## Practice Midterm #1 Answers

### Question 1

What is the frequency associated with radiation of 4.59 x 10^{-8} cm wavelength?

**a. 6.54 x 10 ^{17} s^{-1}**

b. 6.54 x 10

^{15}s

^{-1}

c. 1.53 x 10

^{-8}s

^{-1}

d. 13.8 s

^{-1}

e. 2.18 x 10

^{7}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} mFor all EM radiation, c = vλ, so: (2.998 * 10 ^{8} m/s) = (4.59 * 10^{-10} m) vSolving for v, we find that the frequency of the radiation is 6.54 * 10. ^{17} Hz |

### Question 2

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 * 10 ^{23} photons)E = 9.13 * 10 ^{-19} J/photon = hv = h(c/λ) = (6.626 * 10^{-34} J s) * (2.998 * 10^{8} m/s) / λSolving for λ, we find that the wavelength is 2.17 * 10 ^{-7} m, which is equivalent to 217 nm. |

### Question 3

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. |

### Question 4

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. |

### Question 5

How many electrons would be found in the n=6 level if all possible orbitals were completely filled with electrons?

a. 18

b. 32 **c. 72**

d. 50

e. none of these

Easy way: Number of electrons in a given energy level is given by the formula: # of e ^{-} = 2n^{2} = 2(6)^{2} = 72 electrons.Proof (much longer): Will type up later. |

### Question 6

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. |

### Question 7

The ratio of the radius of the 8 orbit of He^{+} to the radius of the 4 orbit for H is:

a. 1⁄4 **b. 2**

c. 4

d. 1⁄2

e. 1

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 = n ^{2} a/Z for He- r=((8^{2})(52.9))/ 2 = 1692.8 for H r==((4^{2})(52.9))/ 1=846.2 therefore the ratio is 2:1 or 2 |

### Question 8

The ground state electron configuration of Au is:

a. [Xe] 4f^{13} 5d^{9} 6s^{2}

b. [Xe] 5d^{10} 6s^{1}

c. [Xe] 5d^{9} 6s^{2} **d. [Xe] 4f ^{14} 5d^{10} 6s^{1}**

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 6s ^{2}. We hit the lanthanide series (f block), but since Au is past the entire series, we can simply write down 4f^{14}. Finally, counting through the d block, we find that Au contains 5d^{9}, making our initial electron configuration Au = [Xe]4f^{14}5d^{9}6s^{2}.HOWEVER: Au is an exception, as 5d ^{9} is relatively unstable compared to the full orbital of 5d^{10}, 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 (6s^{2}), which becomes 6s^{1}, making our final electron configuration Au = [Xe]4f.^{14}5d^{10}6s^{1} |

### Question 9

[Ar]3d^{10} is the ground state electronic configuration of which species:

a. Zn

b. Cu^{2+}

c. Ge^{2+} **d. Ga ^{3+}**

e. none of these

Easiest way to do this problem is to simply count electrons. [Ar]3d^{10} has 18 (from Ar) + 10 = 28 total electronsA. Zn has 30 electrons. Cu ^{2+} has 29 - 2 = 27 electrons.Ge ^{2+} has 32 - 2 = 30 electrons.Ga has 31 - 3 = 28 electrons, which matches our initial amount and is the correct answer.^{3+}Alternatively, we can count backwards from the ground state of each element. Ga has a ground-state configuration of [Ar]4s ^{2}4p^{1}3d^{10}. The ion Ga^{3+} 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]3d^{10}. |

### Question 10

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]4d ^{10}5s^{2}5p^{3}Noting the last electron is in 5p ^{3}, 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 m^{}l^{} value = -2, which is impossible for a l value of 1 (as m^{}l^{} can thus only be -1, 0, or 1). Thus, n = 5, l = 1, m.^{}l^{} = 0, and m^{}s^{} = 1/2 |

### Question 11

The ion represented as ^{48}Ti^{4+} (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: Ti ^{4+} = [Ar], so it has 0 3d electrons.E. Since electron configuration Ti ^{4+} = [Ar], it has 0 4s, and this is the correct answer.^{2} electrons |

### Question 12

The orbital in which the last electron of Tantalum (Ta) is placed will have a total of how many nodes

(both radial and angular)?

a) 5

b) 3 **c) 4**

d) 6

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]6s ^{2}4f^{14}5d^{3}, so the last electron is the one placed on 5d^{3}.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 |

### Question 13

Which species is the most paramagnetic?

a) Cr^{+} **b) Mn ^{+}**

c) Ag

d) Fe

e) As

The most paramagnetic species will have the most unpaired electrons. Thus, going through the list: A. Cr ^{+} = [Ar]3d^{5} → 5 unpaired electronsB. Mn = [Ar]3d^{+}^{5}4s^{1} → 6 unpaired electronsC. Ag = [Kr]4d ^{10}5s^{1} → 1 unpaired electronD. Fe = [Ar]3d ^{6}4s^{2} → 4 unpaired electronsE. As = [Ar]3d ^{10}4s^{2}4p^{3} → 3 unpaired electrons |

### Question 14

Which of the following electron transitions in the He+ ion results in absorption of light of the longest

wavelength?

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 = Z ^{2}hcRh[(1/ni^{2}) - (1/nf^{2})], 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.) |

### Question 15

What would be the ionization energy for a B^{4+} ion on another planet where a high surface

temperature resulted in the third principle energy level being the ground state for the B^{4+} ionic under

these extreme extraterrestrial conditions?

