Chemical Bonds
1.
Important points about Lewis Dot:
a. Duet Rule: 2 electrons needed to satisfy valence shell.
i. What follows this rule? Hydrogen and Helium
b. Octet Rule: 8 electrons needed to satisfy valence shell.
i. What follows this rule?
C, N, O and F are the only elements that pretty much always follow the
rule.
ii. What can exceed the octet?
Any element whose valence shell is in n=3 or higher have the ability to
carry more than 8 electrons in their valence. This is because at n = 3,
there are additional d orbitals available for bonding.
iii. What can have less than the octet?
Boron is happy with 6 electrons in its valence.
2. What are the two types of electron pairs?
a. Bonding Pair – 2 electrons shared between two atoms in a molecule. These
electrons are confined to the area between the two atoms sharing them.
b. Lone Pair – 2 unshared electrons that are found on one atom. These electrons
take up more room than a bonding pair and tend to push bonding pairs closer
together.
3. Steps for drawing a Lewis Dot structure
a. Add up all valence electrons available using the periodic table.
b. Draw a skeleton structure.
c.
Give all atoms in the compound the need duet or octet.
d. Add up all electrons in use. Compare the number of electrons used to the
number available. Remember you cannot exceed or go below the valence
available. You must use the exact value.
e. If you have used too many electrons you will need to make a double bond
between two atoms. When you add the double bond pull a lone pair off each of
the atoms participating in the double bond. Continue this process until you have
used the correct number of electrons.
f. If you have too few electrons, you will typically add them to the central atom.
4. Draw the Lewis Dot for
a. CH2O
Step 1 – Count up valence electrons
Based on the periodic table we can see that:
C – 4 valence, H – 1 valence, O – 6 valence
Adding them all up we get:
Step 2 – Build a skeleton
If carbon is in a structure it is typically the central atom.
Step 3 – Give all atoms octet/duet
Step 4 – Add up electrons in Lewis
Each bond counts for 2 electrons and each lone pair counts for 2 electrons. In
this structure there are 3 bonds and 4 lone pairs:
3(2e-) + 4(2e-) = 14e- in use
This value exceeds the 12e- available so we are going to have to make additional
bonds.
Step 5 – replace lone pairs with bonds
The only place we can add additional bonds is between carbon and oxygen
(hydrogen does not make more than one bond). When we make this bond we
will have to remove one lone pair from oxygen and one lone pair carbon.
We needed to remove a lone pair from each because they share the bond and
thus the bond counts as 2 electrons for both.
We now recount the number of electrons in use to see if it matches the number
of electrons available.
There are a total of 4 bonds and 2 lone pairs:
4(2e-) + 2(2e-) = 12e- in use
As this value matches the original valence count – this is a valid Lewis structure.
5. What do you do if there is more than one viable Lewis Structure?
You need to look at the formal charges on the atoms in the compound and see which
Lewis structure gives the “best” formal charges.
a. How do you determine the formal charge?
(# of valence electrons) – (# of actual electrons in compound) = formal charge
“# of valence” is the value you get from the periodic table.
When counting the “actual electrons” remember that:
lone pair = 2 ebonding pair = 1eb. What are “favorable” formal charges?
1.
2.
3.
Values close to zero.
Negative formal charges on the most electronegative atom.
Positive formal charges on the least electronegative atom.
6. Draw the Lewis Dot with the best formal charges for
a. OCNStep 1 – Count up valence electrons
Based on the periodic table we can see that:
C – 4 valence, N – 5 valence, O – 6 valence
Adding them all up we get:
Step 2 – Build a skeleton
If carbon is in a structure it is typically the central atom.
Step 3 – Give all atoms octet/duet
Step 4 – Add up electrons in Lewis
Each bond counts for 2 electrons and each lone pair counts for 2 electrons. In
this structure there are 2 bonds and 8 lone pairs:
2(2e-) + 8(2e-) = 20e- in use
This value exceeds the 16e- available so we are going to have to make additional
bonds.
