Atomic Orbitals - Comprehensive Notes

In-class Topics: Atomic Orbitals Overview

• CHM 117, Prof. Jason Khoury, Fall Semester 2025, ASU
• Focus: Quantized energy levels, atomic orbitals, and hydrogen-like systems
• Key takeaway: Orbitals are defined by quantum numbers n, l, and ml; energy levels in hydrogen depend only on n; orbitals within a shell are degenerate in energy for hydrogen; the shapes and orientations of orbitals follow from l and ml.

Weekly schedule and exam scope

• Homework 2 and Quiz 2 due this Friday at 4:59 PM
• Homework 3 and Quiz 3 posted; due Wednesday, September 17th at 4:59 PM
• Exam will cover content from Homework assignments 1, 2, and 3

In-class demonstration: The Flame Test

• Demonstration relevance: flame colors relate to electronic transitions in atoms/ions
• Concept link: emission spectra arise from electron transitions between quantized energy levels

Notable Problems from Homework 1

Problem 1: Black hole density (massive compact object)

• Given: a class of black holes with masses 100 to 10,000 times solar mass; one example with mass =
  $1\times 10^4$ suns and radius equal to one-half the radius of the Moon
• Known/used values:
  • Radius of Sun: R_{\odot} = 7.0 \times 10^5\ \text{km}
  • Sun density: \rho_{\odot} = 1.4 \times 10^3\ \text{kg/m}^3 = 1.4 \times 10^6\ \text{g/m}^3
  • Moon diameter: d_{\text{moon}} = 2.16 \times 10^3\ \text{miles}
  • Conversion: 1\ \text{km} = 0.6214\ \text{miles}
• Steps used (as in the transcript):
  1. Convert Moon diameter to cm for radius and then BH radius:
  • Moon diameter in cm: d_{\text{moon}} = 3.476 \times 10^8\ \text{cm}
  • Moon radius: R{\text{moon}} = d{\text{moon}}/2 = 1.738 \times 10^8\ \text{cm}
  • BH radius: R{\text{BH}} = \tfrac{1}{2} R{\text{moon}} = 8.69 \times 10^7\ \text{cm}
  1. Mass of BH: using the factor of 10^4 solar masses and a solar mass in grams (as used in the transcript):
  • Mass BH: M{\text{BH}} \approx (1 \times 10^4) \times M{\odot}
  • The transcript reports a mass value of roughly M_{\text{BH}} \approx 2.011 \times 10^{37}\ \text{g} (from the computation: volume × density × mass factor)
  1. Volume of BH: V{\text{BH}} = \frac{4}{3}\pi R{\text{BH}}^3 = 2.749 \times 10^{24}\ \text{cm}^3
  2. Density: \rho{\text{BH}} = \frac{M{\text{BH}}}{V_{\text{BH}}} = \frac{2.011 \times 10^{37}}{2.749 \times 10^{24}} \approx 7.3 \times 10^{12}\ \text{g/cm}^3
• Final answer: \rho_{\text{BH}} \approx 7.3 \times 10^{12}\ \text{g/cm}^3
• Notes: The calculation demonstrates extremely dense compact objects; used moon radius as a scale reference; mass and volume inputs dominate the final density.

Problem 2: Alloy composition from atom counts

• System: 67.2 g sample of Au-Pd alloy contains 3.00 \times 10^{23} atoms total
• Aim: compute mass percentage of gold (Au)
• If all gold:
  • Atoms if all Au: 67.2\text{ g} \times \left(\frac{1\ \text{mol Au}}{196.97\text{ g}}\right) \times \left(\frac{6.02 \times 10^{23}}{1\ \text{mol}}\right) = 2.06 \times 10^{23}\ \text{atoms}
• If all palladium:
  • Atoms if all Pd: 67.2\text{ g} \times \left(\frac{1\text{ mol Pd}}{106.42\text{ g}}\right) \times \left(\frac{6.02 \times 10^{23}}{1\ \text{mol}}\right) = 3.80 \times 10^{23}\ \text{atoms}
• Let x be the fraction of Au in the alloy by atoms; 1-x is Pd fraction. Then total atoms balance:
  (2.06 \times 10^{23})\,(x) + (3.80 \times 10^{23})\,(1 - x) = 3.00\times 10^{23}
• Solve: x = 0.459 \quad\text{and}\quad 1 - x = 0.541
• Mass percentages: Au = 45.9%, Pd = 54.1%

Quantized energy levels and the basics of excitation/emission

• Energy levels are quantized and discrete.
• Excitation: promoting an electron from a ground state to an excited state.
• Emission: an electron dropping from a higher energy level to a lower one emits a photon.
• The discreteness of energy levels explains why emission spectra consist of lines at specific energies.

