Covalent Bonding: Orbitals
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Some atoms will form
hybridized orbitals when
bonding with other atoms.
There are three that are
included on the AP exam.
sp, sp2 and sp3
To form the molecule
methane, carbon must bond
with four hydrogens. If the
electron configuration for
carbon is 1s22s22p2 then that
only leaves 2 electrons to
bond with 2 hydrogens but
its been shown by
experiment that carbon
bonds with 4 hydrogens.
How? Carbon forms a
hybridized sp3 orbital
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sp2 hybridization occurs
with ethylene. A double
bond can act as one
effective pair. The
ethylene carbon forms a
trigonal planar
arrangement with bond
angles of 120o. sp3
hybridization forms
109.5o angles so that
won’t work. Combining
one s and two p orbitals
allows for this
arrangement.
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This is the most common
type of hybridization and
it is found in molecules
whose central atom is
surrounded by a stable
octet. CH4 is a classic
example.
Each hydrogen has an
electron in the 1s while
carbon has 2 in the s and 2
in the p lobe.
These s and p orbitals
combine to make an sp3
hybrid orbital.
Hybridization
# bonds
# lone pairs
Molecular shape
Bond angle
Example
sp
2
0
Linear
180
CO2
sp2
3
0
Trigonal planar
120
SO3
sp2
2
1
Angular
<120
SO2
sp3
4
0
Tetrahedral
109.5
CH4
sp3
3
1
Trigonal pyramidal
<109.5
NH3
sp3
2
2
Bent (angular)
<109.5
H2O
sp3d
5
0
Trigonal pyramidal
120, 90
PCl5
sp3d
4
1
Seesaw
<120
SF4
sp3d
3
2
T-Shaped
<90
CF3
sp3d
2
3
Linear
180
XeF2
sp3d2
6
0
Octahedron
90
SF8
sp3d2
5
1
Square pyramidal
<90
IF5
sp3d2
4
2
Square planar
90
XeF4
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1. Determine Lewis structure
2. Determine the number of
electron domains around the
central oxygen atom.
The 4 electron domains are
arranged in a tetrahedron with
2 lone pairs.
Four sp3 hybrid orbitals are
oriented in a tetrahedral array.
Two of the sp3 orbitals are
occupied by bonding electron
pairs and two are occupied by
non-bonding pairs leading to
an angular or bent
arrangement.
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Four electron domains will be most likely to
exist as four sp3 hybrid orbitals.
Due to the large interelectron repulsion of the
lone-paired electrons, the usual 109.5o bond
angle found in a tetrahedron will be reduced
and closer to 104.5o between hydrogen atoms.
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Pi bonds are weaker than sigma
bonds because the side-on
overlap of p orbitals is less
effective than end-on overlap.
In a σ bond, the electrons are in
orbitals between the nuclei of the
bonding atoms. Electron density
is greatest between the nuclei.
The electrons attract the nuclei
and form a σ bond — the
strongest type of covalent bond.
In a π bond, the p orbitals
overlap side-on. The overlap is
less efficient, because the
electron density is off to the sides
of the σ bond. The electrons are
not as effective in attracting the
two nuclei. Thus, a π bond is
weaker than a σ bond.
ChemisNate 8:05
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The steric number of a
molecule is the number
of atoms bonded to the
central atom of a
molecule plus the
number of lone pairs on
the central atom. It is
often used in VSEPR
theory (valence shell
electron-pair repulsion
theory) in order to
determine the particular
shape, or molecular
geometry, that will be
formed.
Formal Charge
Sometimes atoms will have extra or not enough
electrons. This imbalance of electrons is denoted with a
formal charge. A negative formal charge means there are too
many electrons on atom. (Remember electrons are
negatively charged.) A positive formal charge means there
are not enough electrons on an atom. One confusing thing
about formal charges is that we do not simply count up all of
the electrons around an atom. Different types of electrons
are counted differently. Non-bonding electrons are counted
individually. However, electrons in bonds are counted as
being shared and so each pair of electrons in a bond counts
as only one electron.
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Formal charge is the difference between the
number of valence electrons on the free element
and the number of electrons assigned to the atom
in the molecule.
