Lecture 9_Atomic s- and p-orbitals

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Monday Reading: Chapter 6, Sections 5 – 6 ( Use this ) Wednesday Reading: Chapter 6, Sections 5 – 6 ( Use this ) Friday Reading: Chapter 7, Sections 1 – 2 The plan: Orbitals of the hydrogen atom (energy, structure, quantum numbers) Announcements for September 20 th , 2023 • Homework 3 due September 25 th by 11:59 PM • Recitation 3 tomorrow, September 21 st • Exam 1 is scheduled for Wednesday, September 27 th 7:00 pm – 9:00 pm • Coverage: Lectures 1 – 10 • SDS Accommodations? Schedule with SDS asap • Conflict? Alternative exam 8 am – 10 am on 9/27 • Email me by 9/20 to reserve a seat • Permitted one 8.5 in x 11 in sheet of notes (front and back) • 2 Practice Exams posted • Review Session 4:00 pm – 6:00 pm in Chem102 • Quiz 2 Average = 72% ( → B)

• Particle’s radial position is unrestricted and governed by probabilities – e - explores all space • Quantum mechanics predicts total energy is quantized (discrete or single-valued) The energy of electronic states are discrete, but positions are “continuous” If E is quantized, but e - is stabilized by close approach to nucleus, then why doesn’t the atom collapse? Electron bounces in and out of energy “well” 𝐸 𝑛 = 𝐾𝐸 + 𝑃𝐸 = −𝑅 𝐻 𝑍 2 1 ? 2 r = 0 (Nucleus) r → ∞ Energy n = 2 n = 4 n = 1 n = 3 n = 5 n = 6 n = ∞ FREEDOM! ⋮

Why does the electron avoid falling into the nucleus? How does the quantum mechanical picture avoid this anomaly? (?∆𝑣)∆𝑥 ≥ ℎ 4𝜋 Quantum mechanics MUST be consistent with the Heisenberg Uncertainty Principle Heisenberg’s Uncertainty Principle is a consequence of waviness How is this 3D picture “wavy”? Z+ New atomic picture r = 0 (Nucleus)

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Wavefunctions = ORBITAL • One set of quantum numbers ({ n , ℓ, m ℓ }) corresponds to one unique orbital Orbitals contain all information we want to know about e - : probabilities, energy, and waviness of matter The Principal Quantum Number: n n = 1, 2, 3, 4… s-orbitals n = 1, 2, 3, … 𝐸 𝑛 = −𝑅 𝐻 𝑍 2 1 ? 2 As n increases: We predict: (1) The e - is, on average, _____________________ (2) The e - being further away from the nucleus, on average, means that _________________ Radius increases as n increases! ℓ = ___ and m ℓ = ___ Shape: Spherical Let’s dissect the wavefunction! The energy of the orbital increases Is our prediction consistent?____________ Why?________________________

Orbitals are wavefunctions and squaring it ( Ψ 2 ) gives the probability of finding the electron 3D orbital picture (Wavefunction Ψ ) Slice open Nucleus 1D-probability Ψ What do you think Ψ 2 of 1D-probability looks like? Ψ 2 1D segment of 3D picture! How many nodes does the 1s-orbital contain? How far do we have to travel from the nucleus before probability is ultra tiny?? Ψ decays exponentially from nucleus

• Ψ 2 = Probability of finding e - at a POINT • Ψ 2 approaches 0 as r → ∞ (but never = 0), so it’s possible to find an e - ANYWHERE!!! • Relative orbital sizes based on arbitrary cutoff So, what’s the deal with 3D orbital pictures? Following this dotted line traces Ψ 2 plot Volume enclosed by red shell represents a 90% probability of finding e - Edge of orbital picture is the surface of red- dotted “line” (“boundary surface diagram” or BSD) ~1.38 Å radius for H-atom 1s orbital

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Since orbitals are 3D , we need a convenient way to express probability on a graph Most probable point Most probable radial “shell” • Probability within a small, enclosed volume (point-like) • Gray cube = volume within orbital • Probability of finding electron some distance ‘r’ from nucleus (~ “infinitely thin shell”) • Shell is independent of angles • RDF collapses 3D picture into 2D graph by summing all Ψ 2 values at same r RDF → “Onion probability” Probability density ( Ψ 2 ) Radial distribution function (RDF) • All points with same r’s share same probability of finding the electron • Ψ 2 vs r is ‘incomplete’

Consider a set of discrete points that map a discrete probability distribution with a radial dependence. Each point tells us the probability of occupying the point , not the circle. What is the probability of sitting on a circle that joins the set of points sharing a common radial distance from the center? 𝑃 = 18% 𝑃 = 8% 𝑃 = 7% 𝑃 = 2% 𝑷?𝒊??? 𝑪𝒊?𝒄𝒍𝒆 𝑃 = ____ 𝑃 = ____ 𝑃 = ____ 𝑃 = ____

RDF is probability of finding electron within a thin spherical shell a distance r from nucleus Max is trade off between probability and shell size Ψ 2 is big! Shell does not exist at point! Ψ 2 is ___________! Shell is _________! Ψ 2 is __________! Shell is _________! → RDF is 0 Since the RDF is 0 at r = 0, does that mean the probability of finding the electron at the nucleus is 0? • The RDF collapses a 3D plot into a 2D plot • When all probabilities at fixed r are summed , then r = ______ is most probable _______ distance of finding e - from nucleus As r → ∞: Ψ 2 decreases (exponential decay) Shell size increases (bigger sphere) → RDF increases with bigger shell size → RDF decreases due to Ψ 2

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2s orbital introduces a node and shows e - is further from nucleus, on average, than 1s + - Waviness occurs! Sign of Ψ is called phase • Positive and negative • When Ψ changes phase, a node occurs • Phase plays a major role in chemical bonding (later) What do you think Ψ 2 of 1D-probability looks like? How many nodes does the 2s-orbital contain? How far do we have to travel from the nucleus before probability is ultra tiny??

