AP Chemistry Quiz #2

AP/IB Chemistry Name: Ch 8/9 AP Test

Multiple Choice – Circle the letter of the best response:

1. The electron pair in a C–F bond could be considered

a. Closer to C because carbon has a larger radius and thus exerts greater control over the shared electron pair.

b. Closer to F because fluorine has a higher electronegativity than carbon

c. Closer to C because carbon has a lower electronegativity than fluorine

d. An inadequate model since the bond is ionic

e. Centrally located directly between the C and F

2. Which of the following has the greatest bond strength?

a. B2 2– b. O – c. CN + d. O2 – e. NO

3. Which of the following molecules is non-polar overall?

a. SF4

b. SF2

c. CCl4

d. H2S

e. OCl2

2 4. Atoms which are sp hybridized can form _____ pi bond(s).

For 5-6 consider the skeletal structure shown below for the compound C2N2:

N – C – C – N

Complete the Lewis structure above and answer the following:

5. How many of the atoms are sp hybridized?

6. How many pi bonds does the molecule contain?

e. 7 Answer Question 7 in the space provided below:

Work the following problem. Answer neatly, as if you were working the problem for an AP/IB Exam.

- + 7. GeCl4 SeCl4 ICl4 ICl4 The species represented above all have the same number of chlorine atoms attached to the central atom. (a) Draw the Lewis structure (electron-dot diagram) of each of the four species. Show all valence electrons in your structures.

(b) On the basis of the Lewis structures drawn in part (a), answer the following questions about the particular species indicated.

(i) What is the Cl-Ge-Cl bond angle in GeCl4?

(ii) Is SeCl4 polar? Explain.

- (iii) What is the hybridization of the I atom in ICl4 ?

+ (iv) What is the geometric shape formed by the atoms in ICl4 ?