Transcript Expanded Octet, pi and sigma bonds, bonl length and strength

Today’s Class
-Exceptions to the octet rule
-Bond Length and Strength
-Pi and Sigma bonds
-Linking activity
Exceptions to the Octet rule
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The rules that we have used for Lewis structures
apply to most molecules
However, there are cases where the importance
of an octet of electrons is called into question
There are three types of exceptions to the rule:
1) Incomplete octet
2) Expanded octet
3) Odd-electron molecules
Incomplete octet
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Example Boron-group 3A row 2
Tends to form compounds where it has less than 8
electrons surrounding it. (incomplete octet)
BF3-24 electrons
In this structure, Boron only has 6 electrons around it.
The octet rule can be satisfied be drawing a structure
with a double bond
Incomplete octet
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Since fluorine is so much more electronegative
than boron, the structure seems doubtful.
So it is more characteristic for boron to form
molecules where it lacks 8 valence electrons.
On the other hand the elements that are in the
same row as boron obey the octet rule(carbon,
nitrogen, oxygen, and fluorine)
Other examples of elements who do not obey
octet, Beryllium and hydrogen
Expanded Octet
Some atoms appear to exceed the octet
rule
 Behavior observed for elements in period
3 of the periodic table and beyond
 Consider the compound SF6
 Has 48 valence electrons
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Example continued
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We used 12 electrons to form the S-F bonds, which leaves 36
electrons
Since fluorine always follows the octet rule, we completed octets
to give the structure.
Sulfur has 12 electrons around it, therefore it exceeds the octet
rule. How can this happen?
Involves using the empty 3d orbitals on the third-period elements
3rd row elements have 3s, 3p, and 3d orbitals.
The 3s and 3p fill with electrons but the 3d orbitals remain empty
So those 3d orbitals in sulful can be used to accommodate extra
electrons used the 3s and 3p to hold 8 electrons, with the 4 extra
in the formerly empty 3d orbitals
3s
3p
3d- once empty
Summary
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The second-row elements C, N, O, and F should be
assumed to obey octet rule
The second-row elements BE, and B often have fewer
than 8 electrons around them in their compounds
The second-row elements never exceed the octet rule
(valence orbitals can only accommodate only 8
electrons)
Third row and heavier elements often satisfy octet rule,
but can exceed rule by using their empty valence d
orbitals
When writing Lewis structures for a molecule first draw
single bonds between all bonded atoms, and then satisfy
the octet rule for all the atoms. If electrons remain, place
them on the elements having available d orbitals(3rd
period or beyond
Odd-Electron Molecules
Relatively few molecules contain an odd number
of electrons
 Because we need an even number of electrons
for complete pairing the octet rule clearly can’t
be satisfied. (if there is an odd number)
 Example NO-nitric oxide or NO2-nitrogen dioxide
N has five valence electrons
O has six valence electrons
Lewis structures can be drawn for these molecules
but to treat them accurately, a different model is
needed (won’t be discussed)
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Practice
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Explain the bonding in:
ICl4BeF2
PCl5
Bond Length
Bond length is the distance between two bonded
atoms in a molecule
 Bond length is inversely related to bond
order(number of bonds between atoms)
 When more electrons participate in bond
formation the bond will get shorter.
Example: C-C bond length: 154pm
C=C bond length:134pm
C=C bond length:120pm
Bond length is also inversely related to bond
strength, a stronger bond is a shorter bond
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Example bonding and resonance
Ozone-O3 Lewis Structure
 As we’ve seen before, choosing
which O to take the extra electrons
is arbitrary. Therefore Ozone is a resonance structure:
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We know from experimental data that the bonds between the Os are
the same length, therefore neither of these structures are correct
because we know a double bond is shorter than a single
Instead when drawing the structure of ozone it is better like this:
Now each bond can be displayed correctly
Being the same length
Bond Strength
Bond energy is a measure of the strength
of a chemical bond. The larger the bond
energy, the stronger the bond.
 The strength of the bond is determined by
the amount of energy required to break the
bond.
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Sigma and Pi Bonds
Pi bonds- a covalent bond in which parallel p
orbitals share an electron pair occupying
the space above and below the line joining
the atoms
Sigma bonds- a covalent bond in which the
electron pair is shared in an area centered
on a line running between the atoms
Example of Pi and Sigma Bonds
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Take N2 for example, has a linear structure
that looks like this:
N
N
So the middle bond will be a sigma bond, and the two
outer bonds will be pi bonds
Activity
Use the words given to create links between
words and describe the correlation
between them.