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Thread: How can Chlorine make 7 bonds in the perchlorate ion?

  1. #1 How can Chlorine make 7 bonds in the perchlorate ion? 
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    Chlorine being a Halogen should surely have an affinity for just the one bond, right?

    Potassium perchlorate - Wikipedia, the free encyclopedia


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    Thanks so much! I know quite a bit about Orbitals but clearly not enough. Can you link me to a good video or something?


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    exchemist
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    Quote Originally Posted by 09jnewington View Post
    Thanks so much! I know quite a bit about Orbitals but clearly not enough. Can you link me to a good video or something?
    If you don't want to get into MO theory there is a simple way to understand "expanding the octet".

    Since Cl is in the 3rd short period, its outer shell has 3s, 3p and 3d orbitals available. The 3d are of higher energy than 3s and 3p and are unfilled by the end the 3rd period. But they are there. Cl can be thought of as expanding its octet by"unpairing" some or all of its paired electrons in the 3s and 3p promoting some into these unfilled 3d orbitals. If it unpairs all of them it ends up with 7 unpaired electrons, one in 3s, 3 in 3p and 3 more in 3d. So then it can form 7 covalent bonds. It is said to be "sp3d3 hybridised". There is of course an energy "cost" to promoting 3 unpaired electrons to higher energy 3d orbitals, but if the covalent bonds that are formed are strong enough they lead to a lower final energy configuration than the atom would have if these electrons were not participating in bonding. So it is still energetically "worthwhile" - and therefore it happens.

    The "octet rule" is really an exception in chemistry, as it only strictly works for atoms in the 2nd period, for which the outer atomic orbitals are 1 x 2s and 3 x 2p (no d), i.e. 4 orbitals, each of which can accept 2 electrons, so 8 in total. But of course a lot of the more common elements are in this 2nd period, giving rise to the "rule" - which isn't really one, in the grand scheme of things.
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