7th edition 2A.11

Moderators: Chem_Mod, Chem_Admin

Sophia Fox 4B
Posts: 29
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

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.

Return to “Electron Configurations for Multi-Electron Atoms”

Who is online

Users browsing this forum: No registered users and 2 guests