Quantum Mechanics
1. What is an electromagnetic wave?
An electromagnetic wave is one in which there are oscillating magnetic and electric fields in
planes perpendicular to one another.
2. What do the following represent?
a. λ – Wavelength (maximum to maximum or minimum to minimum, in meters)
b. ν – Frequency (# of waves passing through a point per second, in Hertz, s-1)
c. c – 2.998 x 108 m/s (speed of light).
3. What equation relates λ, c and ν?
So the longer the wavelength, the shorter the frequency (or vice versa).
4. In the macroscopic viewpoint, energy is described as being massless and delocalized. Meaning
we don’t ask someone to give us 5 pounds of light (which is a form of energy) nor can we go and
pick up a “piece” of light.
5. In the macroscopic viewpoint, matter is described as having a mass and being located at a
particular point of space.
6. How are the following frequently classified?
a. Ball - Matter
b. Light - Energy
c. Electron – Most people view an electron as a tiny ball, and thus would classify it as
matter. This is an idea we will come to see is not the most complete/correct
perspective.
7. Why did quantum mechanics become necessary?
There are a number of experiments/observations, blackbody radiation (energy can only be
absorbed in integer multiples), double slit experiment (wave formation of electrons),
photoelectric effect (development of the photon) that clearly muddy up the line between
matter and energy.
It was from this incongruity that quantum mechanics was born.
8. What was demonstrated in the double slit experiment?
In the double slit experiment an electron gun was fired through 2 slits onto a screen. When the
gun was fired single points corresponding to an electron hitting the screen where noted. Over
time the following pattern was observed:
From this you can see that there is a wave-like pattern that is observed. The electrons, though
having distinct point of impact on the screen – create a pattern of constructive and destructive
interference (a property of waves – not matter).
9. What is the photoelectric effect?
The photoelectric effect is a phenomenon that Einstein won a Nobel Prize for. What Einstein
showed, was that electrons could be emitted from the surface of a metal when light (at a
minimum frequency) struck the metal surface.
10. Four ideas to know about it
a. There is a minimum frequency needed to get electrons to emit.An increase in intensity
has no effect on electrons if the frequency of the light is below the threshold frequency.
b. Oncethe threshold frequency has been reached, an increase in light intensity will cause
an increase in the number of electrons emitted.
c. The greater the frequency of light, the greater the kinetic energy of the emitted
electron(s).
d. Demonstrated that light has a duality in nature. Exhibiting both particulate and energy
properties. The only way that these electrons could be knocked out is if the light has a
particulate property to it that was able of “bumping” the electrons out.
11. The big discovery of the photoelectric effect was the existence of photons. Photons are little
packets of energy contained in light… one very simplistic way to view it is as though it were a
skittles rainbow:
12. Equation for the energy of a photon
Where h = 6.626 x 10-34 J s (Planck’s constant)
13. Equation for kinetic energy of an electron
Where, ν = incident frequency (meaning the frequency of light that you actually used) and ν0 =
threshold frequency (meaning the minimum energy required to eject electron).
14. How do E, ν, and λ relate?
15. What is really important about E=mc2?
This equation demonstrates the relationship between energy and matter.
16. What is the DeBroglie wavelength equation?
This equation highlights the duality of nature; having both particulate and energy properties.
17. Microwave radiation has a wavelength on the order of 1.0 cm. Calculate the frequency and the
energy of a single photon of this radiation. Calculate the energy of an Avogadro’s number of
photons of the electromagnetic radiation.
This value that we have obtained is the amount of energy per photon of light at this frequency.
In order to get an Avogadro’s number of photons (or mole of photons), we will have to multiply
the value obtained by 6.022 x 1023.
18. The work function of an element is the energy required to remove an electron from the surface
of the solid. The work function for lithium is 279.7 kJ/mol. What is the maximum wavelength of
light that can remove an electron from an atom in lithium?
“work function” = E0 (the energy required to emit a mol of electrons) = 279.7 kJ/mole eWe are trying to figure out the wavelength associated with emitting a single electron from the
lithium metal. The first thing we need to do is determine how much energy is required per
electron:
Now that we know the amount of energy per electron we can solve for λ using:
19. It takes 208.4kJ f energy to remove one mole of electrons from the atoms on the surface of
rubidium metal. If rubidium metal is irradiated with 254-nm light, what is the maximum kinetic
energy the released electron can have?
For this you will need to use the equation:
Eelectron = E – E0
Once again, make sure that you note that the question is asking the amount of energy an
ejected electron will have (not a mole of ejected electrons). This means that you will have to
convert your units for the threshold energy from kJ/mol to J/electron.
