© 2015 L.S. Brown (18 pts) 1. Fill in the blanks in each

(18 pts) 1. Fill in the blanks in each statement below. (You should only need one or two words for each.
You may use the same word more than once if appropriate.)
(For Grading)
(a). A gas is more likely to behave ideally if the temperature is
(b). ____Metals__________________ have relatively low ionization energies
1 (18)
and tend to form positively charged ions.
2 (18)
(c). In any atom, the electrons in the highest occupied shell are called
3 (9)
_____valence electrons_________________.
4 (6)
(d). A photoelectron spectrum for silicon should show a total of
______five (5)_________________________ distinct peaks.
5 (10)
(e). Wavelength and
6 (15)
______frequency_______________________ are two
different ways to specify the color of light.
7 (12)
(f). ______Electronegativity__________________ is a measure of the relative
8 (12)
tendency for an atom to attract the shared electrons in a chemical bond.
(g). An element whose electron configuration contains one or more unpaired
electrons is said to be _______paramagnetic__________________.
(h). The ground state electron configuration for ______iodine (I)_______________________ is
[Kr]5s24d105p2. (Name or symbol is OK.)
(i). Elements in the same _____column (or group)______________ in the periodic table tend to
have similar chemical properties.
We tried to watch for any reasonable synonyms on these.
© 2015 L.S. Brown
2. Answer each of the following questions as indicated.
On b & c: Full credit for the correct sequence. Half credit if you could make it correct by
moving just one of them. No credit if it is more wrong than that.
(4 pts) (a). Write the electron configuration for the monatomic ion most likely to be formed by
nitrogen. (Write out all of the electrons; do not use any shorthand notations.)
1s2 2s2 2p6 (Maybe a point if they just write the configuration for the atom.)
(4 pts) (b). Place the following species in order of increasing radius: I, I–, Xe, Kr. (Please answer by
filling in the blanks below.)
___Kr____ < ___Xe____ < __I_____ < __I–_____
(smallest radius)
(largest radius)
(4 pts) (c). Place the following species in order of increasing ionization energy: Rb, Sr2+, Se, Kr.
(Please answer by filling in the blanks below.)
___Rb____ < ___Se____ < ___Kr____ < ___Sr2+____
(smallest IE)
(largest IE)
(3 pts) (d). Place the following bonds in order of increasing polarity: Li–F, I–I, Si–F. (Please answer by
filling in the blanks below.)
__ I–I__ < ___ Si–F ____ < _ Li–F ____
(least polar)
(most polar)
This one we did as just right or wrong.
(3 pts) (e). All of the following atoms are paramagnetic. Which one should be most strongly
paramagnetic? (Please circle your choice.)
© 2015 L.S. Brown
3. Consider a particular electron in a silicon (Si, Z = 14) atom. The electron has the following
quantum numbers: n = 2, l = 1, ml = 0.
(Please write your answers on the lines provided.)
(3 pts) (a). What type of orbital (1s, 4d, etc.) does this electron occupy?
(3 pts) (b). Assuming the atom is in its ground state electron configuration, what is the maximum
number of electrons in any one silicon atom that could have this set of quantum numbers?
(3 pts) (c). List all the possible values for the ms quantum number for this electron.
_+1/2, –1/2_
4. Consider the following samples of gas, and then answer the questions below by circling the
correct choice. (No need to show work here.)
1.0 L of CH4 at 25°C and 1 atm
1.0 L of F2 at 25°C and 2 atm
(2 pts) (a). Which sample’s molecules have the highest average speed?
the F2
(2 pts)
the CH4
both the same
(b). Which sample’s molecules have the highest average kinetic energy?
the F2
the CH4
both the same
(2 pts) (c). Which sample contains the most molecules?
the F2
the CH4
both the same
© 2015 L.S. Brown
(10 pts) 5. A 1.50-L flask contains a sample of helium gas. When 1.608 g of krypton (Kr, Z = 36) is
injected into the flask, the total pressure of the helium and krypton mixture is found to be 948
torr at 24.0°C. What mass of helium is in the flask? (The volume of the flask and the amount of
helium in the flask remain the same throughout; the krypton is simply being added to the same
flask that already contains the helium.)