Step 5 – replace lone pairs with bonds
We needed to remove a lone pair from each because they share the bond and
thus the bond counts as 2 electrons for both.
We now recount the number of electrons in use to see if it matches the number
of electrons available.
There are a total of 3 bonds and 6 lone pairs:
3(2e-) + 6(2e-) = 18e- in use
We are still over our valence electron limit – this means that we are going to
have to make another bond.
`
Once again, I need to add up the electrons in use – there are four bonds and 4
lone pairs.
4(2e-) + 4(2e-) = 16eWe are using the correct number of valence electrons – so this is a valid Lewis
structure… but is it the best?
Couldn’t we, just as easily, have drawn the structure as:
or
At this point we have 3 valid Lewis structures (valid meaning that octet/duets are
satisfied and proper number of electrons have been used).
Step 6 – Compare formal charges
Looking at the formal charges, structure b is definitely not the correct structure,
a formal charge of -2 on nitrogen would be too high energy. Comparing
structure a and c the difference between their formal charges is the atom on
which the negative charge is located.
Remember that oxygen is more electronegative than N. This means that it
would be preferential for the negative charge to be located on oxygen. This
leads us to the conclusion that structure c is the best Lewis structure. Thus our
final answer would be:
When your compound has an overall charge it is important to remember to
bracket the structure and place the charge outside.
b. NO
Step 1 – Count up valence electrons
Based on the periodic table we can see that:
N – 5 valence, O – 6 valence
Adding them all up we get:
Step 2 – Build a skeleton
If carbon is in a structure it is typically the central atom.
Step 3 – Give all atoms octet/duet
Step 4 – Add up electrons in Lewis
Each bond counts for 2 electrons and each lone pair counts for 2 electrons. In
this structure there is 1 bond and 6 lone pairs:
1(2e-) + 6(2e-) = 14e- in use
This value exceeds the 11e- available so we are going to have to make additional
bonds.
Step 5 – replace lone pairs with bonds
We now recount the number of electrons in use to see if it matches the number
of electrons available.
There are a total of 2 bonds and 4 lone pairs:
2(2e-) + 4(2e-) = 12e- in use
We are still over our valence electron limit – but only by one electron – this
means that we cannot make a bond. We need to remove one electorn from
either O or N and form a radical ( a radical is a compound that contains an
unpaired electron). This means we can form
or
At this point we have 2 valid Lewis structures.
Step 6 – Compare formal charges
Looking at the formal charges, structure b is the best Lewis dot. Al formal
charges are equal to zero.
7. What is resonance?
Resonance is a situation in which there are more than one valid structure for a molecule
such that the actual compound is some combination of all resonance structures.
a. How do you know if you should “pick the best structure” or draw all resonance
structures?
In a case where all formal charges are equal you will have to draw each
resonance structure out. When there is a case where there are variations in
formal charges (as in the example above) you will pick the “best” structure.
8. Draw the Lewis Structure for
a. NO3Step 1 – Count up valence electrons
Based on the periodic table we can see that:
N – 5 valence, O – 6 valence
Adding them all up we get:
Step 2 – Build a skeleton
If carbon is in a structure it is typically the central atom.
Step 3 – Give all atoms octet/duet
Step 4 – Add up electrons in Lewis
In this structure there is 3 bond and 10 lone pairs:
3(2e-) + 10(2e-) = 26e- in use
This value exceeds the 24e- available so we are going to have to make additional
bonds.
Step 5 – replace lone pairs with bonds
We now recount the number of electrons in use to see if it matches the number
of electrons available.
There are a total of 4 bonds and 8 lone pairs:
4(2e-) + 8(2e-) = 24 e- in use
We are using the correct number of electrons. But now we have to consider that
there is no reason why the N has to be double bonded to one oxygen over
another. This is a resonance case. That double bond is equally valid between
the N and any of the oxygens. There is no diffenece in formal charges – so all
structures must be drawn:
9. How do you determine if a molecule is polar?
Determining the electronegativity difference between the two atoms within the bond.
Remember that electronegativity is the ability of an atom to pull electrons towards
oneself.