20. What is an emission spectra?
You can think of them as atomic.molecular “fingerprints”. When energy is pumped into a
molecule/atom and then released, the released energy is often the same value as energy of light
found in the visible spectrum (ROYGBIV).
It is important to note that when the energy id released – it does not emit the entire light
spectrum; rather, it shows discreet values – creating a “line spectrum”
21. What did the line spectra show us?
The allowable energy levels to which an electron can jump within a given element.
22. Is the planetary model for atoms valid?
Absolutely not. If a negative charge were to circle around a positively charged center, the
opposing charges would attract each other and the electron would move in an inward spiral
fashion… ultimately this would lead to implosion.
23. What is the Bohr model based on?
The Bohr model is used the hydrogen emission spectrum to create an equation that was built on
the idea that electrons could only occupy certain areas of space relative to the nucleus.
24. What is the equation?
Which is more frequently used in the form:
This equation can only be used to determine how much energy is absorbed or released when an
electron changes energy orbits.
25. What is the only applicable case to use the Bohr model?
This equation is valid only for atoms with 1 electron, i.e. H, Li2+, C5+, etc.
Another problem with the Bohr model is that it cannot account for bonding…. Obviously a big
issue in chemistry.
26. An electron is excited from ground state to n=3 state in a hydrogen atom. Which of the
following statements are true? Correct any false statements.
a. It takes more energy to ionize (remove) the electron from n=3 than from ground state.
False. At n=3 it would be easier as the electrons are farther away from the nucleus (to
which they are attracted). Further away something is from what it is attracted to…
easier it is to pull them apart.
b. The electron is farther from the nucleus on average in the n=3 state than in ground
state.
True
c. The wavelength of light emitted if the electron drops from n=3 to n=2 is shorter that the
wavelength of light emitted if the electron falls from n=3 to n=1
False. It would take more energy (therefore a shorter wavelength) to transition from
n=3 to n=1
d. The wavelength of light emitted when the electron returns to the ground state from n=3
is the same as the wavelength of light absorbed from n=1 to n=3.
True
27. Does a photon of visible light (λ=400-700nm) have sufficient energy to excite an electron in a
hydrogen atom from the n=1 to the n=5 energy state?
In order to answer this question we are going to have to calculate the minimum energy required
to excite the electron from n=1 to n=5. Remember that because of the inverse relationship
between E and λ… the λ that we solve for (based on the E value) will pertain to the maximum
wavelength possible.
Because we are dealing with hydrogen (which has one electron), we are able to use the Bohr
equation:
Plugging in, we get:
This represents the maximum wavelength of light that would have sufficient amount of energy
to excite an electron from n=1 to n=5. So visible light’s (400-700nm) range it too great. Visible
light would not be able to excite the electron in this way.
28. An excited hydrogen atom emits light with a wavelength of 397.2 nm to reach the energy level
for which n=2. In which principal quantum level did the electron begin?
In this problem we have to determine the energy associated with the wavelength of light that
was emitted. This will allow us to use the Bohr equation to solve for ni.
We can now plug into the Bohr Equation. One very important point to note is that the energy
here is emitted. This means that the electron is releasing this amount of energy upon descent.
So when we use the E value, we just calculated, in the Bohr equation – it must be negative.
Plugging in we get:
29. Consider an electron for a hydrogen atom in an excited state. The maximum wavelength of
electromagnetic radiation that can ionize the electron from the H atom is 1460 nm. Determine
the initial excited state for the electron.
This is the same procedure as normal. Determine the energy associated with this wavelength of
light. Then use this energy to determine the n value using the Bohr equation.
The important thing to realize about this question is what it means to ionize. Ionization is the
complete removal of an electron from an atom. This would require energy to be put in.
Additionally this means that the final n value would equal infinity, the electron would be
infinitely far away from the nucleus if it was removed.
Now we can use the Bohr equation:
Plugging in:
30. As, Bohr’s model, doesn’t really help us understand too many useful things about substances
with more than one electron or bonding. Schrodinger’s equation steps in, to fill that gap.
31. Schrodinger’s wave equation describes orbitals, energy levels and the square of a wave function
describes the 90% probability of where an electron can be found.
32. What is the Heisenberg uncertainty principle?
Tells us that we cannot know both the speed and the location of a particle to the same degree
of accuracy – the better known one value is the less known the second. This is caused by
measurement techniques.
33. What is the equation?
∆x ∆p > h/4π
∆p = uncertainty in momentum = m∆v
∆x = uncertainty in position
34. Quantum Numbers (stemming from Schrodinger)
a. Principal Quantum Number (Shell)
i.
