Unit 6: Electronic Structure of Atoms – Study Notes

Electronic Structure of Atoms

The Wave Nature of Light

Light exhibits both wave-like and particle-like properties. Understanding these characteristics is essential for describing the behavior of electrons in atoms.

• Electromagnetic Radiation (EMR): A form of energy exhibiting wave-like behavior as it travels through space. Includes visible light, X-rays, ultraviolet, infrared, microwaves, and radio waves.

Wave Characteristics of Light

• Wavelength (\( \lambda \)): The distance between two consecutive peaks (or troughs) of a wave. Measured in meters (m), nanometers (nm), or centimeters (cm).

• Frequency (\( \nu \)): The number of wave cycles that pass a given point per second. Measured in hertz (Hz) or s-1.

• Speed of Light (c): The constant speed at which all electromagnetic waves travel in a vacuum: 3.00 × 10^8 m/s.

The relationship between wavelength, frequency, and speed of light is given by:

$\lambda \nu = c$

• Electromagnetic Spectrum: The range of all types of EMR, ordered by increasing wavelength (\( \lambda \)) or frequency (\( \nu \)).

Order (increasing \( \lambda \)): X-rays → Ultraviolet → Visible → Infrared → Microwaves → Radio & TV

Order (increasing \( \nu \)): Radio & TV → Microwaves → Infrared → Visible → Ultraviolet → X-rays

Example 1

Problem: Red light has a wavelength of 725 nm. What is the frequency?

Solution:

• Convert 725 nm to meters: $725 \text{ nm} = 7.25 \times 10^{-7} \text{ m}$

• Use $\lambda \nu = c$:

$\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{7.25 \times 10^{-7}} = 4.14 \times 10^{14} \text{ s}^{-1}$

Example 2

Problem: X-rays have a frequency of $1 \times 10^{18}$ s-1. What is the wavelength?

Solution:

$\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8}{1 \times 10^{18}} = 3.00 \times 10^{-10} \text{ m}$

Particle Characteristics of Light

Light also behaves as if it is made of particles called photons. The energy of these photons is quantized.

• Quantized Energy: Energy can only be absorbed or emitted in discrete amounts called quanta.

• Photon: A quantum of electromagnetic radiation.

The energy of a photon is given by:

$E = h \nu$

where $h$ is Planck's constant ($6.626 \times 10^{-34}$ J·s).

Example 3

Problem: Find the energy of a photon with a frequency of $5.09 \times 10^{14}$ s-1.

Solution:

$E = h \nu = (6.626 \times 10^{-34}) \times (5.09 \times 10^{14}) = 3.37 \times 10^{-19} \text{ J}$

Example 4

Problem: Find the energy of a photon with a wavelength of $5.0 \times 10^{-5}$ cm.

Solution:

• Convert to meters: $5.0 \times 10^{-5}$ cm = $5.0 \times 10^{-7}$ m

• Find frequency: $\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8}{5.0 \times 10^{-7}} = 6.0 \times 10^{14}$ s-1

• Calculate energy: $E = h \nu = (6.626 \times 10^{-34}) \times (6.0 \times 10^{14}) = 3.98 \times 10^{-19}$ J

The Photoelectric Effect

The photoelectric effect demonstrates the particle nature of light. When light of sufficient frequency strikes a metal surface, electrons are ejected. This effect can only be explained if light energy is quantized in photons.

Line Spectra and the Bohr Model

Atoms emit light at specific wavelengths, producing a line spectrum. The Bohr model explains these lines by proposing quantized energy levels for electrons in atoms.

• Bohr's Postulates:

  1. Only certain orbits with specific energies are permitted for electrons in a hydrogen atom.

  2. An electron in a permitted orbit does not radiate energy.

  3. Energy is emitted or absorbed only when an electron changes orbits, with $E = h \nu$.

Example 5

Problem: Calculate the energy change when an electron falls from $n=5$ to $n=3$ in hydrogen. Is light emitted or absorbed?

Solution:

The energy levels of hydrogen are given by:

$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \text{ J}$

Calculate $\Delta E$:

$\Delta E = E_{n=3} - E_{n=5} = \left(-\frac{2.18 \times 10^{-18}}{3^2}\right) - \left(-\frac{2.18 \times 10^{-18}}{5^2}\right)$

$= (-2.42 \times 10^{-19}) - (-8.72 \times 10^{-20}) = -1.55 \times 10^{-19} \text{ J}$

Negative sign indicates energy is emitted.

