Detailed Solution:
Step 1: Pauli's Exclusion Principle states that no two electrons in an atom can have the exact same set of four quantum numbers ($n, l, m_l, m_s$).
Step 2: If two electrons occupy the exact same orbital, they already share the same principal ($n$), azimuthal ($l$), and magnetic ($m_l$) quantum numbers.
Step 3: Therefore, to satisfy the principle, they must differ in their fourth quantum number, which is the spin quantum number ($m_s$). One will have a spin of $+1/2$ and the other $-1/2$.

Detailed Solution:
Step 1: In 1924, Louis de Broglie proposed that just as light exhibits both wave and particle properties, all matter should also exhibit wave-particle duality.
Step 2: He formulated the equation $\lambda = \frac{h}{p}$ (where $\lambda$ is wavelength, $h$ is Planck's constant, and $p$ is momentum).
Step 3: This hypothesis applies universally to all material particles (electrons, protons, atoms, and even macroscopic objects, though for macroscopic objects the wavelength is too small to be observable).

Detailed Solution:
Step 1: According to Molecular Orbital (MO) Theory, atomic orbitals combine to form molecular orbitals. The lowest energy atomic orbitals are the $1s$ orbitals.
Step 2: When two $1s$ orbitals combine in-phase, they form the lowest energy bonding molecular orbital, designated as $\sigma_{1s}$.
Step 3: When they combine out-of-phase, they form the anti-bonding orbital $\sigma_{1s}^*$, which is higher in energy.
Step 4: Therefore, the very first (lowest energy) bonding orbital in the molecule is $\sigma_{1s}$.

Detailed Solution:
Step 1: Aromatic hydrocarbons containing one or more benzene rings are collectively known as arenes.
Step 2: Compounds containing strictly more than one benzene ring fused together (like naphthalene, anthracene) are specifically called polycyclic aromatic hydrocarbons (PAHs) or polyarenes, which falls under the broader classification of Arenes.
Step 3: "Aryls" are functional groups derived from arenes. "Acyls" are derived from carboxylic acids. Therefore, "Arenes" is the correct classification.

Detailed Solution:
Step 1: In an octahedral complex, there are 6 ligands interacting with the central metal ion's d-orbitals directly along the axes.
Step 2: In a tetrahedral complex, there are only 4 ligands, and they do not point directly at any of the d-orbitals, reducing the magnitude of the electrostatic repulsion.
Step 3: Assuming the same metal ion, same ligands, and same metal-ligand distances, Crystal Field Theory calculates that the splitting in a tetrahedral field ($\Delta_t$) is approximately $\frac{4}{9}$ of the splitting in an octahedral field ($\Delta_o$).

Detailed Solution:
Step 1: A $d^6$ metal ion has 6 d-electrons.
Step 2: In an octahedral crystal field, the 5 d-orbitals split into two sets: lower energy $t_{2g}$ (3 orbitals) and higher energy $e_g$ (2 orbitals).
Step 3: "Low spin" implies that the crystal field splitting energy ($\Delta_o$) is greater than the pairing energy ($P$). Therefore, electrons prefer to pair up in the lower energy orbitals rather than jump to the higher ones.
Step 4: The 6 electrons will completely fill the 3 $t_{2g}$ orbitals (2 electrons per orbital), resulting in an electronic configuration of $t_{2g}^6\ e_g^0$.
Step 5: Since all 6 electrons are paired, there are 0 unpaired electrons.

Detailed Solution:
Step 1: Different molecular transitions correspond to different regions of the electromagnetic spectrum depending on their energy gaps.
Step 2: Pure rotational transitions require very little energy and are observed in the Microwave region.
Step 3: Vibrational transitions require more energy and are observed in the Infrared (IR) region.
Step 4: Because rotational energy levels are closely spaced within vibrational energy levels, high-resolution IR spectroscopy reveals both vibrational and rotational transitions simultaneously (vibrational-rotational spectra) in the IR/Near IR region.

Detailed Solution:
Step 1: The quantum number $J$ represents the rotational energy level of a molecule.
Step 2: According to the quantum mechanical selection rules for a rigid diatomic rotor, a molecule must possess a permanent dipole moment to interact with the electromagnetic field and exhibit a rotational spectrum.
Step 3: The strict selection rule for allowed transitions is $\Delta J = \pm 1$. This means a molecule can only transition to the immediately adjacent rotational level above or below its current level.

