# Molecular Geometry Lab Report

Received 5/4/17Lewis Structure Exercise
A Lewis structure shows how the valence electrons are arranged and indicates the bond i ng between atoms in a molecule.
We represent the elements by their symbols. The shared electron pair is shown as a line/bond between the two atoms. All
the other valence electrons are shown as dots or lines around the symbol of the element.
For Example: The Lewis structure for Cl2 is
Let us now see how to draw the Lewis structure f or CO2
Steps f or drawing Lewis Structures
Example with CO2
1. Sum the valence electrons of all the atoms.
The total number of valence electrons is 16 (4 from carbon
• For anions, add one electron for each negative
and 6 from each oxygen).
charge.
• For cations, subtract one electron for each positive
charge.
2. Choose the least electronegative element (other
than H) as the central atom.
• For a molecule of type XVn, X is always the central
atom.
Choose carbon as the central atom since it is less
electronegative and follows the XYn rule.
3. Use single bonds (lines) to connect the central atom
to the surrounding atoms.
Attach the two oxygens to carbon by single bonds.
O
C
O
4. Subtract 2 electrons for each bond from the original
total number of electrons.
We subtract 4 electrons (2 for each bond) from 16, leaving 12
electrons to distribute to the remaining atoms.
5. Complete the octet for all outer atoms (other than
H). Each bond to an atom will count as 2 electrons
for that atom’s octet.
Complete the octet for each of the oxygen atoms which use
the last 12 electrons. (We need to add 6 electrons to each
oxygen.)
O
C
O
6. Complete the octet for the central atom.
There are no more electrons, so we cannot add more to
the central atom to complete its octet.
7. If you run out of electrons before the octet for the
central atom can be completed, start forming
multiple bonds until the central atom has a
complete octet. When you form a multiple bond,
remember to remove an electron pair from the
outer atom.
Since carbon does not have an octet, form multiple bonds
by taking one of the lone pairs:
O
C
O
O
C
O
The central atom is still electron deficient, so share
another pair:
O
8. Make sure the correct number of electrons has
been used and that every atom’s octet is complete.
The exceptions are atoms of group 2 and 3 which
can have incomplete octets and atoms in period 3
or greater which can have expanded octets.
C
O
O
C
O
4 bonds + 2 electron pairs equal 16 electrons. All atoms
have an octet.
O
C
O
Page 1 of 5
Molecular Geometry‐ the Valence Shell Electron Pair Repulsion (VSEPR) Theory
The electron groups around the central atom repel each other and therefore prefer to be as far apart from each other as
possible. This is the main idea of the VSPER theory. We can apply the VSEPR theory to predict the molecular
shape/geometry of a molecule.
1. Draw the Lewis structure for the molecule in question.
2. Count the total number of electron groups on the central atom. Add the number of atoms bonded to the central
atom and the number of lone pairs on the central atom ‐ this is the total number of electron groups. Note that
multiple bonds to one outer atom still count as one electron group.
3. The arrangement of the electron groups is determined by minimizing the repulsions between them.
4. Remember that lone pairs require more space than bonding pairs. Therefore, choose an arrangement that gives
lone pairs as much room as possible.
The attached table shows the relationship between the number of electron pairs and the molecular geometry.
Polarity of a molecule
A covalent bond is polar if there is a difference in electronegativity between the bonded atoms. A molecule like HCI has a
polar covalent bond since there is a difference in electronegativity between the two atoms (the difference is greater than
1.4 and less than 1.8). Thus, HCI possesses a permanent dipole moment because the molecule has a distinct negative end
and a distinct positive end. However, just because there is a polar bond present in a molecule does not necessarily mean
that the molecule is polar. If all the dipoles in the molecule cancel each other out, then the molecule will be non‐polar. For
example, C02 has two polar bonds, but they point in opposite directions and cancel each other out.
Formal Charges
The formal charge of any atom in a molecule is the representation of electron distribution on the atom. Remember that the
formal charge does not represent the real charge on the atom. It is a fictitious charge assigned to each atom that helps in
finding the best Lewis structure for a molecule.
The best Lewis structure will
• Have the lowest possible formal charge on each atom
• Put the negative formal charge on the most electronegative atom (and a positive formal charge on the least
electronegative atom)
Calculating formal charges
• Sum all the electrons in the lone pairs belonging to that atom
• Add to this half of the bonding electrons
• Subtract this total from the number of valence electrons for that atom to get its formal charge.
Valence electrons:
‐(Electrons assigned to atom):
Formal charge:
O
C
O
O
C
O
6
6
0
4
4
0
6
6
0
6
7
‐1
4
4
0
6
5
+1
In the above example for C02, the Lewis structure on the Left is the better one as it places a formal charge of zero on each
atom. Recall that the formal charges sum to zero for a molecule and to the charge for an ion.
