NAME _____________________________
NOTES: UNIT 9 (1): BONDING
I) Why do bonds form? … It ain’t really all that simple of an answer….but generally,
A) Bonds form because the process of bonding tends to *lower the potential energy between the
charged particles that compose atoms.
Much of it goes back to lowering potential energy due to coulombic forces of attraction.
With any luck that may sound a bit like my metaphor of the ladder of energy…. and the idea,
in thermodynamics that reactions tend to occur so as to decrease enthalpy, and to increase the
entropy (the distribution of the energy).
1) Bonding theories should help us to predict the circumstances under which bonds form and
also the properties of the resultant molecules
a) there are three (really) broad classifications of chemical bonds … each is
dependent upon the type of atoms (species) involved in the bonding
Species or Type of
Atom
Type of Bond
Characteristic of the Bond
metal & nonmetal
ionic
electrons are transferred completely, creating
an electrostatic charge attraction with an
electronegativity difference ≥ 1.7
nonmetal and nonmetal
covalent (2 basic types)
 polar covalent
 nonpolar covalent
metal and metal
metallic
electrons are shared
 unequally with an
electronegativity diff. < 1.7 but > 0.4
 equal sharing with an electronegativity
difference of 0 to 0.4
electrons are pooled
For a fairly high caliber set of visuals & explanations, try:
Bozeman Chemistry #19: Covalent https://www.youtube.com/watch?v=Mo4Vfqt5v2A
Bozeman Chemistry #20: Ionic
https://www.youtube.com/watch?v=hiyTfhjeF_U
592
TRY THIS!!! Based solely upon the information of the chart on the prior page, identify the predominate bond
type using the following choices
a) metallic bond(s)
b) ionic bond(s)
c) polar covalent bond(s)
d) nonpolar covalent bond(s)
As you work through this review, keep in mind the idea of what is bonded to what … For instance, in SO2,
are the oxygen bonded to each other? Are the oxygen bonded to the sulfur? (Yeah, go with the latter…)
___1) N2O5
___8) Cl2
___15) OF
___2) KCl
___9) Cu
___16) CaO
___3) Fe
___10) CuCl2
___17) LiF
___4) O2
___11) Fe2O3
___18) HCl
___5) CH4
___12) ICl
___19) N2
___6) CO2
___13) Br2
___20) SO3
___7) NH3
___14) H2O
___21) NO2
___22) PI3
Answers; 1) c 2) b 3) a 4) d 5) d 6) c 7) c 8) d
9) a 10) b 11) b 12) c 13) d 14) c 15) c 16) b
17) b 18) c 19) d 20) c 21) c 22) c
593
Types of Bonds ... A VERY GENERAL SUMMARY
Bond Type
Rough /
Test ID
Metallic
between
metal
species
m
Electronegativity
Difference
Substances
Comments
N/A
Electronegativity
difference does not
play a role in metallic
bonding.
Found in metal
crystals
and between
atoms of
mercury in Hg(l)
key term:  sea of mobile electrons
 pooled or delocalized eThe valence and inner d-orbital e- tend
to be mobile. This means, they are not
bound to any one specific nucleus, and
thus are "mobile". This mobility accounts
for the range of melting points of metals,
the luster, ductility, tenacity, malleability,
and the ability to conduct electricity as in
the solid AND liquid (fused) phase.
Ionic
m+ [nm−]
E.D.  1.7
...because the ions are
of species with widely
different
electronegativity
values … There are a
number of exceptions.
Responsible
for binary and
ternary ionic
compounds
Made by the complete transfer of valence
e-. This transfer produces full + and - ,
oppositely charged ions which then
attract each other. Hence, the ionic bond
is an electrostatic bond, in that the
different charges (electro) are localized
(static) on different species.
Polar
Covalent
nm - nm'
E.D. 0.4  1.7
Responsible
for
molecular
compounds
Tends to most commonly exist between
two nonmetal atoms of different
electronegativity values, in which the
bonding pair of e- is UNequally shared.
One nucleus attracts the bonding e- more
strongly than the other ... producing a
partial negative charge () while the
other atom becomes (+).
Nonpolar
Covalent
...because the atoms
are of different
nonmetal elements,
with only slightly
different
electronegativity
values.
nm - nm
•bond(s) of the
diatomic
The atoms tend to be of
elements,
the same element, with
absolutely no
•C-C bond of
difference in
the parent
electronegativity
chain , ...& the
values. Some species
are
•C – H bond of
so close, the small
organic
difference is negligible
compounds
(e.g. C – H bond is
generally a nonpolar
covalent bond).
E.D.= 0 to 0.4 (-ish)
May be single, double or triple covalent
bonds
These most commonly exist between two
nonmetal atoms of the same
electronegativity values, in which the
bonding pair of e- is EQUALLY shared
Neither nucleus really dominates the epair. The bond is not considered to have
a dipole moment. The C – H bond is
generally considered to be a nonpolar
covalent bond, in spite of a very slight
difference.
May be single, double, triple bonds
594
Examples for the determination of bond type using electronegativity difference:
NaCl
H2O
SO3
CH4
595
II) Review: Lewis Structures of Atoms, Ions and the Octet Rule
A) I have been saying for a while now, that the Octet Rule (the tendency to gain/lose or share valence
electrons so as to get to 8 valence electrons [a stable octet] … is not so much a rule, as it is
a good idea…. There are a number of species that do not achieve a stable valence octet.
H2 starts the list. He, Li, Be, B gets us down the line. Many of the monatomic ions of the
transition metals never get there … So, don’t get too hung up on “the rule” … Just realize
that it has its role … and it helps us simplify the initial work on bonding.
B) The Lewis Dot Diagram (Lewis Structure for an atom)
1) 2 parts: The kernel = the nucleons and all inner electrons
The dot(s) = valence electron(s)
written in their promoted form
when inner electrons are raised to a higher sublevel, thus
creating extra bond sites.
II) Ionic Bonding and Lattice Energy
A) The Lewis Model allows us to represent an ionic bond
Consider a reaction between sodium and sulfur:
Na •
Na •
..
·S:
̇
+
= 2 Na
2 Na + S → Na2S
..
:S:
̇ ̇
2-
The Lewis model
 predicts the formula’s stoichiometric ratio… as each sodium will lose one electron
(for a total of 2 electrons) to the electronegative sulfur.
The sodium atoms at [Ne] 3s1 become isoelectronic with neon at: [Ne] or rather, the
electron configuration of Na+ has the same number of electrons as Ne0 …but not the
same number of protons!!!!! Isoelectronic configurations just imply a common number of eThe sulfur atom at [Ne] 3s2 3p4 gains the two electrons to become isoelectronic with
argon at: [Ar] or rather, the electron configuration of Cl- has the same number of
electrons as Ar0… but not the same number of protons!!!!!

