Practice Midterm #1 Answers
What is the frequency associated with radiation of 4.59 x 10-8 cm wavelength?
a. 6.54 x 1017 s-1
b. 6.54 x 1015 s-1
c. 1.53 x 10-8 s-1
d. 13.8 s-1
e. 2.18 x 107 s-1
First, we must convert the given wavelength back into the standard units of meters to allow us to plug it into the wave equation:
4.59 * 10-8 cm * (1 m / 100 cm) = 4.59 * 10-10 m
For all EM radiation, c = vλ, so:
(2.998 * 108 m/s) = (4.59 * 10-10 m) v
Solving for v, we find that the frequency of the radiation is 6.54 * 1017 Hz.
What is the wavelength, in nm, of light with an energy content of 550 kJ/mol?
a. 0.217 nm
b. 0.419 nm
c. 157 nm
d. 217 nm
e. 6579 nm
Converting the given energy content in kJ/(moles of photons) into J/photon:
(550 kJ / 1 mol ) * (1000 J / 1 kJ) * (1 mol / 6.023 * 1023 photons)
E = 9.13 * 10-19 J/photon = hv = h(c/λ) = (6.626 * 10-34 J s) * (2.998 * 108 m/s) / λ
Solving for λ, we find that the wavelength is 2.17 * 10-7 m, which is equivalent to 217 nm.
Select the correct statement...The photoelectric effect:
a. was discovered by Max Planck.
b. describes the phenomenon of producing a beam of electrons by shining light on any metal surface.
c. contradicted the view that light energy was dependent upon intensity only.
d. was explained by classical physics wave theory
e. results in a beam of electrons which increases in number, but not velocity, as the wavelength of
incident light decreases.
Going through each of the statements:
A is incorrect as the photoelectric effect was first observed by Heinrich Hertz.
B is incorrect as electrons are only ejected if the frequency of the light overcomes the metal's threshold frequency.
C is correct, as the photoelectric effect showed that light energy was dependent on frequency of the light, contrary to Maxwell's wave theory.
D is incorrect, as the theory contradicted classical wave physics.
E is incorrect as decreasing wavelength increases frequency, and will increase the KE of the electrons and increase their velocity. Only intensity of the light increases number of electrons ejected.
In comparing UV and microwave electromagnetic radiation, which statement is correct?
a. Photon energy for microwave > Photon energy for UV
b. ν for UV > ν for microwave
c. λ for UV > λ for microwave
d. UV travels faster than microwave radiation in a vacuum
e. two of these statements are correct
Going through each of the statements:
A is incorrect as UV has higher frequency than microwaves, and thus UV has more energy.
B is correct (see table of EM waves).
C is incorrect as if UV has higher frequency, then microwaves will have the larger wavelength.
D is incorrect as they are both EM waves and travel at the speed of light.
E is incorrect as only one statement is correct.
How many electrons would be found in the n=6 level if all possible orbitals were completely filled with electrons?
e. none of these
Number of electrons in a given energy level is given by the formula: # of e- = 2n2 = 2(6)2 = 72 electrons.
Proof (much longer): Will type up later.
The possible values of the orbital angular momentum quantum number of a 3p electron are:
a. +1/2, -1/2
b. -1, 0, +1
c. 0, 1, 2
d. 1, 2, 3
e. none of these
We can determine from the given information that the principle quantum number 'n' = 3, as we are in the '3'p orbital.
This makes the possible orbital angular momentum quantum numbers (l): 0, 1, 2
But since we already know we are already in the p orbital, we can only use the corresponding 'l' quantum number (0 = s, 1 = p, 2 = d for your reference). Thus, the answer is none of the above, as the only possible answer is l = 1.
