# VSEPR/Hybirdization/diagrams HELP

1. Nov 22, 2004

### Roxy

VSEPR/Hybirdization/diagrams HELP!!

Sorry this is so long but I don't get any of it.

For these 3 molecules:
C2H4
NH2OH
H2SO4

1. Using VSEPR, how to u predict the shape?
2. Indicate hybridization of the central atom???
3. Sketch 3-D diagram an indicate bond angles

2. Nov 22, 2004

### Roxy

if you have time can u help me with these (I tried them but I'm wrong):
2-
SnF
6

and

CF3Cl

3. Nov 23, 2004

### chem_tr

VSEPR is not a hard issue. It is just how you couple the electrons. Select the most appropriate electropositive one and place it in the center, and surround it with electronegative ones.

Secondly, for atoms other than hydrogen, use octet rule (i.e., try to sum the electron count to 8), for hydrogen, use 2. This will give the maximum number of electrons.

Thirdly, calculate all neutral-state electronic configurations to learn how many electrons are present. The difference will be the electrons used for covalent bonding.

With the same method, you can find how many n electrons are present (non-bonding).

Remember that if the central atom has a non-bonding electron on it, the structure will surely be deformed (deviated) from ideality.

4. Nov 28, 2004

### Warwick

1. Calculate the dense areas around the central atom in the bond. This includes lone pairs. Count each grp as 1. For example, h2c2 has 4 total bonds around each c atom. But theres a triple bond, so there is NOT 4 dense areas, there is 2. One on each side. Thus, it would be linear, 180 degrees, there should be a table of all this in your book.

Lets do another example. Lets say you had 4 dense areas around an Xe atom. 2 are lone pairs and 2 are bonds. 4 dense areas = tetrahedral, but with 2 lone pairs it is tetrahedral/bent. Tetrahedral shapes have 109.5 degree angles and if there is lone pairs the angle will be reduced because lone pairs need room. So the angle would be slightly less. For gen chem classes just remember your standard angles for each shape and if it has lone pairs say slightly less than ....

2. Hybridization is confusing to understand, this is a trick I use do get the answer. Count the dense areas. Lets say you get 4, the hybridization would then = sp3
count s as 1 and p as 3, which = 4, the number of dense areas
if you have 2 dense areas it would be sp. count s as 1 and p as 1.
if you have 5 dense areas sp3d
Notice anything? p can only have 3 orbitals, and s only 1. Thus, when you have 5 dense areas it goes to the next subshell d.

If you have not read your text and try to understand what I have said, you will not understand. I'm assuming you know what subshells are, what molecular shapes are, etc.