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Postby joanneyseung22 » Sun Oct 28, 2018 10:04 pm

I just needed some clarification on Professor Lavelle's lecture notes because I am not quite understanding the concepts between the examples of cations and anions.
For example, Professor Lavelle used Indium as an example for a cation. In = [Kr](4d^10)(5s^2)(5p^1), and In+= [Kr](4d^10)(5s^2). In this case why is the entire (5p^1) removed?
Secondly, for anions, the example was Nitrogen, and N:[He](2s^2)(2p^3). Why is that for the N^3-, there is an added p^3 to make (2p^6)?

Kathryn 1F
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Joined: Fri Sep 28, 2018 12:19 am

Re: Cations/Anions

Postby Kathryn 1F » Sun Oct 28, 2018 10:58 pm

Hi! For your first example, the whole 5p1 is removed because that represents just one electron. One electron must be removed to have a charge of +1. The opposite is true for the second example as you need 3 extra electrons

Yvonne Du
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Joined: Fri Sep 28, 2018 12:23 am

Re: Cations/Anions

Postby Yvonne Du » Sun Oct 28, 2018 11:04 pm

For your first example, the whole 5p1 represents one electron only. In the future, you can know the number of electrons in each shell by looking at their superscript. For example, 5p5 means there are 5 electrons in the 5p shell.

For your second example, p shells contain 6 electrons in total. In order to advance to the next shell(d), p shell has to be filled up first. Since you want 3 more electrons for Nitrogen, you add the electrons to the p shell.

Hope this helps.

Rehan Chinoy 1K
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Re: Cations/Anions

Postby Rehan Chinoy 1K » Sun Oct 28, 2018 11:06 pm

Yes, as stated above, 5p^1 represents one electron in the outermost 5p subshell of Indium, so when Indium loses an electron to become the cation In+ it will lose the outermost electron, or 5p^1. When Nitrogen gains 3 electrons to become the anion N^3-, 3 electrons are added to the outermost energy level and subshell, which in this case is 2p, so the configuration turns into 2p^6.

Germar G 4F
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Re: Cations/Anions

Postby Germar G 4F » Sun Oct 28, 2018 11:17 pm

5p1 represents an electron. Since an electron is negatively charged, and with In+ you are losing an electron, taking away the 5p1 accomplishes that in electron configuration notation. Same concept applies to N: if it is N^-3, then making 5p3 to 5p6 represents the three electrons that are gained by N.

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