Question

Consider the ionization process as the reaction Qz → Qz+1 + e–, where the ionization potential is the minimum energy needed to remove the electron. This yields the relation

IP = E(Qz) – [E(Qz+1) + E(e–)], with the minimum energy of the free electron, E(e–) being zero, and the energy of each individual species, E(Qz) or E(Qz+1), defined as the total

energy of all the electrons bound in Qz or Qz+1, respectively.

a)  Using this second approach, apply Slater’s rules to calculate the binding energy of the 2s electron in atomic boron.

b). How do the two approaches compare against the experimental observation, and why do they yield different results?

c) Using the orbital energies (find online) and the first (orbital energy) approach in problem 2, calculate Zeff for the valence s orbitals of the group 1 and group 18 elements through period 6. What is the general trend as a group is descended?

d). How do first ionization potentials trend as a group is descended? What is the physical origin of this trend?

Answer 1

a)

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the effective nuclear charge can be calculated using the Slater's rule, which states that

• Step 1: Write the electron configuration of the atom in the following form:

(1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) . . .

• Step 2: Identify the electron of interest, and ignore all electrons in higher groups (to the right in the list from Step 1). These do not shield electrons in lower groups
• Step 3: Slater's Rules is now broken into two cases:
  • the shielding experienced by an s- or p- electron,
    • electrons within same group shield 0.35, except the 1s which shield 0.30
    • electrons within the n-1 group shield 0.85
    • electrons within the n-2 or lower groups shield 1.00
  • the shielding experienced by nd or nf valence electrons
    • electrons within same group shield 0.35
    • electrons within the lower groups shield 1.00

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the electronic configuration of boron is (1s²)(2s² 2p¹)

we need to find the binding energy of the electron in 2s orbital. so this electron will experience shielding from the 1s orbital electron only.

the shielding constant where n_i is the number of electrons in a specific shell and subshell and S_i is the shielding of the electrons subject to Slater's rules

S = 1.00(0) + 0.85(2) + 0.30(2) = 2.3

0r Z_eff =  5 - 2.30 = 2.70

where Z = Atomic No. of Boron

The binding energy of the valence electron of a non-hydrogen type atom can be found out using the expression

where R_H = Rydberg's Constant = 2.18*10^-18 J and n= Principal Quantum no.

En = 2.70² * 2.18*10^-18/ 2² = 3.973*10^-18 J

b) WHAT IS THE FIRST APPROACH ?? IT IS NOT MENTIONED IN THE QUESTION.

c) 2s orbital energy of boron is 14.0 eV or 22.4*10^-19 J

2p electron is 8.30 eV or 13.28*10^-19 J

Zeff = 2.60

As one moves down the group the no of electron screening the effective pull on the electron increases, thus the effective pull on the electron decreases and so the Zeff values decreases. but as we move across a period no new orbitals/ shells are added thus the effective screening remains constant and so with the increasing atomic no the pull on the electrons increases. thus zeff remains constant across a period. but when the electron fill up in f or d subshell there is somewhat increase in the screening constants because the penultimate d- and antipenultimate f-orbitals are filled before the valence s and p orbital. but since the screening capacity of the d- and f- orbitla electrons are very less they will not have a dominating effect.

d) ionisation potential is the amount of energy required to remove a valence electron from an  isolated atom. it is a function of the distance of electron from the core of the atom. the greater the distance of the electron from the core of the atom, the lesser will be ionisation potential and so easier will be the extraction of the electron from the atom. As we descend down a group we have new shells inserted into the electronic configuration, which in turn increases the distance of the valence electron fro the core of the atom. so, the resulting pull on the valence electron will be diminished and hence lesser energy will be required to pull of the electrons. Thus, the first ionisation energy decreases on moving down a group.

First period has only 1s orbital, second period has 2s and 2p, third period has 3s, 3p and 3d, etc. As the no. of period increases the valence electron enters a position that is farther away compared to the previous filling pattern. i.e. first period electron enters the 1s, in second period it enters into 2s or 2p, etc. thus effective pull on this electron also decreases as the valence electron moves away from the nucleus. thus the ionisation potential decreases on moving down a group.