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1.4 Estimating the effective nuclear charge: Slater Type Orbitals (STO)

NOTE: This section does not appear in your textbook. For further understanding, consult Miessler and Tarr.

PREMISE: We can calculate an approximate value for the effective nuclear charge “felt” by an electron in a particular orbital of a particular atom or ion. We can do this by focussing on a single electron and lumping all the shielding effects from the other electrons into one shielding parameter S.

• Using Slater Type Orbitals (STO), instead of more realistic orbitals, is a reasonable and simple approximation that makes this calculation much easier.

Q. What is a STO?

• A STO assumes radial functions of the general form:

Nr = normalization constant for different orbitals, Nr = f(Z*,n*)

• These orbitals do not depend on the quantum numbers n, l, etc.

• i.e., s, p, and d orbitals all look the same and have only one node at r = 0.

• STO functions work well in approximating true radial function at distances r in the range of covalent bonds, but fail to describe electron distribution in inner shells.

• Slater devised empirical rules to determine the values of Z* and n* (see next page).

• The effective nuclear charge Z* can be calculated as (Z – S).

• The effective nuclear charge Z* can then be used for such purposes as calculating ionic radii (above equation) and electronegativities, hardness, HOMO/LUMO energy levels, etc. of the elements (see notes to come).

• More refined calculations can be used when greater accuracy is needed, however gross trends are nicely reproduced by this oversimplified model.

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Source: Purcell & Kotz, Inorganic Chemistry, 1977.

• For a series of elements with the same (ns,p) or (nd) group electrons, Z* regularly increases by 0.65: Z*n+1 = Z*n + 0.65

• Application of STO’s: Ideally, Slater’s rules permit the calculation of the energy of any electron in any atom or ion by the formula:

E = -13.6 (Z*/n*)2 (Q. Where does 13.6 eV come from?)

Example 1: Predict the energy of a 4s electron in Ca.

Solution: The electron configuration for Ca is 1s22s22p63s23p64s2.

The ten electrons with n=1 and n=2 contribute 1.00 each to the shielding, the eight electrons with n=3 each contribute 0.85, and the other 4s electron contributes 0.35.

S = 10(1.00) + 8(0.85) + 1(0.35) = 17.15 and Z* = 20 – 17.15 = 2.85

E4s = -13.6[(2.85)2 / (3.7)2] = -8.07 eV

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Example 2: Calculate the first ionization energy (IE) of a fluorine atom in the gas phase:

Solution: IE = EF

+ - EF0 …this is the difference in TOTAL energies…

Z = 9; n = 2

F0 (1s2)(2s22p5) → Z* for one e- in (2s, 2p) = 9 - 6(0.35) - 2(0.85) = 5.20

F+ (1s2)(2s22p4) → Z* for one e- in (2s, 2p) = 9 - 5(0.35) - 2(0.85) = 5.55

We are using a “Slater orbital assumption”. In other words, we are assuming that the 2s and 2p orbitals “look the same”…i.e., spherical electron density distribution.

We are ionizing (removing an e-) from the (2s, 2p) orbital group.

Since the electrons in the 1s orbital “see” the same nuclear charge before and after ionization (according to this approximation), their contributions to the energies of F0 and F+ are the same and will cancel when the difference is taken…

THEREFORE:

IE = -13.6[{6(Z*/n*)22s,2p}F+ - {7(Z*/n*)2

2s,2p}F0]

= -13.6[{6(5.55/2)2} – {7(5.20/2)2}]

= -13.6[46.20 – 47.32]

= +15.2 eV

Compare to experimental +17 eV.

Q. How do you define electron affinity (EA)?

HOMEWORK: Find the EA of a fluorine atom in the gas phase.

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Using much more complicated quantum mechanical approaches and powerful computers, very accurate Zeff can be calculated:

Source: Shriver & Atkins, Inorganic Chemistry, 3rd Edition.

The above values originate from a HF-SCF calculation using the true radial functions as starting points for solving the Schrödinger equation with the complete Hamiltonian of the system (see CHEM 386).

Note the difference in Z* for an electron in the 2s versus the 2p orbital…no longer using Slater-type orbitals!

