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12 February, 00:47

Consider the two electron arrangements for neutral atoms A and B. What is the difference between atom A and atom B?

A - 1s22s22p63s1

B - 1s22s22p65s1

Atom B has lost some inner electrons. The outer electron of atom B has moved to a lower energy state. The outer electron of atom B has moved to a higher energy state.

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  1. 12 February, 00:56
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    outer electron moved to a higher energy state

    Explanation:

    To know this, let's write again the electron arrangements:

    [A] = 1s² 2s² 2p^6 3s^1

    [B] = 1s² 2s² 2p^6 5s^1

    Now, let's discart the options.

    First option cannot be, because if you count the inner electrons, which are the numbers uppering (besides the s and p), in both cases, you have the same number of electrons, which is 11 in both cases.

    Second option cannot be either, because electron configuration, always go from lower to higher state. In the case, that one electron move to a lower state, it should move fro 3s to the 2p level. In this case, it was not, mainly because the lower level of 2, has already maxed out it's capacity to hold electrons.

    Therefore, option 3 it's the more accurate option, you can see that from 2p6, it jump to the energy level 5, skipping 3 and 4th level. Therefore this is the correct option.
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