Detailed Solution:
Step 1: The photoelectric effect is governed by Einstein's equation: $K.E. = h\nu - \Phi$ (or $K.E. = E_{photon} - W$).
Step 2: Here, $K.E.$ is the maximum kinetic energy of the ejected electron, $h\nu$ is the energy of the incoming (impinging) photon, and $\Phi$ (or $W$) is the work function (binding energy) of the target metal.
Step 3: While $\Phi$ is a constant for a specific metal, the actual kinetic energy of the ejected electron varies directly with the energy of the impinging photon ($h\nu$).

Detailed Solution:
Step 1: Write down the intercepts of the plane along the x, y, and z axes: $\frac{1}{2}, 1, \infty$.
Step 2: Take the reciprocals of these intercepts: $\frac{1}{1/2}, \frac{1}{1}, \frac{1}{\infty}$.
Step 3: Calculate the reciprocals: $2, 1, 0$.
Step 4: Since these are already small integers, no further clearing of fractions is necessary. The Miller indices are denoted as $(2 1 0)$.

Detailed Solution:
Step 1: Bragg's equation relates the wavelength of X-rays to the interplanar spacing and diffraction angle: $n\lambda = 2d \sin \theta$.
Step 2: Rearranging to solve for the interplanar spacing $d$: $d = \frac{n\lambda}{2 \sin \theta}$.
Step 3: To find the absolute smallest possible spacing ($d_{min}$) that can produce a reflection, we must maximize the denominator. The maximum possible value of $\sin \theta$ is 1 (which occurs at a glancing angle of $90^\circ$).
Step 4: Substituting $\sin \theta = 1$ into the equation yields $d_{min} = \frac{n\lambda}{2(1)} = \frac{n\lambda}{2}$.

Detailed Solution:
Step 1: The first ionization energy is the energy required to remove an electron from a neutral atom.
Step 2: After the first electron is removed, the resulting species is a positively charged cation ($M^+$).
Step 3: In this cation, the number of protons remains the same, but there is one less electron. This decreases electron-electron repulsion and increases the effective nuclear charge felt by the remaining electrons.
Step 4: As a result, the electron cloud contracts, and the remaining electrons are pulled closer and bound much more tightly to the nucleus, requiring significantly more energy to remove a second electron.

Detailed Solution:
Step 1: The First Law of Thermodynamics is the law of conservation of energy. It states that the change in internal energy ($\Delta U$) of a closed system is equal to the heat added to the system ($q$) minus the work done by the system ($w$).
Step 2: The convention historically used in physics and many older engineering textbooks states that $w$ is work done by the system, leading to the formula $\Delta U = q - w$.
Note: The modern IUPAC chemical convention defines $w$ as work done on the system, which changes the formula to $\Delta U = q + w$. However, given the options provided, the historical convention (A) is the intended answer.

Detailed Solution:
Step 1: The E2 (Elimination Bimolecular) mechanism is a single-step, concerted process.
Step 2: In a concerted mechanism, the breaking of the C-H bond by the base and the breaking of the C-LG (Leaving Group) bond occur simultaneously.
Step 3: Because the bonds break and form at the same time, no distinct reactive intermediate (such as a carbocation, carbanion, or free radical) is ever formed.
Step 4: Instead, the reaction passes through a single, highly energetic transition state directly to form the alkene product.

Detailed Solution:
Step 1: The relationship between the electrochemical potential of a cell under non-standard conditions ($E$) and standard conditions ($E^\circ$) is derived from thermodynamics.
Step 2: This foundational equation in electrochemistry is formulated as $E = E^\circ - \frac{RT}{nF} \ln Q$, where $Q$ is the reaction quotient (dependent on ion concentrations).
Step 3: This equation is universally known as the Nernst equation, named after the physical chemist Walther Nernst.

Detailed Solution:
Step 1: Corrosion is the deterioration of a metal as a result of chemical reactions between it and the surrounding environment.
Step 2: The most common form of corrosion (like rusting of iron) happens in the presence of moisture and involves the transfer of electrons from one part of the metal (acting as an anode) to another part (acting as a cathode).
Step 3: Because this process fundamentally relies on the flow of electrons between distinct anodic and cathodic sites through an electrolyte (water), it is classified as an electrochemical reaction.

Detailed Solution:
Step 1: Stereoisomers are molecules that have the same molecular formula and sequence of bonded atoms, but differ in their 3D spatial arrangement.
Step 2: A specific sub-class of stereoisomers occurs when two molecules are exact mirror images of one another (like a left and right hand).
Step 3: If these mirror-image molecules cannot be perfectly overlapped or superimposed on each other, they represent two distinct, chiral compounds.
Step 4: This exact pair of non-superimposable mirror images is defined as enantiomers.

