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Can Berrylium show +6 state?

  1. Apr 16, 2015 #1
    1. The problem statement, all variables and given/known data
    While studying alkaline earth metals of the s block elements I found this equation :
    Be (OH)2+2HCl +2H2O →[Be (OH)4]Cl2

    It is clear that in the complex, the oxidation state of Berrylium is +6. But how is even that possible?

    2. Relevant equations
    None

    3. The attempt at a solution
    We know that Be has only 2 electrons in its s orbital and moreover has a total of only 4 electrons, so how is even that possible?
     
  2. jcsd
  3. Apr 16, 2015 #2

    Borek

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    Staff: Mentor

    Wouldn't hurt to check if the equation is balanced. If it is not, it is wrong.
     
  4. Apr 16, 2015 #3
    Well it's an amazing question.
    Don't know about you but find that reaction in my NCERT book.
    The only reason I can guess is OH donating its electrons to empty p orbitals of Berylium and therefore Berylium having that additional six electrons.
    It is related to ligand theory.
     
  5. Apr 16, 2015 #4
    Yeah it is not balanced. Don't know why in my book it is given as same as the mooncrater's equation.
    Can you, mooncrater give the reference of taking the reaction?
     
  6. Apr 16, 2015 #5

    Borek

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    Mistake is quite obvious here.

    Hint: water is a good ligand.
     
  7. Apr 16, 2015 #6
    Is that intended for me? Well I am not a expert in ligand theory. This thread brought an interest in me and I wanted to be in discussion.
    So then my reason is incorrect for OH donating electrons to empty p orbitals of Berylium?
     
  8. Apr 16, 2015 #7

    Borek

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    You don't have to be an expert to understand that the reaction that is not balanced is wrong. Beryllium compound - as written - doesn't exist. And I gave you both a hint to what does the real compound look like.
     
  9. Apr 16, 2015 #8
    Yeah, the equation should be,
    ## Be(OH)_2 + 2HCl + H_2O → [Be(H_2O)_3] Cl_2 ##
    Thanks Borek Sir.
    The question remains for mooncrater that how in book that he is reading, wrong equation is written?

    As in my book also same equation is given and our teacher ask to rote memorise it without proper logic.

    Though I have passed my 11th grade, I never realized it.
    Once again thanks Borek Sir
     
  10. Apr 16, 2015 #9

    Borek

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    Why Be(H2O)32+?
     
  11. Apr 16, 2015 #10
    Because H2O has a zero charge and Berylium would get +2 charge, which is its normal state.
     
  12. Apr 16, 2015 #11

    Borek

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    But why 3 water molecules? Why not 2 or 4?
     
  13. Apr 16, 2015 #12
    For balancing purpose. Otherwise reaction would not be balanced.
    Like the way, you are making sure that I get every concept of this question. :smile:
     
  14. Apr 16, 2015 #13
    It is from NCERT 11TH
     
  15. Apr 16, 2015 #14
    As I was suspecting.
     
  16. Apr 16, 2015 #15

    Borek

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    Be(OH)2 + 2HCl + 75H2O → [Be(H2O)77]Cl2

    Balanced?

    Google for beryllium aqua complex.
     
  17. Apr 16, 2015 #16
    Main reason for not balancing it:
    It is in our course book.
    Now it's purely impossible for the complex to exist.
    According to me [Be (H2O)3]Cl2(as Raghav said)
    should exist as Be has 2 electrons in its outermost orbital and needs 6 more for inert gas configuration. For that 3 H2Ogroups are needed.
    Is it correct?
     
  18. Apr 16, 2015 #17
    Searched it and got that Be is more stable when 4 molecules of H2O are bonded to it as Ligands.
    Reaction should be
    Be(OH)2 + 2HCl + 2H2O → [Be(H2O)4]Cl2

    mooncrater I realized that I was wrong.
    The compound should be [Be(H2O)4]Cl2.
    The reason is that , the coordination entity is having Be2+ not simply Be
    So 2s orbital , 2py,2px, 2pz orbitals are empty which hybridise to accommodate lone pairs of 4 water Ligands.
     
    Last edited: Apr 16, 2015
  19. Apr 16, 2015 #18
    Okay I got it now. Thanks.
     
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