**a) 25/9 hcRH**

b) -25/9 hcRH

c) 5/9 hcRH

d) -1/9 hcRH

e)hcRH

Ionization energy is given by the following general equation: E = (Z ^{2}hcRH)/(n^{2})Plugging in Z = 5 (the atomic number of boron) and n = 3 for the third principle energy level, we obtain (5 ^{2}/3^{2}) hcRH = (25/9) hcRH. |

### Question 16

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) Li^{2+}, n=4 **d) Li ^{2+}, n=5**

e) Be

^{3+}, n=5

The radius of a Bohr atom is given by the equation r_{n} = (n^{2}a_{o})/Z and the energy of that orbit is given by E = -Z^{2}hcR_{H}/n^{2}.First, taking the radius equation and plugging in the given radius: (441 pm) = [(n ^{2})(53 pm)]/ZSolving for n ^{2}, we obtain the formula:n ^{2} = (441 pm / 53 pm) * Z = 8.32 * ZNext, using the second equation, and plugging in the given energy: -7.84 * 10 ^{-19} J = -Z^{2}[2.179 * 10^{-18}/(n^{2})]Substituting the first equation (solved for n ^{2}) into the second equation yields: -7.84 * 10 ^{-19} J = -Z^{2}[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 Li (Lithium has an atomic number of 3).^{2+}Plugging back Z = 3 into n ^{2} = 8.32 * Z = 8.32 * 3, and solving yields the solution n = 5. |

### Question 17

For the ion Be^{3+}, 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

a) 1

b) 4

c) 5 **d) 7**

e) 9

We must set the Planck postulate and Balmer equation equal in order to solve this question: E = hv = hc/λ E = Z ^{2}hcRH[(1/ni^{2}) - (1/nf^{2})]Equating and eliminating common terms yields: 1/λ = Z ^{2}RH[(1/ni^{2}) - (1/nf^{2})]Plugging in all values: -[1/(5.816 * 10 ^{-9} m)] = (4)^{2}(1.097 * 10^{7}) [(1/ni^{2}) - (1/1^{2})] (note: energy here is negative as energy is being emitted) Solving for ni, we get ni = 7. |

### Question 18

The energy of a laser beam that excites a two-photon absorption in a particular element is 1.22x10^{5}

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 * 10^{5} J/(mol of photons) * (1 mol of photons / 6.023 * 10^{23} photons) = 2.03 * 10^{-19} J/photonUsing E = hv = hc/λ: 2.03 * 10 ^{-19} J = (6.626 * 10^{-34} J s) (2.998 * 10^{8} 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. |

### Question 19

What would be the atomic number for the undiscovered element which would have four electrons in

its 6f orbitals?

a) 120

b) 122 **c) 125**

d) 157

e) 154

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. |

### Question 20

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

field?

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 detector.**

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]3d ^{3}4s^{2}) 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. |

### Question 21

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 10^{2} 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 * 10 ^{2} km / 1 hour) * (1 hour / 3600 seconds) * (1000 m / 1 km) = 33.3 m/sp = 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 |

### Question 22

If light of a frequency of 1.50 x 10^{15} s^{-1} directed at a copper surface, electrons are ejected from the

copper and are measured to have a velocity of 7.40 x 10^{5} 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)mu ^{2} = (1/2)(9.11 * 10^{-31} kg)(7.40 * 10^{5})^{2} = 2.49 * 10^{-19} JSince KE = E - Eo = hv - ϕ: 2.49 * 10 ^{-19} = (6.626 * 10^{-34}) (1.50 * 10^{15}) - ϕSolving yields ϕ = 7.45 * 10 ^{-19} J = 4.65 eV. |

### Question 23

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 * 10^{8} m/s) / (414 * 10^{-9} m) = 4.80 * 10^{-19} J = 3.00 eVSince 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). |

### Question 24

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. |

### Question 25

What would be the ground state electronic configuration for In+?

a. [Kr]4d^{9}5s^{2}5p^{1}

b. [Kr]4d^{10}5s^{1}5p^{1}

c. [Kr]4d^{10}5s^{2}5p^{2}

d. [Kr]4d^{10}5s^{2}5p^{1} **e. none of these**

The standard electron configuration for In = [Kr]4d^{10}5s^{2}5p^{1}.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]4d.^{10}5s^{2} |

### Question 26

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 * 10^{8} m/s)]Solving the equation yields λ = 162 pm. |

### Question 27

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]5s ^{2}4d^{10} → No unpaired electronsCl ^{-} = [Ar] → No unpaired electronsY ^{+} = [Kr]4d^{1}5s^{1} → 2 unpaired electronsTi ^{2+} = [Ar]3d^{2} → Two unpaired electronsAg ^{+} = [Kr]4d^{10} → No unpaired electrons|Thus, there are four diamagnetic species. |

### Question 28

What ionic species with a 3+ charge would you expect to have the following electronic

configuration: [Kr]4d^{5}

**a) Ru ^{3+}**

b) Fe

^{3+}

c) Pd

^{3+}

d) Ni

^{3+}

e) Zr

^{3+}

A species with a 3+ charge means that the original element has lost three electrons. Taking a look at each element: A. Ru = [Kr]4d^{3+}^{5}B. Fe ^{3+} = [Ar]3d^{5}C. Pd ^{3+} = [Kr]4d^{7}]D. Ni ^{3+} = [Ar]3d^{7}E. Zr ^{3+} = [Kr]4d^{1} |