1b. Which element below is the least likely to adopt an oxidation state of +5?

Ti has only 4 valence electrons. Removing 4 electrons is extremely unfavorable since this requires removing a 3p electron.

2b. The chemistry of V²⁺ is dominated by which electrons?

The 4s have been removed in forming the cation. Only 3d are left at the valence level.

3b. Which element below will have reactivity most like that of Cr?

Cr is [Ar]4s¹3d⁵ Mo is [Kr]5s¹4d⁵. - Same valence configuration

4b. Which species below has the same number of valence electrons as Fe³⁺?

Fe³⁺ is [Ar]3d⁵ / Mn²⁺ is [Ar]3d⁵ / Fe²⁺ is [Ar]3d⁶ / Mn¹⁺ is [Ar]4s¹3d⁵ / Ru²⁺ is [Kr]4d⁶ / Co²⁺ is [Ar]3d⁷

Both Fe³⁺ and Mn²⁺ have 5 valence electrons.

5b. Which of the following is the proper electronic configuration for the neutral O atom?

(a) (b) (c)

O has 6 valence electrons. Fill them in to maximize unpaired spins.

6b. Which of the following is the proper electronic configuration for the O^2- ion?

(a) (b) (c)

O has 6 valence electrons. Adding 2 makes 8. Fill them in.

7b. From what you know so far, which ion below would you expect to have the largest number of unpaired spins?

Fe³⁺ has an electronic configuration 3d⁵, with 5 unpaired spins

8b. Among the following elements, which has the largest radius?

Si is at a higher n-level than Ne and F, and has a smaller effective nuclear charge than P or Cl. Alternatively, it’s below and to the left of all of the others in the periodic table..

9b. Among the following elements, which has the largest ionization energy?

Ne is at a lower n-level than Cl, P, or Si, and has a larger effective nuclear charge than F. Alternatively, it’s above and to the right of all of the others in the periodic table..

10b. Which of the following representations is not correct?

(a) P = [Ne]3s²3p³ (b) C = [He]2s²2p² (c) Mg = [Ne]3s²

Cl should be [Ne]3s²3p⁷

11b. Which of the following is not correct?

(a) The first IE of Na is higher than the first IE of K.

(b) The second IE of Mg is higher than the first IE of Mg.

(c) The first IE of Ca is greater than the first IE of K.

(e) The second IE of Sc is higher than the second IE of Ca

This was discussed in class and was explicitly covered in a question raised at the review session. To remove a second electron from Ca, one removes a 4s electron. To remove a second electron from K, one must go up an n-level and remove a 3p electron (in other words, we’re dipping below the valence level). This is VERY costly in energy!

12b. Which of the following molecules is polar?

All the individual bonds have some polarity, but all but NO^2- have a symmetry which cancels out the individual polarities, leading to a nonpolar molecule. Remember that in situations with multiple equivalent resonance forms, you must average in all of them.

13b. Which of the following bonds is the most polar?

(d) H₂B–F (e) CH₃S–SCH₃

14b. Acrylonitrile, shown at right, is the building block of the synthetic fiber Orlon. Which of the following statements is not true of this molecule?

15b. The azide anion N₃^- can be represented by a number of resonance structures. Regarding this anion, which statement below is false?

(a) The average formal charge on the central nitrogen is positive.

(b) The average formal charges on each of the terminal nitrogens are the same.

(d) The average bond order of at least one of the N-N bonds is greater than 1.

(e) The average formal charge on at least one of the terminal nitrogens is negative.

Averaging the 3 structures on the left, the average bond order for both bonds is >1.

The average formal charge on the central N is positive.

The average formal charges on the terminal nitrogens are the same, and negative.

The average bond strengths/orders of the two bonds are the same (symmetry!).

Carbon monoxide, a toxic biproduct of incomplete combustion, is a colorless, odorless gas which can bind to Fe-containing respiratory proteins in the body. The following three questions concern the lowest energy Lewis structure for this simple molecule.

16b. In CO, the number of lone pair electrons on O is:

The question asks for lone pair electronS, not for the number of lone pairS.

There was an error on the original answer key. Grades will be adjusted accordingly.

17b. In CO, the oxidation number of C is:

For oxidation number, we give all of the shared (bonding) electrons to the more electronegative element. Oxygen ends up with 8 electrons, carbon with only 2.

18b. In CO, the formal charge on C is:

19b. Another inhibitor of respiration is the cyanide anion, CN^-. In the cyanide anion, the formal charge on C is:

CN^- is isoelectronic with CO. Same as above.

20b. We have previously seen that H₂ is a good fuel in combustion.

From the bond energies above, what is �H for this reaction in kJ/mol?

�H = [ 2(436) + 498 ] — [ 4(464) ] = 1370 - 1856 = -486