Problem Set 5_Sol

Problem Set 5_Sol - Problem One Ionization Energies 15...

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5 Points 5 Points 5 Points Problem One – Ionization Energies – 15 Points Important Point One: The large differences in ionization energy correspond to transitions from closed shell noble gas configurations to alkali metal configurations. The eight steps indicate the 8 electron sp shell (s 2 p 6 ) because element A has a similar ionization energy to element I. Important Point Two: There are two pairs of noble gases that are 8 electrons apart, Helium-Neon and Neon- Argon. The rest have d electrons in between. This suggests that element A is either Lithium or Sodium, so element L is either Silicon or Titanium. Important Point Three: To distinguish between these two pairs we need to look at elements J and K. If L is Silicon, then J is Magnesium and K is Aluminum. If L is Titanium, then J is Calcium and K is Scandium. The transition from Magnesium to Aluminum has a decrease in ionization energy (as opposed to the general trend) because of the relatively stable [Ne]3s 2 configuration. The transition from Calcium to Scandium does not have this anomaly. The element must therefore be Titanium. Answer: Element L is Titanium
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Problem Two – Atomic Radii Periodic Trend – 15 Points, 5 Points Each a. Cs + and Sr 2+ have the following electronic configurations: [Xe]6s 0 , [Kr]5s 0 . Therefore this is a simple comparison of shell sizes (n=6 > n=5) and effective nuclear charges. While Sr 2+ has a higher effective nuclear charge, the addition of an additional shell of electrons is a much larger effect Answer: Sr 2+ is smaller b. Sr 2+ has an electronic configuration [Kr]5s 0 while Nb 2+ has an electron configuration
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This note was uploaded on 03/30/2010 for the course CH 1a taught by Professor Lewis during the Fall '08 term at Caltech.

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Problem Set 5_Sol - Problem One Ionization Energies 15...

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