CHEM 251 1st Edition Lecture 3 Outline of Last Lecture I. Clicker QuestionII. Periodic Trends (Part 2)III. Overview of Global EnergyOutline of Current Lecture IV. Clicker QuestionV. Quantum NumbersVI. Orbital StructureVII. Electron ConfigurationCurrent LectureIV. Clicker Questiona. Which of the following is in order of increasing ionization energy? (Note: Ca+ denotes the second ionization energy of Ca. All others are first ionization energy)i. Ca+ < Na < Mg < Pii. Na < Ca+ < P < Mgiii. P < Mg < Na < Ca+iv. Na < Mg < P < Ca+ b. (iv) is correct because Ca+’s second ionization energy is much higher than the firstc. What orbital is represented by the radial distribution function below?i. 1sii. 2siii. 2piv. 3dd. (iii) is correct because the node at zero tells us it is a p-orbitalV. Quantum Numbers: define orbital structureThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.a. n = principal quantum number (1 => infinity) b. l = angular momentum quantum number (0 => n-1)c. ml = magnetic quantum number (-l => +l)d. ms = spin quantum number (-1/2 or +1/2)VI. Orbital Structurea.b. n = 1, 1 orbitalc. n = 2, l = 1 exists, leading to ml states and p-orbitalsd. Energy orderinge. Aufbau principle: electrons fill lowest energy orbitals firstf. Relationship to periodic tablei. Periods defined by principle quantum number (n)ii. Blocks reflect sub-shell (l)iii. Group reflects # of electrons in shellg. Shapesi. S-orbital1. non-zero probability of finding electron inside the nucleus2. bigger n gives broader distributions w/ max probability farther from nucleusii. P-orbitals1. Each p-orbital has one node2. Three equivalent orbitals3. Two lobes w/ different wavesign functions4. We use simplified representationsiii. D-orbitals1. Each has 2 nodes2. Five equivalent orbitals3. Not all look the sameVII. Electron Configurationa. Electrons want to fill lowest energy stateb. No 2 electrons can have same 4 quantum #sc. Example: carbon (6 electrons)i. 1s22s22p2 ii. Shorthand/Noble gas configuration: [He] 2s22p2 d. Clicker questioni. Noble gas electronic configuration of Nbii. Answer: [Kr]5s14d4iii. Why? 1. 2nd electron added to 4s-orbital is higher in energy2. 3d orbital energies fall more sharply than 4s orbital energiesAcross periodiv. Electron-electron repulsion vs. populating higher energy orbital1. Costs pairing energy to populate two electrons in single orbital2. Costs energy to populate high energy
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