By: Jonah Landau, David Miron, and Julie Baldassano

  Quantity of energy that an isolated, gaseous atom must absorb to discharge a second electron  Energy required to remove an electron from an atom with a +1 charge  Mg + (g)→ Mg 2+ (g) + e - I 2 = 1451kJ/mol  I 1 < I 2 < I 3 <... < I n Definition

  Generally, the 2 nd ionization energy increases from left to right and from bottom to top, with the top right (He) being the highest  Alkali metals are separate, with the same trend of increasing from bottom to top  Description of Trend

  Atomic radius decreases (left to right and bottom to top), causing the amount of energy needed to break an electron out of orbit increase from left to right and bottom to top.  The smaller the atomic radius, the closer the outer electrons are to the nucleus and more attracted to the center.  Generally, Second ionization energy is greater than the first.  This is because after the first ionization, most atoms become more like a noble gas in their configuration. It is extremely difficult to remove an electron from a full valence shell. Reasons for Trend

  2 nd ionization energy is measured in kilo joules (kJ) (or electron volts (eV)).  There were two main methods that scientists used to calculate the ionization energy; the subtraction method, and Koopman’s theory. How 2 nd Ionization Energy is Measured

  The subtraction method: The first step in this method is to find the ion are looking to find. Next, you would have to subtract the energy value of the neutral atom, hence the “subtraction” method. This is the ionization energy for that ion.  Koopman’s Theory: This method involves the HOMO or highest occupied molecular orbital. It states that an atom of molecule’s ionization energy is equal to the energy of the orbital from which the electron is ejected, meaning HOMO is equal to the ionization energy.  Formal equation: I = -E (I is the orbital the e - is ejected from). How 2 nd Ionization Energy is Measured

  Alkali metals are dramatically larger than the next groups going over, which goes against the trend, saying that the elements on the left should be smaller.  It requires extra energy to lose a second electron from a filled s (lithium) or p (all the others) shell. Anomalies in Trend

  Be B should increase (left to right) but actually decreases.  Be = kJ/mole, and B = kJ/mole. Be has a full 2s subshell. It is stable and wants to stay that way. It is easier to remove the 2p electron from the B atom than an electron from the full 2s subshell of Be.  H Li should decrease (up to down) but actually it increases  H is –kJ/mol while Li is 7298 kj/mol. A normal switch would be something like Be to Mg with Be being 1757 kJ/mol and Mg being 1450 kJ/mol. Anomalies in Trend

  "Ionization Energy: 2nd: Periodicity." WebElements Periodic Table of the Elements. WebElements, n.d. Web. 30 Oct  J.A. Dean (ed) in Lange's Handbook of Chemistry, McGraw-Hill, New York, USA, 14th edition,  "Second, Third, Fourth, and Higher Ionization Energies." Ionization Energy and Electron Affinity. Purdue, n.d. Web. 30 Oct  "Periodic Table Trends." Periodic Table Trends. TAM, n.d. Web. 03 Nov  "Ionization Energy." Ionization Energy. N.p., n.d. Web. 03 Nov Works Cited