Solutions to Homework 8 – chem 344 Sp’ 2014

1.

2.

3.

4. All different orbitals means they could all be

parallel spins

5.

Since electrons are in different orbitals any combination is possible paired or unpaired spins

Equivalent to min J lowest for less than half-filled 6.

a. Allowed, b. Allowed,S = 0 & J = 1, c. Forbidden by spin ,S = 1

7.

8.

9.

10. P15.6) Show that the total energy eigenfunctions 210(r,,) and 211(r,,) are orthogonal. Do you have to integrate over all three variables to show that the functions are orthogonal?

∫ ∫ ∫ ∫

√  3  2 sin  This integral is zero because  cos sin0 d 0   0.   0 3 0

It is sufficient to evaluate the integral over 

11. P15.17) Ions with a single electron such as He+, Li2+, and Be3+ are described by the H

atom wave functions with Z/a0 substituted for 1/a0, where Z is the nuclear charge. The 1s 3 2 Zr a wave function becomes  r  1  Z a e 0 . Using this result, compare the    0  mean value of the radius r at which you would find the 1s electron in H, He+, Li2+, and Be3+.

∫ ∫ ∫

∫

Using the standard integral: ∫

3 4 4Z 6a0 3 r  3 4  a0 a0 16Z 2Z

3 3 1 3 r a;;; r  a r23   a r   a H20 He 4 0 Li 2 0 Be 8 0

12. P15.20) Core electrons shield valence electrons so that they experience an effective

nuclear charge Zeff rather than the full nuclear charge. Given that the first ionization energy of Li is 5.39 eV, use the formula in Problem P15.18 to estimate the effective nuclear charge experienced by the 2s electron in Li. From the previous problem, Z 2 nI2 4 5.39eV I  13.60 eV ; Z    1.26 n2 eff 13.60 eV 13.60 eV

13. P15.24) The force acting between the electron and the proton in the H atom is given by F 2 2 = –e /40r . Calculate the average value F for the 1s and 2pz states of the H atom in

terms of e, 0, and a0. e2 1 Fd *     1s    2   40 r 2 2  e 11r// a r a 2 Fd d esin  e   r d r00    1s 32       4 00  ar0 0 0  222   eee 442r /2 ar / a a Fe  dre  00   0 1s 332  44 0aaa  000  00 220 0

e2 1 Fd *     2 pz    2   40 r 2 2 2  er1122 ra/  F dcos  sin  d   e0 r d r 2 pz 32    4 0 32  a 00 0 0  a 0  r  e231 cos   Fr e  dr  2 ra/ 0 2 pz 4 16a5  3  000 0 e2 1  F r2 era/ 0 dr 2 pz 3  400 24 a 0  n! Using the standard integral renr   n1 0  ee221 Fa  2 3   2 pz 520 4 0 24aa 0 48   0 0 14. P15.30) You have commissioned a measurement of the second ionization energy from two independent research teams. You find that they do not agree and decide to plot the data together with known values of the first ionization energy. The results are shown here: The lowest curve is for the first ionization energy and the upper two curves are the results for the second ionization energy from the two research teams. The uppermost curve has been shifted vertically to avoid an overlap with the other new data set. On the basis of your knowledge of the periodic table, you suddenly know which of the two sets of data is correct and the error that one of the teams of researchers made. Which data set is correct? Explain your reasoning.

The data set shown by the dashed (purple) line is correct, the red one is incorrect. Although the alkali atoms have the lowest ionization energies, they must have the highest ionization energy for the second ionization potential because the singly charged positive ions have the rare gas filled shell configuration. The experimenters that produced the data set shown by the gray (red) line had assigned atomic numbers that were too low by one.

15. P15.33) An approximate formula for the energy levels in a multielectron atom is 2 2 2 E  Z e 8 a n , n  1, 2, 3, . , where Zeff is the effective nuclear charge felt n eff 0 0 by an electron in a given orbital. Calculate values for Zeff from the first ionization energies for the elements Li through Ne (SEE www.webelements.com). Compare these values for Zeff with those listed in TABLE 15.1. How well do they compare? Using: 2 2 -12 2 -1 -1 -11 8 Eion n 0 a0 8 Eion 1 (8.8541878 10 C J m )(5.291772 10 m) Zeff  2  -19 2 23 -1 , e N A (1.602177 10 C) (6.022 10 mol ) the effective charges based on the first ionization energies for the elements Li to Ne are calculated as: Element Li Be B C N O F Ne

-1 Eionization in kJ mol 520.2 899.5 800.6 1086.5 1402.3 1313.9 1681.0 2080.7

Zeff 1.26 1.66 1.56 1.82 2.07 2.00 2.26 2.52

A comparison with the data in Table 15.1 shows that the approximation reproduces the effective charges reasonably well for the second-row elements with only 2s electrons, however, fails to predict the charges for elements with 2p electrons. Extra Problems

1.

2.

3.

4.

5.

6.

a. Allowed, b. Allowed, c. Forbidden 7. P15.8) How many radial and angular nodes are there in the following H orbitals?

a.  2 p r, ,  0 radial node and 1 angular nodes x b. 2s(r) 1 radial node and 0 angular node

c.  3d r,,  0 radial node and 2 angular nodes xz d.  r,,  0 radial node and 2 angular nodes 3d 2 2   x  y

8. P15.16) Calculate the mean value of the radius r at which you would find the electron if the H atom wave function is 210(r,,). r 1 2   r dcos25  sin  d  r ea0 dr 5    32 a0 0 0 0 r r r 2 cos3  2  2   1   r  redr5a0  redr 5 a 0  redr 5 a 0 5 5  5  32a0 30 32 a 0 3 0 24 a 0 0

Using the standard integral: ∫

1 6 ra a 5 5!00 5 24a0 9. P15.18) The energy levels for ions with a single electron such as He+, Li2+, and Be3+ are 2 2 2 + given by En = –Z e /80a0n , n = 1, 2, 3, .... Calculate the ionization energies of H, He , Li2+, and Be3+ in their ground states in units of electron-volts (eV). The ionization potential is the negative of the orbital energy. 2 Z –19 2 22 2 1.602 10 C Ze n 1eV I 2  12  1 2  1  11   19 8 00an 8   8.854  10 J C m  5.292  10 m 1.602  10 J Z 2 I 13.60 eV n2 IIII13.60eV;  54.42eV;  122.4eV;  217.7eV H He Li23  Be 