predicts that the two ions do not “touch” but are strongly attracted to each other via
an electrostatic force (an attractive coulombic force) generated between the two
species produced due to the transfer of valence electrons.
596
1) The bond is made via a transfer of electrons which produces cations and anions which are
attracted to each other via a non-directional electrostatic force of attraction. (Oooo! fancy!!)
Once transferred the electron is localized upon the reduced species (the anion)…and does not
have a type of mobility/delocalization as seen in metals.
C) Summary of Ionic Bonding and its Model
Property
melting point
Physical Characteristic
rather high relative to molecular compounds
NH4NO3 (ammonium nitrate) melts around 169°C … but most ionic compounds have
really high melting points . For example: 1975°C (ZnO), 2870°C (WC), 808°C
(NaCl), 349°C (NaOH)
electrical conductivity
Heck No! There are no mobile or delocalized electrons to accomplish this.
as a solid
electrical conductivity Yup! Lots and lots of freely moving oppositely charged ions. Recall that
as a (fused) liquid
melting (fusing) an ionic compound essentially destroys the bond, as the species
separate in order to accommodate the influx of energy. Think: Clausius
& entropy! …Rather, it is (melting) because the net increase in the number of independently moving
species formed upon dissociation is more able to effectively dissipate the system’s internal kinetic energy.
electrical conductivity
Yup!
in water
solubility in water
Hey ...it will vary… but yes…to some extent. Many are highly soluble, others
will dissolve to a very, very, very low extent.
appearance
Often hard, brittle, crystals or powders. Some hydrates, having cations of
transition metals may be colored…Think CuSO4•5 H2O
Given the nature of our work on lattice energy, the model does account for the high melting points
of ionic solids … Given the non-directional coulombic forces of attraction it is no surprise that a
fair amount of energy is required to melt ionic compounds.
The model includes transferred electron – but once transferred, it stays
localized on the anion species. Thus, there are no “free moving” or delocalized
electrons moving about. This and the fixed position of the ions explains the lack
of conductivity as a solid.
But the model does account for free moving ions (and thus conductivity) when
fused or dissolved in water.
Not a bad model…
http://www.personal.kent.edu/~cearley/ChemWrld/compounds/salt.jpg
597
IV) Covalent Bonding
A) The Lewis Model allows us to represent a covalent bond.
H2 + ½ O2 → H2O
H•
H•
..
· O:
̇
→
..
H :O:
Ḣ ̇
By bonding, the oxygen gains a full valence octet (s2p6) and hydrogen completes its
first principal energy level (1s2), which is as good as it will ever get for the atom.
The shared electrons = bonding pair
The un-bonded electrons on oxygen = “lone” pair of un-bonded electrons
The bond is due to an overlap of orbital space, with the bonding electrons being attracted
to both nuclei … although that attraction may be un-equal (polarized).
The bonds have a directional component, given the shared volume is due to an overlapping
s and p set of orbitals.
There is a bond length … and an associated bond energy. Reflect back on Bozeman #19 and
the determination of the bond length as the “sweet spot” where the repulsive nuclear forces
and the attractive electron-nuclear forces balance out.
http://images.tutorvista.com/cms/images/81/potential-energy-curve-H2-molecule.png
B) The amount of energy released by bond formation, is the amount of energy that must be supplied
in order to break the bond. This introduces the idea of bond strength
1) For a given pair of atoms, the bond is stronger, the greater the overlap of atomic orbitals.
598
a) There is a maximum level of overlap for bonding atoms. Once the maximum level of
overlap occurs,* the repulsive forces between the 2 positive nuclei come into play.
b) This maximum or optimal overlap is called the bond length.
c) For example: With dihydrogen we say that the H─H covalent bond has a
bond length of 0.074 nanometers & a bond strength of 4.52 eV or 7.24 x 10-19 Joules
2) Thus! There is a correlation between the ideas of bond strength and bond length.
a) There are tables listing average bond strengths and lengths
http://mgh-images.s3.amazonaws.com/9780073402659/1748-10-95PEI1.png
i) The table lists average values. Absolute values may vary from these. For
instance, the C – H absolute bond energy in methane may be slightly different
than the absolute bond energy of the C – H bond in chloroform
vs.
methane
chloroform
599
C) Covalent Double and Triple Bonds
1) When atoms share 2 pair of electrons as in O2 or in the carbonyl group of an organic
compound, such as a carboxylic acid, aldehyde or ketone … there is a double bond
a) Double bonds tend to be *shorter bond length and are stronger than single bonds
b) The bonds are not equal however. One bond is called a sigma bond (𝜎), the other is
called a pi bond. The pi bond is the first to break in a chemical reaction
c) Thus a double bond has one sigma bond and one pi bond. A triple bond has one
sigma bond and two different pi bonds
Check out: https://www.youtube.com/watch?v=ree49ge4VA4
2) When atoms share 3 pair of electrons as in N2 or alkyne hydrocarbons, there exists a triple
bond
a) Triple bonds have one sigma and two pi bonds
D) Electronegativity … and the Concept of Bond Polarity
1) By now we pretty much grasp that electrons may be shared unequally by the two atoms
of this energy bridge called a bond.
a) When compounds (such as HF) are put into an electric field, the molecules orient
themselves in such a way as to have the fluoride end attracted to the + plate and
the hydrogen end of the molecule attracted to the – plate.
b) The electron density of the bond is greater on the fluorine atom than on the hydrogen
atom.
2) Polar covalent bond: an intermediate bond, somewhere, along a continuum of sharing
with a pure nonpolar covalent bond at one end and an ionic bond
at the other.
In a polar covalent bond, there is a pole established.
δ+ δH–F
The arrow indicates the direction in which e- density is shifted
or H – F
This indicates the bond has a dipole moment
a) The ability to attract the electrons of a bond is our old friend, electronegativity
600
b) A little of The Story of Linus Pauling: Pauling compared the energy to break
bonds (bond energy) of molecules like HF, by studying H2 and F2.
He reasoned that the strength of the HF bond, if it were purely nonpolar covalent
would be 296 kJ/mol …the average of the bond energies of H2 & F2, which are
436 kJ/mol and 155 kJ/mol, respectively.
He found that the actual bond energy of HF = 565 kJ/mol.
He reasoned that the extra bond energy was due to the ionic character of the bond
From such experiments he developed the concept of un-equal sharing and
electronegativity values.
V) Writing Lewis Structures:
A) Lewis structures give us a clue as to the bonding in molecules and ions. There is a process in your text.
I have adapted and adopted the process introduced in Honors … VEDA CaB … is the mnemonic …
The word veda is Sanskrit for “knowledge”, and I like that … since scientia is the Latin, for
“knowledge” There is a symmetry to be found here. That’s nice…
Some species prove to be a bit more challenging when drawing Lewis Structures. When I was your age,
I (personally) became confused with having to draw out carbon monoxide, or ozone, or (shudder) the
polyatomic ions.
But practice will get you a very, very long way.
601
The overall goal is to get ensure that the number of required e- equal the number of drawn
electrons so that every atom has 8 valence electrons.
One way to do this uses a step-by-step approach and to rely on those lovely double and triple bonds....
I use my little mnemonic the VEDA CaB.
PROCESS: (VEDA CaB ....valence e- / draw atoms and single bonds / add e- / count and bond
1) Use the structure’s formula to determine the number of required Valence Electrons taking into
account any overall charge.
2) Draw the central atom (generally the atom in least abundance and/or the least electronegative
element … except for H, which is usually terminal)
3) Draw the other atoms around the central atom
4) Draw in single bonds (2 electrons per bond) to represent the primary chemical union as covalent
5) Add enough electrons to get 8 valence e-, first for each outer atom, and then add enough around the
central atom to get 8 valence e-….and add them to the central atom, even if doing so results in more
than an octet.
6) Count ... to see if the required number of electrons = the # of drawn e- ... ensuring you have 8 val. earound the central atom.
7) When the central atom does NOT have 8 valence e- (but you have accounted for all of the e-) think of
using some outer atom’s electrons to make a double or triple Bond (thus, re-appropriating 1pr of efrom each atom involved in the multiple bond)
8) Complete the task by including brackets and the overall charge
602
For example: Look at the molecule, carbon monoxide or CO
Step 1) C has 4 val e= 4
O has 6 val e= 6
required e- = 10
Use formula to calculate # Valence Electrons
Draw atoms
draw single bonds
Step 2) & 3)
4)
C
Use: VEDA CaB
CO
O
Add to get octets of e5) & 6)
..
:CO:

There is a total of 10 ebut C does not have 8
valence e- ...Think
about multiple bonds
Count and Bond (if needed)
Because C does not yet have 8
valence e-, “re-deploy” the esurrounding oxygen, to make
double or triple bonds ...
having O still with the e- but
sharing them with C
7)
Ans: move the box away...
..
:C=O:
:C ≡ O:
603
1) Lewis Structures for Polyatomic Ions (Goal: To get the # of required e = to the # of drawn eusing the octet rule)
PROCESS: (VEDA CaB ....valence e- / draw atoms and single bonds / add e- / count and bond)
1) Use the structure’s formula to determine the number of required Valence Electrons taking into
account any overall charge.
2) Draw the central atom (generally the atom in least abundance and/or the least electronegative
element … except for H, which is usually terminal)
3) Draw the other atoms around the central atom
4) Draw in single bonds (2 electrons per bond) to represent the primary chemical union as covalent
5) Add enough electrons to get 8 valence e-, first for each outer atom, and then add enough around the
central atom to get 8 valence e-….and add them to the central atom, even if doing so results in more
than an octet.
6) Count ... to see if the required number of electrons = the # of drawn e- ... ensuring you have 8 val. earound the central atom.
7) When the central atom does NOT have 8 valence e- (but you have accounted for all of the e-) think of
using some outer atom’s electrons to make a double or triple Bond (thus, re-appropriating 1pr of efrom each atom involved in the multiple bond)
8) Complete the task by including brackets and the overall charge
(O2) 2-
e.g.) peroxide
Step 1) O has 6 val e- x 2 = 12
2 extra due to charge = 2 Don’t forget to account for those 2 extra electrons of the charge
required e- = 14
Step 2) & 3)
4)
O
5 & 6)
..
..
:OO:


