Nan in Mass Posted May 31, 2009 Share Posted May 31, 2009 In the section about molecular geometry, it says that CHsubscript4 is tetrahedral. I understand that. But it says that GeClsubscript2 is triangular planar. C and Ge both have four empty spots in their outer valence, right? The electron dot pattern for C would have four single dots around it, right? Wouldn't Ge's dot pattern be the same? Wouldn't they both have 4 unpaired electrons? So why does the book say that Ge binds to two Cl's and has a PAIRed set of electrons, making it triangular planar? Ok - next question: F has a greater electronegativity than O, right? Because of inner shell shielding? I understand (or at least I think I do) that effective nuclear charge is calculated by subtracting the number of inner shell electrons from the number of protons, in which case F's would be greater than O's, resulting in a higher effective nuclear charge and a smaller atom, and I understand that this results in F having a greater electronegativity (at least I think I have that right), but why? Why does having more electrons in an outer orbital make an atom have a greater pull on another electron (as long as the orbital isn't full)? Wouldn't it be the other way around moving across the period because the atoms in the shell would repel each other? It seems like it could vary with how many are paired, also, so if none are paired, there might be a greater pull than if 3 are paired and only one electron unpaired. I think I must be missing something. Or perhaps the model just isn't that good? And it looks like N is bigger than O. Why? What is the difference between ionization energy and electronegativity? I don't remember not understanding this the first time I read it. Sigh. -Nan Quote Link to comment Share on other sites More sharing options...
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