The ratio of the radius of the 8 orbit of He+ to the radius of the 4 orbit for H is:
I keep getting 2 for this question, either I don't know how to do ratios or I suck at chem, but could someone else check this out? The answer is in fact b, "2" there is a mistake in the anwsert key
solution: r = n2 a/Z for He- r=((82)(52.9))/ 2 = 1692.8 for H r==((42)(52.9))/ 1=846.2 therefore the ratio is 2:1 or 2
The ground state electron configuration of Au is:
a. [Xe] 4f13 5d9 6s2
b. [Xe] 5d10 6s1
c. [Xe] 5d9 6s2
d. [Xe] 4f14 5d10 6s1
e. none of these
From the periodic table, working through:
Using noble gas abbreviation, we find that the closest one to Au is [Xe]. Working past that, we reach the s block and counting down families, we find that it is 6s2. We hit the lanthanide series (f block), but since Au is past the entire series, we can simply write down 4f14. Finally, counting through the d block, we find that Au contains 5d9, making our initial electron configuration Au = [Xe]4f145d96s2.
Au is an exception, as 5d9 is relatively unstable compared to the full orbital of 5d10, so Au will take an electron from an outer energy level to reach that stability in n = 5 level. Thus, we remove an electron from the furthest energy level (6s2), which becomes 6s1, making our final electron configuration Au = [Xe]4f145d106s1.
[Ar]3d10 is the ground state electronic configuration of which species:
e. none of these
Easiest way to do this problem is to simply count electrons. [Ar]3d10 has 18 (from Ar) + 10 = 28 total electrons
A. Zn has 30 electrons.
Cu2+ has 29 - 2 = 27 electrons.
Ge2+ has 32 - 2 = 30 electrons.
Ga3+ has 31 - 3 = 28 electrons, which matches our initial amount and is the correct answer.
Alternatively, we can count backwards from the ground state of each element. Ga has a ground-state configuration of [Ar]4s24p13d10. The ion Ga3+ loses 3 electrons, which are always removed from the highest principal shell first—the orbitals with a principal quantum number of 4. This leaves us with [Ar]3d10.
The quantum numbers of the last electron of Antimony (Sb) could be:
a. n = 5, l = 0, ml = 0, ms = 1/2
b. n = 4, l = 2, ml = -2, ms = 1/2
c. n = 5, l = 1, ml = 0, ms = 1/2
d. n = 5, l = 1, ml = -2, ms = 1/2
e. n = 4, l = 3, ml = 3, ms = 1/2
We start out by writing out the electron configuration for Sb:
Sb = [Kr]4d105s25p3
Noting the last electron is in 5p3, n = 5, and l = 1 (as l = 1 corresponds to the orbital p). We are left with answers C and D, but a further examination of D shows that the ml value = -2, which is impossible for a l value of 1 (as ml can thus only be -1, 0, or 1). Thus, n = 5, l = 1, ml = 0, and ms = 1/2.
The ion represented as 48Ti4+ (Ti is atomic number 22) has:
a. 22 electrons
b. 48 neutrons
c. 26 protons
d. 4 3d electrons
e. 0 4s electrons
Going through each of the answers:
A. Ti in its uncharged form has 22 electrons, and the 4+ charge means it has 4 less electrons, so it only has 18 electrons at this point.
B. # of neutrons = mass number - atomic number = 48 - 22 = 26 neutrons, not 48.
C. Atomic number 22 means it has 22 protons, not 26.
D. Writing out the electron configuration: Ti4+ = [Ar], so it has 0 3d electrons.
E. Since electron configuration Ti4+ = [Ar], it has 0 4s2 electrons, and this is the correct answer.
The orbital in which the last electron of Tantalum (Ta) is placed will have a total of how many nodes
(both radial and angular)?
e) none of these
This is another question that involves a simple formula but takes time to prove. The proof will be left for later, but basic equations will be given for now:
The electron configuration of Ta = [Xe]6s24f145d3, so the last electron is the one placed on 5d3.
The number of radial nodes on a subshell is given by the formula # of RN = n - l - 1, where n and l are the principle and azimuthal quantum numbers, respectively. Since we are on the 5d orbital, n = 5 and l = 2, so the # of RN = 5 - 2 - 1 = 2 radial nodes.
The number of angular nodes is simply equal to l, the azimuthal quantum number, so # of AN = 2 angular nodes.
