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)