Page 2 of 5
Resonance
Sometimes one Lewis structure is not enough to describe a molecule completely. For example, ozone (O3) can be
represented by the following two structures:
One might expect ozone to have one single bond and one double bond based on either of the above Lewis structures.
However, experimentally it is found that both the bonds are equivalent and intermediate between a single and a double
bond. Thus, the true structure of ozone is a resonance hybrid of the above two Lewis structures. We represent resonance
by drawing the two structures with a double‐headed arrow between them. Remember that although we may draw two
resonance structures, there is actually only one structure that is a hybrid of the two drawings‐ the molecule does not “flip”
back and forth, but rather is permanently somewhere between the two structures. Each bond i n ozone is a combination of
one single and half a double bond.

In the above example, both the Lewis structures are equivalent and contribute equally to the true structure of ozone.
However, sometimes some of the Lewis structures are preferred over others (see the discussion on formal charges above).
In that case, the preferred Lewis structure(s) contribute more to the overall structure of the molecule.
Hybridization
To explain molecular geometries, we assume that orbitals mix together to form new orbitals. This process of mixing atomic
orbitals is called hybridization. The new orbitals are called hybrid orbitals. The number of hybrid orbitals formed will be
equal to the total number of orbitals that are mixing together.
Page 3 of 5
VSEPR Geometries
Number of
electron groups
on central atom
Hybridization
2
sp
3
Sp2
4
Sp3
5
Sp3d
0 lone pair on
central atom
1 lone pair on
central atom
2 lone pair on
central atom
3 lone pair on
central atom
4 lone pair on
central atom
Page 4 of 5
6
Sp3d2
Page 5 of 5
Revised 7/11/17
0Name:
Station #
Date:
CHM 101 Molecular Geometry
Pre-lab Questions
1. Give a short summary of the main points of VSEPR theory.
2. Fill in the following blanks:
a) A molecule that has two atoms and two lone pairs of electrons off of the central atom
has a VSEPR formula of
. Its approximate bond angle is
.
b) A molecule that has four atoms and one lone pair of electrons off of the central atom
has an electron pair geometry of
. Its approximate
bond angles are
.
c) A molecule that has four atoms and two lone pairs of electrons off of the central atom
has a
electron pair geometry and a molecular shape of
.
3. Draw Lewis Dot structures and predict the bond angles in the following molecules:
a) PF3
b) SF6
c) ICl5
d) CO2
e) H2Se
Page 1 of 1
Revised 7/18/17
Name ____________________________ Station # ______ Date___________________
CHM 101 – Molecular Geometry
Post Lab Questions
1. For the following molecules/ions
a. Draw all possible resonance Lewis structures.
b. Assign formal charges for all atoms in each resonance structure.
c. Circle the favored resonance form; if all forms are equivalent, circle none.
N2O (Connectivity is N-N-O)
CO32−
CH3COOH (both O atoms are connected to the second C, last H attached to O)
Page 1 of 2
Revised 7/18/17
2. The following are the two possible Lewis structures for C2H2F2. Will both of them have the
same dipole moment? Explain clearly.
Page 2 of 2
Lewis Structure Exercise
A Lewis structure shows how the valence electrons are arranged and indicates the bond i ng between atoms in a molecule.
We represent the elements by their symbols. The shared electron pair is shown as a line/bond between the two atoms. All
the other valence electrons are shown as dots or lines around the symbol of the element.
For Example: The Lewis structure for Cl2 is
Let us now see how to draw the Lewis structure f or CO2
Steps f or drawing Lewis Structures
Example with CO2
1. Sum the valence electrons of all the atoms.
The total number of valence electrons is 16 (4 from carbon
• For anions, add one electron for each negative
and 6 from each oxygen).
charge.
• For cations, subtract one electron for each positive
charge.
2. Choose the least electronegative element (other
than H) as the central atom.
• For a molecule of type XVn, X is always the central
atom.
Choose carbon as the central atom since it is less
electronegative and follows the XYn rule.
3. Use single bonds (lines) to connect the central atom
to the surrounding atoms.
Attach the two oxygens to carbon by single bonds.
O
C
O
4. Subtract 2 electrons for each bond from the original
total number of electrons.
We subtract 4 electrons (2 for each bond) from 16, leaving 12
electrons to distribute to the remaining atoms.
5. Complete the octet for all outer atoms (other than
H). Each bond to an atom will count as 2 electrons
for that atom’s octet.
Complete the octet for each of the oxygen atoms which use
the last 12 electrons. (We need to add 6 electrons to each
oxygen.)
O
C
O
6. Complete the octet for the central atom.
There are no more electrons, so we cannot add more to
the central atom to complete its octet.
