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

 

The difference is –31.7 – (–16.3) = –15.4 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