Detailed Solution:
Step 1: The crystal field splitting energy ($\Delta$) depends on several factors: the nature of the ligand, the geometry of the complex, the principal quantum number of the d-orbitals, and the charge (oxidation state) on the central metal ion.
Step 2: As the oxidation state of the central metal ion increases, its positive charge density increases.
Step 3: This higher positive charge draws the negatively charged ligands closer to the metal ion.
Step 4: The closer proximity results in a stronger electrostatic repulsion between the ligand electrons and the metal's d-electrons, which directly increases the magnitude of the splitting ($\Delta_o$).

Detailed Solution:
Step 1: Effective nuclear charge ($Z^*$) is calculated using Slater's Rules: $Z^* = Z - S$, where $Z$ is the actual nuclear charge (atomic number) and $S$ is the shielding (or screening) constant.
Step 2: For Helium, the atomic number $Z = 2$. Its electron configuration is $1s^2$.
Step 3: According to Slater's Rules, for an electron in the $1s$ orbital, the shielding effect provided by the other electron in the same $1s$ orbital is $0.30$.
Step 4: Therefore, the screening constant $S = 1 \times 0.30 = 0.30$.
Step 5: Calculate $Z^* = 2 - 0.30 = 1.70$.

Detailed Solution:
Step 1: The complex $[Pt(NH_3)_2Cl_2]$ is a $d^8$ square planar complex of the type $MA_2B_2$ (where M is the central metal Pt, and A, B are monodentate ligands $NH_3$ and $Cl^-$).
Step 2: In a square planar geometry, the ligands can be arranged in two distinct spatial orientations relative to each other.
Step 3: If the two similar ligands (e.g., both $Cl^-$) are adjacent to each other (at a $90^\circ$ angle), it forms the *cis*-isomer (known as cisplatin, an anti-cancer drug).
Step 4: If the two similar ligands are opposite to each other (at a $180^\circ$ angle), it forms the *trans*-isomer.
Step 5: This type of isomerism based on spatial arrangement around the central atom is called Geometrical Isomerism.

Detailed Solution:
Step 1: In the metal carbonyl complex $[Cr(CO)_6]$, Carbon Monoxide ($CO$) is a neutral ligand, so the oxidation state of Chromium ($Cr$) is $0$.
Step 2: The ground state electronic configuration of neutral Chromium ($Z = 24$) is $[Ar]\ 3d^5\ 4s^1$.
Step 3: Carbon Monoxide ($CO$) is a very strong field ligand (synergic bonding). In the presence of a strong ligand field, the $4s$ electron gets forced into the $3d$ subshell, making the effective configuration $3d^6$.
Step 4: Due to the large crystal field splitting ($\Delta_o > P$), all 6 electrons pair up completely in the lower energy $t_{2g}$ orbitals ($t_{2g}^6\ e_g^0$).
Step 5: Number of unpaired electrons ($n$) = 0.
Step 6: Magnetic moment $\mu = \sqrt{n(n+2)} = \sqrt{0(0+2)} = 0$ B.M.

Detailed Solution:
Step 1: The relationship between the standard Gibbs free energy change ($\Delta G^\circ$) and the equilibrium constant ($K$) is given by the fundamental thermodynamic equation: $\Delta G^\circ = -RT \ln K$.
Step 2: Here, $K = 3.2 \times 10^5$, which is much greater than 1 ($K > 1$).
Step 3: Since $K > 1$, the natural logarithm $\ln(K)$ is a positive value.
Step 4: Substituting a positive value into the equation yields a negative $\Delta G^\circ$ (because of the negative sign in the formula). A negative $\Delta G^\circ$ indicates the forward reaction is highly spontaneous under standard conditions.

Detailed Solution:
Step 1: According to the Second Law of Thermodynamics, the total entropy of an isolated system can never decrease over time. It can only remain constant in ideal reversible processes, or increase for spontaneous (irreversible) processes.
Step 2: An isolated system exchanges neither mass nor energy with its surroundings, fulfilling the condition $\Delta U = 0$.
Step 3: For any spontaneous natural process occurring within this isolated system, entropy must increase ($\Delta S > 0$).
Step 4: If the system is already at equilibrium (a reversible state), the entropy is maximized and remains constant ($\Delta S = 0$).
Step 5: Combining these possibilities, the overall universal condition for an isolated system is $\Delta S \geq 0$.