Is represented by n
ii.
Has values from ranging from1 ∞
iii.
Indicatesthe size and energy level in which the electron is housed. The bigger
the the n value, the further the electro is from the nucleus.
1. Degeneracy is whenorbitals have the same value of n.
b. Angular Momentum Quantum Number (Orbital)
i. Is represented byÄ
ii. Has values from ranging from0 (n-1)
iii.
Indicatesthe type of orbital.
1.
Ä= 0
s orbital
2.
Ä= 1
p orbital
3.
Ä= 2
d orbital
4.
Ä= 3
f orbital
c. Magnetic Quantum Number
i. Is represented by ÅÄ
ii.
Has values from ranging from( - Ä+ Ä)
iii.
Indicates# of orbitals.
1.
Ä =0 (the s orbital)ÅÄ=0
This means that there is only one s-orbital.
2.
Ä =1
(the p orbital)ÅÄ= (-1, 0, +1)
Because there are 3 values of for ÅÄ , there would be three p orbitals.
Ultimately, these correspond to different orientations these orbitals can
have in three dimensional space. In the case of the p orbital we call
them px, py, and pz.
3.
Ä =2
(the d orbital) ÅÄ= (-2, -1, 0, 1, 2)
Because there are 5 values forÅÄ, there would be five differently
orientated d orbitals. Labeled dxz, dxy, dyz, dz2,dx2 - y2
4.
Ä = 3 (the f orbital) ÅÄ= (-3, -2, -1, 0, 1, 2, 3)
Because there are 7 values for ÅÄ, there would be seven differently
orientated f orbitals.
d. Magnetic Spin Quantum Number
i. Is represented by Åá
ii.
Has values from ranging from(+½, - ½)
iii.
Indicates spin of an electron.
35. What is the Pauli Exclusion Principle?
Only 2 electrons can be in a particular orbital and they must have opposite spin numbers.
i.e. no 2 electrons can have the same 4 quantum numbers. They can have the same n, Ä?and ÅÄ
but not the same ÅáA
Think of the quantum numbers as assigning an “address” to each electron in an atom. Each
successive quantum numbers having a greater specificity.
n = street, Ä= building, ÅÄ=apartment number, Åá= top bunk or bottom bunk (only one to a
bunk!)☺
36. Which of the following pairing of quantum numbers is invalid, why?
a. n=3, Ä =2, ÅÄ =2, Åá= ½
This set of quantum numbers is valid. All values fit the required parameters.
b. n=4, Ä =3, ÅÄ =4, Åá=- ½
This set is invalid. The value of ÅÄcannot equal 4 if Ä= 3. The maximum value ÅÄ can be
is +3.
c. n=0, Ä =0, ÅÄ =0, Åá= -½
This is an invalid set of quantum numbers because n cannot equal zero. It’s smallest
value is 1.
d. n=2, Ä =-1, ÅÄ =1, Åá= ½
This is an invalid set of quantum numbers because Äcannot equal -1. T’s lowest value is
zero.
37. How many electrons can have the following designation?
When doing these problems it is obviously important to be able to decipher what this type of
notation indicates, let’s take a look at 2px as an example.
2px
n=2
p = Ä-value (which is 1 for p)
Åá= x
So in this case 3 of the four quantum numbers have been specified. There is only the magnetic
spin number remaining. This means that there are only 2 electrons that could have this
designation. One would have + ½ and the other –½ as their final quantum number.
a. 1p
So, break this down to what each value, that was given, relates to.
n = 1 and = p = 1
It is important to take a pause here and determine whether a value of 1 for Äis possible if
n=1. We realize that it is not. The maximum value for Äwhen n = 1 is zero. Thus the
answer to this question is that there are no electrons that could have this designation.
b. 6dxy
In this case we know that n = 6, Ä= 2 and
Ä= 2.
ÅÄ= xy.
If n = 6 it is completely okay that
This is, therefore, a valid orbital designation. If we consider now, that 3 of the 4
quantum numbers have been specified – we realize that there are only 2 electrons that
can have this “address” one with the + ½ spin and one with a – ½ spin.
c. 4f
In this case n = 4 and Ä= f = 3.
This means that there are 2 quantum numbers remaining to be specified. So now what
we want,is to consider the number of orbitals contained within f – based on previos
calculations we determined that there were seven f orbitals. Remember that we can
determine this by looking at the number of ÅÄvalues that correspond toÄ= 3.
Lastly we consider our final quantum number, Åá, which tells us that there are two
allowable electrons per orbital.
