Chapter 11-12 Problems
Solutions
1. cis and trans C2H2Cl2 has the Lewis structures:
a) Each carbon in each structure has a trigonal planar shape and
therefore will be sp2 hybridized.
b) The first structure will have a dipole moment but the second will
not. In the first structure the “center” of negative charge lies between
the Cl's and the center of positive charge lies between the H's. Since
they do not coincide the molecule will have a dipole moment. In the
second structure the center s of posi tive and negative charge coincide
so there is no dipole moment.
2.
The boron in BF3 has orbital hybridization sp2 whereas in the product is
has orbital hybridization sp3. The orbital hybridization on the N stays
sp3 from reactant to product.
3. The Lewis structure of N3 is:
For shape purposes, remember that a double or triple bond counts as
a single pair of electrons. Therefore, around the central nitrogen atom
there are two pairs of electrons, both bonding pairs, which leads to the
conclusion that the electron pair geometry is linear. The linear
geometry is associated with sp orbital hybridization.
4.
Molecule Lewis
Structure
HSO4-
Shape
Class
Molecular
Shape
AX4
tetrahedral
Orbital
Hybridization
sp3
BeH2
AX2
linear
sp
BF4-
AX4
tetrahedral
sp3
XeOF4
AX5E
square planar
sp3d2
SbCl5
5. diazomethane:
formate anion:
sulfur dioxide:
sulfuric acid:
AX5
trigonal bipyramidal sp3d