Detailed Solution:
Step 1: A Pourbaix diagram maps out the most stable phases of a metal in an aqueous solution as a function of potential (E) and pH.
Step 2: A pH of 12 represents a strongly alkaline (basic) environment.
Step 3: A potential of 1.86 V represents a strongly oxidizing environment (very high up on the y-axis of the diagram).
Step 4: Under strongly basic and highly oxidizing conditions, Iron gets oxidized to its highest common oxidation state (+6), forming the soluble ferrate complex ion, $FeO_4^{2-}$. Lower potentials in basic media form the passive $Fe(OH)_3$ or $Fe_2O_3$ layer.

Detailed Solution:
Step 1: The Helmholtz free energy (denoted by $F$ or $A$) is a thermodynamic potential that measures the useful work obtainable from a closed thermodynamic system at a constant temperature and volume.
Step 2: By definition, it is formulated as the internal energy of the system minus the product of absolute temperature and entropy.
Step 3: Mathematically, it is expressed precisely as $F = U - TS$ (where $U$ = internal energy, $T$ = temperature, $S$ = entropy).

Detailed Solution:
Step 1: The molecular formula $C_6H_{14}$ corresponds to hexane, an alkane. We can determine its structural (chain) isomers by arranging the carbon skeleton.
Step 2: Straight chain: n-hexane.
Step 3: 5-carbon chain with a methyl branch: 2-methylpentane and 3-methylpentane.
Step 4: 4-carbon chain with two methyl branches: 2,2-dimethylbutane and 2,3-dimethylbutane.
Step 5: Counting them all yields exactly 5 unique structural isomers.

Detailed Solution:
Step 1: The $S_N1$ mechanism involves a two-step process where the rate-determining first step is the leaving group departing to form a carbocation intermediate.
Step 2: The overall rate of the $S_N1$ reaction heavily depends on the stability of this carbocation.
Step 3: Carbocation stability follows the order: Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$) > Methyl ($CH_3^+$) due to hyperconjugation and inductive electron-donating effects of alkyl groups.
Step 4: Therefore, the reactivity of the parent alkyl halides directly mirrors this stability trend: $3^\circ > 2^\circ > 1^\circ > CH_3X$.

Detailed Solution:
Step 1: Hex-2-ene is an alkene, which typically undergoes electrophilic addition reactions with hydrogen halides (HX).
Step 2: The rate of electrophilic addition depends heavily on the acidity (bond dissociation energy) of the hydrogen halide. The easier it is for HX to yield a proton ($H^+$), the faster the electrophilic attack on the double bond.
Step 3: The bond strength of HX decreases as the halogen atom size increases down the group ($HF > HCl > HBr > HI$).
Step 4: Consequently, HI has the weakest bond, is the strongest acid, donates a proton most readily, and therefore reacts the fastest with hex-2-ene.

Detailed Solution:
Step 1: Cahn-Ingold-Prelog (CIP) priority rules assign priority based on the atomic number of the atoms directly attached to the chiral center.
Step 2: All four groups have a Carbon directly attached. We must look at the next atoms they are bonded to.
- (A) $-CH(OH)CH_3$: The C is attached to (O, C, H).
- (B) $-CH_2CH_3$: The C is attached to (C, H, H).
- (C) $-CHO$: The C has a double bond to O, which counts twice. Attached to (O, O, H).
- (D) $-CH=CH_2$: The C has a double bond to C. Attached to (C, C, H).
Step 3: Comparing the highest atomic number atoms attached to the first carbon:
- (A) and (C) contain Oxygen (atomic number 8), which beats Carbon (atomic number 6).
- (D) contains two Carbons (equivalent via double bond), which beats the single Carbon in (B).
Step 4: Group (B) is only bonded to one C and two H's, making it the lowest priority among the choices.

