Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Oxygen's hybridization in H2O

  1. Jan 12, 2016 #1
    Guys I saw that during formation of H2O the oxygen atoms undergoes sp3 hybridization but my question is that what is the need for that? Yeah I also noted it was for occuring because of minimizd repulsion between p orbitals but why s orbital undergo for hybridization ?
     
  2. jcsd
  3. Jan 13, 2016 #2

    DrDu

    User Avatar
    Science Advisor

    There is no need for hybridization in H2O and you can describe it perfectly well using only p orbitals for bonding. Results will be almost identical.
     
  4. Jan 14, 2016 #3
    Yeah I thought the same but the geometry of the molecule describes the sp3 hybridization
     
  5. Jan 14, 2016 #4

    DrDu

    User Avatar
    Science Advisor

    Your argument is based on the assumption, that the orbitals forming the bond have to point exactly in the direction of the axis formed by the two bonded atoms. This assumption is called principle of maximal overlapp, but like most principles, it is only an approximation.
    When you describe a water molecule using only p orbitals for bonding, the bonds do not directly align with the orbitals. This can be explained with the repulsion of the positively charged hydrogen atoms. While aligning the bonds hybridizing in some s function would certainly increase the bond strength, this also costs energy as the promotion energy to lift an electron from an s orbital to a p orbital in oxygen is very high (much higher than in carbon).
    In effect, the energetic difference between a VB structure built using only p orbitals and one using sp3 orbitals is minute.
     
  6. Jan 14, 2016 #5
    I understood some of your mentioned concept here. But I want to understand what is actually going on here so can you please try to explain in simple terms?
     
  7. Jan 14, 2016 #6

    James Pelezo

    User Avatar
    Gold Member

    Mr Punjabi, Try this...
    Describing molecular geometries using the VSEPR Theory and the Valence Bond Theories...

    The VSEPR Theory requires determination of two factors that define the molecular geometry of a molecule; one, the number of bonded electron pairs and two, the number of non-bonded electron pairs. The total number of electron pairs determine the geometric valence configuration of the central element in the molecule, i.e., oxygen in this case and is referenced to a 'Parent Structure'... The final structural geometry is a 'derived structure' based on the geometry generated by the bonded pairs of electrons. The non-bonded pair do not define the terminology of the derived structure... So, (1) given H2O, the two Hydrogen attached to Oxygen represent two 'Bonded Pair of Electrons'. (2) The 'Non-bonded' pair is calculated from the simple formula = (∑ Valence electrons - ∑Substrate electrons)/2 = V - S / 2. The ∑V = Number of Valence Electrons = 2H + 2O = 2(1) + 1(6) = 8 valence electrons. The Substrate Number is the total number of electrons associated with the Hydrogen atoms when bonded per the 'Octet Rule'. However, Hydrogen has only 2 electrons in the valence shell when bonded. So, the ∑Substrate electrons when bonded = 2H = 2(2) = 4. So, the Number of Non-Bonded electron pair = V - S / 2 = 8 - 4 / 2 = 2 Non-Bonded electron pair associated with the central element oxygen. The total number of electron pairs associated with the central element oxygen = 2 Bonded Pair + 2 Non-Bonded Pair = 4 Electron Pairs. In the vernacular of the VSEPR Theory, this is an AX4 'Parent' structure which is a regular tetrahedron. Applying 2 Bonded Pr and 2 Non-Bonded Pr to the Tetrahedron => AX2E2 geometry. The final structural geometry is base only upon the 'Bonded Pairs' which gives the final geometry of the H2O molecule = Bent Angular. This approach works for binary molecular structures, but when understood can be extended to more complex multi-centered molecules. Try to define the geometry of CH4, NH3 in the same way. You will find both are based upon a Tetrahedral Parent with methane having the same geometry as the parent, but ammonia is an AX3E geometry. Applied to the regular tetrahedron, this translates into a 'Pyrimidal' Geometry.

