Valence bond picture for HF A nice 2060 Lecture 23: VB Theory HF L23-1 Valence bond picture for HF A...

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Transcript of Valence bond picture for HF A nice 2060 Lecture 23: VB Theory HF L23-1 Valence bond picture for HF A...

  • 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: = 1sF2 2sF2 2pxF2 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: = 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 F = N(2pzF + 2sF) (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:

    = 1sH(1) F(2) + 1sH(2) F(1) + (F(1) F(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


    Since part of 2s was mixed with the 2p F = N(2pzF + 2sF) the LONE PAIR is not simply 2s2 ! lp = N(2s - 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:

    F = N(2pzF + 2sF)

    Normalization means that when an orbital is "scanned over all space (i.e., take the integral over all spacearea under the curve) the probability of finding the electron is 1 (i.e., 100%). [Def] For a normalized wavefunction N, the probability of finding the electron in all space (i.e., anywhere in the universe) is exactly 1. N is the normalization constant.

    1 = N 2 + N 2 2 N = 11+ 2

    F2 = N 2 2pzF( )

    2 =1

    + N 2 2 2sF( )

    2 =1

    + 2N 2 (2pzF 2s)

    = 0


  • 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 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--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 haos, each sp3 orbital should have:

    25% s character and 75% p character.

    For our water molecule, our haos cant have exactly s and 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!)

    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.

    = N p + s( )

    2 d =1

    N 2 p+s( )2d =1

    = N 2 p2 d=1

    + N22 s2 d

    =1 + N

    2 2 sp d=0

  • CHEM 2060 Lecture 23: VB Theory HF L23-10

    So Therefore

    N 2 + N 22 =1

    N = 11+2

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

    Amount p character:



    $ % %


    ' ( (



    1+2 as 0, p-character 100%

    Amount s character:



    $ % %


    ' ( (


    2 =2

    1+2 as 1, s-character 50%

    = 1+ 2( ) 12 p + s( )

  • CHEM 2060 Lecture 23: VB Theory HF L23-11

    How do we choose the correct value of ? The mixing coefficient is clearly related to the bond angle . 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 = 1

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

    cos = - 2

  • 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

    = 0.25 = 0.5

    Amount p character:



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

    This should leave us with 20% s-characterlets double check that:

    Amount s character:



    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 . 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 weve used only one s orbital in total.) Angle between lone pairs

    p-character = 1/(1 + 2) = 0.7 so: -

    2 = 1/0.7 1 = -0.42 therefore = 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)