AAMC CBT6 and 6R OFFICIAL Q&A

[Vihsadas]

This is the official Q&A thread for AAMC CBT6 and 6R.

Please post ONLY questions pertaining to AAMC CBT6 and 6R.
Out of respect for people who may not have completed the other exams, do not post questions or material from any other AAMC exam.

Please see this thread for the rules of order before you post.

Good luck on your MCAT!

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[161927]

Electric power for transmission over long distances is “stepped up” to a very high voltage in order:

A) to produce currents of higher density.

B) to produce higher currents in the transmission wires.

C) to make less insulation necessary.

D) to cut down the heat loss in the transmission wires.
Power transmitted to a load at the user’s end of an electrical line is the product of the current I times the voltage V at the user’s end. The power company has the option of using a low voltage or a high voltage for a given amount of power delivered. Using Ohm’s law (I = V/R), the power delivered can also be written as P = V2/R. The power lost in the transmission line is I2R. Since the resistance R is fixed by the material and dimensions of the line, we see that using a high voltage at a lower current maximizes the power to the user and minimizes the heat loss in the transmission line, choice D.

I am not satisfied with this answer. There is no difference between the formulas I2R and V2/R. Does any one understand this explanation? What exactly is the formula for the heat loss?

This question does not require the use of a formula. Know the rationale behind stepping up voltages. By doing this, you decrease the current across the main transmission line, which results in less dissipation of power.

 

[Vihsadas]

This question does not require the use of a formula. Know the rationale behind stepping up voltages. By doing this, you decrease the current across the main transmission line, which results in less dissipation of power.

Deepa, to see this effect mathematically, take a look at the equation P = IV. From the problem, you know that the power transmitted across the wire is constant. So, if you increase the current, the voltage must decrease, and if you increase the voltage, the current must decrease. In both, situations, however, the power transmitted is the same.

Then you have to know that the heat dissipated depends on the current through the wire, since a higher current leads to heating of the wire.

 

[mytoechondriac]

Hi!!

I'm going nuts trying to figure out why #76 on the paper 6R

"In a healthy person standing at rest, a comparison of arterial blood pressure measured in the arm with that measured in the leg shows that the pressure in the leg..."

seems to be a repeat question to me. I've looked everywhere to figure out where i have seen this question before, but I still cannot find out where I saw it, if I ever did. Does anyone else think they have seen this same question on another practice test??

 

[tncekm]

There ARE repeat questions in the AAMC's. I didn't take them all, so I can't confirm I saw this one twice, but its definitely a possibility.

 

[unsung]

This is more of a complaint than anything else, but, I really thought this was a stupid question. I was confident that this question had something to do with Joe's scared body leading to overheating etc, etc, etc, not that Steve was more frightened than Joe. *sigh* Anybody feeling me here? There were a few really stupid questions on AAMC 6, IMO. But this one and the last one in the PS section (outlined earlier in the thread) really ticked me off.

If the heart rates of Steve and Joe were monitored when they entered the burning warehouse to fight the fire, one would expect:

A) Steve's heart rate to increase more, because of activation of the sympathetic division of the autonomic nervous system.
B) Joe's heart rate to increase more, because of activation of the sympathetic division of the autonomic nervous system.
C) Joe's heart rate to increase more, because of activation of the parasympathetic division of the autonomic nervous system.
D) both heart rates to increase, because of activation of the parasympathetic division of the autonomic nervous system.

This seemed more like a psychology Q than anything else.

There were definitely some really weird Qs on this one.

 

[dwc929]

Did anyone else find this verbal hard? I cruised through the first few passages and only got 1 wrong out of the first 3. Then in the last 4 passages, I got destroyed...don't know how to explain it...

 

[Live4Life]

Nevermind

 

[dwc929]

Can someone please explain #101 in Bio for me?

Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?

A) Isopropyl chloride
B) methyl chloride
C) sec-butyl chloride
D) tert-butyl chloride

 

[161927]

Can someone please explain #101 in Bio for me?

Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?

A) Isopropyl chloride
B) methyl chloride
C) sec-butyl chloride
D) tert-butyl chloride

D

The intermediate is a tertiary carbocation which will not rearrange.

 

[MichaelRowle]

Sorry, but could someone help me out on this one. Could someone work this out for me, or explain the math a little better. I don't know what the hell they did here.

 

[minhaj]

Sorry, but could someone help me out on this one. Could someone work this out for me, or explain the math a little better. I don't know what the hell they did here.