The 2s orbital introduces a node and shows e - is further from nucleus, on average, than 1s RDF Max at ~3 Å RDF: Battle of Ψ 2 vs Shell size (1) Phase information LOST! • Square of all #’s = + (2) Node retained (3) 2 peaks due to Ψ • ________________________ ________________________ • Shell size wins until after last peak in Ψ 2 Ψ 2 decreases exponentially Shell size increases ( ∝ r 2 )

+ + - Waviness occurs! The 3s orbital introduces a node and shows e - is further from nucleus, on average, than 2s RDF Max at ~7.5 Å General trends: • As n grows, the size increases and the energy of the orbital increases • Waviness goes up as n goes up (meaning node count increases) • Total number of nodes = _______________ • RDF shows more probable to find e - further away from nucleus with growing n How many nodes does the 3s-orbital contain? How far do we have to travel from the nucleus before probability is ultra tiny?

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Plotted below are Ψ, Ψ 2 , and the RDF for an s-orbital . Identify the orbital, count the total number of nodes, and estimate the most probable point and radial distance of finding an electron confined to this orbital. Compared to a 3s-orbital, is the electron further or closer to the nucleus on average? + - + - - + 1. Count the nodes and use this to ID the orbital 2. Which plot tells us about the most probable point? radius? 3. Which quantum number tells us something about size?

ℓ = 0, 1, 2, 3…..(n -1) The angular momentum quantum number (ℓ) controls shape Boundary surface diagrams – 90% probability of finding electron is enclosed by these surfaces Orbitals contains all information we want to know about electron (energy, probability) • One set of quantum numbers ({n, ℓ , m ℓ }) corresponds to one unique orbital • Energy is not dependent on ℓ for H-atom • The value of ℓ tells us what orbital shapes are allowed • Shell – set of orbitals with a common n • Subshell – set of orbitals within a shell with common ℓ (share same n and ℓ) ℓ = 0 for all s-orbitals ℓ > 0 for all other orbitals Max allowed value of ℓ is defined by n

1 Angular node (Bisect nucleus) • Plane of 0% probability of finding e - • Color change = phase change • Number of angular nodes = value of ℓ The angular momentum quantum number (ℓ) contributes to nodal structure ( Angular nodes) 1 st Shell n – 1 = 1 – 1 = 0 nodes ( ℓ = 0) n = 1 2 nd Shell 2p – subshell ( ℓ = 1) 2s – subshell ( ℓ = 0) n = 2 ℓ = 0 ℓ = 0 or ℓ = 1 1 radial node • Occurs some radial distance away from the nucleus n - 1 = 1 node Each orbital must have 1 node No other orbitals in 1 st shell No other ℓ values allowed! ℓ = angular node count 1s orbital Orbital energies identical!

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Constructive interference in bonding: Why is phase important? It contributes to the structure of chemical bonds! 0 Amplitude High Amplitude Destructive interference in bonding: Constructive interference: Destructive interference:

The magnetic quantum number describes the spatial orientation of the orbital m ℓ = - ℓ… -2, -1, 0, 1, 2... ℓ The third quantum number ( magnetic quantum number = m ℓ ) is related to ℓ and specifies spatial orientation n = 1 ℓ = 0 1 st Shell: m ℓ = 0 2 nd Shell: n = 2 ℓ = 0 m ℓ = 0 ℓ = 1 m ℓ = ___________  _________________ Same shape, different orientation Orbital lobes 𝟐? ? 𝟐? ? 𝟐? ? Named based on lobe orientation xz-plane yz- plane xy-plane Correlation between orientation and m ℓ value is beyond Gen Chem Orbitals contains all information we want to know about electron (energy, probability) • One set of quantum numbers ({n, ℓ , m ℓ }) corresponds to one unique orbital  ______________  ______________ {2, 1 , -1 } 2p orbitals: {2, 1 , 0 } {2, 1 , 1 }

Determine all possible quantum numbers for n = 3. How many orbitals does this correspond to? The Principal Quantum Number: n = 1, 2, 3, … The Angular Quantum Number: ℓ = 0, 1, 2.. .(n -1) The Magnetic Quantum Number: m ℓ = – ℓ, …, 0, … ℓ 1. Determine the range of acceptable quantum numbers 2. Systematically work through acceptable sets of numbers describing an orbital {n, ℓ, m ℓ } = One orbital

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3𝑝 𝑧 2𝑝 𝑧 Radial nodes? Angular nodes? Total nodes? Radial nodes? Angular nodes? Total nodes? Angular node “Peanut” Radial node Angular node “Peanut within a peanut” The 3p-orbitals are larger in size and higher in energy in comparison to 2p-orbitals Most prob radius: ~2.5Å ~7Å As n increases, size and number of radial nodes increases

Sketch the 4p x -orbital cross-section. 1. Recall the general shape of the p-orbital 2. What are the node counts? 3. What is the orientation? 4. Sketch the best set of axes and apply the structural information Just the boundary surface diagram!