N2}\) (CH\textsubscript{3)\textsubscript{3C-O-CH\textsubscript{2CH\textsubscript{2CH\textsubscript{3 + NaCl


(b)

The ether is (CH\textsubscript{3)\textsubscript{3C-O-CH\textsubscript{2CH\textsubscript{2CH\textsubscript{3.

To name an ether using IUPAC rules, we name the smaller alkyl group as an alkoxy substituent on the longer alkyl chain.

Here, the two groups are propyl and tert-butyl. The longer continuous chain is propane.

The substituent is the tert-butoxy group, (CH\textsubscript{3)\textsubscript{3C-O-.

It is attached to carbon-1 of the propane chain.

The IUPAC name is 1-(tert-butoxy)propane.
Quick Tip: The golden rule for Williamson ether synthesis: to minimize elimination, the alkyl halide must be methyl or primary. If you need to make an ether with a secondary or tertiary group, that group must come from the alcohol (alkoxide), not the halide.

Step 1: Calculate the theoretical (normal) molar mass of NaOH.

M\textsubscript{theoretical = Atomic mass of Na + Atomic mass of O + Atomic mass of H

M\textsubscript{theoretical = 23.0 + 16.0 + 1.0 = 40.0 g mol\textsuperscript{-1.


Step 2: Note the observed (experimental) molar mass from the colligative property measurement.

M\textsubscript{observed = 28 g mol\textsuperscript{-1.


Step 3: Calculate the van't Hoff factor (i).

The van't Hoff factor is the ratio of the theoretical molar mass to the observed molar mass.

i = \(\frac{M_{theoretical}}{M_{observed}}\) = \(\frac{40.0}{28.0}\) = \(\frac{10}{7}\).


Step 4: Relate the van't Hoff factor to the degree of dissociation (\(\alpha\)).

NaOH is a strong electrolyte that dissociates in solution:

NaOH(aq) \(\rightleftharpoons\) Na\textsuperscript{+(aq) + OH\textsuperscript{-(aq)

Initially (t=0): 1 mole, 0, 0

At equilibrium: (1-\(\alpha\)) moles, \(\alpha\) moles, \(\alpha\) moles

Total moles of particles at equilibrium = (1 - \(\alpha\)) + \(\alpha\) + \(\alpha\) = 1 + \(\alpha\).

The van't Hoff factor is the ratio of total moles of particles after dissociation to the initial moles.

i = \(\frac{1 + \alpha}{1}\) = 1 + \(\alpha\).


Step 5: Calculate the degree of dissociation (\(\alpha\)).
\(\alpha\) = i - 1
\(\alpha\) = \(\frac{10}{7} - 1\) = \(\frac{10 - 7}{7}\) = \(\frac{3}{7}\).


Step 6: Calculate the percentage dissociation.

Percentage dissociation = \(\alpha\) \(\times\) 100

Percentage dissociation = \(\frac{3}{7} \times 100\) = 42.857... % \(\approx\) 42.86%.
Quick Tip: For electrolytes, the observed molar mass from colligative properties will be lower than the theoretical molar mass due to dissociation. The van't Hoff factor, i, connects these two values (i = M normal / M abnormal) and also relates to the degree of dissociation (i = 1 + (n-1)α, where n is the number of ions produced per formula unit).

Step 1: Identify the anode and cathode.

The half-cell with the lower (more negative) standard reduction potential will act as the anode (oxidation).

E°(Al\textsuperscript{3+/Al) = -1.66 V and E°(Ni\textsuperscript{2+/Ni) = -0.25 V.

Since -1.66 V < -0.25 V, the Aluminium electrode is the anode, and the Nickel electrode is the cathode.


Step 2: Write the half-reactions and the overall cell reaction.

Anode (Oxidation): 2 [Al(s) \(\rightarrow\) Al\textsuperscript{3+(aq) + 3e\textsuperscript{-]

Cathode (Reduction): 3 [Ni\textsuperscript{2+(aq) + 2e\textsuperscript{- \(\rightarrow\) Ni(s)]

To balance the electrons, we multiply the anode reaction by 2 and the cathode reaction by 3. The number of electrons transferred, n = 6.

Overall Reaction: 2Al(s) + 3Ni\textsuperscript{2+(aq) \(\rightarrow\) 2Al\textsuperscript{3+(aq) + 3Ni(s)


Step 3: Calculate the standard cell potential (E°\textsubscript{cell).

