Winter 2025 Unit 3 Electronics in Atoms Annotated Notes PDF

Unit 3 – Electrons in Atoms ©2024 McMaster University Lydia Chen, Linda Davis, Jim Ghoshdastidar & Rodrigo Vargas–Hernandez Chemistry and Chemical Biology McMaster University...

Unit 3 – Electrons in Atoms ©2024 McMaster University Lydia Chen, Linda Davis, Jim Ghoshdastidar & Rodrigo Vargas–Hernandez Chemistry and Chemical Biology McMaster University CHEM 1E03 – General Chemistry for Engineers  3.0 Unit Coverage 3.1 Light ©2024 McMaster University 3.2 Blackbody Radiation 3.5 Orbitals of Hydrogen 3.3 Photoelectric Effect 3.6 The Periodic Table 3.4 Quantum Mechanics and the 3.7 Electron Configurations Hydrogen Atom CHEM 1E03 – General Chemistry for Engineers  3.0 Chemistry Matter is made of atoms (and ions) ©2024 McMaster University Properties of atoms and ions are determined by their electronic structure – the arrangement of the electrons. CHEM 1E03 – General Chemistry for Engineers  3.0 Unit Objectives – Part 1  Apply the relationship between wavelength, frequency, and energy for light.  Describe the photoelectric effect experiment and ©2024 McMaster University apply the relationship between photon energy, threshold energy (work function) of a metal, and kinetic energy or speed of the ejected electron.  Qualitatively and quantitively describe energy transitions in the hydrogen atom, including absorption and emission.  Describe the wave-particle duality of light and of the electron, including the experimental evidence and theoretical basis. CHEM 1E03 - General Chemistry for Engineering I  ©2024 McMaster University MODULE 3.1 – INTRODUCTION TO LIGHT CHEM 1E03 – General Chemistry for Engineers Brainstorm…What do you know about Light?  Type into Box ©2024 McMaster University CHEM 1E03 – General Chemistry for Engineers  3.1 Electromagnetic Spectrum Light is an electromagnetic ©2024 McMaster University wave which carries energy. Different colors of light correspond to different wavelengths (and frequencies).  3.1 Characterization of Waves Frequency, ν, is the number of peaks passing the observation point per unit time. ©2024 McMaster University  The SI unit of frequency is Hz, (1 Hz = 1 s‒1). Wavelength, λ, is the distance between peaks measured in m.  Nanometers (1 nm = 10‒9 m) are often used in the case of visible ©2023 light.  3.1 Speed of Waves In one period, t, one Since frequency, 𝛎𝛎 = 𝟏𝟏/ 𝐭𝐭, wavelength passes an observation point. 𝝀𝝀 = 𝒄𝒄/𝝂𝝂 𝝂𝝂 = 𝒄𝒄/𝝀𝝀 ©2024 McMaster University The wave travels one wavelength, 𝝀𝝀 = 𝐜𝐜 × 𝐭𝐭, 𝒄𝒄 = 𝝀𝝀 × 𝝂𝝂 where c is the speed of ©2023 8 𝑐𝑐 = 2.9979 × 10 𝑚𝑚/𝑠𝑠 light.  3.1 Electromagnetic Spectrum Revisited Frequency and wavelength of light ©2024 McMaster University are inversely proportional The energy of light is proportional to frequency. 𝐸𝐸 = ℎ𝑣𝑣  3.1 Wave Properties of Light ©2024 McMaster University ©2023 Diffraction – the bending of Refraction – the bending of waves around corners. light as its speed changes in passing from one medium to Constructive and destructive another interference patterns. Wave properties of electromagnetic radiation. X-ray diffraction by glucose isomerase.  ©2024 McMaster University ©2023 MODULE 3.2 – BLACKBODY RADIATION CHEM 1E03 – General Chemistry for Engineers  3.2 Blackbody Radiation Blackbody Radiation – all objects emit electromagnetic radiation due to thermal energy (except at 0 K) ©2024 McMaster University This Photo by Unknown Author is licensed under CC BY CHEM 1E03 – General Chemistry for Engineers  3.2 Ultraviolet Catastrophe Classical wave theory could not ©2024 McMaster University predict observed spectra – the ultraviolet catastrophe. This Photo by Unknown Author is licensed under CC BY-SA-NC CHEM 1E03 – General Chemistry for Engineers  3.2 Max Planck and Quantization  Planck explained the observed spectrum – its exact shape – by postulating that light energy is quantized. ©2024 McMaster University  The smallest increment of light energy is the energy of a single particle of light, a photon. Max Planck The energy of one photon of light with frequency, ν, is given by 