Giving us: 7 x 2 = 14eSo, 14 electrons could have the designation 4f.
d. n=3
For this particular problem, only n has been specified. So we need to figure out which
orbitals and the number of each orbital type are available when n = 3.
When:
n=3
Ä = (0, 1, 2)
or s, p, d
Let’s now add up the number of orbitals associated with s, p and d.
s=1 p=3 d=5
1+3+5 = 9 total orbitals (each of which can house 2 electrons)
9 x 2 e- = 18e 38. What is an electron configuration?
A notation for indicating where all of the electrons in a particular atom/ion.
When we write electron configurations, we must fill electrons into orbitals in order of lowest to
highest energy. This cannot be determined by comparing values of n.
39. What is the penetration effect?
As you can see from the graph, though electrons from the 3d orbital are generally closer to the
nucleus, those electrons closest to the nucleus in a 4s orbital are able to nudge in more than 3d
electrons. The closer an electron can get to the nucleus, the lower it’s energy!
40. How can you figure out how to order the orbitals?
41. What is the Aufbau Principle?
The “building up” principle. Start at the lowest energy orbital and move up.
42. What is Hund’s Rule?
This rule basically instructs us to separate electrons out when creating the configuration. For
example in a p orbital electrons should be spread out as:
43. What is long hand vs. short hand electron configuration?
Long hand notation is where every single orbital in the electrons configuration is written out.
The long hand notation for Mg:
1s22s22p63s2
The short hand notation references the closest noble gas (as it would have the same
configuration as that noble gas) plus the additional electrons placement.
The short hand notation of Mg:
[Ne]3s2
44. Write the short hand electron configurations for
a. Fe
[Ar]4s23d6
b. Hf
[Xe]6s24f 145d2
c. Cr* an exception
[Ar]4s13d4
d. Cu* an exception
[Ar]4s13d 10
45. What “families” of elements have an exception to the normal method of writing out electron
configurations?
The elements with 4 electrons in the d orbital (including Cr, Mo and W).
Normally you would create an electron configuration that looked like:
[noble gas] ns2 (n-1)d4 instead it is written [noble gas] ns1 (n-1)d5
The elements with 9 electrons in the d orbital (including Cu, Ag and Au).
Normally you would create an electron configuration that looked like:
[noble gas] ns2 (n-1)d9 instead it is written [noble gas] ns1 (n-1)d10
46. Which configuration indicates an excited state?
An excited state is one in which electrons are in higher energy orbitals than they would normally
be (i.e. not ground state). The way you can tell if this is the case is to look and see if the
electrons fill in the normal pattern (which would indicate ground) or an abnormal pattern (an
indication of excitement).
a. 1s22s23p1
This would be an excited state. The ground state configuration would be:
1s22s22p1
b. 1s22s22p6
This follows ground state configuration – this means it is not excited.
c. 1s22s22p43s1
This corresponds to an excited state. The “normal” configuration would be:
1s22s22p5
d. [Ar]4s13d10
This is ground state – remember that this is the ground state configuration of Cu.
Usually something like this would indicate an excited state – this is just one of the
exceptions.
47. What are valence electrons?
All of the electrons within the outermost n-value. These are the e- involved in bonding. In a
group (or column) in the periodic table, all the atoms have the same valence e- configuration
48. What are core electrons?
All of the other electrons in an atom. These electrons are bound more tightly than the valence
electrons (this makes them much harder to remove).
49. Write the electron configuration for Sc+.
When creating the electron configuration for an ion – begin by removing the electrons from the
highest energy shell (highest n-value) first.
Sc: [Ar]4s23d1 from this we can see that n=4 is the highest energy level. This means that we pull
the first electron from the 4s orbital. This gives us:
Sc+ = [Ar]4s13d1
50. What is shielding?
Shielding is the phenomenon where the positive charge of the nucleus (Z) is felt as a reduced
positive charge (Zeffective) by some electrons.
Ultimately the core electrons are very good at shielding the valence electrons from feeling the
full positive charge from the nucleus. Valence electrons are very bad at shielding each other
from the positive charge.
(If you were in CLAS, consider the dodge ball example)
51. What is ionization energy?
The energy required to remove an electron from a gaseous atom/ion.
X(g) X+(g) + e a. General trend on the periodic table:
52. What can ionization energy tell us about core and valence electrons?
Al (g) Al+(g) + eAl+ (g) Al2+(g) + eAl2+ (g) Al3+(g) + eAl3+ (g) Al4+(g) + e-
I1 = 580 kJ/mol
I2 = 1815 kJ/mol
I3 = 2740 kJ/mol
I4 = 11600 kJ/mol
Why the jump from I3 I4? Because I1 – I3 is the energy related to pulling off valence electrons.