    Describing molecular geometry using the Valence Bond Theory uses the same basic Parent Structures but describes the geometry in terms of 'Hybridized Valence Orbitals'. That is, the valence shell of the ground state electron configuration of the central element must undergo change in order to accommodate the Bonded and Non-Bonded electron pairs associated with the central element of the binary molecule. Applied to water, begin with the electron configuration of Oxygen =>

    O: [He}2s22px2py1pz1. The valence shell of Oxygen can not accommodate bonding in this configuration because of energy differences between the valence shell sub-orbitals. So, the valence orbitals blend together to generate 4 equal energy 'Hybridized Orbitals' made from 1 s-orbital and 3 p-orbitals. Each of the 4 hybrid orbitals are termed sp3 hybrids b/c they are formed by blending 1s and 3p orbitals. Each is an asymmetric figure 8.
    upload_2016-1-14_22-25-7.png
    sp.gif

    Four of the hybrids (green) will join at the apex origin and assume positions in 3D space that minimize repulsion energy. Two of the hybrids have paramagnetic electrons (i.e., a single electron) and two have diamagnetic electrons (i.e., Non-Bonded electron pairs). The Hydrogen 1s1 orbitals pair with the paramagnetic electrons to form two Bonded pair and a Bent Angular Geometry.

    hybrid1.gif

    Two of the hybrids contain diamagnetic electron pairs ( = Non-Bonded Prs ) while the two with paramagnetic electrons bond by paring with the 1s electron in each of the Hydrogen atoms.

    Now, one thing you need to understand about this is, the hybrids are 'pre-bonding' conditions. Once the H bonds to the Oxygen the bond becomes a 'Sigma Bond'. Which is defined by the fact that a line of symmetry can be drawn through the nuclei of the bonding elements. This means that water has two sp3 hybrids with Non-Bonded Diamagnetic electrons and two Sigma bonds accommodating the Hydrogen to Oxygen Bonded Pair.

    206water.gif
    The H - O bonds are Sigma Bonds and the Non-Bonded Pr are in sp3 hybrids. Try this with CH4 and NH3. Methane has 4 Sigma Bonds from H-C bonds and zero Non-Bonded Pr => geometry = Tetrahedron. Ammonia has 3 Sigma Bonds and 1 sp3 hybrid with 1 Non-Bonded Pr => geometry = Pyramidal. Remember, geometries are defined only in terms of bonded pairs. Also, geometries are equivalent to those derived in the VSEPR Theory. :-)
     
  8. Jan 15, 2016 #7

    DrDu

    User Avatar
    Science Advisor

    You could have a look at this article:
    McWeeny, Roy, and F. E. Jorge. "Hybridization in valence bond theory: The water molecule." Journal of Molecular Structure: THEOCHEM 169 (1988): 459-468.
     
  9. Jan 15, 2016 #8

    James Pelezo

    User Avatar
    Gold Member

    I respectfully disagree. The ground state valence shell electron configuration of Oxygen can not accommodate covalent bonding, first, because the s orbitals and p orbitals are at different energy levels, second because the s-orbital and one p-orbital have diamagnetic electron pairs at two different energy levels and third, the configuration can not assume a bent angular geometry with a bond angle of 104.4o. This has been defined many times independent of either VSEPR or Valence Bond considerations by instrumental spectroscopic methods. Even if boding did occur in the ground state p-orbital configuration, the bond angle would be 90o between p-orbital orientations ... NOT!!! Experimental evidence has confirmed 104.4o at 250C. Also, the fundamental energy values for p-orbital + s-orbital overlap are too thermodynamically unstable to accommodate such a bond. The only p-orbital bonding that is found in the literature is the sideways overlap of non-hybridized p-orbitals giving π-bonds in unsaturated 2nd and 3rd order bond systems. Furthermore, the only s-orbital to s-orbital bonding that I know of is in molecular Hydrogen giving a symmetrical Sigma Bond between Hydrogen nuclei.

    Also, I am familiar with McWeeny, Roy and Jorge hybridization concepts. The conclusions are valid as applied to hybridization outcomes, but much simpler methods are available that define the number of hybrids needed for the central element of a binary molecular configuration.
     