# of RN + # of AN = 2 + 2 = 4 total nodes
Which species is the most paramagnetic?
The most paramagnetic species will have the most unpaired electrons. Thus, going through the list:
A. Cr+ = [Ar]3d5 → 5 unpaired electrons
B. Mn+ = [Ar]3d54s1 → 6 unpaired electrons
C. Ag = [Kr]4d105s1 → 1 unpaired electron
D. Fe = [Ar]3d64s2 → 4 unpaired electrons
E. As = [Ar]3d104s24p3 → 3 unpaired electrons
Which of the following electron transitions in the He+ ion results in absorption of light of the longest
a) n = 8 to n = 7
b) n= 4 to n = 5
c) n = 1 to n = 8
d) n = 2 to n = 3
e) n = 1 to n = 2
Energy is absorbed when an electron transitions from a low to high energy level. Immediately, we can eliminate A, as photons will be emitted rather than absorbed.
Since the change in energy during an electron transition is given by ΔE = Z2hcRh[(1/ni2) - (1/nf2)], and energy is inversely proportional to wavelength, we want to find the electron transition that gives us the lowest amount of energy absorbed. The smallest transition farthest from the lower energy levels will thus give the smallest energy absorption, making n = 4 to n = 5 the correct answer. (Alternatively, you can plug in values into the equation and find which one provides the smallest ΔE.)
What would be the ionization energy for a B4+ ion on another planet where a high surface
temperature resulted in the third principle energy level being the ground state for the B4+ ionic under
these extreme extraterrestrial conditions?
a) 25/9 hcRH
b) -25/9 hcRH
c) 5/9 hcRH
d) -1/9 hcRH
Ionization energy is given by the following general equation:
E = (Z2hcRH)/(n2)
Plugging in Z = 5 (the atomic number of boron) and n = 3 for the third principle energy level, we obtain (52/32) hcRH = (25/9) hcRH.
The nth orbit of a hydrogen like species has a radius of 441 pm and an energy of -7.84 x 10-19 J. Identify the correct H-like species and electron orbit.
a) He+, n=3
b) He+, n=4
c) Li2+, n=4
d) Li2+, n=5
e) Be3+, n=5
The radius of a Bohr atom is given by the equation rn = (n2ao)/Z and the energy of that orbit is given by E = -Z2hcRH/n2.
First, taking the radius equation and plugging in the given radius:
(441 pm) = [(n2)(53 pm)]/Z
Solving for n2, we obtain the formula:
n2 = (441 pm / 53 pm) * Z = 8.32 * Z
Next, using the second equation, and plugging in the given energy:
-7.84 * 10-19 J = -Z2[2.179 * 10-18/(n2)]
Substituting the first equation (solved for n2) into the second equation yields:
-7.84 * 10-19 J = -Z2[2.179 * 10-18 / (8.32 * Z) = -Z[2.179 * 10-18 / 8.32]
Solving the equation yields Z = 3, so the species must be Li2+ (Lithium has an atomic number of 3).
Plugging back Z = 3 into n2 = 8.32 * Z = 8.32 * 3, and solving yields the solution n = 5.
For the ion Be3+, an emission line is detected at 5.816 nm. This line is due to an electron transition
from the nth orbit to the ground state orbit of this ionic species. Identify the value of n for the initial orbit
for this emission line
We must set the Planck postulate and Balmer equation equal in order to solve this question:
E = hv = hc/λ
E = Z2hcRH[(1/ni2) - (1/nf2)]
Equating and eliminating common terms yields:
1/λ = Z2RH[(1/ni2) - (1/nf2)]
Plugging in all values:
-[1/(5.816 * 10-9 m)] = (4)2(1.097 * 107) [(1/ni2) - (1/12)]
(note: energy here is negative as energy is being emitted)
Solving for ni, we get ni = 7.
