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Inorganic : polarity
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 Message 1 of 2 in Discussion 
From: dianna  (Original Message)Sent: 10/26/2004 4:13 AM
How do u tell if something is polar or nonpolar just by looking at the Lewis Dot strucutre?


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 Message 2 of 2 in Discussion 
From: MSN Nickname·Steve·Sent: 10/26/2004 5:06 AM
Hi Dianna, we can't say much about the polarity of a molecule from the dot structure unless we know what the shape, or geometry of the molecule is.  For example, we can draw a dot structure of water that is linear in shape:
 
  :
H-O-H   (the colons are supposed to be the two lone pairs on the oxygen!  )
  :

Now if water really had this shape, we wouldn't be here, nor would any liquid water likely exist on the earth, because with this shape water would be a nonpolar molecule.  The relatively high boiling point of water is due to its polar nature which results from its bent geometry.  Attractions between water molecules, responsible for its high boiling point, are called dipole-diple attractions, or since hydrogens are involved, "hydrogen bonding".  Without its "unbalanced" bent shape, the dipole-dipole interactions would be much weaker.

We can speculate for fun about the carbon dioxide molecule also.  The dot structure is

O=C=O   (with two lone pairs on each oxygen)

and the molecule really is staight or linear in shape.  Because of this, CO2 is nonpolar and has such a low boiling point that it is a gas at normal temperatures and pressures.  Attractive forces between nonpolar molecules are much weaker because there are little or no dipole-dipole attractions present.  But if CO2 was a bent molecule like water, oh boy, what a different world we would be in!  Again, much in the way of life as we know it would likely not be able to exist.  CO2 might exist as a liquid like water (can you imagine oceans of liquid CO2?).

So in summary, polarity in molecules is mainly determined by 1)  the shape of the molecule along with  2)  the differences in electronegativities of the atoms in the molecule (not sure if you've covered electronegativity yet!).
 
Hope this helps!
 

Steve