7. If you run out of electrons before the octet for the
central atom can be completed, start forming
multiple bonds until the central atom has a
complete octet. When you form a multiple bond,
remember to remove an electron pair from the
outer atom.
Since carbon does not have an octet, form multiple bonds
by taking one of the lone pairs:
O
C
O
O
C
O
The central atom is still electron deficient, so share
another pair:
O
8. Make sure the correct number of electrons has
been used and that every atom’s octet is complete.
The exceptions are atoms of group 2 and 3 which
can have incomplete octets and atoms in period 3
or greater which can have expanded octets.
C
O
O
C
O
4 bonds + 2 electron pairs equal 16 electrons. All atoms
have an octet.
O
C
O
Page 1 of 5
Molecular Geometry‐ the Valence Shell Electron Pair Repulsion (VSEPR) Theory
The electron groups around the central atom repel each other and therefore prefer to be as far apart from each other as
possible. This is the main idea of the VSPER theory. We can apply the VSEPR theory to predict the molecular
shape/geometry of a molecule.
1. Draw the Lewis structure for the molecule in question.
2. Count the total number of electron groups on the central atom. Add the number of atoms bonded to the central
atom and the number of lone pairs on the central atom ‐ this is the total number of electron groups. Note that
multiple bonds to one outer atom still count as one electron group.
3. The arrangement of the electron groups is determined by minimizing the repulsions between them.
4. Remember that lone pairs require more space than bonding pairs. Therefore, choose an arrangement that gives
lone pairs as much room as possible.
The attached table shows the relationship between the number of electron pairs and the molecular geometry.
Polarity of a molecule
A covalent bond is polar if there is a difference in electronegativity between the bonded atoms. A molecule like HCI has a
polar covalent bond since there is a difference in electronegativity between the two atoms (the difference is greater than
1.4 and less than 1.8). Thus, HCI possesses a permanent dipole moment because the molecule has a distinct negative end
and a distinct positive end. However, just because there is a polar bond present in a molecule does not necessarily mean
that the molecule is polar. If all the dipoles in the molecule cancel each other out, then the molecule will be non‐polar. For
example, C02 has two polar bonds, but they point in opposite directions and cancel each other out.
Formal Charges
The formal charge of any atom in a molecule is the representation of electron distribution on the atom. Remember that the
formal charge does not represent the real charge on the atom. It is a fictitious charge assigned to each atom that helps in
finding the best Lewis structure for a molecule.
The best Lewis structure will
• Have the lowest possible formal charge on each atom
• Put the negative formal charge on the most electronegative atom (and a positive formal charge on the least
electronegative atom)
Calculating formal charges
• Sum all the electrons in the lone pairs belonging to that atom
• Add to this half of the bonding electrons
• Subtract this total from the number of valence electrons for that atom to get its formal charge.
Valence electrons:
‐(Electrons assigned to atom):
Formal charge:
O
C
O
O
C
O
6
6
0
4
4
0
6
6
0
6
7
‐1
4
4
0
6
5
+1
In the above example for C02, the Lewis structure on the Left is the better one as it places a formal charge of zero on each
atom. Recall that the formal charges sum to zero for a molecule and to the charge for an ion.
Page 2 of 5
Resonance
Sometimes one Lewis structure is not enough to describe a molecule completely. For example, ozone (O3) can be
represented by the following two structures:
One might expect ozone to have one single bond and one double bond based on either of the above Lewis structures.
However, experimentally it is found that both the bonds are equivalent and intermediate between a single and a double
bond. Thus, the true structure of ozone is a resonance hybrid of the above two Lewis structures. We represent resonance
by drawing the two structures with a double‐headed arrow between them. Remember that although we may draw two
resonance structures, there is actually only one structure that is a hybrid of the two drawings‐ the molecule does not “flip”
back and forth, but rather is permanently somewhere between the two structures. Each bond i n ozone is a combination of
one single and half a double bond.

In the above example, both the Lewis structures are equivalent and contribute equally to the true structure of ozone.
However, sometimes some of the Lewis structures are preferred over others (see the discussion on formal charges above).
In that case, the preferred Lewis structure(s) contribute more to the overall structure of the molecule.
Hybridization
To explain molecular geometries, we assume that orbitals mix together to form new orbitals. This process of mixing atomic
orbitals is called hybridization. The new orbitals are called hybrid orbitals. The number of hybrid orbitals formed will be
equal to the total number of orbitals that are mixing together.
Page 3 of 5
VSEPR Geometries
Number of
electron groups
on central atom
Hybridization
2
sp
3
Sp2
4
Sp3
5
Sp3d
0 lone pair on
central atom
1 lone pair on
central atom
2 lone pair on
central atom
3 lone pair on
central atom
4 lone pair on
central atom
Page 4 of 5
6
Sp3d2
Page 5 of 5

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