Question

iopyb Part 2. Periodic Trends (CH 9) 3. (a) Write the full electron configuration for Cu. (b) Using Slater's rules, calculate Z for a 4s electron and a 3d electron in a copper atom. Which electron is easier to remove upon forming the Cu' cation? Briefly explain your choice. 4. Using Slater's rules, calculate Zon for the outermost electron in the following species: Al, Al', AP, and Al". Discuss the results relative to the expected ionization energies for these species. 5. Ofo, F, Cl, and Ne, which has the highest (i.e., most negative) electron affinity? Calculate Zetr for the added/incoming electron in each of these species and use it to rationalize your answer.

Answer 1

3. a)The full electronic configurstion for Cu is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d¹⁰.

b) Slater's rule of shielding is process of calculatin the effective nuclear charge on each electron in an atom. The electronic configuration is to be written in the increasing orders of the quantum numbers n and l, in this case as, (1s)² (2s,2p)⁸ (3s, 3p)⁸ (3d)¹⁰ (4s)¹.

Electrons to the right on this list will not shield the ones to their left.

The shielding constant S for each electron is given as:

0.35 for each electron in the same group. 0.85 for each electron in the (n-1) groups. 1.00 for each electron in (n-2) or lower groups. For nd or nf valence electrons, all electrons in (n-1) group or lower will contribute 1.00.

Thus, for Cu, S_4s = 18x0.85 + 18x1.00 = 25.3

Z_eff = Z-S = 29 - 25.3 = 3.7

S_3d = 9x0.35 + 18x1.00 = 21.15

Z_eff = Z-S = 29 - 21.15 = 7.85

For the formation of Cu⁺ cation, since the effective nuclear charge of 4s electron is less (3.7) than 3d electron (7.85), the 4s electron will preferably be removed. This is also beacuse 4s is the outermost electron the Cu atom, and electron lost in an ionization process is always lost from the outemost electron. This is supported by the Z_eff values.