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MODULE 9. MANY ELECTRON ATOMS With the exception of hydrogen the elements of the Periodic Table consist of atoms that are made up of 2 or more electrons

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MANY ELECTRON ATOMS

With the exception of hydrogen the elements of the Periodic Table consist of atoms that are made up of 2 or more electrons

Thus in our Schrödinger equation for the energy we have multiple kinetic energy operators and multiple potential energy operators to contend with.

The simplest many electron atom is that of helium with Z = 2,

i.e., two electrons and one nucleon having +2 charge.

r12

r1

N

2

r2

MODULE 9

As with the H atom we neglect the motion of the atom as a whole through space.

Thus we need to account for the kinetic energy of the two electrons about the center of mass.

Also there are PE terms for the various Coulombic attractions and repulsions present.

The hamiltonian operator in atomic units is

1, 2, ..., n identifies the spatial coordinates of the n electrons.

For He atom n = 2, and the hamiltonian becomes

or

where h(1) and h(2) are one electron hamiltonians.

The third term is a two electron operator (inter-repulsion of e-)

This term prevents the Schrödinger equation from being separable into a pair of single variable equations.

The 3-particle He atom is impossible to solve exactly and we need to resort to approximation methods, such as variation theory.

A very obvious (but drastic) approximation is the one we already used in our variation treatment, viz., to simply ignore the 1/r12 term.

This led to the concept of “effective nuclear charge” and is the “independent electron approximation”

The h(1) and h(2) terms in are hamiltonians for a pair of hydrogenic atoms containing a Z = 2 nucleus.

The wavefunctions are the 1s, 2s, 2p, etc atomic orbitals.

Ignoring the 1/r12 term in the full hamiltonian generates an approximate hamiltonian

Note that the 1s, 2s, etc atomic orbitals are eigenfunctions of ^Happrox but not of the full hamiltonian.

In this approximation (zero order) the ground state of the He atom can be written as 1s(1)1s(2), where the (1) and (2) refer to the individual electrons in the atom.

Both electrons are associated with the same AO

But now we need to worry about distinguishability of electrons

Are Electrons Individually Recognizable?

In the macroscopic world we can distinguish individual objects in a group of identical objects by their positions relative to a fixed point or, if they are in motion, by their individual velocity vectors.

Their different trajectories/positions make them distinguishable.

Electrons are also identical, same rest mass, same intrinsic angular momentum and same charge.

There are no ways to tag individual electrons.

In the sub-microscopic world of the electron, the momentum and position operators fail to commute and therefore these two observables cannot be simultaneously specified with arbitrary precision. If the momentum of an electron is sharply defined, its position is very uncertain, and vice versa.

Thus there is no property of one of the electrons in an atom that we can use to distinguish it from another electron.

If there are several electrons occupying different AOs we cannot state which of the electrons occupies which orbital.

The description of the electronic configuration of He ground state as 1s(1)1s(2) is satisfactory because it states that both electrons are occupying the same AO (1s) and there is no attempt to distinguish between the electrons – 1s2 .

However, to write the zeroth-order wavefunction of the excited state as 1s(1)2s(2) is not allowed, since to do so implies that electron (1) is in the 1s AO and electron (2) is in 2s.

This implies that we can somehow distinguish electron (1) from electron (2); but we cannot.

Instead we must use superposition functions such as

This allows for all combinations of electrons and AOs and does not specify that a given electron is in a particular AO

The wavefunction of a system of particles is a function of all the variables that pertain, both spatial and spin.

For electron (1) in a set of n, the variables are x1, y1 and z1 (space) and s1(spin).

For brevity we denote all of these variables for each of n electrons as qn.

Thus the wavefunction of a system of n identical particles can be written as y (q1, q2, … qn).

Let us define the permutation (or exchange) operator ^Pjk as the operator that exchanges all the space and spin coordinates of the particles (j) and (k) in the n-particle system.

For a two electron system (He atom), writing the wavefunction as y(q1, q2) and the exchange operator as ^P1,2, we see

thus the operator has exchanged the coordinates of the electrons.