Detailed Solution:
Step 1: Draw the structure of 2,2-dimethylbutane: $CH_3-C(CH_3)_2-CH_2-CH_3$.
Step 2: The second carbon (C2) is bonded to carbon 1 (a methyl group), carbon 3 (a methylene group), and two methyl branches. Thus, C2 has three methyl ($CH_3$) groups attached to it.
Step 3: When viewing down the C2-C3 bond in a Newman projection, the front carbon (C2) will have these three methyl groups radiating from its center.
Step 4: The back carbon (C3) is bonded to two hydrogens and one ethyl group (or a methyl group C4). Therefore, the groups on the front carbon are strictly three methyls.

Detailed Solution:
Step 1: The conversion of a ketone ($R-CO-R'$) to a secondary alcohol ($R-CH(OH)-R'$) requires the addition of hydrogen, which is a reduction reaction.
Step 2: We must select a reducing agent from the given options.
Step 3: $O_2$, $KMnO_4$, and $CrO_3$ are all strong oxidizing agents and cannot perform this reduction.
Step 4: Sodium borohydride ($NaBH_4$) is a common, mild hydride donor (reducing agent) that cleanly reduces ketones and aldehydes to their corresponding alcohols.

Radial Wave Function ($R(r)$):
The Schrödinger wave equation for the hydrogen atom can be separated into a radial part and an angular part: $\psi(r, \theta, \phi) = R(r) Y(\theta, \phi)$.
- The radial wave function, $R(r)$, depends only on the distance $r$ from the nucleus and the quantum numbers $n$ (principal) and $l$ (azimuthal).
- It determines the size of the orbital and the probability of finding the electron at a distance $r$ from the nucleus.
- Nodes: A radial node occurs where $R(r) = 0$ (probability is zero). The number of radial nodes is given by the formula $(n - l - 1)$.
- Probability Density: $R^2(r)$ represents the probability density at a point. However, the Radial Probability Distribution Function, $4\pi r^2 R^2(r)$, gives the probability of finding the electron in a spherical shell of thickness $dr$ at distance $r$. This plot shows peaks corresponding to the most probable distance (e.g., Bohr radius $a_0$ for the 1s orbital).

MO Diagram of $O_2$ Molecule:
- Oxygen atomic number is 8. Configuration: $1s^2 2s^2 2p^4$. Total electrons in $O_2$ = 16.
- Due to the large energy gap between 2s and 2p orbitals in Oxygen, there is no s-p mixing. Thus, the $\sigma_{2p_z}$ orbital is lower in energy than the $\pi_{2p_x}$ and $\pi_{2p_y}$ orbitals.
- Filling Order: $\sigma_{1s}^2 < \sigma_{1s}^{*2} < \sigma_{2s}^2 < \sigma_{2s}^{*2} < \sigma_{2p_z}^2 < \pi_{2p_x}^2 \equiv \pi_{2p_y}^2 < \pi_{2p_x}^{*1} \equiv \pi_{2p_y}^{*1}$
- Bond Order: $\frac{N_b - N_a}{2} = \frac{10 - 6}{2} = 2$. Therefore, $O_2$ has a double bond.
- Magnetic Property: The two highest energy electrons occupy the degenerate $\pi_{2p}^*$ antibonding orbitals singly (Hund's rule). Since there are 2 unpaired electrons, $O_2$ is paramagnetic.

Based on Crystal Field Theory (CFT), octahedral complexes can be high spin or low spin depending on the ligand strength (spectrochemical series).

1. High Spin Complexes (Weak Field):
- The ligands produce a small crystal field splitting ($\Delta_0 < P$, pairing energy).
- Electrons prefer to occupy the higher energy $e_g$ orbitals rather than pair up in the $t_{2g}$ orbitals.
- Result: Maximum number of unpaired electrons.
- Example: $[Fe(H_2O)_6]^{3+}$. $Fe^{3+}$ is $d^5$. The configuration is $t_{2g}^3 e_g^2$. It has 5 unpaired electrons.

2. Low Spin Complexes (Strong Field):
- The ligands produce a large crystal field splitting ($\Delta_0 > P$, pairing energy).
- Electrons prefer to pair up in the lower energy $t_{2g}$ orbitals before occupying the $e_g$ orbitals.
- Result: Minimum number of unpaired electrons.
- Example: $[Fe(CN)_6]^{3-}$. $Fe^{3+}$ is $d^5$. The configuration is $t_{2g}^5 e_g^0$. It has only 1 unpaired electron.

