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 2px

2 2py2 2pz

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The simplest VB wavefunction for HF is: ψ = 1sF

2 2sF2 2pxF

2 2pyF2 [1sH(1) 2pzF(2) + 1sH(2) 2pzF(1)]

We can rewrite a simplified version of this wavefunction by ignoring the non-bonding 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 φF = 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 Decrease in e-e repulsion concentrated between

nucleii

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)

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.

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1 = N 2 + N 2γ 2 ⇒ N = 11+γ 2

φF2∫ ∂τ = N 2 2pzF( )2∫ ∂τ

=1

+ N 2γ 2 2sF( )2∫ ∂τ

=1

+ 2N 2γ (2pzF∫ 2s)∂τ

= 0

=1

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). Ground state electronic configuration of atomic O is: 1s2 2s2 2px

2 2py1 2pz

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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!

OH H

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:

N is the normalization coefficient

All this means is that the new orbital contains 1 electron. So the integral of the function squared over all space =1 (orthogonal!)

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.

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φ = N p + γs( )

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φ2∫ dτ =1

N 2 p+γs( )∫2dτ =1

= N 2 p2∫ dτ=1

+ N 2γ2 s2∫ dτ=1 + N 2 2γ sp∫ dτ

=0

CHEM 2060 Lecture 23: VB Theory HF L23-10

So… Therefore…

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N 2 + N 2γ2 =1

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N =11+γ2

hybrid wavefunction is then: The s and p characters are now easy to get… square of that part of φ

Amount p character:

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11+γ2

#

$ % %

&

' ( (

2

=1

1+γ2 …as γ →0, p-character →100%

Amount s character:

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11+γ2

#

$ % %

&

' ( (

2

γ2 =γ2

1+γ2 …as γ →1, s-character →50%

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φ = 1+ γ 2( )− 12 p + γs( )

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:

(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) O=C=O carbon atom in CO2 The bond angle is 180°. cos180° = -1 so γ = 1

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φsp = 12 s+ p( ) REMEMBER: Mix 2 (s & p) → Get 2 (2 x sp)

cos θ = - γ2

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

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γ = 0.25 = 0.5

Amount p character:

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11+γ2

=1

1+ 0.52= 0.80 i.e. 80% p-character

This should leave us with 20% s-character…let’s double check that:

Amount s character:

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γ2

1+γ2=0.52

1+ 0.52= 0.20 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: -γ

2 = 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)