Answer the following questions related to Slater\'s rules and its applications.

Question

Answer the following questions related to Slater's rules and its applications. Predict Z_eff for a 6s, 4f, 5d, and 6p electron in bismuth. From which orbital is an electron removed first in Bi? Look up and list the first five ionization energies of Bi, being sure to cite where you found these ionization energies. Why does ionization energy generally increase as the element becomes further ionized? That is, why is the second ionization energy expected to be larger than the first ionization energy for the same element? Indicate where there is a very large increase in ionization energy (more than a few hundred kj/mol), and explain why this occurs. Consider the electron configuration of the ion that is being ionized.

Explanation / Answer

Accroding to Slater's rule

1) we should write the elecronic configuration first with the form

[Xe] 4f14 5d10 6s2 6p3

(1s2) (2s2 2p6) (3s2 3p6)(3d10) (4s2 4p6) (4d10) ( 4f14) (5s2 5p6) (5d10) (6s2 6p3)

2) We will look for the electron talked about

a) 6s , 5d , 6p

3) the electrons shielding the different electrons are

(i) 6s : Shielded by (1s2) (2s2 2p6) (3s2 3p6)(3d10) (4s2 4p6) (4d10) ( 4f14) (5s2 5p6) (5d10)(6s2 6p3)

(ii) 5d : Shielded by (1s2) (2s2 2p6) (3s2 3p6)(3d10) (4s2 4p6) (4d10) ( 4f14) (5s2 5p6) (5d10)

(iii) 6p : Shielded by (1s2) (2s2 2p6) (3s2 3p6)(3d10) (4s2 4p6) (4d10) ( 4f14) (5s2 5p6) (5d10)(6s2 6p3)

4) let us consider shielding experienced by 6s and 6p electrons

Remember : The shielding contributed by electrons of same group will be multiple of 0.35 nuclear charge

The shielding contributed by electrons with one less value of "n" will be multiple of 0.85

The shielding contributed by electrons with two less value of "n" will be multiple of 1

so 6s and 6p will be shielded by same extent

Calculation will be:

4 X 0.35 + 18 X 0.85 + 60X1 = 76.7

Z* = Shielding = 83 - 76.7 = 6.3 for an electrons

5) let us consider 5d electrons : here all electrons is the same group will contribute 0.35 and other electrons multiple of one

So

9X0.35 + 68 X 1 = 71.15

Z* = 83 - 71.15 = 11.85

6) For 4f electrons:

(1s2) (2s2 2p6) (3s2 3p6)(3d10) (4s2 4p6) (4d10) ( 4f14)

here all electrons is the same group will contribute 0.35 and other electrons multiple of one

13X0.35 + 46X1 = 50.55

Z* = 83-50.55 = 32.45

b) From which orbital electron released from Bi first:

From 6p orbital

c) The first five ionization energies are (from https://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements)

I : 703 II 1610 III 2466   IV 4370 V 5400

The second ionization energy will be energy to remove an electron from a positive ion, and its always difficult to remove and electron from a positive ion as compared to a neutral atom

The large increase is after third ionization energy as it becomes very difficult the remove electrons from an element with completely filled orbitals

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