Woot
Shapes are determined by the Lewis Dot Structures. Do you know how to do those? The basic concept is that the electrons (both free and in bonds) always try to be as far away from each other as possible. The number of electrons swarming around determines which way the bonds/free e- have to be oriented in order to maximize distance.
Linear is when you have only 2 sets of bonds (double bonds, triple bonds, and lone e- pairs act as a single 'unit' when determining structure), such as CO2. O == C == O is as far as the two sets of bonds can be from each other. Tetrahedral is basically the same thing, except with 4 sets of bonds and hence a kinda funky shape.
The strangeness happens when you start factoring lone e- pairs in. They have the same effect on the positioning of other e- and atoms as a normal bond would, except you don't count it when you find geometric structure. For example, H2O has 2 free electron pairs and 2 bonds, hence 4 'units' of repulsion. The structure of this is
actually closer to tetrahedral, because the e- pairs occupy the same places (give a few degrees; free pairs repulse just a little more than a bond) as an atom would. A 4-bond molecule would have the tetrahedral shape in order to maximize distance, no? If you're looking for electronic structure, you're done.
But wait! What if you're finding the geometric structure? (more common question) Two of those 'atoms' are actually electron pairs! What to do.
The answer is: ignore them. Recognize the repulsive effect they have on the other atoms, but only think about actual bonds when finding the geometric structure. Hence, you have a bent molecule (take a tetrahedral molecule and imagine what it would look like without two of the outer atoms. It's bent! Kinda cool, ja?)
www.chem.ufl.edu/~myers/chm2045/shapes.htm has a pretty good tutorial on it.
Polar covalent bonds does not equal a polar molecule. The reason is this:
A molecule is only nonpolar if its dipole moment is equal to 0. The dipole moment is the sum of all the dipole vectors in the molecule.
Take a look at that, tell me if it makes sense. No? Good
Think about a molecule such as CO2 (This one is so easy to type <3)
O == C == O
The oxygen is more electronegative than the carbon, and hence exerts a pull on the electrons for the molecule. So the electrons are wiggling over this way:
O <== C ==> O
Right? Each arrow is called a 'dipole vector,' and represents how hard and in what direction the electrons are getting pulled (determined by electronegativity difference between the two atoms). I'm going to assume you know how to add vectors, but if you don't it's really just common sense.
When you add up all the dipole vectors, you get what's called the 'dipole moment.' If it's equal to 0, the whole molecule is nonpolar (because there is no 'net polar' happening; all the attraction one way or another is negated).
Realize here that the two oxygens (being equally electronegative) are pulling with equal intensity and in opposite directions.
Hence
<== + ==> = 0.
But what if you have the hypothetical and impossible molecule COF?
O == C == F (I know this is completely wrong, but just bear with me).
Then, since fluorine is more electronegative than oxygen:
O <== C ======> F
Get it?
And hence the dipole moment would be:
<== + ======>
=
====>
It's not zero, and hence you have a polar molecule D8
A strange case happens when it's bent, or something other weird shape:
. .
O
/ \
H H
Mmkay?
Imagine those .'s are actually ..'s. I ran out of room.
The two H's are pulling equally strongly
but not in opposite directions. Hence some verreh bad things are happening, because if you think about it, the dipole moment goes DOWN and hence you have a polar molecule D8
All good? ^_________^
Interesting tidbit: The direction of the dipole moment vector determines which side of the molecule is positive, and which side is negative. Since it points to where the electrons are pulled to more, the vector is pointing at the negative side XD
I love chem~