The energy of a laser beam that excites a two-photon absorption in a particular element is 1.22x105
J/(mol of photons). What would be the wavelength of light that would excite a one-photon absorption
between the same states in the material?
a) 980 nm
b) 1960 nm
c) 490 nm
d) 8.11 x 10-31 m
e) none of these
1.22 * 105 J/(mol of photons) * (1 mol of photons / 6.023 * 1023 photons) = 2.03 * 10-19 J/photon
Using E = hv = hc/λ:
2.03 * 10-19 J = (6.626 * 10-34 J s) (2.998 * 108 m/s) / λ, and solving yields λ = 980 nm.
However, since this λ provides the energy that excites a two-photon absorption, all we must do is divide by 2 to obtain the wavelength to excite a one-photon absorption:
980 nm / 2 = 490 nm.
What would be the atomic number for the undiscovered element which would have four electrons in
its 6f orbitals?
I personally believe the answer to this question is wrong, as it doesn't take into consideration the g orbital, which should begin after element 121.
The correct answer to this question should be element 142, Uqb, not element 125.
Assume the Laws of Nature were different and Hund’s rule required that unpaired electrons fill
empty degenerate orbitals singly with all electrons in a spin up configuration for every atom, before a
second electron with opposite spin could be added to each orbital. What would be the outcome of the
Stern Gerlach type experiment where a beam of vanadium (V) atoms are passed through a magnetic
a) The V atoms would pass through the magnetic field undeflected and produce one spot on the detector.
b) The V atoms would be deflected in two separate directions by the magnetic field and produce two
spots on the detector.
c) The V atoms would be deflected in three separate directions by the magnetic field and would produce
three spots on the detector.
d) The V atoms would be deflected backwards by the magnetic field would not reach the detector.
e) The V atoms would be deflected in one direction by the magnetic field producing one spot on the
In the original Stern-Gerlach experiment, silver atoms that were shot at from an electron gun had half deflected upwards by the magnetic field and half deflected downwards by the magnetic field. This is due to the fact that in the silver atom, the one unpaired electron it has will determine the direction of the magnetic field generated, which will determine the direction in which it is deflected.
Looking at the electron configuration for V ([Ar]3d34s2) we can expect a similar outcome due to its unpaired electrons.
However, if hypothetically, the Laws of Nature™ are changed such that each degenerate orbital is filled up first with a spin-up electron, the generated magnetic field will always be in the same direction, and the V atoms will only be deflected in one direction.
A US car company lists the mass of a particular model as 3142 pounds. If the velocity of the car is
measured to be u = 1.20 x 102 km/hour using a radar speed gun with a 2 % uncertainty, what is the
uncertainty (∆x) in the position of the car? (1 pound = 453.59 g)
a) ≥ 5.55 x 10-38 m
b) ≥ 3.08 x 10-43 m
c) ≥ 1.54 x 10-41 m
d) ≥ 5.55 x 10-41 m
e) none of these
The Heisenberg uncertainty principle is expressed in the equation:
∆x∆p ≥ h/4π
However, we must first convert all given information into a usable form first:
m = 3142 pound * (452.49 g / 1 pound) * (1 kg / 1000 g) = 1422 kg
u = (1.20 * 102 km / 1 hour) * (1 hour / 3600 seconds) * (1000 m / 1 km) = 33.3 m/s
p = mu = (1422 kg) * (33.3 m/s) = 47,400 kg m/s
∆p = (0.02) * (47,400 kg m/s) = 948 kg m/s
Substituting back into the original equation:
∆x ≥ h / (4π * ∆p) ≥ (6.626 * 10-34) / (4π * 948) ≥ 5.55 * 10-38 m.
If light of a frequency of 1.50 x 1015 s-1 directed at a copper surface, electrons are ejected from the
copper and are measured to have a velocity of 7.40 x 105 m s-1. What is the work function of copper in
units of eV? (1eV = 1.602 x 10-19 J)
a) 9.94 x 10-19 eV
b) 4.65 eV
c) 6.20 eV
d) 1.19 x 10-37 eV
e) 1.50 eV
Using the mass of the electron and the given velocity, we can find the kinetic energy:
KE = (1/2)mu2 = (1/2)(9.11 * 10-31 kg)(7.40 * 105)2 = 2.49 * 10-19 J
Since KE = E - Eo = hv - ϕ:
2.49 * 10-19 = (6.626 * 10-34) (1.50 * 1015) - ϕ
Solving yields ϕ = 7.45 * 10-19 J = 4.65 eV.
