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


This mechanism occurs in three steps.


Step 1: Protonation of the alcohol. The oxygen atom of the hydroxyl group in the alcohol acts as a Lewis base and donates an electron pair to a proton (H\(^{+}\)) from the sulfuric acid.


This forms a protonated alcohol (an oxonium ion), R-OH\(_{2}^{+}\).


This initial step is fast and reversible.


Its purpose is to convert the poor leaving group (-OH) into a good leaving group (H\(_{2}\)O).


Step 2 is the formation of a carbocation by the loss of a water molecule, and Step 3 is the elimination of a proton to form the alkene.


Therefore, the very first step of the reaction is the protonation of the alcohol molecule.
Quick Tip: In acid-catalyzed reactions involving alcohols, the first step is almost always the protonation of the hydroxyl group. This converts the poor leaving group, OH\(^{-}\), into an excellent leaving group, H\(_{2}\)O.

DDT (Dichlorodiphenyltrichloroethane) is a well-known organochlorine compound.


It was extensively used as an insecticide for controlling pests in agriculture and disease-carrying insects like mosquitoes.


A key chemical property of DDT is its high stability and resistance to decomposition by natural processes in the environment.


Substances that are not easily broken down by microorganisms or other environmental factors are termed non-biodegradable.


This persistence leads to bioaccumulation in the fatty tissues of organisms and biomagnification up the food chain, causing significant ecological harm.


Thus, DDT is correctly classified as a non-biodegradable insecticide.
Quick Tip: Remember the key environmental issue with chlorinated organic compounds like DDT is their stability and persistence. "Non-biodegradable" means it stays in the environment for a very long time.

The reaction between permanganate ion (MnO\(_{4}^{-}\)) and iodide ion (I\(^{-}\)) is a redox reaction whose products depend on the pH of the medium.


In a neutral or weakly alkaline medium, the permanganate ion is a strong oxidizing agent.


It oxidizes iodide ions (I\(^{-}\)) to iodate ions (IO\(_{3}^{-}\)).


In this process, the oxidation state of iodine increases from -1 to +5.


The permanganate ion (MnO\(_{4}^{-}\), with Mn in +7 oxidation state) is itself reduced to manganese dioxide (MnO\(_{2}\), with Mn in +4 oxidation state).


The balanced chemical equation is: 2MnO\(_{4}^{-}\) + I\(^{-}\) + H\(_{2}\)O \(\rightarrow\) 2MnO\(_{2}\) + IO\(_{3}^{-}\) + 2OH\(^{-}\).


Therefore, the oxidation product of I\(^{-}\) in an alkaline medium is IO\(_{3}^{-}\).
Quick Tip: The oxidizing power of KMnO\(_{4}\) is medium-dependent. Remember the reduction products: In acidic medium, MnO\(_{4}^{-}\) \(\rightarrow\) Mn\(^{2+}\). In neutral/weakly alkaline medium, MnO\(_{4}^{-}\) \(\rightarrow\) MnO\(_{2}\). In strongly alkaline medium, MnO\(_{4}^{-}\) \(\rightarrow\) MnO\(_{4}^{2-}\).

The reaction sequence starts with aniline.


Step 1: Aniline (C\(_{6}\)H\(_{5}\)NH\(_{2}\)) is treated with NaNO\(_{2}\) and HCl at 0-5°C.


This reaction is called diazotization, and it converts the primary aromatic amine into a diazonium salt.


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


Step 2: The benzenediazonium chloride (X) is treated with ethanol (C\(_{2}\)H\(_{5}\)OH).


In this context, ethanol acts as a reducing agent.


It reduces the diazonium salt by replacing the entire diazonium group (-N\(_{2}^{+}\)Cl\(^{-}\)) with a hydrogen atom.


The final product Y is benzene (C\(_{6}\)H\(_{6}\)).


The complete reaction is: C\(_{6}\)H\(_{5}\)N\(_{2}^{+}\)Cl\(^{-}\) + CH\(_{3}\)CH\(_{2}\)OH \(\rightarrow\) C\(_{6}\)H\(_{6}\) + N\(_{2}\) + HCl + CH\(_{3}\)CHO.
Quick Tip: Diazonium salts are versatile intermediates. Remember that they can be reduced to benzene using mild reducing agents like ethanol (C\(_{2}\)H\(_{5}\)OH) or hypophosphorous acid (H\(_{3}\)PO\(_{2}\)).

A molecule is chiral if it is non-superimposable on its mirror image. A common feature of chiral molecules is the presence of a chiral center.


A chiral center (or asymmetric carbon) is a carbon atom bonded to four different groups.


Let's analyze the structure of each compound:


(A) 2-Bromobutane: CH\(_{3}\)-CH(Br)-CH\(_{2}\)-CH\(_{3}\). The second carbon atom is bonded to four different groups: -H, -Br, a methyl group (-CH\(_{3}\)), and an ethyl group (-CH\(_{2}\)CH\(_{3}\)). Thus, it has a chiral center and is a chiral molecule.


(B) 1-Bromobutane: CH\(_{2}\)(Br)-CH\(_{2}\)-CH\(_{2}\)-CH\(_{3}\). The first carbon has two identical hydrogen atoms. No carbon is chiral. It is achiral.


(C) 2-Bromopropane: CH\(_{3}\)-CH(Br)-CH\(_{3}\). The second carbon has two identical methyl groups. It is achiral.


(D) 1-Bromopropane: CH\(_{2}\)(Br)-CH\(_{2}\)-CH\(_{3}\). The first carbon has two identical hydrogen atoms. It is achiral.


Therefore, only 2-Bromobutane is chiral.
Quick Tip: To quickly identify a chiral molecule, look for an sp³-hybridized carbon atom bonded to four distinctly different atoms or groups. If you find one, the molecule is chiral (unless it is a special case like a meso compound).

The anomeric carbon is a special type of stereocenter created during the cyclization of a monosaccharide.


In the open-chain form of the sugar, this carbon was part of the carbonyl group (aldehyde or ketone).


A key feature to identify the anomeric carbon in a cyclic hemiacetal or hemiketal is that it is the only carbon atom in the ring that is bonded to two oxygen atoms.


One oxygen is part of the ring (the ether linkage), and the other is part of a hydroxyl group (-OH).


Looking at the provided Haworth structure, the carbon atom numbered '1' is bonded to the oxygen within the ring and also to an -OH group.


No other carbon in the ring has this feature.


Therefore, the anomeric carbon is numbered as 1.
Quick Tip: In aldoses (like glucose and ribose), the anomeric carbon is always C-1. In ketoses (like fructose), the anomeric carbon is always C-2. This is a direct result of the position of the carbonyl group in the open-chain form.

Henry's Law describes the solubility of a gas in a liquid.


The law is mathematically stated as: P = K\(_{H}\) \(\times\) x.


Here, P is the partial pressure of the gas above the liquid.


x is the mole fraction of the gas dissolved in the liquid, which is a measure of its solubility.


K\(_{H}\) is Henry's Law constant, which is specific to a given gas-solvent pair at a particular temperature.


Rearranging the equation to solve for solubility (x), we get: x = P / K\(_{H}\).


This equation shows that at a constant partial pressure (P), the solubility (x) is inversely proportional to Henry's constant (K\(_{H}\)).


Therefore, a gas with a lower solubility will have a greater value of K\(_{H}\), and a gas with a higher solubility will have a smaller value of K\(_{H}\).
Quick Tip: Think of K\(_{H}\) as a measure of a gas's "reluctance" to dissolve. A high K\(_{H}\) value means the gas prefers to stay in the gas phase rather than dissolve, hence it has low solubility.

