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CHEM1100 Lecture Notes - Lecture 2: Atomic Orbital, Covalent Bond, Valence Electron

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20 Jul 2018

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Document Summary

Atomic orbital electron distributions and energies: energy is determined by n principal number, thus, for h, all orbitals with the same value for n have the same energy- they are. Thurs: 1-4pm degenerate: for non-hydrogen atomic orbitals, the electrostatic attraction is different between each orbital (closer orbitals = more) In the hydrogen atom (1 electron) the energy is identical whether the electron is in the 2s or. Consequence of the effects of shielding and penetration: Electron spin quantum number (ms: electrons have a spin", e. g. +1/2 or -1/2 (only these two values: spinning electrons have a magnetic moment (spin up or spin down, electrons have many orbitals, but the outer ones are empty (ready to be filled if electron is excited) The aufbau principle: allows us to write electron configurations electrons in an atom results from placing each successive electron in the most stable (lowest energy) available atomic orbital.

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Related Questions

Slater’s rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the “shielding constant,” S, is calculated. The effective nuclear charge is then the difference between S and the atomic number, Z.

The shielding constant, S, is calculated using the following rules:

(1) The electrons of an atom are grouped as follows: (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d), and so on.

(2) Electrons in higher groups (to the right) do not shield those in the lower groups.

(3) For ns and np valence electrons

a) Electrons in the same ns, np group contribute 0.35 (for 1s 0.30 works better).

b) Electrons in n − 1 groups contribute 0.85.

c) Electrons in n − 2 groups (and lower) contribute 1.00.

(4) For nd and nf electrons, electrons in the same nd or nf group contribute 0.35, and those in groups to the left contribute 1.00.

As an example, let us calculate Z* for the outermost electron of oxygen:

Here is the calculation for a d electron in Ni:

and for an s electron Ni:

(Here 3s, 3p, and 3d electrons are in the (n − 1) groups.)

a) Calculate Z* for F and Ne. Relate the Z* values for O, F, and Ne to their relative atomic radii and ionization energies.

b) Calculate Z* for one of the 3d electrons of Mn, and compare this with Z* for one of the 4s electrons of the element. Do the Z* values give us some insight into the ionization of Mn to give the cation?

Slater’s rules are a way to estimate the effective nuclear charge experienced by an electron. In this approach, the “shielding constant,” S , is calculated. The effective nuclear charge is then the difference between S and the atomic number, Z . (Note that the results in Table 7.2 and Figure 7.2 were calculated in a slightly different way.) Z * = Z ‒ S The shielding constant, S , is calculated using the following rules: (1) The electrons of an atom are grouped as follows: (1 s ) (2 s , 2 p ) (3 s , 3 p ) (3 d ) (4 s , 4 p ) (4 d ), and so on. (2) Electrons in higher groups (to the right) do not shield those in the lower groups. (3) For ns and np valence electrons a) Electrons in the same ns, np group contribute 0.35 (for 1 s 0.30 works better). b) Electrons in n ‒ 1 groups contribute 0.85. c) Electrons in n ‒ 2 groups (and lower) contribute 1.00. (4) For nd and nf electrons, electrons in the same nd or nf group contribute 0.35, and those in groups to the left contribute 1.00. As an example, let us calculate Z* for the outermost electron of oxygen: S = (2 × 0.85) + (5 × 0.35) = 3.45 Z * = 8 ‒ 3.45 = 4.55 Here is a calculation for a d electron in Ni: Z * = 28 ‒ [18 × 1.00] ‒ [7 × 0.35] = 7.55 and for an s electron in Ni: Z * = 28 ‒ [10 × 1.00] ‒ [16 × 0.85] ‒ [1 × 0.35] = 4.05 (Here 3 s , 3 p , and 3 d electrons are in the ( n ‒ 1) groups.) a) Calculate Z * for F and Ne. Relate the Z * values for O, F, and Ne to their relative atomic radii and ionization energies. b) Calculate Z * for one of the 3 d electrons of Mn, and compare this with Z * for one of the 4 s electrons of the element. Do the Z * values give us some insight into the ionization of Mn to give the cation?

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