Chem1a Practice Midterm Ii Free Response Solutions

Chemistry 1A, Spring 2009 Midterm Exam II, Extended Practice Free-Response Answer Key 24) Resonance structures of Ozone (O3) O  O  O 

O 

O 

O

The ozone molecule exhibits resonance in its bonds between the oxygen atoms. This would mean that the electrons involved in the second bond of the double bond would be spread over both bonding regions, which gives the molecule a bonding order of 1½. The increased electron density in the regions will decrease the bond length below a single bond’s length, but it will still be still be longer than a normal double bond. Remember that as the bond length becomes shorter, the bond also becomes stronger, which means a triple bond would have a very short bond length. This explains why the bond length in ozone is between the single and double bond lengths for oxygen atoms. 26) Resonance structures of Carbonate (CO32-) 2‐ 

O 

2‐

O 

C 

C

C  O 

O 

O

O 

2‐

O

O

O

The -2 charge allows for two extra electrons in this system which completes the unpaired electrons in two of the three oxygen atoms. Carbon double-bonds with the last oxygen atom to complete its own octet along with the oxygen atom. If you want to check whether your model is correct or not, you would calculate the formal charge of each atom. In this case, carbon has 0 charge, the doublebonded oxygen atom has 0 charge, while the two single-bonded oxygen atoms have a charge of -1. A total of -2, which matches the charge of the molecule, and the charges are on the most electronegative atoms, which are the oxygen atoms. The overall partial charges on the atoms also seem to be reduced to the fullest extent, so this would be the correct model. The double bond is then represented in the three separate locations. (Remember that bonds are actually “hybrid” combinations between a single and double bond.) 27) Resonance structures of Formate (HCO2-) ‐ 

O  C  O 

‐

O C

H 

O 

H

Basically the same as with the carbonate ion. You’ll have one extra electron in this system due to the -1 charge, which would finish one of the oxygen atom’s octet. The carbon would double-bond with the other oxygen atom to finish off both their octets. The double bond is represented in two different locations, with hydrogen being unable to double-bond due to its duplet.

28) Resonance structures of Oxalate (C2O42-) O 

O  C 

2‐ 

O 

C 

O 

O C

C  O 

2‐

O 

O

The oxalate ion is a bit different. First off, there are only two resonance structures. If you thought there were three or four, then most likely you were thinking of the first one flipped around or the double bonds being pointed upward or downward. The problem with those structures is that the single bond between the two carbon atoms can rotate (unlike double bonds which prevent rotation). So basically, if the carbon atoms rotate once, the third “resonance structure” will be formed, and if only one side rotates, then the fourth “resonance structure” will appear. However, since these were made simply by rotation, they do not count as resonance structures. 60)

Lewis Dot Structures

SF6

BrF5 F

F  F 

F 

F 

F 

F

F  F

F 

F

F 

F

Cl 

Cl

Cl  Cl 

Cl 

Cl P

Cl Cl  F

Cl P 

F F

F 

Cl 

F

Br

S 

Cl 

Cl F

F

F  F 

F

Br

F 

F 

F

F

F  S 

PCl6-

ClF3

Cl

Cl

No Dipole No Dipole These are all octet rule breakers and their structures are shown as such. When finding dipole moments, you should always imagine the molecules in 3-dimensions, in their VSPER molecular shapes. The above structures are pretty much arranged as their 3-D structures would look like, and the dipole moment arrows represent the “net” dipole moment, which is basically the sum of all the individual dipole moment arrows. When you find the dipole moment, draw the molecule roughly in 3-D and draw the corresponding arrows where the point is towards the more electronegative atom. Those that are equal in magnitude (that is the elements involved in the dipole moment are the same for both arrows) and opposite in direction, they cancel each other out. Make sure this is in 3-D and not 2-D because that can really confuse you. For example the SF6 molecule would have two individual dipole moment arrows coming out the top and bottom since F is more electronegative, but since they’re dealing with the same elements (same electronegativity difference, same magnitude) and they’re in opposite directions, they’re cancelled out. Same goes for the two pairs in the square planar region of the molecule. If they’re not cancelled out, like for one of the dipole moments in the square planar region for BrF5, the overall dipole moment would be in that direction. If there are multiple dipole moments that are remaining, then you will average them accordingly, and find out the relative direction if they were added together (in terms of opposite directions means subtraction, same directions means addition).