Let's evaluate each statement:


(A) The electronic configuration of Lanthanum (La, Z=57) is [Xe] 5d\(^{1}\)6s\(^{2}\). Since its last electron enters the d-subshell, it is classified as a d-block element, which are also known as transition elements. This statement is correct.


(B) Zirconium (Zr) is in period 5 and Hafnium (Hf) is in period 6. The intervening filling of the 14 electrons in the 4f-orbitals (the lanthanoids) causes a significant increase in effective nuclear charge, which contracts the atomic size. This "lanthanoid contraction" almost perfectly cancels the expected size increase down a group. Thus, Zr and Hf have nearly identical radii. This statement is correct.


(C) Moving across the lanthanoid series from La to Lu, electrons are added to the same inner 4f subshell. However, the nuclear charge increases by one proton at each step. The poor shielding effect of f-electrons results in a stronger pull from the nucleus, causing a steady decrease in ionic (and atomic) radii. This is the definition of lanthanoid contraction. This statement is correct.


(D) This statement claims that lanthanoids in general are radioactive. This is false. Out of all the lanthanoids, only one element, Promethium (Pm), is radioactive. All other lanthanoids have stable, naturally occurring isotopes. Therefore, this statement is incorrect.
Quick Tip: A common point of confusion is between lanthanoids and actinoids. All actinoids are radioactive. However, for the lanthanoids, only Promethium (Pm) is radioactive. Remember this exception.

The voltage of an electrochemical cell under non-standard conditions is described by the Nernst equation.


E\(_{cell}\) = E\(_{cell}^{\circ}\) - \(\frac{RT}{nF}\)lnQ


For the given reaction, the reaction quotient Q is given by Q = \(\frac{[Mg^{2+}]}{[Ag^{+}]^{2}}\).


The Nernst equation becomes: E\(_{cell}\) = E\(_{cell}^{\circ}\) - \(\frac{RT}{nF}\)ln(\(\frac{[Mg^{2+}]}{[Ag^{+}]^{2}}\)).


As the reaction progresses, the cell operates or "discharges".


Reactants are consumed, so the concentration of Ag\(^{+}\) decreases.


Products are formed, so the concentration of Mg\(^{2+}\) increases.


As [Mg\(^{2+}\)] increases and [Ag\(^{+}\)] decreases, the value of the reaction quotient Q increases.


Since Q is increasing, the term that is being subtracted from E\(_{cell}^{\circ}\) also increases.


Consequently, the overall voltage of the cell, E\(_{cell}\), decreases.


The cell voltage continues to decrease until the reaction reaches equilibrium, at which point E\(_{cell}\) = 0.
Quick Tip: For any spontaneous galvanic cell, as the cell operates (discharges), the concentration of products increases and reactants decreases. This always causes the reaction quotient Q to increase, leading to a decrease in the cell potential until it reaches zero at equilibrium.

The spin-only magnetic moment (\(\mu\)) is calculated using the formula: \(\mu = \sqrt{n(n+2)}\) Bohr Magnetons (B.M.), where 'n' is the number of unpaired electrons.


The magnetic moment increases with the number of unpaired electrons. Therefore, we need to find the ion with the maximum number of unpaired electrons.


Let's determine the electronic configuration and the value of 'n' for each ion:


1. Ti (Z=22) \(\rightarrow\) [Ar] 3d\(^{2}\) 4s\(^{2}\). For Ti\(^{3+}\), the configuration is [Ar] 3d\(^{1}\). Number of unpaired electrons, n = 1.


2. Cr (Z=24) \(\rightarrow\) [Ar] 3d\(^{5}\) 4s\(^{1}\) (anomalous configuration). For Cr\(^{3+}\), the configuration is [Ar] 3d\(^{3}\). Number of unpaired electrons, n = 3.


3. Mn (Z=25) \(\rightarrow\) [Ar] 3d\(^{5}\) 4s\(^{2}\). For Mn\(^{2+}\), the configuration is [Ar] 3d\(^{5}\). This represents a half-filled d-subshell with one electron in each of the five d-orbitals. Number of unpaired electrons, n = 5.


4. Ni (Z=28) \(\rightarrow\) [Ar] 3d\(^{8}\) 4s\(^{2}\). For Ni\(^{2+}\), the configuration is [Ar] 3d\(^{8}\). The d-orbitals will have three filled pairs and two unpaired electrons. Number of unpaired electrons, n = 2.


Comparing the number of unpaired electrons: Mn\(^{2+}\) has the highest value (n=5).


Therefore, Mn\(^{2+}\) will have the highest magnetic moment.
Quick Tip: The maximum number of unpaired electrons possible in a d-subshell is 5 (in a d\(^{5}\) configuration). Ions with a d\(^{5}\) configuration, such as Mn\(^{2+}\) and Fe\(^{3+}\), will have the highest spin-only magnetic moments among the first-row transition metal ions.

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


The general form of the reaction is: R-CONH\(_{2}\) + Br\(_{2}\) + 4NaOH \(\rightarrow\) R-NH\(_{2}\) + Na\(_{2}\)CO\(_{3}\) + 2NaBr + 2H\(_{2}\)O.


The reaction is "given by" the reactant, which must be a primary amide.


Let's examine the options:


(A) CH\(_{3}\)NO\(_{2}\) is nitromethane.


(B) CH\(_{3}\)NH\(_{2}\) is methanamine, which is the product of the reaction if the reactant were ethanamide.


(C) CH\(_{3}\)CONH\(_{2}\) is ethanamide, which is a primary amide and a suitable starting material for this reaction.


(D) CH\(_{3}\)CH\(_{2}\)NH\(_{2}\) is ethanamine.


Therefore, the Hoffmann Bromamide degradation reaction is given by the primary amide, CH\(_{3}\)CONH\(_{2}\).
Quick Tip: The name "Hoffmann Bromamide" itself gives a clue: the reaction involves Bromine (Bromo) and an Amide. It's a "degradation" because the carbon chain is shortened by one carbon atom.

First, we need to write the balanced half-reaction for the oxidation of water (H\(_{2}\)O) to oxygen (O\(_{2}\)).


The oxidation half-reaction is: 2H\(_{2}\)O(l) \(\rightarrow\) O\(_{2}\)(g) + 4H\(^{+}\)(aq) + 4e\(^{-}\).


This balanced equation shows that for the oxidation of 2 moles of H\(_{2}\)O, 4 moles of electrons are transferred.


The question asks for the charge required for the oxidation of 1 mole of H\(_{2}\)O.


From the stoichiometry, if 2 moles of H\(_{2}\)O release 4 moles of electrons, then 1 mole of H\(_{2}\)O will release half that amount, which is 2 moles of electrons.


The charge carried by 1 mole of electrons is defined as one Faraday (1 F).


Therefore, the total charge required for the transfer of 2 moles of electrons is 2 Faradays (2 F).
Quick Tip: When dealing with stoichiometry in electrochemistry, always start by writing the balanced half-reaction. The stoichiometric coefficient of the electrons (e\(^{-}\)) directly tells you how many moles of electrons are involved per mole of reaction as written.

Let's analyze the Assertion and Reason.


The cuprous ion is Cu\(^{+}\).


The atomic number of Copper (Cu) is 29, and its electronic configuration is [Ar] 3d\(^{10}\) 4s\(^{1}\).


To form the cuprous ion (Cu\(^{+}\)), one electron is removed from the outermost shell (4s).


