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JFraser

  • one year ago

Tutorial: Ionization Energy

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  1. JFraser
    • one year ago
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    Ionization energy is defined as the energy input required to remove 1 mole of electrons from 1 mole of gaseous atoms, and is summarized by the general equation:\[X(g) \rightarrow X^{+1}(g) + e^{-1}\]The ease with which this process occurs will vary from element to element, but varies in a regular, periodic fashion. A second ionization may occur, so long as the atom has a second electron to remove (so any atom other than H), and the generalized equation looks similar:\[X^{+1}(g) \rightarrow X^{+2}(g) + e^{-1}\] Successive ionizations are always more difficult, since we are removing electrons from an increasingly unbalanced system. Consider an atom of aluminum, with 3 valence electrons. A neutral atom will have 13 protons pulling on 13 electrons, 3 of which occupy the 3rd principle energy level. Aluminum's electron configuration is\[[Ne]3s^23p^1\] Removal of a valence electron leaves aluminum with 12 electrons, still being attracted by 13 protons. Each electron now "feels" more than a full proton's equivalent attraction, and the electrons in the 3s-orbital are drawn in slightly, decreasing the radius of the Al^{+1} ion. In order to remove a second electron, a larger energy "cost" is needed, and the remaining ion's radius is decreased further still, as the nucleus becomes, in effect, stronger, as valence electrons are removed (see table)

  2. JFraser
    • one year ago
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    As the table shows, repeatedly removing electrons from the same atom gets more and more difficult, but the table also shows a HUGE increase in IE (shown by the yellow line) occurring at a different point for each atom in the second period. In the case of a sodium atom, removing a second electron "costs" nearly 10 times as much as removing the first, where the second IE of magnesium is only twice as much as the first. The huge jump in the ionizations of magnesium occurs when the third electron is removed. This jump is easily explained if we look at WHICH electrons are being removed. Sodium has just one valence electron, sitting in the 3s orbital. If we personalize the sodium atom, we can imagine it being "desperate" to lose that electron, empty that level, and gain a configuration just like neon. So the loss of a single electron from an atom of sodium isn't particularly costly, since the sodium atom benefits from this loss energetically. If a second electron were to be removed, it would come from a different principle energy level.\[Na: 1s^12s^22p^63s^1\]\[Na^{+1}: 1s^22s^22p^6\] Removing an electron from a full principle level would result in a loss of stability, so the cost to remove one of the 2p-electrons will be a several-fold increase in IE. A magnesium atom, with 2 valence electrons to lose, doesn't see a huge increase in IE until the THIRD ionization, once the 2 valence electrons have already been removed. This pattern allows us to use the successive ionizations of an unknown atom to place the unknown atom in a particular group, based on when the jump in IE occurs. We may not be able to identify the atom, but with information about its successive ionizations, we can, at least, identify its group number.

  3. JFraser
    • one year ago
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    This table shows the first ionizations for the main group of the periodic table (s- and p-block only), and we see two reliable trends. Firstly, moving from top to bottom within any group, ionization energy decreases. This trend is most easily explained by looking at the atomic radii of the atoms within the group. Atoms at the top of a group are filling a principle energy level that is fairly close to the nucleus, so the force of attraction of the oppositely charged particles (an inverse-square law) is fairly strong. As we move down the group, the principle level being filled becomes larger, so the orbitals being filled are farther and farther away from the nucleus. As a result, the inverse-square law shows that the force of attraction is weaker, and the valence electrons themselves are also shielded from the nucleus by more and more layers of core electrons. For example:\[Na: 1s^22s^22p^63s^1\]\[Cs: 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^1\] Where the valence level of a sodium atom is shielded by only 2 layers and 10 electrons, the valence level of a cesium atom is shielded by portions of five different layers and 54 core electrons. The 6s-electron in cesium is very weakly-held, and is very easy to remove. Down any group we see the same general decrease, because each atom within a group has an electron configuration similar to those above it, simply within a principle energy level that is farther from the nucleus

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  4. JFraser
    • one year ago
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    When we look across the period, we see a general increase in IE. As each principle level is in the process of being filled, atoms at the left of the period have a low resistance to the removal of a valence electron, where atoms at the right of each period have a decreasing willingness to have one of their valence electrons removed. Where lithium has an electron configuration of \(1s^22s^1\), fluorine's electron configuration is: \(1s^22s^22p^5\), one electron shy of full. The fluorine atom will be in no "mood" to have one of its valence electrons removed, so its IE is the second highest of any in its period, after neon for obvious reasons

  5. quintess
    • one year ago
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    You lost me

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