𝐸𝐸𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = ℎ𝜈𝜈 where h = 6.62606957×10‒34 Js is called Planck’s constant. CHEM 1E03 – General Chemistry for Engineers  iClicker – Light Choose the false statement: ©2024 McMaster University A. The frequency of red light is larger than that of blue light. B. The wavelength of red light is longer than that of blue light. C. The energy of red light is lower than that of blue light. CHEM 1E03 – General Chemistry for Engineers  ©2024 McMaster University MODULE 3.3 – PHOTOELECTRIC EFFECT CHEM 1E03 – General Chemistry for Engineers  3.3 The Photoelectric Effect ©2024 McMaster University Ammeter → no reading A Voltmeter → no reading Albert Einstein V Voltmeter → measures Ammeter → how many e- how fast the e- are going are ejected and travelling (Kinetic Energy) V A (Current) CHEM 1E03 – General Chemistry for Engineers  3.3 The Photoelectric Effect Shine light with a large wavelength (700 nm) on a metal surface 700 nm ©2024 McMaster University Ammeter → no reading A Voltmeter → no reading V Voltmeter → measures Ammeter → how many e- how fast the e- are going are ejected and travelling (Kinetic Energy) V A (Current) Ered light < Ethreshold  3.3 The Photoelectric Effect Let’s kick up the intensity of our 700 nm light! 700 nm Ammeter → no reading ©2024 McMaster University (no change) A Voltmeter → no reading (no change) V Voltmeter → measures Ammeter → how many e- how fast the e- are going are ejected and travelling (Kinetic Energy) V A (Current) Ered light < Ethreshold  3.3 The Photoelectric Effect What if we switch to a different colour light. Let’s try green (550 nm)! 700 nm 550 nm Ammeter → reading ©2024 McMaster University (increases!) A Voltmeter → reading (increases) V Voltmeter → measures Ammeter → how many e- how fast the e- are going are ejected and travelling (Kinetic Energy) V A (Current) Egreen light > Ethreshold  3.3 The Photoelectric Effect Now violet light (400 nm)! Let’s see what changes… 700 nm 550 nm Ammeter → reading ©2024 McMaster University 400 nm (no change compared to green light?) A Voltmeter → reading V (increases compared to green) Voltmeter → measures Ammeter → how many e- how fast the e- are going are ejected and travelling (Kinetic Energy) V A (Current) Eviolet light > Ethreshold  3.3 The Photoelectric Effect What if we increase the intensity of violet light (400 nm)? 700 nm 550 nm Ammeter → reading ©2024 McMaster University 400 nm (2x the previous reading!) A Voltmeter → reading (no change compared to less V intense light) Voltmeter → measures Ammeter → how many e- how fast the e- are going are ejected and travelling (Kinetic Energy) V A (Current) Eviolet light > Ethreshold  3.3 The Photoelectric Effect Photoelectric Effect – Sufficiently high frequency/energetic photons eject electrons from the surface ©2024 McMaster University of a metal.  Ephoton > Ethreshold  Ethreshold = work function of the metal, symbol Φ  Φ changes with different metals CHEM 1E03 – General Chemistry for Engineers  3.3 Work Function If Ephoton > Φ, then the excess energy goes towards the kinetic energy of the ejected electron ©2024 McMaster University 1 𝐸𝐸𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = ℎ𝜈𝜈 = 𝑚𝑚𝑢𝑢2 + 𝜙𝜙𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 2 where m and u are the electron mass and velocity, respectively, and ϕmetal is called the work function of the metal. Exactly one electron is ejected from the metal surface for every one photon of light (provided it is energetic enough).  3.3 Work Function  Alkali metals have the smallest work functions - it ©2024 McMaster University takes less energy to eject an electron from an alkali metal.  According to decreasing work function,  lithium > sodium > potassium > rubidium > cesium. CHEM 1E03 – General Chemistry for Engineers  Example – Threshold Frequency The work function for calcium is 4.64 × 10‒19 J (or 279 kJ mol‒1 – this is the energy needed to eject 1 mol of electrons). a) What is the threshold frequency above which calcium will exhibit the ©2024 McMaster University photoelectric effect? Givens Required Equations 𝜙𝜙𝐶𝐶𝐶𝐶 = 4.64 × 10−19 𝐽𝐽 𝜈𝜈𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = ? [𝐻𝐻𝐻𝐻] 𝜙𝜙𝐶𝐶𝐶𝐶 = 𝐸𝐸𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = ℎ𝜈𝜈 Solution 𝜙𝜙𝐶𝐶𝐶𝐶 4.64 × 10−19 𝐽𝐽 ℎ𝜈𝜈 = 𝜙𝜙𝐶𝐶𝐶𝐶 = 4.64 × 10−19 𝐽𝐽 𝜈𝜈 = = ℎ 6.626 × 10−34 𝐽𝐽 𝑠𝑠 𝜈𝜈 = 7.00 × 1014 𝐻𝐻𝐻𝐻 = 700. 