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This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points Which of the following types of electromag- netic radiation (EMR) has the greatest energy per photon: infra-red, ultra-violet, radio, or visible?
- They all have the same energy because they all travel at the same speed.
- radio
- visible
- infra-red
- ultra-violet correct
Explanation:
002 10.0 points
Which type of electromamgnetic radiation excites inner shell electrons?
- ultraviolet light CORRECT
- visible light
- microwave radiation
- infrared radiation
Explanation:
The other 3 answer choices interract with matter in other ways, not with inner shell electrons.
003 10.0 points The observation and characterization of black body radiation was key to the devel- opment of our quantum mechanical view of the atom. Which of these statments is correct regarding blackbody radiation?
- The emission from a blackbody consists of a series of discrete spectral lines, the colors of which are unique to the type of blackbody used.
- The emission curve for a blackbody is a broad lopsided hump with a peak that shifts to shorter wavelengths as the temperature of the blackbody increases. CORRECT
- Theoreticians when trying to model the blackbody emission were unable to explain why they emitted so much UV light at room temperature, this difficulty became known as the ”UV Catastrophe”.
- The emission of radiation from an in- candescent lightbulb does not resemble the emission of radiation from a blackbody.
Explanation:
A blackbody’s radiation emission curve is broad, lopsided and has a peak which shifts to shorter wavelengths as the temperature of the blackboady increases. Theoreticians tried to model this using classical mechanics and never could get their models to behave be- cuase they predicted vast amounts of UV, gamma, etc. being produced by a blackbody above absolute zero - the ”UV Catastrophe”. A plain old incandescent lightbulb with a dim- mer switch can be used to reproduce the emis- sion curves of a blackbody - and even vary its temperature by changing the current with the dimmer switch.
004 10.0 points
005 10.0 points What is the de Broglie wavelength of a bowl- ing ball rolling down a bowling alley lane? Assume the mass of the ball is 5000 g and it is moving at 6.63 m/s.
- 2 × 10 −^35 m correct
- 1 × 10 −^40 m
- 2 × 10 −^38 m
- 5 × 10 −^32 m
- 5 × 1037 m
Explanation:
006 10.0 points Polonium 210 (^210 Po) is a highly radioactive metal found in high concentrations in most tobacco leaves. It undergoes radioactive de- cay, releasing an alpha particle, which is de- tected as it strikes a metal surface. The device used to measure the position is only sensitive down to 1 ˚A. What is the minimum uncer- tainty in the alpha particle’s velocity upon impact? (An alpha particle has a mass of
64 × 10 −^27 kg. 1 ˚A = 10−^10 m)
160 m · s−^1
800 m · s−^1
80 m · s−^1 correct
8 m · s−^1
Explanation:
∆v ≥
h 4 πm∆x
- 626 × 10 −^34 J · s 4 π (6. 64 × 10 −^27 kg) (10−^10 m) ≥ 80 m · s−^1
007 10.0 points Consider a 1-dimensional system of length 160 nm. If an electron in this ”box” is in principle energy level 4, at what positions along the length of the box will there be zero probability of finding the electron?
- 40 nm, 80 nm, 120 nm correct
- 60 nm, 100 nm
- 20 nm, 60 nm, 100 nm, 140 nm
- 80 nm
Explanation: For principle energy level 4, the particle in a box will have 3 nodes at
and
the length of the box.
008 10.0 points Which of the following statements about the Schr¨odinger equation is incorrect?
- There are both kinetic and potential en- ergy terms in the equation
- The classical standing wave equation so- lution for a guitar string yields the same solu- tion as the particle in a box case.
- The hydrogen atom solution is simplified by changing to spherical coordinates.
- The principle, angular momentum and magnetic quantum numbers yield information about energy (size), shape and orientation of the electron orbits.
- Setting the potential energy term to zero allows the accurate calculation of wave equa- tion solutions for the hydrogen atom. cor- rect
- [Kr] 5s^2 5 d^5 5 p^3
- [Kr] 5s^1 4 d^9
- [Kr] 5s^2 4 d^8
Explanation: In3+^ is isoelectronic with Pd (it has 46 elec- trons). The enhanced stability afforded by a filled d subshell results in a both 5s electron being “borrowed” to fill In3+’s 4d subshell; therefore, In3+^ is [Kr] 5s^0 4 d^10.