The electronic configuration of Cu\(^{+}\) is [Ar] 3d\(^{10}\).


In this configuration, all five 3d-orbitals are completely filled with electrons, meaning all electrons are paired.


A substance with no unpaired electrons is diamagnetic. Therefore, cuprous salts are diamagnetic. Assertion (A) is true.


Now let's examine the Reason (R). It states that in the cuprous ion, 3d-orbitals are partially filled.


This is incorrect. As shown above, the 3d-subshell in Cu\(^{+}\) has a 3d\(^{10}\) configuration, which means it is completely filled, not partially filled.


Therefore, Reason (R) is false.
Quick Tip: Paramagnetism is caused by unpaired electrons, while diamagnetism is caused by paired electrons. To determine the magnetic property of an ion, always write its electronic configuration and check for unpaired electrons. A d\(^{10}\) configuration (like Cu\(^{+}\) and Zn\(^{2+}\)) is always diamagnetic.

Let's analyze the Assertion and Reason.


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


In aniline (C\(_{6}\)H\(_{5}\)NH\(_{2}\)), the lone pair on the nitrogen is delocalized into the benzene ring through resonance, which already reduces its basicity compared to aliphatic amines.


In acetanilide (C\(_{6}\)H\(_{5}\)NHCOCH\(_{3}\)), the lone pair on the nitrogen is involved in resonance with the adjacent electron-withdrawing carbonyl group (-C=O).


This delocalization is more significant than the resonance with the benzene ring, making the lone pair on the nitrogen atom much less available for protonation.


Therefore, acetanilide is less basic than aniline. Assertion (A) is false.


Now let's examine the Reason (R). It states that acetylation of aniline results in a decrease of electron density on nitrogen.


This is true. The electron-withdrawing nature of the acetyl group (-COCH\(_{3}\)) pulls the lone pair of electrons from the nitrogen atom towards the carbonyl oxygen via resonance.


This process decreases the electron density on the nitrogen atom. Therefore, Reason (R) is true.
Quick Tip: Electron-withdrawing groups (like -NO\(_{2}\), -CN, -C=O) attached to the nitrogen atom or the benzene ring decrease the basicity of anilines. Conversely, electron-donating groups (like -CH\(_{3}\), -OCH\(_{3}\)) increase the basicity.

Let's analyze the Assertion (A).


During the electrolysis of aqueous NaCl, at the cathode (reduction), H\(_{2}\)O is reduced to H\(_{2}\) gas in preference to Na\(^{+}\) because water has a higher reduction potential. (2H\(_{2}\)O + 2e\(^{-}\) \(\rightarrow\) H\(_{2}\) + 2OH\(^{-}\)).


At the anode (oxidation), there is competition between the oxidation of Cl\(^{-}\) ions and H\(_{2}\)O. Due to the phenomenon of overpotential (or overvoltage) of oxygen, the oxidation of Cl\(^{-}\) to Cl\(_{2}\) gas is kinetically favored and occurs preferentially. (2Cl\(^{-}\) \(\rightarrow\) Cl\(_{2}\) + 2e\(^{-}\)).


Thus, the products are indeed H\(_{2}\) at the cathode and Cl\(_{2}\) at the anode. Assertion (A) is true.


Now let's analyze the Reason (R).


It states that chlorine has a higher oxidation potential than H\(_{2}\)O.


The standard oxidation potential for water is: 2H\(_{2}\)O \(\rightarrow\) O\(_{2}\) + 4H\(^{+}\) + 4e\(^{-}\), E\(^{\circ}_{ox}\) = -1.23 V.


The standard oxidation potential for chloride is: 2Cl\(^{-}\) \(\rightarrow\) Cl\(_{2}\) + 2e\(^{-}\), E\(^{\circ}_{ox}\) = -1.36 V.


Since -1.23 V is greater (less negative) than -1.36 V, water actually has a higher standard oxidation potential than the chloride ion.


This means, based on thermodynamics alone, water should be easier to oxidize. The reason it doesn't is due to kinetics (overpotential).


Therefore, the statement in Reason (R) is false.
Quick Tip: In the electrolysis of aqueous solutions, remember the concept of overpotential. At the anode, the oxidation of chloride, bromide, and iodide ions is preferred over the oxidation of water, even if thermodynamically less favorable, due to the high overpotential required for oxygen evolution.

Let's analyze the Assertion (A).


The boiling point of alkyl halides depends on the strength of intermolecular forces, which are primarily van der Waals forces.


The strength of van der Waals forces increases with the size and mass of the molecule.


Comparing n-butyl chloride and n-butyl bromide, the alkyl group is the same. The halogen atoms are chlorine and bromine.


Bromine has a higher atomic mass and larger size than chlorine.


Therefore, n-butyl bromide has a larger molecular mass and surface area, leading to stronger intermolecular van der Waals forces than n-butyl chloride.


This means n-butyl bromide has a higher boiling point than n-butyl chloride.


The Assertion (A) states the opposite, so Assertion (A) is false.


Now let's analyze the Reason (R).


Bond polarity depends on the difference in electronegativity between the bonded atoms.


The electronegativity order is F > Cl > Br > I.


Since chlorine is more electronegative than bromine, the electronegativity difference between carbon and chlorine is greater than that between carbon and bromine.


Therefore, the C-Cl bond is indeed more polar than the C-Br bond. Reason (R) is true.


Since Assertion (A) is false and Reason (R) is true, the correct option is (D).
Quick Tip: For alkyl halides with the same alkyl group, the boiling point order is always R-I > R-Br > R-Cl > R-F. This trend is dominated by the increasing strength of van der Waals forces due to increasing molecular mass and size, and not by dipole-dipole interactions.

Essential Amino Acids:

These are the amino acids that are required by the human body for protein synthesis and proper growth but cannot be synthesized by the body itself.


Therefore, they must be supplied through the diet.


Examples include valine, leucine, isoleucine, lysine, etc.


Amphoteric Nature of Amino Acids:

Amino acids are considered amphoteric because they contain both an acidic functional group (the carboxyl group, -COOH) and a basic functional group (the amino group, -NH\(_{2}\)) within the same molecule.


The carboxyl group can donate a proton (acting as an acid), while the amino group can accept a proton (acting as a base).


Because of this dual nature, they can react with both acids and bases.


In aqueous solution, they primarily exist as zwitterions (H\(_{3}\)N\(^{+}\)-CHR-COO\(^{-}\)), which have both a positive and a negative charge.
Quick Tip: Remember the mnemonic "PVT TIM HALL" for the essential amino acids: Phenylalanine, Valine, Threonine, Tryptophan, Isoleucine, Methionine, Histidine, Arginine (conditionally essential), Leucine, Lysine.

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


1. First half-life: The time taken for the concentration to fall from 0.40 M to 0.20 M is (1 - 0) = 1 hour.


2. Second half-life: The time taken for the concentration to fall from 0.20 M to 0.10 M is (2 - 1) = 1 hour.


3. Third half-life: The time taken for the concentration to fall from 0.10 M to 0.05 M is (3 - 2) = 1 hour.


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


The rate law for a first-order reaction is given by the expression:


Rate = k[A] or Rate = k[A]\(^{1}\)


where 'k' is the rate constant and [A] is the concentration of reactant A.
Quick Tip: A constant half-life is the hallmark of a first-order reaction. For a zero-order reaction, the half-life decreases as concentration decreases. For a second-order reaction, the half-life increases as concentration decreases.

The given rate law is: Rate = k[H\(_{2}\)][I\(_{2}\)].


