We’re on to discrete section 1 in the Blueprint MCAT full-length 1 today. We evaluate formal charges, calculate projectile speed, define chiral centers, and more!
We’re joined by Phil from Blueprint MCAT, formerly Next Step Test Prep.
Listen to this podcast episode with the player above, or keep reading for the highlights and takeaway points.
[03:57] Question 14
What are the formal charges on the atoms in the molecule below? (Note: the molecule has not been labeled with overall formal charge and may have an overall formal charge that is 0 or a value other than 0).
I – oxygen with two little dots on the left
II – a triple bond to a nitrogen
III – connected to the nitrogen is another oxygen with six dots around it.
Dots on the left has a lone pair and triple bonded to a nitrogen, which is then single bonded to another oxygen with three lone pairs.
(A) I: +1; II: +1; III: -1
(B) I: -1; II: +1; III: +1
(C) I: +1; II: 0; III: -1
(D) I: +1; II: -1; III: -1
It’s about asking what are the formal charges on each of these atoms? And it tells us that we don’t know what the overall charge is. It doesn’t tell us what it is. So it could be zero. So it’s kind of a nice little hint to say, it doesn’t have to add up to zero. It could be something other than zero.
That’s actually good to note because NO2 is generally nitrite. But this is not the nitrite most people are familiar with. So if you know the charge, you’ve got to throw that out. And then the question here is telling you that you can’t predict the charge. It’s going to be something weird. So go from there. You have to figure out how to calculate a formal charge.
The equation for the formal charge is the number of electrons normally found on the thing on the element, or the atom, minus the number of sticks and dots.
The idea here is the dots are lone pairs of electrons. A bond is two electrons, but they’re being shared sort of unevenly or evenly in this case.
And so we have the oxygen has two lone pairs, and it’s got three electrons, that from those double bonds. Each of those bonds is two electrons, but one of them goes with the nitrogen. One of them goes to the oxygen.
Looking at the periodic table, anything on the far right is going to have an octet. So with the exception of Helium, Neon, Argon, Krypton, Xenon, all of those are going to be like 8 valence electrons. Then you go to the left, they’re missing one. So fluorine is going to have 7. They want electrons and oxygen is going to have 6.
Using the equation, start with 6, and then subtract the number of dots and sticks. So you add them together. A bond counts as just one electron and then any free electron counts as an electron as well.
So 6 minus one is plus one. Plus one gives me answer choices A, C and D. So B is immediately out. Then Nitrogen is 5 – 4 +1. It leaves you with answer choice A without looking at answer choice three.
And once you know that equation, and you’re already going to apply it. This question is actually pretty straightforward. But the challenge is just making sure that you’ve got that little outside discrete knowledge.
[10:44] Question 15
A mass of 10 kg was dropped from a height of 20 m. Ignoring air resistance, what is the maximum speed achieved by the mass? (assume g = 10 m/s2)
(A) 10 m/s2
(B) 20 m/s2
(C) 200 m/s2
(D) 400 m/s2
D is very fast. Putting D’s speed in conceptual understanding, this means that every five seconds, this thing is going two kilometers. That’s very fast. C is also very similar. This thing is 10 kg. If you drop it from 20 meters, it’s not going to be going fast. So this leaves us with A and B.
Now, there are actually two ways to solve this question the way that the MCAT kind of sets this up. You can use your kinematic equations. We’re trying to find our final velocity.
Initial velocity = 0
Distance = 20 m
Acceleration = 10 m/s2
You can use your kinematic equations for that. The other one is kinetic energy. Very interestingly, there are some cases where kinematic equations don’t work where the forces are changing.
For example, as objects get closer, they pull more electrostatics or gravity or things like that. Or springs, the more you compress it, the more they push back. And so in those cases, you can’t use kinematics.
But you can still use energy. Phil suggests trying energy first always. Because in theory, that should work unless there is some friction going on. And in this question, that isn’t the case.
So now, we just have to convert our gravitational potential energy into kinetic energy. The potential energy equation is:
mgh = 1/2 mv2
Masses on both sides, that cancels out.
gh = 1/2 v2
So technically, if we ignore drag, a feather, and a bag of bricks, they’re going to fall at the same speed.
The problem is the drag. And we’re assuming there’s not gonna be any drag in this case, especially because we’re only dropping it from 20 meters. In this case, the mass ends up canceling out and so we end up not needing to worry about that.
gh = 1/2 v2
Then you can just plug in the numbers to find the velocity.
g = 10
h = 20.
So we have two times 10 times 20 is 400. Square root it, and it turns into 20.
The takeaway here is to use kinetic energy first when they asd for velocity.
You can convert gravitational potential energy or electrostatic potential energy or spring potential energy into velocity through kinetic energy. Not only is that does that almost always work compared to the kinematics. It’s also easier in terms of the math because the kinematic equations can be kind of complex at times.
[17:10] Question 16
How many tetrahedral stereocenters are present in cholesterol (pictured below)?
Tetra means 4. Stereo is like right and stereocenter means you have a right and left version of the center, which is another way to say the chiral center. And so tetrahedral chiral center is something that’s connected to four things. It’s tetrahedral, and it’s also got some stereochemistry with the right and the left-hand version of this as well.
Stereocenter means there’s got to be a difference from one side versus the other.
This is one of those questions that students very often end up taking like three or four minutes, but you definitely don’t.
On the MCAT, the only time you would include a dash or a wedge is if it matters if which means there’s a right and left form.
There’s 5 wedges and 3 dashes so the answer is 8.
A lot of times, a student will get this question correct. If you spend five minutes on this question, then you’re going to miss some other questions just because you didn’t get to them. So even if you get a question correct, it’s really important to go back and see how long you spent on it.
If you spent a long time, see if maybe there’s a faster way. See if there’s a shortcut. Look for a backdoor to the question because they exist pretty often.
[23:32] Question 17
The preferred ion configuration of many elements on the periodic table is determined by:
(A) the electron configuration of the nearest noble gas; elements will gain or lose electrons until they have the same core electron configuration as this noble gas.
(B) the electronegativity of the element directly above it, within the same group; elements will gain or lose electrons until they have an electronegativity that matches the period immediately above them.
(C) the electron configuration of the nearest noble gas; elements will gain or lose electrons until they have the same valence electron configuration as this noble gas.
(D) the relative stability of the nearest d subshell; elements will gain or lose electrons until the outermost d subshell available is stable.
Ion is when something’s like gained or lost like.
By changing the electrons in the quick, within the inner shells, that’s way more difficult than adding one onto the outside or taking one off of the outside.
Another way to think about this, what are some ions that you know. So it’s turning into the kind of like the noble gas configuration and that just makes sense. And it’s absolutely the valence shell because we pull or take them off the outer shell. And not the inner shells because that’s difficult and annoying. It’s like trying to paint the inside of a baseball. It’s a lot easier to add something to the outside of it, rather than the inside.
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