81) NO

4σ*

2p

 

 

2π* 

 

 

E

 

2p

3σ 1π 

 

2s

2σ*

 

2s

1σ Nitrogen Atomic Orbitals

Molecular Orbitals

Oxygen Atomic Orbitals

For the molecular orbital energy diagram, you basically need to understand how it’s completed. First off, you should complete the two atomic orbital by filling in the orbital with the number of valance electrons the element has. In this case, N has 5 and O has 6. You also fill them in the same way you did for the boxes in a row, where when there are multiple orbitals (boxes), you fill them up with 1 first before going back to pair them. Now for the molecular orbitals, you take the total number of electrons (11), and fill from the bottom up, using the same methods. Your molecular orbitals will generally not be completely filled up and that’s perfectly fine. You don’t need to memorize the energy diagrams for the elements as it will be provided on the midterm. 82) NO is paramagnetic because there is an unpaired electron in the 2π* orbital. Paramagnetism is defined as having a magnetic characteristic, which is only possible if there are unpaired electrons that are able to provide the magnetic effect. Diamagnetic would be the opposite where there is no magnetic character because all the electrons are paired. 83)    

 

  # 

 

# 

 

       

 

     

#    2 #    2 8 3 5   2 2

     

 

 

 

84)

N 

O 

NO is a radical, which means it has an unpaired electron making it very volatile, since any compound with an unpaired electron can make it spontaneously (on its own) react. If a molecule has all their electrons paired up, then generally it will not exhibit such a reactive behavior. This is generally why NO is found in the form of NO-, where there is one extra electron that pairs to finish nitrogen’s octet. 85) HF

σ* E

1s

 

σ 2s (n.b.) Hydrogen Atomic Orbitals

2p

 

2px,2py (n.b.)

 

n.b. - non-bonding

 

2s Fluorine Atomic Orbitals

Molecular Orbitals

Same as with NO, fill the atomic orbitals with their corresponding number of valence electrons and then fill the molecular orbitals with the total number of valence electrons. Note that nonbonding orbitals still have to be filled, but play no role in the bonding between the two atoms. 86) HF is diamagnetic because it has no unpaired electrons. See problem 82 for definitions. 87)  

2 1 2 2 Note that this questions tests whether you understand how to calculate bond order. Only electrons that contribute to the attraction (bonding) or repulsion (antibonding) of the atoms are counted for the bond order. So those electrons in the non-bonding orbitals will play no role in determining the strength of the bond. Thus, the total bond order is 1, which is expected, since in the lewis dot model, hydrogen is bonded to fluorine with a single bond.  

 

2

0

88) The energy of the atomic orbitals deal with the magnitude of their electronegativity. The more electronegative the element is, the less energy is has. This linked to the atomic radius of electronegative elements, because as electrons are held closer to the nucleus, their kinetic energy (the energy involved in motion) is reduced, because the nucleus exerts a greater attraction on the electrons, which also is the reason why electronegative elements are able to capture electrons so efficiently. Since the kinetic energy is reduced, the overall energy of electronegative atoms are reduced, which is why they contribute most to the lower energy orbitals, which are generally bonding, while the least electronegative atom contributes more to the higher energy orbitals, which are generally antibonding. Another perspective is that electrons are the ones that contribute to the bonding effect between atoms, so the electronegative atom is attempting to attract electrons from the other atom, which effectively contributes to the bonding. The less electronegative atom generally has a lower ionization energy, but since it needs energy to lose an electron, it will not willingly release it, causing the antibonding effect. 89) In ionization, the electron with the higher energy (in terms of becoming less negative, furthest away from the nucleus) will be ejected. So from the energy diagrams, we would remove one electron from the last filled molecular orbital, not the atomic orbitals, because this would imply that the atoms themselves had ionized before they even bonded (or not bonded in the case of 0 bond order). So in NO’s case, it would have loss an electron from its 2π* molecular orbital, which is antiboding. The removal of an antibonding electron strengthens its bond order to 3 (8-2/2). While in HF’s case, an electron is removed from a non-bonding orbital, which already has no effect on the bonding order, so the bond order is not effected. Stability is measured in terms of the strength of the bond order, and hence, NO is more stable after undergoing ionization. 90) (This question was actually improperly worded, I apologize for that) In electron capture, each molecule gains one extra electron. So as above in ionization, an electron is added to the molecular orbitals. So for NO, an electron is added to its 2π* molecular orbital, which decreases its bond order from 2½ to 2, due to the extra antibonding electron. For HF, an electron is added not to its non-bonding orbital, since it’s already filled, but to its σ* molecule orbital. This also decreases its bond order from 1 to ½. Unfortunately, due to the poor wording of this question, it can be interpreted as which molecule has the shorter bond after electron capture, or which molecule has a shorter bond after electron capture, so I will answer both. Bond length reflects how strong the bond actually is, whereas the shorter the bond, the stronger the bonding, which may be counter-intuitive. This is because the shorter distance between the two atoms allows the nuclei to attract the electrons of the other atom more effectively, also increasing the electron density in the bond. This means that a single bond has the longest bond length, followed by a double bond, with a triple bond having the shortest bond length. So the molecule with the shorter and stronger bond will be NO due to its bond order of 2 compared to ½. In terms of the other question, both molecules experienced a lengthening of their bonds since their bond orders both decreased, so it was a trick question. 91-92) I had noticed that these were improper questions to ask since I had forgotten that I had use an molecule with non-bonding orbitals, which makes these comparison a lot more difficult to comprehend. Hopefully, questions of this difficulty will not show up on the midterm.