𝑇𝑇𝑇𝑇𝑇𝑇  Example – Kinetic Energy The work function for calcium is 4.64 × 10‒19 J (or 279 kJ mol‒1 – this is the energy needed to eject 1 mol of electrons). b) What is the kinetic energy of an electron ejected from calcium by a ©2024 McMaster University photon of wavelength, 300. nm? Givens Required Equations 1 −9 𝜆𝜆 = 300 × 10 𝑚𝑚 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘 = ? [𝐽𝐽] 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘 = 𝑚𝑚𝑢𝑢2 = ℎ𝜈𝜈 − 𝜙𝜙𝐶𝐶𝐶𝐶 2 𝜙𝜙𝐶𝐶𝐶𝐶 = 4.64 × 10−19 𝐽𝐽 Solution 𝑐𝑐 2.9979 × 108 𝑚𝑚/𝑠𝑠 𝜈𝜈 = = −9 = 999.3 𝑇𝑇𝑇𝑇𝑇𝑇 𝜆𝜆 300 × 10 𝑚𝑚 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘 = (6.626 × 10−34 𝐽𝐽 𝑠𝑠)(999 × 1012 1/𝑠𝑠) − (4.64 × 10−19 𝐽𝐽) 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘 = 1.98 × 10−19 𝐽𝐽  iClicker – Photoelectric Effect The work function for calcium is 4.64 × 10‒19 J (or 279 kJ mol‒1 – this is the energy needed to eject 1 mol of electrons). c) What is the velocity of an electron ejected from calcium by a ©2024 McMaster University photon of wavelength, 300. nm? 𝑚𝑚𝑒𝑒 − = 9.109 × 10−31 𝑘𝑘𝑘𝑘 A. 1.52 × 10-6 m/s B. 4.66 × 105 m/s C. 6.59 × 105 m/s D. 4.35 × 1011 m/s CHEM 1E03 – General Chemistry for Engineers iClicker – Photoelectric Effect (Solution) The work function for calcium is 4.64 × 10‒19 J (or 279 kJ mol‒1 – this is the energy needed to eject 1 mol of electrons). c) What is the velocity of an electron ejected from calcium by a ©2024 McMaster University photon of wavelength, 300. nm? 𝑚𝑚𝑒𝑒 − = 9.109 × 10−31 𝑘𝑘𝑘𝑘 A. 1.52 × 10-6 m/s 1 2𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘 B. 4.66 × 105 m/s 𝐸𝐸𝑘𝑘𝑘𝑘𝑘𝑘 = 𝑚𝑚v 2 =v 2 𝑚𝑚𝑒𝑒 − C. 6.59 × 105 m/s 𝑚𝑚2 𝑘𝑘𝑘𝑘 2 2(1.98 × 10 −19 𝐽𝐽) 𝑠𝑠 𝑚𝑚 D. 4.35 × 1011 m/s v= = 6.59 × 105 9.109 × 10−31 𝑘𝑘𝑘𝑘 𝑠𝑠 CHEM 1E03 – General Chemistry for Engineers  Unit 3 – Electrons in Atoms ©2024 McMaster University Lydia Chen, Linda Davis, Jim Ghoshdastidar & Rodrigo Vargas–Hernandez Chemistry and Chemical Biology McMaster University CHEM 1E03 – General Chemistry for Engineers  3.0 Unit Coverage 3.1 Light ©2024 McMaster University 3.2 Blackbody Radiation 3.5 Orbitals of Hydrogen 3.3 Photoelectric Effect 3.6 The Periodic Table 3.4 Quantum Mechanics and the 3.7 Electron Configurations Hydrogen Atom CHEM 1E03 – General Chemistry for Engineers  ©2024 McMaster University MODULE 1.4 – THE HYDROGEN ATOM CHEM 1E03 – General Chemistry for Engineers  Dalton’s Atom – single, indivisible entity unique for each element, cannot be altered or destroyed and combines to more complex compounds John Dalton BUT there are these negatively 1803 charged particles called Electrons! Democritus ©2024 McMaster University Need a new theory! 370 BC Plum Pudding Model – negative charged “corpuscles” (electrons) embedded in a sea of positive charge BUT there we fire alpha particles at gold foil, they get deflected! Need a new theory! J.J. Thompson 1906 Nuclear Model – highly concentrated positive charge in Ernest Rutherford a very small volume which contained the bulk of the 1908 atom’s mass surrounded by orbiting electrons  3.4 Emission Spectra of Elements  Each element produces light ©2024 McMaster University with a unique emission spectra  But how is this explained by what we know about the atom? CHEM 1E03 – General Chemistry for Engineers Problem: any charged particle moving in a circular orbit continuously loses energy and will eventually spiral into the ©2024 McMaster University Heisenberg & nucleus. Bohr Bohr’s solution: Combine the model of the atom with Planck’s ideas of quantized energies. CHEM 1E03 – General Chemistry for Engineers  3.4 Discrete Energy Levels Quantum mechanics tells us that an atom has a discrete set of ©2024 McMaster University energy levels  i.e., only certain energies are allowed. Quantization of energy arises from wave properties of matter described by quantum mechanics. CHEM 1E03 – General Chemistry for Engineers  3.4 Absorption of Electrons  Atoms can also absorb photons of n=∞ … light. ©2024 McMaster University  An electron can be ejected from the atom - this is called ionization).  