Asked by Ezra
In the ground-state electron configuration of Fe^{3+}, how many unpaired electrons are present?
An explanation would be great, thanks.
An explanation would be great, thanks.
Answers
Answered by
DrBob222
Electron configuration Fe is
26Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2.
To make the 3+ ion we remove the two 4s electrons and one of the 3d6 to make it 3d5. Remember Hund's that electrons don't pair until the must AND that there are 5 3d orbitals, then all 5 of the 3d electrons must be unpaired.
26Fe = 1s2 2s2 2p6 3s2 3p6 3d6 4s2.
To make the 3+ ion we remove the two 4s electrons and one of the 3d6 to make it 3d5. Remember Hund's that electrons don't pair until the must AND that there are 5 3d orbitals, then all 5 of the 3d electrons must be unpaired.
Answered by
Ezra
Thanks so much, that really helped me solve other problems.
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