Name
CHEM 115 EXAM #2
Practice Fall 2013
Circle the most correct answer for the following. (3 points each)
1.
Which of the following electron configurations is not allowed by the Pauli exclusion
principle?
a. 1s22s22p2
b. 1s22s22p63s23p1
c. 1s22s22p63s23p4
2
2
6
2
6
d. 1s 2s 2p 3s 3p
e. none of the above [an example of a bad e
configuration that violates the Pauli Excl. Prin. would be 1s22s22p63s33p1]
2.
3.
What is the ground state electron configuration of argon (Ar)?
a. 1s22s22p63s3
b. 1s22s22p63s23p1
2
2
6
2
6
d. 1s 2s 2p 3s 3p e. none of the above
c. 1s22s22p63s13p4
How many neutrons, protons, and electrons are present in
?
a.
b.
c.
d.
e.
4.
48 neutrons,
22 protons,
22 electrons
22 neutrons,
48 protons,
22 electrons
24 neutrons,
24 protons,
22 electrons
26 neutrons,
26 protons,
26 electrons
none of the above, the correct answer is: 26 neutrons, 22 protons, 22 electrons
Of the species listed below, which should be the smallest?
a. Ca2+
5.
e. N3-
b. sulfur, S
c. magnesium, Mg
d. fluorine, F
e. potassium, K
b. oxygen, O
c. magnesium, Mg
d. chlorine, Cl
e. potassium, K
Which of the elements below, in the ground electronic state, has one (1) 5s electron?
a. Na
8.
d. Cl-
Which of the following elements has chemical properties MOST SIMILAR to those of sodium,
Na?
a. Iron, Fe
7.
c. S2-
Which of the following elements has chemical properties MOST SIMILAR to those of chlorine,
Cl?
a. Iron, Fe
6.
b. K+
b. K
c. Al
d. Ga
e. none of the above
Which of the following orbitals has the highest energy in a multielectron atom?
a. 3s
b. 3p
c. 4s
d. 3d
e. 4p
(6 points)
9.
Write the chemical symbols for six ions (ex. Mg2+) that are isoelectronic with Ne AND arrange
them in order of increasing size.
Al3+ < Mg2+ < Na+ < Ne < F- < O2- < N3-
(over)
2
Solve the following problems. Show all your work, always show your units, and express your answers in
the proper number of significant figures. (point values as indicated)
(12 points)
10.
The equation below is the Bohr equation for the energy levels in any hydrogen-like atom.
a. Calculate the first three energy levels for the Li2+ ion.
The atomic number (Z) for Li is 3. Substitution of 3 for Z in the above equation yields:
For n = 1 E1 = -19.62 x 10-18 J
En = -9
B
2
/n =
For n = 2, E2 = -4.905 x 10-18 J
-18
-19.62x10
J
/n
2
For n = 3, E3 = - 2.18 x 10-18 J
b. Calculate the energy lost by an electron dropping from level 3 to level 1 in Li2+.
Δ E = Ef – Ei = E1 – E3 = -19.62 x 10-18 J – (-2.18 x 10-18 J) = -17.44 x 10-18 J
c. Calculate the frequency and wavelength of a photon of light that corresponds to the
energy lost by the electron in part “b”. [h = 6.626x10-34 Js and c = 3.00x10+8 m/sec]
(12 points)
11.
a. Write the electron configuration for the following two species (you may use the shorter
version . . . you know, the one that will start with [Kr]): . . . sorry for the error, I was
editing an old question that asked about Pb!
Sn
[Kr] 5s24d105p2
or
[Kr] 4d105s25p2
Sn2+
[Kr] 5s24d10
or
[Kr] 4d105s2
b. Define the terms paramagnetic and diamagnetic AND use Sn and Sn2+ as examples to more
clearly demonstrate the meaning of these terms.
Paramagnetic species have one, or more, unpaired electrons, Sn is paramagnetic b/c the two
electrons in the 5p level are unpaired (per Hund’s Rule of Maximum Multiplicity).
Diamagnetic species have all electrons paired, Sn2+ is diamagnetic b/c the electrons in the
4d, 5s, and all lower E orbitals are paired.
c. Which electron(s) has(have) the highest energy in Sn? The 5 p electrons.
d. Which is larger Sn or Sn2+? Explain your choice. Sn is larger b/c it has 50 protons
attracting 50 electrons. Sn2+ is smaller b/c the 5p level is empty and the same 50
protons are now attracting only 48 electrons.
3
For numbers 12 – 14 answer in concise complete sentences as appropriate and provide the specific
details requested in the other questions. (6 points each)
12.
What are the three types of radioactivity (discussed in this course) that helped to reveal
the structure of the atom. Give the mass number, atomic number, and charge on each.
The three were alpha (a), beta (b), and gamma (g) rays.
Mass number (A)
Atomic number (Z)
alpha (α)
4
2
beta (β)
0
-1
gamma (γ)
0
0
13.
Charge
+2
-1
0
What are cathode rays? Are the cathode rays produced by all metals the same? Explain
why.
Cathode rays are actually streams of electrons that were ejected from the cathode (the
negative electrode) in a vacuum tube. Since all elements possess identical electrons (same
charge and mass), it does not matter what type of metal the electrode was made out of,
the cathode rays had the same properties.
14.
Explain in some detail our current understanding of the nature of isotopes. Providing
specific examples will enhance your explanation.
Isotopes are variants of atoms of a particular chemical element (which by definition must
contain the same number of protons), that have different numbers of neutrons. The number of
nucleons (protons and neutrons) in the nucleus, known as the mass number, is not the same
for two isotopes of any element. (example: C-12 has 6 protons and 6 neutrons while C-13 has
6 protons and 7 neutrons).
15.
Identify each of the following as either molecular or ionic compounds and provide the
name of the compounds (0.5 points each blank)?
Compound
Ionic or Molecular
Compound name
CuBr2
ionic
copper (II) bromide
Fe2(SO4)3
ionic
iron (III) sulfate
P2 O5
molecular
diphosphorus pentoxide
FeCl3·6 H2O
ionic
iron (III) chloride hexahydrate
CO
molecular
carbon monoxide
Fe2O3
ionic
iron (III) oxide
Ni3(PO4)2
ionic
nickel (II) phosphate
H2 O
molecular
water (or dihydrogen monoxide)
SF6
molecular
sulfur hexafluoride
KNO3
ionic
potassium nitrate
4
(6 points)
16. Use dot symbols to show the formation of NaCl(g) and MgO(g).
(12 points)
17. Use the following reaction diagram (a Born-Haber cycle) to find the lattice energy for MgO(s).
Sum of E loop = 0 = (+411 kJ + 107 kJ + 502 kJ + 121 kJ – 355 kJ + lattice E)
Lattice E = -786 kJ / mole NaCl
5
BONUS (for up to ten points)
Avogadro’s number is huge. To try to understand how big this number really is consider the following:
An average M&M has a volume of 1.1 cm3. Calculate the volume of a mole of M&Ms.
You may want to calculate this in m3 for easier use in the next step.
1.1 cm3 = 1.1x10-6 m3
One mole of these occupies a volume = (6.022x1023) X(1.1x10-6 m3) = 6.62x1017 m3
The surface area of the 48 contiguous states is 7.825 x 1012 m2. Calculate the depth (in meters) to which
one mole of M&Ms would cover the 48 contiguous states.
Volume / area = depth
6.62x1017 m3 / 7.825 x 1012 m2 = ~84,654 m
Finally, to put this into a number which might be more meaningful to a U.S. citizen, convert this depth to
miles (1 mile = 1609.3 meters).
~52.6 miles !!!!