Detailed Solution:
Step 1: Polymers are classified into thermoplastics (which soften on heating and can be reshaped) and thermosetting polymers (which undergo permanent chemical cross-linking upon heating and set into a hard, infusible mass).
Step 2: Polystyrene, PVC (Polyvinyl Chloride), and Polyethene are long-chain polymers with weak intermolecular forces. They melt upon heating and are thermoplastics.
Step 3: Bakelite (Phenol-formaldehyde resin) is heavily cross-linked in its three-dimensional network structure. Once heated and molded, it cannot be remelted, making it a classic thermosetting polymer.

Detailed Solution:
Step 1: Polytetrafluoroethylene (PTFE), widely known as Teflon, possesses unique properties due to the strong Carbon-Fluorine bonds shielding the carbon backbone.
Step 2: Due to its non-polar nature and highly electronegative fluorine atoms, it is an excellent electrical insulator.
Step 3: Furthermore, PTFE has one of the lowest coefficients of friction of any solid material, making it highly suitable for use as a dry/solid lubricant in bearings, gears, and non-stick coatings.

Detailed Solution:
Step 1: Materials used for hemodialysis (blood purification) tubing must be highly biocompatible, flexible, chemically inert, and capable of being sterilized repeatedly.
Step 2: Silicone rubber (polydimethylsiloxane) fits all these criteria perfectly. It does not react with blood, prevents platelet adhesion, and remains highly flexible.
Step 3: While PVC is also occasionally used for generic blood lines, Silicone is the standard premium material explicitly for biocompatible hemodialysis peristaltic pump tubing.

Detailed Solution:
Step 1: Crystallinity in polymers depends on how tightly and regularly the polymer chains can pack together.
Step 2: Atactic and random polymers have irregular substituent placement, preventing close packing, making them highly amorphous.
Step 3: Both Isotactic (substituents all on one side) and Syndiotactic (regular alternating) allow for regular packing.
Step 4: However, Isotactic polymers generally permit the most highly ordered and dense crystal lattice packing (forming strong helical structures like in isotactic polypropylene), generally yielding the highest degree of crystallinity among tacticity types.

Detailed Solution:
Step 1: In a standard structural composite, the matrix is usually a polymer, and the filler consists of strong, stiff fibers (like carbon or glass).
Step 2: As load is applied, the more ductile matrix transfers the stress into the stiffer, stronger fibers.
Step 3: When pushed to the limit, the matrix generally yields or undergoes micro-cracking first due to its lower tensile strength.
Step 4: The embedded fibers bridge these micro-cracks and continue to bear the overwhelming majority of the structural load. The composite completely fails only when the critical fibers finally snap (fail last).

Detailed Solution:
Step 1: In a typical engineering stress-strain curve, the initial linear region is where Hooke's Law applies. This ends at the Proportionality Limit.
Step 2: Immediately following the proportionality limit, the curve is no longer strictly linear, but the material can still return to its original shape if the load is removed. This specific point is the Elastic Limit.
Step 3: If loading continues past the elastic limit, the material hits the Upper Yield Point, indicating the onset of permanent (plastic) deformation. Thus, the elastic limit immediately follows the proportionality limit.

Detailed Solution:
Step 1: Bragg's law for X-ray diffraction is given by $n\lambda = 2d \sin \theta$.
Step 2: For diffraction to occur, we must observe at least the first-order spectrum, meaning $n_{min} = 1$. The equation becomes $\lambda = 2d \sin \theta$.
Step 3: Rearranging for $d$, we get $d = \frac{\lambda}{2 \sin \theta}$.
Step 4: To find the absolute minimum possible value for $d$, we must maximize the denominator. The maximum possible value for $\sin \theta$ is 1 (at $\theta = 90^\circ$).
Step 5: Substituting $\sin \theta = 1$, we get the minimum spacing condition: $d_{min} = \frac{\lambda}{2(1)} = \frac{\lambda}{2}$.

Detailed Solution:
Step 1: Young's Modulus ($Y$ or $E$) is defined by Hooke's Law as the ratio of tensile stress to tensile strain in the elastic region of a material.
Step 2: Formula: $Y = \frac{\text{Stress}}{\text{Strain}}$.
Step 3: Substitute the given values: Stress = $4\text{ N/m}^2$ and Strain = $0.15$ (Strain is dimensionless).
Step 4: Calculate: $Y = \frac{4}{0.15} = \frac{400}{15} = \frac{80}{3}$.
Step 5: $80 \div 3 = 26.666...\text{ N/m}^2$. This matches option (A).

