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# Chapt 10 Rev Qu

### Chem Chap 10 Rev question answers

Question | Answer |
---|---|

Why is molecular geometry important? Cite some examples. | Molecular geometry is very important as it tells us about the shapes of molecules. This helps us account for the properties of molecules also. |

According to VSEPR theory, what determines the geometry of a molecule? | The geometry of a molecule is determined by: 1. Number of electron groups around the central atom. 2. Number of bonding groups and lone pairs among the electron groups. |

Describe linear electron geometry. | This geometry is observed when two electron groups are present. The bond angle is 180 degrees. All bonds are sorted in a line. NO LONE PAIRS ON CENTRAL ATOM. |

Describe trigonal planar electron geometry | Trigonal planar geometry has a bond angle of 120 degrees and all three groups are arranged in one plane. |

Describe tetrahedral electron geometry | This geometry is observed when four electron groups are present. two on equatorial, one above, and one in the z plane. The bond angle is 109.5 |

Describe trigonal bipyramidal electron geometry. | When 5 electron groups are present. The bond angle between the equatorial positions are 120 degrees (3 equatorial, 2 above). The 2 above, or axial positions have 90 degrees between axial and eq bond. |

Describe octahedral electron geometry. | When six electron groups are present. The bond angle is 90 degrees between all bonds. |

Explain the difference between electron geometry and molecular geometry? | E geo is the geo arrangement of the e groups while M geo is the geo arrangement of atoms. |

Under what circumstances are electron and molecular geometries NOT the same? | When a molecule contains one or more lone pairs on a centry atom, then the molecular geometry differs from the electron geometry. This because the lone pair exerts slightly greater repulsion than a bonding pair. |

Give E and M geo and angle for the following info: 2 bonding groups 0 lone pairs | E geo = linear, M geo = Linear, angle = 180 |

Give E and M geo and angle for the following info: 3 bonding groups 0 lone pairs | E geo = Trigonal Planar, M geo = Trigonal Planar,, angle = 120 |

Give E and M geo and angle for the following info: 2 bonding groups 1 lone pairs | E geo = Trigonal Planar, M geo = Bent,, angle = <120 degrees |

Give E and M geo and angle for the following info: 4 bonding groups 0 lone pairs | E geo = Tetrahedral, M geo = Tetrahedral,, angle = 109.5 degrees |

Give E and M geo and angle for the following info: 3 bonding groups 1 lone pairs | E geo = Tetrahedral, M geo = Trigonal pyramidal,, angle = <109.5 degrees |

Give E and M geo and angle for the following info: 2 bonding groups 2 lone pairs | E geo = Tetrahedral, M geo = bent, angle = <109.5 degrees |

Give E and M geo and angle for the following info: 5 bonding groups 0 lone pairs | E geo = Trigonal bipyramidal, M geo = Trigonal bipyramidal,, angle = 120 degrees (equatorial) 90 degrees (axial) |

Give E and M geo and angle for the following info: 4 bonding groups 1 lone pairs | E geo = Trigonal bipyramidal, M geo = Seesaw, angle = <120 degrees (equatorial) <90 degrees (axial) |

Give E and M geo and angle for the following info: 3 bonding groups 2 lone pairs | E geo = Trigonal bipyramidal, M geo = T-shaped, angle = 90 degrees |

Give E and M geo and angle for the following info: 2 bonding groups 3 lone pairs | E geo = Trigonal bipyramidal, M geo = Linear angle = 180 degrees |

Give E and M geo and angle for the following info: 6 bonding groups 0 lone pairs | E geo = Octahedral, M geo = Octahedral = 90 degrees |

Give E and M geo and angle for the following info: 5 bonding groups 1 lone pairs | E geo = Octahedral, M geo = Square pyramidal <90 degrees |

Give E and M geo and angle for the following info: 4 bonding groups 2 lone pairs | E geo = Octahedral, M geo = Square planar 90 degrees |

How do you apply VSEPR theory to predict the shape of a molecule with more than one interior atom? | With more than one interior atom, the same principles apply for predicting m geometries. Apply to each central atom and then take each geoemtry and build the three dim shaps of the molecules being determined. |

How do you determine if a molecule is polar? Why is polarity important? | Draw lewis diagram structure, determine the m geometry, determine if m geo has polar bonds based on the difference in electro negativities of atoms. determine if molecule is polar if net dipole moment forms. |

Polar or nonpolar: Linear e geometry | nonpolar |

Polar or nonpolar: Bent e geometry | polar |

Polar or nonpolar: Trigonal planar e geometry | nonpolar |

Polar or nonpolar: Tetrahedral | nonpolar |

Polar or nonpolar: Trigonal pyramidal | polar |

Polar or nonpolar: Trigonal planar | nonpolar |

What is a chemical bond, according to valence bond theory? | When the energy of two atoms is lowered due to the interactions between them, a chemical bond is said to be formed. When two atoms approach each other, the electrons and nucleus of an atom interact with electrons and nucleus of the other atom. |

In valence bond theory, what determines the geometry of a molecule? | The geometry of a molecule, according to VB theory, can be determined by using the hybridzation concept. |

What is hybridization? | Hybridization is the procedure in which standard atomic orbitals combine to form new atomic orbitals called hybrid orbitals. |

In valence bond theory, the interaction energy between the electrons and nucleus of one atom with the electrons and nucleus of another atom is usually negative (stabilizing) when ___________. | a bond forms between the two atoms. |

Why is hybridization necessary in valence bond theory? | Hybridization is necessary for valence bond theory to adequately explain bonding in some molecules. The concept of half filled orbitals overlapping is not sufficient to explain the bonding in all molecules. |

How does hybridization of atomic orbitals in the central atom of a molecule help lower the overall energy of the molecule. | atomic orb. combine to form hybrid orb. which have different shapes & energies from those of standard atomic orbs.. In these hybrid orbs., the e probab. density is more concentrated in a single directional lobe. hybrid lower energy max orb overlap in bon |

How is the number of hybrid orbitals related to the number of standard atomic orbitals that are hybridized? | The number of hybrid orbitals formed is always equal to the number of standard atomic orbitals that are hybridized. |

In the lewis model, the two bonds in a double bond look identical. However, valence bond theory shows that they are not. Describe a double bond according to valence bond theory. | VB T: Double bond consists of 2 diff kinds. One is omega, the other is pi. The omega bond is formed by end to end overlap of orbitals. Pi is formed by side by side overlap of p. |

Linear hybridization scheme. | sp |

Trigonal planar hybridization scheme is | sp^2 |

Tetrahedral hybridization scheme is | sp^3 |

Trigonal pipyramidal hybridization scheme is | sp^3d |

Octahedral hybridization scheme is | sp^3d^2 |

Created by:
aiur100