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AAMC Chemistry Practice Set Q.72
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Anastasia_6156
#1 Posted : Monday, May 17, 2021 9:54:38 PM
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Question 72 asks about the least polar compound. I do not understand why is it CBr4 - I don't think it was on a lecture?..
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Youssef_6546
#2 Posted : Tuesday, May 18, 2021 5:18:12 AM
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Hi Anastasia,

For question 72, AAMC is assessing our understanding of polariity and what makes a compound more polar than the other. Yes, we didn’t specifically discuss CBr4 in lecture, but we did learn how to determine a compound’s polarity based on the net dipole of the compound’s atoms (page 51 of ChemOchem Class Companion). CBr4 has a tetrahedral shape like CCl4 (as shown in page 51). With this shape, the individual dipoles cancel out which gives a net dipole moment of zero. Hence CBr4 is non-polar and the least polar among the other three options in the question

Hope these help!
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INSTR_Katrina_128
#3 Posted : Thursday, May 20, 2021 10:18:32 PM
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Great explanation, Youssef!

I want to mention that the OVERALL polarity of the molecule CBr4 is 0.

However, each bond of C---Br is (slightly) polar.

What's the difference or distinction in these statements?

Stretch your two arms out beside you, and imagine that you have two friends (wow... friends... remember those? 😥) tugging on your arms. If they are both EQUALLY strong, you will feel an equal-and-opposite balanced force on your arms, and you won't move anywhere. This is what C-Br4 feels. It has FOUR ARMS, and feels an equal and opposite force in 4 different directions. It's not moving anywhere (i.e., it's not a polar molecule), but each of its individual arms FEELS a pulling force, so each individual ARM is (slightly) polar.

I say *slightly* polar because the electronegativity difference (a way that we can quantify polarity) between C and Br is small (about 0.3).
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