Generally:
∆EN >1.8
0.4<∆EN<1.8
∆EN< 0.4
indicates
indicates
indicates
Ionic Bond
Polar Covalent
Non Polar Covalent
10. What is a handy way to remember electronegativity ordering?
F >O>N>Cl>Br>I>S>C>H (pronounced “FONClBrISCH” ☺ )
Fluorine, Oxygen, Nitrogen, Chlorine, Bromine, Iodine, Sulfur, Carbon, Hydrogen
(Fluorine being the most electronegative)
The further apart 2 atoms are the greater the difference in their electronegativities and
more polar bond they would form. Notice that metals are not part of this list as metals
tend to lose electrons, meaning they would have very weak electronegativity.
11. What is the premise of VSEPR theory?
Electron pairs orientate themselves in a manner which minimizes repulsion (i.e.
maximizes distance between electron pairs).
12. VSEPR Chart
13. How do electronic and molecular structure differ?
Electronic structure is the shape that a molecule would take based solely on the number
of areas of electron density (also known as steric number) – in this case we do not need
to differentiate between bonding pairs and lone pairs. Meaning we are looking at how
ALL of the electrons lone/bonding will be arranged
Molecular structure is the arrangement that focuses on how the bonding electrons will
be orientated in space.
14. What are the bond angles of the non-lone pair bearing shapes?
a.
b.
c.
d.
e.
Linear (2 areas of e- density, no lone pairs) – 180o
Trigonal Planar (3 areas of e- density, no lone pairs) – 120o
Tetrahedral (4 areas of e- density, no lone pairs) – 109.5o
Trigonal bipyramidal (5 areas of e- density, no lone pairs) – 120o and 90o
Octrahedral (6 areas of e-, no lone pairs) – 90o
15. Do lone pairs affect bond angles?
Because lone pairs take up more room they essentially push the bonding pairs closer
together, thus altering typical bond angles.
For example in NH3 (which has one lone pair) the bond angle is 107.3o and in H2O (which
has 2 lone pairs) the bond angle is 104.5o. In both cases however, there are 4 areas of
electron density around the central atom.
16. How can shape affect polarity?
Symmetry can cancel out a polar bond. If identical atoms are attached to an atom and
the molecule is symmetric, it will be non-polar overall.
Let’s say that B is more electronegative than A, such that they form a polar bond. The
following is a list of some examples of how symmetry can cancel out polarity (keep in
mind, this list is not exhaustive).
Basically what you are looking for is a symmetry in which the electrons are being pulled
with equal strength but in opposing directions.
If you picture the pull on electrons as a tug of war… imagine that the opposing teams
are identical in strength and are pulling at angles that perfectly counter each other. If
this were the case the electrons would not be pulled in any one direction more strongly
than the other – thus they would stay where they were.
17. Determine the electronic and molecular structures around each of the central atoms
and whether the molecule is polar.
In order to do this we will first have to draw the Lewis Dot structure.
a. CHF3
Following the rules described in preceding problems we would get the structure:
In this case, the central atom, carbon, has 4 areas of electron density.
That means that electronic configuration around carbon would be tetrahedral
configuration. Additionally, because there are no lone pairs on the carbon, the
molecular structure would also be tetrahedral.
The molecular structure would look like:
In a tetrahedral the bond angles are 109.5o. This would be a polar molecule as F
is more electronegative than C.
b. I3-
In this case, the central I has 5 areas of electron density. This means that the
electronic configuration would be trigonal bipyramidal.
Because three of the areas of electron density are lone pairs, the molecular
configuration would be a linear configuration.
With all atoms in the bond being identical, and it being symmetric, this would be
a nonpolar molecule.
c. BrF5
In this case, the central atom, bromine, has 6 areas of electron density. This
means that the electronic configuration would be octahedral.
As one of these areas corresponds to a lone pair – the molecular configuration
would be square pyramidal. Thus the molecule around bromine would look like:
This would be a polar molecule. Due to the lone pairs, there would not be
a canceling of polar bonds due to symmetry.
