Download Chemistry 1000 (Fall 2011) Problem Set #2: Orbitals and Electrons

Chemistry 1000 (Fall 2011)
Problem Set #2: Orbitals and Electrons
Solutions
Answers to Questions in Olmsted (only those w/out answers at the back of the book)
93.
In the real periodic table, screening effects only compete with the effect of increasing
quantum number when n > 2 (i.e. with d and f orbitals). As such, the first two rows of the
periodic table would not change if the effect of increasing n was always more significant
than the effect of screening. We would, however, see the 3d subshell filled before the 4s
subshell (and the 4d before 5s, 5d before 6s, etc.) we would see the 4f subshell filled
before the 5s (and the 5f before 6s). In effect, the order in which electrons would fill the
subshells would be:
1s, 2s, 2p, 3s, 3p, 3d, 4s, 4p, 4d, 4f, 5s, 5p, 5d, 5f, 5g(!), 6s, 6p, etc.
(a)
The periodic table would therefore look something like:
H
Li
Na
K
Rb
He
Be
Mg
Ca
Sr
(b)
(c)
(d)
97.
(a)
(b)
(c)
B
Al
Ga
In
C
Si
Ge
Sn
N
P
As
Sb
O
S
Se
Te
F
Cl
Br
I
Ne
Ar
Kr
Xe
Sc
Y
La
Ti
Zr
Hf
V
Nb
Ta
Cr
Mo
W
Mn
Tc
Re
Fe
Ru
Os
Co
Rh
Ir
Ni
Pd
Pt
Cu
Ag
Au
Zn
Cd
Hg
lanthanides
actinides
g-block!!!!
This is, of course, assuming that the name of each element is the same based on the
identity and number of electrons in the last subshell. In the crazy world described by this
question, who knows if that would be the case!!!
Note that it is due to the competition between the effects of increasing quantum number
and screening that we have the periodic table we do (without a g-block!)
If you used atomic numbers for your periodic table (or our current elements in order of
atomic number), that’s fine too. It’s the overall shape change that’s important.
Given the table above, the elements we’d expect to behave as “halogens” (adding an
electron easily) would be those one electron away from a complete valence shell: H, F,
Cu, the second last lanthanide and the second last element in the period 5 g-block.
Given the table above, the elements we’d expect to behave as “noble gases” (chemically
unreactive) would be those with a complete valence shell: He, Ne, Zn, the last lanthanide
and the last element in the period 5 g-block.
Any opinion with a reasoned rationale is acceptable.
Morspin element 18 would be expected to have a larger first ionization energy than
Morspin element 30. They are both in the same Group and therefore have analogous
valence electron configurations (3s3 for Morspin element 18 and 4s3 for Morspin element
30 ); however, the valence electrons in Morspin element 18 have a lower n quantum
number which means that they are closer to the nucleus and therefore more strongly
attracted to it. Thus, a removal of a valence electron from Morspin element 18 will
require more energy.
Morspin element 15 would be expected to have a larger radius than Morspin cation 172+.
They both have the same electron configuration (since they both have 15 electrons);
however, Morspin cation 172+ has two more protons. As such, the electrons in Morspin
cation 172+ are more strongly attracted to the nucleus and its radius is therefore smaller.
Morspin element 47 would be expected to have a more negative enthalpy of electronic
attraction (i.e. “larger electron affinity”) than Morspin element 48. The valence electron
configuration for Morspin element 47 is 4s3 4p2 while the valence electron configuration
for Morspin element 48 is 4s3 4p3. Since the quantum number laws *other* than that for
electron spin appear to be the same on Morspin, it is reasonable to conclude that there are
still three p orbitals in each subshell (as evidenced by there being nine Groups in the
Morspin p-block). Adding one electron to the third (empty) 4p orbital in Morspin
element 47 should therefore release more heat than adding a second electron into one of
the three (partially filled) 4p orbitals in Morspin element 48. This is because there will
be enough electrostatic repulsion between the two electrons in the same p-orbital to
counteract the fact that Morspin element 48 has one more proton than Morspin element
47 (and therefore a greater nuclear charge).
