chapter 6 electronic structure

Question 1

Which of the following has the highest frequency?

a. microwaves
b. radio waves
c. gamma rays
d. infrared
e. ultraviolet

Answer 1

gamma rays

This is pure memorization.

Question 2

Which of the following has the shortest wavelength?

a. visible light
b. microwaves
c. radio waves
d. infrared
e. ultraviolet

Answer 2

ultraviolet

This is pure memorization.

Question 3

Arrange the following in order of increasing energy:
x-rays…infrared…ultraviolet…radio waves

a. UV…infrared…radio waves…x-rays
b. radio waves…infrared…UV…x-rays
c. x-rays…UV…infrared…radio waves
d. x-rays…infrared…radio waves…UV
e. radio waves…UV…infrared…x-rays

Answer 3

radio waves…infrared…UV…x-rays

Question 4

Arrange the following in order of increasing wavelength:
visible…gamma rays…UV…microwaves

a. gamma…UV…micro…visible
b. microwaves…gamma…UV…visible
c. microwaves…UV…gamma…visible
d. gamma…UV…visible…micro
e. UV…gamma…visible…micro

Answer 4

d.

Question 5

Arrange the following in order of increasing frequency:
x-rays…gamma rays…visible…infrared

a. infrared…visible…x-rays…gamma
b. gamma…visible…infrared…x-rays
c. gamma…visible…x-rays…infrared
d. gamma…x-rays…visible…infrared
e. infrared…gamma…x-rays…visible

Answer 5

a.

Question 6

What is the energy of a photon with a frequency of 1.0 x 10^15 Hz
(see study guide for notation definitions)
λν = c
E = hν
c = 3.0 x 108m/s
h = 6.63 x 10-34J∙s

a. 3.0 x 10^23 J
b. 3.33 x 10^6 J
c. 6.63 x 10^-19 J
d. 2.0 x 10^-25 J
e. 3.0 x 10^-7 J

Answer 6

6.63 x 10^-19

This gives you more information than you need. Just focus on the E = hv
v is the frequency which is in the question.
h is Planck’s constant.
Just multiply them.

Question 7

What is the frequency of a photon with a wavelength of 750 nm?
λν = c
E = hν
c = 3.0 x 108m/s
h = 6.63 x 10-34J∙s

a. 4.0 x 10^5 Hz
b. 2.3 x 10^11 Hz
c. 4.0 x 10^14 Hz
d. 2.5 x 10^-15 Hz
e. 2.5 x 10^16 Hz

Answer 7

4.0 x 10^14 Hz

The tricky part here is nano means 9 zero’s. I always think it’s 6 for some reason so it messes me up.
You just use the equation for
wavelength x frequency = speed of light

Question 8

What is the energy of a photon with a wavelength of 300 nm?
λν = c
E = hν
c = 3.0 x 108m/s
h = 6.63 x 10-34J∙s

a. 6.63 x 10^-19 J
b. 9.0 x 10^10 J
c. 1.0 x 10^15 J
d. 2.0 x 10^-40 J
e. 90 J

Answer 8

6.63 x 10^-19 J

In this one, you have to incorporate two of the equations. First, figure out the frequency and then plug that into the
E = hv

Question 9

What is the frequency of a photon with an energy of 6.0 x 10^-20 J ?
λν = c
E = hν
c = 3.0 x 108m/s
h = 6.63 x 10-34J∙s

a. 1.1 x 10^-14 Hz
b. 2.0 x 10^-28 Hz
c. 5.0 x 10^27 Hz
d. 2.5 x 10^-15 Hz
e. 9.0 x 10^13 Hz

Answer 9

9.0 x 10^13 Hz

You can just use your given in the equation:
E = hv

Question 10

What is the wavelength of a photon with an energy of 5.0 x 10^-18 J
λν = c
E = hν
c = 3.0 x 108m/s
h = 6.63 x 10-34J∙s

a. 6.0 x 10^25 m
b. 7.5 x 10^15 m
c. 2.5 x 10^7 J
d. 4.0 x 10^-8 m
e. 7.5 x 10^-6 m

Answer 10

4.0 x 10^-8 m

This is another one where you have to incorporate two of the equations. First, find you frequency.