Is it on CBT or paper? if CBT can you post the question?

 

[MichaelRowle]

Assuming equal masses, how would the detection times of 3H+ and 3He+ compare? A
) 3H+ would have a longer flight time than 3He+. B
) 3H+ would have a shorter flight time than 3He+. C
) 3H+ would have the same flight time as 3He+.
Detection times (time-of-flight) depends of the charge to mass ratio of the ion being accelerated. This fact is deduced from the energy equation given above. The singly-ionized isotopes of 3H and 3He have the same charge to mass ratio and therefore the same detection times for the same accelerating potential. Answer C is correct.

D
) The radioactive 3H+ would always decay before
detection.


This doesn't seem to make sense to me wouldn't H have 1 proton and 2 neturons while He would have 2 protons and 1 neutron?

 

[MichaelRowle]

God I hate this test. Here's another one if anyone has any insight. Thanks.

Item 49

Solution

Mark Hooke’s law relates stress (force/unit area) and strain (elongation/unit length) with Young’s modulus Y by the expression, F/A = YΔL/L. Suppose a mass M suspended by a wire of length L and radius R stretches the wire by an amount ΔL. By how much will M stretch a wire of the same material with double the length and double the radius? A
) ΔL/4 B
) ΔL/2
Rearrange the Hooke’s law equation to give ΔL1 = FL/YA. Replace L by 2L and A by 4A (because doubling the radius increases the area by four). Now the new ΔL2 = F2L/Y4A, which can be written as ΔL2 = 2ΔL1/4. The new stretch is only half the old stretch because the increase in wire area more than compensates the increase in wire length. Answer B is correct.

C
) 2ΔL D
) 4ΔL

 

[bluemonkey]

They both have the same charge, +1. As you said, H+ must have one proton, two neutrons, and zero electrons while He+ has two protons, one neutron, and one electron. Since they both have the same charge and essentially the same mass (He+ has an extra electron), the mass/charge is the same.

Assuming equal masses, how would the detection times of 3H+ and 3He+ compare? A
) 3H+ would have a longer flight time than 3He+. B
) 3H+ would have a shorter flight time than 3He+. C
) 3H+ would have the same flight time as 3He+.
Detection times (time-of-flight) depends of the charge to mass ratio of the ion being accelerated. This fact is deduced from the energy equation given above. The singly-ionized isotopes of 3H and 3He have the same charge to mass ratio and therefore the same detection times for the same accelerating potential. Answer C is correct.

D
) The radioactive 3H+ would always decay before
detection.


This doesn't seem to make sense to me wouldn't H have 1 proton and 2 neturons while He would have 2 protons and 1 neutron?

 

[pullandbear]

Did anyone else find this verbal hard? I cruised through the first few passages and only got 1 wrong out of the first 3. Then in the last 4 passages, I got destroyed...don't know how to explain it...

yeah i thought it was really hard. I got an 11 on AAMC 7's verbal and got cocky because everyone said it was hard. Then I ran into this verbal.....very humbling lol. That damn ice sheet passage killed me. And i wanted to stab myself after reading the "formalist critique"....holy crap. And the passage about brand advertistemtn in movies was ridiculous. The questions were written by a 4 year old kid....no logic whatsoever.

 

[engineeredout]

God I hate this test. Here's another one if anyone has any insight. Thanks.

Item 49

Solution

Mark Hooke's law relates stress (force/unit area) and strain (elongation/unit length) with Young's modulus Y by the expression, F/A = YΔL/L. Suppose a mass M suspended by a wire of length L and radius R stretches the wire by an amount ΔL. By how much will M stretch a wire of the same material with double the length and double the radius? A
) ΔL/4 B
) ΔL/2
Rearrange the Hooke's law equation to give ΔL1 = FL/YA. Replace L by 2L and A by 4A (because doubling the radius increases the area by four). Now the new ΔL2 = F2L/Y4A, which can be written as ΔL2 = 2ΔL1/4. The new stretch is only half the old stretch because the increase in wire area more than compensates the increase in wire length. Answer B is correct.

C
) 2ΔL D
) 4ΔL

ΔL = FL/AY, and with doubling the length and doubling r (A = pi*r^2, 2r^2 = 4, so new A = 4A) you get ΔL = 2FL/4AY = FL/2AY.

What a lot of people get wrong on these kinds of questions is going oh look, if I move the 2 to the other side then I see that it equal 2ΔL so the answer is C. This is incorrect.

