Homework 2A.9

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KTran 1I
Posts: 47
Joined: Fri Aug 30, 2019 12:15 am

Homework 2A.9

Postby KTran 1I » Sun Nov 03, 2019 7:45 pm

Which M^2+ ions (where M is a metal) are predicted to have the following ground-state electron configurations?
(a) [Ar]3d^7
(b) [Ar]3d^6

The answer in the solutions manual is (a) Co^2+ and (b) Fe^2+, but can someone explain how to get these answers?

KarineKim1L
Posts: 47
Joined: Fri Aug 30, 2019 12:16 am

Re: Homework 2A.9

Postby KarineKim1L » Sun Nov 03, 2019 7:54 pm

When removing electrons from an atom to form an ion, you remove from the 4s orbital first. Thus, the 2+ ion electron configuration for Co^2+ and Fe^2+ would be a) and b) respectively, as the electrons in the 4s orbital are removed.

ckilkeary 1H
Posts: 45
Joined: Fri Aug 09, 2019 12:16 am

Re: Homework 2A.9

Postby ckilkeary 1H » Sun Nov 03, 2019 7:57 pm

We know that these are ions that have had two electrons taken away because their charge is 2+. So we can add those ions back to the electron configuration, but we have to keep in mind that the two electrons were likely taken from the 4s subshell since 4s has to be filled before we start adding to the 3d orbitals but 4s is a lesser energy orbital once 3d starts filling up so its the outermost shell. So the neutral element's atom would be a) [Ar] 3d^7 4s^2 and b) [Ar] 3d^6 4s^2.
Counting on the periodic table we know that Co corresponds to a) and Fe corresponds to b). So the M^2+ ions are a) Co^2+ and b) Fe^2+


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