Additional Problems for practice:
1. Use the VSEPR rules to determine the shape of each of the following
molecules:
(a) CBr4
With four substituents around carbon, according to VSEPR theory, each of
the bonds should be 109.5° apart and the overall molecule has a tetrahedral geometry
(b)(CH3)3N Although there are 3 atoms bonded to Nitrogen, the steric number of this
compound is 4 since there is also a lone pair on nitrogen. With four bonds/lone pairs
around nitrogen, the overall molecule has a tetrahedral geometry (each methyl group is
also tetrahedral), with the bond angles compressed slightly from their ideal value of
109.5° due to the larger steric requirement of a lone pair. This molecule may also be
thought of as pyramidal or a trigonal pyramid if the lone pair is discounted in drawing the
molecule
(c)(CH3CH2)2O This molecule has two bonds and two lone pairs on the central oxygen,
and so tetrahedral geometry around oxygen is expected (of course, each carbon is
tetrahedral as well!). The “bent” geometry at oxygen is obvious when the lone pairs are
discounted in drawing the molecule.
(d)BCl3 : there are three bonds and no lone pairs around B atom, so we expect a trigonal
planar geometry.
(e) CH3C
N The central carbon atom has two atoms bonded to it, and thus a linear
geometry is expected (both carbons and nitrogen are in a straight line), bond angles 180°.
The CH3 group is tetrahedral at carbon. The nitrogen has one lone pair and one other
atom bonded to it, so it also has a linear geometry, with the lone pair oriented 180° away
from the C-N triple bond.
a
Br
c
b
C
Br
Br
H3C
O
H3C
N
Br
CH3
CH3
d
h
CH3
H
H
H
Cl
B
Cl
Cl
H
e
C
H
C
N
H
2. Indicate the kind of hybridization you might expect for each carbon
atom in these molecules:
Draw out the Lewis structure for each molecule
H
H
C1
H
C2
C3
C4
(a) CH2=CHC
CH
H
Carbon 1 has 3 sigma bonds, no lone pairs, (remember pi bonds are
formed from p orbitals, which are not hybridized) and thus displays sp2
hybridization. Carbon 2 has 3 sigma bonds, no lone pairs, and thus has sp2
hybridization. Carbons 3 and 4 each have 2 sigma bonds, no lone pairs,
and thus each has sp hybridzation
O
H
H
H
H
H
H
(b) CH3OCH3 Both carbon atoms have four sigma bonds and thus sp3
hybridization; oxygen has 2 sigma bonds and 2 lone pairs (4 total) and
thus has sp3 hybridization.
H3C
H
C3
C4
H
H3C
(c) (CH3)2C=CH2 The two methyl groups have sp3 hybridization, since
they each have four sigma bonds; C3 and C4 have three sigma bonds,
no lone pairs, and thus each has sp2 hybridization.
H
C
H
H
H
H
H
H
H
(d) CH3CH2CH3 All carbon atoms have four sigma bonds, and thus each
has sp3 hybridization.
3. What shape would you expect these species to have?
(a) NH4+ :four sigma bonds, no lone pairs, tetrahedral
H
N
H
H
H
(b) (CH3)3P :three sigma bonds, one lone pair, tetrahedral (pyramidal)
P
H3C
CH3
CH3
(c) (CH3)3B : three sigma bonds, no lone pairs, trigonal planar geometry
H3C
B
H3C
CH3
(d) H2C=O: carbon has three sigma bonds, no lone pairs, trigonal planar
geometry; oxygen has two lone-pairs, one sigma bond, also trigonal
planar geometry.
H
C
O
H
4. Draw the structure of the following molecules, indicating the types of
bonds involved and the hybrid orbitals on each atom:
(a) CH3C
N
H
C
H
C
N
H
σ−Csp-Csp3
σ−Csp-Nsp
σ−Csp3-Hs
H 3C
H2
C
NH2
Nsp
π bonds
(b) CH3CH2CH=CH2
σ−Csp3-Csp3
σ−Csp2-Csp2
H H
H H
H
H3C
C
C
σ−Csp2-H1s
H
H
σ−Csp3-Hs
σ−Csp3-Hs
(c) CH3C
CCH3
π bond
H
H
C
H
C
C
C
H
H
H
σ−Csp-Csp3
σ−Csp-Csp
σ−Csp3-Hs
H 3C
H2
C
π bonds
(d) CH2=C=CH2
H
H
H
H
σ−Csp-Csp2
σ−Csp-Csp2
σ−Csp2-H1s
σ−Csp2-H1s
π bonds
(e)
(f) CH2=NH
σ−Csp2-Nsp2
H
H
C
σ−Csp2-H1s
N
σ−Nsp2-H1s
H
π bond
5. Draw an orbital diagram for the following species, showing the hydrid
orbitals that combine to form bonds. Indicate the geometry of each:
(a) CH3+ planar:
side view
empty
p orbital
top view
σ-Csp2-Hs
planar molecule
with p-orbital
perpendicular to
plane of hybrid orbitals
perpendicular p-orbital
-
(b) H3C: ; tetrahedral
C
H
H
H
(c) CO2 linear
O
C
σ−Csp-Osp2
σ−Csp-Osp2
O
Οsp2
Οsp2
π bonds
6.
In which of the following molecules do you expect to
observe cis/trans isomerism? Draw Lewis structures for the
different possibilities.
(a) CH3CH2CH2CH3 No double bond, no cis/trans iomerism
(b) (CH3)2C=CH2 Only one possibility, no cis-trans isomerism possible
(c) CH2=CHCl Again, only one possibility, no cis-trans isomerism
possible
(e) CH3CH2CH=CHCH2CH3 yes, cis,trans possible:
H
H
H
cis
H
trans