Chemical Bonding Test Review

Be sure you have read the notes carefully and thoroughly. Study your worksheets and practice problems.

Part 1

1. What are sigma and pi bonds?

2. Determine which type of bond (ionic, polar covalent, or nonpolar covalent) would form between the following atoms. Show the polarity of the bond if the bond is polar.

a. Ca and Nb Mg and Oc. C and Cl

d. Cl and Cle. N and Cf. S and I

3. The bonds between the following pairs of atoms are covalent. Arrange them according to polarity, naming the most polar bond first. H-Cl, H-C, H-F, H-O, H-H

4. For each of the following, answer these questions:

How many polar bonds? How many nonpolar bonds? Is the molecule polar or nonpolar?

a. H2Sb. CO2c. CHO2

d. CCl4e. CH2Cl2f. NH3

g.BrClh. Cl2j. SF6 (exception, S will have 12 ve-)

k. PF5 (exception, P will have 10 ve-)

5. What is the shape and bond angle for the following? How many sigma and pi bonds are present in each molecule? What is the hybridization (or hybrid orbital)? Is the molecule polar or nonpolar? (Draw them out!!!)
a. AsI3b. BBr3c. SF2

e. COf. SiCl4g. H2O

h. HCli. PH3

6. The phosgene molecule, COCl2, has the oxygen and chlorines attached directly to carbon. Write the Lewis dot structure and predict the shape. (Hint: Cl wll not form double bonds)

7. Draw the Lewis dot structure for the HCO31- ion. Carbon is the central atom and hydrogen is bonded to the oxygen.

8. Draw the nitrate ion, NO31-. Are there any resonant structures? If so, draw them.

Part 2

1. Explain the following terms: valence electrons and the Octet Rule

2. Draw electron dot structures for the following atoms or ions:

Rb atom, Rb ion, Be atom, Be ion, Kr atom, Br atom, Br ion, C atom

3. Transition metals often form ______ions because they have valence electrons in the s and d sublevel.

4. Name 5 ions that have the same electron configuration:

5. Ionic compounds are formed when electrons are ______. (shared or transferred)

6. List 5 properties of ionic compounds.

7. Draw electron dot structures showing the formation of the following ionic compounds:

a. calcium nitrideb. magnesium oxidec. aluminum chlorided. rubidium sulfide

Chemical Bonding Test Review Answer Key

Part 1 Answers:

  1. single bond = sigma bond, double bond = 1 sigma bond and 1 pi bond, triple bond = 1 sigma bond and 2 pi bonds
  2. a. ionicb. ionic c. polar (Cl has higher EN) d. nonpolar e. polar (N has higher EN) f. nonpolar
  3. H-F (1.9), H-O (1.4), H-Cl (0.9), H-C (0.4), H-H (0)
  4. a. 2 polar bonds, 0 nonpolar bonds, polar moleculeb. 2 polar bonds, 0 nonpolar bonds, nonpolar moleculec. CHO2, 3 polar bonds, 0 nonpolar bonds, polar molecule d. 4 polar bonds, 0 nonpolar bonds, nonpolar molecule e. 4 polar bonds, 0 nonpolar bonds, polar molecule f. 3 polar bonds, 0 nonpolar bonds, polar molecule g. 0 polar bonds, 1 nonpolar bond, polar molecule h. 0 polar bonds, 1 nonpolar bond, nonpolar molecule j. 6 polar bonds, 0 nonpolar bonds, nonpolar molecule k. 5 very polar bonds (would be ionic, but P is a not a metal and will not lose its ve-), 0 nonpolar bonds, nonpolar molecule
  5. a. Trigonal pyramidal, <109.5°, 3 single bonds (3 sigma bonds), sp3, polar molecule b. (exception B will have 6ve-) Trigonal planar, 120°, 3 single bonds (3 sigma bonds), sp2, nonpolar molecule c. bent, <109.5°, 2 single bonds (2 sigma bonds), sp3, polar molecule e. linear, 180°, 1 triple bond (1 sigma bond and 2 pi bonds), sp, polar molecule f. tetrahedral, 109.5°, 4 single bonds (4 sigma bonds), sp3, nonpolar molecule g. bent, <109.5°, 2 single bonds (2 sigma bonds), sp3, polar molecule h. linear, 180°, 1 single bond (1 sigma bond), sp, polar molecule i. Trigonal pyramidal, <109.5°, 3 single bonds (3 sigma bonds), sp3, polar molecule
  6. trigonal planar (1 double bond between C and O, 1 single bond between C and Cl, 1 single bond between C and other Cl)
  7. add the additional 1 ve- to the C, show 1 double bond between C and an O, 1 single bond between C and an O, and a single bond between C and the last O and the H will single bond to that O)
  8. add the additional 1 ve- to the N, show 1 double bond between N and an O, 1 single bond between N and an O, and 1 single bond between N and the last O. There are 3 resonant structures that can be drawn (the double bond can be between the N and any of the O’s)

