1 / 29

HL STANDARD E

HL STANDARD E. LEWIS STRUCTURES MOLCULAR GEOMETRY & HYDRIDIZATION. 14.1 MOLECULES w/ EXPANDED OCTETS. WHEN THE CENTRAL ATOMS IS AN ELEMENT FROM THE 3 rd PERIOD OR BELOW, WE SOMETIMES FIND COMPOUNDS IN WHICH THERE ARE MORE THAN EIGHT ELECTRONS AROUND THE CENTRAL ATOM.

tayte
Download Presentation

HL STANDARD E

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. HL STANDARD E LEWIS STRUCTURES MOLCULAR GEOMETRY & HYDRIDIZATION

  2. 14.1 MOLECULES w/ EXPANDED OCTETS • WHEN THE CENTRAL ATOMS IS AN ELEMENT FROM THE 3rd PERIOD OR BELOW, WE SOMETIMES FIND COMPOUNDS IN WHICH THERE ARE MORE THAN EIGHT ELECTRONS AROUND THE CENTRAL ATOM. • THIS ARRANGEMENT IS POSSIBLE BECAUSE THE d ORBITALS IN THE VALANCE SHELL OF THESE ATOMS HAVE ENERGY VALUES CLOSE TO THOSE OF THE p ORBITALS

  3. TYPES OF COVALENT BONDS • SINGLE BONDS • LONGEST OF THE 3 TYPES • WEAKEST OF THE 3 TYPES • CONTAINS ONE PAIR OF ELECTRONS (2 ELECTRONS) • DOUBLE BONDS • LENGTH IS GREATER THAN TRIPLE BUT LESS THAN SINGLE • STRENGTH IS GREATER THAN SINGLE BUT LESS THAN TRIPLE • CONTAINS TWO PAIRS OF ELECTRONS (4 ELECTRONS) • TRIPLE BONDS • SHORTEST OF THE 3 TYPES • STRONGEST OF THE 3 TYPES • CONTAINS THREE PAIRS OF ELECTRONS (6 ELECTRONS) • LONGEST BONDS(SINGLE) ARE THE WEAKEST • SHORTEST BONDS(TRIPLE) ARE THE STRONGEST

  4. VSERP THEORYVALENCE SHELL ELECTRON REPLUSION THEORY VSERP THEORY states that electron pairs found in the outer energy level or valence shell of atoms repel each other and thus position themselves as far apart as possible

  5. The VSERP theory helps to predict the shape of the molecule. The following points should be considered in predicting shape (molecular geometry): • Double or triples bonded electron pairs are oriented together and so behave in terms of repulsion as a single unit know as a ‘negative charged center’. • The repulsion applies to both bonding and non- bonding (lone) pairs of electrons • The total number of charge centers around the central atom will determine the geometrical arrangement of the electrons • The shape of the molecule is determined by the angles between the bonded atoms • Because non-bonded (lone) pairs are not shared between two atoms they cause more repulsion than bonded pairs • The repulsion decrease in the following order: lone-lone pair>lone-bonding pair>bonding-bonding pair

  6. FOR EXAMPLE, IN THE LEWIS STRUCTURE OF WATER, THERE ARE TWO NON-BONDED (LONE) PAIRS OF ELECTRONS ON THE OXYGEN ATOM. THUS THE NON-BONDED (LONE) PAIR OF ELECTRONS REPEL MORE THAN THE LONE-BONDED PAIR BETWEEN THE OXYGEN AND HYDROGEN AND THE BONDED-BONDED PAIR OF ELECTRONS BETWEEN THE TWO HYDROGENS HAVE THE LEAST REPULSION. NON-BONDED (LONE)& NON-BONDED (LONE) PAIR OF ELECTRONS NON-BONDED(LONE) & BONDED PAIR OF ELECTRONS BONDED-BONDED PAIR OF ELECTRONS

