<p><a href="http://i45.tinypic.com/34e3ty8.jpg%5B/url%5D">http://i45.tinypic.com/34e3ty8.jpg</a></p>
<p>Q: Give the electron-domain geometry on which the molecular geometry is based..</p>
<p>Answer: first one is octahedral, second tetrahedral, and third is trigonal bipyramidal..</p>
<p>I am a little confused on this answer.
For the first one, I know that octahedral has 6 bonding and 0 nonbonding but i dont see so in the picture? I don't even get how they got the third answer either..</p>
<p>To me they all look tetrahedral, what am I missing?</p>
<p>they stated the electron domain geometries for the answer. the only way to get those “molecular geometries” would be with those e- domain geometries. the first is square planar obviously, but you need 6 e- domains (sp3d2) to form it, which makes it octahedral. the second, is the same for both e- domain and molecular geometries. the last is see-saw (molecular geom that is) but its e- domain geom is trig bipyramidal.</p>
<p>i think you misunderstood it.<br>
the answer is not included in the question, the question is just the picture and the question.
I’m posting the answer so someone can help me figure out why those are correct heh</p>
<p>I don’t think they want you to focus on the number of bonds, but on the geometry of each. You can obviously tell that the second one is a tetrahedral because of its shape.</p>