The measured values for the work function of some metals are Al = 4.19 eV, Zn= 4.33 eV, Mg = 3.66
eV, Na = 2.7 eV and Cs = 2.14. If light having a wavelength of 414 nm is directed on samples of these
metals, what would you expect to observe experimentally? (KE means Kinetic Energy in the answers;
1 eV = 1.602 x 10-19 J)
a) electrons would be ejected by Na and Cs with KE e- Cs > KE e- Na
b) electrons would be ejected from Al, Zn, and Mg with KE e- Zn > KE e- Al > KE e- Mg
c) electrons would be ejected from Al, Zn with KE e- Al > KE e- Zn
d) electrons would be ejected from Mg, Na, and Cs with KE e- Cs > KE e- Na > KE e- Mg
e) none of these would occur
Electrons will be ejected with a kinetic energy (KE) if the energy of the light (E) shone on the metal is greater than its workfunction (∅). Thus:
E = hv = hc/λ = (6.626 * 10-34 J s) * (2.998 * 108 m/s) / (414 * 10-9 m) = 4.80 * 10-19 J = 3.00 eV
Since the energy of the shone light is greater than the workfunctions of Na and Cs, but lower than Al, Zn, and Mg, only Na and Cs will have electrons ejected.
Since KE = E - ∅, then KE (Cs) > KE (Na).
Which is not a valid set of quantum numbers for an electron?
a. n = 5, l = 2, ml = -2, ms = 1/2
b. n = 3, l = 0, ml = 0, ms = 1/2
c. n = 10, l = 2, ml = 2, ms = 1/2
d. n = 5, l = 4, ml = 4, ms = -1/2
e. n = 6, l = 3, ml = -4, ms = -1/2
|The value of ml can only be integers between the value of l and - l. Thus, E is not valid, as -4 does not fall in the range -3 ≤ ml ≤ 3.|
What would be the ground state electronic configuration for In+?
e. none of these
The standard electron configuration for In = [Kr]4d105s25p1.
With the positive charge, one electron is removed from the furthest energy level in the furthest shell, so the correct ground state electronic configuration would be [Kr]4d105s2.
What would be the wavelength of an electron that is moving at 1.5% of the speed of light?
a) 74 pm
b) 162 pm
c) 52.9 pm
d) 1.62 pm
e) 7.4 pm
λ = h/mu = (6.626 * 10-34 J s) / [(9.11 * 10-31 kg) * (.015 * 2.998 * 108 m/s)]
Solving the equation yields λ = 162 pm.
How many of the following species would you expect to be diamagnetic?
Be, Cd, Cl- , Y+, Ti2+ Ag+
a) one diamagnetic species
b) two diamagnetic species
c) three diamagnetic species
d) four of them
e) all of them
A species is diamagnetic if it has no unpaired electrons. Going through each of the species:
Be = 1s2 2s2 → No unpaired electrons
Cd = [Kr]5s24d10 → No unpaired electrons
Cl- = [Ar] → No unpaired electrons
Y+ = [Kr]4d15s1 → 2 unpaired electrons
Ti2+ = [Ar]3d2 → Two unpaired electrons
Ag+ = [Kr]4d10 → No unpaired electrons|
Thus, there are four diamagnetic species.
What ionic species with a 3+ charge would you expect to have the following electronic
A species with a 3+ charge means that the original element has lost three electrons. Taking a look at each element:
A. Ru3+ = [Kr]4d5
B. Fe3+ = [Ar]3d5
C. Pd3+ = [Kr]4d7]
D. Ni3+ = [Ar]3d7
E. Zr3+ = [Kr]4d1