Change in Rate of Reaction:

The rate of the reaction is directly proportional to the concentration of H\(_{2}\) (i.e., Rate \(\propto\) [H\(_{2}\)]).


When 1 mole of H\(_{2}\)(g) is added to the sealed container, the number of moles of H\(_{2}\) increases.


This leads to an increase in the concentration (or partial pressure) of H\(_{2}\).


Since the rate depends directly on [H\(_{2}\)], the rate of the reaction will increase.


Change in Rate Constant:

The rate constant (k) for a reaction depends only on the temperature and the presence of a catalyst.


The problem states that the temperature was kept constant.


Changing the concentration of reactants or products does not affect the value of the rate constant.


Therefore, the rate constant (k) will remain unchanged.
Quick Tip: Remember the key difference: reaction rates depend on concentration and temperature, but the rate constant (k) depends only on temperature (and catalyst). This is a frequent point of confusion in exam questions.

The fact that the complex does not give a precipitate of AgCl upon addition of AgNO\(_{3}\) solution indicates that there are no free chloride ions (Cl\(^{-}\)) in the solution.


This means that all the chloride ions are inside the coordination sphere, bonded directly to the central metal atom, Platinum (Pt).


The components are one Pt, four Cl from PtCl\(_{4}\), and two Cl from 2KCl, making a total of six chloride ligands. The two K\(^{+}\) ions act as counter-ions outside the coordination sphere.


Structural Formula: The formula of the coordination compound is K\(_{2}\)[PtCl\(_{6}\)].


This shows two potassium ions as counter-ions and a complex anion [PtCl\(_{6}\)]\(^{2-}\).


IUPAC Name:

1. Name the cation first: Potassium

2. Name the ligands in the anion: There are six chloride ligands, so it is hexachlorido.

3. Name the central metal atom. Since the complex is an anion, the metal name ends in '-ate': platinate.

4. Determine the oxidation state of the metal. Let the oxidation state of Pt be x. 2(+1) + x + 6(-1) = 0 \(\Rightarrow\) x = +4. The oxidation state is written in Roman numerals: (IV).


Combining these parts, the full IUPAC name is Potassium hexachloridoplatinate(IV).
Quick Tip: The AgNO\(_{3}\) test is a classic method to determine which ions are inside and outside the coordination sphere. If a precipitate forms, the ion is outside. If no precipitate forms, the ion is inside, acting as a ligand.

Definition of Fuel Cell:

A fuel cell is an electrochemical cell (a galvanic cell) that converts the chemical energy of a fuel (such as hydrogen) and an oxidizing agent (such as oxygen) directly into electrical energy through a continuous electrochemical process.


Unlike conventional batteries, fuel cells require a continuous supply of fuel and oxidant to operate.


Two Advantages of Fuel Cell over Ordinary Cell (Battery):

1. High Efficiency: Fuel cells have a very high efficiency of converting chemical energy into electrical energy, typically around 60-70%. This is significantly higher than the efficiency of thermal power plants or internal combustion engines (around 30-40%).


2. Pollution-Free Operation: The most common type, the hydrogen-oxygen fuel cell, produces only water as its byproduct. This makes them non-polluting and environmentally friendly, as they do not emit greenhouse gases (like CO\(_{2}\)) or other harmful pollutants (like SO\(_{x}\) and NO\(_{x}\)).


(An additional advantage is that they can run continuously as long as fuel is supplied, whereas ordinary batteries need to be recharged or replaced after they are discharged.)
Quick Tip: The key difference between a fuel cell and a battery is that a fuel cell has its reactants supplied externally and continuously, whereas a battery has a fixed amount of reactants stored internally.

(a) 6 R\(_{3}\)C-OH + 2Al \(\rightarrow\)


This reaction shows a tertiary alcohol reacting with aluminum, which is an active metal. Alcohols are weakly acidic and react with such metals to liberate hydrogen gas and form a metal alkoxide.


Since aluminum has a valency of +3, one aluminum atom reacts with three alcohol molecules. The balanced equation requires 2 moles of Al and 6 moles of alcohol.


The product is aluminum tri-tert-alkoxide and hydrogen gas.


The structure of the main product is Aluminum tri-tert-alkoxide, [(R\(_{3}\)C-O)\(_{3}\)Al].



(b) Reaction with NaBH\(_{4}\)


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


NaBH\(_{4}\) (sodium borohydride) is a mild and selective reducing agent. It reduces aldehydes and ketones to their corresponding alcohols but does not typically reduce esters, carboxylic acids, or amides under normal conditions.


Therefore, the ketone group (C=O) in the cyclohexane ring will be reduced to a secondary alcohol group (-OH), while the ester group (-COOCH\(_{3}\)) on the side chain will remain unaffected.


The structure of the main product is:
Quick Tip: Remember the selectivity of common reducing agents: NaBH\(_{4}\) reduces only aldehydes and ketones. LiAlH\(_{4}\) is much stronger and reduces aldehydes, ketones, esters, carboxylic acids, and amides.

(a) Preparation of t-butyl methyl ether:


No, sodium methoxide (CH\(_{3}\)ONa) and t-butyl bromide ((CH\(_{3}\))\(_{3}\)CBr) cannot be used for the preparation of t-butyl methyl ether via the Williamson synthesis.


Reason: The reaction involves a strong nucleophile/base (methoxide) and a tertiary alkyl halide (t-butyl bromide).


In this case, the methoxide ion acts as a strong base and preferentially abstracts a \(\beta\)-hydrogen from the t-butyl bromide.


This leads to an elimination reaction (E2 mechanism) being the major pathway, producing 2-methylpropene as the main product, instead of the desired substitution product (ether).


(CH\(_{3}\))\(_{3}\)CBr + CH\(_{3}\)ONa \(\xrightarrow{Elimination}\) CH\(_{2}\)=C(CH\(_{3}\))\(_{2}\) + CH\(_{3}\)OH + NaBr


Appropriate Starting Materials:

To prepare t-butyl methyl ether successfully, the roles of the reactants should be reversed.


We should use a primary alkyl halide and a tertiary alkoxide.


The appropriate starting materials are sodium tert-butoxide ((CH\(_{3}\))\(_{3}\)CONa) and methyl bromide (CH\(_{3}\)Br) or methyl chloride.


Here, the primary halide (CH\(_{3}\)Br) readily undergoes S\(_{N}\)2 reaction, and elimination is not possible.


(CH\(_{3}\))\(_{3}\)CONa + CH\(_{3}\)Br \(\xrightarrow{S_{N}2}\) (CH\(_{3}\))\(_{3}\)C-O-CH\(_{3}\) + NaBr


(b) IUPAC Name:

The structure of t-butyl methyl ether is (CH\(_{3}\))\(_{3}\)C-O-CH\(_{3}\).


For IUPAC naming of ethers, the smaller alkyl group plus oxygen is named as an alkoxy substituent on the longer alkyl chain.


The parent alkane is the three-carbon chain with a methyl group at position 2, which is 2-methylpropane.


The smaller group is methyl, so the substituent is methoxy (-OCH\(_{3}\)).


The methoxy group is attached to carbon-2 of the propane chain.


Therefore, the IUPAC name is 2-methoxy-2-methylpropane.
Quick Tip: For a successful Williamson ether synthesis, always choose the combination with the primary (or methyl) alkyl halide. Using secondary or tertiary halides often leads to elimination as the major competing reaction.

(a) Increasing order of pK\(_{b}\) values:

Increasing pK\(_{b}\) corresponds to decreasing basic strength. We need to arrange the amines from most basic to least basic.

