CHEM 2060 Lecture 23: VB Theory HF L23-1
Valence bond picture for HF… A nice example
H atom ground state electronic configuration: 1s1
F atom ground state electronic configuration: 1s2 2s2 2px2 2py2 2pz1
The simplest VB wavefunction for HF is:
ψ = 1sF2 2sF2 2pxF2 2pyF2 [1sH(1) 2pzF(2) + 1sH(2) 2pzF(1)]
We can rewrite a simplified version of this wavefunction by ignoring the nonbonding electrons of the F atom:
ψ = 1sH(1) 2pzF(2) + 1sH(2) 2pzF(1)
This first guess does not adequately describe the bond energy
of HF.
We can improve the stability of the wavefunction by allowing
the 2s orbital of the F atom to participate in bonding.
CHEM 2060 Lecture 23: VB Theory HF L23-2
Improvements made by including 2s on F atom:
We do this by hybridizing the fluorine 2s and 2pz atomic orbitals to get a set of
two sp hybrid atomic orbitals, one of which is oriented towards the H nucleus.
i.e.
2pzF is replaced by φ F = N(2pzF + γ2sF)
(new sp hao)
(recall: should be sp3)
• Ionic terms can also be introduced into the wavefunction.
• Good Idea! …based upon electronegativity considerations.
H+ F- will contribute more than H- F+
New wavefunction is now:
ψ = 1sH(1) φF(2) + 1sH(2) φF(1) + λ(φF(1) φF(2))
Both electrons on F
“Ionic Term”
Ionic term contributes ~50% (compare only 6% in H2).
CHEM 2060 Lecture 23: VB Theory HF L23-3
Advantages in Hybridization
Electron Density
concentrated between
nucleii
Decrease in e-e repulsion
Since part of 2s was mixed with the 2p
φF = N(2pzF + γ2sF)
the LONE PAIR is not simply 2s2 !
φlp = N(γ2s - 2p)
This decreases e-e repulsion for one of the F lone pairs and makes it more
directional.
Question: Which orbitals describe the other two lone pairs?
CHEM 2060 Lecture 23: VB Theory HF L23-4
Hybrids for HF
- formation of bonding hybrid (sp)
- overlap with H 1s
- formation of lone pair hybrid (sp)
- electron density in of lone pair points
away from bond
⇐ HF BOND
⇐ “LONE PAIR”
CHEM 2060 Lecture 23: VB Theory HF L23-5
A Note on Normalization:
φF = N(2pzF + γ2sF)
2
2
φ
∂
τ
=
N
∫ F
2
2
2 2
2
2
p
∂
τ
+
N
γ
2s
∂
τ
+
2N
γ ∫ (2 pzF 2s)∂τ = 1
(
)
(
)
∫zF
∫F 



=1
=1
2
2 2
1 = N +N γ ⇒ N =
=0
1
1+ γ 2
Normalization means that when an orbital is "scanned” over all space (i.e., take
the integral over all space…area under the curve) the probability of finding the
electron is 1 (i.e., 100%).
€
[Def] For a normalized wavefunction Nψ, the probability of finding the electron
in all space (i.e., anywhere in the universe) is exactly 1.
N is the normalization constant.
CHEM 2060 Lecture 23: VB Theory HF L23-6
Water: VB Model (hybrid orbitals)
Goal: To describe the bonding in H2O and
account for the structure, (i.e., appropriate
bond angle and 2 lone pairs).
O
H
H
Ground state electronic configuration of atomic O is: 1s2 2s2 2px2 2py1 2pz1
What would the water molecule look like if we simply used the 2py and 2pz
orbitals of oxygen as our bonding orbitals?
⇒ We would end up with 2 bonds BUT an H-Ô-H bond angle of 90°
…obviously not a good model for the bonding!
VB approach:
Since the 2s orbital is spherical, mixing some 2s character in can adjust the bond
angle!
CHEM 2060 Lecture 23: VB Theory HF L23-7
Question: What is the hybridization of the O atom in water?
We learn in 1st year that the
hybridization of the oxygen atom is sp3.
Bonding hybrids
Lone pair hybrids
4 orbital lobes in (“tails”)
4 orbital lobes out (“bodies”)
MIX 4 (s, px, py, pz) ⇒ GET 4 (4 x sp3)
CHEM 2060 Lecture 23: VB Theory HF L23-8
How do we get a better bond angle?
• Water is almost 4 x sp3 but not perfectly!
• The 4 lobes are not quite a perfect tetrahedron.