This effect can be observed on the real periodic table (Earth version). Nitrogen
(analogous to Morspin element 48) is considered to have no electron affinity (enthalpy of
electronic attraction = 0) whereas carbon does release heat when adding an electron to
make C-.
99.
(a)
(b)
The first three noble gases on Morspin would be elements 3, 15 and 27 (each having the
same number of electrons as the atomic number).
Morspin element 4: valence electron in 2s therefore:
2s
n = 2, l = 0, ml = 0, ms = 0
ms = +½ or -½ would also be acceptable instead of ms = 0
Morspin element 7: valence electrons in 2s (3 e-) and 2p (1 e-) therefore:
2s
n = 2, l = 0, ml = 0, ms = 0
n = 2, l = 0, ml = 0, ms = +½
n = 2, l = 0, ml = 0, ms = -½
2p
n = 2, l = 1, ml = 0, ms = 0
ml = +1 or -1 would also be acceptable instead of ml = 0
ms = +½ or -½ would also be acceptable instead of ms = 0
Morspin element 32: valence electrons in 4s (3 e-) and 3d (2 e-) therefore:
4s
n = 4, l = 0, ml = 0, ms = 0
n = 4, l = 0, ml = 0, ms = +½
n = 4, l = 0, ml = 0, ms = -½
3d
n = 3, l = 2, ml = 0, ms = 0
n = 3, l = 2, ml = +1, ms = 0
values for ml may be +1, 0 or -1; however, the two *must* be different!
values for ms may be +½, 0 or -½; however, the two *must* be the same!
Additional Practice Problems
1.
Give the orbital label (1s, 2p, 6f, etc.) for one orbital that corresponds to each set of
quantum numbers. Also, sketch the orbital using Cartesian axes to show its orientation.
Note that you can rotate the axes if it makes your picture clearer/easier to draw.
See your notes/text for pictures of the different types of orbitals.
(a)
n = 4, l = 2, ml = -2
4dxy, 4dxz, 4dyz, 4dx2-y2, or 4dz2
(c)
n = 3, l = 1, ml = +1
3px, 3py, or 3pz
(b)
n = 6, l = 0, ml = 0
6s
(d)
n = 2, l = 1, ml = 0
2px, 2py, or 2pz
2.
Sketch the full set of 3d orbitals and label them with their Cartesian labels.
The orbitals are 3dxy, 3dxz, 3dyz, 3dx2-y2, and 3dz2.
See your notes/text for pictures.
3.
What is the maximum number of electrons (in a single atom) that can be associated with
each of the following combinations of quantum numbers?
(a)
n=2
8
(b)
n = 3, l = 2
10
2
(c)
n = 2, l = 0, ml = 0
(d)
n = 6, l = 3, ml = -3, m s = -½
1
4.
Does each of the following set of quantum numbers describe a possible atomic orbital? If
so, give the label for this orbital. If not, explain why an electron with that set of quantum
numbers isn’t possible.
(a)
n = 4, l = 3, ml = -3, m s = ½
YES 4f
(b)
n = 1, l = 0, ml = 0, m s = -1
NO
m s must be either +½ or -½
(c)
n = 0, l = 0, ml = 0, m s = -½
NO
n cannot be zero
(d)
n = 2, l = 3, ml = 0, m s = ½
NO
l cannot be greater than n
(no 2f orbital exists)
(e)
n = 3, l = -2, ml = 3, m s = -½
NO
l cannot be negative and, since ml
cannot be greater than l and l cannot
be greater than or equal to n, ml
cannot be equal to n
5.
(a)
Explain why each of the following statements is true.
Hydrogen has a more negative enthalpy of electronic attraction than helium.
H-1 = 1s2
He = 1s2
He-1 = 1s22s1
Electron configurations:
H = 1s1
Enthalpy of electronic attraction is the enthalpy change when a neutral gaseous atom
gains an electron.
When gaining an electron, hydrogen goes from having an incomplete valence shell (1s1)
to a full valence shell (1s2). Thus, H-1 is relatively stable and can form.
If helium were to gain an electron, it would go from having a full valence shell (1s2) to
having an incomplete valence shell (1s22s1). Thus, He-1 does not form because it is much
less stable than He.
Since H-1 can form and He-1 cannot, hydrogen has a more negative enthalpy of electronic
attraction than helium (which is one of the few elements to have a positive enthalpy of
electronic attraction).
(b)
(c)
(d)
6.
(a)
(b)
(c)
7.
(a)
Potassium has a greater second ionization energy than calcium.
Second ionization energy is the energy required to remove an electron from X+1 where X
is the element of interest.
For calcium, this is the energy required to form Ca2+ from Ca+. Ca2+ has a complete
valence shell whereas Ca+ does not. As such, Ca2+ is relatively stable and forms readily
from Ca+.
For potassium, this is the energy required to form K2+ from K+. K+ has a complete
valence shell whereas K2+ does not. As such, K+ is more stable and very difficult to
convert to K2+.
The atomic radius of Br -1 is larger than the atomic radius of Br.
Br and Br -1 have the same number of protons, but Br -1 has one more electron. This extra
electron partially shields the other electrons in the valence shell from the positive charge
of the nucleus. As such, the effective nuclear charge felt by the valence electrons of Br -1
is smaller than that felt by the valence electrons of Br, and they are not pulled as strongly
toward the nucleus (increasing the atomic radius).
The atomic radius of nitrogen is smaller than the atomic radius of boron.
Nitrogen has two more protons in its nucleus than boron does. It also has two more
electrons, but these do not completely shield the valence electrons from the increased
positive charge of the nucleus. As such, the effective nuclear charge felt by the valence
electrons of nitrogen is higher than that felt by the valence electrons of boron. The
valence electrons of nitrogen are therefore pulled more strongly toward the nucleus
(decreasing the atomic radius).
Rank the following sets of atoms/ions according to the number of valence electrons (from
least to most).
Sn2+ (2)
Sb (5)
Se (6)
S2- (8)
(i)
S2-, Sb, Se, Sn2+
(ii)
F-, Fe2+, Fe3+, Fr
Fr (1)
Fe3+ (5)
Fe2+ (6)
F- (8)
(iii)
C, Ca, Cl, Co
Ca (2)
C (4)
Cl (7)
Co (9)
List all of the atoms/ions from part (a) that are paramagnetic.
Se
[Ar]4s23d104p4
Sb
[Kr]5s24d105p3
Fr
[Rn]7s1
Fe3+ [Ar]3d5
Fe2+ [Ar]3d6
C
[He]2s22p2
Cl
[Ne]3s23p5
Co
[Ar]4s23d7
List all of the atoms/ions from part (a) that are diamagnetic.
Sn2+ [Kr]5s24d10
S2[Ne]3s23p6
F[He]2s22p6
Ca
[Ar]4s2
Consider H and He+ in the ground state.
Which would you expect to have a higher ionization energy?
He2+ (Both are one-electron atoms/ions, but He2+ has two protons in the nucleus so the
electron should be more strongly attracted to the nucleus therefore more difficult
to remove.)
(b)
Calculate the ionization energy for H.
First ionization energy is the energy required to remove an electron from an atom (far
enough away that it no longer feels any attraction to the nucleus. Mathematically, we can
approximate this as the energy required to excite an electron from n = 1 to n = ∞ (where
∞ stands for infinity). Recall that 1/0 = ∞ therefore 1/∞ = 0.
For a hydrogen atom, Z = 1 (1 proton in the nucleus), n1 = 1 and n2 = ∞
I1 ( H ) = En =∞ − En =1
⎛
⎛ 1
Z2 ⎞ ⎛
Z2 ⎞
1 ⎞
1⎞
⎛ 1
I1 ( H ) = ⎜⎜ − RH 2 ⎟⎟ − ⎜⎜ − RH 2 ⎟⎟ = − RH ⋅ Z 2 ⎜⎜ 2 − 2 ⎟⎟ = ( −2.179 × 10 −18 J )(1) 2 ⎜ 2 − 2 ⎟ = 2.179 × 10 −18 J
n2 ⎠ ⎝
n1 ⎠
n1 ⎠
⎝∞ 1 ⎠
⎝
⎝ n2
(c)
Calculate the ionization energy for He+.
For a helium cation, Z = 2 (2 protons in the nucleus), n1 = 1 and n2 = ∞
I 1 ( He) = E n =∞ − E n =1
⎛
⎛ 1
1 ⎞
1⎞
Z2 ⎞
Z2 ⎞ ⎛
⎛ 1
I 1 ( He) = ⎜⎜ − RH 2 ⎟⎟ − ⎜⎜ − RH 2 ⎟⎟ = − RH ⋅ Z 2 ⎜⎜ 2 − 2 ⎟⎟ = ( −2.179 × 10 −18 J )(2) 2 ⎜ 2 − 2 ⎟ = 8.716 × 10 −18 J
n1 ⎠
n1 ⎠
n2 ⎠ ⎝
⎝∞ 1 ⎠
⎝
⎝ n2
(d)
Did your calculations in parts (b) and (c) support your answer to part (a)?
Yes. The ionization energy for He+ was four times higher than that for H.
8.
Give the name and symbol for the neutral element matching each of the following
descriptions.
“I have 8 neutrons and 6 valence electrons.”
oxygen (O)
6 valence electrons = group 6 or 16; the only element in either
group likely to have only 8 neutrons (i.e. mass of about 2×8 = 16
g/mol) is oxygen
“I have a full octet and belong to the same period as nickel.”
krypton (Kr)
full octet = noble gas (group 18); nickel is in period 4
“Half of my electrons are valence electrons.”
beryllium (Be)
If half the electrons are valence, then half are core. There are only
a few possible values for the number of core electrons. Any
element in period 1 has 0 core electrons; any element in period 2
has 2 core electrons (1s2); any element in period 3 has 10 core
electrons (1s22s22p6); any element in period 4 has 18 core electrons
(1s22s22p63s23p6); etc. Beryllium has 2 core electrons plus 2
valence electrons. Note that there is no other element that meets
these criteria (calcium has 20 electrons, but 18 are core – not 10).
“I have twice as many valence electrons as core electrons.” see logic for 6(c)
carbon (C)
2 core electrons and 4 valence electrons.
(a)
(b)
(c)
(d)
(e)
“I have twice as many core electrons as valence electrons.” (bonus: name all three
elements fitting this description) see logic for 6(c)
lithium (Li)
2 core electrons + 1 valence electron
phosphorus (P)
10 core electrons + 5 valence electrons
cobalt (Co)
18 core electrons + 9 valence electrons
***note that xenon (Xe) has 8 valence electrons – not 18!***
9.
(a)
For a neutral ground state nickel atom,
Write the complete electron configuration.
1s2 2s2 2p6 3s2 3p6 4s2 3d8
Write the electron configuration using the noble gas abbreviation.
[Ar] 4s2 3d8
Draw an orbital occupancy diagram showing the valence electrons. Label each subshell.
(b)
(c)
3d
4s
(d)
Use the table below to list a set of quantum numbers describing the valence electrons.
Use as many rows as necessary; the correct answer may include one or more empty
rows.
Electron
ms
n
l
ml
1 (4s)
4
0
0
+½
2 (4s)
4
0
0
–½
3 (3d)
3
2
+2
+½
4 (3d)
3
2
+1
+½
5 (3d)
3
2
0
+½
6 (3d)
3
2
-1
+½
7 (3d)
3
2
-2
+½
8 (3d)
3
2
+2
–½
9 (3d)
3
2
+1
–½
10 (3d)
3
2
0
–½
11
12