- Aliphatic amines ((C\(_{2}\)H\(_{5}\))\(_{2}\)NH, C\(_{2}\)H\(_{5}\)NH\(_{2}\)) are stronger bases than aromatic amines (C\(_{6}\)H\(_{5}\)NHCH\(_{3}\), C\(_{6}\)H\(_{5}\)NH\(_{2}\)) because the lone pair on nitrogen in aromatic amines is delocalized into the benzene ring.

- Between (C\(_{2}\)H\(_{5}\))\(_{2}\)NH (secondary) and C\(_{2}\)H\(_{5}\)NH\(_{2}\) (primary), the combined effect of +I effect and solvation makes the secondary amine more basic.

- Between C\(_{6}\)H\(_{5}\)NHCH\(_{3}\) and C\(_{6}\)H\(_{5}\)NH\(_{2}\), the +I effect of the methyl group in N-methylaniline makes it slightly more basic than aniline.

- Order of basicity: (C\(_{2}\)H\(_{5}\))\(_{2}\)NH > C\(_{2}\)H\(_{5}\)NH\(_{2}\) > C\(_{6}\)H\(_{5}\)NHCH\(_{3}\) > C\(_{6}\)H\(_{5}\)NH\(_{2}\).

- Order of increasing pK\(_{b}\) (decreasing basicity):

C\(_{6}\)H\(_{5}\)NH\(_{2}\) < C\(_{6}\)H\(_{5}\)NHCH\(_{3}\) < C\(_{2}\)H\(_{5}\)NH\(_{2}\) < (C\(_{2}\)H\(_{5}\))\(_{2}\)NH


(b) Decreasing order of boiling point:

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

- C\(_{2}\)H\(_{5}\)OH (ethanol) can form strong intermolecular H-bonds due to the highly electronegative oxygen atom.

- C\(_{2}\)H\(_{5}\)NH\(_{2}\) (ethylamine, 1°) and (CH\(_{3}\))\(_{2}\)NH (dimethylamine, 2°) can also form H-bonds, but they are weaker than those in alcohols as N is less electronegative than O.

- Between the two amines of the same molar mass, the primary amine (ethylamine) can form more extensive H-bonds than the secondary amine (dimethylamine).

- The order of decreasing boiling point is:

C\(_{2}\)H\(_{5}\)OH > C\(_{2}\)H\(_{5}\)NH\(_{2}\) > (CH\(_{3}\))\(_{2}\)NH


(c) Decreasing order of solubility in water:

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

- C\(_{2}\)H\(_{5}\)NH\(_{2}\) has a small hydrophobic part (ethyl group) and can form H-bonds, making it highly soluble.

- (C\(_{2}\)H\(_{5}\))\(_{2}\)NH has a larger hydrophobic part (two ethyl groups), which decreases its solubility compared to ethylamine.

- C\(_{6}\)H\(_{5}\)NH\(_{2}\) (aniline) has a large, bulky, hydrophobic phenyl group, which makes it sparingly soluble in water despite its ability to form H-bonds.

- The order of decreasing solubility is:

C\(_{2}\)H\(_{5}\)NH\(_{2}\) > (C\(_{2}\)H\(_{5}\))\(_{2}\)NH > C\(_{6}\)H\(_{5}\)NH\(_{2}\)
Quick Tip: For amines, basicity is increased by electron-donating groups (+I) and decreased by electron-withdrawing groups and resonance. Boiling points and solubility are highest for primary amines > secondary > tertiary (of comparable mass) due to better H-bonding ability.

Step 1: Identify the anode and cathode.

The half-cell with the higher reduction potential acts as the cathode. E°\(_{Ni^{2+}/Ni}\) (-0.25 V) > E°\(_{Al^{3+}/Al}\) (-1.66 V).

Therefore, Nickel is the cathode (reduction) and Aluminum is the anode (oxidation).


Anode: Al(s) \(\rightarrow\) Al\(^{3+}\)(aq) + 3e\(^{-}\)

Cathode: Ni\(^{2+}\)(aq) + 2e\(^{-}\) \(\rightarrow\) Ni(s)


Step 2: Write the balanced overall cell reaction.

To balance the electrons, multiply the anode reaction by 2 and the cathode reaction by 3.

Overall reaction: 2Al(s) + 3Ni\(^{2+}\)(aq) \(\rightarrow\) 2Al\(^{3+}\)(aq) + 3Ni(s)

The number of electrons transferred, n = 6.


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

E°\(_{cell}\) = E°\(_{cathode}\) - E°\(_{anode}\) = E°\(_{Ni^{2+}/Ni}\) - E°\(_{Al^{3+}/Al}\)

E°\(_{cell}\) = (-0.25 V) - (-1.66 V) = 1.41 V


Step 4: Use the Nernst equation to find the cell voltage (E\(_{cell}\)).

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

The reaction quotient, Q = \(\frac{[Al^{3+}]^{2}}{[Ni^{2+}]^{3}}\)

Given [Al\(^{3+}\)] = 0.001 M = 10\(^{-3}\) M and [Ni\(^{2+}\)] = 0.001 M = 10\(^{-3}\) M.

Q = \(\frac{(10^{-3})^{2}}{(10^{-3})^{3}} = \frac{10^{-6}}{10^{-9}} = 10^{3}\)


Step 5: Calculate E\(_{cell}\).

E\(_{cell}\) = 1.41 V - \(\frac{0.0591}{6}\) log(10\(^{3}\))

E\(_{cell}\) = 1.41 - \(\frac{0.0591}{6} \times 3\)

E\(_{cell}\) = 1.41 - \(\frac{0.0591}{2}\)

E\(_{cell}\) = 1.41 - 0.02955

E\(_{cell}\) \(\approx\) 1.38 V
Quick Tip: Remember the Nernst equation: E\(_{cell}\) = E°\(_{cell}\) - (0.0591/n)logQ (at 298K). Always correctly identify the anode and cathode to calculate E°\(_{cell}\), and be careful with the stoichiometry when writing the expression for Q.

(i) Allyl chloride vs n-propyl chloride hydrolysis:

Allyl chloride hydrolyzes more readily because it can proceed via an S\(_{N}\)1 mechanism, forming a highly stable intermediate carbocation.

When the C-Cl bond breaks, it forms the allyl carbocation (CH\(_{2}\)=CH-CH\(_{2}^{+}\)).

This carbocation is stabilized by resonance, delocalizing the positive charge over two carbon atoms.

n-Propyl chloride, being a primary halide, would form a very unstable primary carbocation, so it reacts much more slowly via an S\(_{N}\)2 mechanism. The stability of the intermediate in the allyl chloride pathway makes its hydrolysis faster.


(ii) Alkyl halides with silver cyanide:

Silver cyanide (AgCN) is a predominantly covalent compound, with a covalent bond between Ag and C.

When it acts as a nucleophile, the carbon end is not readily available to donate electrons.

The nitrogen atom has a lone pair of electrons which is available for donation, making it the nucleophilic site.

Therefore, the nitrogen atom attacks the alkyl halide, forming a C-N bond and resulting in the formation of an alkyl isocyanide (R-NC).


(iii) S\(_{N}\)2 reactivity of Methyl chloride vs t-butyl chloride:

The S\(_{N}\)2 mechanism involves a backside attack by the nucleophile on the carbon atom bearing the leaving group.

The rate of this reaction is highly dependent on steric hindrance.