18. Explain why CF4 and XeF4 are non-polar yet SF4 is polar.
In order to explain this, we will have to look at the shape as predicted by VSEPR.
We’ll start with our Lewis Dots.
19. Indicate the symmetry around all non-hydrogen atom
The carbon has three areas of electron density, no lone pairs. This would mean that
around the carbon the arrangement would be trigonal planar. The oxygen oxygen in the
C-O-H bond has four areas of electron density (with 2 lone pairs) so the atoms would be
arrange in a bent (or v-shaped) fashin around that oxygen.
Putting all of this together the VSEPR structure would look like:
20. Consider CH4
a. What orbitals are available for bonding for each atom involved in the molecule?
Each hydrogen has a 1s orbital available for bonding.
Carbon has one 2s and 3 2p orbitals available for bonding.
This was determined by looking at the valence of each atom. Rememebr that
only valence electrons participate in bonding.
b. Based on this would you expect all the C-H bonds in methane (CH4) to be
identical?
To answer this let’s look at the valence electron configuration for carbon.
In it’s current format, it doesn’t look like carbon can make more than 2 bonds, as
it has only 2 unpaired electrons. What winds up happening is that one of the 2s
electrons gets promoted into the 2p shell so that we get:
When this happens, we are now able to bond carbon 4 times, as there are four
unpaired electrons.
Next let’s consider this in conjunction with the 4 hydrogen atoms.
From this we can see that each one of the hydrogens can pair up with one of the
unpaired electrons from the carbon atom.
Looking at these results we can now say that based on this initial information we
would expect 3 of the four bond to be identical. This is the case because 3 of the
bonds are made from a 1s/2p combination and one bond is made from a 1s/2s
combination. Different orbitals used = different bonds.
c. What have experiments shown?
Experiments have shown that all bonds are actually equivalent (i.e. identical to
one another)..
d. What does this indicate?
This means that carbon cannot be using its atomic orbitals (2s and 2p). If all
bonds are identical is must be using 4 equivalent orbitals to bond with the
hydrogens.
21. This means that when an atom patricpiates in a bond, its atomic orbitals (regular
valence shell orbitals) are combined to form a hybrid. This new hybrid orbital is what
particpates in the bond.
22. How does the energy of the hybrids compare to the atomic orbitals?
The hybrid orbital has an energy level somewhere between the energy level of the
composite atomic orbitals.
23. The number of hybrid orbitals formed is equal to the number of atomic orbitals used.
24. What should you do first, when trying to determine the type of hybrid orbitals needed?
Draw out your Lewis Dot structure.
a. Why?
It will help you to determine the number of areas of electron density (or steric
number) of each of atom in the molecule. The type of hybrid used depends on
the steric number.
b. How many areas of electron density do the following account for?
i. Lone Pairs - one
ii. Single bond - one
iii. Double Bond - one
iv. Triple Bond – one
25. Fill In the Following
26. Draw the following
a. CH4
Step 1 – Lewis Dot
Step 2 – Determine the Steric Number for Atoms in Molecule
Carbon
Steric Number = 4
This means we will need to create a hybrid that has 4 orbitals.
This means we need to combine 4 atomic orbitals. One s and three p orbitals –
which come together to form 4sp3 hybrids.
Hydrogen
doesn’t create hybrid orbitals because it only has a 1s electron available
Thus, the molecule would look like:
b. CH2O
Step 1 – Lewis Dot
Step 2 – Determine the Steric Number for Atoms in Molecule
Carbon
Steric Number = 3
This means that we need to create 3 hybrid orbitals.
This means we need to combine 3 atomic orbitals. One s and two p orbitals –
which come together to form 3 sp2 hybrids and one unhybridized p is left over.
Oxygen
Steric Number = 3
This means that we need to create 3 hybrid orbitals.
This means we need to combine 3 atomic orbitals. One s and two p orbitals –
which come together to form 3 sp2 hybrids and one unhybridized p is left over.
Hydrogen
Does not form hybrids
Thus the molecule would look like:
27. What are sigma and pi bonds?
Sigma bonds are bonds that occur when orbitals have a head on overlap.
