Page 1 of 1

### 2A.11

Posted: Mon Oct 28, 2019 12:35 am
Which M31 ions (where M is a metal) are predicted to have the following ground-state electron configurations: (a) [Ar]3d6; (b) [Ar]3d5; (c) [Kr]4d5; (d) [Kr]4d3?

### Re: 2A.11

Posted: Mon Oct 28, 2019 12:52 am
What part of the question are you having trouble with?

### Re: 2A.11

Posted: Mon Oct 28, 2019 8:40 am
In order to find what metal with a 3+ state would have the ground state configuration, I find it helpful to first write what configuration the metal would have if it were not an ion. For the first one, for example, if the 3+ state is [Ar] 3d6, then the configuration for the unionized version would be [Ar] 3d7 4s2. Then, you can use the periodic table to see that this corresponds to Co, so then it would be Co3+.

### Re: 2A.11

Posted: Mon Oct 28, 2019 9:33 am
To find the 3+ ions of the specified electron configurations you need to consider that metals lose their s electrons first. So in the case of all these metals you would just be moving one to the right on the periodic table as the other two electrons are lost from the s orbital of the shell above.

### Re: 2A.11

Posted: Mon Oct 28, 2019 3:51 pm
For part a, [Ar]3d^6 has a total of 24 electrons, which is chromium. However, since the unknown metal has an additional 3 electrons, you would find the element that has 27 electrons, which is cobalt. The answer to part a would be Co^3+. You can apply the same process to parts b, c, and d.