OO
O
with a total of 14 e-
7)
No need to
change anything
since the 14 eare so arranged
as to give octets.
8)
..
..
:OO:


2..
..
:OO:


14 e-
604
2) Lewis Structures for Polyatomic Ions (Goal: To get the # of required e = to the # of drawn e- using
the octet rule)
PROCESS: (VEDA CaB ....valence e- / draw / add / count and bond)
1) Use the structure’s formula to determine the number of required Valence Electrons taking into account any
overall charge.
2) Draw the central atom (generally the atom in least abundance and/or the least electronegative element …
except for H, which is usually terminal)
3) Draw the other atoms around the central atom
4) Draw in single bonds (2 electrons per bond) to represent the primary chemical union as covalent
5) Add enough electrons to get 8 valence e-, first for each outer atom, and then add enough around the central
atom to get 8 valence e-….and add them to the central atom, even if doing so results in more than an octet.
6) Count ... to see if the required number of electrons = the # of drawn e- ... ensuring you have 8 val. e- around
the central atom.
7) When the central atom does NOT have 8 valence e- (but you have accounted for all of the e-) think of using
some outer atom’s electrons to make a double or triple Bond (thus, re-appropriating 1pr of e- from each
atom involved in the multiple bond)
8) Complete the task by including brackets and the overall charge
e.g.) nitrate (NO3)1step 1) N has 5 val eO has 6 val e- x (3)
one extra due to charge
required esteps 2) & 3)
= 5
= 18
= 1
= 24
4)
O

ON

O
O
O
. .
:O:
.. 
:ON
5) & 6)
N
O
˙˙

:O:
˙˙
There ae 24 e- but the central
atom does NOT have 8 val.e-
Used 2 e- from the O atom to get
8 val. e- around the central atom
with the new double bond
Inserted
a double bond,
because step 6 showed
that N (the central atom)
did not have 8 valence e-
7)
8)
:O:
. . 
:ON
˙˙ 
:O:
˙˙ 24e-
:O:
. . 
:ON
˙˙ 
:O:
-1
(Resonance
structures exist)
˙˙
605
3) Lewis Structures for Polyatomic Ions (Goal: To get the # of required e = to the # of drawn eusing the octet rule)
PROCESS: (VEDA CaB ....valence e- / draw / add / count and bond)
1) Use the structure’s formula to determine the number of required Valence Electrons taking into account any
overall charge.
2) Draw the central atom (generally the atom in least abundance and/or the least electronegative element …
except for H, which is usually terminal)
3) Draw the other atoms around the central atom
4) Draw in single bonds (2 electrons per bond) to represent the primary chemical union as covalent
5) Add enough electrons to get 8 valence e-, first for each outer atom, and then add enough around the central
atom to get 8 valence e-….and add them to the central atom, even if doing so results in more than an octet.
6) Count ... to see if the required number of electrons = the # of drawn e- ... ensuring you have 8 val. e- around
the central atom.
7) When the central atom does NOT have 8 valence e- (but you have accounted for all of the e-) think of using
some outer atom’s electrons to make a double or triple Bond (thus, re-appropriating 1pr of e- from each
atom involved in the multiple bond)
8) Complete the task by including brackets and the overall charge
e.g.) cyanide (CN)1step 1) N has 5 val e= 5
C has 4 val e=
4
one extra due to charge = 1
required e- = 10
*remove box
steps 2) & 3)
4)
C
N
5) &6)
..
: CN :

