1 / 45

VSEPR THEORY:

VSEPR THEORY:. How do we determine the shapes of molecules and ions?. VSEPR THEORY. What does VSEPR stand for? V alence S hell E lectron P air R epulsion. VSEPR THEORY. Why is this important to know? It explains how molecules and ions behave. VSEPR THEORY: Basic procedure.

trapper
Download Presentation

VSEPR THEORY:

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. VSEPR THEORY: How do we determine the shapes of molecules and ions?

  2. VSEPR THEORY • What does VSEPR stand for? • Valence Shell Electron Pair Repulsion

  3. VSEPR THEORY • Why is this important to know? • It explains how molecules and ions behave.

  4. VSEPR THEORY: Basic procedure • 1) Determine the central atom (usually the atom with the lowest subscript and/or the atom capable of forming the most bonds).

  5. VSEPR THEORY: Basic procedure • 2) Draw the electron dot structure and bar diagram

  6. VSEPR THEORY: Basic procedure • 3) Determine the molecular geometry using ALL electron pairs AND atoms around the central atom.

  7. VSEPR THEORY: Basic procedure • 4) Modify the geometry to determine the molecular shape (if non-bonding electron pairs exist by ignoring them)

  8. VSEPR THEORY: Example: BeH2 • 1) Central Atom? Be (only 1 atom)

  9. H Be H VSEPR THEORY: Example: BeH2 • 2) Electron Dot? • 2) Bar Diagram? Note that Be violates the octet rule—this is an exception! H—Be—H

  10. Be H H VSEPR THEORY: Example: BeH2 • 3) Geometry? Hint: What is the furthest apart you can spread two atoms attached to a central atom?

  11. Be H H VSEPR THEORY: Example: BeH2 • 4) Shape? Ignore any unbonded pairs of electrons —not necessary in this case.  LINEAR

  12. VSEPR THEORY: Example: BF3 • 1) Central Atom? B (only 1 atom)

  13. F B F F VSEPR THEORY: Example: BF3 • 2) Electron Dot? • 2) Bar Diagram? Note that B violates the octet rule—this is an exception! F—B—F F

  14. VSEPR THEORY: Example: BF3 • 3) Geometry? Hint: What is the furthest apart you can spread three atoms attached to a central atom? F B F F

  15. F B F F VSEPR THEORY: Example: BF3 • 4) Shape? Ignore any unbonded pairs of electrons —not necessary in this case.  trigonal planar

  16. VSEPR THEORY: Example: CH4 • 1) Central Atom? C (only 1 atom)

  17. H H C H H H H—C—H H VSEPR THEORY: Example: CH4 • 2) Electron Dot? • 2) Bar Diagram?

  18. H C H H H VSEPR THEORY: Example: CH4 • 3) Geometry? Hint: What is the furthest apart you can spread four atoms attached to a central atom? Think in 3D!

  19. H C H H H VSEPR THEORY: Example: CH4 • 4) Shape? Ignore any unbonded pairs of electrons —not necessary in this case.  tetrahedral

  20. VSEPR THEORY: Example: NH3 • 1) Central Atom? N (only 1 atom)

  21. H N H H H—N—H H VSEPR THEORY: Example: NH3 • 2) Electron Dot? • 2) Bar Diagram?

  22. ~109.5o H N H H VSEPR THEORY: Example: NH3 • 3) Geometry? Hint: What is the furthest apart you can spread three atoms plus one unbonded pair of electrons attached to a central atom? Think in 3D!

  23. ~109.5o H N H H VSEPR THEORY: Example: NH3 • 4) Shape? Ignore any unbonded pairs of electrons —it IS necessary in this case.  trigonal pyramidal

  24. VSEPR THEORY: Example: H2O • 1) Central Atom? O (only 1 atom)

  25. O H H O—H H VSEPR THEORY: Example: H2O • 2) Electron Dot? • 2) Bar Diagram?

  26. ~109.5o O H H VSEPR THEORY: Example: H2O • 3) Geometry? Hint: What is the furthest apart you can spread two atoms plus two unbonded pairs of electrons attached to a central atom? Think in 3D!