Find P for the krypton:
⎛ 1.608 g ⎞
L torr
)(297.15 K)
mol K
n Kr RT ⎝ 83.80 g/mol ⎠
PKr =
=236.98 torr
1.50 L
Subtract from total P to get PHe = 948 – 236.98 = 711 torr
Solve for moles then mass of helium:
nHe =
(711 torr)(1.50 L)
= 0.05755 mol , and that means 0.2304 g
L torr
)(297.15 K)
mol K
nKr =
1.608 g
= 0.01919 mol
83.8 g/mol
nTotal =
(948 torr)(1.50 L)
= 0.07673 mol
L torr
)(297.15 K)
mol K
Subtract those to get nHe = 0.07673 – 0.01919 = 0.05755 mol
Which again is 0.2304 g
© 2015 L.S. Brown
6. An atom originally in its ground state absorbs a photon (we can call it ‘A’) with a
wavelength of 100.0 nm as it makes a transition to a higher energy state. The atom then emits
two photons: one (B) with wavelength 500.0 nm to reach an intermediate energy level, and then
a second (C) to return to the ground state.
(5 pts) (a). In the box to the right, sketch an energy
level diagram depicting this process. Label
the various transitions as A, B, and C to
show which one corresponds to each of the
photons. (λA = 100 nm, λB = 500 nm, and λC
is unknown.)
(10 pts) (b). Find the frequency of photon C (the second emitted photon).
I’ll start by finding the E’s for the 2 given wavelengths:
hc (6.626 × 10 –34 Js)(2.9979 × 108 m/s)
= 1.986 × 10 –18 J
100 × 10 m
hc (6.626 × 10 Js)(2.9979 × 108 m/s)
EB =
= 3.973 × 10 –19 J
500 × 10 m
EA =
The diagram shows us that C = A – B, so we subtract those 2 E’s to get
EC = (1.986 – 0.3973) ×10–18 J = 1.589 ×10–18 J
Convert that to frequency:
E 1.589 × 10 –18 J
= 2.40 × 1015 s –1
h 6.626 × 10 –34 Js
© 2015 L.S. Brown
(12 pts) 7. Nickel metal will react with CO gas to form a compound called nickel tetracarbonyl
(Ni(CO)4), which is a gas at temperatures above ~45°C.
A 1.50-L glass bulb is filled with CO gas to a pressure of 1.20 atm at 73.0°C, and then 0.5869 g
of pure Ni is added. If the reaction described above occurs and goes to completion at constant
temperature, what will the final total pressure in the bulb be? (You can assume that any volume
occupied by the solid nickel is negligible compared to the volume of the flask.)
This is a limiting reagent problem. So I’ll start by writing the equation for the reaction.
Ni(s) + 4 CO(g) → Ni(CO)4(g)
Next find moles of each reactant.
(1.20 atm)(1.50 L)
= 0.0634 mol
RT (0.08206 L atm )(346 K)
mol K
0.5869 g
= 0.01 mol
58.69 g/mol
So the nickel is the limiting reagent. At the end of the reaction, the flask will contain the
Ni(CO)4 product PLUS the left over CO that did not react.
We will get 0.01 mol of the product, because we have 1:1 mole ratio with Ni.
Find the amount of CO that DOES react with the nickel, then subtract from the initial
4 mol CO
= 0.0400 mol CO used up
1 mol Ni
0.0634 mol CO initially – 0.0400 mol CO used up = 0.0234 mol CO left
0.01 mol Ni ×
So total moles of gas is 0.01 + 0.0234 = 0.0334. Finally, use gas law to find P:
L atm
nRT (0.0334 mol)(0.08206 mol K )(346 K)
= 0.632 atm (or 480 torr)
(1.50 L)
© 2015 L.S. Brown
(12 pts) 8. When a photoelectric effect experiment was carried out using a metal 'M' and light at
wavelength λ1, electrons were emitted with a kinetic energy of 1.9 × 10–19 J. The wavelength
was reduced to 1/2 of its original value and the experiment was repeated (still using the same
metal target). This time electrons were emitted with a kinetic energy of 8.1 × 10–19 J. Find the
electron binding energy for metal M. (The actual wavelengths used were not recorded, but it is
still possible to find the binding energy.)
There are a few ways to start. I think the easiest is to see that the difference in the KE’s
given is the difference in the photon energies (because the binding E doesn’t change). Then
also see that the photons with λ 2 will have twice the energy of those with λ 1. So the
difference in energies is equal to the energy of the original (λ 1) photons. Then I can solve
for the binding E:
8.1 × 10 –19 J – 1.9 × 10 –19 J = 6.2 × 10 –19 J
6.2 × 10 –19 J = photon energy at λ1
6.2 × 10 –19 J = BE + 1.9 × 10 –19 J
so BE = 4.3 × 10 –19 J
More in equations:
= BE + KE1 and
= BE + KE2
But we know that λ 2 =
λ 1, so:
= BE + KE1 and
= BE + KE2
Subtract the 1st of those from the 2nd and get
= KE1 – KE2 = 6.2 × 10 –19 J
Put that into
= BE + KE1 and get BE = 4.3 × 10–19 J
© 2015 L.S. Brown