For lower energy photons (i.e., lower frequency), an excited state of the atom can be produced. CHEM 1E03 – General Chemistry for Engineers  3.4 Transition Between Energy Levels absorption emission  Electrons do not lose energy n=∞ … when at a given energy level. ©2024 McMaster University  They can transition from one energy level by absorbing or releasing a packet of energy  Electrons cannot exist in- between the allowed energy levels. CHEM 1E03 – General Chemistry for Engineers  3.4 Hydrogen Atom  The hydrogen atom – the simplest atom with only one ©2024 McMaster University electron – has the simplest set of energy levels, 𝑅𝑅𝐸𝐸 𝐸𝐸𝑛𝑛 = − 2 , 𝑛𝑛 = 1,2,3, … , ∞ 𝑛𝑛 where Re = 2.17987 x 10–18 J is the Rydberg Energy and n is the principal quantum number/energy level. CHEM 1E03 – General Chemistry for Engineers  3.4 Energy of Electron Transition The difference between two of these energies determines a transition energy equal to the energy of the photon emitted or absorbed. ©2024 McMaster University ∆E > 0: absorption of a photon ∆E < 0: emission of a photon 1 1 ∆𝐸𝐸 = 𝐸𝐸𝑓𝑓 − 𝐸𝐸𝑖𝑖 = −𝑅𝑅𝐸𝐸 2 − 2 = ℎ𝑣𝑣 𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖 Ephoton = hν =│∆E│, always positive CHEM 1E03 – General Chemistry for Engineers  Example – Hydrogen Atom Determine the frequency of the emission lines resulting when an excited state with ni = 3 drops to the (i) nf = 2 , or (ii) nf = 1 level. Are either of these emissions in the visible range? ©2024 McMaster University Givens Required Equations 𝑛𝑛𝑖𝑖 = 3 𝑣𝑣 = ? [𝐻𝐻𝐻𝐻] 𝑅𝑅𝐸𝐸 1 1 𝑛𝑛𝑓𝑓 = 2 𝑣𝑣 = 2 − 2 ℎ 𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖 2.17987 × 10−18 𝐽𝐽 1 1 𝑣𝑣 = 4.57 × 1014 1/𝑠𝑠 Solution 𝑣𝑣 = − 2 6.626 × 10 −34 𝐽𝐽 𝑠𝑠 22 3 𝑣𝑣 = 457 𝑇𝑇𝑇𝑇𝑇𝑇 𝑐𝑐 2.9979 × 108 𝑚𝑚/𝑠𝑠 𝜆𝜆 = = 14 = 656 𝑛𝑛𝑛𝑛 Balmer Series 𝜈𝜈 4.57 × 10 1/𝑠𝑠  Example – Hydrogen Atom Determine the frequency of the emission lines resulting when an excited state with ni = 3 drops to the (i) nf = 2 , or (ii) nf = 1 level. Are either of these emissions in the visible range? Equations ©2024 McMaster University Givens Required 1 1 1 𝑛𝑛𝑖𝑖 = 3 𝜆𝜆 = ? [𝑛𝑛𝑛𝑛] = 𝑅𝑅∞ 2 − 2 𝜆𝜆 𝑛𝑛2 𝑛𝑛1 𝑛𝑛𝑓𝑓 = 1 𝑛𝑛1 > 𝑛𝑛2 1 1 1 Solution 7 = 1.097 × 10 𝑚𝑚 −1 − 2 1 𝜆𝜆 12 3 = 9.754 × 10−6 𝑚𝑚−1 𝜆𝜆 𝜆𝜆 = 1.0251 × 10−7 𝑚𝑚 = 103 𝑛𝑛𝑛𝑛 Lyman Series  Try this at Home! – Relationship between 𝑅𝑅∞ and 𝑅𝑅𝐸𝐸 We used two different equations to solve for wavelength of a photon emitted – one that involves Rydberg’s Equation (RE) and another than uses Rydberg’s Constant (𝑅𝑅∞ ) ©2024 McMaster University 𝑅𝑅𝐸𝐸 1 1 1 1 1 𝑣𝑣 = 2 − 2 = 𝑅𝑅∞ 2 − 2 ℎ 𝑛𝑛𝑓𝑓 𝑛𝑛𝑖𝑖 𝜆𝜆 𝑛𝑛2 𝑛𝑛1 ∗ 𝐹𝐹𝐹𝐹𝐹𝐹 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑛𝑛1 > 𝑛𝑛2 Using these two equations can you determine the relationship between these two constants? How does 𝑛𝑛1 > 𝑛𝑛2 relate to nf and ni? CHEM 1E03 – General Chemistry for Engineers  Example – Hydrogen Atom Of all the absorption frequencies associated ©2024 McMaster University with transitions between all levels up to n = 4, which transitions have the largest and the smallest frequencies? CHEM 1E03 – General Chemistry for Engineers  iClicker – Bohr Model Which requires greater energy, an electron promoted from n=1 to n=2, or an electron promoted from n=3 to n = 4? ©2024 McMaster University Which releases a photon with greater frequency, an n = 5 to n = 4 transition or an n = 3 to n = 2 transition? Which releases a photon with greater wavelength, an n = 5 to n = 4 transition or an n = 3 to n = 2 transition? CHEM 1E03 – General Chemistry for Engineers  iClicker – Bohr Model (Solutions) Which requires greater energy, an electron promoted from n=1 to n=2, or an electron promoted from n=3 to n = 4? ©2024 McMaster University Which releases a photon with greater frequency, an n = 5 to n = 4 transition or an n = 3 to n = 2 transition? Which releases a photon with greater