CN
All 10 e- are used, but
the N does NOT have
8 valence -
7)
8)
1:C≡N:
:C≡N:
ANSWER
10 e-
Created a triple bond scenario by drawing in 2 more bonds, requiring
the re-appropriation of 2 pr of carbon’s e- to be shared with N
606
4) Draw the Lewis Structure for PCl3
Try using VEDA CaB
count valence e1) 5 + (3 x 7) = 26 edraw atoms, draw single bonds
get octets & maybe multiple bonds
5) – 6)
2) - 4)
Cl – P – Cl
|
Cl
5) Draw the Lewis Structure for HCN
..
..
:Cl – P – Cl:
˙˙ | ˙˙
:Cl:
˙˙
7)
..
.. ..
:Cl – P – Cl:
˙˙ | ˙˙
ANSWER
:Cl:
˙˙
You have the example for CN-1 try the molecule HCN
H – C ≡ N:
ANSWER
6) Draw the Lewis Structure for BrO31-
.. .. .. -1
:O– Br – O:
˙˙ | ˙˙
ANSWER
:O:
˙˙
607
7) NO1+ has a multiple bond. What is the number of pi bonds? One or Two
To answer the questions …draw the Lewis Structure.
[ : N ≡ O : ]+1
there is a triple
bond,ANSWER
so there are
2 pi bonds
8) How many multiple bonds are in ClO21-?
a) zero
b) one
c) two
..
..
..
1:OANSWER
– Cl – O:
˙˙ ˙˙ ˙˙
a) there are zero
multiple bonds
9) Draw the Lewis structure for the polyatomic ion nitrite: (NO2)1*
Go online ...and move this box for the answer … There is
a resonance structure… thus either for these purposes
would be deemed correct.
10) Draw the Lewis structure for the polyatomic ion carbonate: (CO3)2Go online ...and move
* this box for the answer
608
VI) Molecular Shape / Molecular Polarity / VSEPR Theory / Hybridization
CHECK THIS OUT! From Crash Course: http://www.youtube.com/watch?v=a8LF7JEb0IA
A) Lewis Structures for molecules : When atoms bond with each other, they often produce
structures with specific shapes. The three-dimensional angles and
spacings between the bonded species of a molecule can influence the
behavior (properties) of the molecule. Cobb & Fetterolf The Joy Of Chemistry p 178
••
H ― Cl :
••
••
H―O
H
••
H―N―H
|
H
These three molecules have different angles between the atoms. This difference gives the
molecules different shapes. Different shapes can help account for different properties.
1) For our course there are: 5 basic shapes of simple molecules:
tetrahedral, linear, pyramidal, trigonal planar, and bent
a) Recall that molecular shape has a huge influence upon chemical activity and
physical characteristics. This idea that shape confers activity can be seen throughout
the study of molecules as in enzymes and Lock and Key Theory, solubility in water,
and phase changes.
2) There are a number of theories used to explain/predict molecular bonding and shape.
One theory is, VSEPR Theory. This essentially attempts to explain how the shape of the
molecule is


dependent upon how many bonding orbitals are occupied AND
how the shape is dominated by pairs of unbonded electrons (lone pairs of e-) on
the central atom

Recall that there are three p orbitals (px, py and pz), each organized
in a different spatial
direction
http://2012books.lardbucket.org/books/principles-of-general-chemistry-v1.0/s13-molecular-geometry-and-covalen.html
609
B) VSEPR Theory: Valence Shell Electron Pair Repulsion Theory
Edited Reading: http://www.chem1.com/acad/webtext/chembond/cb05.html or http://tinyurl.com/bsxboaz
TEXT REFERENCE: p. 232-233
The covalent model of chemical bonding assumes that the electron pairs responsible for bonding are
concentrated into the region of space between the bonded atoms.
Premise 1) The fundamental idea of VSEPR theory is that these regions of negative electric charge will
repel each other, causing them (and thus the chemical bonds
that they form) to stay as far apart as possible.
Thus the two electron clouds contained in a simple tri-atomic
molecule AX2 will extend out in opposite directions; an angular
separation of 180° places the two bonding orbitals as far away
from each other they can get.
We therefore expect the two chemical bonds to extend in opposite directions, producing
a linear molecule.
Premise 2) When the central atom also contains one or more pairs of NONbonding electrons, these
additional regions of negative charge will behave very much like those associated with the
bonded atoms.
The NONbonding electrons of the central
oxygen extend out into space and essentially
repel each other.
The Hydrogen each have a
POSITIVE partial charge ( ),
indicating that their bonding
electrons are dislocated from
their nuclei toward the nucleus of
the oxygen ....Thus creating a
polar covalent bond between each
H and the central O.
However, their repulsion is of greater
magnitude than the repulsion due to the 
of the hydrogen.
Thus, the hydrogen are pushed closer to each
other than would be predicted.
These  species repel each other