• The H-O-H bond angle is SMALLER than 109.5° (It is closer to 104.5°.)
We can rationalize this by thinking about the s and p characters of the hybrids...
In a perfectly sp3 hybridized set of hao’s, each sp3 orbital should have:
25% s character
and 75% p character.
For our water molecule, our hao’s can’t have exactly ¼ s and ¾ p…
…in other words there is uneven distribution of s and p character between the 4
hybrid orbitals.
First we will write down the wavefunction and see what this means and then we
will rationalize it.
CHEM 2060 Lecture 23: VB Theory HF L23-9
In this course we will write sp hybrids in the general form:
φ = N ( p + γs)
N is the normalization coefficient
γ tells us how much s is added to the p.
It must
€ be related to the s character of the hybrid.
In order to get at this we need to normalize the wavefunction.
All this means is that the new orbital contains 1 electron.
So the integral of the function squared over all space =1
∫ φ 2 dτ = 1
2
2
∫ N ( p + γ s ) dτ = 1
= N 2 ∫ p 2 dτ + N 2γ 2 ∫ s 2 dτ + N 2 2γ ∫ spdτ



=1
=1
=0
(orthogonal!)
CHEM 2060 Lecture 23: VB Theory HF L23-10
So…
Therefore…
2
2 2
N + N γ =1
€ hybrid wavefunction is then:
N=
(
φ = 1+ γ
€
1
2 − 2
)
1
1+ γ 2
( p + γs)
The s and p characters are now easy to get… square of that part of φ
€
#
2
&
1
1
(
%
=
Amount p character: %
2
2(
1+
γ
1+
γ
$
'
2
# 1 &2
γ
(( γ 2 =
%%
€
Amount s character:
2
2
1+
γ
1+
γ
$
'
…as γ →0, p-character →100%
…as γ →1, s-character →50%
CHEM 2060 Lecture 23: VB Theory HF L23-11
How do we choose the correct value of γ ?
The mixing coefficient γ is clearly related to the bond angle θ.
Using some simple trigonometric relationships, it can be proven that:
2
cos θ = - γ
(If you really want to know how this was derived, see DeKock & Grey
“Chemical Structure and Bonding”, pp.146-147)
For example a 50:50 mix (sp hybrid orbital)
The bond angle is 180°.
φ sp =
1 s+
(
2
p)
cos180° = -1
O=C=O carbon atom in CO2
so γ = 1
REMEMBER: Mix 2 (s & p) → Get 2 (2 x sp)
CHEM 2060 Lecture 23: VB Theory HF L23-12
Applying the bond angle equation to H2O
Water: The bond angle is 104.5°.
cos104.5° = -0.25
so γ = 0.25 = 0.5
1
1
Amount p character:
i.e. 80% p-character
=
= 0.80
2
2
1+ γ
1+ 0.5
€
This should leave us with 20% s-character…let’s double check that:
γ2
0.5 2
=
= 0.20
Amount s character:
2
2
1+ γ
1+ 0.5
€
i.e. 20% s-character
The two hybridized atomic orbitals of oxygen involved in bonding are each
80% p and 20% s
€
What are the remaining atomic orbitals of oxygen that contain the 2 lone pairs?
CHEM 2060 Lecture 23: VB Theory HF L23-13
Water: p-character of lone pair atomic orbitals of O atom
Without a bond angle to start from, we cannot derive γ.
But we do know that the O atom has 3 p orbitals.
So the TOTAL p-character count must be 3.
Let x be the p-character in the lone pairs:
0.8 + 0.8 + x + x = 3
(Note: we are assuming the lone pairs are identical.)
Solving for this, x = 0.7
(i.e. 70% p and 30% s)
(Double check for yourself that we’ve used only one s orbital in total.)
Angle between lone pairs
p-character = 1/(1 + γ2) = 0.7
so: -γ2 = 1/0.7 – 1 = -0.42
therefore θ = 115o
CHEM 2060 Lecture 23: VB Theory HF L23-14
Water: conclusions
The angle between the lone pairs is greater (115°) than the bond angle (104.5°).
The sp3 hybrid atomic orbitals of the lone pairs have > 25% s-character.
- less directional
- held more tightly to the O atom
The sp3 hybrid atomic orbitals of the bonding pairs have < 25% s-character.
- more directional (more p-character)
- electron density found in the bonding region between O and H
HOMEWORK: Can you now write down the full wavefunctions for the O atom
hybrid orbitals of water? (bonding & lone pair)