1.5 Atomic configurations

Pauli’s rule: No two electrons in the same atom/molecule can have the same set of quantum numbers

n l ml ms

Two electrons in the same orbital must be spin correlated, i.e. have opposite spins ± ½.

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• Because of Coulomb repulsion, pairing of electrons in one orbital costs energy: → Charge Correlation Energy

• Exchange energy: Electrons of like spin in energy degenerate orbitals can “exchange” leading to an overall lower energy (quantum mechanical resonance phenomenon.)

• Another way of looking at this is to consider a diagram of the px and py orbitals of an atom. Although these orbitals are orthogonal (zero overlap integral), there is nevertheless a region of actual orbital overlap in space (shown in red).

• Consider one electron in each orbital. If they have opposite spin, their spatial wave function can put them in exactly the same space at the same time (high electrostatic repulsion). If they have the same spin, there must be a constraint on their spatial wave function so as to not put them in exactly the same space at the same time (therefore lower electrostatic repulsion if spin is the same)

• Leads to Hund’s Rule of maximum multiplicity: Energy degenerate orbitals are filled with electrons in such a way that the maximum possible total spin is realized.

• Consequences: As a general rule, the most stable oxidation states are those with - completely filled/empty (sub)shells → octet rule - exactly half-filled (sub)shells → maximizes exchange energy

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Applications:

• Using these concepts we can qualitatively/empirically predict the stability of oxidation states!

Examples:

Phosphorus

P0 [Ne] 3s23p3 stable in at least three different allotropic forms

White phosphorus is composed of P4 units. These are highly strained units and ignite (i.e. oxidize in air) at 30 °C. White P is used as a chemical warfare agent.

Red phosphorus can be amorphous or can be crystalline. In its crystalline form it forms chains that can be described as P4 units joined by breaking one bond of the tetrahedron and making a bond with the adjacent tetrahedron. Red P does not readily ignite in air but it thermally decomposes to white P upon heating to 260 °C. This is what happens when you strike a match.

Black phosphorus is thermally stable. It is a flaky solid that resembles carbon. In the crystal structure of black P is shown.

CONCLUSION: Relative stability also depends on other factors such as structure and bonding.

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P+ [Ne] 3s23p2 very rarely observed

P2+ [Ne] 3s23p1 very rarely observed

P3+ [Ne] 3s23p0 very common: e.g., organophosphines PR3

2,2'-Bis(diphenylphosphino)-1,1'-binaphthyl, commonly known as BINAP, is a chiral ligand used to design enantioselective homogeneous catalysts.

Ryoji Noyori won the Nobel Prize in Chemistry in 2001 for asymmetric hydrogenation catalysis, primarily using Rh and Ru complexes of BINAP as catalysts.

P4+ [Ne] 3s13p0 very rarely observed

P5+ [Ne] 3s03p0 very common: phosphates (e.g. P2O5)

Phosphorus pentoxide is an anhydride of phosphoric acid. As such, it is a powerful desiccant (drying agent). We use it to dry acetonitrile in our laboratory stills. **Caution: P2O5 reacts violently with water.

Phosphates are very common in biological chemistry (ADP, ATP, DNA, hydroxyapatite, etc.)

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Phosphazenes (compounds with –N=PR2– linkages) provide a good example of stable species in which N and P are in different oxidation states.

The electronegativity difference (χN = 3.04; χP = 2.19) is less than in the siloxane link, but some multiple bond character strengthens the framework.

Phosphazene polymers can be biodegradable with harmless decomposition products.

Here are two figures from an article describing the use of a tube made of PEIP as a prosthetic nerve guide for repairing nerve damage in a living organism.

Ref: F. Langone, S. Lora, et al., Biomaterials, 16 (1995) 347-353.

Sulfur

S2- [Ne]3s23p6 sulfides (e.g. H2S)

S0 [Ne]3s23p4 S4+ [Ne]3s23p0 sulfites (e.g. Na2SO3)

S6+ [Ne]3s03p0 sulfates (e.g. Na2SO4)

What are the most likely oxidation states of Ga, Sn, Sr, Br, Xe ?

HOMEWORK: Use Slater’s rules to calculate Z* for highest occupied atomic orbitals of the elements Li through F and compare the trend in Z* to the trend in electronegativity. Textbook problems: 1.19-1.22, 1.26, 1.27, 1.29.