Quantum mechanical description of atomic orbitals: n, l, ml

• Schrödinger’s equation is solvable exactly for hydrogen; the solutions yield atomic orbitals defined by three quantum numbers:
  • Principal quantum number: n\in{1,2,3,\dots}
  • Physical significance: defines the size and energy of an orbital; also designates the electron shell.
  • Angular momentum quantum number (Azimuthal): l\in{0,1,2,…,n-1}
  • Physical significance: defines the shape and angular momentum of an orbital; corresponds to s, p, d, f, … orbitals.
  • Magnetic quantum number: m_l\in{-l,-l+1,…,l-1,l}
  • Physical significance: defines the orientation of an orbital in space; how many distinct orbitals exist within a subshell.
• Key definitions:
  • An electron shell is a collection of orbitals with the same principal quantum number n.
  • A subshell is a collection of orbitals with the same n and l.

Principal, angular momentum, and magnetic quantum numbers (detailed)

• n defines size and energy level of an orbital
  • For example, the 2s orbital exists in shell n=2 and is smaller than the 3s orbital (n=3).
  • The principal quantum number also defines the electron shell.
• l defines the shape and identity of the orbital
  • allowed values: 0 ≤ l ≤ n-1
  • corresponds to orbital types: l=0\rightarrow s,\ l=1\rightarrow p,\ l=2\rightarrow d,\ l=3\rightarrow f
  • shapes: s (spherical), p (dumbbell), d (clover), f (more complex)
• m_l defines orientation and count within a subshell
  • m_l ranges from -l to +l in integer steps
  • Example: for l=1 (p subshell), m_l = -1, 0, +1 → 3 orbitals (px, py, pz orientations)
• Example statement from transcript: if you can go up to l=3, that means the shell corresponds to n=4 (since l ≤ n-1)

How many orbitals are in a shell? (n-shell counts)

• n=1: only s orbitals → 1 orbital total (1s)
• n=2: s and p orbitals → 1s + 3p = 4 orbitals total
• n=3: s, p, and d orbitals → 1s + 3p + 5d = 9 orbitals total
• General pattern: each subshell contributes (2l+1) orbitals; total per shell n is sum_{l=0}^{n-1} (2l+1) = n^2 orbitals
• Example: n=4 shell → 4 subshells (l=0,1,2,3) with orbital counts 1, 3, 5, 7 → total 1+3+5+7 = 16 orbitals

Visualizing the atomic orbitals of hydrogen

• Cross-sectional images show orbitals for various n and l; principal quantum number n is shown on the right, and angular momentum quantum number l is labeled across the top
• Observations:
  • There is more than one orbital per hydrogen atom (one electron can occupy different orbitals)
  • Orbitals grow larger with larger n (size scales with n)
  • Orbital shapes are determined by l (s, p, d, f)
• Probability density (electron density):
  • Coloring in images represents probability density distribution in space
  • Probability density is proportional to the squared wavefunction: \rho(\mathbf{r}) \propto |\Psi(\mathbf{r})|^2 = \Psi^2(\mathbf{r})
  • Units: probability per unit volume; density interpretation of \Psi^2
• Each orbital has a characteristic energy and electron density distribution, defined by its quantum numbers

Hydrogen energy levels and their implications

• The energies of hydrogenic orbitals are negative relative to a free electron (negative binding energy)
• Energy levels are determined by the principal quantum number n only (in the hydrogen atom): all orbitals with the same n have the same energy (degenerate within a given shell)
• Hydrogen emission spectrum reflects transitions between these energy levels: each spectral line corresponds to a transition between two levels with energies E{ni} to E{nf}
• Conceptual link: the energy ladder convergence toward 0 eV explains why higher-n levels approach the ionization limit
• If needed, the hydrogen energy formula (typical filling in notes): E_n = -\frac{13.6\text{ eV}}{n^2}
  • Where 13.6 eV is the Rydberg energy for hydrogen
• Relation to potential energy: the negative sign indicates the bound state energy is lower than a free electron; the electrostatic interaction lowers the energy

Energies of hydrogen atomic orbitals and degeneracy

• En is the energy for all orbitals with a given n (independent of l and ml in pure hydrogen)
• The spectrum lines demonstrate transitions between different n levels (ΔE = E{nf} - E{ni}) and emit photons with corresponding frequencies/frequencies via:
  • Photon energy: E{photon} = h\nu = E{nf} - E{n_i}
  • This provides spectroscopic evidence for the quantized energy levels

Quick concept checks (Clicker-style Questions) – answers included

Clicker: Which quantum number defines the energy and size of hydrogen atomic orbitals?

