Module 1
· Question 1
0 out of 1 points
Consider the following sets of quantum numbers. Circle all of the allowed sets of quantum numbers for an electron in a hydrogen atom. The electron may be either in a ground state or an excited state. Mark all that apply.
Recommended time: 1 minute
Selected Answers:
[None Given]
Answers:
n=4, l=3, ml =-3, ms=+1/2
n=3, l=-1, ml =-1, ms=-1/2
n=0, l=0, ml =0, ms=+1/2
n=2, l=1, ml =-1, ms= -1/2
n=3, l=0, ml =1, ms=-1/2
n=2, l=0, ml =0, ms=+1/2
Response Feedback:
Look at the rules for electron quantum numbers in Table 6.1.
n=3, l=-1, ml =-1, ms=-1/2 --Violates the rules because l cannot be negative.
n=3, l=0, ml =1, ms=-1/2 --Violates the rules because ml cannot be greater than l.
n=0, l=0, ml =0, ms=+1/2 --Violates the rules because n must be 1 or greater, and l cannot be greater than n-1.
· Question 2
0 out of 2 points
What is the best estimate for the C-O-H bond angle in ethanol, CH3CH2OH?
Recommended time: 2 minutes
Selected Answer:
[None Given]
Answers:
Slightly less than 109.5o
109.5o
Slightly greater than 109.5o
Slightly less than 120o
120o
Slightly greater than 120o
Slightly less than 180o
180o
Slightly greater than 180o
Response Feedback:
If you do the Lewis structure, you find that that there are two lone pairs of electrons on the oxygen as well as the C-O and O-H bonds, so four electron regions. The region geometry is tetrahedral, so the angles are near 109.5. But the electron pairs are more repulsive than the bonds, so the C-O and O-H bonds are pushed away from the electron pairs and toward each other. That makes the angle a bit less than 109.5 degrees.
· Question 3
0 out of 2 points
Which of the following molecules possess trigonal planar molecular geometry? Check all that apply.
Recommended time: 3 minutes
Selected Answers:
[None Given]
Answers:
H2S
NH3
CH2S
C2H2
H3O+
None of these.
Response Feedback:
H2S has the two S-H bonds and two lone pairs on the S. That means there are four electron regions on the S, so the electron region geometry is tetrahedral and the molecular geometry is bent.
NH3 has three N-H bonds and one line pair on the N. That makes four electron regions again, so the electron region geometry is tetrahedral and the molecular geometry is trigonal pyramidal.
CH2S has a C=S double bond and two H-C bonds from the ca
on. That makes three electron domains, and trigonal planar geometry. You may be confused, thinking the fact that one of them is a double-bond, but it still counts as trigonal planar. The double bond is more repulsive than the single bonds, so the H-S-H angle is a little less than 120o and the S-C-H angle is a bit more than 120o, but the label "trigonal planar" still applies.
C2H2 is acetylene. The Lewis structure goes H-C=C-H, meaning there's a triple bond between the C's. There are only 2 electron regions around each ca
on, so they are linear. In fact the whole molecule turns out to be linear.
H3O+ has a lone pair of electrons on the O, along with the three bonds. The electron region geometry is tetrahedral, and the geometry is trigonal pyramidal. The ion is actually isoelectronic with NH3.
· Question 4
0 out of 1 points
Which ion below has the largest radius?
Recommended time: 2 minutes
Selected Answer:
[None Given]
Answers:
Cl -
K +
Br -
F -
Na +
Response Feedback:
F-, Cl-, and Br- are all elements in the same group with the same charge. The one with the most electrons will have the largest radius, so Br- is largest.
Na+ and K+ have the same relationship, and K+ has more electrons, so it is larger.
Comparing K+ and Br-, K+ is isoelectronic with Ar and Br- is isoelectronic with Kr. That means the valence shell for K+ has n=3, and for Br- it is n=4, so Br- is larger.
Note that if you compared neutral K and Br, K would be larger because of its smaller effective nuclear charge. But that's not what was asked in this question.
· Question 5
0 out of 2 points
Which of the answers given below is the best estimate of the angle theta in the figure below?
Recommended time: 2 minutes
Selected Answer:
[None Given]
Answers:
Slightly more then 109.5o
Exactly 120o
Slightly more than 120o
Slightly less than 120o
None of these is close.
Slightly less then 109.5o
Exactly 109.5o
Response Feedback:
There are three bonding domains and zero nonbonding domains around the central ca
on, meaning it has a trigonal planar geometry. Since one of them is a double bond containing four electrons, it is more strongly repulsive than the single bonds, which each contain only two. That means the single bonds will be pushed away from the double-bond, forcing them closer to each other. Since the angle specified is between the two single bonds, the total angle will be slightly less than 120 degrees.