wavelength, an n = 5 to n = 4 transition or an n = 3 to n = 2 transition? CHEM 1E03 – General Chemistry for Engineers  Demonstration – Bowls of Flame Why do the flames ©2024 McMaster University of different elements generate different colours of light? CHEM 1E03 – General Chemistry for Engineers  Unit 3 – Electrons in Atoms ©2024 McMaster University Lydia Chen, Linda Davis, Jim Ghoshdastidar & Rodrigo Vargas–Hernandez Chemistry and Chemical Biology McMaster University CHEM 1E03 – General Chemistry for Engineers  3.0 Unit Coverage 3.1 Light ©2024 McMaster University 3.2 Blackbody Radiation 3.5 Orbitals of Hydrogen 3.3 Photoelectric Effect 3.6 The Periodic Table 3.4 Quantum Mechanics and the 3.7 Electron Configurations Hydrogen Atom CHEM 1E03 – General Chemistry for Engineers  3.0 Unit Objectives  Use the four quantum numbers to describe atomic orbitals, electrons in orbitals, and electron configurations. ©2024 McMaster University  Define and apply the three basic principles governing electron configuration: the Aufbau Principle, Hund’s Rule, and the Pauli Exclusion Principle.  Describe how orbital energies are modified when more than one electron is present in an atom.  Determine the number of unpaired electrons in a main group (s- or p-block) atom or ion from its electron configuration; use the terms “paramagnetic” or “diamagnetic” to describe the atom or ion. CHEM 1E03 - General Chemistry for Engineering I  Demonstration – Bowls of Flame Explained  We are observing electronic ©2024 McMaster University transitions.  The emitted light consists of certain discrete wavelengths. CHEM 1E03 – General Chemistry for Engineers 3.4 Wait…electrons are waves now? ©2024 McMaster University Electron diffraction pattern in Ni2MnIn The Thomsons alloy Together J. J. Thomson and his son G. P. Thomson demonstrated the wave-particle duality of electrons. J. J. Thomson won the Nobel Prize in Physics in 1906 and G. P. won the award in 1937.  3.4 Wavelength of Matter  Electrons & nuclei, & light are all treated as wave-like & particle-like. ℎ ©2024 McMaster University  de Broglie first suggested that 𝜆𝜆 = particles, such as electrons, are wave- 𝑚𝑚𝑚𝑚 like, characterized by a wavelength … Louis de Broglie  The wavelength of matter (in meters) is equal to the ratio of Planck’s constant over the product of the mass (in kg) times the velocity (in m/s). CHEM 1E03 – General Chemistry for Engineers  3.4 Heisenberg Uncertainty Principle  There is uncertainty in the position of an electron in an atom.  One can only speak of the probability of finding the electron in a ©2024 McMaster University volume within the atom. ℎ ∆𝜒𝜒Δ𝜌𝜌 ≥ 4𝜋𝜋 Werner Heisenberg CHEM 1E03 – General Chemistry for Engineers  Example – de Broglie Wavelength In a photoelectric effect experiment, photons with a wavelength of 40.0 nm strike a gold metal surface (Ethreshold = 8.15 x 10-19 J), causing photoelectrons to be ejected with a speed of 3.02 x 106 m/s. What is the ©2024 McMaster University de Broglie wavelength (in nm) of the emitted photoelectrons? Mass of electron = 9.11 x 10-31 kg 𝑚𝑚2 𝑘𝑘𝑘𝑘 2 𝑠𝑠 ℎ 6.626 × s 10−34 J λ= = A. 0.241 nm 𝑚𝑚𝑚𝑚 (9.11 × 10−31 kg)(3.02 × 106 m/s) B. 2.46 nm C. 4.15 nm λ = 2.41 × 10−10 m = 0.241 nm D. 338 nm Notice there are two different formula for wavelength: λ=ν 𝑐𝑐 applies to light (any electromagnetic radiation) ℎ λ = 𝑚𝑚𝑚𝑚 applies to matter  AN ELECTRON ©2024 McMaster University MODULE 3.5 – ORBITALS OF HYDROGEN CHEM 1E03 – General Chemistry for Engineers  3.5 Wave Functions  The solutions to Schrodinger’s Equation correspond to a wave function, Ψ, which (together with the spin) describes the state of the electron in the atom as a wave ©2024 McMaster University  The square of the wave function, Ψ2, at each position gives the likelihood of finding the electron in that location.  