E°\textsubscript{cell = E°\textsubscript{cathode - E°\textsubscript{anode

E°\textsubscript{cell = (-0.25 V) - (-1.66 V) = -0.25 + 1.66 = 1.41 V.


Step 4: Apply the Nernst equation to find the cell voltage (E\textsubscript{cell).

E\textsubscript{cell = E°\textsubscript{cell - \(\frac{0.0591}{n}\) log(Q)

The reaction quotient Q = \(\frac{[Al^{3+}]^2}{[Ni^{2+}]^3}\) = \(\frac{(0.002)^2}{(0.002)^3}\) = \(\frac{1}{0.002}\) = 500.


Step 5: Substitute the values and calculate.

E\textsubscript{cell = 1.41 - \(\frac{0.0591}{6}\) log(500)

log(500) = log(5 \(\times\) 100) = log(5) + log(100) = 0.6990 + 2 = 2.6990.

E\textsubscript{cell = 1.41 - (0.00985) \(\times\) 2.6990

E\textsubscript{cell = 1.41 - 0.02658 \(\approx\) 1.3834 V.

The cell voltage is approximately 1.38 V.
Quick Tip: When using the Nernst equation, follow these steps systematically: 1. Identify anode/cathode from E° values. 2. Write the balanced overall reaction to find 'n'. 3. Calculate E°\textsubscript{cell}. 4. Set up the reaction quotient Q correctly (products over reactants, raised to stoichiometric powers). 5. Substitute and solve.

(a)

An example of a liquid pair showing negative deviation from Raoult's law is a mixture of Chloroform (CHCl\textsubscript{3) and Acetone (CH\textsubscript{3COCH\textsubscript{3).

The reason for this deviation is that the intermolecular forces between chloroform and acetone molecules are stronger than the forces within pure chloroform or pure acetone.

Specifically, a hydrogen bond is formed between the hydrogen atom of chloroform and the oxygen atom of acetone.

This stronger attraction (A-B > A-A, B-B) reduces the escaping tendency of the molecules, leading to a lower vapour pressure than predicted by Raoult's law.


(b) (i)

Raoult's law for a solution containing volatile components states that for each component, its partial vapour pressure in the solution is equal to the product of its mole fraction in the solution and its vapour pressure in the pure state.

Mathematically, for a component 'A', P\textsubscript{A = P°\textsubscript{A \(\times\) x\textsubscript{A.


(b) (ii) OR

Raoult's Law is P = P° \(\times\) x. Henry's Law is P = K\textsubscript{H \(\times\) x.

Both laws express a linear relationship between the partial pressure of a volatile component and its mole fraction in the solution.

Raoult's law can be seen as a special case of Henry's law where the Henry's constant, K\textsubscript{H, becomes equal to the vapour pressure of the pure component, P°.

This holds true for the solvent in an ideal solution or in any very dilute solution.


(c)

Two main characteristics of an ideal solution are:

1. There is no change in enthalpy upon mixing the components, meaning no heat is absorbed or evolved. (\(\Delta H_{mix} = 0\)).

2. There is no change in volume upon mixing the components. The final volume is the sum of the volumes of the individual components. (\(\Delta V_{mix} = 0\)).
Quick Tip: Remember the deviations from Raoult's Law by the strength of intermolecular forces. Stronger A-B interactions lead to negative deviation (less vapor pressure). Weaker A-B interactions lead to positive deviation (more vapor pressure). Ideal solutions have equal A-A, B-B, and A-B interactions.

(a)

On complete hydrolysis, DNA yields three components:

1. A pentose sugar: 2-deoxyribose.

2. Phosphoric acid (H\textsubscript{3PO\textsubscript{4).

3. Nitrogenous bases: Adenine (A), Guanine (G), Cytosine (C), and Thymine (T).

Structural differences between DNA and RNA are:

1. Sugar: DNA contains 2-deoxyribose, whereas RNA contains ribose.

2. Base: DNA contains the base Thymine (T), whereas RNA contains Uracil (U) in its place.

3. Strand: DNA is typically a double-stranded helix, while RNA is generally a single-stranded molecule.


(b)

A nucleoside is a molecule formed by the attachment of a nitrogenous base to the 1' carbon of a pentose sugar (ribose or deoxyribose).

A nucleotide is formed when a phosphate group is attached to the 5' carbon of the sugar in a nucleoside.

Therefore, Nucleotide = Nucleoside + Phosphate group.


(c) (i)

Two important functions of nucleic acids are:

1. DNA (Deoxyribonucleic acid): It is the chemical basis of heredity and serves as the storage of genetic information, which is passed from one generation to the next.

