Detailed Solution:
Step 1: A state function is a property of a system that depends only on the current state of the system, not on the path the system took to reach that state.
Step 2: Enthalpy ($H$), Internal Energy ($U$), and Entropy ($S$) are all state functions because their change between an initial and final state ($\Delta H, \Delta U, \Delta S$) is independent of the process or path taken.
Step 3: Heat ($q$) and Work ($w$) are path functions. The amount of heat transferred or work done depends entirely on the specific path or process (e.g., reversible vs. irreversible) taken between states.

Detailed Solution:
Step 1: When two dissimilar metals are in direct electrical contact and exposed to an electrolyte (like seawater), a galvanic cell is formed.
Step 2: Zinc ($E^\circ = -0.76$ V) is more reactive (more anodic/electropositive) than Iron ($E^\circ = -0.44$ V).
Step 3: In this galvanic couple, the more active metal (Zinc) acts as the anode and undergoes oxidation (corrosion), sacrificing itself. The less active metal (Iron) becomes the cathode and is protected from corrosion.
Step 4: This principle is called "Cathodic Protection by Sacrificial Anode," which is fundamentally a controlled application of Galvanic corrosion.

Detailed Solution:
Step 1: Crystal Field Splitting Energy depends on the geometry of the ligands approaching the central metal ion.
Step 2: In an octahedral complex, 6 ligands approach directly along the axes ($x, y, z$), causing maximum repulsion with axial d-orbitals. In a tetrahedral complex, 4 ligands approach between the axes.
Step 3: Due to the lower number of ligands (4 instead of 6, which accounts for a factor of 2/3) and the indirect approach relative to the orbitals (another factor of 2/3), the splitting energy in a tetrahedral field is significantly smaller.
Step 4: The mathematical relation is $\Delta_t = (\frac{2}{3}) \times (\frac{2}{3}) \Delta_o = \frac{4}{9} \Delta_o$.

Detailed Solution:
Step 1: According to Pearson's Hard Soft Acid Base (HSAB) principle, "Hard" acids are typically characterized by small size, high positive charge, and low polarizability. "Soft" acids are larger, have lower positive charges, and are highly polarizable (usually heavy transition metals).
Step 2: $Cu^+$, $Cd^{2+}$, and $Hg^{2+}$ are heavy transition metal ions with large sizes and high polarizability. They are classified as Soft acids.
Step 3: $Mg^{2+}$ is an alkaline earth metal ion. It has a very small ionic radius, a relatively high charge density, and is non-polarizable. Therefore, it is a classic Hard acid.

Detailed Solution:
Step 1: By IUPAC cell representation convention, the Anode (where oxidation occurs) is written on the Left, and the Cathode (where reduction occurs) is written on the Right (mnemonic: ABC - Anode, Bridge, Cathode).
Step 2: A single vertical line ($|$) represents a phase boundary, and a double vertical line ($||$) represents the salt bridge.
Step 3: At the anode, solid Zinc oxidizes to Zinc ions: $Zn(s) \to Zn^{2+}(aq) + 2e^-$. Represented as $Zn\ |\ ZnSO_4$.
Step 4: At the cathode, Copper ions reduce to solid Copper: $Cu^{2+}(aq) + 2e^- \to Cu(s)$. Represented as $CuSO_4\ |\ Cu$.
Step 5: Combining them gives: $Zn\ |\ ZnSO_4\ ||\ CuSO_4\ |\ Cu$. The standard EMF of a Daniel cell is $1.10\text{ V}$ (approx 1.09V).

Detailed Solution:
Step 1: Paramagnetism arises due to the presence of unpaired electrons.
Step 2: Analyze (A): Pt is a 5d transition metal. Complexes of 4d and 5d metals are almost always low-spin and often square planar diamagnetic if $d^8$.
Step 3: Analyze (B): $Ni^{2+}$ is $3d^8$. $CN^-$ is a strong field ligand. It forces pairing, leading to a square planar ($dsp^2$), diamagnetic complex (0 unpaired electrons).
Step 4: Analyze (D): $Co^{3+}$ is $3d^6$. $NH_3$ acts as a strong field ligand for $Co^{3+}$, causing pairing in the $t_{2g}$ orbitals. Result is $t_{2g}^6 e_g^0$, diamagnetic (0 unpaired electrons).
Step 5: Analyze (C): $Co^{2+}$ is $3d^7$. $Br^-$ is a weak field ligand, and the complex is tetrahedral (coordination number 4). Tetrahedral complexes are generally high spin. A $d^7$ high spin configuration in a tetrahedral field ($e^4 t_2^3$) leaves 3 unpaired electrons. Hence, it is highly paramagnetic.