The region of space where we find an electron 90% of the time defines the boundaries of the orbital CHEM 1E03 – General Chemistry for Engineers  3.5 Principal Quantum Number, n  The Principal Quantum Number describes the energy and distance of the electron from the nucleus ©2024 McMaster University  It is analogous to the energy levels in the Bohr Model of the Atom  The greater the value of n, the further the electron is from the nucleus and the more energy it has n = 1, 2, …, ∞ CHEM 1E03 – General Chemistry for Engineers  3.5 Angular Momentum Number, ℓ  Because of the spherical symmetry of the hydrogen atom (and electron spin), there is more than one state for each energy level. ©2024 McMaster University  The electron in hydrogen has an orbital angular momentum associated as it orbits about the proton nucleus.  Angular momentum (azimuthal) has a magnitude, with associated quantum number - angular momentum quantum number ℓ = 0, 1, 2, …, n ‒1 CHEM 1E03 – General Chemistry for Engineers  3.5 Magnetic Quantum Number, mℓ  Angular momentum has an orientation-associated quantum number - magnetic quantum number ©2024 McMaster University  Orbitals with the same angular momentum number are degenerate in energy mℓ = ‒ℓ, ‒ℓ + 1, …, ℓ ‒1, ℓ. CHEM 1E03 – General Chemistry for Engineers  3.5 s orbitals  The ℓ = 0 orbitals have spherical shape. mℓ = 0 ©2024 McMaster University  Between nodes is a nodal sphere  The ns orbital has n – 1 nodal spheres. CHEM 1E03 – General Chemistry for Engineers  3.5 p orbitals  The ℓ = 1 orbitals have dumbell shape. ©2024 McMaster University  Nodal plane between positive and negative lobes  Three degenerate np  The np orbital has n – 2 orbitals mℓ = -1, 0, +1 nodal spheres. CHEM 1E03 – General Chemistry for Engineers  3.5 d orbitals  The ℓ = 2 orbitals have clover shape. ©2024 McMaster University  Nodal plane between positive and negative lobes  Five degenerate np orbitals mℓ = -2, -1, 0, +1, +2 CHEM 1E03 – General Chemistry for Engineers  3.5 Specifying Orbitals Specifying the (n, ℓ, mℓ) values determines an orbital of the hydrogen atom, which determines the state of the electron with respect to its motion about the nucleus. ©2024 McMaster University For n = 1, there is only one ℓ value, ℓ = 0. For ℓ = 0, there is only one mℓ value, mℓ = 0. Therefore, the n = 1 energy level has only one orbital – the 1s orbital. CHEM 1E03 – General Chemistry for Engineers  3.5 Specifying Orbitals Specifying the (n, ℓ, mℓ) values determines an orbital of the hydrogen atom, which determines the state of the electron with respect to its motion about the nucleus. ©2024 McMaster University For n = 2, there are two ℓ values, ℓ = 0 and 1 For ℓ = 0, we have only (n, ℓ, mℓ) = (2, 0, 0) (the 2s orbital). For ℓ = 1, we have (2, 1, ‒1), (2, 1, 0) and (2, 1, 1), the three 2p orbitals. CHEM 1E03 – General Chemistry for Engineers  3.5 Spin Quantum Number In addition to the orbital quantum numbers already introduced, the electron has a spin quantum number, ©2024 McMaster University  ms = ½ or ‒½, spin angular momentum pointing up or down, respectively. There are two states associated with the lowest energy level:  (n, ℓ, mℓ, ms) = (1, 0, 0, ½) and (1, 0, 0, ‒½) CHEM 1E03 – General Chemistry for Engineers 3.5 Hydrogen Energy Levels 𝑁𝑁𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑛𝑛2 ©2024 McMaster University CHEM 1E03 – General Chemistry for Engineers  3.5 Orbitals and States of Hydrogen Number of For every shell n, n ℓ Number of States Orbitals total total 𝑁𝑁𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑛𝑛 2 ©2024 McMaster University 1 0 1 1 2 2 𝑁𝑁𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 𝑛𝑛 0 1 2 2 2 1 3 4 6 8 𝑁𝑁𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 2𝑛𝑛 0 1 2 3 1 3 6 For every subshell ℓ, 2 5 9 10 18 0 1 2 𝑁𝑁𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 = 2ℓ+1 1 3 6 4 2 5 10 𝑁𝑁 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 = 2(2ℓ+1) 3 7 16 14 32 CHEM 1E03 – General Chemistry for Engineers  Example – Quantum Numbers (a) Which of the following sets of quantum numbers, (n, ℓ, mℓ, ms), corresponds to an actual state of a hydrogen atom? ©2024 McMaster University (i) (1, 0, 1, ½) (ii) (2, 1, ‒1, ‒½) (iii) (2, 2, 0, ‒½) (iv) (3, 1, ‒1, 0) (v) (3, ‒2, ‒2, ½) CHEM 1E03 – General Chemistry for Engineers  Example – Quantum Numbers (a) Which