Chemistry 2000 (Spring 2010)
Problem Set #3: Valence Bond Theory
Solutions
Answers to Questions in Petrucci (only those w/out answers at the back of the book)
11.12.
(a)
In HCN, C is sp hybridized.
(b)
In CH3OH (methanol), C is sp3 hybridized.
(c)
In (CH3)2CO (acetone), C is sp2 hybridized.
(d)
In carbamic acid, C is sp2 hybridized.
11.14.
(a)
H
σ:C(sp3) –Cl(3p)
..
: Cl
..
C
H
σ:C(sp3) –H(1s)
(b)
σ:C(sp3) –H(1s)
-
..
: O..
..
Cl
.. :
σ:C(sp3) –Cl(3p)
N:
C
σ:C(sp) –N(2pz)
σ:C(sp) –O(2pz)
π:C(2px) –N(2px)
π:C(2py) –N(2py)
..
: F:
(c)
2
σ:B(sp ) –F(2p)
σ:B(sp2) –F(2p)
.. B ..
: F..
..F :
11.16.
..
O
..
C
σ:B(sp2) –F(2p)
..
O
..
σ:C(sp) –O(2pz)
σ:C(sp) –O(2pz)
π:C(2px) –O(2px)
π:C(2py) –O(2py)
11.18.
(a)
σ:N(sp2) –O(2p)
- :O..
..
N
..
O:
..
σ:N(sp2) –O(2p)
π:N(2pz) –O(2pz)
(c)
σ:C(sp2) –O(2p)
: O:
σ:C(sp2) –O(2p)
-
π:C(2pz) –O(2pz)
.. C
:O
C
..
.. O
.. :
σ:C(sp2) –O(2p)
: O:
σ:C(sp2) –O(2p)
π:C(2pz) –O(2pz)
(d)
σ:C(sp2) –O(2p)
: O:
σ:C(sp2) –O(2p)
-
π:C(2pz) –O(2pz)
.. C .. H
:O
O
..
..
σ:O(sp3) –H(1s)
σ:C(sp2) –O(sp3)
11.24. The central C is sp hybridized while the two outer C are sp2 hybridized:
H
C
σ:C(sp) –C(sp2) H
π:C(2px) –C(2px)
C
C
H
H
σ:C(sp) –C(sp2)
π:C(2py) –O(2py)
The C-H bonds were omitted from the diagram for clarity. They are all σ:C(sp2)–H(1s).
Note that the two pi bonds are perpendicular to each other.
11.58.
: O:
H
.. C .. H
N
N
H
H
..
At first glance, it would appear that the C is trigonal planar
while the Ns are both trigonal pyramidal. If the molecule is
completely planar, though, that cannot be. Instead, the C
and both Ns must all be trigonal planar. As such, they will
all be sp2 hybridized.
We can draw resonance structures in which the lone pair on
a nitrogen atom is used to make a C=N double bond and
the pi electrons in the C=O bond become a lone pair on O:
-
: O:
H
C
.. : O:
: O:
.. H
H
.. C .. H
.. C
H
H
+N
N
N
N
N
N+
H
H
H
H
H
H
It is now easy to see why the Ns should also be trigonal planar.
So, if each N is sp2 hybridized, the lone pair on each N is in the 2pz orbital.
Each N-H bond is σ:N(sp2)–H(1s).
Each C-N bond is σ:C(sp2)–N(sp2).
The C-O bond is σ:C(sp2)–O(2p) and π:C(2pz)–O(2pz). .
Additional Practice Problems
1.
The structural formula of glycine is shown below. Name the hybrid orbital set used by
each central atom when VB theory is applied to glycine and indicate how many
unhybridized 2p orbitals remain on each central atom.
sp3 (no leftover 2p)
sp3 (no leftover 2p)
H : O:
..
H
sp3 (no leftover 2p)
..
N
C
H
H
C
O
..
Suggested Approach for This Type of Question:
1. Add lone pairs to convert structural formula to Lewis dot structure.
2. Determine electron group geometry for each 'central' atom.
3. Choose hybrid orbital set based on electron group geometry.
H
sp2 (+1 leftover 2p)
2.
