Hix3r
Regular
Joined: 24 Apr 2007
Posts: 43
Location: Hungary
|
 Posted: Sun Jun 03, 2007 9:58 pm Post subject: |
 |
|
Hmm.. okay, I'm not the best in these kind of questions, but
Sb usually takes the +3 state, because it has the last orbital with three unpaired electrons. in this compound Sb has +3 charge(or whatever you call it) SbCl5(2-). Each chloride has -1 charge, 5 makes -5. because the compound has -2 charge two of the chloride drops out. Then we have three minus charges for one Sb-ion which then has +3 charge.
Anyway this tells us that in this ion Sb does not form a hibrid orbital, because it only took the three unpaired electrons.
In the case of SbCl5 with no overall charge, the question is different. You see the atom has only three unpaired electrons. To make more, it destroys another pair of electrons to make more unpaired. I don't really know this with atoms having the d-orbital, but I think this one still breaks the pair in the s orbital and puts one electron into the 6s-orbital, because that is the orbital that comes next energetically. (You know that after 3p the order of the orbitals becomes a mess  ) So it looks like this 5s1 5p3 6s1 and with now 5 unpaired electrons it can form an Sb+5 ion, to have 5 chloride. I don't really know how is this called, I have a hunch that these names come that it indicates the orbital the electron came from, and the orbital it went including the other orbitals that are int he hibrid state, like carbon has an sp3 hibridization so the electron came from the s-orbital and went to the p-orbital which has three unpaired slots and the entire p is in hibrid state. So with this I would name it sps1 or sp3s or I don't know... I really don't know for sure. Anyway thank god your question doe not involved this.
Again please if anyone sees an error please correct me. I am often wrong.
|
|