Solution for 21(a): CFT for $\text{Mn}^{3+}$ ($d^4$)

Step 1: Understand the Ion and Field
$\text{Mn}^{3+}$ has an electronic configuration of $[Ar] 3d^4$. In an octahedral field, the five degenerate d-orbitals split into a lower energy $t_{2g}$ set (3 orbitals) and a higher energy $e_g$ set (2 orbitals).

Step 2: Weak Field (High Spin Complex)
- The splitting energy ($\Delta_0$) is less than the pairing energy ($P$). $\Delta_0 < P$.
- Electrons fill according to Hund's rule before pairing.
- Configuration: $t_{2g}^3 e_g^1$
- Number of Unpaired Electrons ($n$): 4
- CFSE Calculation: CFSE $= [-0.4x + 0.6y]\Delta_0$
CFSE $= [-0.4(3) + 0.6(1)]\Delta_0 = [-1.2 + 0.6]\Delta_0 = -0.6\Delta_0$
- Magnetic Moment ($\mu$): $\mu = \sqrt{n(n+2)} = \sqrt{4(4+2)} = \sqrt{24} = 4.89 \text{ B.M.}$

Step 3: Strong Field (Low Spin Complex)
- The splitting energy ($\Delta_0$) is greater than the pairing energy ($P$). $\Delta_0 > P$.
- Electrons pair up in the lower energy orbitals before occupying the higher energy ones.
- Configuration: $t_{2g}^4 e_g^0$
- Number of Unpaired Electrons ($n$): 2
- CFSE Calculation: CFSE $= [-0.4(4) + 0.6(0)]\Delta_0 + 1P = -1.6\Delta_0 + P$
- Magnetic Moment ($\mu$): $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} = 2.83 \text{ B.M.}$

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Solution for 21(b): Isomerism in Transition Metal Complexes

Isomerism is divided into two broad categories: Structural Isomerism and Stereoisomerism.

1. Structural Isomerism:
- Ionization Isomerism: Exchange of ligands between the coordination sphere and ionization sphere.
Example: $[\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4$ (red-violet) and $[\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}$ (red).
- Linkage Isomerism: Occurs with ambidentate ligands (e.g., $\text{NO}_2^-$ bonding through N or O).
Example: $[\text{Co}(\text{NH}_3)_5(\text{NO}_2)]^{2+}$ (yellow) and $[\text{Co}(\text{NH}_3)_5(\text{ONO})]^{2+}$ (red).
- Coordination Isomerism: Exchange of ligands between complex cation and complex anion.
Example: $[\text{Co}(\text{NH}_3)_6][\text{Cr}(\text{CN})_6]$ and $[\text{Cr}(\text{NH}_3)_6][\text{Co}(\text{CN})_6]$.
- Hydrate Isomerism: Exchange between water inside the coordination sphere and water of crystallization.
Example: $[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$ (violet) vs $[\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2\cdot\text{H}_2\text{O}$ (green).

2. Stereoisomerism:
- Geometrical Isomerism: Distinct spatial arrangements (Cis vs Trans).
Example: $[\text{Pt}(\text{NH}_3)_2\text{Cl}_2]$ - Cis-platin (adjacent) and Trans-platin (opposite).
- Optical Isomerism: Non-superimposable mirror images (enantiomers). Present in complexes lacking a plane of symmetry.
Example: $[\text{Co}(\text{en})_3]^{3+}$ exists as $d$ (dextro) and $l$ (laevo) isomers.

Solution for 22(a): Applications of Nernst Equation

Nernst Equation (General Form):
$E = E^0 - \frac{RT}{nF} \ln Q = E^0 - \frac{0.0591}{n} \log_{10} \frac{[\text{Products}]}{[\text{Reactants}]}$ (at 298 K)

1. Application in Redox Reactions:
The Nernst equation allows the calculation of cell potential under non-standard conditions.
Example: Daniel Cell
Cell Reaction: $\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightleftharpoons \text{Zn}^{2+}(aq) + \text{Cu}(s)$
$n = 2$ electrons transferred.
$E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{2} \log \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]}$
This is used to determine the concentration of an unknown ion or the spontaneity of the redox process at arbitrary concentrations.