Question 11

If a photon with an energy of 10eV strikes a metal plate resulting in the ejection of an electron with a maximum kinetic energy of 7 eV, then what is the ionization energy (or work function) of the metal?
K.E.electron = Ephoton minus the work function

a. -3 eV
b. 3 eV
c. 7 eV
d. 10 eV
e. 13 eV

Answer 11

b. 3 eV

Question 12

Which of the following electronic transitions would involve an absorption of a photon of the highest energy?

a. n=1 to n=2
b. n=3 to n=1
c. n=1 to n=3
d. n=4 to n=1
e. n=1 to n=4

Answer 12

e. n=1 to n=4

When an electron absorbs energy, it hangs onto that energy by being promoted to a higher energy level. The outermost shells of an electron become higher and higher in energy. So…
The electron at n=1 receives energy and absorbs it which transfers it to a higher (outer) shell. The most drastic of these transitions to choose from is n1 to n4

Question 13

Which of the following electronic transitions would involve an absorption of a photon of the highest frequency?

a. n=2 to n=5
b. n=3 to n=1
c. n=2 to n=3
d. n=2 to n=1
e. n=2 to n=4

Answer 13

a. n=2 to n=5

The represent moving 3 levels.
In answer c., the electron is only moving 1 level and in answer e., the electron is only moving up in energy 2 levels so…

Question 14

Which of the following electronic transitions would involve the emission of a photon of the longest wavelength?

a. n=2 to n=5
b. n=3 to n=5
c. n=3 to n=4
d. n=2 to n=1
e. n=5 to n=1

Answer 14

d. n=2 to n=1
Energy is being released (emitted) not absorbed so the electron is going to drop from a higher energy level (outer shell) down to a lower level. The only two answer choices that do this and d. and e.
But there is a trick to the question. The ‘longest wavelength’ means the weaker of the two transitions. If it would have said ‘the highest frequency’ then it would have been e.

Question 15

What is the wavelength of an electron moving with a velocity of 1.5 x 10^6 m/s?
wavelength = h/mv
h = 6.63 x 10-34J∙s
mass of electron = 9.11 x 10-31kg

a. 2.1 x 10^9 m
b. 1091.7 m
c. 2.0 x 10^-11 m
d. 4.9 x 10^-10 m
e. 3.2 m

Answer 15

4.9 x 10^-10 m

I actually got 5.1 x 10^-10 but I think this is from rounding

Question 16

The Heisenberg Uncertainty Principle quantifies limits on the certainties in which of the following?
a. mass and velocity
b. density and volume
c. position and momentum
d. velocity and acceleration
e. position and acceleration

Answer 16

c. position and momentum

Question 17

Which of the following could be a set of quantum numbers [n, l, m(l), m(s)] for an electron in a 3d orbital?
a. [3, 2, -1, 1]
b. [3, 1, 1, -1/2]
c. [2, 2, 0, -1/2]
d. [3, 2, -1, -1/2]
e. [3, 2, 3, 1/2]

Answer 17

d. [3, 2, -1, -1/2]

So if it’s 3d, your first number is right there: 3
The next number pertains to the gibven letter so… 0 = s, 1 = p, 2 = d, 3 = f
So now you have [3, 2,
Next, if l = 2, that means that m(l) could equal -2, -1, 0, 1, or 2
so this leaves you with answers a. and d.
and since a. has a spin of 1, that can’t be it because spins can only be 1/2 or -1/2

Question 18

What is the value of the principal quantum number for a valence electron in a fluorine atom?
a. 0
b. 1
c. 2
d. 3
e. 4

Answer 18

c. 2
fluorine has 9 electrons which means electrons would fill up to 2p

Question 19

How many possible sets of quantum numbers [n, l, m(l), m(s)] are possible for an electron in 2p?
a. 1
b. 2
c. 3
d. 5
e. 6

