Answers to Practice midterm 2
  1. D
  2. A
  3. B
  4. B
  5. C
  6. B
  7. A
  8. C
  9. D
  10. A
  11. A
  12. C
  13. B
  14. D
  15. C
  16. A
  17. A
  18. C
  19. D
  20. D
  21. A
  22. C
  23. D
  24. B
  25. C
  26. A
  27. D
  28. C
  29. C
  30. A  : triple bond
  31. A
  32. C
  33. D
  34. E
  35. B

 

 

I.  A)  1&2. polar covalent bonds   3.  O-N=O arrangement with 3 pairs of nonbonding electrons around the left oxygen, one pairs of nonbonding electrons on nitrogen, and two nonbonding pairs of electron on the right oxygen.   4.  EG: trigonal planar  5. MG: bent   6.  dipoles exist pointing towards each oxygen and one points to the nonbonding pair of electrons on Nitrogen:  this is a polar molecule.   7.  the bond angle is ~ 120o  8.  resonance: O=N-O

 

B)  1&2. polar covalent bonds   3.  Br is in the middle with 4 pairs of dots around it in the four quadrants and each oxygen is in a quadrant with 4 pairs of electrons (7+4(6)+1=32 electrons total).   4.  EG: tetrahedral  5. MG: tetrahedral   6.  dipoles exist pointing towards each oxygen, it is symmetric:  this is a nonpolar polyatomic ion.   7.  the bond angle is ~ 109.5o  8.  NA

 

C)  1&2. polar covalent bonds   3. Si is in the middle with 4 pairs of dots around it in the four quadrants and each hydrogen is in a quadrant with 2 pairs of electrons (8 electrons total).   4.  EG: tetrahedral  5. MG: tetrahedral   6.  dipoles exist pointing towards each hydrogen (electronegativity value of H is 2.1 and Si is 1.8; a hydride):  this is a nonpolar molecule.   7.  the bond angle is ~ 109.5o  8.  NA

 

I.  D)  1&2. polar covalent bonds   3.  C=C central arrangement with an F (3 lone pairs of electrons) and Cl (3 lone pairs of electrons) attached to a C.   4.  EG: trigonal planar  5. MG: trigonal planar   6.  dipoles exist pointing towards each fluorine atom and each chlorine atom:  this is a polar molecule.   7.  the bond angle is ~ 120o  8.  isomers exist:  you could arrange the  fluorine atoms all on one carbon with the chlorine atoms all on the other carbon  or  geometric isomers (cis & trans).

F         Cl                                     F           F

   C=C                                             C=C

Cl        F                                     Cl         Cl

 Trans                                             cis

 

 

 

II.  A)  approximately 1.60 angstroms.   B)  the weakest X-H bond happens with N-H because it is the longest bond with the greatest difference in electronegativity.

 

III.  A.

(#1) :N=N=O: with one other lone pair on left N and on right O

 

(#2) :N≡N-O: with two more lone pairs on O (note the triple N to N bond)

 

(#3) :N=O=N: with another lone pair on each N

 

B.  only #1 is both linear and polar, #2 is bent & polar while #3 is linear & nonpolar.

 

C.  Using formal charge structure # 1 has formal charges of –1, +1, 0, while structure #2 has formal charges of 0, +1, -1 and structure #3 has formal charges of –1, +2, -1.  The most stable structure is #2 because it has the most electronegative atom (O) with the most negative formal charge.

 

IV. 

Molecular: Ag2SO4 (aq) + 2 HCl (aq) →  2 AgCl(s) + H2SO4(aq)

Net ionic: Ag+ + Cl- → AgCl

MM (Ag2SO4) = 312 g/mol it is the limiting reactant

mass (AgCl) = 0.44 g

 

V.    BaN  &  Ba3N2