2. Application in Acid-Base Reactions (pH measurement):
The Nernst equation is fundamental in pH meters, typically using a Glass Electrode coupled with a reference electrode (e.g., Calomel).
Example: Hydrogen Electrode
Reaction: $H^+ + e^- \rightarrow \frac{1}{2}H_2$
$E_{H^+/H_2} = E^0_{H^+/H_2} - \frac{0.0591}{1} \log \frac{1}{[H^+]}$
Since $E^0 = 0$ V and $\log(1/[H^+]) = -\log[H^+] = \text{pH}$:
$E_{H^+/H_2} = -0.0591 \times \text{pH}$
By measuring the potential of the electrode against a standard, the exact pH of the solution is determined.

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Solution for 22(b): Gibbs-Helmholtz Equation

Derivation:
Step 1: Start with the fundamental definitions.
$G = H - TS$
For a process at constant temperature, the change is: $\Delta G = \Delta H - T\Delta S \quad \text{--- (1)}$

Step 2: We know the relation between Free Energy and Entropy is given by the partial derivative:
$\left( \frac{\partial \Delta G}{\partial T} \right)_P = -\Delta S \quad \text{--- (2)}$

Step 3: Substitute equation (2) into equation (1):
$\Delta G = \Delta H + T\left( \frac{\partial \Delta G}{\partial T} \right)_P$
Rearranging yields the Gibbs-Helmholtz equation:
$\Delta H = \Delta G - T\left( \frac{\partial \Delta G}{\partial T} \right)_P$

Alternative form (often more useful):
Divide by $T^2$ and use the quotient rule: $\left( \frac{\partial (\Delta G / T)}{\partial T} \right)_P = -\frac{\Delta H}{T^2}$

Applications:
1. Calculating Enthalpy ($\Delta H$): If the free energy change ($\Delta G$) at different temperatures is known, $\Delta H$ for a reaction can be calculated without direct calorimetric measurement.
2. Electrochemistry: Using $\Delta G = -nFE$, the equation becomes: $\Delta H = -nFE + nFT \left( \frac{\partial E}{\partial T} \right)_P$. Here, $\left( \frac{\partial E}{\partial T} \right)_P$ is the temperature coefficient of the cell. This allows calculating $\Delta H$ and $\Delta S$ exclusively from cell potential measurements.
3. Predicting $\Delta G$: If $\Delta H$ is known (assumed constant over a small range) and $\Delta G$ is known at one temperature, $\Delta G$ at another temperature can be determined.

Solution for 23(a): $S_N1$ vs $S_N2$ Reactions

$S_N1$ Reaction (Substitution Nucleophilic Unimolecular):
- Kinetics: First-order. Rate depends only on substrate: $\text{Rate} = k[\text{Substrate}]$.
- Mechanism: Two steps. Step 1 forms a carbocation intermediate (slow, rate-determining). Step 2 is nucleophile attack (fast).
- Stereochemistry: Results in racemization (both retention and inversion) due to planar carbocation.
- Substrate Reactivity: $3^\circ > 2^\circ > 1^\circ > \text{Methyl}$ (stabilized by hyperconjugation).
- Example: Hydrolysis of tert-butyl bromide in water.
$(CH_3)_3C-Br \rightleftharpoons (CH_3)_3C^+ + Br^- $
$(CH_3)_3C^+ + H_2O \rightarrow (CH_3)_3C-OH + H^+$

$S_N2$ Reaction (Substitution Nucleophilic Bimolecular):
- Kinetics: Second-order. Rate depends on both: $\text{Rate} = k[\text{Substrate}][\text{Nucleophile}]$.
- Mechanism: Concerted (single step). Bond breaking and forming occur simultaneously via a transition state.
- Stereochemistry: Complete inversion of configuration (Walden inversion).
- Substrate Reactivity: $\text{Methyl} > 1^\circ > 2^\circ > 3^\circ$ (due to steric hindrance).
- Example: Reaction of methyl bromide with hydroxide ion.
$HO^- + CH_3-Br \rightarrow [HO \cdots CH_3 \cdots Br]^\ddagger \rightarrow HO-CH_3 + Br^-$

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Solution for 23(b): Conformational Analysis of n-Butane

Conformational analysis studies the energy changes occurring during rotation around the C2-C3 single bond in n-butane ($CH_3-CH_2-CH_2-CH_3$). Using Newman projections looking down the C2-C3 bond, we observe extreme conformations.