2. RNA (Ribonucleic acid): It plays a crucial role in the process of protein synthesis within the cell. Different types of RNA (mRNA, tRNA, rRNA) have specific roles in this process.


(c) (ii) OR

The linkage that joins two nucleotides together to form a polynucleotide chain is called a phosphodiester linkage.

This bond connects the 3'-hydroxyl group of one sugar to the 5'-phosphate group of the next sugar.

The nitrogenous base that is found in RNA but not in DNA is Uracil (U).
Quick Tip: Remember the components: Base + Sugar = Nucleo\textbf{s}ide. Base + Sugar + Phosphate = Nucleo\textbf{t}ide. The 't' in nucleotide can remind you of the "tri" in "triphosphate". The key structural differences (DNA vs RNA) are Deoxyribose vs Ribose and Thymine vs Uracil.

The atomic number of Fe is 26. The electronic configuration of Fe\textsuperscript{3+ is [Ar] 3d\textsuperscript{5.


In an octahedral complex, the d-orbitals split into two sets: t\textsubscript{2g (lower energy) and e\textsubscript{g (higher energy).


(I) With a Strong field ligand:

A strong field ligand causes a large crystal field splitting (\(\Delta_o\)), which is greater than the pairing energy (P).


Electrons will pair up in the lower energy t\textsubscript{2g orbitals before occupying the higher energy e\textsubscript{g orbitals.


For d\textsuperscript{5, the configuration will be t\textsubscript{2g\textsuperscript{5 e\textsubscript{g\textsuperscript{0.


This configuration has 1 unpaired electron. This is a low spin complex.


(II) With a Weak field ligand:

A weak field ligand causes a small crystal field splitting (\(\Delta_o\)), which is less than the pairing energy (P).


Electrons will occupy all orbitals singly before pairing up, according to Hund's rule.


For d\textsuperscript{5, the configuration will be t\textsubscript{2g\textsuperscript{3 e\textsubscript{g\textsuperscript{2.


This configuration has 5 unpaired electrons. This is a high spin complex.
Quick Tip: Remember the rule for filling d-orbitals in octahedral complexes: Strong field ligand \(\rightarrow\) \(\Delta_o\) > P \(\rightarrow\) electrons pair up first (low spin). Weak field ligand \(\rightarrow\) \(\Delta_o\) < P \(\rightarrow\) electrons occupy all orbitals singly first (high spin).

The reaction is given to be of the first order.


The integrated rate law for a first-order reaction is:

k = \(\frac{2.303}{t}\) log \(\frac{[A]_0}{[A]_t}\)


Here, the given values are:

Initial concentration, [A]\textsubscript{0 = 1.2 \(\times\) 10\textsuperscript{-2 mol L\textsuperscript{-1.


Concentration at time t, [A]\textsubscript{t = 0.2 \(\times\) 10\textsuperscript{-2 mol L\textsuperscript{-1.


Time, t = 60 minutes.


First, calculate the ratio of concentrations:
\(\frac{[A]_0}{[A]_t}\) = \(\frac{1.2 \times 10^{-2}}{0.2 \times 10^{-2}}\) = 6.


Now, substitute the values into the rate law equation:

k = \(\frac{2.303}{60 min}\) log(6).


We are given that log(6) = 0.778.

k = \(\frac{2.303 \times 0.778}{60}\) min\textsuperscript{-1.


k = \(\frac{1.791734}{60}\) min\textsuperscript{-1.


k = 0.02986 min\textsuperscript{-1 \(\approx\) 0.0299 min\textsuperscript{-1.
Quick Tip: For first-order kinetics problems, the integrated rate law `k = (2.303/t) log([A]0/[A]t)` is essential. Ensure your units for the rate constant 'k' are consistent with the units of time used in the calculation (e.g., s\textsuperscript{-1}, min\textsuperscript{-1}, hr\textsuperscript{-1}).

(I)

The order of a reaction represents the dependence of the reaction rate on the concentration of reactants and can only be determined experimentally.


A balanced chemical equation shows the overall stoichiometry of the reaction, i.e., the molar ratios of reactants and products.


It does not provide information about the reaction mechanism, which is the sequence of elementary steps by which the reaction occurs.


The rate of a complex reaction is determined by the slowest step in its mechanism (the rate-determining step), and the order reflects the molecularity of this step, not the overall stoichiometry.


(II)

A bimolecular reaction, for example, A + B \(\rightarrow\) Products, would typically have a rate law of Rate = k[A][B], making it second order.


