Bond Polarity

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pilotdentist

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I understand that polar bonds can be included in an overall nonpolar molecule.
CF4 is tetrahedral and a nonpolar molecule with polar bonds.
What about SO3? It should be Sulfer double bonded to Oxygen in a trigonal planar arrangement.

Am I missing something? To me both A/D are correct.

13. Which of the following is an example of a compound that has polar bonds but a non-polar shape?
  • 1. A. SO3
  • 2. B. H2O
  • 3. C. N2
  • 4. D. CF4
  • 5. E. NH3
Correct
There are two different types of polarity that have to be understood to answer this question. They are bond polarity and molecular polarity. For bond polarity, if there is a large difference in the electronegativities of the elements that make up a bond, then the elements are most likely sharing the electrons in the bond unequally, resulting in a polar bond. However, molecular polarity deals with the shape of the molecule, looking for symmetry. Symmetry leads to a non-polar molecule. The only molecules in the answer options that have a symmetrical shape are CF4 and N2. However, N2 would have non-polar bonds as the N-N bond would have the same electronegativity and would be sharing the electrons equally. Therefore, CF4 must be the answer as the electronegativities vary with the C and the F making the C-F bonds polar.
Topic: Liquids and Solids

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I believe it's due to the fact that SO3 has resonance structures, where one of the pi bonds move onto the O, making the O negative and S positive --> making the molecule polar.
SO3_meso2.png
 
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Maybe, but the "net" (-) charge would be distributed evenly around the outside of the molecule since the drawn structures don't really "exist."

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Unrelated follow up question about molar solubility. I get confused as to when to use (3x)^3 vs just (x)^3 when solving these types of problems.

1-What is the maximum concentration of fluoride ions that could be present in 0.032M Ba(NO3)2 (Ksp,BaF2 = 3.2x10-8?
3.2X10^-8 = .32 * X^2

2-What is the molar solubility of a salt of the formula AB3 (Ksp=3.0x10-19)?
3.0X10^-19 = X * (3X)^3

Why does problem 1 use X^2 and not 2X^2 similar to question 2? Does it have to do with the question asking for IONS present vs molar solubility? Does anyone have a good set of rules for evaluating these problems?

Thank you so much!
 
I understand that polar bonds can be included in an overall nonpolar molecule.
CF4 is tetrahedral and a nonpolar molecule with polar bonds.
What about SO3? It should be Sulfer double bonded to Oxygen in a trigonal planar arrangement.

Am I missing something? To me both A/D are correct.

13. Which of the following is an example of a compound that has polar bonds but a non-polar shape?
  • 1. A. SO3
  • 2. B. H2O
  • 3. C. N2
  • 4. D. CF4
  • 5. E. NH3
Correct
There are two different types of polarity that have to be understood to answer this question. They are bond polarity and molecular polarity. For bond polarity, if there is a large difference in the electronegativities of the elements that make up a bond, then the elements are most likely sharing the electrons in the bond unequally, resulting in a polar bond. However, molecular polarity deals with the shape of the molecule, looking for symmetry. Symmetry leads to a non-polar molecule. The only molecules in the answer options that have a symmetrical shape are CF4 and N2. However, N2 would have non-polar bonds as the N-N bond would have the same electronegativity and would be sharing the electrons equally. Therefore, CF4 must be the answer as the electronegativities vary with the C and the F making the C-F bonds polar.

I think that NH3 has a polar shape! Here molecular geometry and electron geometry isn't the same. There is a lone pair on the N!!
Topic: Liquids and Solids
 
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I think that NH3 has a polar shape! The electron geometry and molecular geometry isn't the same! The shape of NH3 is Trigonal Pyramidal. There is a partial negative on the Nitrogen!!!!!!!
 
SO3 is also a non-polar molecule with polar bonds (either way you draw it, with all double bonds or the form with resonance structures above). You are correct, both answers are valid.
 
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