  7. BOND ANGLES Due to the fact that lone pair of electrons repel more than bonded pairs, the bond angles will be different in different shaped molecules BEND(with 2 lone pair of e-) <TRIGONAL PYRAMIDAL<TETRAHEDRAL< BEND(with 1 lone pair of e-)<TRIGONAL PLANAR<LINEAR ****KNOW THE BOND ANGLES OF THIS SHAPES

  8. TETRAHEDRAL 109.5o TRIGONAL PLANAR LINEAR 120o 180o .. S // \ :O: :O: .. BENT (w/ 1 lone pairs of e-) BENT (w/ 2 lone pairs of e-) TRIGONAL PYRAMIDAL 109.5 o (104.5o) 109.5o (107.5o ) 120o (119o)

  9. ETHANOIC ACID CH3COOH BENT 104.5o TETRAHEDRAL TRIGONAL PLANAR

  10. CAN A MOLECULE BE NONPOLAR AND STILL HAVE POLAR BONDS WITHIN THE MOLECULE??? FIRST THINK ABOUT WHAT MAKES A BOND POLAR! NOW THINK ABOUT WHAT MAKES A MOLECULE NONPOLAR!

  11. YES THE DIFFERENCE IN ELECTRONEGATIVITY VALUES (CHARGE SEPARATION) DETERMINE IF A BOND WILL BE POLAR OR NONPOLAR. 0-.3 difference is NONPOLAR .3-1.7 difference is POLAR IF A MOLECULE IS SYMMETRICAL OR IF THERE IS NO NET DIPOLE MOMENT THEN IT IS NONPOLAR EXAMPLE: CCl4 THE C-Cl BOND IS POLAR BUT THE MOLECULE IS TETRAHEDRAL SO IT IS SYMMETRICAL AND THUS NONPOLAR

  12. WHICH OF THE FOLLOWING BONDS IS THE MOST POLAR B-C or C-O or N-O or O-F ANSWER: C-O b/c they are further apart on the periodic table which means that they have the greatest charge separation.

  13. HYBRIDIZATION • HYBRIDIZATION IS THE MIXING OF ORBITALS • THE NEW HYBRID ORBITALS FORMED MAY BE sp, sp2, sp3, (sp3d, sp3d2 not covered)

  14. Formation of sp hybrid orbitals The combination of an s orbital and a p orbital produces 2 new orbitals called sp orbitals. These new orbitals are called hybrid orbitals The process is called hybridization What this means is that both the s and one p orbital are involved in bonding to the connecting atoms

  15. EXAMPLES OF MOLECULES WITH sp HYBRIDIZATION ALL LINEAR MOLECULES

  16. Formation of sp2hybrid orbitals

  17. EXAMPLES OF sp2 HYBRIDIZATION ALL TRIGONAL PLANAR AND BENT(one lone pair of e-) .. S // \ :O: :O: ..

  18. Formation of sp3hybrid orbitals

  19. EXAMPLES OF sp3 HYBRIDIZATION ALL TETRAHEDRAL TRIGONAL PYRIMIDAL & BENT (w/ two lone pair of e-)

  20. Hybrid orbitals can be used to explain bonding and molecular geometry

  21. Multiple Bonds Everything we have talked about so far has only dealt with what we call sigma bonds Sigma bond (s)  A bond where the line of electron density is concentrated symmetrically along the line connecting the two atoms.

  22. Pi bond (p)  A bond where the overlapping regions exist above and below the internuclear axis (with a nodal plane along the internuclear axis).

  23. Example: H2C=CH2

  24. Example: H2C=CH2

  25. Example: HCCH

  26. Delocalized p bonds When a molecule has two or more resonance structures, the pi electrons can be delocalized over all the atoms that have pi bond overlap.

  27. Example: C6H6 benzene Benzene is an excellent example.  For benzene the p orbitals all overlap leading to a very delocalized electron system In general delocalized p bonding is present in all molecules where we can draw resonance structures with the multiple bonds located in different places.

More Related