  27. ~109.5o O H H VSEPR THEORY: Example: H2O • 4) Shape? Ignore any unbonded pairs of electrons —it IS necessary in this case.  bent

  28. VSEPR THEORY • In conclusion: Since water (also called the universal solvent) is bent it is able to dissolve ionic substances:

  29. O H H O side tends to be – (the electron pairs hybridize into one group) H sides tend to be +

  30. O H H This negative side tends to attract positive ions These positive ends tend to attract negative ions

  31. What is the VSEPR Theory? • The VSEPR Theory is used to predict the shapes of molecules based on the repulsion of the bonding and non-bonding electrons in the molecule. • The shape is determined by the number of bonding and non-bonding electrons in the molecule. • In order to determine the shape, the Lewis diagram must be drawn first. • When determining the shape of a molecule with multiple bonds, treat the multiple bonds as if they were single bonds (i.e. one bonding pair)

  32. Molecules with the central atom surrounded by four bonding pairs (i.e. four atoms) • If the central atom is placed at the center of a sphere, than each of the four pairs of electrons will occupy a position to be as far apart as possible. • This will result in the electron pairs being at the corners of a regular tetrahedron, therefore these molecules are said to have a TETRAHEDRAL SHAPE. • The angle between each bond will be 109.5°

  33. Example CCl4

  34. Molecules with the central atom surrounded by 3 bonding pairs and 1 non-bonding pair • Four pairs of electrons will always arrange themselves tetrahedrally around the central atom. • The shape of the molecule is determined by the arrangement of the atoms not the electrons. • As a result such molecules will have a TRIANGULAR (TRIGONAL) PYRAMIDAL shape. • Due to the repulsion, a non-bonding electron pair requires more space than a bonding pair, the angles in these molecules are 107° not 109.5° as in the tetrahedral molecules.

  35. Example NH3

  36. Molecules with the central atom surrounded by 2 bonding pairs and 2 non-bonding pairs • The four pairs of electrons will be arranged tetrahedrally but since only 2 pairs are bonding electrons, the surrounding atoms are at 2 corners of the tetrahedron. • As a result these molecules will have a BENT OR V-SHAPE. • The repulsion between the non-bonding pairs will result in a bond angle of 104.5°. • For each pair of non-bonding electrons, the bond angle decreases by 2.5°

  37. Example H2O

  38. Molecules with the central atom surrounded by 5 pairs of bonding pairs • As with the tetrahedral molecules, the electron pairs will arrange themselves as far apart as possible. • To achieve this, the atoms will arrange themselves in a TRIANGULAR (TRIGONAL) BIPYRAMIDAL SHAPE which consists of 2 pyramids sharing the same base. • In this type of molecule, the 3 atoms making the base will lie in the same plane with the central atom in the middle of it. The other atoms will be positioned above and below this plane. • The bond angles within the base will 120° and the bond between the other atoms and the base will be 90°.

  39. Example PF5

  40. Molecules with the central atom surrounded by 6 pairs of bonding pairs • In order for the 6 pairs of electrons to be as far apart as possible in this case, each bonding pair will be at one corner of a REGULAR OCTAGON • The central atom is at the center of a square plane made up of 4 atoms and the other 2 atoms will be placed above and below the plane. • All bond angles will be 90°

  41. Example SF6

  42. Molecules with the central atom with an incomplete octet • Molecules that only have 2 bonding pairs on the central atom will have a LINEAR SHAPE with a bond angle of 180° • Molecules that only have 3 bonding pairs on the central atom will have a TRIANGULAR (TRIGONAL) PLANAR SHAPE with bond angles of 120°

  43. Example BeF2 Example BF3

  44. Things to Remember • In order to predict the shape of a molecule you must draw the Lewis Dot Diagram for the molecule, determine the number of bonding and non-bonding electron pairs and compare this with the chart you have been given (the general shapes must be memorized).

  45. Thank You Arunava Das, PGT Chemistry, GDGPS Siliguri, Darjeeling, West Bengal Ex – Research Scholar UGC CSIR DBT JRF II@ Molecular Biology Lab, I.I.Sc., Bangalore and Soil Biology Lab, GKVK, Bangalore

More Related