• Answer: Principal quantum number n

Clicker: Which pair of quantum numbers corresponds to three possible atomic orbitals?

• Options:
  • n = 1, l = 0
  • n = 2, l = 1
  • n = 3, l = 0
  • n = 3, l = 2
• Answer: n=2,\ l=1 (p subshell has 3 orbitals aligned in 3 orientations)

Clicker: Which of the following statements about the energy of electrons is true?

• Options:
  • A free electron has lower energy than an electron associated with a nucleus.
  • The energy levels of hydrogen atomic orbitals depend on the orbital angular momentum.
  • The energy of an electron in an atomic orbital is negative because it is lower in energy than a free electron.
  • None of these
• Answer: The energy of an electron in an atomic orbital is negative because it is lower in energy than a free electron.

Conceptual question: Which pair of hydrogen atomic orbitals do not have the same energy?

• Options:
  • 2s and 2p
  • 2s and 3p
  • 3p and 3d
  • 4s and 4f
• Answer: 2s and 3p (different n values give different energies; within the same n, orbitals have equal energy in the hydrogen atom)

Summary of key relationships and rules to memorize

• Quantum numbers:
  • n \in {1,2,3,…} determines size and energy level
  • l \in {0,1,2,…,n-1} determines orbital shape (s, p, d, f, …)
  • m_l \in {-l,-l+1,\dots, l-1,l} determines orbital orientation
• Orbitals per shell:
  • Total orbitals in shell n: N_{\text{orbitals}} = n^2
  • Breakdown per subshell: for each l, number of orbitals is 2l+1
  • Example: n=4 → 16 orbitals (1,3,5,7 from l=0,1,2,3)
• Energy structure:
  • For hydrogen, energy depends only on n: all orbitals in shell have same energy
  • Bound-state energy is negative: E_n = -\frac{13.6\text{ eV}}{n^2}
• Visual interpretation:
  • Increasing n means larger orbital size and more interior nodes (for higher n, more radial nodes exist in the wavefunction)
  • Probability density is given by \rho(\mathbf{r}) = |\Psi(\mathbf{r})|^2 with units of 1/volume
• Practical connections:
  • The quantization explains line spectra (flame tests, hydrogen emission lines)
  • Real-world relevance to bonding, chemical properties, and spectroscopy

Quick reference numbers and formulas

• Moon diameter: d_{\text{moon}} = 2.16 \times 10^3\ \text{miles}
• Sun radius: R_{\odot} = 7.0 \times 10^5\ \text{km}
• Sun density: \rho_{\odot} = 1.4 \times 10^3\ \text{kg/m}^3 = 1.4 \times 10^6\ \text{g/m}^3
• BH radius (given): R{\text{BH}} = \frac{R{\text{moon}}}{2} = 8.69 \times 10^7\ \text{cm}
• BH mass (example): M_{\text{BH}} \approx 2.011 \times 10^{37}\ \text{g}
• BH volume: V{\text{BH}} = \frac{4}{3}\pi R{\text{BH}}^3 = 2.749 \times 10^{24}\ \text{cm}^3
• BH density: \rho{\text{BH}} = \frac{M{\text{BH}}}{V_{\text{BH}}} \approx 7.3 \times 10^{12}\ \text{g/cm}^3
• Atomic counts in alloy problem:
  • Avogadro: N_A = 6.02 \times 10^{23}\ \text{atoms/mol}
  • Gold: \text{Au atomic mass} = 196.97\ \text{g/mol}
  • Palladium: \text{Pd atomic mass} = 106.42\ \text{g/mol}
• Hydrogen energy: E_n = -\frac{13.6\text{ eV}}{n^2}
• Transition energy relation: E{photon} = h\nu = E{nf} - E{n_i}

End of notes