March 28, 2024, 07:25:21 PM
Forum Rules: Read This Before Posting


Topic: How electron configuration is related to the oxidation state of an element?  (Read 733 times)

0 Members and 1 Guest are viewing this topic.

Offline Blueberries116

  • Regular Member
  • ***
  • Posts: 32
  • Mole Snacks: +0/-1
  • Gender: Male
  • Freelance scientist and learner
The question arises from trying to get where does the values of [itex]-2[/itex], [itex]0[/itex], [itex]+2[/itex], [itex]+4[/itex] and [itex]+6[/itex] from the electron configuration.

The electron configuration for sulphur [itex](Z=16)[/itex] is [itex][Ne]3s^23p^4[/itex].
It makes sense that to complete the octet sulphur needs to gain two electrons hence it will have [itex]-2[/itex], but it can lose all the electrons [itex]2+4=+6[/itex]. But I have no idea what sort of explanation can be made for obtaining the other values?. Does it exist a rule or something that am I unware of?.
Have a nice day!

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4041
  • Mole Snacks: +304/-59
The octet rule is over-simplified, alas. Complicated world.

Other theories can predict additional valences more or less, but when you arrive at transition metals, all simple rules and theories fail. At best, heavy software computations can try a prediction.

Sponsored Links