3712BH MINING BRO 04Title: 3712BH_MINING_BRO_04 Author: Justin Created Date: 7/10/2017 2:27:41 PM
PowerPoint Presentationchemistry.oregonstate.edu/courses/ch411/… · PPT file · Web view ·...
Transcript of PowerPoint Presentationchemistry.oregonstate.edu/courses/ch411/… · PPT file · Web view ·...
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CH2. Molecules and covalent bonding Lewis Structures VSEPR MO Theory
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Lewis structure H3PO4 • Skeleton is:
• Count total valence electrons:1 P = 53 H = 34 O = 24Total = 32 e- or 16 valence e- pairs.
• 7 e- pairs needed to form skeleton.
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Lewis structure H3PO4 • Add remaining e- pairs:
• Left has a formal charge of +1 on P and -1 on one O, right has 5 e- pairs around P (hypervalence)
• Analysis of phosphoric acid shows purely Td phosphate groups, which requires something beyond either simple Lewis model.
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Resonance in NO3-
experimental data - nitrate is planar with 3 equivalent N-O bonds
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VSEPR model• Count e- pairs about the central atom (draw Lewis
structure if needed). Include non-bonding pairs, but not multiple bonds.
• Geometry maximizes separation:# e pairs geometry example
2 linear HF2-
3 equilateral triangular BF3
4 tetrahedral (Td) CF4
5 trigonal bipyramidal (TBP) PF5
6 octahedral (Oh) SF6
7 pentagonal bipyramidal IF7
8 square antiprismatic TaF83-
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Drawing Oh and Td molecules
It's often useful to draw octahedra and tetrahedra with a cubic framework
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Deviations from ideal geometries:
unshared pairs and multiple bonds require larger biteex: CH4, NH3, H2O
<H-C-H = 109.5°,<H-N-H = 107.3,<H-O-H = 104.5
ex: ICl4-
6 e pairs around I, 2 lone pairs and 4 e pair bonds to ClOh coordination, and geometry is square planar (lone pairs are trans, not cis)
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POCl3
based on Td geometry< ClPCl = 103.3° due to repulsion by multiple bond
note that in :PCl3 the <ClPCl = 100.3, the lone pair is more repulsive towards other ligands than the multiple bond !
Ligands move away from multiple bond
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XeF5+
5 Xe-F bonds and 1 lone pair on Xe geometry based on Oh coordination lone pair repulsion gives < FeqXeFeq = 87°
< FaxXeFeq = 78°
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Fajan’s rule
bond polarization is towards ligands with higher , decreasing repulsive effect. Lone pairs are the most repulsive. ex: NH3 vs NF3
< HNH = 107.3°
< FNF = 102.1°
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Inert pair effect• VSEPR geometries require
hybridization (valence bond term) or linear combinations (MO term) of central atom orbitals. For example, Td angles require sp3 hybrid orbitals. More on this in MO theory section.
• Period 5 and 6 p-block central atoms often show little hybridization (ex: they form bond with orbitals oriented at 90° as in purely p orbitals). This can be ascribed to the weaker bonding of larger atoms to ligands.
In Sn Sb Te
Tl Pb Bi
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Inert pair effect - evidence• Bond angles near 90°:
NH3 107.2 H2O 104.5AsH3 91.8 H2Se 91SbH3 91.3 H2Te 89.5
• Increased stability of lower oxidation statesex: Si, and Ge are generally 4+, but Sn and Pb are common as 2+ ions (as in stannous fluoride SnF2)ex: In and Tl both form monochlorides, B, Al, Ga form trichlorides.
• Vacant coordination sites where the lone pair resides
ex: PbOPbO unit cell
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Fluxionality
• PF5 if TBP has 2 types of F ligands (equatorial and axial).
• 19F NMR spectra at RT show only a single peak (slightly broadened).
• PF5 is fluxional at RT, i.e. the F ligands exchange rapidly, only a single "average" F ligand is seen by NMR.
• Only occurs if ligand exchange is faster than the analytical method. IR and Raman have shorter interaction times and show 2 types of P-F bonding at RT.
• Even low temp NMR studies cannot resolve two F environments
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Berry pseudo-rotation
Sequences of the MD-Simulation of PF5 at 750K (Daul, C., et al, Non-empirical dynamical DFT calculation of the Berry pseudorotation of PF5, Chem. Phys. Lett. 1996, 262, 74)
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Molecular Orbitals Use linear combinations of atomic orbitals to
derive symmetry-adapted linear combinations (SALCs).
Use symmetry to determine orbital interactions. Provide a qualitative MO diagram for simple
molecules. Read and analyze an MO diagram by sketching
MO’s / LCAO’s, describing the geometric affect on relative MO energies.
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H2
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Some rules The number of AO’s and MO’s must be equal. This
follows from the mathematics of independent linear combinations.
More on symmetry labels later, but they come from the irreducible representations for the point group. MO’s are symmetric about bond axis, MO’s are not. Subscipt g is gerade (has center of symmetry), u is ungerade. Antibonding orbitals are often given a * superscript.
The bond order = ½ (bonding e- - antibonding e-). The bond energy actually depends on the energies of the filled MO’s relative to filled AO’s.
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O2
• MO theory predicts 2 unpaired e-, this is confirmed by experiment.• Bond order = ½ (8-4) = 2, as in Lewis structure.• MO indicates distribution and relative energies of the MO's, Lewis structure says only bonding or non-bonding.
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I and Ea for atoms and diatomics
species I (kJ/mol) Ea
N 1402
O 1314 142 O2 1165 43 NO 893 F 1681 F2 1515 C 1086 123 C2 300
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Li2 – F2 MO’s
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Some diatomic bond data
bond order r0 in pm D0 in kJ/mol
O2 2 121 494
O2- 1 ½ 126
O22- 1 149
F2 1 142 155 O2
+ 2 ½ 112 NO 2 ½ 115 NO+ 3 106 N2 3 110 942
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Spectroscopic data for MO’s
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HF
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Ketalaar triangle
HF
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Hybridization• Linear combinations of AO’s from same atom
makes hybrid orbitals.
• Hybridization can be included in the MO diagram.
• In MO theory, any proportion of s and p can be mixed (the coefficients of the AO’s are variable). sp and sp3 hybrids are specific examples.
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H3+
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BeH2
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Correlation diagram for MH2
M < HMHBe 180°
B 131
C 136
N 103
O 105
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Bonding MO’s in H2O
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NH3
Use triangular H3 MO’s from above as SALC's of the H ligand orbitals. Must relabel to conform with lower symmetry pt group C3v. They become a1 and e.
Combine with N valence orbitals with same symmetry.
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NH3 --calculated MO diagram
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SF6
See textbook Resource Section 5 for SALCs