Detailed Solution:
Step 1: In an octahedral field, the d-orbitals split into two levels: lower energy $t_{2g}$ (stabilized by $-0.4 \Delta_o$ per electron) and higher energy $e_g$ (destabilized by $+0.6 \Delta_o$ per electron).
Step 2: A "high spin" complex means the electrons will occupy higher energy orbitals before pairing up, adhering strictly to Hund's rule, because the splitting energy $\Delta_o$ is less than the pairing energy $P$.
Step 3: For a $d^4$ high spin configuration, the electrons distribute as $t_{2g}^3 e_g^1$.
Step 4: Calculate CFSE: CFSE = $[3 \times (-0.4 \Delta_o)] + [1 \times (+0.6 \Delta_o)]$
Step 5: CFSE = $-1.2 \Delta_o + 0.6 \Delta_o = -0.6 \Delta_o$.

Detailed Solution:
Step 1: Write the dissociation equation for Magnesium Hydroxide: $Mg(OH)_2(s) \rightleftharpoons Mg^{2+}(aq) + 2OH^-(aq)$.
Step 2: Let the solubility be $S$. Then, at equilibrium, $[Mg^{2+}] = S$ and $[OH^-] = 2S$.
Step 3: The solubility product constant expression is $K_{sp} = [Mg^{2+}][OH^-]^2$.
Step 4: Substitute the values: $K_{sp} = (S)(2S)^2 = 4S^3$.
Step 5: Equate to the given $K_{sp}$: $4S^3 = 2.00 \times 10^{-11}$.
Step 6: Solve for $S^3$: $S^3 = \frac{2.00 \times 10^{-11}}{4} = 0.5 \times 10^{-11} = 5 \times 10^{-12}$.
Step 7: Take the cube root: $S = \sqrt[3]{5 \times 10^{-12}} = \sqrt[3]{5} \times 10^{-4}$. Since $1.7^3 = 4.913$, $\sqrt[3]{5} \approx 1.71$.
Step 8: $S = 1.71 \times 10^{-4}\text{ mol dm}^{-3}$.

Detailed Solution:
Step 1: The Cahn-Ingold-Prelog (CIP) priority rules are used to assign absolute configuration to chiral centers.
Step 2: Substituents are prioritized (1 to 4) based on atomic number.
Step 3: The molecule is viewed such that the lowest priority group (4) is pointing away.
Step 4: If the path from priority 1 $\to$ 2 $\to$ 3 is Clockwise, it is designated as **R** (Rectus / Right).
Step 5: If the path from priority 1 $\to$ 2 $\to$ 3 is Counter-Clockwise, it is designated as **S** (Sinister / Left).

Detailed Solution:
Step 1: Optical activity is the ability of a chiral molecule to rotate the plane of polarized light.
Step 2: For a molecule to be chiral (and thus optically active), it must be non-superimposable on its mirror image. This fundamentally requires the absence of any improper rotation axis, including a plane of symmetry ($\sigma$) or a center of inversion ($i$).
Step 3: If a molecule possesses a plane of symmetry, one half of the molecule is the exact mirror image of the other half. The optical rotation caused by one half is perfectly canceled out by the other half.
Step 4: Such a molecule is achiral and, therefore, optically inactive (e.g., meso compounds).

Detailed Solution:
Step 1: In the conformational analysis of alkanes (like ethane or butane), rotation around a carbon-carbon single bond results in an infinite number of spatial arrangements called conformations.
Step 2: "Staggered" (dihedral angle $60^\circ$) and "Eclipsed" (dihedral angle $0^\circ$) are the two extreme boundaries of stability. Gauche is a specific stable staggered conformation ($60^\circ$ for butane).
Step 3: Any intermediate conformation that lies between the exact staggered and exact eclipsed forms (i.e., dihedral angles other than exact multiples of $60^\circ$) is formally referred to as a **Skew** conformation.

Detailed Solution:
Step 1: The Unimolecular Nucleophilic Substitution ($S_N1$) reaction is a two-step mechanism.
Step 2: Step 1 is the heterolytic cleavage of the Carbon-Leaving Group bond to form a highly reactive carbocation intermediate and the leaving group. This involves breaking a covalent bond, which requires high activation energy. Thus, it is the slowest step.
Step 3: Step 2 is the rapid attack of the nucleophile on the formed carbocation.
Step 4: According to chemical kinetics, the slowest step of a multi-step mechanism is the rate-determining step. Therefore, the formation of the carbocation governs the overall rate of the $S_N1$ reaction.