Solution for 28(a): Particle in a 1D Box

1. The Model:
Consider a particle of mass $m$ trapped in a one-dimensional box of length $L$ (from $x=0$ to $x=L$).
Potential Energy inside the box: $V(x) = 0$ (for $0 < x < L$)
Potential Energy outside the box: $V(x) = \infty$

2. Schrödinger Equation:
The time-independent equation is: $\frac{-h^2}{8\pi^2 m} \frac{d^2\psi}{dx^2} + V\psi = E\psi$
Inside the box ($V=0$): $\frac{d^2\psi}{dx^2} + \frac{8\pi^2 m E}{h^2} \psi = 0$. Let $k^2 = \frac{8\pi^2 m E}{h^2}$.
The general solution is: $\psi(x) = A \sin(kx) + B \cos(kx)$

3. Boundary Conditions:
- At $x=0$, $\psi(0) = 0 \implies B = 0$. Therefore, $\psi(x) = A \sin(kx)$.
- At $x=L$, $\psi(L) = 0 \implies A \sin(kL) = 0$. Since $A \neq 0$, $\sin(kL) = 0 \implies kL = n\pi \implies k = \frac{n\pi}{L}$.
(where $n = 1, 2, 3...$)

4. Energy Levels:
Substitute $k$ back into $k^2 = \frac{8\pi^2 m E}{h^2}$:
$\left(\frac{n\pi}{L}\right)^2 = \frac{8\pi^2 m E}{h^2} \implies E_n = \frac{n^2 h^2}{8mL^2}$
This shows that the energy is quantized. The lowest energy state ($n=1$) is the Zero Point Energy: $E_1 = \frac{h^2}{8mL^2} > 0$.

Solution for 29(a): Crystal Field Theory (CFT)

(i) Salient Features of CFT:
1. The central metal ion is considered a positive point charge, and the ligands are considered negative point charges (or point dipoles).
2. The interaction between the metal ion and the ligands is purely electrostatic (ionic). There is no mixing of orbitals (no covalency).
3. In a free, isolated gaseous metal ion, the five d-orbitals are degenerate (have the same energy).
4. When ligands approach the metal ion, the spherically symmetric field becomes asymmetric, causing the degeneracy of the d-orbitals to be lifted (Crystal Field Splitting).

(ii) Splitting in an Octahedral Complex:
- In an octahedral complex, six ligands approach the central metal ion along the x, y, and z axes.
- The d-orbitals that lie directly along these axes ($d_{x^2-y^2}$ and $d_{z^2}$, the $e_g$ set) experience greater electrostatic repulsion from the incoming ligands.
- The d-orbitals that lie between the axes ($d_{xy}$, $d_{yz}$, $d_{zx}$, the $t_{2g}$ set) experience less repulsion.
- As a result, the degenerate d-orbitals split into two sets: the $e_g$ orbitals are raised in energy by $+0.6\Delta_0$, and the $t_{2g}$ orbitals are lowered in energy by $-0.4\Delta_0$ relative to the barycenter (average energy level).
- The energy gap between these two sets is called the Crystal Field Splitting Energy ($\Delta_0$).

Solution for 30(b)(ii): Bragg's Law of X-Ray Diffraction

Principle:
When X-rays hit a crystal, they are scattered by the atoms in the crystal lattice. Because the atoms are arranged in regular, periodic planes, the scattered waves can interfere constructively or destructively. Constructive interference occurs only at specific angles, producing diffraction peaks.

Derivation of Bragg's Law:
1. Consider parallel atomic planes in a crystal separated by an interplanar distance '$d$'.
2. Two parallel X-ray beams of wavelength '$\lambda$' strike the adjacent planes at an angle of incidence '$\theta$'.
3. The beam striking the lower plane travels an extra distance compared to the beam striking the upper plane.
4. By drawing perpendiculars from the point of incidence on the top plane to the lower beam, we can form two right-angled triangles.
5. The extra path difference = $AB + BC$.
6. From trigonometry, $AB = d \sin\theta$ and $BC = d \sin\theta$.
7. Total path difference = $2d \sin\theta$.
8. For constructive interference (a sharp diffraction peak), this path difference must be an integer multiple of the wavelength ($n\lambda$).

Equation:
$n\lambda = 2d \sin\theta$
Where:
$n$ = Order of reflection (1, 2, 3...)
$\lambda$ = Wavelength of incident X-rays
$d$ = Interplanar spacing
$\theta$ = Bragg angle (angle of diffraction)