Key Conformations:
1. Anti (Staggered): Dihedral angle ($\theta$) = $180^\circ$. The two bulky $-CH_3$ groups are farthest apart. Lowest steric strain, lowest energy (most stable).
2. Eclipsed (minor): $\theta = 120^\circ$. $CH_3$ group is eclipsed by an $H$ atom. Higher energy due to torsional strain.
3. Gauche (Staggered): $\theta = 60^\circ$. The two $-CH_3$ groups are adjacent. More stable than eclipsed, but less stable than anti due to steric repulsion between the methyl groups.
4. Fully Eclipsed (Syn): $\theta = 0^\circ$. The two bulky $-CH_3$ groups perfectly overlap. Maximum steric and torsional strain. Highest energy (least stable).

Energy Profile:
- $0^\circ$ (Fully Eclipsed) = Energy Maximum (~19 kJ/mol)
- $60^\circ$ (Gauche) = Energy Valley (~3.8 kJ/mol)
- $120^\circ$ (Eclipsed) = Energy Peak (~14 kJ/mol)
- $180^\circ$ (Anti) = Global Minimum (0 kJ/mol)

(Examiner expects a sine-wave-like energy curve mapping Energy vs Dihedral Angle corresponding to these points.)

Solution for 24(a): Teflon and PVC

1. Teflon (Polytetrafluoroethylene, PTFE)
- Synthesis: Synthesized via free-radical addition polymerization of tetrafluoroethylene ($CF_2=CF_2$) gas under high pressure and temperature in the presence of a peroxide catalyst (e.g., benzoyl peroxide).
$n CF_2=CF_2 \xrightarrow{\text{Polymerization}} -(CF_2-CF_2)_n-$
- Properties: Highly unreactive, chemically inert, extremely high melting point, excellent electrical insulator, extremely low coefficient of friction.
- Applications: Non-stick coatings for cookware, gaskets, valves, insulation for high-voltage cables, solid lubricants.

2. PVC (Polyvinyl Chloride)
- Synthesis: Synthesized by the addition polymerization of vinyl chloride monomer ($CH_2=CHCl$) in the presence of a catalyst (e.g., benzoyl peroxide) at 40-70°C.
$n CH_2=CHCl \xrightarrow{\text{Polymerization}} -(CH_2-CHCl)_n-$
- Properties: Rigid, hard, and flame-retardant (due to chlorine). Can be made flexible by adding plasticizers (e.g., DOP).
- Applications:
- Rigid PVC: Water pipes, window frames, door panels.
- Plasticized PVC: Wire insulation, synthetic leather, medical tubing, flooring.

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Solution for 24(b): Doping in Conducting Polymers

Intrinsically conducting polymers (e.g., Polyacetylene, Polyaniline) possess a conjugated backbone of alternating single and double bonds. However, pristine polymers are usually insulators. Their conductivity is increased by orders of magnitude via doping (adding/removing electrons).

1. p-Doping (Oxidative Doping):
- The polymer is treated with an oxidizing agent (electron acceptor) such as $I_2$, $Br_2$, or $AsF_5$.
- The oxidizing agent removes an electron from the $\pi$-backbone of the polymer, creating a positively charged site (hole) called a polaron.
- Further oxidation creates a bipolaron (two positive charges).
- These positive charges are mobile along the conjugated chain, facilitating electrical conductivity.
- Reaction: $(Polymer)_n + \frac{3}{2}x I_2 \rightarrow (Polymer)^{+x}_n + x(I_3)^-$

2. n-Doping (Reductive Doping):
- The polymer is treated with a reducing agent (electron donor) such as alkali metals (Li, Na, K).
- The reducing agent donates an electron into the conduction band ($\pi^*$ orbital) of the polymer chain.
- This creates a negatively charged mobile site.
- Reaction: $(Polymer)_n + x Na \rightarrow (Polymer)^{-x}_n + x Na^+$
- n-doped polymers are generally highly sensitive to oxygen and moisture, making them less stable in air compared to p-doped polymers.