93) Note that temperature and the moles of gas stay constant, so deriving from the ideal gas law.   ,           Thus, the product of the pressure and volume in the initial system, and the product of the pressure and volume in the final system must equal the same constant. So… 1.21

4.35

6.54 1.82

94) In this problem, the moles of gas were given, but since the gas is in a closed container, there is no change in moles. The volume is also constant. So deriving from the ideal gas law again:         Thus, the ratio of the pressure to the temperature of the gas is related to a constant, in which the ratio of the final pressure and temperature must equal the same constant, so…

1.32

0.95 253 253

1.32 0.95 352

95) In this problem, we have three variables, pressure, volume, and temperature. Only the moles of gas stay constant, so deriving from the formula.    

0.932

168

1.01

142

24° This is incorrect! The temperature must be converted to Kelvins (K) in all chemistry related problems. The relation is K=˚C+273.15 1.01 142 0.932 168 24 273.15 272 96) Similar to the problem above, except this time you’ll be solving for the final volume rather than temperature.

Convert temperature in Kelvins!!! 1.21 1.02 3.23 35 273.15 20 273.15 2.86

98) In this problem, you’ll be approximating the extinction coefficient with Beer’s Law with only one data point. This is obviously inaccurate compared to the multiple readings, but in this context of the problem, you can assume that it’s only an estimation. ·ℓ·       ,         , ℓ      ,       You should have this formula memorized since it may not be on your formula sheet. So you’re given the absorbance of a sample along with its concentration. The length is assumed to be 1cm, thus you already have your numbers. 0.43 0.37 · 1 · 0.86 Note that absorbance is unitless and the extinction coefficient has the units that cancel out the units of length and concentration. 99) In the experiment, you had set the spectrometer to measure the absorbance at 310nm, which is in the region of ultraviolet light, perfect for testing sunscreen. However at 580nm, you’re in the region of visible light, to be precise, yellow light. While sunlight is generally colored yellow in drawings, especially in kindergarten, you’re not testing for whether the sunscreen absorbs visible light. Basically the answer is that you’ll get the absorbance of visible light, rather than ultraviolet light. 100) This question mainly asks, “Were you paying attention to what you were doing in lab?” When you used the spectrometer, you were asked to put a blank solution of just the solute 2-propanol into the spectrometer. Why was this done? A spectrometer works by emitting a set amount of photons of the specific wavelength through the samples and recording how many actually went through, which is your absorbance value. However, what if the light was not “absorbed”, but rather blocked? Although liquids seem to be transparent, they’re still made out of solid particles, which can indeed block the photons. Another characteristic is that the liquid can actually bend the path of photons so that they never reach the recording region. By placing a sample of the blank solution, you’re basically calibrating the spectrometer to make up for all those errors. If the absorbance of the blank solution isn’t 0, which means all light passed through, then the measurement now becomes the “0” point to make up for the light lost through various sources. This dramatically improves the accuracy of your results, especially when using solutes that already have a bit of absorbance. 101) O 

O  102) In the experiment “How the Nose Knows”, you were investigating the reasons why molecules smelled as they do. What you should have found out what that functional groups and atomic orientation in the molecules play a big role in determining their smells. For this question, an ester was present, which is basically a double bonded oxygen along with a oxygen bonded in the chain nearby. Esters generally do smell fruity, and are often in perfumes. Depending on what is given, try to find the compound that matches the particular function group. Don’t be fooled! There are functional groups that look very similar!