Thus we have the Repulsion of Valence Shell e- determining the shape of the molecule
(VSEPR ... Valence Shell Electron Pair Repulsion Theory)
The orbitals containing the various bonding and nonbonding pairs in the valence shell will extend out
from the central atom in directions that minimize their mutual repulsions.
You will note how the
extra NONbonding electrons occupy space and help to influence the shape of the molecule.
Premise 3) If the central atom possesses partially occupied d-orbitals, it may be able to accommodate five or
six electron pairs, forming what is sometimes called an “expanded octet” (This is essentially a
violation of the (albeit) relatively arbitrary Octet Rule. But, our course work does, for now, value the Octet
Rule, and thus, a VSEPR discussion of the expanded octet is reserved for AP Chemistry.
For students interested in the variations of bond theory, you may wish to look into the formation of complex
ions, with transition metals, at: http://tinyurl.com/aak2fou or http://alevelchem.com/aqa_a_level_chemistry/unit3.5/s354/02.htm)
It is another AP Chemistry topic, but it is interesting, using coordinate covalent bonding and quantum.
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C) Lewis Structures for molecules (and polyatomic ions)
1) Basic Shapes
180º bond angle due to repulsion of bonded species to a central species
Linear A - Z or A - Z - A
* 2 unbonded (lone) pairs of eBent
* bond angle approximates 104.5
* Only 1 lone pair of e- with the central atom
at the “top”
Pyramidal
Tetrahedral
* bond angle approximates 107
* bond angle approximates 109.4º
* central atom is in the middle (no unbonded pairs of e-)
http://www.epa.gov/globalmethane/basicinfo.htm
http://encarta.msn.com/media_461544861/tetrahedral_bonding_in_methane.htmlPRACTICE
:
Trigonal Planar: *bond angle approximates 120°
There are 0 lone pairs of eon the central atom
Checkout: http://phet.colorado.edu/sims/html/molecule-shapes/latest/molecule-shapes_en.html
Prepare for tomorrow’s work by checking out the site. Click on “MODEL”. Then, on the new
site, in the lower left corner, click on molecular geometry and electron geometry. Investigate
what happens as you add lone pairs of electrons, or new bonds (found in the upper right corner)
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VII) Hybridization of atomic orbitals …
A) Hybridization: The mixing of 2 or more atomic orbitals of similar energies (e.g. s and p) on the
same atom (often, but not always, the central atom) … so as to produce new orbitals
of different energies of the original, but equal energies when hybridized, facilitating
bonding, and the lowering of energy, in valence electron bond theory.
1) Hybrid orbitals: Orbitals of equal energy produced by the blending of two or more orbitals
on the same atom
a) evidence comes from experimental results dealing with methane (for instance)
atomic orbitals of carbon
hybrid orbitals found in methane
We can also witness sp and sp2 hybridization with carbon
…. But first … some of the basics
B) Essentially, the bonding orbitals will be “blended” (hybridized) and the shape and energy of the
orbital will be altered … merged into a new shape and probably lower (averaged) energy.
Be
1s
2s


2p
no classically identifiable bond sites seem to exist
1 electron is promoted:



For Be, promotion seems to explain the formation of two bond sites. The electron clouds of
the bonding Be and 2 Cl atoms will thus be re-shaped (hybridized) so that the 3p electron
of one Cl atom can make a bond with the 2s of Be … and the 3p electron of the other Cl
atom will bond with the 2p electron of Be….And both bonds will end up being identical
in both bond length and bond energy!!!
Since the central atom of Be uses an s and p sublevel to bond, chemists say that a new (in
shape and energy), sp hybrid orbital is produced. The higher electronegative chlorine
𝛿atoms are each 𝛿- and repel each other, producing a linear molecule. 𝛿-
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1) Sigma and Pi bonds … Just a bit more so we can speak of them….
a) Sigma Bond: A bonding electron pair, localized in the area centered along the line
between the two nuclei …
i) the stronger covalent bond … sometimes described as the one which is
“head on” between the 2 nuclei
b) Pi Bond: a bonding pair of electrons, which utilize a p orbital which is NOT involved
in the hybridization process …
i) of considered to be the weaker of the bonds in a multiple bond system. It is
a pathway NOT centered upon the two nuclei.
Check out: https://www.youtube.com/watch?v=ree49ge4VA4 …. Look at 3ˈ15ˈˈ for pi bonds
C) Hybridization theory helps explain why the experimental results (such as bond angles, bond energy
and molecular geometry) are different from the theorized values. Something must be going on to
cause the experimental results to differ from the theoretical … That “something” is hybridization.
1) Hybridization theory finds its use mainly in organic chemistry, although it is used pretty
extensively to explain compounds made with Be, B, N, O, and S, as well, for example.
2) Hybridization happens when differing atomic orbitals mix to form new “hybrid” orbitals.
The new hybrid orbitals have the