Answer 19

e. 6
This is a little tricky. I always want to say 5 because…
n won’t change. It will always be 2.
l won’t either. It will always be 1 for p. Since l = 1, that means that m(l) can be one of three different numbers. -1, 0, or 1. So that’s three possbile quantum numbers right there. The spin can be one of two possible numbers. -1/2 or 1/2
This gives 5 options. But when you combine (-1)(-1/2), (-1)(1/2), (0)(-1/2), (0)(1/2), (1)(-1/2), (1)(1//2)
So that’s 6 total combinations.
It’s honestly kind of a stupid question. It’s not like it’s ever going to come up in real life.

Question 20

How many values of m(l) quantum numbers are possible for an electron in a 3p orbital?
a. 0
b. 1
c. 3
d. 5
e. 77

Answer 20

c. 3
the l number will come from the p shell (given). the s shell equals 0, the p shell equals 1, the d shell equals 2
So you’re dealing with l = 1
This means that your numbers for m(l) can be -1, 0, 1
That’s 3 possible numbers

Question 21

What is the value of the azimuthal number for a valence electron in a potassium atom?
(azimuthal = l) like the letter l, not 1
a. 0
b. 1
c. 2
d. 3
e. 4

Answer 21

a. 0
K has one valence electron. It’s all the way out in 4s. The 4 is pretty irrelevant in this question. But the fact that it’s in an s subshell makes l equal to 0

Question 22

The following electron configuration is a violation of which of the following?
2s has one electron in it
1s has one electron in it
a. Hess’s law
b. Heisenberg Uncertainty Principle
c. Hund’s rule
d. Aufbau principle
e. Pauli Exclusion principle

Answer 22

d. Aufbau principle
electrons fill orbitals in order of increasing energy
Just use the study guide for these questions as I haven’t really come up with a good way (yet) for memorizing all these.

Question 23

The following electron configuration is a violation of which of the following?
1s is full
2 s is full
but 2p has one in the first square, none in the second square, and two in the third square
(it may help to draw this)
a. Hess’s Law
b. Heisenberg Uncertainty principle
c. Hund’s rule
d. Aufbau principle
e. Pauli Exclusion principle

Answer 23

c. Hund’s rule
degenerate orbitals each get an electron before pairing

Question 24

The following electron configuration is a violation of which of the following…
1s is filled
2s is filled but both electrons are flipped up
a. Hess’s law
b. Heisenberg Uncertainty principle
c. Hund’s rule
d. Aufbau principle
e. Pauli Exclusion principle

Answer 24

e. Pauli Exclustion principle
No two electrons in an atom have the same 4 quantum numbers.
In this case, one electron in the 2s shell must be flipped up and the other must be flipped down or (-1/2)(1/2)

Question 25

What is the ground state electron configuration of oxygen? a. 1s2 2s2 2p4 b. 1s2 2s2 2p6 c. 1s2 2s2 2p2 3s2 d. 1s2 2p6 e. 1s2 2s2 2p3

Answer 25

a. 1s2 2s2 2p4 Remember, oxygen might be in row 16 but it only has 8 total electrons

Question 26

What is the ground state electron configuration for Sc? a. 1s2 2s2 2p6 3s2 3p6 4s1 3d2 b. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d1 c. 1s2 2s2 2p6 3s2 3p6 4s2 3d1 d. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s1 4d2 e. 1s2 2s2 2p6 3s2 3p6 4s2 4p1

Answer 26

c. It's a little weird that 4s2 is filled before 3d1 but it's because they overlap a little. It's easier to see when looking at the study guide. It can also be written as [Ar] 4s2 3d1

Question 27

What is the ground state electron configuration for Ag? a. [Kr] 5s2 4d9 b. [Ar] 5s2 4d9 c. [Kr] 5s1 4d10 d. [Ar] 4s2 3d9 e. [Kr] 4s2 4d9

Answer 27

c.

