Two more chem questions... sigh... I am so stupid about this...

[Nan in Mass]

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

[Jane in NC]

This is Jane's son conscripted to answer questions. With the CH4 the number of atoms affect shape (you need 4 atoms for a tetrahedron while a triagular planar has 3). The Cl may also be double bonded to the Ge which would fill Ge's shell (not sure but from my memory this should work). I can't help with the size issue as far as I remember my textbook gave a table with an explanation and I was working through other elements in the chapter to think about why (or my memory is failing).

[ChemMommy]

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?

 

The mistake in thinking here was starting with an assumption about how carbon and germanium will be bonded before assembling the molecule. Here's the order to assemble the molecule:

 

1. Count valence electrons (4 for Ge, 7 for each Cl, for a total of 18)

2. Choose the central atom (this will be the atom farthest to the left in the periodic table, except for H), in this case Ge.

3. Attach two Cl's to the Ge with single bonds. At this point, you've used up 4 of the 18 valence electrons. The 14 remaining electrons will go on in pairs, starting with the outermost atoms.

4. Each Cl gets 3 sets of electrons. That uses up 12 of the 14 remaining electrons.

5. The last two go on as a pair of electrons on Ge.

 

(To make you feel better, this is an unusual molecule. Usually, things in this column make four bonds-sigma or pi.)

 

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

 

Here's a short answer. The large effective nuclear charge means that the valence shell electrons are pulled hard down toward the nucleus. This results in a smaller atom. Atom size, in general, decreases as you go across a period because the Zeff increases.

 

Now, if the NUCLEAR (which is positive) charge is increasing, it makes sense that the atom is more interested in attracting electrons (which are negative).

 

Finally, there is a tremendous stabilizing influence when the p-orbital is "completed". The entire subshell moves down toward the nucleus and becomes more stable. Adding just one electron to F does increase the amount of repulsion between valance electrons, BUT all six electrons in the p-orbital are stabilized as a result of the filled shell.

 

 

Let me know if this is still confusing.....

[Miss Marple]

Here's a link that might help you visualize the GeCl2 molecule. Is it the word "Paired" set that is confusing? It's basically a leftover pair of electrons. I believe unbound electrons take up a bit more space than those that are bonded which causes the molecule to bend some.

 

Electronegativity increases as one moves from left to right across the periodic table and as one goes down the periodic table. Atomic size (radius) is the inverse of this. As one moves from left to right the radius decreases due to increasing protons pulling harder on the electrons - bringing them closer to the nucleus. So, yes, nitrogen would be slightly larger than oxygen because it has fewer protons in the nucleus. Electronegativity is a measure of the ability of an atom to attract electrons towards itself. So since Fluorine has more protons than oxygen it will attract its electrons more forcefully which will also make it a smaller atom.