Chemistry 181
Professor S. Alex Kandel
Fall 2015
Due 9/16/2015
Problem Set 3
We will discuss this in class (3:30) on Wednesday the 16th. Be prepared to present
answers to and to discuss any of these up at the blackboard.
1. The densities of the alkali metals (in g/cm3 ) are
Li
0.534
Na
0.97
K
0.86
Rb
1.532
Cs
1.879
The arrangement of atoms is the same in all of these metals (that is, they
have the same crystal structure). Given that lithium has an atomic radius of
approximately 152 pm (1 pm = 10−12 m), estimate the atomic radii of the
other alkali metals. Does this make sense given your understanding of periodic
trends?
The mass of a single lithium atom is the molar mass divided by Avogadro’s
number:
1 mol
6.941 g Li
= 1.153 × 10−23 g
mol
6.022 × 1023 atoms
The volume of a sphere of radius 152 pm is
V
4 3
πr
3
3
4 =
π 152 × 10−12 m
3
= 1.471 × 10−29 m3
=
If we divide this mass by this volume, we will get a density for a single atom
(this is only really meaningful with the very simple description of volume we’re
using here):
1.153 × 10−23 g
= 7.835 × 105 g m−3 = 0.784 g cm−3
1.471 × 10−29 m3
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This number is larger than the actual density because bulk lithium metal isn’t
made up of cubic atoms that pack together densely. Instead, spheres must leave
gaps when packed together, so the actual density is only
0.534 g cm−3
= 0.681
0.784 g cm−3
that is, only 68% of the volume is taken up by atoms-as-spheres. That’s the
packing fraction for a bcc lattice, which is what the crystal structure is for
lithium. Once you’ve got this packing fraction, you can run the calculation for
all of the other elements; for sodium, e.g.:
1 mol
1 cm3 22.99 g
·
·
= 3.936 × 10−23 cm3 /atom
23
0.97 g 1 mol 6.022 × 10 atoms
Again, that’s the amount of space per atom if the atoms are densely packed;
using the 0.68 packing fraction:
V
= 3.936 × 10−23 cm3 · 0.68
4 3
πr = 2.676 × 10−23 cm3
3
r3 = 6.389 × 10−24 cm3
r = 1.85 × 10−8 cm
= 185 pm
If you wanted to take a shortcut (not necessary), you could ratio out the other
alkali metals to lithium, figuring:
density ∝
atomic mass
(atomic radius)3
and so
atomic radius ∝
atomic mass
density
!1/3
so
atomic radius Na
=
atomic radius Li
atomic mass Na density Li
·
atomic mass Li density Na
2
!1/3
22.99 g mol−1 0.534 g cm−3
6.941 g mol−1 0.97 g cm−3
=
!1/3
= 1.8231/3
atomic radius Na = 1.221 · atomic radius Li
= 185 pm
My numbers (possibly a little sloppy with significant figures) in pm were
Li
152
Na
185
K
230
Rb
246
Cs
267
The general trend is towards increasing size as you go down the column. Looking more carefully, it’s interesting that these numbers “roll off”, with Cs being
only slightly larger than Rb, which is only slightly larger than K. (Theoretical calculations for Fr, which is too radioactive to make enough to measure
experimentally, show it as only minutely larger than Cs.) The valence orbital
increases from 2s to 6s, which is the reason for the increase in size, but working
opposite to this trend is a steady increase in effective nuclear charge as you
move down the column: the core electrons do not shield perfectly, and so the
valence electron sees a little bit of the (increasingly positive) core.
As mentioned in class, the size of the s orbitals grows quadratically with n: in
hydrogen (where the nuclear charge stays as 1), the 6s orbital to be 9 times
bigger than the 2s orbital. So the story is less that potassium is bigger than
expected if you are simply considering orbital size, and more that rubidium and
cesium are much smaller than expected.
2. The transition metals have the following densities and crystal structures:
Sc
hcp
2.989
Ti
hcp
4.5
V
bcc
5.96
Cr
bcc
7.20
Mn
bcc
7.47
Fe
bcc
7.86
Co
fcc
8.92
Ni
fcc
8.90
Cu
fcc
8.94
Zn
hcp
7.14
Again, estimate the radii and discuss how they relate to periodic trends.