of the following sets of quantum numbers, (n, ℓ, mℓ, ms), corresponds to an actual state of a hydrogen atom? ©2024 McMaster University (i) (1, 0, 1, ½) violates mℓ ≤ ℓ. (ii) (2, 1, ‒1, ‒½) (iii) (2, 2, 0, ‒½) violates ℓ ≤ n – 1 (iv) (3, 1, ‒1, 0) violates ms = ±½ (v) (3, ‒2, ‒2, ½) violates 0 ≤ ℓ CHEM 1E03 – General Chemistry for Engineers  Unit 3 – Electrons in Atoms ©2024 McMaster University Lydia Chen, Linda Davis, Jim Ghoshdastidar & Rodrigo Vargas–Hernandez Chemistry and Chemical Biology McMaster University CHEM 1E03 – General Chemistry for Engineers  3.0 Unit Coverage 3.1 Light ©2024 McMaster University 3.2 Blackbody Radiation 3.5 Orbitals of Hydrogen 3.3 Photoelectric Effect 3.6 The Periodic Table 3.4 Quantum Mechanics and the 3.7 Electron Configurations Hydrogen Atom CHEM 1E03 – General Chemistry for Engineers Brainstorm…What did you learn about orbitals?  Type student answers here ©2024 McMaster University CHEM 1E03 – General Chemistry for Engineers  Example – Quantum Numbers (b) How many orbitals are there with energy up to and including n=4 or E4? How many electron states are there with energy up to this value? ©2024 McMaster University Energy Level (n) # Orbitals Total Orbitals Electron States n=1 1s 1 2 n=2 1s + 3p 4 8 n=3 1s + 3p + 5d 9 18 n=4 1s + 3p + 5d + 7f 16 32 Total 30 Orbitals 60 Electron States 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇 𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂𝑂 𝑓𝑓𝑓𝑓𝑓𝑓 𝐸𝐸𝑛𝑛 = 𝑛𝑛2 3.5 Radial Probability Distributions ©2024 McMaster University  Spotlight in Engineering – Electron Corral The corral is an artificial structure created from 48 iron atoms (the sharp peaks) on a copper surface. The wave patterns in this scanning tunneling microscope image are formed by copper electrons confined by the iron atoms. ©2024 McMaster University Don Eigler and colleagues created this structure in 1993 by using the tip of a low-temperature scanning tunneling microscope (STM) to position iron atoms on a copper surface, creating an electron-trapping barrier. This was the first successful attempt at manipulating individual atoms and led to the development of new techniques for nanoscale construction. Question – What is the energy level of this electron? Don Eigler, IBM Almaden Research Center  Spotlight in Engineering – Electron Corral Question – What is the energy level of this electron? ©2024 McMaster University  STM detects electron density which is the square of the electron wave’s amplitude. But nodes will still be present at 0.  The total nodes is n – 1 (dark regions)  There are four nodes/dark regions in the image  Therefore, the energy level of the electron is n = 5 Don Eigler, IBM Almaden Research Center  ©2024 McMaster University MODULE 3.6 – PERIODIC TABLE CHEM 1E03 – General Chemistry for Engineers  3.6 One Electron Systems  Ions with only one electron – i.e., He+, Li2+,…, have the same orbitals as hydrogen, except that they held more ©2024 McMaster University closely to the nucleus by the larger nuclear charge. 𝑍𝑍 2 𝑅𝑅𝐸𝐸 𝐸𝐸𝑛𝑛 = − 2 𝑛𝑛 = 1,2,3, … , ∞ 𝑛𝑛  The energy levels of He+ are four times deeper than those of H because the nuclear charge is doubled. CHEM 1E03 – General Chemistry for Engineers  3.6 Multielectron Systems  Electron repulsion makes calculating the states and energy levels difficult – computers approximate this ©2024 McMaster University  Each electron is assigned to an orbital  Pauli Exclusion Principle – two electrons with the same spin cannot occupy the same orbital.  Ground State Electron Configuration – set of occupied orbitals filled in order of increasing energy CHEM 1E03 – General Chemistry for Engineers  3.6 Ground State Electron Configurations Aufbau’s Principle – the occupation of electrons in order of increasing ©2024 McMaster University orbital energy 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p 7s Famous Chemists – Prof Jaime Keller (Mexico) ©2024 McMaster University Prof. Jaime Keller (1936 – 2011) CHEM 1E03 – General Chemistry for Engineers  3.6 Orbital Energies But why do the 2s and Energy 2p orbitals have different energies? ©2024 McMaster University 2s 2p 2p 2s 2p 1s 2s 2p has higher energy H ground state than 2s because the two 1s electrons screen 1s a 2p electron better He ground state than a 2s electron. 