## 7th edition 2A.9 and 2A.11

Alyssa Bryan 3F
Posts: 61
Joined: Fri Sep 28, 2018 12:26 am
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### 7th edition 2A.9 and 2A.11

I am a little confused how to predict the ions for these two problems.
2A.9 Which M^2+ ions (M=metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d7 (b) [Ar]3d6
2A.11 Which M^3+ ions (M=metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d6 (b) [Ar]3d5

Jessica Castro 2H
Posts: 60
Joined: Fri Sep 28, 2018 12:29 am

### Re: 7th edition 2A.9 and 2A.11

All this question is asking is which atom (of ^+2, which will make it an ion-cation) has the given configuration? However, remember: each orbital contains a pair of (2)electrons, so if the ion contains a charge greater than 2 (i.e. +3 charge), it will likely be referring to the atom in the next orbital.

Hope this helps!

ryanhon2H
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### Re: 7th edition 2A.9 and 2A.11

One way to figure the questions out is to just add up the number of electrons.

For example, 2A.9 a) is [Ar]3d7. Ar has 18 electrons, plus the 7 given in 3d7, for a total of 25 electrons. The question tells you that it is a +2 charge ion, which means the neutral atom lost 2 electrons. 25 electrons is the number of electrons for the ion, so just add 2 to 25 to get 27 for the neutral atom. Since a neutral atom has the same atomic number as the number of electrons, you know the element is atomic number 27, which is Cobalt, so the answer is Co^2+.

You can do these for the rest of the questions.