3.9

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Kelly Kiremidjian 1C
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Joined: Fri Sep 29, 2017 7:04 am

3.9

Postby Kelly Kiremidjian 1C » Sun Nov 05, 2017 9:29 am

Can anyone explain 3.9 from the homework problems? I am confused on what the question is asking for. Also for the given electron configurations
a.) [Ar] 3d7
b.[Ar]3d6

are these possible? Doesn't the 4s orbital need to be filled for to have an electron in 3d?

thanks

Sarah_Stay_1D
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Re: 3.9

Postby Sarah_Stay_1D » Sun Nov 05, 2017 10:19 am

Kelly Kiremidjian wrote:Can anyone explain 3.9 from the homework problems? I am confused on what the question is asking for. Also for the given electron configurations
a.) [Ar] 3d7
b.[Ar]3d6

are these possible? Doesn't the 4s orbital need to be filled for to have an electron in 3d?

thanks


So The confusing part about this question is that in this case M represents an unknown element (it acts as a variable almost). This question want you to look at the electron configurations given, and determine which metal ions could have those configuration. (Now note that in the syllabus Lavelle said we only needed to do a and b). We know that all metals form cations, which means that an electron is removed from an atom. If you look at configurations from a and b, you were correct in noticing that the 4s orbital is not there. This is because for the 3d metals, the 4s orbital is actually higher in energy than the 3d orbital, meaning the 4s electrons will be removed first. Now we know that two electrons have definitely been removed from our mystery atom. Now all you need to do is note that for a, there are 7 electrons in the d orbital, which corresponds to cobalt. This means a is Co2+. For b, note that there are 6 electrons in the d orbital, this corresponds to iron. This means b is Fe2+. Remember these configurations are possible because the 4s electrons were removed from both atoms, creating two cations.

miznaakbar
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Re: 3.9

Postby miznaakbar » Sun Nov 05, 2017 10:22 am

From what I understand, when an electron configuration includes electrons in both the d and s orbitals, such as [Ar] 3d^7 4s^2, the 4s level is higher in energy and thus electrons must be removed from it first. Therefore, when in question 3.9 it asks for which M^(2+) ions have the following electron configurations, you need to take into account the fact that electrons will be removed from 4s before they are removed from 3d.

In part a, they give you the configuration [Ar]3d^7 and they say that the ion whose configuration this is has a 2+ charge (meaning the original atom lost 2 electrons). Since we know any regular atom would need to have 4s filled before 3d, we can assume the original atom before it was ionized had the configuration [Ar] 3d^7 4s^2. We can see that this is the configuration of the cobalt atom. Thus, [Ar] 3d^7 is the configuration of the Co^2+ ion (it became +2 charged when it lost its 4s electrons).

When it comes to ions, the 4s orbital does not need to be full for the electrons to be in the 3d. Hope this helps!


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