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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: Blueberries116 on February 22, 2020, 04:15:02 AM

Title: How electron configuration is related to the oxidation state of an element?
Post by: Blueberries116 on February 22, 2020, 04:15:02 AM

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?.

Title: Re: How electron configuration is related to the oxidation state of an element?
Post by: Enthalpy on February 24, 2020, 01:18:16 PM

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.