However, if one reactant (say, B) is taken in a very large excess compared to the other reactant (A), its concentration [B] will not change significantly as the reaction proceeds.


In this specific condition, the concentration [B] can be considered constant and can be merged with the rate constant k to give a new constant k' (where k' = k[B]).


The rate law then simplifies to Rate = k'[A], which is the rate law for a first-order reaction.


Such reactions are called pseudo-first-order reactions. A common example is the hydrolysis of an ester in an aqueous solution.
Quick Tip: Remember the key distinction: Molecularity is a theoretical concept applying to elementary steps, while Order is an experimental quantity applying to the overall reaction. For pseudo-order reactions, identify the reactant in large excess (often the solvent).

(I)

This is the reaction of a ketone (cyclohexanone) with a derivative of ammonia (semicarbazide).


It is a nucleophilic addition-elimination reaction. The -NH\textsubscript{2 group of semicarbazide attacks the carbonyl carbon, followed by the elimination of a water molecule.


The product is a semicarbazone.



(II)

This reaction involves an organometallic compound, dialkylcadmium (specifically, dimethylcadmium), reacting with an acid chloride (acetyl chloride).


This is a standard method for the preparation of ketones. The reaction produces a ketone and cadmium chloride.


(CH\textsubscript{3)\textsubscript{2Cd + 2CH\textsubscript{3COCl \(\rightarrow\) 2CH\textsubscript{3-CO-CH\textsubscript{3 + CdCl\textsubscript{2.


The main organic product is acetone (propanone).


(III)

This is the Rosenmund reduction.


This reaction involves the catalytic hydrogenation of an acid chloride (benzoyl chloride) to an aldehyde (benzaldehyde).


The catalyst used is palladium on a barium sulfate support (Pd/BaSO\textsubscript{4), which is partially poisoned (e.g., with sulfur or quinoline) to prevent the further reduction of the aldehyde to an alcohol.


C\textsubscript{6H\textsubscript{5COCl + H\textsubscript{2 \(\xrightarrow{Pd/BaSO_4}\) C\textsubscript{6H\textsubscript{5CHO + HCl.


The product is benzaldehyde.
Quick Tip: Recognize named reactions: Carbonyl + Ammonia derivative \(\rightarrow\) Imine/Oxime/Semicarbazone etc. Organocadmium + Acid chloride \(\rightarrow\) Ketone. Acid chloride + H\textsubscript{2}/Pd-BaSO\textsubscript{4} \(\rightarrow\) Rosenmund Reduction (aldehyde).

(I) Ethyl benzoate (an ester) and Benzoic acid (a carboxylic acid)

Test: Sodium Bicarbonate Test.


Procedure: Add a spatula of sodium bicarbonate (or a few mL of its aqueous solution) to both compounds.


Observation:

- With Benzoic acid, brisk effervescence will be observed due to the liberation of carbon dioxide gas.

C\textsubscript{6H\textsubscript{5COOH + NaHCO\textsubscript{3 \(\rightarrow\) C\textsubscript{6H\textsubscript{5COONa + H\textsubscript{2O + CO\textsubscript{2(g) \(\uparrow\).


- With Ethyl benzoate, there will be no reaction and no effervescence.


(II) Propanal (an aldehyde) and Propanone (a ketone)

Test: Tollens' Test (Silver Mirror Test).


Procedure: Add Tollens' reagent (ammoniacal silver nitrate solution) to both compounds and warm gently in a water bath.


Observation:

- With Propanal, a bright silver mirror is formed on the inner walls of the test tube, as the aldehyde is oxidized to a carboxylate ion and Ag\textsuperscript{+ is reduced to Ag(s).

CH\textsubscript{3CH\textsubscript{2CHO + 2[Ag(NH\textsubscript{3)\textsubscript{2]\textsuperscript{+ + 3OH\textsuperscript{- \(\rightarrow\) CH\textsubscript{3CH\textsubscript{2COO\textsuperscript{- + 2Ag(s) + 4NH\textsubscript{3 + 2H\textsubscript{2O.


- With Propanone, there will be no reaction and no formation of a silver mirror, as ketones are not oxidized by mild oxidizing agents like Tollens' reagent.
Quick Tip: Master the key distinguishing tests: Carboxylic acids vs other groups \(\rightarrow\) NaHCO\textsubscript{3} test. Aldehydes vs Ketones \(\rightarrow\) Tollens' test or Fehling's test. Methyl ketones vs other ketones \(\rightarrow\) Iodoform test.