These are all examples of sigma bonds.
Pi bonds overlapping parallel p orbitals.
These are two ways to represent the pi bond.
28. What are each type of bond composed of?
Single = 1 sigma bond
Double = 1 sigma bond + 1 pi bond
Triple = 1 sigma bond + 2 pi bonds
29. Predict the shape, hybridization of central atom and polarity for the following
a. OF2
Shape – Bent (4 areas of e- density and 2 lone pairs)
sp3 hybridization (Steric Number = 4, combination of 4 atomic orbitals)
Polar
b. TeF4
Shape – See Saw ( 5 areas of e- density and 1 lone pair)
sp3d hybridization (Steric Number = 5, combination of 5 atomic orbitals)
Polar
c. BF3
Shape – Trigonal Planar (3 areas of e- density)
sp2 hybridization (Steric Number = 3, combination of 3 atomic orbitals)
Non-polar (symmetry cancels out polar bonds)
30. Label the hybridization of C, O, and N in the following molecules. Also count total
number of sigma bonds and total number of pi bonds.
31. Are all the atoms in the same plane?
a. C2H2
This would be a linear molecule (2 areas of e- density around each carbon, no lone
pairs)
b. CH2CCH2
All of the atoms are not in the same plane. In order for the central carbon to
double bond to each of the outer carbons they must form a pi bond. Pi bonds are
made up of parallel p orbitals. This means that one of the carbons will have to
bond with the py and the other will bond with the px. The sigma bonds occur on
the z plan.
32. How can CO32- help us understand the short falls of LE model?
If you look at the Lewis Dot for CO32-:
This is a resonance structure. As you can see based on where we position the double
bond, the hybridization of the oxygen atoms change. We cannot label it sp2 or sp3. Our
model is falling short here because we cannot describe resonance structures. This means
that we need to revise out understanding a bit.
33. Orbitals are wave functions. One property of waves is that they can constructively or
destructively interfere with each other.
34. What is constructive interference?
When two waves constructively interfere the resultant wave has an increased amplitude.
35. What is destructive interference?
When two waves destructively interfere they basically cancel each other out.
36. How does this relate to our understanding of orbital interactions?
Just as waves have a positive phase
s orbital phases:
and negative phase
- so do orbitals.
p orbitals phase
* a node is a location where there can be no electrons.
When orbitals of the same phase interact with each other a bond forms. The electrons are
located, primarily, between the two nuclei of each atom.
When orbitals of different phases interact with each an “anti-bond” occurs. There is a
node between the two nuclei of the atoms.
37. What is Molecular Orbital Theory?
In this theory the atomic orbitals of different atoms are combined to make new bonding
orbitals called molecular orbitals.
38. Facts about molecular orbitals?
a.
Molecular orbitals can hold up to 2e- with opposite spin.
b.
(Molecular Orbital)2 = 90% probablility of finding e- within orbital.
c. Only Molecular Orbitals exist after combining, no atomic orbitals.
d.
Molecular orbitals can be used to electron configuration for molecules.
e.
# of atomic orbitals used = # of molecular orbitals formed.
f. When atomic orbitals interact they can do so constructively and destructively.
g. Constructive interference produces bonding molecular orbitals that are lower in
energy than the atomic orbitals contributors.
h. Destructive interference produces anti-bonding molecular that are higher in
energy than the atomic orbital contributors.
i. Generally sigma bonds are lower in energy than pi bonds.
39. Why do bonds occur?
The energy of the atoms combined is lower than the sum of all their energies independent
from one another.
40. Fill in the following chart
Remember that degeneracy is when orbitals are at the same energy level.
41. What is a Molecular Orbital Diagram?
A diagram that shows which atomic orbitals are being contributed to by the atoms and
what molecular orbitals they will form. Additionally it indicates the relative energy level
of all orbitals.