Try looking at it this way: Rearrange the formula so everything but the number is on top: ΔL = ((FL)*(AY)^-1)/2. Now its somewhat easier to see that with the new values for length and area, the ΔL is decreased by a factor of 2, so the answer is ΔL/2

 

[engineeredout]

Electric power for transmission over long distances is "stepped up" to a very high voltage in order:

A) to produce currents of higher density.

B) to produce higher currents in the transmission wires.

C) to make less insulation necessary.

D) to cut down the heat loss in the transmission wires.
Power transmitted to a load at the user's end of an electrical line is the product of the current I times the voltage V at the user's end. The power company has the option of using a low voltage or a high voltage for a given amount of power delivered. Using Ohm's law (I = V/R), the power delivered can also be written as P = V2/R. The power lost in the transmission line is I2R. Since the resistance R is fixed by the material and dimensions of the line, we see that using a high voltage at a lower current maximizes the power to the user and minimizes the heat loss in the transmission line, choice D.

I am not satisfied with this answer. There is no difference between the formulas I2R and V2/R. Does any one understand this explanation? What exactly is the formula for the heat loss?

Actually when I did this question I just remember that the whole advantage of using AC power is that it runs at a lower current so there are less power losses. DC power on the other hand runs at a higher current, so there is a lot more power loss therefore DC energy cannot be transmitted over great distances.






As an aside, I got 11PS/11VR/9BS on this. I was totally unprepared for all the orgo that was on the BS section (although I think I did all right considering). Damn I never wanna see that much orgo on the real thing.


And what the hell was with the passage with one question? I thought my internet connection died in the middle of the test or something.

 

[madame99]

What about an entire passage in the BS section dedicated to ONE question, i had to check that the questions were numbered correctly to make sure my computer wasnt messing up

 

[engineeredout]

VR question #76:

The author's major thesis is that:

A) Rational people can be expected to base their decisions on self interest.
--While acknowledging the seemingly rationality of acting solely on the basis of self-interest, the passage author argues that this policy would now lead to ruin for all.

B) The unregulated use of resources that are available to all is detrimental to society

It says B is correct, but I don't get why A isn't correct. The answer doesn't even really seem to explain why its wrong.

 

[jdla]

Can someone help me with question 34?

I don't understand why the answer is 9/4.

 

[engineeredout]

Can someone help me with question 34?

I don't understand why the answer is 9/4.

You're going to have to post the question. The CBT question #34 has no numbers in it so you're probably getting it from the paper version which not everyone has.

 

[tigress139]

Which of the following compounds has the same geometry as methane?

B. H2S D. SiCl4

I understand that SiCl4 has a tetrahedral geometry and is the correct answer, but why not H2S? The answer to this question states that "Geometry/shape of a molecule refers only to positions of atoms, not e- pairs." I thought this definition only applied to SHAPE, and that geometry takes atoms and lone pairs into account.

Would it be correct to say that H2S has tetrahedral geometry, but a bent shape?

 

[futuredoctor10]

Which of the following compounds has the same geometry as methane?
B. H2S D. SiCl4

I understand that SiCl4 has a tetrahedral geometry and is the correct answer, but why not H2S? The answer to this question states that "Geometry/shape of a molecule refers only to positions of atoms, not e- pairs." I thought this definition only applied to SHAPE, and that geometry takes atoms and lone pairs into account.

Would it be correct to say that H2S has tetrahedral geometry, but a bent shape?

Ah I see why you thought it was H2S. I think they should have edited this question to say: "Which of the following has the same MOLECULAR geometry as methane"

Electronic geometry refers to the # of electron groups or electron regions. Both H2S and SiCl4 have tetrahedral electron geometries because both have 4 electron regions (H2S, the S has 2 H + 2 lone pairs, while SiCl4 the Si has 4 Cl atoms attached).

Since both these answer choices are tetrahedral, we have to look at molecular geometries to distinguish. H2S has 2 bonding groups and 2 lone pairs which is bent, while SiCl4 has 4 bonding and 0 lone pairs which is tetrahedral. Since CH4 is also tetrahedral molecular geometry, the correct answer choice was SiCl4.

So by process of elimination you know it has to be SiCl4, but I agree they probably should have said "MOLECULAR" geometry in the question to avoid any ambiguity. Hope this helps

 

[tigress139]

You're right, I think it was just the unfamiliar terminology. Thanks a lot!