Methyl chloride (CH\(_{3}\)Cl) has three small hydrogen atoms, which offer minimal steric hindrance, allowing easy access for the attacking OH\(^{-}\) ion.

t-Butyl chloride ((CH\(_{3}\))\(_{3}\)CCl) is a tertiary halide with three bulky methyl groups surrounding the central carbon. These groups sterically block the path for a backside attack, making the S\(_{N}\)2 reaction extremely slow or impossible.
Quick Tip: Remember the factors affecting S\(_{N}\)1 and S\(_{N}\)2 reactions. S\(_{N}\)1 is favored by stable carbocations (3° > 2° > allyl/benzyl) and polar protic solvents. S\(_{N}\)2 is favored by less sterically hindered substrates (methyl > 1° > 2°) and strong nucleophiles.

(a) Mn\(^{3+}\) as an oxidizing agent and Cr\(^{2+}\) as a reducing agent:

Both Mn\(^{3+}\) and Cr\(^{2+}\) have the d\(^{4}\) electronic configuration.

Mn\(^{3+}\) acts as an oxidizing agent because it has a strong tendency to accept an electron to achieve a more stable configuration.

Mn\(^{3+}\)(d\(^{4}\)) + e\(^{-}\) \(\rightarrow\) Mn\(^{2+}\)(d\(^{5}\)). The resulting Mn\(^{2+}\) ion has a half-filled d-subshell (d\(^{5}\)), which is exceptionally stable.

Cr\(^{2+}\) acts as a reducing agent because it has a strong tendency to lose an electron.

Cr\(^{2+}\)(d\(^{4}\)) \(\rightarrow\) Cr\(^{3+}\)(d\(^{3}\)) + e\(^{-}\). The resulting Cr\(^{3+}\) ion has a half-filled t\(_{2g}\) level (t\(_{2g}^{3}\)) in an octahedral field, which provides extra stability according to Crystal Field Theory.


(b) Actinoid contraction vs Lanthanoid contraction:

Actinoid contraction is greater than lanthanoid contraction because the electrons being filled are in the 5f orbitals for actinoids, compared to 4f orbitals for lanthanoids.

The 5f orbitals are more diffuse and extend further in space than the 4f orbitals.

This means the shielding effect of 5f electrons is even poorer than that of 4f electrons.

As a result, the outer electrons experience a stronger effective nuclear pull as the atomic number increases across the actinoid series, leading to a more significant size reduction from one element to the next.


(c) Formation of Interstitial Compounds:

Transition metals have a regular crystalline lattice structure with empty spaces between the metal atoms, called interstitial sites or voids.

Small non-metallic atoms like Hydrogen (H), Boron (B), Carbon (C), and Nitrogen (N) have small atomic radii.

These small atoms can easily be trapped or fitted into the interstitial voids of the transition metal lattice.

This forms non-stoichiometric compounds known as interstitial compounds, without disturbing the parent metal lattice significantly.
Quick Tip: Stability of electronic configurations (especially half-filled d\(^{5}\) and filled d\(^{10}\) subshells, and half-filled t\(_{2g}^{3}\) orbitals) is a key factor in explaining the redox properties of transition metal ions.

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

M\(_{normal}\) = Atomic mass of Na + Atomic mass of O + Atomic mass of H

M\(_{normal}\) = 23 + 16 + 1 = 40 g mol\(^{-1}\).


Step 2: Calculate the van't Hoff factor (i).

The van't Hoff factor is the ratio of the normal molar mass to the observed molar mass for solutes that dissociate or associate.

i = \(\frac{Normal Molar Mass}{Observed Molar Mass}\)

i = \(\frac{40 g mol^{-1}}{25 g mol^{-1}}\) = 1.6


Step 3: Relate the van't Hoff factor to the degree of dissociation (\(\alpha\)).

NaOH dissociates in solution as: NaOH \(\rightleftharpoons\) Na\(^{+}\) + OH\(^{-}\).

The number of particles produced from one formula unit is n = 2.

The formula relating i, n, and \(\alpha\) for dissociation is: i = 1 + (n - 1)\(\alpha\).

1.6 = 1 + (2 - 1)\(\alpha\)

1.6 = 1 + \(\alpha\)
\(\alpha\) = 1.6 - 1 = 0.6


Step 4: Calculate the percentage dissociation.

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

Percentage dissociation = 0.6 \(\times 100\)% = 60%.
Quick Tip: For colligative properties of electrolytes, always remember to use the van't Hoff factor 'i'. It's defined as i = (Observed value of colligative property) / (Calculated value) = (Normal molar mass) / (Observed molar mass).

1. Compound 'A' (C\(_{4}\)H\(_{8}\)O) gives a positive 2,4-DNP test, which indicates the presence of a carbonyl group (C=O). So, 'A' is either an aldehyde or a ketone.


2. 'A' does not react with Tollen's or Fehling's reagent. This confirms that 'A' is a ketone, not an aldehyde.


3. 'A' gives a yellow precipitate ('B') with sodium hypoiodite (iodoform test). This is a positive test for methyl ketones (compounds containing the CH\(_{3}\)-CO- group).


4. From these clues, 'A' must be a methyl ketone with 4 carbon atoms. The only possible structure is CH\(_{3}\)-CO-CH\(_{2}\)-CH\(_{3}\).

Structure of A: \hspace{1cm CH\(_{3}\)-CO-CH\(_{2}\)-CH\(_{3}\)

IUPAC Name of A: \hspace{0.5cm Butan-2-one


5. The yellow precipitate 'B' formed in the iodoform test is iodoform.

Structure of B: \hspace{1cm CHI\(_{3}\)

IUPAC Name of B: \hspace{0.5cm Triiodomethane


6. Drastic oxidation of 'A' (Butan-2-one) with KMnO\(_{4}\) cleaves the C-C bond adjacent to the carbonyl group. According to Popoff's rule, the C=O group stays with the smaller alkyl group. Cleavage of the C2-C3 bond occurs.

CH\(_{3}\)-CO-|-CH\(_{2}\)-CH\(_{3}\) \(\xrightarrow{KMnO_{4}}\) CH\(_{3}\)-COOH + CO\(_{2}\) + H\(_{2}\)O (from the ethyl group being oxidized). A major product is ethanoic acid.

Structure of C: \hspace{1cm CH\(_{3}\)-COOH

IUPAC Name of C: \hspace{0.5cm Ethanoic acid
Quick Tip: Key tests for carbonyl compounds: 2,4-DNP test confirms a carbonyl group. Tollen's/Fehling's test distinguishes aldehydes (positive) from ketones (negative). The iodoform test identifies methyl ketones or alcohols that can be oxidized to methyl ketones.

(a) Products of DNA hydrolysis and DNA vs RNA structure:


When DNA undergoes complete hydrolysis, it yields three main components:

1. A pentose sugar called 2-deoxyribose.

2. Phosphoric acid (or phosphate group).

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


DNA is structurally different from RNA in two main ways:

1. Sugar Unit: DNA contains 2-deoxyribose sugar, while RNA contains ribose sugar.

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


(b) Nucleotide vs Nucleoside:

Nucleoside: A nucleoside is formed when a nitrogenous base is attached to the 1' position of a pentose sugar (ribose or deoxyribose). (Base + Sugar).

Nucleotide: A nucleotide is formed when a phosphate group is attached to the 5' position of the sugar unit in a nucleoside. It is a phosphate ester of a nucleoside. (Base + Sugar + Phosphate).


(c) (i) Two important functions of nucleic acids:

1. DNA (Deoxyribonucleic Acid): It is the chemical basis of heredity and is responsible for storing and transferring genetic information from one generation to the next.