It gets successively larger because as the Al develops the positive charge the electrons are more
and more attracted to the ion. The cause for the massive jump from I3 I4 is that I4 represents
the amount of energy required to pull off a core electron. Core electrons are much more tightly
bound to the nucleus than are valence electrons.
53. Why does phosphorus have a first ionization energy of 1060 kJ/mol and sulfur have a first
ionization energy of 1005 kJ/mol?
P: [Ne]3s23p3
S: [Ne]3s23p4
As you can see from the electron configuration of phosphorus and sulfur, the electrons in the
phosphorus configuration are fully spread out over the 3p orbitals. In the sulfur, on the other
hand, there would be some repulsion between the electrons sharing the orbital. Because of this
repulsion is would be easier to pull the sulfur electron out.
54. Which configuration would have the highest 1st ionization energy and which would have the
lowest 2nd ionization energy?
a. 1s22s22p6 – this would have the highest first ionization energy because the electrons are
in a lower energy level and therefore – closer to the nucleus.
b. 1s22s22p63s1
c. 1s22s22p63s2 – this would have the lowest first ionization energy.
55. What is electron affinity?
The energy change when an e- is added to a gaseous atom.
X(g) + e- X – (g)
a. What element has the greatest electron affinity?
Chlorine has the greatest electron affinity. The reason that chlorine has a greater
electron affinity than fluorine us that fluorine small size leads to some electron-electron
repulsion – making it less favorable.
b. General trends on the periodic table:
56. Trends in atomic radius
57. Trends in ionic radius
Cations are smaller than their atomic counterpart. This is because; as the atoms lose electrons,
the positive charge (Zeff) gets larger. The more positivity the electrons feel, the closer the
electrons pull in.
Anions are larger than their atomic counterparts. This is because; as the atoms gain electrons,
the greater the negativity felt (smaller Zeff). The increased negativity causes the electrons to
repel one another – pushing each other further apart.
58. Arrange the following in order of increasing size
a. Ba, F, Si
Ba is in the 6th row. F is in the 2nd row. Si is in the 3rd row. Atomic size increases going
down the periodic table so…
F< Si<Ba
b. K, Br, Ni
K, Br and Ni are all in the 4th row. Thus we will arrange these knowing that the radius
decreases going from group 1 group 8. Thus…
Br < Ni < K
59. Which in each set has the smallest 1st ionization energy?
It is, generally, easiest (lowest ionization energy) to pull electrons off of bigger atoms and
negatively charged ions. Though there are occasion where it may be necessary to look at the
electron configurations to fully determine lowest first ionization energy.
a. Ca, Sr, Ba
Based on placement in the periodic table, the size ordering of the atoms of each
element would be as follows:
Ca < Sr < Ba
Because Ba is the largest atom, it would be the easiest to pull an electron from, thus it
would have the lowest first ionization energy.
i. N, O, F
Based on size: F < O < N, one would assume that N would have the lowest first
ionization energy. However, it would be necessary to double check this by
looking at the electron configurations:
N : [He]2s22p3
O: [He]2s22p4
F: [He]2s22p5
What do you notice about these configurations? In the 2p orbital of nitrogen,
the electrons are spread out over the 3 p orbitals. In the oxygen configuration
there is one orbital (within 2p) that has a set of electrons together. In the
fluorine configuration, there are 2 orbital (within the 2p) with electrons paired
up. What can you glean from this… which atom has the most to gain from the
removal of one electron?
Oxygen. If it loses one electron all the electrons in the 2p shell will be spread
out – biog drop in energy because their will not be a shared pair of electrons
that experience repulsion.
ii. S2-, S, S2+
In this case we are focused on the charges - it is easiest to remove an electron
from a negatively charged ion due to all the repulsion felt by the electron.
This means that S 2- would have the lowest first ionization energy.
60. Order in terms of increasing radius
a. Na-, Na, Na+
Remember that because of positive/negative attraction – cations are smaller than the
neutral atom and because of negative/negative repulsion - anions are larger than the
neutral atom.
This leads to:
Na+ < Na < Na 61. How would you determine the electron affinity for Mg2+
Look at the second ionization energy of magnesium.
Remember that electron affinity means adding an electron Mg 2+ + e - Mg+
This reaction is the reverse of the second ionization energy of Mg.
So to determine the energy associated with the flip reaction – just change the sign of the
ionization energy.
62. To which family would the following element belong?
This jump means that the first 2 electrons that were removed valence and the 3 started pulling
from core. This means that this substance has 2 valence electrons and therefore comes from
group 2.