Transcript 14.2 Hybridization

14.2 Hybridization
(AND QUICK REVIEW OF INTERMOLECULAR
FORCES-4.3)
Warm-up 10/6
 Draw the Lewis Dot Structure, determine the
molecular shape and determine the polarity of the
following:
a. ICl2b. ICl4-
Now to 14.2-Hybridization
Hybridization
 CH4 - 1s2 2s2 2p2


How can we explain the 4 equivalent bonds?
Bonding electrons are in different orbitals

2 in the 2p orbitals and 2 in the 2s orbital
Hybrid orbitals are created
 Must be 4 equivalent orbitals
 CH4 = four sp3 hybrid orbitals
CH4
 4 sp3 orbitals
 sp3 orbitals look
like half p orbitals
http://www.mhhe.com/physsci/chemistry/essentialchemistry/
flash/hybrv18.swf
Key Ideas
 When atoms join to form molecules (except H)
 Outer atomic orbitals produce hybrid orbitals
 Focus only on the central atom
 Results in same number of hybrid orbitals as original
orbitals involved

Now they all have the same energy and are arranged
symmetrically
Hybridization Demonstration
Shapes to Hybrid Orbitals
sp = 2 charge centers (linear)
sp2 = 3 charge centers (trigonal
planar)
sp3 = 4 charge centers (tetrahedral)
Lewis Dot Structures to Hybrid Orbitals
 From Lewis Dot Structure:
 Number of orbitals around central atom = number of charge
centers
Example: H2O
-2 single bonds and 2 non-bonding pairs 
4 orbitals so sp3 hybridization
Practice IB Question
 Explain the meaning of the term hybridization. State the
type of hybridization shown by the carbon atoms in
carbon dioxide, diamond, graphite and the carbonate
ion.
 mixing/combining/merging of (atomic) orbitals to form
new orbitals (for bonding);
Allow molecular or hybrid instead of new.
Do not allow answers such as changing
shape/symmetries of atomic orbitals.
 Carbon dioxide: sp;
Diamond: sp3;
Graphite: sp2;
Carbonate ion: sp2;
5
Sigma and Pi Bonds

Sigma (σ) Bonds
End-to-end overlap

Pi bonds (π) Bonds
Sideways overlap
http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp
/bom5s2_6.swf
Single, Double, and Triple Bonds
Explaining the bond strengths/lengths
 Single bond is longest/weakest
 Least area of overlap (just sigma)
 Double bond shorter/stronger
 3 areas of overlap (one sigma and one pi)
 Pulls two nuclei closer together
 Triple bond shortest/strongest
 5 areas of overlap (1 sigma, 2 pi)
 Two nuclei even closer together
Homework
Alternative text Ch. 4
 Pg. 117

#1, 2
 Pg. 118
 #1, 2