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Topic: Who doesn't enjoy VSEPR?  (Read 3571 times)

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Offline nickthegolfer

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Who doesn't enjoy VSEPR?
« on: January 20, 2007, 10:52:49 PM »
Well, you guys seem to have such an efficient and articulate response time that I'll ask another question.  My grade in my AP Chem class is teetering between B- and C, so I'd like to use all the resources available to make sure my understanding is solid. Okay, this is about the Valence Shell Electron Pair Repulsion Theory, of which I have a limited amount of knowledge.  Here it is:

Use the principles of bonding and molecular structure to explain the following statements.

a.) The angle between the N-F bonds in NF3 is smaller than the angle between the B-F bonds in BF3 .

b.)  I2 (s) is insoluble in water, but it is soluble in carbon tetrachloride.

c.) Diamond is one of the hardest substances on Earth.

d.)  HCl has a lower boiling point than either HF or HBr.

So for (A) shouldn't the answer be that although the Lewis Structures of NF3 and BF3 look alike, in practice the shape of the central atoms are different, so therefore the angles between the bonds are different.

And for (B), well, I assume I2 is immune to water's strange dipole, but the shape of carbon tetrachloride allows it to penetrate the bonds holding I2 together... there's probably a much more articulate way to say that.

(C) seems simple, I mean, diamonds are pure carbon atoms that are connected in tight clusters, their geometric shape makes them extremely dense, aka hard

(D) someone bail me out on this one...

So there we have it.  If someone could tell me where I'm going wrong and what my answers should resemble, I'd owe them immensely

Offline english

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Re: Who doesn't enjoy VSEPR?
« Reply #1 on: January 20, 2007, 11:09:36 PM »
a)  The bond angles are dissimilar because boron in BF3 hybridizes to form 3 atomic orbitals, each one bonded to fluorine, so you have a trigonal planar geometry.  Boron is a..."special" case.  If you promote a 2s electron from its electron configuration to an empty 2p oribtal, you will see it has 3 unpaired electrons, and one empty p orbital.  Boron is what we call an electron deficient atom.  It doesn't normally attain an octet after bonding.  NF3, on the other hand, has a tetrahedral geometry.  Now what do you know of the bond angles of tetrahedral angles and trigonal planar angles?

b)  This is just trying to get you to think in terms of dipole moments.  Is I---I polar or nonpolar?  I'll give you a hint:  iodine's electronegativity is 2.5.  Compare that to CCl4.  Carbon's electronegativity is 2.5, but chlorine's is 3.0. 

c)  Diamond forms something called a network solid.  You are right.  Don't doubt yourself.  You know more than you give yourself credit for. 

d)  You know that HF can form hydrogen bonds, what we call a dipole-dipole force.  HBr cannot, nor can HCl.  So you can predict that HF should have a higher boiling point that HBr or HCl.  HBr has a higher boiling point than HCl.  What trend did we just talk about do you think would result in this?  Hint:  boiling requires intermolecular forces (dipole-dipole forces for example) to be overcome.


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