Solution for 25(a): Stress-Strain Plot

The Stress-Strain curve graphically represents the behavior of a material when subjected to load.

(i) Elastic Region:
- This is the initial, linear portion of the curve where Stress is directly proportional to Strain (obeys Hooke's Law).
- If the load is removed in this region, the material fully recovers its original shape with no permanent deformation.
- The slope of this line determines Young's Modulus ($E$).
- It ends at the Yield Point (or Elastic Limit).

(ii) Plastic Region:
- This region begins past the yield point.
- Here, stress increases non-linearly with strain, and the material undergoes permanent (irreversible) deformation.
- If the load is removed, the material will not return to its original length (permanent set).
- This region includes the Ultimate Tensile Strength (UTS) (the maximum peak of the curve) and ends at the Fracture Point (where the material breaks).

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Solution for 25(b): Matrix Composites

A composite material consists of a matrix phase (continuous) and a reinforcement phase (dispersed). They are categorized by their matrix material.

1. Metal Matrix Composites (MMC):
- Matrix: Ductile metals like Aluminum, Titanium, Magnesium, or Copper.
- Reinforcement: High-strength ceramics like $SiC$, $Al_2O_3$, or Carbon fibers.
- Properties: High specific strength, stiffness, excellent high-temperature resistance, good thermal/electrical conductivity, and wear resistance.
- Example: Aluminum matrix reinforced with Silicon Carbide ($SiC$) particles. Used in automotive engine blocks, aerospace components, and brake rotors.

2. Ceramic Matrix Composites (CMC):
- Matrix: Brittle ceramics like Silicon Carbide ($SiC$), Alumina ($Al_2O_3$), or Glass.
- Reinforcement: Ceramic fibers (e.g., Carbon fibers, $SiC$ fibers) to provide fracture toughness.
- Properties: Extremely high thermal stability, high wear and corrosion resistance, and significantly improved fracture toughness (prevents catastrophic brittle failure characteristic of pure ceramics).
- Example: Carbon-Carbon composites (Carbon matrix with Carbon fiber reinforcement) or $SiC/SiC$ composites. Used in space shuttle heat shields, rocket nozzles, and high-performance jet engine turbine blades.

Solution: Pourbaix Diagram for Iron

A Pourbaix Diagram (Potential-pH diagram) maps out the thermodynamically stable phases of an aqueous electrochemical system (like Iron in water) as a function of electrode potential ($E$) and pH.

1. Construction and Axes:
- Y-Axis: Redox Potential ($E$, in Volts) vs Standard Hydrogen Electrode (SHE). High $E$ means oxidizing conditions; low $E$ means reducing conditions.
- X-Axis: pH (0 to 14). Low pH means acidic; high pH means alkaline.

2. Three Main Regions of Iron:
The Pourbaix diagram defines three critical zones for corrosion engineering:
(i) Immunity Region:
- Located at the bottom of the diagram (very negative potential, highly reducing environment).
- The thermodynamically stable species is solid Iron ($Fe_{(s)}$).
- Corrosion cannot occur because oxidation to iron ions is not thermodynamically favored.
- Application: Cathodic protection forces the metal into this region.

(ii) Corrosion Region:
- Located at moderate potentials and acidic to slightly neutral pH.
- The stable species are soluble ions like $Fe^{2+}$ (ferrous, pale green) and $Fe^{3+}$ (ferric, yellow/brown).
- Since the oxidation products dissolve into the electrolyte, active corrosion and material loss occur continuously.

(iii) Passivation Region:
- Located at moderate to high potentials and neutral to alkaline pH.
- The stable species are solid oxide/hydroxide films like $Fe(OH)_2$, $Fe_2O_3$ (hematite), or $Fe_3O_4$ (magnetite).
- These solid products form a tight, continuous layer on the metal surface, acting as a barrier that slows down or stops further corrosion (passivity).