  10. Jan 16, 2016 #9

    DrDu

    User Avatar
    Science Advisor

    As I explained before, the bond angle has not to equal the angle between the orbitals. The difference between 104.5 and 90 is only 15 degrees so that each p orbital is only 7.5 degrees off from the C-H bond axis. The reduction in overlap is therefore quite small. The description of the VB structure for H2O I found also in the book by Ira Levine, Quantum Chemistry.
     
  11. Jan 17, 2016 #10

    James Pelezo

    User Avatar
    Gold Member

    Never said it was ... My emphasis is on the bond angles being the angle(s) between the Sigma Bonds. The orbitals with non-bonded electron pairs are not referenced in the geometry. For water, the non-bonded pairs repel the bonded pairs to form the 104.4o angle between the Sigma Bonds. If you wish to compare the bond angles of the series methane, ammonia and water, then one must consider the 4 hybrid orbitals in which all three cases join together to form a regular tetrahedrons. For CH4 => AX4 geometry with all Sigma bonds having bond angles of 109.4o. All have equivalent repulsion effects and assume positions in space of the regular tetrahedron. For NH3 the geometry is AX3E with the non-bonded pair providing a slightly higher torque repulsion energy and gives three Sigma bonds at 107o. Water is a bent angular AX2E2 configuration with a H - O - H bond angle of 104.4o. The repulsion energy of two non-bonded electron pairs is ~2x that of a single non-bonded pair. The ground state s and p orbitals in the 2nd energy level can not arbitrarily form covalent bonds without hybridization. The repulsion effects are just too unstable. Visualize a stiff coiled spring distorted to 90o angle. Then (visualize) releasing a bit of applied torque. The torque force to sustain a 104.4o angle is less than the torque energy needed to sustain a 90o angle. Electron pair repulsions function in the same way; i.e., repel to positions in space that minimize repulsion potential energy. The structural model of molecular water based upon the ground state electron configuration of Oxygen doesn't take into account the repulsion effects of the diamagnetic s and px orbitals. These just can't be ignored. Also, I am familiar with Levine's work and do not agree with his conclusions on the topic for the same reasons. Brilliant mind, but I have some issues with some of the fundamental assumptions. Let's not go there as the debate will become intractable and more confusing than enlightening.
     
  12. Jan 17, 2016 #11

    DrDu

    User Avatar
    Science Advisor

    It would be nice to do some ab initio calculations. Should be easy using e.g. Gamess US which includes several implementations of Valence Bond code. However, I don't have the possibility to do so.
     
  13. Jan 17, 2016 #12

    DrDu

    User Avatar
    Science Advisor

    I just found the time to look up what Linus Pauling, the great inventor of hybridization, has to say on bonding in water:
    On page 111 of "The nature of the chemical bond, third edition, Cornell University Press, 1960", he says:
    "The conclusion that p bond tend to be at right angles to one another is verified to some extent by experiment (Table 4-1). In water, with the structure HOH [printed with a right angle] the bond angle is 104.5 deg. We expect the bond to be p bonds rather than s bond for the followign reason: A 2s electron of oxygen is more stable than a 2p electron by about 200 kcal /mole; and if the s orbital were used in bond formation (being then occupied effectively by only one electron) rather than for an unshared pair the molecule would be made unstable to this extent. The difference of 14.5 deg between the observed value of the bond angle and the expected value of 90 deg is probably to be attributed in the main to the partial ionic character of the O-H bonds , estimated in the preceding chapter to be 39 percent. This would give a resultant positive charge to the hydrogen atoms, which would repel one another an thus cause an increase in the bond angle. This effect is discussed in the more detailed treatment of bond angles that is given in Section 4-3. The large value for ammonia, 107 deg, may be attributed to the same cause."

    In section 4-3 he estimates the ammount of hybridization in H2O based on the equilibrium of hybrid orbitals forming somewhat stronger bonds and the promotion energy. For water, he finds a value of 5% s character of the bond orbitals. Compare this to 25% s-character of an sp3 hybrid orbital.
     