Formal charge = valence e- - nonbinding e- bonding e- ÷ 2.
OR: Formal Charge = (valence number) - (number
of bonds) - (non-bonding e's)
Calculate the formal charges for the practice
problems in the NMSI handout.
Formal charge for XeO3 explained.
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Bond order indicates bond strength. A high bond
order indicates more attraction between electrons.
A higher bond order also means that the atoms are
held together more tightly. With a lower bond
order, there is less attraction between electrons and
this causes the atoms to be held together more
loosely. Bond order also indicates the stability of
the bond. The higher the bond order, the more
electrons holding the atoms together, and therefore
the greater the stability.
Bond order = (number of electrons – number of
antibonding electrons) ÷ 2
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Bond order is the number of chemical bonds
between a pair of atoms. For example, in
diatomic nitrogen N≡N the bond order is 3, in
acetylene H−C≡C−H the bond order between
the two carbon atoms is also 3, and the C−H
bond order is 1. Bond order gives an indication
of the stability of a bond.
Bond order and length are inversely
proportional to one another. ie: the longer the
bond length the lower the bond order (less
attraction/strength)
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Draw the Lewis structure.
Count the total number of
bonds.
4
Count the number of bond
groups between
individual atoms.
3
Divide the number of
bonds between individual
atoms by the total number
of bonds.
4 ÷3 = 1.33
Bond order is 1.33
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Determine the bond
order for a nitronium
ion- NO2 +.
Bond total =
4
Bond groups =
2
Divide =
4÷2=2
Bond order is 2
Write electron configuration.
Homo vs heteronuclear. If
homonuclear the diagram is
symmetrical. If not, the more
electronegative atom goes
lower in the diagram (extra
electrons= more stability)
Fill in orbitals from bottom
to top.
Edvantage pages 440, 442
More examples (9:00)
Crash Course summary (10:51)
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Explain the following in terms of the electronic
structure and/or bonding of the compounds involved.
At ordinary conditions, HF (normal boiling point =
20oC is a liquid, whereas HCl normal boiling point = 114oC) is a gas.
1. The hydrogen is bonded to the highly
electronegative F atom and exhibits a special type of
dipole-dipole IMF called hydrogen bonding.
2. H-Cl molecules are held together with only dipoledipole forces (not FON) therefore they do not have
hydrogen bonding. It takes more energy to separate
the HF molecules than the HCl molecules due to the
added H-bonding of the HF molecule.
1 pt IMF, 1 pt energy
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Molecules of ASF3 are polar, whereas molecules of
AsF5 are nonpolar, why?
1. First draw out the Lewis structure to see what
you’re dealing with.
2. AsF3 is trigonal pyramidal and AsF5 is trigonal
bipyamidal.
3. AsF3 is polar because it has a net dipole moment
due to the presence of a lone pair of e- with no other
lone pair (of e-) oriented to cancel out the dipole.
4. AsF5 has all of its dipole moments cancelling each
other.
1 pt for AsF3 having a lone pair of electrons making it
polar, and 1 pt for AsF5 being nonpolar because its
dipole moments cancel.
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The N-O bonds in the NO2- ion are equal in
length, whereas they are unequal in HNO2.
Draw structures first. When drawing acids, the
hydrogen will always bond to the oxygen.
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NO2- has resonance structures with a bond
order of 1.5 for each N-O bond. This is because
there are two single and one double bond for 2
structures. HNO2 does not have resonance and
has one single bond which is longer than the
other double bond which has a shorter length.
1 pt resonance structure (you don’t have to
have bond order).
1 pt for HNO2 having two distinctly different
bonds-you can even go into sigma, pi if you
want to.
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For sulfur, SF2, SF4 and SF6 are known to exist
whereas for oxygen only OF2 is known to exist.
Sulfur has valence electrons in the 3rd principle
energy level (n=3). But since its on the 3rd level,
those electrons can expand its octet into the d
sublevel and form 6 bonds around the central
atom (SF6). Oxygen has valence electrons in
the 2nd energy level (n=2) and cannot expand
into the d orbitals because they are not
available. Therefore, oxygen cannot expand its
octet.