This will not affect the state of the system since the electrons are not distinguishable (an electron is an electron is an electron…).

It serves only to change the labels we placed on the electrons.

Application of the exchange operator a second time will bring the system back to the original description, so we can say

where ^1 is the unit operator.

It can be shown that the eigenvalues of any operator who’s square is the unit operator are +1 and –1.

Thus when a wavefunction is an eigenfunction of ^P1,2 then

If the exchange operation retains the sign of the function then the function is symmetric wrt interchange of electrons (1) and (2).

If the operation changes the sign of the function we say that the function is antisymmetric wrt the interchange.

The conclusion applies in the general case:

“A wavefunction for a system of n identical particles must be symmetric OR antisymmetric with respect to every possible interchange of any two of the particles.”

Since the particles are identical it is not conceivable that some interchanges would be symmetric and others antisymmetric.

The Periodic Table of elements provides abundant evidence (see later) that for systems of electrons (atomic states) only the antisymmetric result is found and we can formulate an important quantum mechanical postulate that

“The wavefunction of a system of electrons must be antisymmetric with respect to the interchange of any two electrons.”

Wolfgang Pauli employed relativistic quantum field theory to show that fermions (electrons and other particles with half-integral spin; s = 1/2, 3/2,…) require antisymmetric wavefunctions, whereas bosons (s = 0, 1, 2, and so on) require symmetric wavefunctions.

To demonstrate the effect of the Pauli principle for identical fermions we start with the recognition ( from the postulate) that the antisymmetry requirement is

Now assign the same four coordinates (x, y, z, s) to two (identical) electrons, i.e. two electrons occupy the same AO.

Thus, x1 = x2, y1 = y2, z1 = z2, and s1 = s2; thus q1 = q2.

2y = 0

and the probability of finding the two electrons in the orbital defined by y is zero.

Thus two electrons having the same spin function (a, say) have zero probability of being found with the same values of space coordinates, in other words they cannot co-exist in the same region of space (AO).

This “Pauli repulsion” forces electrons of the same spin to keep apart from one another, e.g., in different orbitals.

In other words the Pauli principle requires that no two electrons can have the same four quantum numbers, n, l, ml, and s

The Helium Atom

The ground state of He in the zeroth order can be written as 1s(1)1s(2), i.e., both electrons occupying the same orbital.

Now we must consider the spin state of the two electrons and at first sight there are 4 spin combinations

The indistiguishability of identical particles is clearly upheld in the first pair, since both electrons have the same spin function.

The second two combinations show a particular electron with a particular spin function, and this violates the principle.

the first and second are symmetric (+) and the third and fourth are neither + nor -, with respect to interchange.

These last two then are unacceptable.

Moreover, even though the first and second spin combinations are acceptable in themselves, when coupled with the symmetric 1s(1)1s(2) spatial function, we find the Pauli principle forbids the combinations

and

since they are both overall symmetric to electron interchange thereby violating the Pauli rule.

The way to proceed is to seek a spin function that is an antisymmetric superposition, e.g.

(no conflict with electron distinguishability)

These are the symmetric (+) and antisymmetric (-) eigenfunctions of ^P1,2, the exchange operator

we combine the antisymmetric one with the symmetric spatial function 1s(1)1s(2) to make an overall antisymmetric total function

This is the total wave function for He ground state (in zero order) including spatial and spin contributions.

It is a single state -- no degeneracies.

He in an excited electronic state

A 1s electrons is promoted to the n = 2 level which could be described as 1s12s1 or 1s12p1x/y/z

In non-hydrogenic systems the 2s orbital is at a slightly lower energy than any of the three 2p orbitals.

The two-electron combination might be written as 1s(1)2s(2) or 1s(2)2s(1), but these descriptions imply that we can identify electrons (1) and (2) as being in the designated orbitals. This clearly violates the concept of indistiguishability.

Moreover, the two product functions are neither symmetric nor antisymmetric on electron exchange and thus cannot contribute to a total wavefunction that must beoverall antisymmetric with respect to electron exchange.