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My values were
Sc
164
Ti
146
V
132
Cr
124
Mn
125
Fe
124
Co
124
Ni
124
Cu
127
Zn
139
The big story here is that these are the radii of the atoms calculated from how
they bond in a metal—and how they bond in a metal depends both on size
and on how the electrons can interact with neighboring atoms. Other measures
of size (calculation, van der Waals) show Ni and Cu and Zn as progressively
smaller. So part of this trend is that half-full d orbitals overlap to make metallic
bonds much more easily than full d orbitals.
Graded problem:
This should be written up and handed in before class begins on Wednesday the 16th.
This problem may also be discussed in class.
1. The first ionization energy of cesium (Cs) is 0.624 aJ. The first electron affinity
of neon (Ne) is not known, but it is certainly greater than 0 and likely equal
to or greater than 0.2 aJ—for this problem, assume it is 0.2 aJ. (Why is this
statement made so uncertainly?)
(a) Consider the formation of a gas-phase molecule CsNe through the reaction
Cs(g) + Ne(g) −→ Cs+ Ne− (g)
How short would the Cs–Ne bond have to be for this to be an energetically
favorable process?
(b) Look up the size of the Cs+ ion, and give your best estimate of the size of
Ne− . Is the bond length you calculated above a reasonable one?
The best estimate of the size of Ne− comes from looking at Na, which is
isoelectronic—you don’t want to compare to F− , because there’s a huge
difference between adding a 2p electron to a fluorine and adding a 3s
electron to a neon. Na and Ne− have the same electron configuration, but
Na has the more positively charged nucleus, so the size of Na is an absolute
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minimum size for Ne− .
If you wanted to refine your guess more, you could try to find a number
for the radius of Na− ; the thinking would be that with Na− you’re still
looking at 3s electrons and so the same basic orbital size, but trying to find
a system where there are more electron-electron repulsions. In any case,
while it’s hard to track down a number for Na− , the ion does exist and has
been measured, and has a radius between 200 and 270 pm (depending on
who you ask.)
Working with sodium, you could take the metallic radius of 186 pm, or
the van der Waals radius of 227 pm. In either case, combined with the
167 pm diameter of Cs+ , there’s no way to get the ions within the 280 pm
you calculated as being necessary for stability.
(c) The Kapustinskii equation estimates the energy of making an ionic solid
from gas-phase ions:
−10
E = 4.0 × 10
3.45 × 10−11 m
z+z−
·
1
−
aJ m · +
r + r−
r+ + r−
!
where z + and z − are the charges and r+ and r− are the radii of the cation
and anion, respectively. (Note that z + z − is negative—that is, making
a solid from gas-phase ions releases energy.) Use this to estimate how
short the Cs–Ne bond needs to be to make the solid stable relative to the
separated elements. Is this a reasonable length for this bond?
I calculate 448 pm. Remember that this is the distance that, if the ions
can be brought this close together in the lattice, then it’s energetically
favorable to form them. We expect this number to be larger than part
(a): in an ionic solid, each positive charge interacts with many negative
charges, and vice versa. So, the solid doesn’t have to pack the ions as
closely together in order to stabilize them.
You also get a second root when solving the quadratic here. The Kapustinskii only will give a good answer for for ionic lengths of hundreds
of picometers, as it was developed based on a quadratic fit to experimental data. At very short distances, it predicts the energy should go back
up, but this is not a problem because that rise occurs at shorter distances
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than you’d find in actual physical materials. You do end up with that
non-physical second root when you solve the equation, but it can (should)
be ignored.
So, could CsNe(s) actually be a stable compound? This number says yes,
provided:
• the Ne− is as small as Na, or not too much bigger,
• the 0.2 aJ number for the Ne electron affinity is not too low.
In actuality, if we consider entropy as well as energy (we will discuss how
the two combine later in the class), the answer is probably: “it’s not a
totally crazy idea, but it probably doesn’t exist, either.”
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