Quick G-chem Question

[DAT-Scared]

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Hey Guys,
just a quick question about G-chem.

How would you write a Lewis structure for a compound that has an odd number of valence electron for example ClO2...I think I ve forgotten something about this,
It would be great if you guys shed some light on this....
Thanks a lot, 😳

 

[Djapprentice]

DAT-Scared,

One thing I've noticed about the Lewis Structures on the practice tests is that all you gotta do is count the total number of valence electrons in the entire molecule, & you can pretty much get the answer based on that alone. Calculating the formal charges aren't really necessary, unless you get a tie between like 2 or 3 molecules. In this case, ClO2 has a total of 19 electrons, Cl - 7 & O = 2 x 6 = 12 Place a bond between the Chlorine and Oxygen, so its now 19-2 = 7 electrons remaining, so you want to make sure the chlorine, being the center atom since it is the least EN & there's more Oxygens, satisfies the octet rule. It doesn't because it needs two more electrons, so you can take 2 lone pairs from each oxygen, & voila, there's your L. Structure for Cl02.

 

[Djapprentice]

in case you didn't get my lingo, EN = electronegative

 

[DAT-Scared]

in case you didn't get my lingo, EN = electronegative

Thanks Djapprentice, but I thougth you have two oxygen to be added. maybe i didn't understand your explanation perfectly!!!

 

[Djapprentice]

Thanks Djapprentice, but I thougth you have two oxygen to be added. maybe i didn't understand your explanation perfectly!!!

No problem Dat-Scared, but here's thing thing the 2 x 6, part, look at it for a second, do you see, that 2, that 2 takes into account, the 2 Oxygens, then you place the oxygens between the chlorine, so you make your little chlorine sandwich. I hope that helps, let me know if you're still having trouble. I'm still working on ways to do these problems quicker, but counting the total valence electrons is the best way in my opinion.

 

[DAT-Scared]

No problem Dat-Scared, but here's thing thing the 2 x 6, part, look at it for a second, do you see, that 2, that 2 takes into account, the 2 Oxygens, then you place the oxygens between the chlorine, so you make your little chlorine sandwich. I hope that helps, let me know if you're still having trouble. I'm still working on ways to do these problems quicker, but counting the total valence electrons is the best way in my opinion.

Thanks Djapprentice, i think it made sense.

 

[dontbam]

For CLO2, you look at Cl first and recognize there are 7 valence electrons. Then you look to see how many Cl-O bonds there are, since there are 2 for two oxygens you normally add 2 e' to get the total to 9, however oxygen bonds are an exception to this rule, i forget exactly why but it may be due to effective nuclear charge between the bonds.

So now you have 7 valence electrons, you add one electron to get the number to even to 8 if your molecular is ClO2- which i'm expecting you mean. Divide by 2 to get 4 pairs/bonds. 4 pairs is usually a tetrahedral symmetry but there are exceptions and O=Cl-O /w 4 lone electrons (and negative charge) is what you're looking for

 

[DAT-Scared]

For CLO2, you look at Cl first and recognize there are 7 valence electrons. Then you look to see how many Cl-O bonds there are, since there are 2 for two oxygens you normally add 2 e' to get the total to 9, however oxygen bonds are an exception to this rule, i forget exactly why but it may be due to effective nuclear charge between the bonds.

So now you have 7 valence electrons, you add one electron to get the number to even to 8 if your molecular is ClO2- which i'm expecting you mean. Divide by 2 to get 4 pairs/bonds. 4 pairs is usually a tetrahedral symmetry but there are exceptions and O=Cl-O /w 4 lone electrons (and negative charge) is what you're looking for

Thanks dontbam, but the molecule is ClO2 without any charge if you go to kaplan book page 810 practice problem 12 has this compound. that's why I was comfused!!!!!

 

[tom_servo_dds]

Thanks dontbam, but the molecule is ClO2 without any charge if you go to kaplan book page 810 practice problem 12 has this compound. that's why I was comfused!!!!!

On page 798 under "Exceptions to the octet rule" it says that atoms in the third period and beyond can have more than eight valence electrons. Cl in ClO2 has more than eight. Of its seven valence electrons, two are involved in sigma bonds to the oxygen and two are involved in pi bonds as well to form double bonds to the oxygens. Three valence electrons remain unbonded and give the molecule a bent configuration (incidently, unbound electron gives it a paramagnetic nature as well). With the highly electronegative oxygens attached, the molecule becomes slightly polar which is what the question was referring to. Because it does not follow the octet rule it sometimes throws people off. Remember that the octet rule isn't really a hard and fast "rule" as much as it is a guideline for molecules. Just be careful of those in the third period or beyond. For more practice on this, try the problem on pg. 808 6c.

If you would like more information about ClO2 and to see a 3d electric potential map of the molecule, go here:

http://en.wikipedia.org/wiki/Chlorine_dioxide

Best of luck on your preparations! 🙂

 

[DAT-Scared]

On page 798 under "Exceptions to the octet rule" it says that atoms in the third period and beyond can have more than eight valence electrons. Cl in ClO2 has more than eight. Of its seven valence electrons, two are involved in sigma bonds to the oxygen and two are involved in pi bonds as well to form double bonds to the oxygens. Three valence electrons remain unbonded and give the molecule a bent configuration (incidently, unbound electron gives it a paramagnetic nature as well). With the highly electronegative oxygens attached, the molecule becomes slightly polar which is what the question was referring to. Because it does not follow the octet rule it sometimes throws people off. Remember that the octet rule isn't really a hard and fast "rule" as much as it is a guideline for molecules. Just be careful of those in the third period or beyond. For more practice on this, try the problem on pg. 808 6c.

If you would like more information about ClO2 and to see a 3d electric potential map of the molecule, go here:

http://en.wikipedia.org/wiki/Chlorine_dioxide

Best of luck on your preparations! 🙂

Thanks tom_servo_dds, that was a great explanation.