42. Draw the Molecular Orbital Diagram
a. H2
When filling in the diagram always fill in lowest available molecular orbital first.
b. H2+
In this case the molecule being formed is H2+. The positive charge indicates that
one electron has been lost – thus only one electron will be contributed. It does not
matter which hydrogen you leave as the electron contributor.
c. H2-
In this case the molecule being formed is H2-. The negative charge indicates that
an extra electron has been gained – thus three electrons are being contributed. It
doesn’t matter which hydrogen you add the electron to.
43. What is bond order?
Bond order indicates the relative strength/stability of a bond.
The formula for determining the bond order of a compound is:
44. Some facts about bond order:
a. A bond order of zero indicates that the compound does not exist.
b. The bond order typically corresponds to the type of bond present.
i.e.
Bond order = 1 indicates a single bond
Bond order = 2 indicates a double bond
Bond order = 3 indicates a triple bond
45. What do the bond orders for #41 tell us about relative stabilities of H2 , H2+ ,H2-?
H2 = (2 – 0) / 2 = 1
H2+ = (1-0)/2 = ½
H2- = (2-1)/(2) = ½
Based on these bond orders we can see that H2 is the most stable compound.
46. Paramagnetic substances have unpaired electrons and are therefore attracted toward a
magnetic field.
Diamagnetic substances have paired electrons and are therefore repelled by a magnetic
field.
47. What is the order of molecular orbitals for B2, C2 and N2?
σ2s σ*2s π2p σ2p π*2p
Remember that there are 2 degenerate orbitals in both the π and π*. Thus there are a total
of 4 electrons that are contained within the π orbitals.
48. What is the order of molecular orbitals for O2 and F2?
σ2s σ*2s σ2p π2p π*2p
49. Draw the Molecular orbital diagrams for N2 and O2.
50. Why do B2, C2 and N2 have a different MO order compared to O2 and F2?
π2p is lower in energy for B2, C2 and N2 because of e- repulsion σ2p has with σ2s.
Additionally there is some s/p orbital mixing that changes energies.
σ2p is lower in energy because as protons are added the 2s orbital diminishes in size such
that the e- repulsion is minimalized.
51. Write out the MO electron configuration for each.
N22-, O22-, F22N2 2-:
O22-:
F22-:
a. Which of the following is predicted to be the most stable diatomic species?
In order to determine this we will have to look at the bond orders of each
molecule. You can look at all the electrons or just the valence electrons to
determine the bond order – you will get the same answer using either method.
N22-: using all electrons (10-6)/2 = 2 , gives the same answer as looking only at
the valence electrons (8-4)/2 = 2.
O22-: (8-6)/2 = 1
F22-: (8-8)/2 = 0
Based on these bond order values we can determine that N22- is the most stable of
all the species. We can also determine that F22- doesn’t exist as it has a bond
order equal to zero.
b. Indicate whether each is paramagnetic or diamagnetic.
In order to determine this, you will need to determine if the electrons in the last
molecular orbital are paired up or not. If they are paired up, it is diamagnetic; if
not, paramagnetic.
N22- - paramagnetic
O22-: diamagnetic
F22-: paramagnetic
52. Use the MO model to predict the bond order and magnetism for Ne2 and P2.
Ne2:
Because Ne comes after O and F on the periodic table, we will use their molecular orbital
configuration.
Bond Order = (8-8)/2 = 0
This compound does not exist.
P2 :
Because P is in the same family as N, we will use the B2, C2, N2 configuration.
Bond Order = (8-2)/2 = 3
Diamagnetic
53. Complete the MO diagram for NO.
When creating the MO diagram for a molecule that has elements has B, C or N and O or
F – default to the B, C, and N MO order. Notice that when using molecular orbital
theory, there is absolutely no issue with an odd number of electrons, unlike Lewis.
54. How do we reconcile these two theories (i.e. LE and MO models)?
For simplicities sake we will view σ bonds as being localized (hybrid orbitals) and we
will view π bonds through molecular orbital theory, in which the electrons are
delocalized.
To better understand this, we will consider benzene, C6H6, a resonance structure.
In the combined view of localized σ bonds and delocalized π, the new depiction looks
like: