Chapter 81 Atomic Electronic Configurations and Chemical Periodicity Chapter 8

2 Electron Spin Effective Nuclear Charge Effective nuclear charge - The charge experienced by an electron in a many-electron atom.

Chapter 83 Electron Spin Effective Nuclear Charge -Electrons are attracted to the nucleus, but repelled by the electrons that screen it from the nuclear charge. -The nuclear charge experienced by an electron depends on its distance from the nucleus and the number of core electrons. -As the average number of screening electrons (S) increases, the effective nuclear charge (z eff ) decreases. -As the distance from the nucleus increases, S increases and z eff decreases. z eff = z - S

Chapter 84 Electron Spin Pauli Exclusion Principle - Pauli’s Exclusion Principle - no two electrons can have the same set of 4 quantum numbers. -Therefore, two electrons in the same orbital must have opposite spins. -One electron has a spin of +½, the other –½.

Chapter 85 Electron Spin Magnetism Diamagnetic – Property of a substance that is repelled by a magnetic field. Paramagnetic – Property of a substance that is attracted by a magnetic field. Whether a substance is diamagnetic or paramagnetic depends on its electron configuration. -Atoms without “paired” spins result in a paramagnetic substance.

Chapter 86 Electron Assignment Orbitals and Quantum Numbers -Orbitals can be ranked in terms of energy to yield an orbital energy diagram. -As n increases, note that the spacing between energy levels becomes smaller. -As shielding increases the energy of an orbital increases.

Chapter 87 Electron Assignment

Chapter 88 Electron Configuration Electron configurations tells us in which orbitals the electrons for an element are located. Three rules: -Electrons fill orbitals starting with lowest n and moving upwards. -no two electrons can fill one orbital with the same spin (Pauli Exclusion Principle). -for degenerate orbitals, electrons fill each orbital singly before any orbital gets a second electron (Hund’s rule).

Chapter 89 Electron Configuration

Chapter 810 -There is a shorthand way of writing electron configurations -Write the core electrons corresponding to the filled Noble gas in square brackets. -Write the valence electrons explicitly. Example, P: 1s 2 2s 2 2p 6 3s 2 3p 3 but Ne is 1s 2 2s 2 2p 6 Therefore, P: [Ne]3s 2 3p 3. Electron Configuration

Chapter 811 Electron Configuration of Ions Cations: To form a cation, an electron from the highest principle quantum number and angular momentum quantum number is removed. Mg: [ 1s 2 2s 2 2p 6 3s 2 ]  Mg 2+ : [1s 2 2s 2 2p 6 ] + 2e- Anions: To form an anion, an electron is added to the highest principle quantum number and angular momentum quantum number. Cl: [1s 2 2s 2 2p 6 3s 2 3p 5 ] + 1e-  Cl-: [1s 2 2s 2 2p 6 3s 2 3p 6 ]

Chapter 812 Atomic Properties and Periodic Trends

Chapter 813 Atomic Properties and Periodic Trends Electron Shells in Atoms -Elements in the same column have the same electron configuration. Consider:Ne: 1s 2 2s 2 2p 6 Ar: 1s 2 2s 2 2p 6 3s 2 3p 6 Both elements have the same electron configuration: [Element]ns 2 np 6 -Therefore, the elements in the periodic table should exhibit regular variations in there physical properties.

Chapter 814 -Atomic size varies consistently through the periodic table. -As we move down a group, the atoms become larger. -As we move across (left to right) a period, atoms become smaller. -There are two factors at work: -principal quantum number, n -the effective nuclear charge, z eff Atomic Properties and Periodic Trends Atomic Size

Chapter 815 -As the principle quantum number increases, the distance of the outermost electron from the nucleus becomes larger. Hence, the atomic radius increases. - As we move across the periodic table, there is an increased attraction between the nucleus and the outermost electrons. This attraction causes the atomic radius to decrease. Atomic Properties and Periodic Trends Atomic Size

Chapter 816 Atomic Properties and Periodic Trends Atomic Size

Chapter 817 First Ionization Energy - The first ionization energy, I 1, is the amount of energy required to remove an electron from a gaseous atom: Na(g)  Na + (g) + e - -The larger ionization energy, the more difficult it is to remove the electron. -There is a sharp increase in ionization energy when a core electron is removed. Atomic Properties and Periodic Trends Ionization Energy

Chapter 818 Atomic Properties and Periodic Trends Ionization Energy

Chapter 819 -Ionization energy decreases down a group. -As the atom gets bigger, it becomes easier to remove an electron from the most spatially extended orbital. - Ionization energy generally increases across a period. -Two exceptions: removing the first p electron and removing the fourth p electron have a lower energies. -This indicates that half-filled and completely filled subshells are more stable. Atomic Properties and Periodic Trends Ionization Energy

Chapter 820 The s electrons are more effective at shielding than p electrons. Therefore, forming the s 2 p 0 becomes more favorable. When the p subshell has four electrons, one orbital has two electrons. When this electron is removed, the resulting s 2 p 3 configuration is more stable than the starting s 2 p 4 configuration (the final state has a much lower electron-electron repulsion). Atomic Properties and Periodic Trends Ionization Energy

Chapter 821 Atomic Properties and Periodic Trends Ionization Energy

Chapter 822 Electron affinity – The energy required to add an electron to an atom in the gaseous state: Cl(g) + e -  Cl - (g) -Electron affinity can either be exothermic (as the above example) or endothermic. Atomic Properties and Periodic Trends Electron Affinity

Chapter 823 Atomic Properties and Periodic Trends Electron Affinity

Chapter 824 -Cations are smaller than the parent atom. -Anions are larger than the parent atom. Atomic Properties and Periodic Trends Ion Sizes

Chapter 825 -Since the representative (main group) elements have the same valence electron configuration, their chemical properties should be similar. Example: 2 Li(s) + Cl 2 (g)  2 LiCl(s) 2 Na(s) + Cl 2 (g)  2 NaCl(s) 2 K(s) + Cl 2 (g)  2 KCl(s) Periodic Trends and Chemical Properties

Chapter 826 4, 20, 22, 32, 34, 44Homework