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For which of the following atoms is the 2+ ion paramagnetic?


The answer is Mn. How do you figure this question out?

  • Chemistry -

    Write out the electron configuration for each. Unpaired electrons will cause an atom or ion to be paramagnetic. For example, for Mn, it is
    1s2 2s2 2p6 3s2 3p6 3d5 4s2 and all 3d electrons are unpaired (remember Hund's rule that electrons don't pair until an orbital is at least half filled). If Mn were measured it would should it to paramagnetic AND that it had 5 unpaired electrons.

  • Chemistry -

    So where does the 2+ ion part of the question come in? Or does that even matter, do you just work them like you would if it would have just asked which atom is paramagnetic?

  • Chemistry -

    Sorry, I didn't notice the +2 part of the question. Everything I said above still holds; however, the electron configuration changes. For the +2 ion, remove the two 4s electrons so we have remaining
    1s2 2s2 2p6 3s2 3p6 3d5 (23 electrons) and you still have 5 unpaired electrons. Hund's rules still holds. (By the way, Hund's rule holds in MOST circumstances but now and then there is an exception)

  • Chemistry -

    Thank you very much i understand now. When you do the electron configuration do you just have to know what order it goes in? Like memorize the aufbau chart or is there a trick?

  • Chemistry -

    The Aufbau Principle works most of the time but not all the time. If you haven't seen it, here is a way to help count up which orbital fills next.

    However, there are exceptions to that, too. Here are a few tips that I use.
    First, I don't have all of the electron configurations memorized; personally, I think that is a waste of my time when I can go to and look each one up. Second, it IS convenient to know some of them. Before I retired, I asked my students to know the first 38. You can use the chart from the above site (the 1s 2s 2p 3s 3p etc chart---not the web elements chart) to correctly configure all of the elements from 1 through 38 EXCEPT for two elements. You memorize those. They are Cr and Cu (24 and 29).
    Second, since so much attention is paid to the 3d transition series, there is a trick to knowing how many electrons are in the 3d orbital--You know how many are in the others from the chart. The number of electrons in the 3d orbital is the same as the second number of the atomic number for the +2 ion. (I call this the DrBob222 rule).
    Sc is 1 (21), Ti is 2(22), V is 3(23) Ti is 4(24), Mn is 5(25), Fe is 6(26), Co is 7(27), Ni is 8(28) Cu is 9(29) and Zn is 10(30) (OK you must remember that 30 is 10 and not zero---the zero was at Ca (zero d electrons)(20) so the next zero must be 10. AND get this--there are no exceptions to that rule through the 3d series. Why you ask (I'm glad you asked), since Cr and Cu are exceptions that I mentioned above. The answer is that the exception for Cr (24) for the element is that we expect 3d4 4s2 but it REALLY is 3d5 4s1 BUT when we remove the two outside electrons to make the +2 ion we remove those two "pesky" electrons that are causing the trouble so the +2 ion is ok by the DrBOb222 rule. Same thing for Cu. We expect Cu to be 3d9 4s2 but it REALLY is 3d10 4s1 for the element. Again, removing the two outside electrons for the +2 ion gives us the 3d9 configuration we expect. Neat, huh?
    This may be more than you ever wanted to know about electron configuration, especially the 3d series, but such is life. There are a few tricks here and there as one goes through the table. Again, I think its a waste of ones time to memorize all of them.

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