## Solutions to Exam 4 for CHM 1311 July 26, 1999

Open Periodic Tables and Closed Everything Else.
Answer any 10 of the 11 questions

• Answers are bulletted and in san serif font.

1. As complex as nitrogen-containing molecules can become, which of the following molecular shapes about a central nitrogen atom will never be observed? Why?

1. bent
2. octahedral
3. trigonal planar
4. trigonal pyramidal
5. trigonal bipyramidal

• The simple answer is that (b) and (e) can never be observed since they call for the nitrogen atom to be bonded to 6 or 5 other atoms, respectively. That violates nitrogen's octet, which it cannot permit.

The more thoughtful answer would be to provide examples of the ones that can exist, but that wasn't required. For completeness, let X, Y, and Z be other atoms and : be lone pairs of electrons; then

```
:             Z            X
/             /            /
(a) X = N     (c) X = N      (d) : N - Y
\             \            \
Y             Y            Z
```

2. Give the most likely Lewis structure(s) and describe the shape of the thiosulfate ion, S2O32-, around each sulfur if its skeletal structure is:
```                  2-
O   S   O   S   O```

• Both S and O have 6 valence electrons each; that's 30. But the anionic charge means there's two more for a total of 32 electrons. In order to link all 5 atoms together, we'll need at least 4 bonding pairs, reducing the number left to distribute to 32-8=24. So each of the 5 atoms can take 4 more lone pairs, leaving 24-20=4 electrons to be split as additional lone pairs on the end atoms.
```
(-1) ..   ..  ..  ..  .. (-1)
:O - S - O - S - O:
¨   ¨   ¨  ¨   ¨
```
It is true that sulfur will accept more than an octet, so we could imagine additional resonance structures that would involve using the extra lone pairs on the end oxygens to make a double bond to one or both sulfurs, but that would then put the (-1) formal charge on the sulfur rather than the oxygen! And of the two, oxygen would far rather have it due to its higher electronegativity. So the 3 structuress like the following would contribute rather little to extra stability (see question #11) in the ion.
```         (-1)
..   ..  ..  ..  .. (-1)
:O = S - O - S - O:
¨   ¨  ¨   ¨
```
With or without the alternate resonance structures, the electronic structure about each of the central atoms, SOS, involves 4 directions in space whose electrons must avoid one another. So all will be bent, and the entire molecule will zig-zag like /\/\.

3. The thiocyanide anion is shown in three possibilities below, each with a different choice for the central atom. They are all the best structures that can be made for that central atom choice. Only one of these structures is correct. Which is the real thiocyanide ion and why?
```      ..             -        ..             -        ..         ..  -
(a) [ :S : N ::: C: ]   (b) [ :S : C ::: N: ]   (c) [ :C :: S :: N: ]
¨               ¨
(-1) (+1) (-1)          (-1) (0)   (0)          (-2) (+2)  (-1)
```

• The (formal charges) are atrocious in (a) and (c), so (b) must be the SCN - ion.

4. At right is the skeletal structure of vitamin C. The multiple bonds are not shown. Determine the hybridization of the non-hydrogen atoms in its ring and from that, show where the multiple bonds, if any, must be. There are no multiple bonds beyond any that show up in or on the ring.

(Red is oxygen, gray is carbon, and white is hydrogen. The "ring" is that cycle of atoms at the right end of the molecule comprised of one oxygen and four carbons with other stuff dangling off. So we don't get confused, let's number the ring atoms starting with the oxygen as number 1 and proceeding clockwise around the ring to number 5, the carbon just below #1.)

• Atom #1, the ring's only oxygen (not counting danglers) is sp3 hybridized to accommodate bonds to the adjacent two carbons and oxygen's pair of lone pairs. So atom 1's bonds are all single.

Atom #2, the topmost carbon, is hybridized sp2 to make three s bonds to the atoms attached to it. But that leaves carbon's extra p electron to double bond with the top-right oxygen. That oxygen must also be hybridized sp2 to accommodate its pair of lone pairs and its double bond to atom #2.

Atom #3, the rightmost carbon, is also hybridized sp2 since he too is bonded to three other atoms. But his bond to his oxygen is only single as we can see since his oxygen is also bonded to a hydrogen (the rightmost). His oxygen is therefore sp3 hybridized to accommodate two single bonds and its pair of lone pairs. That means atom #3 has no choice but to use its extra p electron to double bond to atom #4, the bottomost carbon.

Fortunately, atom #4 is also sp2 hybridized to accept 3's double bond. And 4's oxygen is an OH too. So it too is sp3.

Finally, #4 has only a single bond left over for #5, the carbon atom that closes the ring back to that (#1) oxygen. So atom #5 is single bonded to four other atoms and must be sp3 hybridized.

So a Lewis structure for vitamin C would be sort of