DNA consists of three types of molecules connected together. The “coding” molecules
are called nitrogenous bases (because they are bases that contain nitrogen). The molecule
shown below is adenine, one of the four nitrogenous bases in DNA.
H
..
H
N
sp2 (+1 leftover 2p)
:N
..
C
N
C
C
..
C
C
H
sp2 (+1 leftover 2p)
H
N
N
..
H
sp2 (+1 leftover 2p)
(a)
Name the hybrid orbital set used by each of the three bolded atoms when VB theory is
applied to adenine. Also, indicate the number of unhybridized 2p orbitals remaining on
each bolded atom.
(b)
How many σ bonds are there in one molecule of adenine? 16
(c)
How many π bonds are there in one molecule of adenine? 4
3.
Acetyl chloride (CH3COCl) has the connectivity shown:
(a)
Draw the best Lewis structure for acetyl chloride.
(b)
Indicate the molecular geometry at each central atom.
C (of CH3) = tetrahedral
(c)
H
..
O:
H C C ..
Cl
.. :
H
C (of COCl) = trigonal planar
What is the hybridization of the following atoms when VB theory is applied to acetyl
chloride:
C (of CH3) = sp3
C (of COCl) = sp2
(d)
How many sigma bonds are there in one molecule of acetyl chloride? 6
(e)
How many pi bonds are there in one molecule of acetyl chloride? 1
4.
Acetonitrile (CH3CN) has the connectivity shown:
(a)
Draw the best Lewis structure for acetonitrile.
(b)
Indicate the molecular geometry at each central atom.
H
C (of CH3) = tetrahedral
(c)
H C C N:
H
C (of CN) = linear
What is the hybridization of the following atoms when VB theory is applied to
acetonitrile:
C (of CH3) = sp3
C (of CN) = sp
(d)
How many sigma bonds are there in one molecule of acetonitrile? 5
(e)
How many pi bonds are there in one molecule of acetonitrile? 2
5.
The Lewis structure for N2H2 is shown below. Consider this molecule according to
valence bond theory.
H
..
N
N
..
H
(a)
Name the set of hybrid atomic orbitals used by each N atom.
(b)
Clearly indicate which orbitals contribute to each σ bond in N2H2.
σ:N(sp2)–H(1s)
H
..
N
N
..
Clearly indicate which orbitals contribute to each π bond in N2H2.
H
..
N
H
6.
σ:N(sp2)–H(1s)
σ:N(sp2)–N(sp2)
H
(c)
sp2
N
..
π:N(2p)–N(2p)
Use diagram(s) to explain how sp orbitals are formed. Clearly indicate the number, type
and geometry of all orbitals involved.
An s orbital and a p orbital in the same shell of the same atom combine to make two sp
orbitals:
7.
Use diagrams in your answers to the following questions.
(a)
What is the main difference between a hybrid atomic orbital and a molecular orbital?
Hybrid atomic orbitals are formed by mixing orbitals on the same atom.
Molecular orbitals are formed by mixing orbitals on different atoms.
(In both cases, the total number of orbitals is not changed. Mixing two atomic
orbitals gives two new orbitals – be they hybrid atomic orbitals or be they
molecular orbitals.)
(b)
What is the main difference between a σ bond and a π bond?
A sigma (σ) bond has electron density along the line connecting the nuclei of the two
bonded atoms.
A pi (π) bond has a node along the plane of the molecule.
Sigma:
Pi:
8.
Use diagram(s) to explain why it is not possible to rotate about the double bond in
molecule A to make molecule B:
F
F
C
H
H
F
X
C
H
C
C
H
F
molecule B
molecule A
A π bond has a node along the plane of the molecule. Rotating one of the atoms in the π
bond will rotate the orbitals on that atom and break the symmetry required for the π bond.
The bond will be broken by twisting it. Thus, twisting a π bond requires enough energy
to break the bond (and is therefore difficult to do).
H
F
H
C
C
F
molecule A
(p orbitals have correct
symmetry to overlap)
rotate
X
rotate
H
C
F
C
H
F
X
pi bond broken!
(p orbitals do not have correct
symmetry to overlap)
F
H
H
C
C
F
molecule B
(p orbitals have correct
symmetry to overlap)