  14. Jan 17, 2016 #13

    James Pelezo

    User Avatar
    Gold Member

    Yes that was the early Pauling supposition for methane; i.e., 3 p-orbital bonds at right angles along with a weak s-orbital bond in an arbitrary direction. However, he recanted this later by stating in reality the bonding configuration consists of a blend of s and p-orbital hybrid orbitals (sp3) having equivalent character and assuming positions in space of 104.4o. This is maintained if all 4 hybrids are involved in equivalent covalent bonds with same substituent. However, if the central element hybrid orbital was a diamagnetic non-bonded pair or a sigma bond with a highly electronegative element, the bond angles would be compressed due to stronger electron pair repulsion energies. The O - H electropositive repulsions can 'seem' logical for the 107o H - O - H bond angle, but find difficulty in application to hypervalent structures such as PCl5 and SF6 and especially in binary derivatives such as SF4, ICl5. Sulfur has the same ground state valence configuration as does Oxygen, but try to apply the right angle bonding concept in p-orbitals without hybridization to SF6. Here Sulfur must accommodate 6 equivalent sigma bonds to Fluoride atoms to form a regular octahedron. The most practical approach ( and in my humble opinion ) is to determine the number of bonded and non-bonded electron pairs needed to accommodate a given binary molecular system. The total number of electron pairs defines the saturated parent geometry which when substituted with bonded and non-bonded pairs define the related derived geometry from the parent structure.

    Octet Centers:
    CH4 (Tetrahedral parent AX4) => 4 bonded pair & 0 non-bonded pair => regular tetrahedron => θb = 109.5o
    NH3 (Pyramidal derivative of AX4 => AX3E) => 3 bonded pair & 1 non-bonded pair => θb = 107o
    H2O (Bent angular derivative of AX4 => AX2E2) => 2 bonded pair & 2 non-bonded pair => θb = 104.4o

    Hypervalent Centers:
    PCl5 (Trigonal Bipyramidal Parent AX5) => 5 bonded pair & 0 non-bonded pair => Trigonal Bipyramidal => Planar θb = 120o and Axial θb = 90o
    SF4 (See Saw derivative of AX5 parent => AX4E geometry) => 4 bonded pair & 1 non-bonded pair => See Saw geometry... etc...
     
  15. Jan 18, 2016 #14

    DrDu

    User Avatar
    Science Advisor

    I have given 3 I think very sound references to support my point. I am not referring to very early work of Pauling, but his book from 1960, where he uses sp3 hybrids to describe bonding in Methane. Most of your replies are some general statements based on VSEPR ideas. So I would kindly ask you to name some serious sources or calculations that purport your view.
     
  16. Jan 18, 2016 #15

    James Pelezo

    User Avatar
    Gold Member

    Well, here are about 300 current College Chemistry textbooks all of which support the hybridization theory => https://www.google.com/search?q=col...hUKEwiQ-JTTw7TKAhUGTCYKHbdCDGQQ7AkIWA&dpr=0.9 ... As for 'general statements based on the VSEPR ideas' my last post was quite specific with respect to Valence Bond Theory and the need for hybrid orbitals.

    As for your references, I don't refute your supporting references, they are very legit and do support your position, but that does not mean that I have to agree with the conclusions. Actually, I didn't mean for the debate to extend into a justification for bonding theory, all I wanted to say is I don't agree with the contention that the geometry of molecules (i.e., the water reference in your original post reply to Mr Punjabi) can be done without using hybridization of the ground state valence shell electrons. Please forgive if I've left any inappropriate implications.
     
  17. Jan 19, 2016 #16

    DrDu

    User Avatar
    Science Advisor

    I have heard and later on given theoretical chemistry classes myself. I got the feeling that most of these college level books tend to ignore the insights gained into bonding acquired since the advent of computers (so the last 50 years or so). E.g. they still discuss bonding in main group elements using spxdy hybrid orbitals although one knows least since the end of the 70's that d orbitals play at most a role as polarisation functions in main group compounds.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Oxygen's hybridization in H2O
  1. Formation of H2O? (Replies: 42)

  2. O2 + H2O is? (Replies: 4)

  3. H2O Formation (Replies: 2)

  4. H2O in Equilibria (Replies: 8)

Loading...