```
H            :O:
\     ..    //
H     H  :O:   :O - C
\    |   |    /    |   ..
:O - C - C - C     C - O:
"   |   |   | \ //    |
H   H   H  C      H
|
:O:
/
H
```
and everybody has zero formal charge.

5. Which of the following molecules are polar?

1. carbon oxysulfide, OCS

• No choice but to be polar. No geometry could cancel out unequal bond polarities!
2. boron tribromide, BBr3

• Nonpolar since B, having no lone pairs, makes a trigonal planar shape, and the B-Br bond polarities all cancel vectorially.
3. nitrogen tribromide, NBr3

• Ah, but N has a lone pair, so the Br's are pushed down in a trigonal pyramid structure. The nonplanar component of the N-Br polarity leaves this molecule polar.
4. chloroform, HCCl3

• Here "H" takes the place of a lone pair forcing the Cl's down to near tetrahedral bonding. So since C-H polarity is radically less than C-Cl, chloroform is a polar molecule
5. arsenic pentachloride, AsCl5

• Like PCl5, it's Periodic Table brother, AsCl5 must be trigonal bipyramidal with 3 of the chlorines equatorial (with vector cancelled polarity) and 2 chlorines axial (whose As-Cl polarity also cancels). Like PCl5, it is nonpolar.

6. It may seem to you that our obsession with second row diatomic molecules is ludicrous, considering the enormous variety of all other molecules. We do it because the molecular orbitals are particularly simple in these diatomic molecules; so they make good introductory examples. But we can broaden even this simple use of MO theory to a few more examples by taking advantage of the fact that molecules with the same number of electrons (isoelectronic) have similar electronic structures!

Take CO, carbon monoxide, for example. With what second row homonuclear diatomic, E2, is CO isoelectronic? And what does that tell you about the relative magnetic properties of CO and CO+ ?

• With 10 valence electrons (4+6), CO is isoelectronic with N2 (5+5). So CO is expected to have completely filled the three bonding 2p overlapped molecular orbitals just like N2. Makes it a triple bond as the Lewis structure also predicts. But since the electrons are all paired, it is diamagnetic.

Remove one electron to make CO+ and its buddy is unpaired! So CO+ would be paramagnetic.

7. Toward which side (reactants or products) does equilibrium lie in the following and why?

NH3 + PH4+ NH4+ + PH3

• Both NH4+ and PH4+ are binary acid (actually conjugate acids to the weak bases NH3 and PH3). But binary acids get stronger as one goes down the Periodic Table even though the nonmetal's electronegativity decreases. That's because its size increases, and the proton finds itself further away from the nonmetal.

So NH4+ is a weaker acid than PH4+, and equilibrium lies to the right with the weaker acid.

If you remain unconvinced, the weaker conjugate acid (NH4+) is associated with the stronger conjugate base (NH3). So again the weaker base (PH3) makes equilibrium lie to the right! The weaker conjugate acid and the weaker base are both on the same side.

8. With which pairs below (if any) would you expect to see some sort of chemical reaction? Balance any appropriate reaction(s).

1. Au(s) and Cu2+(aq)

2. Au3+(aq) and Cu(s)

• Since Cu is a more active metal than Au, the former will displace the latter but not vice versa. So the only reaction would come from (b) as

3 Cu(s) + 2 Au3+(aq) 3 Cu2+(aq) + 2 Au(s)

9. (Warning: This is one of Dr. Parr's "creative" questions.)

Molecular fluorine is made by the electrolysis of the molten salt, KHF2. The existence of that salt suggests a bizarre HF2- ion! Lewis would absolutely choke on such a species, but MO makes it simple to understand. That ion would hold together only if there was a bond between hydrogen's 1s orbital and the highest occupied molecular orbital of F2-. Since F2- has an odd number of electrons, that highest MO orbital (s2px* ) has only one electron, and it stands to reason that it can bond with the single electron in hydrogen atom.

Remember the origin and hence the shape of that s2px* orbital to prove that HF2- must be linear. (The key is orbital overlap. The x axis runs between the fluorine atoms.)

• s2px* is the MO that is the antibonding combination of the 2px atomic orbitals on each fluorine. Those atomic orbitals are cylindrically symmetric around the F-F bond direction, x. So must be their combination. So the s2px* juts beyond both F atoms in line with their bond. The electron density is therefore concentrated along the x-axis in the s2px* (and, of course, the s2px as well). For the best overlap, H must line itself (and its 1s orbital) along the x-axis. So all 3 nuclei are collinear. The overlap is less as the hydrogen moves off the x-axis; so the most stable configuration is linear.

There's no hope for triangular at all since the sign of the s2px* MO as we look along the F2- bond direction varies as

+ F -|+ F -

for the antibonding s2px*. That | is a node in the MO exactly between the fluorines. Were the 1s orbital of hydrogen to attempt to straddle the ion, that change in sign across the node would give exactly as much positive as negative overlap, meaning no overlap at all! So triangular geometries are nonbonding. Bent geometries are weakly overlapping and so weakly bonding, but linear geometries are strongly overlapping hence strongly bonding.

10. What is the shape of the ion and bond order of the N O bonds in the nitrite ion, NO2- ?

• The two Lewis resonance structures below (18 valence electrons) show that NO2- is bent and has bond order 1½.
```
(-1)                  (-1)
..                    ..
:O: :O:          :O: :O:
\ //   <---->   \\ /
N                N
¨          ¨
```

11. It is a fact that the more (legitimate) Lewis resonance structures a chemist can imagine for a molecule, the more stable that molecule turns out to be. Clearly the molecule isn't reading the chemist's mind. So what accounts for that greater stability?

• Resonance structures, like that in #10 above, are Lewis's way of delocalizing electrons, letting them spread out around the molecule, a foretaste of the molecular orbitals concept. If the electrons are less confined than in the local electron theories, their average wavelength will be longer (lowering their momenta by deBrolie's equation p=h/l) and letting them get near more than just two nuclei (which lowers their potential energy as well). It's a win-win bargain for the electrons. If they are thus less energetic, they've made stronger bonds, and the overall stability of the molecule is improved.