362 Lecture 6 and 7. Spring 2017 Monday, Jan 30

Transcription

1 362 Lecture 6 and 7 Spring 2017 Monday, Jan 30

2 Quantum Numbers n is the principal quantum number, indicates the size of the orbital, has all positive integer values of 1 to (infinity) l is the angular momentum quantum number, represents the shape of the orbital, has integer values of (n 1) to 0 m l is the magnetic quantum number, represents the spatial direction of the orbital, can have integer values of -l to 0 to l m s is the spin quantum number, has little physical meaning, can have values of either +1/2 or -1/2 l (angular momentum) orbital 0 s 1 p 2 d 3 f Other terms: electron configuration, noble gas configuration, valence shell Pauli Exclusion principle: no two electrons can have all four of the same quantum numbers in the same atom (Every electron has a unique set.) Hund s Rule: when electrons are placed in a set of degenerate orbitals, the ground state has as many electrons as possible in different orbitals, and with parallel spin. Aufbau (Building Up) Principle: the ground state electron configuration of an atom can be found by putting electrons in orbitals, starting with the lowest energy and moving progressively to higher energy.

3 Energy Levels for Electron Configurations Guiding Principles for Electron Assignment: The Aufbau The Pauli Exclusion Principle Hund s Rules

4 And Why are s, p, d orbitals of different energy? Need both radial and angular functions # Angular Nodes = l Both radial and angular functions have nodes

5 Energies of the electrons Nodes, (not toads) Region of space of zero probability Summary: Total # Nodes = n 1 # Angular Nodes = l # Radial Nodes = n - l - 1 Where n = principal quantum number; l = angular momentum quantum number

6 Radial Wave Functions and Nodes # Radial Nodes = n - l - 1 Copyright 2014 Pearson Education, Inc.

7 Radial Probability Functions and Nodes # Radial Nodes = n - l - 1 Copyright 2014 Pearson Education, Inc.

8 Screening: The 4s electron penetrates Inner shell electrons more efficiently than does 3d in neutral atoms. Reverses in positive ions.

9 How to handle atoms larger than H? Effective Nuclear Charge or Z eff

10 Electrons Characterized by a) Principal energy level, n b) Orbital or angular momentum, l = # of angular nodes c) Z eff In the presence of a magnetic field of l is oriented and composed of m l components. d) Spin-spin and spin-orbital coupling

11 Slater s Rules for Calculating Z eff 1) Write the electron configuration for the atom as follows: (1s)(2s,2p)(3s,3p) (3d) (4s,4p) (4d) (4f) (5s,5p) 2) Any electrons to the right of the electron of interest contributes no shielding. (Approximately correct statement.) 3) All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units 4) If the electron of interest is an s or p electron: All electrons with one less value of the principal quantum number shield to an extent of 0.85 units of nuclear charge. All electrons with two less values of the principal quantum number shield to an extent of 1.00 units. 5) If the electron of interest is an d or f electron: All electrons to the left shield to an extent of 1.00 units of nuclear charge. 6) Sum the shielding amounts from steps 2-5 and subtract from the nuclear charge value to obtain the effective nuclear charge.

12 Slater s Rules: Examples Calculate Z eff for a valence electron in fluorine. (1s 2 )(2s 2,2p 5 ) Rule 2 does not apply; therefore, for a valence electron the shielding or screening is (0.35 6) + (0.85 2) = 3.8 Z eff = = 5.2 Calculate Z eff for a 6s electron in Platinum. (1s 2 )(2s 2,2p 6 )(3s 2,3p 6 ) (3d 10 ) (4s 2,4p 6 ) (4d 10 ) (4f 14 ) (5s 2,5p 6 ) (5d 8 ) (6s 2 ) Rule 2 does not apply, and the shielding is: (0.35 1) + ( ) + ( ) = Z eff = = 4.15 for a valence electron.

13 Trends in Atomic Properties Size (atomic, ionic, covalent, van der Waals radii) Ionization Potential (A 0 (g) + I.E. A + + e - ) Electron Affinity Energies (A 0 (g) + e - A - + E.A.E.) Electronegativity: Ability of an atom, within a molecule to attract electrons to itself.

14 Inorganic Chemistry Chapter 1: Figure W.H. Freeman

15 Inorganic Chemistry Chapter 1: Table W.H. Freeman

16 Lanthanide Contraction: particulary large decrease in ionic radii size due to Particularly poor shielding by electrons in f orbitals.

17 Inorganic Chemistry Chapter 1: Figure W.H. Freeman

18 Metallic Single-Bond Distances (useful for M-M bonding and intermetallic compounds)

19 Inorganic Chemistry Chapter 1: Table 1.4 Cations are smaller than Neutral atom Anions are Larger than Neutral atom 2009 W.H. Freeman

20 Trends in Atomic Properties Size (atomic, ionic, covalent, van der Waals radii) Ionization Potential energy (A 0 (g) + I.E. A + + e - ) Electron Affinity Energy (A 0 (g) + e - A - + E.A.E.) Electronegativity: Ability of an atom, within a molecule to attract electrons to itself.

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22 Ionization Energies

23 Inorganic Chemistry Chapter 1: Figure W.H. Freeman

24 Ionization Energies kj/mol Transition Metals - some complications

25 Inorganic Chemistry Chapter 1: Table W.H. Freeman

26 Copyright 2014 Pearson Education, Inc.

27 The answer to why? Z effective! Figure 1.7 The variation of the radial density distribution function with distance from the nucleus for electrons in the 1s, 2s, and 3s orbitals of a hydrogen atom.

28 2p 3p 3d

29 nd vs. (n + 1)s in the Transition Metals

30 BDE: kj/mol

31 Linus Pauling BDE H 2 = 436 kj/mol BDE Cl 2 = 239 BDE HCl = 427 Pauling: If strictly covalent: BDE HCl should be average of H 2 and Cl 2 Which would be ½ ( ) = 338 kj/mol. The extra stability is Due to electronegativity difference, and electrostatic attraction. But... BDE H 2 = 436 kj/mol BDE I 2 = 153 BDE HI = 299 Average ½ ( ) = 295 kj/mol Note that χ H = 2.2 and χ I = 2.6 and we expect that because HI isn t stable!

32 Electronegativity Pauling Mulliken Rochow

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34 Inorganic Chemistry Chapter 1: Figure W.H. Freeman

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37 Inorganic Chemistry Chapter 1: Figure W.H. Freeman

38 Properties of Hydrogen Halides: polar covalent bonds, bond dipoles, electric dipole moments, µ = q x r Why??? Electric dipole moment is the product of magnitude of charges and the distance of separation between the charges. Magnitude of charges will depend on difference in electronegativity of the atoms. Now in this case, F has highest electronegativity hence HF will have highest magnitude of charges. But, as we go down the group, the atomic radii increases, hence the distance of separation (bond length) will increase. So, we would expect the two factors to balance each other. Here, it seems the difference in electronegativity out weighs the distance (bond length) and that is why Dipole moment decreases,