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A problem in hybridization

  1. Aug 17, 2013 #1
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    How could I know the hybridization type in such a question
     
  2. jcsd
  3. Aug 18, 2013 #2

    Borek

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    How do you determine hybridization in general?
     
  4. Aug 18, 2013 #3
    If the atom have one bond with another, the hybridization type is SP3. If it's double-bonded, the hybridization type is SP2. If it's triple-bonded with another one, the hybridization type is SP.
    Can I use this in less general cases?
     
  5. Aug 18, 2013 #4

    Borek

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  6. Aug 18, 2013 #5
    I find that the simplest approach to determine hybridization is just to count the number of regions containing electrons. To clarify what I mean, a single bond, a lone pair, a double bond, a triple bond etc would each count as one. It helps to draw lewis structures of each compound, that way you can make sure you get the right number of lone pairs.

    After you know the number of regions containing electrons, you know the hybridization! Just use the table below.

    2 regions = sp
    3 regions = sp2
    4 regions = sp3
     
  7. Aug 22, 2013 #6
    Thanks all, that was so useful. obviously letter B is the right answer
     
  8. Aug 22, 2013 #7

    DrDu

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    Why not also E?
    Is BeCl2 in the gas phase or solid?
    Does neutral ClO4 really exist?
    I wonder who poses these brain dead exercises?
     
  9. Aug 22, 2013 #8
    In E. hybridization type is sp3 in NH3 & it is sp2 in SO2 so it is wrong
    If BeCl2 is gas or solid, I don't think there would be a difference in the answer
    ClO4, I think it is so hard for such a molecule to be found
     
  10. Aug 23, 2013 #9

    DrDu

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    You can argue that in SO2 S is also sp3 hybridized and that the somewhat larger bond angle is due to electrostatic repulslion of the oxygens. So you can equally well describe it with sp2 or sp3.
    In the case of BeCl2, which is a solid at room temperature, Be is coordinated by 4 Cl atoms, so that a sp3 hybridization is also most natural.
     
  11. Aug 24, 2013 #10
    Step 1: Count the number of bonding regions on the atom in question plus the number of unshared pairs of electrons.

    Step 2: Beginning with the s orbital and working your way up, keep adding orbitals to the hybridization until the superscripts add to the number of bonding regions.

    Ex. for methane, carbon has 4 bonding areas. Add the s orbital which gives you 1, and then all three p orbitals for 3, so you have sp3 hybridization.

    Ex.2 For ammonia, you have 3 bonding areas, plus an unshared pair on the Nitrogen, so again, it is sp3.

    Ex.3 For PCl5, the phosphorous has 5 bonding areas of electrons and 0 unshared pairs, so you add in the s orbital, 3 p orbitals, and 1 of the d orbitals for sp3d orbital.
     
    Last edited: Aug 24, 2013
  12. Aug 24, 2013 #11
    For carbon you can use this as a general rule. You can also use it for Nitrogen if you consider its unshared pair a bond.
     
  13. Aug 24, 2013 #12

    DrDu

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    These rules are more a variant of VSEPR theory.
    Involution of d orbitals in main group elements has been disprooven since at least 50 years ago!
     
  14. Aug 24, 2013 #13
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