Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

lgbasallote

  • 3 years ago

For which of the following species are the intermolecular interactions entirely due to dispersion forces? A) C2H6 B) CH3OCH3 C) NO2 D) H2S E) CaNO3

  • This Question is Closed
  1. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    im thinking it would be the nonpolar molecule...so A?

  2. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yup its A ...... because its non polar ..... we have hydrogen bonding in B, C and D

  3. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    thus instantaneous-induced dipole forces are present between non polar molecule :D

  4. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    c d and e dont have h-bonding o.O

  5. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    sorry //// C , D , and E are POLAR

  6. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    i misread the question :( //// but the answer is A

  7. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    when flourine , chlorine, oxygen and nitrogen are present in a non symetrical molecule ... then it is polar

  8. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    and polar molecule have permanent dipole

  9. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    THUS A IS 110 % correct

  10. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    idk how to prove c d and e are polar though

  11. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    how do i prove they are polar?

  12. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you know something about dipole moment >????

  13. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    all i know is drawing lewis...and it seems tedious

  14. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you know something about dipole moment >????

  15. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yes i know dipole moment...somehow

  16. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    or relative electronegativity of elements

  17. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    when the directions of the electronegativity in the lewis cancel out then zero dipole moment

  18. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    FLORINE, CHLORINE AND OXYGEN .. are the MOST ELECTRONEGATIVE ELEMENT

  19. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    mmhmm

  20. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    NO2 is polar because it has a lone electron

  21. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    unbonded

  22. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah...i figured that one out already :D

  23. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    H2S is polar because of 2 lone pairs

  24. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    H2S i think looks something like |dw:1343732893300:dw|

  25. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yup its shape is bent

  26. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    calcium nitrate...idk how that looks

  27. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you dont need the structure ..... just look at the oxygen ... secondly ... it is non symmetircal ... thus it must be polar

  28. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm i see

  29. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1343733005262:dw|

  30. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    lol enough with H2S we've already confirmed it's polar :p

  31. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    C2H6 looks like this right.. |dw:1343733047139:dw| that's why it's nonpolar?

  32. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    lgbasallote ..its a basic question :D

  33. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    well im not chemistry inclined -_-

  34. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    i can call laplace transforms basic but you wont right? ;)

  35. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1343733096534:dw|

  36. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm bottomline they cancel hehe

  37. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yup :D

  38. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    so wait....if there are no lone pairs then i's nonpolar?

  39. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    it's*

  40. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    not necessarily ...

  41. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oh

  42. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    its must be symmetrical

  43. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1343733217757:dw|

  44. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1343733238479:dw|

  45. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    you see the difference :D

  46. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ohhhh

  47. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    anyway ..we are done with this question

  48. lgbasallote
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah i think so too

  49. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    CLOSSSSSSSEEEEEEEEEE IIIIITTTTT

  50. Hashir
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 2

    |dw:1343733296219:dw|

  51. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy