Chem 6A 2011 (Sailor)
Name:
Student ID Number:
Section Number:
QUIZ #7
VERSION A KEY
Some useful constants and relationships:
Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45
1 atm = 760 Torr
1J = 1kg.m2/s2
1 eV = 1.6022 x 10-19J
101.325 J = 1 L.atm
.
.
-1 . -1
.
-1 . -1
R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K
Avogadro constant: 6.022 x 1023 mole-1
Planck's constant = h = 6.6261 x 10-34 J.s
8
RH = 1.097 x 10-2 nm-1
c = speed of light: 3.00 x 10 m/s
-2
.
C2 = second radiation constant = 1.44 x 10 K m
q = CpmΔT
1
Emitted power (W)
Tλmax = C2
= (constan t)T 4
e = mc 2
c = λν
5
Surface area (m2 )
⎛ 1
⎛ Z 2 ⎞
1 ⎞
hc
1
= R H ⎜ 2 − 2 ⎟
E=
E(in Joules) = −2.18 ×10−18 ⎜ 2 ⎟
E = hν
λ
λ
⎝ n ⎠
⎝ n1 n 2 ⎠
€
Chem 6A 2011 (Sailor)
QUIZ #7
1. Rank each set of elements in order of decreasing size. The first one is done for you as
an example. (10 pts)
(a) Li, Na, K K > Na > Li
(b) Sr, Ba, Ca Ba > Sr > Ca (5 pts if correct, all or nothing)
(c) O, O-, O2-
O2- > O- > O (5 pts if correct, all or nothing)
2. Fill in the electrons in the valence level orbital diagrams below to provide a ground state
electron configuration for each gas phase atom or ion that is consistent with the magnetic
property given and label the valence s, d, and p shells with the correct principle quantum
number. The first one is done as an example. (20 pts)
Paramagnetic Br
↑↓
↑↓
↑↓
↑↓
2+
Diamagnetic
€ Cd €
€
€
↑↓
↑↓
€
€
€
↑↓
↑↓
€
↑↓
€
↑↓
↑
€
€
↑↓
4d
Paramagnetic
€ Cu €
4s
↑↓
4p
↑↓
5s
↑
↑↓
3d
4s
€
↑↓
5p
€
↑↓
↑↓
3d
↑↓
4p
2 pts for correct
quantum
€
€
€shell numbers,
€
€all or nothing. 8 points for correct electron
configuration, all or nothing This was homework problem 8.61 in the book.
3. Identify each element from the given electron configuration for its neutral atom. Indicate the
number of unpaired electrons in each. The first one is done as an example. (20 pts) This was
homework problem 8.30 in the book.
element
Electron configuration
# unpaired electrons
P
[Ne] 3s23p3
3
Sr
(3 pts)
Si
(3 pts)
Ne(3 pts)
Fe
(3 pts)
[Kr] 5s
2
0 (2 pts)
[Ne] 3s23p2
2 (2 pts)
[He] 2s22p6
0 (2 pts)
[Ar] 4s23d6
4 (2 pts)
Chem 6A 2011 (Sailor)
Name:
Student ID Number:
Section Number:
QUIZ #7
VERSION B KEY
Some useful constants and relationships:
Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45
1 atm = 760 Torr
1J = 1kg.m2/s2
1 eV = 1.6022 x 10-19J
101.325 J = 1 L.atm
.
.
-1 . -1
.
-1 . -1
R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K
Avogadro constant: 6.022 x 1023 mole-1
Planck's constant = h = 6.6261 x 10-34 J.s
8
RH = 1.097 x 10-2 nm-1
c = speed of light: 3.00 x 10 m/s
-2
.
C2 = second radiation constant = 1.44 x 10 K m
q = CpmΔT
1
Emitted power (W)
Tλmax = C2
= (constan t)T 4
e = mc 2
c = λν
5
Surface area (m2 )
⎛ 1
⎛ Z 2 ⎞
1 ⎞
hc
1
= R H ⎜ 2 − 2 ⎟
E=
E(in Joules) = −2.18 ×10−18 ⎜ 2 ⎟
E = hν
λ
λ
⎝ n ⎠
⎝ n1 n 2 ⎠
€
Chem 6A 2011 (Sailor)
QUIZ #7
1. Rank each set of elements in order of decreasing size. The first one is done for you as
an example. (10 pts)
(a) Li, Na, K K > Na > Li
(b) N, As, P As > P > N (5 pts if correct, all or nothing)
(c) S, S-, S2-
S2- > S- > S (5 pts if correct, all or nothing)
2. Fill in the electrons in the valence level orbital diagrams below to provide a ground state
electron configuration for each gas phase atom or ion that is consistent with the magnetic
property given and label the valence s, d, and p shells with the correct principle quantum
number. The first one is done as an example. (20 pts)
Paramagnetic Br
↑↓
↑↓
↑↓
↑↓
2+
Paramagnetic
€ Ru €
€
↑↓
€
↑
5s
€
↑↓
↑↓
↑↓
€
↑↓
€
€
↑
€
↑↓
↑
4p
↑
€
€
↑
4d
Paramagnetic
€ Ge €
↑↓
↑↓
3d
4s
€
↑↓
5p
€
↑↓
↑↓
↑
↑
4s
3d
4p
2 pts for correct quantum shell numbers, all or nothing. 8 points for correct electron
configuration,
all€or nothing
€
€
€
€
€
€
This was homework problem 8.61 in the book.
3. Identify each element from the given electron configuration for its neutral atom. Indicate the
number of unpaired electrons in each. The first one is done as an example. (20 pts) This was
homework problem 8.30 in the book.
element
Electron configuration
# unpaired electrons
P
[Ne] 3s23p3
3
Rb
(3 pts)
Al
(3 pts)
Cl
(3 pts)
Co
(3 pts)
[Kr] 5s1
1 (2 pts)
[Ne] 3s23p1
1 (2 pts)
[Ne] 3s23p5
1 (2 pts)
[Ar] 4s23d7
3 (2 pts)
Chem 6A 2011 (Sailor)
Name:
Student ID Number:
Section Number:
QUIZ #7
VERSION C KEY
Some useful constants and relationships:
Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45
1 atm = 760 Torr
1J = 1kg.m2/s2
1 eV = 1.6022 x 10-19J
101.325 J = 1 L.atm
.
.
-1 . -1
.
-1 . -1
R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K
Avogadro constant: 6.022 x 1023 mole-1
Planck's constant = h = 6.6261 x 10-34 J.s
8
RH = 1.097 x 10-2 nm-1
c = speed of light: 3.00 x 10 m/s
-2
.
C2 = second radiation constant = 1.44 x 10 K m
q = CpmΔT
1
Emitted power (W)
Tλmax = C2
= (constan t)T 4
e = mc 2
c = λν
5
Surface area (m2 )
⎛ 1
⎛ Z 2 ⎞
1 ⎞
hc
1
= R H ⎜ 2 − 2 ⎟
E=
E(in Joules) = −2.18 ×10−18 ⎜ 2 ⎟
E = hν
λ
λ
⎝ n ⎠
⎝ n1 n 2 ⎠
€
Chem 6A 2011 (Sailor)
QUIZ #7
1. Rank each set of elements in order of decreasing size. The first one is done for you as
an example. (10 pts)
(a) Li, Na, K K > Na > Li
(b) Mg, Be, Ca Ca > Mg > Be (5 pts if correct, all or nothing)
(c) Se, Se-, Se2-
Se2- > Se- > Se (5 pts if correct, all or nothing)
2. Fill in the electrons in the valence level orbital diagrams below to provide a ground state
electron configuration for each gas phase atom or ion that is consistent with the magnetic
property given and label the valence s, d, and p shells with the correct principle quantum
number. The first one is done as an example. (20 pts)
Paramagnetic Br
↑↓
↑↓
↑↓
↑↓
Paramagnetic
€ Se €
↑↓
€
€
↑↓
↑↓
↑↓
↑↓
€
€
↑↓
↑↓
€
€
↑↓
↑
3d
2+
Paramagnetic
€ Mo €
€
↑
€
↑
↑
4p
↑↓
4s
€
↑↓
3d
4s
€
↑↓
↑
4p
€
↑
€
€
€
↑
5s
4d
5p
2 pts for correct quantum shell numbers, all or nothing. 8 points for correct electron
configuration,
or nothing
€ all €
€
€
This was homework problem 8.61 in the book.
3. Identify each element from the given electron configuration for its neutral atom. Indicate the
number of unpaired electrons in each. The first one is done as an example. (20 pts) This was
homework problem 8.30 in the book.
element
Electron configuration
# unpaired electrons
P
[Ne] 3s23p3
3
K
(3 pts)
S
(3 pts)
I
(3 pts)
Ni
(3 pts)
[Ar] 4s1
1 (2 pts)
[Ne] 3s23p4
2 (2 pts)
[Kr] 5s24d105p5
1 (2 pts)
[Ar] 4s23d8
2 (2 pts)