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Homework 5

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  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    To which second period element do these ionization values belong?

    • \(IE_{1}=899\;kJ/mol\)
    • \(IE_{2}=1,757\;kJ/mol\)
    • \(IE_{3}=14,849\;kJ/mol\)
    • \(IE_{4}=21,006\;kJ/mol\)

    In order to understand and solve this problem, you must first understand the definition of ionization energy. Ionization energy is the minimum amount of energy required in order to successively remove an electron from an element. You also need to know which elements belong to which period. A period is a horizontal row on a periodic table. The second period elements include the following: lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine, and neon. It is also helpful to know the ionization trends for the periodic table. From left to right, ionization energy increases, which makes sense because metals(left side) lose electrons easily while nonmetals(right side) do not. With that being said each ionization energy an element has means we have lost one electron.The question gives you 4 values meaning this particular element can lose 4 electrons. This eliminates lithium since it only has 3 all together. Also, when there is a large jump from one value to another, this signifies that you have transitioned from valence electrons to core electrons. This is because core electrons are harder to remove. Notice that between the second and third values there is a significant jump in energy. This means that after the removal of two valence electrons, we have entered core electron territory. The only element that has 2 valence electrons in the second period is Beryllium.

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