7th edition 2A.11

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Sophia Fox 4B
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Joined: Fri Sep 28, 2018 12:27 am

7th edition 2A.11

Postby Sophia Fox 4B » Sun Nov 04, 2018 1:24 am

7th edition 2A.11: Which M^3+ ions (where M is a metal) are predicted to have the following ground state electron configurations: (a) [Ar]3d^6, (b) [Ar]3d^5.
I thought the answer to a was Cu^3+, and that the answer for b was Ni^3+, but the answer key says that a is Co^3+ and b is Fe^3+. I really can't figure out why. Can someone explain this to me please

armintaheri
Posts: 68
Joined: Fri Sep 28, 2018 12:26 am

Re: 7th edition 2A.11

Postby armintaheri » Sun Nov 04, 2018 1:04 pm

Short answer: transition metals are the spawn of satan and they are out to ruin your life.
Long answer: During ionization, 4s electrons are lost before 3d electrons. So if you removed three electrons and ended up with 3d6, that means you removed 2 electrons from 4s and 1 electron from 3d. So the original configuration was 4s23d7, which is Cobalt. I think your mistake was assuming 4s was filled, so you just added 3 electrons to 3d and ended up with copper. But if you look at the question, it gives the electron configuration as [Ar]3d6. That meant the electron configuration of Argonne, plus 6 electrons in 3d. But Argonne has no electrons in 4s, so you should know your 4s subshell is empty, even if you don't remember that 4s electrons are ionized first.


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