AP CHEM HELP: Ionization Energies

<p>Hi, I've been going through old CB free response questions and I got stuck on this and some other ones. If you could help, that's great.</p>

<p>2007 AP CHEMISTRY FREE-RESPONSE QUESTION - FORM B; # 6</p>

<p>Element, First Ionization Energy, Second Ionization Energy, Third Ionization Energy
Element 1: 1,251 2,300 3,820
Element 2: 496, 4560, 6910
Element 3: 738, 1450, 7730
Element 4: 1,000 2,250, 3,360</p>

<p>The table above shows the first three ionization energies for atoms of four elements from the third period of the periodic table. The elements are numbered randomly. Use the information in the table to answer the following: </p>

<p>a) Which element is most metallic in character? Explain</p>

<p>b) Identify element 3. Explain.</p>

<p>c) Write the complete electron configuration for an atom of element 3.</p>

<p>d) What is the expected oxidation state for the most common ion of element 2? </p>

<p>e) What is the chemical symbol for element 2? </p>

<p>f) A neutral atom of which of the four elements has the smallest radius?</p>

<p>Thanks in advance!</p>

<p>a) Metallic character increases as you go further left across a period (which makes sense intuitively if you think of where metals and nonmetals are on the periodic table). Since first ionization energy decreases as you go left across a period (since there are fewer protons to attract the electrons), the most metallic element will be the one with the lowest first ionization energy. In this case that is element 2.</p>

<p>b) Element 3 has a very large gap between its second and third ionization energies. When an ionization energy is significantly higher than the previous one, you can deduce that it's because it represents the energy required to strip an electron from a species that has attained the electron configuration of a noble gas. Since the third ionization energy in element 3 is very large, we know that the element must be magnesium.</p>

<p>c) 1s^2 2s^2 2p^6 3s^2</p>

<p>d) You can use the same logic used in part (b) to realize that element 2 is sodium. The most common ion of sodium is Na+. The expected oxidation state is +1.</p>

<p>e) We already figured that the element is sodium. The chemical symbol is Na.</p>

<p>f) Atomic radius decreases as you move to the right across a period because there are more protons to draw the electrons in. First ionization energy increases as you move to the right across a period, so the one with the smallest radius will also have the largest first ionization energy. That element is element 1.</p>

<p>Lemme know if any of that reasoning is shaky or if I made a mistake. Now my question to you: Why on earth are you doing this review stuff AFTER the AP???</p>