 

[byeh2004]

Can someone give a better explanation to number 28 of the PS section than what AAMC gave us? Its the one with the molecule set up to be 2 springs. Thanks!

 

[byeh2004]

On number 36 on the PS section, how do we know the comet is going away from the sun? Because I thought the comet was going towards the sun so I picked the answer that had the perturbing forces going towards it.

 

[futuredoctor10]

Can someone give a better explanation to number 28 of the PS section than what AAMC gave us? Its the one with the molecule set up to be 2 springs.

#28. Identify... the correct formula from which period can be calculated.

I did this one differently (not using dimensional analysis that is suggested in the soln):

f = 1/(2pi) x sqrt(k/m) have to know period or frequency formula
Since T = 1/f:
T = 2 pi sqrt (m/k)
T^2 = 2^2 pi^2 [sqrt (m/k)]^2
T^2 = 4 pi^2 m/k
(T/pi)^2 = 4 m/k

Hope this helps.

 

[futuredoctor10]

On number 36 on the PS section, how do we know the comet is going away from the sun? Because I thought the comet was going towards the sun so I picked the answer that had the perturbing forces going towards it.

I missed this one too. I have no idea how we were supposed to get this one! "Shadowing effects" seems very vague (their solution explanation makes sense, but their answer choice should have been more clear like the solution is!)

I picked "No, because the sun radiates in all directions"

And the "coma dust" question -- that one tripped me up too on this passage set.

 

[Lukkie]

did anyone else find this test to have the easiest VR section of all the AAMC?

 

[futuredoctor10]

did anyone else find this test to have the easiest VR section of all the AAMC?

I thought it was on par/similar with the later AAMC VR sections - because my VR score has been consistent since 6.

 

[Piston95]

on the question about the galvanic cell, with Pb, Ag and Zn....did anyone else think it should have only been B.)lead only instead of C.) lead and Zinc? I get that they should have negative reduction potentials, so that they will be oxidized and copper will be reduced...but if it has to be galvanic, doesnt the EMF have to be positive for spontaneous reaction? If thats the case shouldnt it just be lead..that would give -.13 + .52= +.39 for EMF, if it was zinc it would be -.76+.52=-.24,...so wouldnt that give a negative EMF, so it wouldnt make a the cell galvanic right? or did i miss something?

 

[Lukkie]

on the question about the galvanic cell, with Pb, Ag and Zn....did anyone else think it should have only been B.)lead only instead of C.) lead and Zinc? I get that they should have negative reduction potentials, so that they will be oxidized and copper will be reduced...but if it has to be galvanic, doesnt the EMF have to be positive for spontaneous reaction? If thats the case shouldnt it just be lead..that would give -.13 + .52= +.39 for EMF, if it was zinc it would be -.76+.52=-.24,...so wouldnt that give a negative EMF, so it wouldnt make a the cell galvanic right? or did i miss something?

making it way too complicated and i think that's why you got it wrong. just look at the numbers, and what E0 means. the way i think about it, E0 is how much something WANTS to get reduced. so Cu sorta wants to get reduced, Ag REALLY wants to get reduced, but Pb and Zn don't want to get reduced (actually, Pb and Zn want to get OXIDIZED). i don't even know what a galvinic cell and electromotive force and all that jazz is and i was able to get this question simply based on that.

 

[Schwann]

did anyone find the verbal hard? I did very well on the previous verbal sections, but on this verbal i for some reason did badly by 3 points!

 

[Lukkie]

._.

the answer selections would have excluded a better answer. for MCAT you have to choose the BEST answer. there might be 2 right answers; one will be better than the other.

in this case, you had to choose between "A, not B or Z" or "Z, not A or B". Z is obviously the best choice since its the closest to the required #. choosing the "A, not B or Z" option would have excluded Z. choosing "Z, not A or B" would have excluded a POSSIBLE answer (A) but retains the BEST answer (Z).

 

[Ari1584]

Can someone please explain a question to me? It is from BS #15 for the discrete part.

#15- Which of the following would be LEAST likely to prevent bacterial synthesis of the superantigen protein?

A) Adding tRNA nucleotides that can bind to mRNA and bacterial ribosomes
B) Adding a repressor protein to bind to operator site of the superantigen gene
C) Adding specific complementary nucleotide sequence that can bind to mRNA transcribed from the superantigen
D) Adding a stop codon within the bacterial superantigen gene

The answer is A, and i got it right, but now that i look at the question again, why couldnt it be C? What makes A a better answer than C, because wouldnt adding a tRNA nucleotides prevent the mRNA from being translated? Just not sure why A is the better answer here...thanks a lot!