2. RNA (Ribonucleic Acid): It is primarily responsible for the process of protein synthesis in the cell.


OR


(c) (ii) Linkage and unique base in RNA:

The linkage that joins two nucleotides together in a nucleic acid chain is called a phosphodiester linkage.

The base that is found in RNA but not in DNA is Uracil (U).
Quick Tip: Remember the fundamental difference: Nucleo\textbf{s}ide = \textbf{S}ugar + Base. Nucleo\textbf{t}ide = Sugar + Base + Phospha\textbf{t}e. The 't' in tide can help you remember the phosphate.

(a) Negative Deviation from Raoult's Law:

Example: A mixture of chloroform (CHCl\(_{3}\)) and acetone (CH\(_{3}\)COCH\(_{3}\)).


Reason: Negative deviation occurs when the intermolecular forces of attraction between the molecules of the two components (A-B interactions) are stronger than the intermolecular forces in the pure components (A-A and B-B interactions).


In the chloroform-acetone mixture, a hydrogen bond is formed between the hydrogen atom of chloroform and the oxygen atom of acetone.


This new, stronger interaction reduces the escaping tendency of the molecules from the solution, leading to a lower vapor pressure than predicted by Raoult's law.


(b) (i) Raoult's Law for volatile components:

For a solution of volatile liquids, Raoult's law states that the partial vapor pressure of each component in the solution at a given temperature is directly proportional to its mole fraction in the solution.

Mathematically, P\(_{A}\) = P°\(_{A}\) \(\times\) x\(_{A}\), where P\(_{A}\) is the partial pressure of component A, P°\(_{A}\) is its vapor pressure in the pure state, and x\(_{A}\) is its mole fraction.


OR


(b) (ii) Raoult's law as a special case of Henry's law:

Yes, this is correct. According to Henry's law, P = K\(_{H}\)x. According to Raoult's law, P = P°x.

The two equations are similar in form. Raoult's law can be considered a special case of Henry's law in which the Henry's law constant, K\(_{H}\), becomes equal to the vapor pressure of the pure component, P°.

This holds true for the solvent in a very dilute solution, which behaves ideally.


(c) Two characteristics of an ideal solution:

1. The enthalpy of mixing is zero (\(\Delta H_{mix}\) = 0). No heat is evolved or absorbed when the components are mixed.

2. The volume of mixing is zero (\(\Delta V_{mix}\) = 0). The total volume of the solution is exactly the sum of the volumes of the pure components.

(A third characteristic is that it must obey Raoult's law over the entire range of concentrations.)
Quick Tip: Remember the signs for deviations: \textbf{Negative deviation} \(\rightarrow\) Stronger A-B forces, \(\Delta H_{mix}\) < 0 (exothermic), \(\Delta V_{mix}\) < 0. \textbf{Positive deviation} \(\rightarrow\) Weaker A-B forces, \(\Delta H_{mix}\) > 0 (endothermic), \(\Delta V_{mix}\) > 0.

(a) (i) Calculation of Rate Constant:

For a first-order reaction, the integrated rate law is:

k = \(\frac{2.303}{t}\) log \(\frac{[R]_{0}}{[R]_{t}}\)


Given:

Initial concentration [R]\(_{0}\) = 1.2 \(\times\) 10\(^{-2}\) mol L\(^{-1}\)

Final concentration [R]\(_{t}\) = 0.2 \(\times\) 10\(^{-2}\) mol L\(^{-1}\)

Time, t = 60 minutes


Substituting the values:

k = \(\frac{2.303}{60 min}\) log \(\frac{1.2 \times 10^{-2}}{0.2 \times 10^{-2}}\)

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

k = \(\frac{2.303}{60} \times 0.778\)

k = 0.03838 \(\times\) 0.778

k \(\approx\) 0.02986 min\(^{-1}\)


(ii) (I) Order from balanced equation:

The order of a reaction is an experimental quantity that depends on the reaction mechanism.

It represents the sum of the powers of the concentration terms in the rate law.

A balanced chemical equation only shows the stoichiometry of the overall reaction and gives no information about the intermediate steps or the slowest (rate-determining) step.


(ii) (II) Bimolecular reaction becoming first order:

This can occur in a pseudo-first-order reaction.

If a bimolecular reaction involves two reactants, and one of them is present in a very large excess, its concentration remains practically constant throughout the reaction.

The reaction rate then becomes dependent only on the concentration of the reactant present in the smaller amount, making the reaction kinetically of the first order.


OR


(b) (i) Calculation of Activation Energy:

The relationship between rate constants at two different temperatures is given by the Arrhenius equation:

log \(\frac{k_{2}}{k_{1}}\) = \(\frac{E_{a}}{2.303 R}\) [\(\frac{T_{2} - T_{1}}{T_{1} T_{2}}\)]


Given:

The rate doubles, so \(\frac{k_{2}}{k_{1}}\) = 2.

T\(_{1}\) = 298 K

T\(_{2}\) = 298 + 10 = 308 K

2.303 R = 19.15 JK\(^{-1}\) mol\(^{-1}\)


Substituting the values:

log 2 = \(\frac{E_{a}}{19.15}\) [\(\frac{308 - 298}{298 \times 308}\)]

0.3 = \(\frac{E_{a}}{19.15}\) [\(\frac{10}{91784}\)]

E\(_{a}\) = \(\frac{0.3 \times 19.15 \times 91784}{10}\)

E\(_{a}\) = 52719.5 J mol\(^{-1}\) or 52.72 kJ mol\(^{-1}\).


(b) (ii) Rate Law, Order and Molecularity:

(1) Rate Law: The rate law is determined by the slowest step in the reaction mechanism.

The slow step is: H\(_{2}\)O\(_{2}\) + I\(^{-}\) \(\rightarrow\) H\(_{2}\)O + IO\(^{-}\).

Therefore, the rate law is: Rate = k[H\(_{2}\)O\(_{2}\)][I\(^{-}\)]


(2) Overall Order: The order is the sum of the powers of the concentration terms in the rate law.

Order = 1 (with respect to H\(_{2}\)O\(_{2}\)) + 1 (with respect to I\(^{-}\)) = 2.


Molecularity: Molecularity is defined for elementary steps, not for the overall complex reaction.

The molecularity of the slow step (step I) is 2 (bimolecular). The molecularity of the fast step (step II) is also 2 (bimolecular). The molecularity of the overall reaction is not defined.
Quick Tip: The rate of a multi-step reaction is governed by its slowest step, known as the rate-determining step (RDS). The rate law is written directly from the stoichiometry of this elementary step.

(a) (i) Completing Reactions:

(I) Ketone + Semicarbazide \(\rightarrow\) Semicarbazone. The product is cyclohexanone semicarbazone.


(II) Dimethyl cadmium + Acetyl chloride \(\rightarrow\) Ketone. This reaction prepares ketones. The product is propanone.

(CH\(_{3}\))\(_{2}\)Cd + 2CH\(_{3}\)COCl \(\rightarrow\) 2CH\(_{3}\)COCH\(_{3}\) + CdCl\(_{2}\). Product: CH\(_{3}\)COCH\(_{3}\)


(III) Rosenmund Reduction: Acyl chloride is reduced to an aldehyde using H\(_{2}\)/Pd-BaSO\(_{4}\). The product is benzaldehyde.



(a) (ii) Distinguishing Tests:

(I) Ethyl benzoate and Benzoic acid: Add aqueous sodium bicarbonate (NaHCO\(_{3}\)) solution to both. Benzoic acid, being an acid, will produce brisk effervescence due to the release of CO\(_{2}\) gas. Ethyl benzoate (an ester) will not react.