same total electron capacity as the old ones. And,

The properties and energies of the new, hybridized orbitals are an 'average'
of the original un-hybridized orbitals
The hybrid sp3 orbitals are an
average, in terms of energy of the
original atomic orbitals. Due to the
energy differences, hybridized
orbitals must also have a different
shape, and “look” different, from
the original atomic orbitals
http://www.chemicool.com/definition/hybridization.html
a) It is important to grasp that when atomic orbitals are blended into hybridized
orbitals, the number of these new hybrid orbitals must be equal to the number of
electron domains (numbers of atoms and non-bonded electron pairs) surrounding
the central atom! https://chemistry.boisestate.edu/richardbanks/inorganic/bonding%20and%20hybridization/bonding_hybridization.htm
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For instance: In CH4 … carbon uses (1) s and (3) p orbitals
s + p + p + p = sp3 This is a total of 4 hybridized orbitals capable of making up
to 4 bonds.
In the molecule NH3, nitrogen also uses one s and 3 p orbitals (sp3 hybridization)
That is: N has 4 orbitals involved in the
hybridization. NH3 has 3 bonds and
one lone pair of un-bonded electrons.
This connects electron domain geometry with hybridization… For instance
with water, all four domains are hybridized thus the electron domain geometry
tetrahedral. Due to the nonbonding electrons the molecular geometry is bent.
http://www.chemistry-assignment.com/wp-content/uploads/2013/01/1103.png
b) Not every orbital is “automatically hybridized” as we try to explain what is going on.
Remember, the number of hybrid orbitals = the number of bonded species
&/or lone pair of electrons (electron domains). Thus, we use the idea that the
number of bonded groups to the central atom dictates the number of hybridized
orbitals.
The un-hybridized orbitals can help explain the evolution of pi (double bonds), as in
aldehydes, ketones, and organic acids.
c) Hybrid orbitals can bond with other hybrid orbitals … or just about any other variation
of which you can think.
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★★d) When you are asked questions about “the type of hybridization”, you can:
 count up the # of electron domains (total number of bonded species and/or unbonded electron pairs) Count “multiple bonds” as 1…since only sigma bonds
represent hybridized orbitals
e.g. in CO2
e) Summary of hybridization in CO2 …for both C and O
Atom
Hybridization
C
sp
Reasoning
The number of hybridized orbitals is dependent upon the number
of species to which an atom is bonded, and/or the number of lone
pairs of electrons.
Thought to occur when Carbon has only 2 species bonded to it, and
there no un-bonded pairs of electrons (lone pairs) on the carbon.
each O
sp2
Thus, only 2 sets of hybridized orbitals are produced, using one s
and one p producing sp hybridization.
Again, the number of hybridized orbitals is dependent upon the
number of species to which an atom is bonded, and/or the number
of lone pairs of electrons.
Since the O atoms are identical … the analysis of just one will due.
Notice that the O has two lone pair of electrons and the sigma bond of
the double bond to carbon. This makes 3 regions resulting in the need for
3 hybridized orbitals or sp2 hybridization.
We only count one of the bonds (the sigma) of a double bond pair,
because only orbitals making sigma [head to head bonds, in line with the
two nuclei] are susceptible to hybridization.
Check Out: http://www.youtube.com/watch?v=JKdVaN-q4AM Hybridization in CO2
d) For carbon, sp3, sp2 and sp hybridization can be summarized in a presentation by
Dr. Sarah A. Sheeley, at the University of Illinois, upon whose work I built my
powerpoint, which is online.
at: http://www.chem.uiuc.edu/CLCtutorials/104/Hybridization/SeeIt.html
or: https://tinyurl.com/carbon-hybridization
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SUMMARY FOR HYBRIDIZATION FOR CARBON
Hybridization
Concept
3
sp
one s and three p orbitals are
blended
Molecular Shape
tetrahedral (trigonal pyramidal in the case of
nitrogen-based compounds.
Examples
CH4
CCl4
C2H6
trigonal planar … There are no un-bonded electrons on
CH2O
BCl3
All the bonds are sigma bonds
(single) and that means that
there is free rotation around.
sp2
sp
an s and two p orbitals are
blended leaving 2 unhybridized p orbitals capable
of producing double bonds.
an s and a p orbital are
blended, leaving 2 unhybridized p orbitals capable
of producing double bonds.
the central atom. This can apply to many compounds, but
especially to Group 13 compounds. All the atoms are in
the same plane.
linear
CO2
D) Hybridization for other atoms … N, O, S …
From: http://faculty.otterbein.edu/DJohnston/chem220/tutorial1/nitrogen.html
1) Nitrogen - sp3 hybridization: The nitrogen atom also hybridizes in the sp3 arrangement, but differs
from carbon in that there is a "lone pair" of electron left on the nitrogen that does not participate in the
bonding. The geometry about nitrogen with three bonded ligands is therefore trigonal pyramidal.
e.g. ammonia
methylamine
2) Nitrogen -sp2 hybridization: Nitrogen will also hybridize sp2 when there are only two atoms bonded
to the nitrogen (one single and one double bond). Just as for sp3 nitrogen, a pair of electrons is left on
the nitrogen as a lone pair. The resulting geometry is bent with a bond angle of 120 degrees.
e.g. pyridine
3) Oxygen - sp3hybridization: Oxygen bonded to two atoms also hybridizes as sp3. There are two
lone pairs of electrons located on the oxygen atom (not shown) and the resulting geometry is bent
with a bond angle ~109 degrees. Note that in acetic acid one of the oxygen atoms is bonded to only
one atom. We therefore do not have to consider the geometry (or hybridization) around that
particular atom.
e.g water
e.g. methanol
e.g. acetic acid
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4) Sulfur - sp3 hybridization: The geometry of sulfur compounds is essentially the same as for
oxygen compounds with sp3 hybridization found with two atoms bonded to sulfur.
e.g) dihydrogen sulfide
methanethiol
So, here is a question posed by a student:
Why does NH3 have sp3 hybridization? It doesn’t have four bonds. I read that it has sp3 hybridization, but I don't
understand why. Ammonia (NH3) seems to me to not require sp3 hybridization because all of its bond lengths &
energies are already equal. It has 3 hydrogens bonded to the p orbitals. Why can't the lone electron pair in
nitrogen's 2s shell stay put where it is? Someone told me that one of the proofs for sp3 hybridization is found in the
equivalent bond energy (and length) when ammonia become ammonium (NH4+) …but I don’t see how.
Here is my response to the student …to be used sort of an alternative summary, in terms of:




why we believe hybridization occurs,
bond angles,
energy
electron domain geometry and molecular geometry or
There are some experimental data suggesting that hybridization may not be all that we believe … especially
when involving d sublevel orbitals. However, most first year courses still embrace sp, sp2 and sp3
hybridization. We do so because hybridization theory helps to explain why the experimentally obtained
information for such issues as bond angles, equality of bond energies and basic molecular geometry, may be
different from the theorized results, when using only VSEPR Theory.
Justification for the existence of sp3 hybridized orbitals for nitrogen in ammonia can be found in:
Bond angle: If the molecule did not have hybrid orbitals, and instead had just “atomic” p-orbitals taking part in
the bond formation, then the bond angle between the orbitals would be 90 degrees. But in reality, the bond
angle is nearly 107 which makes the molecule more stable, decreasing the bond pair-bond pair and bond pairlone pair repulsion. This supports VSEPR theory, and helps to explain the experimental results regarding 4
equivalent polar covalent bonds, when ammonia is converted via a coordinate covalent bond into the
polyatomic ion, ammonium. Hybridization seems to answer how 4 bonds of equal length and energy can be
made from on s and 3 p sublevels. All four domains must be hybridized ….thus equalized.
Energy: Recalling the definition of hybridization, it is the mixing of the atomic orbitals having slightly different
energies to form new orbitals that have equal energies. This stabilizes the molecule. As for ammonia molecule
due to hybridization the energies of the lone pair of electrons and the bond pair of electrons becomes almost
equal, thereby increasing the stability of the molecule.
Geometry: Thinking about the spatial arrangement of the atoms in the molecule of ammonia if there is no
hybridization in the molecule then the size of the orbital containing lone pair of electrons would be different
from that of the orbitals containing bond pair of electrons. Also if hybridization were taking place, explaining
the geometry of the molecule’s pyramidal shape, would not be possible.
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PRACTICE: Try the following questions regarding hybridization for only the central atom in each molecule.
★★★ RECALL … identification of the type of hybridization, can often be determined by the number of electron domains are around
the central atom.
Substance
Central
Atom
# of
bonded
species
# of un-bonded
e- on the
central atom
Total domains
around the
central atom
Hybridization
methane
C
4
0
4
sp3
ammonia
N
3
1
4
*sp3
boron trifluoride
B
*3
0
*3
sp2
methanal
C
*3
0
*3
*sp2
carbon dioxide
C
*2
*0
2
*sp
beryllium difluoride
Be
2
*0
*2
*sp
water
O
*2
*2
*4
*sp3
sulfur trioxide
S
*3
0
*3
*sp2
Bonus Round: Identify the
hybridization for each O in a
molecule of ozone.
1
2
3
Oxygen 1 has * 3
domains so it has * sp2 hybridization
Oxygen 2 has * 3
domains so it has * sp2
hybridization
Oxygen 3 has * 4
domains so it has * sp3
hybridization
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