 

[Lukkie]

Can someone please explain a question to me? It is from BS #15 for the discrete part.

#15- Which of the following would be LEAST likely to prevent bacterial synthesis of the superantigen protein?

A) Adding tRNA nucleotides that can bind to mRNA and bacterial ribosomes
B) Adding a repressor protein to bind to operator site of the superantigen gene
C) Adding specific complementary nucleotide sequence that can bind to mRNA transcribed from the superantigen
D) Adding a stop codon within the bacterial superantigen gene

The answer is A, and i got it right, but now that i look at the question again, why couldnt it be C? What makes A a better answer than C, because wouldnt adding a tRNA nucleotides prevent the mRNA from being translated? Just not sure why A is the better answer here...thanks a lot!

C would do a VERY GOOD job of preventing synthesis, since the mRNA would not be able to be read, and tRNA would not be able to bind, etc. however, what good would adding tRNA do? very little, if anything. lets assume these tRNA can't pick up AAs. they'd go into the ribosome (a, p sites and all that jazz) and then come off. sooner or later the real tRNA would come in wtih the AA and add to the chain like it normally does.

 

[kish180]

This is a little belated dwc929, but I thought this wasnt explained properly so here it is:

Can someone please explain #101 in Bio for me?

Which of the following alkyl halides is most readily prepared by a reaction between the corresponding alcohol and concentrated hydrochloric acid?

A) Isopropyl chloride
B) methyl chloride
C) sec-butyl chloride
D) tert-butyl chloride

I got this answer wrong too awhile ago and couldn't figure it out at all so i actually emailed the orgo tutor for TPR and what he answered was basically this:
The mechanism for the rxns are Sn1/2. Sn2 is fastest as methyl and Sn1 is fastest as tertiary, so answer is between B and D. The reason why tert-butyl is faster is b.c the sn1 mechanism involves the formation of a planar (sp2) carbocation, which can get attacked from essentially any angle other than the plane its in. On the other hand, Sn2 always has to occur from backside attack. For this reason, the Sn1 mechanism is faster. I know, its pretty ridic for the AAMC to expect you to know or posit this. It is what it is, hope this helped

 

[jayel]

why did this exam have 8 passages in the Bio section? After I finished my seventh section, I was like, wow, I finished early, and clicked next. I almost crapped in my pants when I found out that there was an 8th section.

Will the real thing have 7 passages only? or can they change it up whenever they feel like it? I have a feeling that it's the latter

 

[BerkReviewTeach]

In the dirty-snowball model, does the perturbing force on the comet due to sublimation act in any preferred direction?

A) No, because the nucleus tends to have a roughly spherical surface

B) No, because the sun radiates with equal intensity in all directions

C) Yes, more or less outward from the sun because of shadowing effects
Sunlight is absorbed on one side of the comet. The sublimating gases are preferentially ejected toward the Sun because the solids block the other direction. The force on the comet is away from the Sun. Choice C states this conclusion.


D) Yes, more or less toward the sun because of the temperature gradient

I thought the answer would be B. Where in the passage does it say that the "solids block the other sirection"??

Choice B is a classic wrong answer in that it is a completely true statement that is absolutely irrelevant. They love to include those on most questions, especially in verbal reasoning. The sun may radiate in all directions, but the comet is only struck on the side facing the sun.

In general, I think they expect a certain amount of background knowledge on many of their passages. It is reasonable to expect a student to conclude that something as large as a comet would be opaque (objects of that size are never transparent just due to impurities in the structure adding up over such a large scale). They likely expect that test-takers are aware that there is a "dark side of the moon." By combining the idea of a dark side of a comet with conservation of momentum generating recoil, it is completely reasonable that a student should be able to choose C without the passage.

Besides, that passage was one of the more cumbersome ones, so you probably need to use more intuition than information from the passage to do well. It's a thinking test.

 

[banister]

I found this verbal to be pretty hard. I just took the paper version of this (9 passages) and got destroyed at the end when I realized I was running out of time. Ended up getting an 8 on it, after getting a 12 on CBT3 a few days ago and consistently getting 10-11 on EK101 =(.

did anyone find the verbal hard? I did very well on the previous verbal sections, but on this verbal i for some reason did badly by 3 points!