(II) Propanal and Propanone: Use Tollen's test. Add Tollen's reagent to both compounds and warm. Propanal, being an aldehyde, will give a silver mirror or a black precipitate of silver. Propanone, a ketone, will not give a positive test.


OR


(b) (i) Completing Synthesis:

(I) The product is benzoic acid. This is formed by the strong oxidation of the ethyl group on ethylbenzene. The missing reagent is an alkaline solution of potassium permanganate followed by acidification. Reagent: (i) KMnO\(_{4}\), KOH, Heat (ii) H\(_{3}\)O\(^{+}\).


(II) The product is cyclohexanecarbaldehyde formed from an alkene. This is an ozonolysis reaction. The starting material must be methylenecyclohexane. Starting Material: Methylenecyclohexane. Reagent: (i) O\(_{3}\) (ii) Zn/H\(_{2}\)O.


(III) This shows a ketone reacting with Tollen's reagent [Ag(NH\(_{3}\))\(_{2}\)]\(^{+}\). Standard ketones do not react. However, the reactant shown is an alpha-hydroxy ketone which does give a positive Tollen's test. Let's assume there is a typo in the question and the reactant is cyclohexanecarbaldehyde. The product would be the cyclohexanecarboxylate ion and a silver mirror. Product: Cyclohexanecarboxylate ion and Ag(s).


(b) (ii) Conversions:

(I) Benzaldehyde to Benzophenone:

Step 1: Oxidize benzaldehyde to benzoic acid using an oxidizing agent like Tollen's reagent or KMnO\(_{4}\).


C\(_{6}\)H\(_{5}\)CHO \(\xrightarrow{KMnO_{4}}\) C\(_{6}\)H\(_{5}\)COOH

Step 2: Convert benzoic acid to benzoyl chloride using thionyl chloride (SOCl\(_{2}\)).


C\(_{6}\)H\(_{5}\)COOH + SOCl\(_{2}\) \(\rightarrow\) C\(_{6}\)H\(_{5}\)COCl

Step 3: React benzoyl chloride with benzene in a Friedel-Crafts acylation reaction.

C\(_{6}\)H\(_{5}\)COCl + C\(_{6}\)H\(_{6}\) \(\xrightarrow{Anhyd. AlCl_{3}}\) C\(_{6}\)H\(_{5}\)COC\(_{6}\)H\(_{5}\) (Benzophenone)


(II) Benzaldehyde to 3-phenyl propanol:

Step 1: Perform a crossed aldol condensation between benzaldehyde and ethanal (acetaldehyde).

C\(_{6}\)H\(_{5}\)CHO + CH\(_{3}\)CHO \(\xrightarrow{dil. NaOH}\) C\(_{6}\)H\(_{5}\)CH=CH-CHO (Cinnamaldehyde)


Step 2: Reduce both the double bond and the aldehyde group using a strong reducing agent like H\(_{2}\) with a Ni catalyst under pressure and heat.

C\(_{6}\)H\(_{5}\)CH=CH-CHO + 2H\(_{2}\) \(\xrightarrow{Ni, Heat, Pressure}\) C\(_{6}\)H\(_{5}\)CH\(_{2}\)CH\(_{2}\)CH\(_{2}\)OH (3-phenyl propanol)
Quick Tip: For multi-step conversions, work backward from the product (retrosynthesis). Think about what reaction forms the final functional group and what starting material that would require.

(a) (i) Reasons:

(I) In [Ni(CO)\(_{4}\)], Ni is in the 0 oxidation state (3d\(^{8}\)4s\(^{2}\)). CO is a strong field ligand, causing the pairing of all electrons, including the 4s electrons which move to the 3d orbitals. The configuration becomes 3d\(^{10}\). With no unpaired electrons, it is diamagnetic. In [NiCl\(_{4}\)]\(^{2-}\), Ni is in the +2 state (3d\(^{8}\)). Cl\(^{-}\) is a weak field ligand and does not cause pairing. The geometry is tetrahedral (sp\(^{3}\)), and the 3d\(^{8}\) configuration has two unpaired electrons. Hence, it is paramagnetic.


(II) CO is a stronger ligand than NH\(_{3}\) because, in addition to being a sigma donor, CO is also a pi-acceptor. It accepts electron density from the filled d-orbitals of the metal into its vacant \(\pi\)* anti-bonding orbitals. This synergic bonding strengthens the metal-ligand bond significantly. NH\(_{3}\) is only a sigma donor.


(III) The trans isomer of [Co(en)\(_{2}\)Cl\(_{2}\)]\(^{+}\) is optically inactive because it is superimposable on its mirror image. This is due to the presence of a plane of symmetry that bisects the molecule.


(a) (ii) Unpaired electrons in Fe\(^{3+}\) complexes:

Fe\(^{3+}\) has the electronic configuration [Ar] 3d\(^{5}\).


(I) Strong field ligand: The crystal field splitting is large (\(\Delta_{o}\) > P). Electrons pair up in the lower energy t\(_{2g}\) orbitals first. The configuration is t\(_{2g}^{5}\)e\(_{g}^{0}\). The number of unpaired electrons is 1.


(II) Weak field ligand: The crystal field splitting is small (\(\Delta_{o}\) < P). Electrons fill all orbitals singly before pairing. The configuration is t\(_{2g}^{3}\)e\(_{g}^{2}\). The number of unpaired electrons is 5.


OR


(b) (i) Isomerism:

(I) [Co(NH\(_{3}\))\(_{6}\)] [Cr(CN)\(_{6}\)]: Exhibits Coordination Isomerism, which arises from the interchange of ligands between the cationic and anionic coordination spheres. Its isomer is [Cr(NH\(_{3}\))\(_{6}\)] [Co(CN)\(_{6}\)].


(II) [Co(en)\(_{3}\)]\(^{3+}\): Exhibits Optical Isomerism. The complex is chiral and exists as a pair of non-superimposable mirror images (enantiomers).



(III) [Co(NH\(_{3}\))\(_{3}\)(NO\(_{2}\))\(_{3}\)]: Exhibits Geometrical Isomerism (specifically facial-meridional isomerism). The isomers are the fac isomer (where the three identical ligands occupy the corners of one face of the octahedron) and the \textit{mer isomer (where they occupy positions around the meridian of the octahedron).



(b) (ii) Weak vs Strong Field Ligands:


Difference: Ligands are classified based on their ability to cause splitting of the d-orbitals of the central metal ion. This ability is summarized in the spectrochemical series. Ligands that cause a large splitting are called strong field ligands (e.g., CN\(^{-\), CO). Ligands that cause a small splitting are called weak field ligands (e.g., I\(^{-}\), Br\(^{-}\), Cl\(^{-}\)).


Influence on Spin: The strength of the ligand determines the magnitude of the crystal field splitting energy (\(\Delta_{o}\)). If \(\Delta_{o}\) is greater than the pairing energy (P), as with strong field ligands, it is energetically more favorable for electrons to pair up in the lower t\(_{2g}\) orbitals. This results in a low-spin complex. If \(\Delta_{o}\) is less than P, as with weak field ligands, it is energetically more favorable for electrons to occupy the higher e\(_{g}\) orbitals before pairing up. This results in a high-spin complex.
Quick Tip: For octahedral d\(^{4}\), d\(^{5}\), d\(^{6}\), and d\(^{7}\) configurations, the ligand field strength (weak vs. strong) determines whether the complex will be high-spin or low-spin. For other configurations (d\(^{1,2,3,8,9,10}\)), there is only one possible electron arrangement.