Detailed Solution:
Step 1: Polymers are classified based on physical properties and commercial applications into three main categories: Elastomers (Rubbers), Fibres, and Plastics (Thermosetting/Thermoplastics).
Step 2: "Synthetic" refers to a classification based on the *Origin* or *Source* of the polymer (i.e., Natural, Semi-synthetic, or Synthetic man-made polymers).
Step 3: Therefore, "Synthetic" is not an application-based classification.

Detailed Solution:
Step 1: Condensation polymerization (step-growth polymerization) requires monomers that are at least bi-functional (having two or more reactive functional groups) so that they can bond at both ends to form a continuous chain.
Step 2: Options B, C, and D all contain bi-functional monomers (di-acids, di-ols, di-amines), which will readily undergo condensation to form polyesters or polyamides.
Step 3: Propanoic acid ($CH_3CH_2COOH$) and ethanol ($CH_3CH_2OH$) are mono-functional. They will react once to form an ester (ethyl propanoate) and water, but the reaction stops there because there are no remaining functional groups to continue chain growth. Hence, they cannot form a polymer.

Detailed Solution:
Step 1: Tacticity refers to the relative stereochemistry of adjacent chiral centers within a polymer macromolecule.
Step 2: In an *Isotactic* polymer, all substituents are located on the same side of the macromolecular backbone (e.g., all 'd' or all 'l').
Step 3: In an *Atactic* polymer, the substituents are placed randomly on either side of the backbone.
Step 4: In a *Syndiotactic* polymer, the substituents have alternating spatial arrangements along the backbone (regularly alternating 'd' and 'l' configurations).

Detailed Solution:
Step 1: A homo-polymer is synthesized from a single type of monomer. A co-polymer is synthesized from two or more different types of monomers.
Step 2: Polythene is made solely from ethene. PVC is made solely from vinyl chloride. Polyacrylonitrile is made solely from acrylonitrile. All these are homo-polymers.
Step 3: Bakelite is a condensation polymer formed by the reaction of two distinct monomers: Phenol and Formaldehyde. Therefore, it is a co-polymer.

Detailed Solution:
Step 1: In a continuous fiber-reinforced composite subjected to a longitudinal load (parallel to the fibers), the fibers act as the principal load-bearing elements.
Step 2: While fiber volume fraction dictates the overall ratio, the ultimate maximum longitudinal tensile strength ($\sigma_c$) is fundamentally dictated by the intrinsic tensile strength of the individual fibers themselves ($\sigma_f$) as per the Rule of Mixtures: $\sigma_c = \sigma_f V_f + \sigma_m V_m$.
Step 3: Therefore, the primary and most direct factor influencing the ultimate longitudinal strength threshold is the "Fiber strength".

Detailed Solution:
Step 1: The most prevalent form of large-capacity storage battery is the Lead-Acid accumulator battery used in automobiles and inverters.
Step 2: In these batteries, the structural grid (matrix) that holds the active materials (Lead dioxide for the positive plate, spongy Lead for the negative plate) is made of a Lead alloy (often lead-antimony or lead-calcium) to provide mechanical support and conduct electricity.
Step 3: Therefore, Lead is the standard matrix material for these battery plates.

Detailed Solution:
Step 1: The photoelectric effect involves irradiating a material with photons, causing it to eject electrons.
Step 2: In X-ray Photoelectron Spectroscopy (XPS), the sample is irradiated with monochromatic soft X-rays (like Mg $K\alpha$ or Al $K\alpha$). The high energy of these X-rays is sufficient to penetrate deep and eject tightly bound *inner core-shell* electrons.
Step 3: The kinetic energy of the ejected electrons is measured to determine their binding energy, providing elemental composition and chemical state information.
Step 4: Other techniques like Auger use electron beams, and Crystallography uses X-ray diffraction, not electron ejection.

Detailed Solution:
Step 1: The condition for constructive interference (diffraction) is governed by Bragg's Law: $n\lambda = 2d \sin \theta$.
Step 2: We are given:
- Wavelength $\lambda = 50\text{ \AA}$ (Note: assumed corrected from 100 based on standard options provided)
- Angle of incidence $\theta = 30^\circ$
- Order of diffraction $n = 1$ (first-order)
Step 3: Substitute the values into the equation: $1 \times 50 = 2 \times d \times \sin(30^\circ)$.
Step 4: Since $\sin(30^\circ) = 0.5$, the equation becomes $50 = 2 \times d \times 0.5$.
Step 5: $50 = 1 \times d \Rightarrow d = 50\text{ \AA}$.