Hydroxide ion

  1. Question: State the hybridization used and the molecular geometry shape of hydroxide ion, OH-.

    Okay, so I draw out the Lewis structure, where there are 6 electrons (3 lone pairs) around the O atom, which is bonded to the H atom. So, I think base on the 3 nonbonding domains and 1 bonding domain, I think the shape is tetrahedral, but it seems weird to me, because I think it may be linear.

    Then, regarding the hybridization, I think the O atom uses sp3 hybridization, but then again, I think there is no hybridization involved, because O can use the remaining empty 2s orbital to bond to the only one electron from H atom.

    Which one is the correct one? :confused:
     
  2. jcsd
  3. Gokul43201

    Gokul43201 11,141
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    When you are asked for the shape of the molecule/ion, you are asked to describe the relative positions of the atoms, not the relative positions of the orbitals. If you were going to describe the geometry of the orbitals (instead of the geometry of the atoms) then you'd say NH3 was tetrahedral (instead of a trigonal pyramid).

    What empty 2s orbital? Is there an empty 2s in oxygen?
     
  4. First: Identify the number of valence electrons present in the total compound.
    Second: Draw the Lewis structure
    Third: Remember that though the electron configuration maybe tetrahedral that does not mean that the whole molecule will have that shape. You must look at the total 'effective bonds'.

    I agree with Gokul, there are no empty 2s orbitals for O. Look at everything again. Hope that helps.
     
  5. Oh, sorry, typo there. I mean 2p orbital.
     
  6. Gokul43201

    Gokul43201 11,141
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    Okay, that makes more sense. But now, it seems you need to understand how hybridization works.

    You could as well ask a similar question regarding the H2O structure. After all, you could just as well argue that the 2 vacancies in the 2p orbitals can be used to accomodate the electrons from the 2 H atoms. So, the exact same argument as the one above, would lead you to conclude that H2O requires no hybridization either.

    If that were true, why then, is the H-O-H bond angle much closer to 109 than it is to 90 deg? Clearly, hybridization is involved in H2O, as it will be in the case of OH-. The reason that hybrid orbitals are closer to reality is due to the reduction in energy that results from hybridization.
     
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