VSEPR Thoery and Polarity

[saku]

Hi! I just a few questions that i'm not sure of, if theyre correct or not. If anything is incorrect could it be corrected.

 

VSEPR

 

 

Shape: trigonal planar

 

 

N/A - i don't need to find out the shape

 

 

trigonal pyramid

 

 

N/A - i don't need to find out the shape

 

 

octahedral (square pyramid)

 

Are those correct ?? esp the lewis structure diagrams

 

Polarity *note* the limit given by my teacher is non-polar 0-0.5; Polar 0.5-1.7

 

CH20 - Polar; EN (C-O=1.0, C-H=1.4) Would the double effect the polarity like when you do the vector thing.

 

CO2 - Polar; EN=1.0. The ends aren't different, all negative, so is it still polar?

 

ClO2 - Non-polar; EN=0.5

 

H2S2 - Non-polar; EN=0.4. Ends have different charges.

 

Thanks

[UncleAl]

You had best go back and practice counting electrons. Sulfite is pyramidal (Advanced Inorganic Chemistry, 6th Edtion, Cotton, Wilkinson, et al., p. 526) with its electron lone pair on sulfur. A square pyramid is not an octahedron. Saying linear symmetric O=C=O is polar is a terrible utterance for any of a large number of reasons. ClO_2 has a dipole moment of 1.792 debye. That's kinda fat for a "nonpolar" molecule (J. Mol. Spect. 98 425 (1983)). ClO_2 has an odd number of electrons. What does that mean in the great scheme of things? Inorg. Chem. 31 4740 (1992).

 

http://www-theory.chem.washington.edu/prezhdo/oleg/Pubs_pdf/OClO/oclo_solvents.pdf

especially footnotes 23,29

 

 

The ClO_2 dipole moment isn't news, either: Zeitschrift Fur Physikalische Chemie-Abteilung B-Chemie Der Elementarprozesse Aufbau Der Materie 28 (1): 17-30 (1935)

[TeleMad]

I'll use CN- as an example of how we were taught to draw Lewis structures (this is not the precise language we used, by the way).

 

1) Determine how many valance electrons there are.

 

 

2) Determine which atoms will be terminal and which will be central.

 

a. hydrogens are terminal atoms

 

b. central atom(s) should be the element(s) with the lowest electronegativity.

 

c. in oxoacids, acidic hydrogens are usually bonded to an oxygen

 

d. molecules and polyatomic ions usually have a compact and symmetrical shape (organic molecules are often exceptions to this rule of thumb)

 

 

3) Draw a molecular skeleton using only single bonds. Each single bond uses up 2 of the electrons from step (1).

 

 

4) Using the remaining number of valence electrons from step (1), complete terminal atoms' valence shells. It is okay if you have some valence electrons left over, but you can't use more than the number determined in step (1).

 

 

5) If there are any valence electrons remaining, assign them to the central atom(s)

 

 

6) If there were not enough valence electrons to complete all atoms' valence shells, then convert lone pairs on a terminal atom into a higher order bond with a cental atom. Continue until the number of valence electrons is sufficient on all atoms.

 

 

7) If more than one possible Lewis structures arise, choose the one with the best formal charges. Use the following rules to make that determination.

 

a. the magnitude of formal charges should be as small as possible

 

b. if negative formal charges must be used, they should be on the atoms with the greatest electronegativies

 

c. no two adjacent atoms should have formal charges of the same sign.

 

 

 

So here's how CN- would be done.

 

1) Carbon has 4 valence electrons and nitrogen has 5; the cyanide molecule carries a -1 electric charge so there is an extra electron, for a total of 10 valence electrons.

2) Here we don't have a choice as to which should be the central atom, since there are only 2 atoms

3) C-N

     ..
4) : C - N :
     .. 

5) there are no valence electrons left over

6) Nitrogen does not have a full valence shell so we have to convert some of the lone pairs around C into higher order bonds with N.

Converting one lone pair on C into a higher order(double) bond gives...
  ..
: C = N :

Nitrogen is stil short, so... converting another lone pair on C into a higher order (triple) bond gives...

: C /// N :   (/// represents a triple bond here: sorry for the crudeness)

7) Only one Lewis structure resulted, so there's no need to compare formal charges for different possibilities.


At this point, all has been taken care of.  The final Lewis structure has C triple bonded to N, with both C and N having one lone pair each.

 

As far as your other set of questions...

Polarity *note* the limit given by my teacher is non-polar 0-0.5; Polar 0.5-1.7

 

...

 

CO2 - Polar; EN=1.0. The ends aren't different, all negative, so is it still polar?

 

It's not enough to note that a molecule has polar bonds; polar bonds and polar molecules are not the same thing. You have to consider both the bonds and the molecular geometry in order to figure out if a molecules is polar.

 

In CO2, which has a simplified Lewis structure of O=C=O, the carbon in the middle is not as electronegative as the oxygens surrounding it, so yes, the two individual double bonds are polar. But the molecule is linear (all valence electrons on the central C are bonding electrons, so the electron group geometry is the same as the molecular geometry). So the magnitude that the left O pulls the electon-charge distribution towards itself is exactly counterbalanced by the magnitude that the right O pulls the electron-charge distribution towards itself, ending up with the two vectors cancelling each other out.

[TeleMad]

Sorry, temporary post. As soon as I can post in the Physics and Mathematics section again (I've tried half a dozen times tonight and I keep getting a "Web Site Not Responding" error) I will 'move' this there.

 

It dawned on me that I could leverage my 50-million row lookup table of prime/non-prime numbers. That is, using my already-precalculated 50-million known prime/nonprime values, I can rather quickly calculate whether or not any number up to a theoretical 50-million squared ... that's up to 2,500,000,000,000,000 which is 2.5 quadrillion ... is prime.

 

All it required was a simple change to an existing function I had. Here's the update algorithm.

 

FUNCTION IsPrime(lnNumber, lcSieveAlias)
LOCAL llIsPrime, lnSelect, lnLcv, lnMax
lnSelect = SELECT()
SELECT (lcSieveAlias)
IF (lnNumber <= RECCOUNT())
	GO (lnNumber)
	llIsPrime = prime
ELSE
	llIsPrime = .T.	
	lnMax = CEILING(SQRT(lnNumber))
	SCAN FOR prime
		IF (RECNO() > lnMax)
			EXIT
		ENDIF
	
		IF (lnNumber % RECNO() = 0)
			llIsPrime = .F.
			EXIT
		ENDIF
	ENDSCAN
ENDIF
SELECT (lnSelect)
RETURN llIsPrime
ENDFUNC

 

The only cleaning up that would need to be done is that IF someone specifies a number that it soo large - a number greater than 2.5 quadrillion - the function doesn't throw an exception. A simple fix, but one I will ignore for now.

 

I'll time test this function tonight using 10,000 pseudo-randomly generated numbers between 1 and a quadrillion, posting the MIN, MAX, and AVG time intervals when it completes.