exercices resolves del atkins
TRANSCRIPT
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Exercise 7.3
Determine the symmetry elements and assign the point group of (a) NH2Cl, (b) CO32–, (c) SiF4, (d)
HCN, (e) SiFClBrI, (f) BF4–.
Answer
Use VSEPR to find the structure and then assign the point group and identify the symmetry elements.
(a) NH2Cl
Point group = Cs
Symmetry elements: E,
(b) CO32–
Point group = D3h
Symmetry elements: E, C3, C2, h, v, S3
(c) SiF4
Point group = Td
Symmetry elements: E, C3, C2, d, S4
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(d) HCN
Point group = C v
Symmetry elements: E, C , v
(e) SiFClBrI
Point group = C1
Symmetry elements: E
(f) BF4–
Point group = Td
Symmetry elements: E, C3, C2, d, S4
Determine the point group of SnF4, SeF4, and BrF4–.
Answer
Use VSEPR to find the structure and then assign the point group.
SnF4
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Point group = Td
SeF4
Point group = C2v
BrF4–
Point group = D4h
Problem 7.1Consider a molecule IF3O2 (with I as the central atom). How many isomers are possible? Which is likely to have the lowest energy? Assign point group designations to each isomer.
Answer
The Lewis dot structure is:
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The electron distribution is based on 5 sites, so a trigonal bipyramidal distribution is used. This leads to 3 different isomers (both O axial, both O equatorial, and 1 O axial and 1 O equatorial), as shown below with the appropriate point group.
The C2v structure is most likely the lowest energy because this minimizes the number of 90o repulsions between the I=O double bonds and the I-F single bonds.
Group Theory Solutions: #1
1. Identify the point group of each of the following pictures. Which are chiral? Which have dipole moments?
A. five-pointed star
Hay un eje de rotación 5 veces perpendicular al plano de la página. Hay 5 C2 ejes perpendiculares al eje de rotación principales, por lo que este es un grupo D. Hay un plano de simetría en el plano de la página (perpendicular al eje de rotación principio), por lo que este grupo es D5h, que no es quiral o polar. Nótese que el uso de los criterios de búsqueda simplificado, que no tuvimos que hacer una lista de todos los elementos de simetría que la imagen tiene el fin de decidir sobre un grupo de puntos (claramente hay rotaciones inadecuadas, etc.).
B. Star of David
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Hay un eje de rotación 6 veces perpendicular al plano de la página. Hay 6 C2 ejes perpendiculares al eje de rotación principales, por lo que este es un grupo D. Hay un plano de simetría en el plano de la página (perpendicular al eje de rotación principio), por lo que este grupo es d6h. Tenga en cuenta que a diferencia de la imagen en la parte a, esta imagen tiene un centro de inversión. Un interesante hecho de trivia es que todos los grupos Dnh donde n es siquiera tienen un centro de inversión; los grupos Dnh donde n es impar, no. En cualquier caso, ninguna molécula grupo Dnh es polar o quiral.
C. baseball
Given this baseball (and assuming it looks the same from behind), we have the following symmetry elements:
We also have mirror planes containing all three axes. Therefore the point group of this picture is D2h, which is neither chiral or polar.
D. pencil
This linear picture has a C-infinity rotation axis along the axis of the pencil and no inversion center. Therefore the pencil is C-infinity-v. This is not a chiral group, but it is polar.