3. Types of Boundaries (Lines in the Diagram):
- Horizontal Lines: Represent purely redox reactions (electron transfer). Independent of pH. Example: $Fe^{2+} + 2e^- \rightleftharpoons Fe$.
- Vertical Lines: Represent purely acid-base reactions (proton transfer). Independent of potential. Example: $Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$.
- Slanted Lines: Represent reactions involving both electron and proton transfer (redox + acid-base). Example: $Fe_2O_3 + 6H^+ + 2e^- \rightleftharpoons 2Fe^{2+} + 3H_2O$.

4. Limitations:
- It is derived strictly from thermodynamics; it yields no information on the kinetic rate of corrosion.
- It does not account for localized corrosion (like pitting) caused by specific ions like chlorides.
- It applies only to pure metals in pure water at standard temperature (25°C).

Solution for 27(i): E2 Mechanism

Elimination Bimolecular (E2) Mechanism:
- The E2 reaction is a one-step, concerted elimination process where a strong base removes a proton from the $\beta$-carbon, the $C=C$ double bond forms, and the leaving group departs from the $\alpha$-carbon simultaneously.
- Kinetics: Second-order. $\text{Rate} = k[\text{Substrate}][\text{Base}]$.
- Stereochemistry: It strictly requires an anti-periplanar geometry. The leaving group and the $\beta$-proton being removed must be $180^\circ$ apart (anti) to each other to allow the developing p-orbitals to overlap perfectly and form the $\pi$-bond.
- Example: Dehydrohalogenation of 2-bromopropane using sodium ethoxide.
$CH_3-CHBr-CH_3 + C_2H_5O^- \rightarrow CH_3-CH=CH_2 + C_2H_5OH + Br^-$
Mechanism step: The ethoxide ion ($C_2H_5O^-$) attacks a $\beta$-hydrogen, pushing electrons to form a double bond between C1-C2, which simultaneously ejects the bromide ($Br^-$) ion.

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Solution for 27(ii): X-ray Photoelectron Spectroscopy (XPS)

XPS, also known as Electron Spectroscopy for Chemical Analysis (ESCA), is a powerful surface-sensitive quantitative spectroscopic technique.

Principle:
XPS is based on the Photoelectric Effect. A sample is irradiated with a beam of X-rays (typically Al $K\alpha$ or Mg $K\alpha$). The X-ray photon energy ($h\nu$) is absorbed by a core-level electron of an atom in the sample. If the photon energy is greater than the binding energy ($E_B$) of the electron, the electron is ejected (emitted as a photoelectron).
The kinetic energy ($E_K$) of the emitted electron is measured by the spectrometer.
The Binding Energy is calculated using the equation: $E_B = h\nu - E_K - \Phi$
Where:
$h\nu$ = Energy of the incident X-ray photon
$E_K$ = Kinetic energy of the emitted electron (measured)
$\Phi$ = Work function of the spectrometer
Since the core-level binding energies are highly characteristic of specific elements and their chemical environments (oxidation states), XPS can determine elemental composition and chemical state.

Instrumentation:
XPS systems must operate under Ultra-High Vacuum (UHV) ($10^{-9}$ mbar) to prevent electrons from colliding with gas molecules and to keep the sample surface free of contaminants. The major components are:
1. X-ray Source: Generates the incident X-rays. Typically uses a magnesium (Mg $K\alpha$ = 1253.6 eV) or aluminum (Al $K\alpha$ = 1486.6 eV) anode. A monochromator is often used to narrow the X-ray bandwidth.
2. Sample Chamber: A UHV chamber where the sample is mounted and manipulated.
3. Electron Energy Analyzer: Usually a Concentric Hemispherical Analyzer (CHA). It acts as a filter, using an electric field to allow only electrons of a specific kinetic energy to pass through to the detector at any given time.
4. Detector: Typically an electron multiplier (or array of multipliers). It counts the number of electrons arriving at each kinetic energy, producing the XPS spectrum (Counts per second vs Binding Energy).
5. Data System: A computer that controls the voltages on the analyzer, records the data, and processes the final spectrum.