Chapter
9: Krane’s Modern
Physics
Question
1:
Why does H2 have a smaller radius
and a greater binding energy than H2+?
On
page 274 Krane states: “Comparing Figs. 9.4 and 9.5
you can immediately see the effect of adding an electron to H2+: The binding energy is greater (the molecule
is more tightly bound), and the nuclei are drawn closer together. Both of the effects are due to the presence
of the increased electron density in the region between the protons.”
One
might also add to this the magnetic coupling of the antiparallel
electron spins in the bonding state.
Moreover, since the electrons are in an antisymmetric
singlet (Fermi exclusion) spin state, the spatial wave functions are in a
symmetric state. Thus a Bose
condensation of the electron wave function between the two protons further
enhances the negative charge density there.
Question
2:
The
molecule LiH has a simple electronic structure. The H atom would like to gain an electron to
fill its 1s subshell, but the Li atom would similarly
like to fill its 2s subshell.
Based on the atomic structure of H and Li, which would you expect to
dominate in the desire for the additional electron? Are the electronegativity
values of Table 9.6 consistent with this?
The binding energy of the outermost electron of a fully screened atom
13.6 eV/n2. For H 1s this is 13.6 eV, for Li 2s this is 3.4 eV. Although the screening of the 2s in Li is not
complete, the larger 2s orbital is nonetheless less tightly bound than the 2s
in Li. Thus the H tends to steal the 2s
electron from the Li. Table 9.6 confirms
this, as the electro-negativities are 2.1 for H, and
1.0 for Li.
Question
4:
Explain why the bond angles of the sp directed
bonds (Table 9.3) approach 90o as the atomic number of the central atom
increases.
On
page 278 Krane states that directed bonds deviate
from 90o because of repulsion of two H atoms which spread the angle.
As indicated at the top of page 279, a large charge on the central atom makes
the H repulsion less dominant. Thus “the
bond angle does indeed approach 90o … as the Z of the central atom
increases.”
Problem
1:
Calculate the ionization energy of H2.
Fig.
9.6 – Depth of the well for H2 is 2(–13.6) – 4.5 = – 31.7 eV
Page
271 – Depth of the well for H2+ is –16.3 eV
Problem
3:
The
ionization energy of potassium is 4.34 eV; the
electron affinity of iodine is 3.06 eV. At what separation distance will the KI
molecule gain enough Coulomb energy to overcome the energy needed to form K+
and I – ions?
K
needs 4.34 eV, I gives 3.06 eV,
must borrow 4.34 – 3.06 = 1.28 eV = ke2/r
r = (1.44 eV-nm)/(1.28 eV) = 1.125 nm