103) Acid-Base titrations were done in this lab, which have the goal of finding the concentration of an unknown acid or basic solution. In this particular problem, you’re attempting to find the concentration of an unknown acidic solution, just as you had done in lab with HCl. The unit of concentration is molarity, which is defined as the ratio of moles to liters. So basically, you’re attempting to solve for the moles of the acidic solution over the liters of the acidic solution. In a titration, the endpoint is defined as the moment when the solution is neutralized, where the moles of acid is the same as the moles of base. This moment is generally visually apparent due to an indicator, which was BCG. This indicator is yellow in the presence of an acidic solution, green in a neutral solution, and blue in a basic solution. So from the information, this point occurred when 26.35mL of the unknown acidic solution was added to the basic solution made from 0.2431g of TRIS and 45.63mL of distilled water. That’s your first piece of information, the mL of acid can be changed into liters of acid by dividing by 1000. Your second piece of information, the moles of acid, relies of the purpose of an acid-base titration. As I said before, at the endpoint, the moles of acid is equivalent to the moles of base, so the moles of TRIS = moles of acid. By calculating how many moles of TRIS you used to create the basic solution, you can find out the moles of acid at the endpoint. Note that this doesn’t depend on the amount of distilled water you used! That’s why the experiment only required you to put about 50mL rather than exactly 50mL. The amount of distilled water does not affect the moles of TRIS. It’s only there to let you clearly see the color change. So the numbers are all here, the calculation is shown below: 26.35       0.02635       0.2431   0.0020    121.14   0.002 

 

:

 

:

 

    0.002    0.0020 0.02635

      0.0760

0.0760

104) The question is basically asking, if you increased the moles of TRIS, what happens to the calculated concentration of the acidic solution? Thus, recalculate the moles of TRIS 0.3431   0.0028    121.14 0.0028 0.106 0.02635 Wow, that’s quite an error. Hopefully you didn’t make the same mistake!  

 

 

:

0.1063

105) This is basically the same concept as above, except you’re trying to find the concentration of the basic solution rather than an acidic solution. The titration problem is calculated the same both ways. You’ll need to find the moles of base and the liters of base. The latter was given to be 42.23mL, so the remaining value is the moles of base. As stated before, moles of acid=moles of base at the endpoint. So if you solve for the moles of acid, you’ll be able to find the moles of base at the endpoint. This is a slightly different calculation since you’re given the concentration of the acid, which you basically multiply with the liters of acid you used to find the moles of acid that were present. Calculations shown below: 42.23       0.04223       0.43 0.43 0.43   0.03 0.0129        0.0129 

 

 

:

 

:

    0.0129    0.0129 0.04223

 

 

  0.3055

0.3055

106) If you had titrated incompletely as described in the problem, your volume of the basic solution would have been smaller than its true amount. So if you had used calculated the concentration with a slightly higher volume of basic solution, you’ll get a smaller concentration. So what you calculated initially was higher than the true value. 107) This is a calculation you did in your lab report as well. What you can take advantage of is that the moles of the base would not change no matter what its volume is. The only thing that changes with volume is the concentration of the basic solution. So we use the diluted base concentration calculated before: 0.3055M, and use the following formula: What this basically means is the moles of the stock solution = the moles of the dilute solution, since M, concentration, is measured in moles per liter, while volume is measured in liters that cancel that particular unit out. We’re solving for Mstock, so we still need Vdilute and Vstock. Vdilute is basically how much of the diluted solution you made, which is limited by the volumetric flask, which has a volume of 300mL. Vstock is the volume of the stock solution you used to make that diluted solution, which was given to be 7.5mL. Solve! 300 7.5 0.3055 1000 1000 0.3055 300 12.22 1000 12.22 7.5 1000 Wow, that’s a very concentration stock solution. 108)   ∑ #   

| 3.243

|

100%

3.302 3.221 3 |3.243 3.255|  1 %  100% 3.255 |3.302 3.255| 100%  2 %  3.255 |3.221 3.255|  3 %  100% 3.255 That’s not bad in terms of precision, but it could be better!

3.255 0.37%  1.44% 1.04%