013 10.0 points What is the effective nuclear charge expe- rienced by the 2p and (3p) electrons of an Chlorine atom (Cl), respectively?
- 17, 10
- 11, 7
- 17, 7
- 15, 7 correct
- 7, 2
Explanation: Electrons within the same energy level do not shield one another. Therefore the 2s elec- trons are not shielding the 2p electrons. To calculate effective nuclear charge the number of shielding electrons (all electrons in lower energy levels) are subtracted from the actual nuclear charge (number of protons in the nu- cleus). Chlorine, atomic number 17, has 17 protons. All of its electrons in principal en- ergy level 2 are shielded only by the electrons in principal energy level 1, i.e. the 2 1s elec- trons. 17 - 2 = 15 and for 3p 17 - 10 =
014 10.0 points From the data below
Element First Ionization Energy 1 1310 kJ/mol 2 1011 kJ/mol 3 418 kJ/mol 4 2080 kJ/mol 5 947 kJ/mol
which element is likely to be a metal?
- 3 correct
- 1
- 2
- 5
- 4 Explanation: Metals form positive ions; therefore they are likely to have relatively low first ionization energies. (It will be easy to remove the first electron.)
015 10.0 points For an isoelectronic series of ions, the ion that is the smallest is always
- the most negatively charged ions.
- the ion with the greatest number of p orbitals.
- the ion with the highest atomic number. correct
- the ion with the fewest protons.
- the ion with the most electrons.
- the ion with the most neutrons. Explanation: Isoelectronic ions all have the same num- ber of electrons. However, the one with the highest atomic number will have the largest effective nuclear charge.
016 10.0 points
Which of the following groups of atoms is correctly arranged in order of decreasing first ionization energy?
- Rh > Ag > Pd > Cd
- Cd > Ag > Pd > Rh
- Cd > Pd > Ag > Rh correct
- Rh > Pd > Ag > Rh
Explanation: Ionization energy increases as one moves from left to right on a given row and increases from bottom to top as one moves up a given family. In the case of Rh, Pd, Ag, and Cd, Pd has a higher ionization energy because re- moving an electron moves the electron config- uration from the stable filled d configuration.
017 10.0 points Write the ground-state electron configuration of a chromium atom.
- [Ar] 4s^1 4 d^5
- [Ar] 4s^2 4 d^4
- [Ar] 3d^5 4 s^1 correct
- [Ar] 3d^6
- [Ar] 3d^4 4 s^2
Explanation: The Aufbau order of electron filling is 1s, 2 s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f , 5 d, 6p, etc. s orbitals can hold 2 electrons, p orbitals 6 electrons, and d orbitals 10 electrons. Note some exceptions do occur in the electron con- figuration of atoms because of the stability of either a full or half-full outermost d-orbital, so you may need to account for this by ‘shuffling’ an electron from the (n − 1) s orbital. Finally use noble gas shorthand to get the answer: [Ar] 3d^5 4 s^1.
018 10.0 points
Which atom, selenium (Se) or Arsenic (As), has the higher electron affinity?
- Neither, because selenium and arsenic have the same electron affinity.
- Arsenic, because it has a stable, half-filled p subshell and adding an electron increases its stability.
- Arsenic, because it has a lower effective nuclear charge.
- Selenium, because it has a higher effective nuclear charge.
- Selenium, because it does not have a sta- ble, half-filled p subshell and adding an elec- tron does not decrease its stability. correct
Explanation: Arsenic (electron affinity = 79) has a a half-filled p subshell and one additional elec- tron does not increase its stability, in fact it perturbs it from a relatively stable configu- ration. Selenium (electron affinity = 195) is moved slightly close to a filled shell and no- ble gas configuration by addition of a single electron and sacrifices no half-filled stability in the process.
019 10.0 points Use electron-dot notation to demonstrate the formation of ionic compounds involving the elements Ca and I.
- Ca b+ b b bb
b b Ib −→
Ca+^ + bb bb
b b Ibb
− −→ CaI
- Ca b+ Ca b+ b bb
b b Ib −→
Ca+^ + Ca+^ + bb bb
b b Ibb
2 − −→ Ca 2 I
- Ca bb+ b bb
b b Ib −→
Ca2+^ + bb bb
b b bb
I
2 − −→ CaI
The Lewis formula for formaldehyde
(CH 2 O) is H C
O
b^ b bb
H
021 10.0 points
How many resonance forms are possible for the formate ion? Formate ion is CHOO−.