Question 28

What is the ground state electron configuration of P^3- ? a. [Ar] 4s2 3d10 4p6 b. 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p3 c. 1s2 2s2 2p6 3s2 3p3 d. 1s2 2s2 2p6 3s2 3p6 e. 1s2 2s2 2p6 3s2

Answer 28

d. P^3- is going to have 18 because of the 3 extra electrons.

Question 29

What is the ground state electron configuration of Ca^2+ a. [Ar] b. 1s2 2s2 2p6 3s2 3p6 4s2 3d2 c. [Ar] 4s2 d. [Ar] 4s2 4p2 e. [Ar] 4s2 4d2

Answer 29

a. [Ar] Ca^2+ is going to have 18 electrons because two were removed. When removing electrons, I think they're removed from the highest energy level first. So, even though 4s2 is slightly below 3d on the diagram (study guide), they will be removed first because 4 is a higher energy level than 3. So basically, you're left with something with no electrons in the 4 energy level at all. And the only answer that doesn't mention '4' is [Ar]

Question 30

What is the ground state electron configuration of Ti^2+ a. [Ar] 4s2 3d2 b. [Ar] 4s2 3d4 c. [Ar] 4s2 4d4 d. [Ar] 4s2 e. [Ar] 3d2

Answer 30

e. The 2 electrons removed are going to be removed from the highest energy level (4) first.

Question 31

How many valence electrons does Na have? a. 0 b. 1 c. 2 d. 3 e. 4

Answer 31

b.

Question 32

How many valence electrons does Se have? a. 2 b. 3 c. 4 d. 5 e. 6

Answer 32

e. The column on the periodic table basically tells you but to count them is a little weird. The first ring has 2 electrons like every element. The 2nd ring has 8. But then the 3rd ring has 18.

Question 33

How many valence electron does vanadium have? a. 2 b. 3 c. 4 d. 5 e. 6

Answer 33

d.

Question 34

Which of the following has the greatest number of valence electrons? a. Na b. Ca c. Si d. Sc e. Cs

Answer 34

c.

Question 35

Which of the following shows a nitrogen atom in an excited state? a. 1s2 2s2 2p6 b. 1s2 2s2 2p4 c. 1s2 2s2 2p3 d. 1s2 2s2 2p2 4s1 e. 1s2 2s2 2p1 5d1

Answer 35

d.

Question 36

Which of the following shows an iron atom in an excited state? a. [Ar] 4s2 3d6 b. [Ar] 3d6 c. [Ar] 4s1 3d6 6p1 d. [Ar] 3d5 e. [Ar] 4s2 3d7

Answer 36

c.

Question 37

How many upaired electrons does a single nitrogen atom have? a. 0 b. 1 c. 2 d. 3 e. 4

Answer 37

d. 3 To guess b. 1 would be a common mistake. The electrons will fill up a shell one by one before pairing off so there will be 3 unpaired electrons in the 2p shell. This would be Hund's rule

Question 38

Which of the following is paramagnetic? 'paramagnetic' describes a substance that contains unpaired electrons a. Ne b. Be c. Zn d. P e. Ca

Answer 38

d. P A quick hint is it's also the only element there with an odd number of electrons

Question 39

Which of the following is paramagnetic? 'paramagnetic, describes a substance that contains unpaired electrons I. Cd II. O III. Cl a. I b. III c. I and III d. II and III e. I, II, and III

Answer 39

d. Cd is not paramagnetic. It has 48 electrons which fill up all orbitals completely.

Question 40

How many orbitals are there in the 2nd shell? a. 2 b. 4 c. 6 d. 8 e. 9

Answer 40

b. The 2nd shell includes 2s and 2p There is one orbital in 2s (one square) and 3 orbitals in 2p (three squares) And them together to get 4

Question 41

What is the maximum number of electrons that could occupy the 3rd shell? a. 2 b. 4 c. 9 d. 12 e. 18

Answer 41

e. 18 The 3rd shell includes 3s 3p and 3d 3s has one orbital that can fit 2 electrons 3p has 3 orbitals that can fit 6 electrons 3d has 5 orbitals that can fit 10 electrons