Chem 1320 1st Edition Lecture 20Outline of Last Lecture I. Electron Configurations ContinuedII. Periodic PropertiesOutline of Current Lecture I. Periodic Properties ContinuedII. Underlying ConceptIII. Orbital EnergiesIV. Ionization EnergiesCurrent LectureI. Periodic Properties ContinuedThere are different types of configurations. The arrangement of valence electrons is related to the chemical and physical properties of an element. The periodic table is organized which reflects common trends. II. Underlying Concept This is referring to the effective nuclear charge and orbital size. It starts with the attraction to the nucleus. Then remember that the orbital size and distance is important. Lastly it will experience screening or shielding. Zeff= effect nuclear charge (charge felt by valence electrons) O= shielding constant (unique to an element) Zeff (H) = 1.28 Zeff (Cs) = 6.63O = - number o = + numberThese 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.The lower you go in the periodic table, the greater the Zeff is. The more to the left an element is on the periodic table, the greater the Zeff is. The more across and left an element is on the periodic table, the greater the Zeff is. III. Orbital EnergiesThis is the probability of finding the electron at distance r form nucleus as a function of r. IV. Ionization EnergiesThis is also referred to as ionization potential. This defines the reactions. I stands for ionization energy. X(9) + energy X(9) + electrons. This measures how tightly an electron is held onto an atom. Example: Mg[Ne] 3s2; Mg(9) + energy Mg(9)+ + electronsI= 738.1 KJ/mol[Ne] 3s1 ; Mg(9)+ + energy Mg(9) 2+ + electrons I= 1450 KJ/molThis means that it took less energy per electron with I2 because it goes to a stable state The general trend I1 < I2 As the periodic table goes up, so does the ionization energy. The farther left the element is on the periodic table, the greater the ionization energy. Example: B 1s2 2s2 2p1I1 = 801 KJ/molI2 = 2430 KJ/ mol trying to remove electrons from filled shellI3 = 3600
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