Example 6

Problem: What is the wavelength of the light associated with the above transition?

Solution:

$\lambda = \frac{hc}{|\Delta E|}$

Plug in values to find $\lambda$.

Limitations of the Bohr Model

• Only accurately describes hydrogen-like atoms (single electron).

• Cannot explain spectra of multi-electron atoms.

• Does not account for electron-electron interactions or quantum mechanical principles.

The Wave Behavior of Matter

Louis de Broglie proposed that particles, like electrons, can exhibit wave-like properties. The wavelength of a particle is given by:

$\lambda = \frac{h}{mv}$

• Where $m$ is mass (kg), $v$ is velocity (m/s), $h$ is Planck's constant.

Example 7

Problem: Calculate the wavelength of a baseball (mass = 146 g, velocity = 44.1 m/s).

Solution:

• Convert mass to kg: 146 g = 0.146 kg

• $\lambda = \frac{6.626 \times 10^{-34}}{0.146 \times 44.1} = 1.03 \times 10^{-34}$ m

(Extremely small, not observable.)

Example 8

Problem: Calculate the wavelength of an electron (mass = $9.11 \times 10^{-31}$ kg, velocity = $3.0 \times 10^6$ m/s).

Solution:

$\lambda = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-31} \times 3.0 \times 10^6} = 2.42 \times 10^{-10} \text{ m}$

(Comparable to atomic dimensions.)

Heisenberg's Uncertainty Principle

It is impossible to know both the exact momentum and exact position of an electron simultaneously. This principle sets a fundamental limit on measurement at the quantum scale.

Quantum Mechanics and Atomic Orbitals

Schrödinger developed a wave equation to describe the behavior of electrons in atoms. Solutions to this equation are called orbitals, which are regions in space with a high probability of finding an electron.

Shapes and Orientations of Orbitals

• s orbitals: Spherical shape

• p orbitals: Dumbbell shape, oriented along x, y, z axes

• d and f orbitals: More complex shapes

Quantum Numbers

Each electron in an atom is described by a unique set of four quantum numbers:

Quantum Number	Symbol	What it Describes	Allowed Values
Principal	n	Energy level, size of orbital	1, 2, 3, ...
Angular Momentum	l	Shape of orbital	0 to n-1 (0=s, 1=p, 2=d, 3=f)
Magnetic	ml	Orientation in space	-l to +l
Spin Magnetic	ms	Spin direction	+1/2, -1/2

• Pauli Exclusion Principle: No two electrons in an atom can have the same set of four quantum numbers.

Electron Configurations and Orbital Diagrams

Electron configurations show the arrangement of electrons in an atom's orbitals. Orbital diagrams use arrows to represent electron spins.

• Hund's Rule: Electrons occupy degenerate orbitals singly before pairing up.

Examples:

• H: 1s1

• He: 1s2

• C: 1s2 2s2 2p2

• K: [Ar] 4s1

• Ni: [Ar] 4s2 3d8

• Tc: [Kr] 5s2 4d5

• Ca2+: [Ar]

Anomalous Electron Configurations

• Cr: [Ar] 4s1 3d5 (not 4s2 3d4)

• Cu: [Ar] 4s1 3d10 (not 4s2 3d9)

Relative Energies of Orbitals

• 3s < 3p < 3d < 4s (increasing energy)

Condensed Electron Configurations

Use the noble gas core to simplify notation. For example, C: [He] 2s2 2p2

Valence and Core Electrons

• Valence electrons: Electrons in the outermost energy level (highest n).

• Core electrons: Electrons in inner energy levels.

Examples:

• Number of valence electrons for C: 4 (2s2 2p2)

• Number of unpaired electrons for C: 2

• Valence electrons in Ba: n = 6

• Element with 7 valence electrons in the 5th energy level: Iodine (I)

Additional info: The above examples and explanations provide a comprehensive overview of the quantum mechanical model of the atom, including the historical development from classical to quantum concepts, and their application to electron configurations and periodic properties.