C HAPTER 9: M OLECULAR GEOMETRY AND BONDS. J ANUARY 28 TH, 2013 Do Now: Calculate the ∆H: C 2 H 4 + HCN  CH 3 CH 2 CN.

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Presentation transcript:

C HAPTER 9: M OLECULAR GEOMETRY AND BONDS

J ANUARY 28 TH, 2013 Do Now: Calculate the ∆H: C 2 H 4 + HCN  CH 3 CH 2 CN

M OLECULE SHAPE How do we determine the shape of a molecule? How are bonds related to the location of electrons? Draw CO 2, how many electron domains are located on the central carbon?

VSEPR T HEORY States: “Best arrangement of a given number of electron domains is the one that minimizes the repulsions among them” Predict the difference between electron-domain geometry and molecular geometry. How do we use VSEPR Theory: Draw lewis structure of molecule or ion and count number of electron domains around central atom. Determine electron-domain geometry Using table determine (eventually *memorize*) molecular geometry

P RACTICE Predict the molecular geometry of: O 3 SnCl 3 - Compare and contrast: H 2 O CH 4 NH 3 Explain the ideal bond angles.

W ORK IT OUT How are all the bond angles related to one another? With a partner or small group, discuss why it is possible that all these angles are so large, similar. In fancy terms: Non-bonding electrons give off greater repulsive forces therefore compressed bond angles. ** NOTE: double bonded atoms tend to exert a greater repulsive force as well.

P RACTICE : Determine the following molecular geometries: SF 4 IF 5 BrF 3 ICl 4 -

L ARGER THAN LIFE … Large molecules need to be broken down into their smaller components: Draw CH3COOH (use VSEPR for each central atom)

P OLARITY Why is polarity important? How do we determine polarity? Bonds: Molecules: Predict the polarity of the following: BrCl SO 2 SF 6

E XPLANATIONS How do we explain bonding? How is energy related to distance of covalent bonds? Why is there a sharp increase with shorter distance? All in all: bond length is distance at which the attractive forces between unlike charges are balanced by repulsive forces within the molecule