Chapter 3: Bonding

[bradley.hu5547]

In question 3 HBr has a dipole moment of 0.75 Debye. The solution to this question does not take into account the 3.34x10^-30 if charge is in coulombs and distance is in meters. Is this a mistake or am I missing something.

As a side note question 6 has 3 answers which refer to chlorine instead of bromine. Is this a mistake?

[goldstanda3269]

In question 3 HBr has a dipole moment of 0.75 Debye. The solution to this question does not take into account the 3.34x10^-30 if charge is in coulombs and distance is in meters. Is this a mistake or am I missing something.

You are correct. And strangely, the answer was also basically correct. The problem was that the question stem had an inflated number for the charge. This has been updated. You can try the question again to see if you get the correct answer. The worked solution has also been updated.

As a side note question 6 has 3 answers which refer to chlorine instead of bromine. Is this a mistake?

Corrected! This was one of our brand new units so there were a couple of bumps. Thanks for the alert.

[elena_333]

Foundational MCQs Q2: Why can't N have 4 bonds and be neutral (as in B)? If it has 5 valence e-, couldn't it form 5 bonds? And similarly, why doesn't it have a charge when bonded only 3 times (as in C)?

Foundational MCQs Q15: I thought it would be formal charge = 5 - (1+4) = 0, as N has 5 valence e- and is bonded 4 times so has 1 non-bonding e-. Why is it +1?

Thanks

[goldstanda3269]

Foundational MCQs Q2: Why can't N have 4 bonds and be neutral (as in B)? If it has 5 valence e-, couldn't it form 5 bonds? And similarly, why doesn't it have a charge when bonded only 3 times (as in C)?

Of course, I'm keeping this explanation at GAMSAT level because this matter can get complex quickly and no longer relate to this exam. So, these are the key points: 1) a covalent bond includes 2 electrons, in a normal covalent bond each atom contributes 1 electron to the bond; 2) many common atoms follow the octet rule, the atom is more stable when we can draw a Lewis structure surrounded by 8 electrons; 3) so, as you noted, N has 5 valence electrons so it needs 3 more for the octet and that directly relates to the 3 bonds (recall C has 4 valence and bonds 4 times to be neutral, O has 6 valence and bonds 2 times to be neutral like H2O).

Foundational MCQs Q15: I thought it would be formal charge = 5 - (1+4) = 0, as N has 5 valence e- and is bonded 4 times so has 1 non-bonding e-. Why is it +1?

I believe this is explained above. Please consider looking at the videos on gamsat-prep.com like the 2 for the Periodic Table, and then . . .
Chemical Bonds
Hydrogen Bonds
Ionic and Covalent Bonds
***Lewis Dot Structures
Multiple Bonds

[elena_333]

Continuing from my last question about Q2, For A why is N positive and O negative though? If N is neutral at 3, when it has 4 bonds hasn't it got more than a full outer shell so shouldn't it be negative?

[goldstanda3269]

Continuing from my last question about Q2, For A why is N positive and O negative though? If N is neutral at 3, when it has 4 bonds hasn't it got more than a full outer shell so shouldn't it be negative?

Here is the easy answer:

If N is bonded 3 times, it is neutral; 4 times, it is positively charged; 2 times, it is negative. This is the routine for the foundational understanding for this exam. During your GAMSAT practice, you will see ammonium (NH4+) dozens of times, it is nitrogen bonded 4 times and thus it is positive.

If N needs to add an extra bond (beyond 3) it needs to give up an electron for that bond and so it becomes positive: https://i.stack.imgur.com/Z2n1k.png

Now for the explanation that you were seeking (which is really more detail than you need for this exam): keeping in mind that nitrogen has 5 valence electrons and oxygen has 6 valence electrons, you can watch the following video (again, this is a detail beyond the GAMSAT): https://www.youtube.com/watch?v=3gRFffBhdrQ

[horbowiec.7527]

This is regarding Worked solution for GEN CHM Chapter 3, Question 13 (CHM-83):

I am not understanding your answer for question 13. You say "only shape consistent with 5 other bonds from Table 1 is trigonal planar". Isn't it trigonal bipyramidal since that is included as the correct answer?

Additionally, I chose seesaw because trigonal bipyramidal is under "electron geometry" so I assumed "molecular geometry" with seesaw as my answer is what is referred to as "shape of phosphorus pentachloride". Where in the passage states that electron geometry = shape of molecule and thus making answer C the only correct answer?

Regards,
Ania

[goldstanda3269]

This is regarding Worked solution for GEN CHM Chapter 3, Question 13 (CHM-83):

I am not understanding your answer for question 13. You say "only shape consistent with 5 other bonds from Table 1 is trigonal planar". Isn't it trigonal bipyramidal since that is included as the correct answer?

Correct, "trigonal bipyramidal" was the intent. The typo will be corrected within 24 hours.

Additionally, I chose seesaw because trigonal bipyramidal is under "electron geometry" so I assumed "molecular geometry" with seesaw as my answer is what is referred to as "shape of phosphorus pentachloride". Where in the passage states that electron geometry = shape of molecule and thus making answer C the only correct answer?

Side note: It is suspicious that both seesaw and T-shaped are in the Molecular Geometry column and potential answer choices. However, in neither case is the central atom connected to 5 atoms like PCl5.

From the passage: “The presence of lone pair electrons can affect molecular geometry.” The implication is that in the absence of lone pair electrons (i.e. PCl5), the molecular geometry is unaffected from the original geometry (i.e. the electron geometry).