· Question 6
0 out of 2 points
What is the best estimate for the C-N-C bond angle in the molecule shown below? Note: Bonds are shown, but lone pairs are not included.
Recommended time: 2 minutes
Selected Answer:
[None Given]
Answers:
Slightly less than 109.5o
109.5o
Slightly greater than 109.5o
Slightly less than 120o
120o
Slightly greater than 120o
Slightly less than 180o
180o
Slightly greater than 180o
Response Feedback:
The N has a lone pair of electrons on it, so four electron regions. That means the angle has to be near 109.5 for the bonds, except that they are pushed slightly away from the more repulsive lone pair. So, their angles are a bit less than 109.5.
A ring like this looks intimidating, but the truth is that rings play by the same bond-angle rules as other types of molecules except in very special circumstances we'll learn about later.
· Question 7
0 out of 1 points
Which electron configuration represents a violation of Hund's rule for an atom in its ground state?
Recommended time: 1 minute
Selected Answer:
[None Given]
Response Feedback:
Recall that Hund's rule requires that electrons be put into o
itals with their spins aligned until it is impossible to do so. For something like an s-o
ital, the first one goes in spin-up and the other spin-down (or vice versa). But for degenerate o
itals, like p- or d-o
itals, you need to fill them all with spins aligned before adding any electrons with opposite spin.
· Question 8
0 out of 1 points
An electron cannot have the quantum numbers n = __________, l = __________, m l = __________.
Recommended time: 2 minutes
Selected Answer:
[None Given]
Answers:
2, 0, 0
2, 1, -1
3, 1, -1
1, 1, 1
3, 2, 1
Response Feedback:
Only certain combinations of quantum numbers for an electron are allowed. Have a look at Table 6.1 in the Flowers text to see how the quantum numbers are connected to each other.
· Question 9
0 out of 1 points
Which electron configuration represents a violation of the Pauli exclusion principle?
Recommended time: 1 minute
Selected Answer:
[None Given]
Response Feedback:
The Pauli exclusion principle fo
ids two electrons from having the same set of quantum numbers. Two atoms can sit in the same o
ital if they are spin-up and spin-down, since that means their spin quantum numbers are different (+1/2 and -1/2). But in the violation above, there are two spin-up electrons in the same o
ital, meaning they have the same o
ital quantum numbers (n, l, ml) and the same spin quantum number.
· Question 10
0 out of 1 points
Of the following species, __________ has the largest radius.
Recommended time: 2 minutes
Selected Answer:
[None Given]
Answers:
Rb +
Sr 2+
Br -
K
A
Response Feedback:
Kr and Ar are both noble gases, but Kr is in a lower period so it must have a larger radius.
The remaining atoms and ions are all isoelectronic, meaning they have the same number of electrons. Since the number of electrons is the same, the only thing determining their difference in radius is how strongly the nucleus pulls on them. The stronger the pull, the lower the radius. Since the Br nucleus has the lowest charge (fewest protons), it draws the electrons the most weakly and so has the largest radius.
· Question 11
0 out of 1 points
The __________ o
ital is degenerate with 5py in a many-electron atom.
Recommended time: 1 minute
Selected Answer:
[None Given]
Answers:
5s
5p x
4p y
5d xy
5d 2
Response Feedback:
In a many-electron atom, o
itals with the same values of n and l are degenerate (meaning equal in energy), but different values of n or l lead to different values of energy. The 5p-o
itals both have n=5, l=1 (if you are not sure why this is true, look back at Section 6.3). Since the other o
itals in the list have different n,l values, they are not degenerate.
· Question 12
0 out of 2 points
Which of the following molecules are polar? Check all that apply.
Recommended time: 3 minutes
Selected Answers:
[None Given]
Answers:
CS2
CCl4
C2Cl4
F2
XeF4
None of these are polar.
Response Feedback:
It is important to do Lewis structures for each molecule.
CS2 has only two electron domains around the ca
on, meaning it is linear. That means the dipole moments of the bonds point in opposite directions.
CCl4 has 4 single bonds, so is tetrahedral. That means that it is nonpolar for the same reason methane is (see course notes).
C2Cl4 has a double bond between the ca
ons and a trigonal planar geometry for each ca
on center. That means each Cl-C-Cl a
angement makes a triangle with a dipole pointing from the ca
on to a point halfway between the two chlorines. Since there are two triangles with their points in