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### Electron Configurations

Posted: **Wed Nov 01, 2017 5:34 pm**

by **Abby Ellstrom 1I**

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}

The answer to

(a) is Co^{3+}

(b) is Fe^{3+}

I am just very confused as to where these answers came from.

### Re: Electron Configurations

Posted: **Wed Nov 01, 2017 8:36 pm**

by **Joyce Lee 1C**

The answer to part (a) is Co3+ because Co has the electron configuration: [Ar] 3d7 4s2

However, Co3+ loses 3 electrons (2 from 4s and 1 from the 3d orbitals). The new configuration is [Ar] 3d6

Similarly, for part (b), if two electrons from the 4s orbital and one electron from the 3d orbital are lost, you get the given configuration: [Ar] 3d5. Thus the answer must be Fe3+

### Re: Electron Configurations

Posted: **Wed Nov 01, 2017 8:51 pm**

by **Isabella Sanzi 2E**

So for this problem, you have to look at the order of removal based on an electron's configuration. In the case of part a, the book asks for the metal (M) that would have the configuration for M3+: [Ar]3d^6. We know that when electrons are removed from an atom, they are removed from the highest energy level, so because we are working backwards since we are starting from the 3+ state, we can assume that the 2 out of 3 of electrons that would be added back on to make it neutral would first fill the 4s orbital (as these would be the first to be removed). Then we can assume that the d orbital would resume being filled to make it 3d^7 with the remaining electron. The configuration [Ar]3d^7 4s^2 corresponds to cobalt as the unknown metal. We can apply a similar principle to part b. Because [Ar]3d^5 is given, we can assume that 2 of the 3 electrons required to make the atom neutral would fill the 4s orbital and then the remaining would go to the 3d orbital to make it 3d^6. The configuration [Ar]3d^6 4s^2 corresponds to Fe. I hope this helps!