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CHEM 150 Chapter Notes - Chapter 4: Valence Electron, Ion, Atomic Radius

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6 Aug 2019

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Emory University

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Explaining the physical and chemical properties of the elements. Nuclear charge (z) = # of protons in the nucleus of an atom. E ective nuclear charge (ze ) : actual magnitude of positive charge that is. * hydrogen = nuclear charge = e ective nuclear charge. Shielding: electrons are simultaneously attracted to the nucleus & repelled by one another. An electron in a multi electron atom is partially shielding from the positive charge of the nucleus by the other electrons in the atom. Easy to remove the 1st electron but the 2nd electron is more di cult to remove, require more energy. When the charges have opposite signs = negative f = attractive force. When the charges have like signs = positive f = repulsive force. = distance between the nucleus and valence shell of a cation/anion. When an atom loses an electron and becomes a cation. When an atom gains an electron and becomes an anion.

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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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