1s CHEM 1E03 – General Chemistry for Engineers Li ground state 3.6 Nuclear Shielding and Penetration  For one electron species, the greater the nuclear charge (Z), the lower the energy of ©2024 McMaster University the electron.  For multielectron species, electrons repulsions destabilize the electrons (shielding – S) and increase their potential energy CHEM 1E03 – General Chemistry for Engineers 3.6 Nuclear Shielding and Penetration  Inner electrons shield outer electrons from the full attractive charge from the ©2024 McMaster University nucleus (effective nuclear charge) Zeff = Z – S  Orbitals that penetrate closer to the nucleus experience a higher effective nuclear charge and reduced shielding (s > p > d > f) CHEM 1E03 – General Chemistry for Engineers  3.6 Hund’s Rule Hund’s Rule – Energy electrons ©2024 McMaster University occupy 2s 2p 2p 2s 2p degenerate 2s orbitals singly 1s first to avoid Be ground state repulsions 1s (pairing energy) B ground state 1s C ground state CHEM 1E03 – General Chemistry for Engineers  Example – Electron Configurations 1s2 2s2 2p3 A. N ©2024 McMaster University B. Ne C. P D. Ca E. V CHEM 1E03 – General Chemistry for Engineers  Example – Electron Configurations 1s2 2s2 2p63s2 3p3 A. N ©2024 McMaster University B. Ne C. P D. Ca E. V CHEM 1E03 – General Chemistry for Engineers  Example – Electron Configurations A. N ©2024 McMaster University 1s2 2s2 2p63s2 3p64s2 3d3 B. Ne C. P D. Ca E. V CHEM 1E03 – General Chemistry for Engineers  Example – Electron Configurations To which element does ©2024 McMaster University the following atomic ground state electron configuration correspond? 1s22s22p63s23p64s23d7 There are 27 electrons in this atom. Therefore, it is cobalt (Z =  3.6 Electron Configuration Exceptions These are abbreviated Electron Configurations  Cr: 1s22s22p63s23p64s13d5 [Ar]4s13d5 ©2024 McMaster University where the core electrons are represented by the electron configuration of a Noble Gas  half-filled 3d subshell  Cu: 1s22s22p63s23p64s13d10 [Ar] 4s13d10  filled 3d subshell CHEM 1E03 – General Chemistry for Engineers  3.6 Diamagnetic vs Paramagnetic  Paramagnetic – atoms with unpaired electrons that are attracted to magnetic fields 1s 2s 2p ©2024 McMaster University  C:[He] 2s2 2p2  F: [He] 2s2 2p5 We can also use box diagrams to represent electron configurations.  Diamagnetic – atoms with entirely paired electrons that are weakly repelled by magnetic fields  Be: [He] 2s2 and Ne: [He] 2s2 2p6 = [Ne] CHEM 1E03 – General Chemistry for Engineers  3.6 Periodic Table Blocks ©2024 McMaster University s Block p Block d Block f Block 3.6 Development of the Periodic Table Dmitri Mendeleev (1834 – 1907 CE) formulated the Periodic Law and the Periodic Table of Elements which predicted elements yet discovered ©2024 McMaster University Dmitri Mendeleev’s periodic table from 1871. Mendeleev organized the known elements into eight groups, in order of increasing mass. He left gaps for undiscovered elements and predicted their properties  Dalton’s first list of elements from 1803 consisted of 5 elements. His second list in 1808 featured 20 elements with their atomic weights. John Dalton Berzelius sought to accurately measure the atomic ©2024 McMaster University weights of the elements publishing his list in 1828. He established the system of chemical symbols. Jons Berzelius Döbereiner tried to find order in the chemical Johann properties of elements. He identified groups of three Döbereiner chemically similar elements called triads. John Newlands ordered the 56 elements by atomic weights and found recurring properties every eight elements known as the Law of Octaves. John Newlands  Unit 4 – Periodic Trends ©2024 McMaster University Lydia Chen, Linda Davis, Jim Ghoshdastidar & Rodrigo Vargas–Hernandez Chemistry and Chemical Biology McMaster University CHEM 1E03 – General Chemistry for Engineers