LEWIS STRUCTURES
MOLCULAR GEOMETRY
&
HYDRIDIZATION
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
WHEN THE CENTRAL ATOMS IS AN ELEMENT
FROM THE 3rd PERIOD OR BELOW, WE
SOMETIMES FIND COMPOUNDS IN WHICH
THERE ARE MORE THAN EIGHT ELECTRONS
AROUND THE CENTRAL ATOM.
THIS ARRANGEMENT IS POSSIBLE BECAUSE
THE d ORBITALS IN THE VALANCE SHELL OF
THESE ATOMS HAVE ENERGY VALUES CLOSE
TO THOSE OF THE p ORBITALS
TYPES OF COVALENT BONDS
SINGLE BONDS
•LONGEST OF THE 3 TYPES
•WEAKEST OF THE 3 TYPES
•CONTAINS ONE PAIR OF ELECTRONS (2 ELECTRONS)
DOUBLE BONDS
•LENGTH IS GREATER THAN TRIPLE BUT LESS THAN SINGLE
•STRENGTH IS GREATER THAN SINGLE BUT LESS THAN TRIPLE
•CONTAINS TWO PAIRS OF ELECTRONS (4 ELECTRONS)
TRIPLE BONDS
•SHORTEST OF THE 3 TYPES
•STRONGEST OF THE 3 TYPES
•CONTAINS THREE PAIRS OF ELECTRONS (6 ELECTRONS)
LONGEST BONDS(SINGLE) ARE THE WEAKEST
SHORTEST BONDS(TRIPLE) ARE THE STRONGEST
VSERP THEORY states that electron pairs found in
the outer energy level or valence shell of atoms repel
each other and thus position themselves as far apart
as possible
The VSERP theory helps to predict the shape of the molecule. The following points
should be considered in predicting shape (molecular geometry):
lone-lone pair>lone-bonding pair>bonding-bonding pair
FOR EXAMPLE, IN THE LEWIS STRUCTURE OF WATER, THERE ARE TWO NONBONDED (LONE) PAIRS OF ELECTRONS ON THE OXYGEN ATOM. THUS THE
NON-BONDED (LONE) PAIR OF ELECTRONS REPEL MORE THAN THE LONEBONDED PAIR BETWEEN THE OXYGEN AND HYDROGEN AND THE BONDEDBONDED PAIR OF ELECTRONS BETWEEN THE TWO HYDROGENS HAVE THE
LEAST REPULSION.
NON-BONDED (LONE)&
NON-BONDED (LONE)
PAIR OF ELECTRONS
NON-BONDED(LONE)
& BONDED PAIR
OF ELECTRONS
BONDEDBONDED PAIR OF
ELECTRONS
Due to the fact that lone pair of
electrons repel more than bonded
pairs, the bond angles will be different
in different shaped molecules
BEND(with 2 lone pair of e-) <TRIGONAL PYRAMIDAL<TETRAHEDRAL<
BEND(with 1 lone pair of e-)<TRIGONAL PLANAR<LINEAR
****KNOW THE BOND ANGLES OF THIS SHAPES
TETRAHEDRAL
109.5o
TRIGONAL PLANAR
LINEAR
180o
120o
..
S
// \
:O: :O:
..
BENT (w/
TRIGONAL PYRAMIDAL
109.5o (107.5o )
1 lone pairs
of e-)
120o
(119o)
BENT (w/ 2 lone pairs of e-)
109.5 o (104.5o)
ETHANOIC ACID
CH3COOH
BENT
104.5o
TETRAHEDRAL
TRIGONAL
PLANAR
CAN A MOLECULE BE NONPOLAR
AND STILL HAVE POLAR BONDS
WITHIN THE MOLECULE???
FIRST THINK ABOUT WHAT MAKES A BOND POLAR!
NOW THINK ABOUT WHAT MAKES A MOLECULE NONPOLAR!
YES
THE DIFFERENCE IN ELECTRONEGATIVITY VALUES (CHARGE
SEPARATION) DETERMINE IF A BOND WILL BE POLAR OR NONPOLAR.
0-.3 difference is NONPOLAR
.3-1.7 difference is POLAR
IF A MOLECULE IS SYMMETRICAL OR IF THERE IS NO NET DIPOLE
MOMENT THEN IT IS NONPOLAR
EXAMPLE: CCl4
THE C-Cl BOND IS POLAR BUT THE MOLECULE IS
TETRAHEDRAL SO IT IS SYMMETRICAL AND THUS
NONPOLAR
WHICH OF THE FOLLOWING BONDS IS THE MOST POLAR
B-C or C-O or N-O or O-F
ANSWER:
C-O
b/c they are further apart on the periodic table which means that
they have the greatest charge separation.
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
HYBRIDIZATION IS THE MIXING OF
ORBITALS
THE NEW HYBRID ORBITALS FORMED
MAY BE sp, sp2, sp3, (sp3d, sp3d2 not covered)
Formation of sp hybrid orbitals
The combination of an s orbital and a p orbital produces 2
new orbitals called sp orbitals.
These new orbitals are called hybrid orbitals
The process is called hybridization
What this means is that both the s and one p orbital are
involved in bonding to the connecting atoms
EXAMPLES OF MOLECULES WITH sp HYBRIDIZATION
ALL LINEAR MOLECULES
Formation of sp2 hybrid orbitals
EXAMPLES OF sp2 HYBRIDIZATION
ALL
TRIGONAL PLANAR
AND
BENT(one lone pair of e-)
..
S
// \
:O: :O:
..
Formation of sp3 hybrid orbitals
EXAMPLES OF sp3 HYBRIDIZATION
ALL
TETRAHEDRAL
TRIGONAL PYRIMIDAL
&
BENT (w/ two lone pair of e-)
Hybrid orbitals can be used to explain bonding and molecular geometry
Multiple Bonds
Everything we have talked about so far has only dealt with
what we call sigma bonds
Sigma bond (s)  A bond where the line of electron density
is concentrated symmetrically along the line connecting the
two atoms.
Pi bond (p)  A bond where the overlapping regions exist
above and below the internuclear axis (with a nodal plane
along the internuclear axis).
Example: H2C=CH2
Example: H2C=CH2
Example: HCCH
Delocalized p bonds
When a molecule has two or more resonance structures, the pi
electrons can be delocalized over all the atoms that have pi
bond overlap.
Example: C6H6 benzene
Benzene is an excellent example. For benzene the p orbitals
all overlap leading to a very delocalized electron system
In general delocalized p bonding is present in all molecules
where we can draw resonance structures with the multiple
bonds located in different places.