Part 2 Answers: 1. valence electrons = electrons involved in bonding, Octet Rule = atoms will form bonds in order to obtain 0 or 8 valence electrons. Metals will lose electrons and have 0 of their original valence electrons. Nonmetals will gain or share valence electrons and will end up with 8 valence electrons. 2. Rb atom 1 dot, Rb ion 0 dot with a +1 charge, Be atom 2 dots, Be ion 0 dots and +2 charge, Kr atom 8 dots, Br atom 7 dots, Br ion 8 dots with a -1 charge, C atom 4 dots. 3. multiple 4. oxide ion, fluoride ion, sodium ion, magnesium ion, aluminum ion 5. transferred 6. hard, brittle solids, high melting and boiling points, crystalline in nature 7a. Ca3N2 7b. MgO 7c. AlCl3 7d. Rb2S (come in to look at key for the dot structures)

Chemical Bonding Test Review Answer Key

Part 1 Answers:

  1. single bond = sigma bond, double bond = 1 sigma bond and 1 pi bond, triple bond = 1 sigma bond and 2 pi bonds
  2. a. ionicb. ionic c. polar (Cl has higher EN) d. nonpolar e. polar (N has higher EN) f. nonpolar
  3. H-F (1.9), H-O (1.4), H-Cl (0.9), H-C (0.4), H-H (0)
  4. a. 2 polar bonds, 0 nonpolar bonds, polar moleculeb. 2 polar bonds, 0 nonpolar bonds, nonpolar moleculec. CHO2, 3 polar bonds, 0 nonpolar bonds, polar molecule d. 4 polar bonds, 0 nonpolar bonds, nonpolar molecule e. 4 polar bonds, 0 nonpolar bonds, polar molecule f. 3 polar bonds, 0 nonpolar bonds, polar molecule g. 0 polar bonds, 1 nonpolar bond, polar molecule h. 0 polar bonds, 1 nonpolar bond, nonpolar molecule j. 6 polar bonds, 0 nonpolar bonds, nonpolar molecule k. 5 very polar bonds (would be ionic, but P is a not a metal and will not lose its ve-), 0 nonpolar bonds, nonpolar molecule
  5. a. Trigonal pyramidal, <109.5°, 3 single bonds (3 sigma bonds), sp3, polar molecule b. (exception B will have 6ve-) Trigonal planar, 120°, 3 single bonds (3 sigma bonds), sp2, nonpolar molecule c. bent, <109.5°, 2 single bonds (2 sigma bonds), sp3, polar molecule e. linear, 180°, 1 triple bond (1 sigma bond and 2 pi bonds), sp, polar molecule f. tetrahedral, 109.5°, 4 single bonds (4 sigma bonds), sp3, nonpolar molecule g. bent, <109.5°, 2 single bonds (2 sigma bonds), sp3, polar molecule h. linear, 180°, 1 single bond (1 sigma bond), sp, polar molecule i. Trigonal pyramidal, <109.5°, 3 single bonds (3 sigma bonds), sp3, polar molecule
  6. trigonal planar (1 double bond between C and O, 1 single bond between C and Cl, 1 single bond between C and other Cl)
  7. add the additional 1 ve- to the C, show 1 double bond between C and an O, 1 single bond between C and an O, and a single bond between C and the last O and the H will single bond to that O)
  8. add the additional 1 ve- to the N, show 1 double bond between N and an O, 1 single bond between N and an O, and 1 single bond between N and the last O. There are 3 resonant structures that can be drawn (the double bond can be between the N and any of the O’s)

Part 2 Answers: 1. valence electrons = electrons involved in bonding, Octet Rule = atoms will form bonds in order to obtain 0 or 8 valence electrons. Metals will lose electrons and have 0 of their original valence electrons. Nonmetals will gain or share valence electrons and will end up with 8 valence electrons. 2. Rb atom 1 dot, Rb ion 0 dot with a +1 charge, Be atom 2 dots, Be ion 0 dots and +2 charge, Kr atom 8 dots, Br atom 7 dots, Br ion 8 dots with a -1 charge, C atom 4 dots. 3. multiple 4. oxide ion, fluoride ion, sodium ion, magnesium ion, aluminum ion 5. transferred 6. hard, brittle solids, high melting and boiling points, crystalline in nature 7a. Ca3N2 7b. MgO 7c. AlCl3 7d. Rb2S (come in to look at key for the dot structures)