Chemical Bonding
Chapter 12
GCC
CHM 130
12.1 Chemical Bonding
Atoms want to be like noble gases (stable and happy)
 Goal = 8 outer valence electrons =Octet Rule (except H, He)

 Metals lose electrons and become (+) cations
Nonmetals gain electrons and become (-) anions
 A metal / nonmetal compound is IONIC w/ IONIC bond
(the + and – attract each other)
 Who makes a good partner for Ca? Br? Li?
 Ionic Examples: KCl, CaBr2
Nonmetals can also share electrons with each other
 A nonmetal / nonmetal compounds is COVALENT
w/COVALENT bonds.
 Example: diatomic elements (H2, N2, F2…)
 Covalent Examples: H2O, CO2
Why is the formula CaBr2? Why one Ca
per two Br? Any ideas?
Because Ca ion is +2 and Br ion is -1. So
need two Br -1 ions to balance with one Ca
+2 ion. +2-1-1=0
(Br-1Ca2+Br-1)
The answer is NOT because Br is diatomic
Note, Br is diatomic BY ITSELF (Br2) but
when in a compound the Br ‘s break apart to
bond with other atoms! The diatomic
elements are NOT diatomic anymore once
bonded with others.
Which compounds are Ionic?
•
•
•
•
•
•
KBr
SO3
HCl
Br2
CO2
MgCl2
Answer: The ones with a
metal and a nonmetal.
KBr and MgCl2
12.2 Ionic Bonding
Electrons are completely transferred from metal to
nonmetal.
Draw electron dot structures for
Mg and S atoms then Mg2+ and S2ions in MgS. How many protons and
electrons in Mg2+ and S2- ions?
Notice that Mg2+ and S2- are “like” noble gases. They are
isoelectronic with Ne and Ar, and that is what makes
them happy and stable. While the number of electrons
changed, the number of protons did NOT. # Protons
never change in chemical reactions.
Ionic Radius
• Cations have lost electrons, so there are
more protons, so pull the electron orbits in
closer to nucleus (smaller than the atom)
• Anions have gained electrons, so there
are more electrons, they repel and push
orbits farther from nucleus (larger than the
atom)
True or False regarding an ionic bond
between aluminum and iodine?
1. The aluminum atom loses electrons, and the
iodine atom gains electrons.
2. The aluminum atom is larger in radius than the
True
True
aluminum ion.
3. The iodine atom is smaller in radius than the
iodine ion.
True
4. The aluminum and iodine ions form a bond by
attraction.
True
12.3 Covalent Bonding
…is when nonmetals share electrons.
Single = 2, double = 4, triple = 6 e- shared
Note the bond length is less than r1+r2 due to orbital overlap
Bond Energy
•=Energy required to break a bond.
•
•Breaking bonds always requires E.
• E is a reactant, it is absorbed. Endothermic
•Forming bonds always releases E. Exothermic
• E is a product, it is produced.
HCl (g) + heat
H (g) + Cl (g)
H (g) + Cl (g)
HCl (g) + heat
True or False regarding H2S?
1. Electrons are shared in H2S.
True
2. The bond between H and S is ionic.
False
3. The H-S bond length is less than
the sum of the two atomic radii.
True
4. Breaking the H-S bond releases
False
energy.
12.4 Draw electron dot
structures for H2 and HCl
12.4 Electron Dot Structures
1. Add up the total number of
valence electrons.
2. Surround the central atom with
the other atoms and draw single
bonds to them.
3. All atoms want octet 8e- except H
wants 2 e-.
4. Final Check: Make SURE you use
the total # of e-, no more or less.
bonding e-= shared elone pairs = unshared e-
If single bonds don’t work, try
double, then triple.
H2O
Total = 8e-
..
H :O
. .:
H
Examples
to put on
board:
•
HCN
•
CHCl3
•
CO2
•
NH3
The central atom is in bold.
12.5 Electron Dots of Polyatomic Ions
+
NH4
Add electrons for
anions and
+1
charge
means
one
less
e
subtract electrons
Total =
for cations. Put
5 + 4(1) – 1 = 8 ebrackets around
the ion and charge
in the right corner.
Examples for the board:
•
BrO3-
•
SO42-
•
CN-
The central atom is in bold.
12.10 Valence Shell Electron Pair Repulsion Theory
VSEPR
• Electron pairs (bonded and lone pairs) repel
each other and move as far away from each
other as possible.
• Molecular Shape or Geometry – the 3 D
arrangement of the atoms.
.Print out Shape Table from the web page
A = Central Atom
B = Outer Atom
E = Lone Pair on central atom
Linear – AB and AB2
Examples: H2, HCl, CO2
Bond
Angle is
180
Trigonal Planar – AB3
Example: Formaldehyde, CH2O
O
C
H
H
120°
Tetrahedral – AB4
Example: CH4, CF4, CH2F2
Bent – AB2E
Example: SO2
S
O
O
< 1 20 °
Trigonal Pyramidal – AB3E
Example: ammonia, NH3
Bent – AB2E2
Example: water, H2O
Summary
• Given any molecule or polyatomic ion you
should be able to
– Draw the Electron dot structure
– Determine the shape and bond angles
Get used to the Table of Shapes online – you
will get it on the exam over this chapter!!!
Practice: PH3 and ozone O3
12.6&7 Polar and Nonpolar Covalent Bonds
• A covalent bond where electrons are
shared equally is a nonpolar bond. (no
poles, no magnet)
• A covalent bond where electrons are
shared unequally is a polar bond. (has
poles like a magnet)
Symbols used to indicate polarity:
d+ = Partially positive atom
d- = Partially negative atom
points toward more EN atom
What does partial charge mean?
• The atoms in ions are completely +1, +2,
-3, -2 and such
• Polar bonds make the atoms just a little bit
+ and -, like maybe +0.001 and -0.001
• So ions are WAY more + and – than polar
covalent bonded atoms
• Ionic bond Na-Cl is completely +1 and -1
• Polar Covalent bond N-F is a little bit d+
and d-
Electronegativity (EN) is the ability of a
BONDED atom to attract electrons.
Noble Gases don’t have an EN
Why? Any ideas?
• EN = ability of an atom to pull BONDED
electrons close
Well noble gases don’t BOND! So they can’t
pull bonded electrons close. So what is the
atom with the highest EN???
Yep, F. F pulls electrons closer than anything! F is an
electron hog. Nothing holds electrons tighter than F.
Nonpolar covalent bonds
•When an atom is bonded to itself, that bond is nonpolar
because the electrons are shared equally between them.
•Diatomic molecules have nonpolar covalent bonds.
Examples:
H2, N2, F2, O2, I2, Cl2 , Br2
•Note C and H are about the SAME in EN so
also make nonpolar covalent bonds.
Example:
C-H bond in CH4
Polar covalent bonds
•In general, when two different nonmetal atoms are
bonded, the bond is polar because the more EN atom
pulls the electrons closer so they are shared unequally.
Examples of polar bonds:
C-O, H-F, S-F, C-N
Examples:
(a) Add the delta notation
(b) Add the polarity arrow
C-F
O-C
C-H
C-Cl
O-H
Ionic, polar covalent, or nonpolar covalent???
C=O bond
Cl-Cl bond
Na-O bond
C=C bond
Polar covalent
Nonpolar cov
Ionic
Nonpolar cov
Polarity - Review
• This is really important and will come up
later again and again and again.
• Think of this as a tug-of-war for bonded
electrons. The more EN atom pulls them
closer, and since e- are negative, that
makes that atom a little bit d-. By default
the other atom is a little bit d+. The bond is
thus polar. If the atoms have the same EN,
like C and H, then it is a tie (nonpolar).
Ionic, Polar, or Nonpolar Bonds?
•
•
•
•
•
•
•
•
Na-Cl
H-Cl
H-H
Cl-C
C-H
O=O
K-O
P-F
Ionic
Polar Covalent
Nonpolar Covalent
Polar Covalent
Nonpolar Covalent
Nonpolar Covalent
Ionic
Polar Covalent
Metallic Bonding
•
•
•
Pure metals have a freely moving “sea of
electrons”.
The electrons are shared among all the
metal atoms.
This is why they conduct heat and
electricity so easily.
12.10 Polarity of Molecules
• All nonpolar bonds = nonpolar molecule.
• Polar bonds that don’t cancel out = polar
molecule.
• Polar bonds that do cancel out = nonpolar
molecule.
Polar Bonds BUT Nonpolar Molecule
Polar bonds cancel out = nonpolar molecule.
Summary
Draw the electron dot structure, determine
shape, bond angle, determine if bonds are
polar and if molecule is polar.
• Water
• Ammonia
• Carbon dioxide
Yes, we skipped sections 8 and 9
For Fun if time
• Why care about molecular shape? Well
look at what cis-platin can do thanks to its
shape! http://www.youtube.com/watch?v=Wq_up2uQRDo&feature=related
• Antioxidants and free radicals – what
are they? Free radicals are when a
molecule has an unpaired electron.
They are considered bad for you.
http://www.youtube.com/watch?v=KVyjmt10CH0&feature=related
Self Test
• Page 352
• Try 1-6, 10, 12, 14 (shape only), 16-17
(don’t worry about when it says electron
geometry, worry about shape)
• Answers in Appendix J