- 3
- 1
- 2 correct
- 4
- 5
Explanation: The Lewis structures are
C
O
bbb bb b
H O
bb bb
−
C
bOb b^ b
bb H
(^) bb O bb
−
022 10.0 points Which of the following is the correct Lewis structure of dimethyl ether (CH 3 OCH 3 )?
1. C
H
H
H
O
bb C
H
H
H
2. C
H
H
H
O C
H
H
H
3. C
H
H
H
C
bb O
H
H
H
4. C
H
H
H
O
bb^ bb^ C
H
H
H
correct
5. C
H
H
H
C
H
H
O H
bb^ bb
Explanation: All of the other choices have octet rule vi- olations, and because carbon and oxygen are Period 2 elements, they cannot form hyperva- lent compounds.
023 10.0 points Which of the following hypervalent com- pounds cannot exist even in theory?
- XeF 2
- I− 3
- SBr 6
- OCl 6 correct
- XeF 4 Explanation:
024 10.0 points The dot structure for BeCl 2 has how many lone pairs on the central atom?
- 5
- 3
- 2
- 0 correct
Explanation: Beryllium brings in 2 valence electrons and each chlorine brings in 7 valence electrons, for a total of 16 valence electrons: Cl Be Cl and the octet rule is satisfied for each Cl. Be is a known exception to the octet rule in that it can form stable compounds with only 4 valence electrons.
025 10.0 points Which of the species NO, BrO, CH+ 3 , BF− 4 are radicals?
- NO, CH+ 3 and BF− 4 only
- BrO, CH+ 3 and BF− 4 only
- NO and CH+ 3 only
- NO and BF− 4 only
- BrO and BF+ 4 only
- NO and BrO only correct
- BrO and CH+ 3 only
Explanation: The Lewis structures are N O bbb bb
bb Brbb O bbb bb
bb
B F
F
F
bbF bb bb
bb bb
bb
bb
bbbb bb bb
bb
−
H C H
H
Radicals are species with an unpaired elec- tron, so only NO and BrO are radicals.
026 10.0 points Rank the crystal lattice energy of the salts
Al 2 O 3 , CaCl 2 , CaO, NaF, Mg 3 (PO 4 ) 2 from least to greatest:
- NaF < CaO < CaCl 2 < Mg 3 (PO 4 ) 2 < Al 2 O 3
- CaO < NaF < CaCl 2 < Al 2 O 3 < Mg 3 (PO 4 ) 2
- NaF < CaCl 2 < CaO < Mg 3 (PO 4 ) 2 < Al 2 O 3 correct
- Mg 3 (PO 4 ) 2 < NaF < CaCl 2 < CaO < Al 2 O 3
- Mg 3 (PO 4 ) 2 < NaF < < CaO CaCl 2 < Al 2 O 3
Explanation: Sodium fluoride has the least lattice energy because of its small charges, then calcium chloride and calcium oxide with incremen- tally increasing charges. Both magnesium phosphate and aluminum oxide have identi- cal magnitudes for their charges, but both of the latter are smaller than than their counter- parts in the former and consequently have a greater lattice energy.
027 10.0 points What is the correct order of increasing elec- tronegativity?
- F, Si, P, S, N, C, O
- Si, P, S, C, N, O, F correct
- F, O, N, C, S, P, Si
- O, C, N, S, P, Si, F
Explanation: Generally speaking electronegativity in- creases from left to right across the periodic table and from bottom to top for elements within a group. Si(1.9) < P(2.2) < S(2.6) ≈ C(2.6) < N(3.0) < O(3.4) < F(4.0)
028 10.0 points
Explanation: Dipole moments result because of the differ- ence in electronegativity (EN) between bond- ing atoms. The bonds with the largest dipole will have the largest difference in EN. EN in- creases from bottom to top and left to right on the periodic table. C C and N N have the same EN, so there is no dipole moment. Between N, O, and F, F is the most EN, and has the largest difference in EN with C. Therefore, the C N bond has the largest dipole.
