Hi Nicole!

Q31: Agreed. It would be A BIT less than 109.5°. However, the question is asking us to choose the angle that BEST describes the C-S-C bond, and 90° is WAAAAAAY LESS than 109.5°.

In reality, the bond angle of C-S-C varies between 101 to 107° (you don't need to know this), which is less than 109.5°, but not as drastic as 90°.

From our basic geometry, we see that the central atom is bonded to FOUR GROUPS: two other atoms, and two lone pairs. Therefore, we have a tetrahedral arrangement, and an approximate bond angle of 109.5°.

The best answer is B.

Q25 EK and Q33 ICE: Awesome question! These concepts seem like they are opposing each other, but amazingly, these are both asking the same thing!

Use the Newman projection diagram on pg. 129 in the CC to help you out. 😀

Imagine that there is a horizontal line drawn from Eclipsed I (left) to Eclipsed I (right).

The distance from that horizontal line DOWN to another conformation is the energy REQUIRED to break apart that molecule. For example, notice that the energy "well" is VERY DEEP for Gauche I, while the energy "well" for Eclipsed II is very SHALLOW.

Hey, wait a minute...

Using this knowledge, we can then understand Q25 EK a bit better. A more stable molecule has a higher ENERGY WELL, while a less stable molecule has a lower ENERGY WELL.

Conversely, the distance from the BOTTOM of the axis tells us the amount of energy that a molecule will RELEASE when it is broken apart. Notice that the more stable molecule is closer to the bottom axis, while the less stable molecule is FURTHER from the bottom axis. Therefore, a more stable molecule will release LESS ENERGY than a less stable molecule. Using this, we can understand Q31 ICE 2 better! Since the trans isomer is more stable, it will RELEASE less energy than the cis isomer.