 

[Deepa100]

Deepa, to see this effect mathematically, take a look at the equation P = IV. From the problem, you know that the power transmitted across the wire is constant. So, if you increase the current, the voltage must decrease, and if you increase the voltage, the current must decrease. In both, situations, however, the power transmitted is the same.

Then you have to know that the heat dissipated depends on the current through the wire, since a higher current leads to heating of the wire.

Thanks, this makes sense.

 

[FutureScaresMe]

Wow, who said verbal was easy? I wanted to shoot myself after reading some of those passages. And then the ice plate one? argggh. At least PS and BS carried me through...

14 PS 9 VR 14 BS

Hopefully I can get that verbal up, but I think I'm pretty much capped at 10.

Question: Does the MCAT have passages with just one question following? I guessed and skipped it over, coming back later. Wasn't worth the time for a passage that looked pretty complicated after reading 2 sentences for just one question...

 

[futuredoctor10]

Question: Does the MCAT have passages with just one question following? I guessed and skipped it over, coming back later. Wasn't worth the time for a passage that looked pretty complicated after reading 2 sentences for just one question...

I have not seen a post on SDN of someone having a one-question passage on a real MCAT, so probably not... but who knows with the AAMC.

 

[ohsmurfyea]

how the uniform acceleration equation does not apply so that t=sqrt 2h/g. instead it seems like it was just Vf = Vi + at to derive time
b/c of that i picked B instead of A.

 

[ksmi117]

how the uniform acceleration equation does not apply so that t=sqrt 2h/g. instead it seems like it was just Vf = Vi + at to derive time
b/c of that i picked B instead of A.

Ok I'm trying to figure out what you're asking here. The answer does come down to a ratio involving the "vf=vi + at" equation. What I'd suggest trying is just setting vi equal to some arbitrary number. I picked 1. We know that vf will be 0 at the top of the projectory. (Although it won't affect this problem, remember it's t/2 to get to the top). Then solve for t for each case. t = 6/5 s when we have g/6 and t = 1/5 s when we have g. So when g/6 the ball takes 6 times as long to get back to its original height.

Just a note: the two balls do not go to the same height, because the acceleration pulling it back down is not the same. So you can't use that first equation.

Does this even answer your question???

 

[ohsmurfyea]

Ok I'm trying to figure out what you're asking here. The answer does come down to a ratio involving the "vf=vi + at" equation. What I'd suggest trying is just setting vi equal to some arbitrary number. I picked 1. We know that vf will be 0 at the top of the projectory. (Although it won't affect this problem, remember it's t/2 to get to the top). Then solve for t for each case. t = 6/5 s when we have g/6 and t = 1/5 s when we have g. So when g/6 the ball takes 6 times as long to get back to its original height.

Just a note: the two balls do not go to the same height, because the acceleration pulling it back down is not the same. So you can't use that first equation.

Does this even answer your question???

Yea thanks. I just assumed that the heights would be the same which is why I used the wrong equation and got the wrong answer!

 

[favored]

Ok, so I'm having some difficulty understanding this question on AAMC Test 6:

Q: Which of the following has the same geometry as methane.

a. H2S
b. CO2
c. XeF4
d. SiCl4 (correct answer)

Methane has a tetrahedral GEOMETRY because geometry considers the number of bonded groups as well as electron pairs, SO my question is why couldn't H2S be the correct answer choice being that there are four groups of electron density surrounding it????

Help!

 

[Marjan Islam]

Yeah, this was a bad question. You take into account electron densities when they're asking electron geometry, but you DON'T take them into account during MOLECULAR geometry. I'm guessing they meant to ask for MOLECULAR geometry. I went with D, because it fits both electronically (there isn't even any electrons on the central atom!) AND molecular.

 

[Vihsadas]

Please do not post questions from the AAMC exams in the main forum. It's unfair to your fellow members who may not have taken the exam yet, as it would ruin the testability of the question. Only post these questions in the relevant OFFICIAL thread in the Study Q&A Forum. Thanks.

http://forums.studentdoctor.net/showthread.php?t=516243

I'll merge this thread with the relevant official one now.

 

[LaCasta]

I am wondering if my CBT is messed up for this one

For passage II of BS, there is only one question showing. It is numbered 99 and then it jumps to 100 on the next passage

I had to pause my test, if anyone could give me a fast reply I would appreciate it