As before we use a superposition that can satisfy the symmetry requirements.

The sum function is symmetric and the difference function is antisymmetric with respect to electron exchange

The symmetric spatial combination can conform to the Pauli principle of He (1s2s) in a product with an antisymmetric spin term

The antisymmetric spatial combination can conform to the Pauli principle in a product with a symmetric spin term.

Thus, four total wavefunctions satisfy these requirements for the 1s2s excited state of He.

And the configuration 1s2s shows 4 distinct states.

To obtain the state energies we need to operate on the individual wavefunctions with the full hamiltonian.

But we have used the independent electron approximation (ignored the inter-electron repulsion term 1/r12)

Therefore, the functions are not eigenfunctions of ^H(1,2) and we cannot simply compare the eigenvalues

However, we can calculate the average energies by evaluating the expectation values of with the four functions listed.

Note that the usual denominator is missing since both space and spin parts are normalized.

Nothing in the operator we are using will interact with any of the spin contributions in the four wavefunctions, and since the spin terms are normalized, their integrals are unity.

Therefore, the average energies will be determined by the spatial contributions, and since three of the functions contain the same (antisymmetric) spatial part, they form a three-fold degenerate set.

Only ys,a has a different spatial part and therefore we can anticipate this function to have a different energy from the three ya,s functions.

The excited state of He (1s2s) will therefore have two energy states, one of which is triply degenerate.

To find what the energy values are we need to grind through some integrations – see Appendix I

There we find the conclusion that the two energies are given by

where the first two terms on the RHS are the (exact) energies of the 1s and 2s states for He+ (Z=2), respectively.

The integrals J (the Coulomb integral) and K (the Exchange integral) are

J is always positive because it describes the Coulombic interaction between the two electrons (1s*1s and 2s*2s).

The interaction is repulsive, and thus the energy of the combination is raised (less negative).

The integral K has the product functions in the integrand differing by exchange of electrons, hence its name.

As Appendix I explains K has negative and positive contributions but is positive overall and not as large as J.

The procedure in the Appendix shows that the triply degenerate level is lower in energy than the non-degenerate level and the separation is 2K.

We can think of the J term as resulting from homogeneous/time-averaged electron densities in the charge-clouds of the orbitals

We can think of the K term as being a correction arising from the inhomogeneities that occur as a consequence of electrons wanting to avoid each other.

For example, if we approximate the two linear combination spatial terms in our wavefunctions as follows

where a is a function of the radius vector r1 and b is a function of the radius vector r2, then as r1 approaches r2 so y-tends to vanish, i.e., the electrons tend to avoid each other in the difference combination.

This leads to a Fermi hole.

In a magnetic field the triply degenerate level is split into three discrete energy states-it is a triplet of states, or a triplet state.

Slater Determinants

With just two electrons (ground and excited states of He) it is straightforward to write down antisymmetric total wavefunctions as multiples of spatial and spin parts.

It is another matter to construct an asymmetric wave function for N electrons, by inspection.

In 1930’s Slater introduced the concept of using determinants to construct the required wavefunctions.

For example take the He ground state wavefunction that we have shown to be

This wavefunction is composed of a symmetric space function and an antisymmetric spin function so it is overall antisymmetric with respect to interchange of the pair of electrons-as Pauli requires.

As a determinant this becomes

[Show this by expanding the determinant and rearranging]

The individual terms in the determinant (e.g. 1sa) are spinorbitals that describe the spatial and spin state of the associated electron.

The determinant has the individual spinorbitals along the rows and the associated electrons in the columns (others reverse this procedure, but this has no effect since interchanging columns and rows leaves the determinant unchanged).

Determinants such as the above are called Slater determinants

The wavefunction they represent is a determinantal wavefunction

If the labels (1) and (2) are interchanged in the determinant this places electron (1) in column 2 and electron (2) in column 1

This interchanges two columns of the determinant and changes its sign (see Barrante, chapter 9)

The wavefunction must also change sign and thus the determinant represents an antisymmetric function.

Furthermore, suppose that our two electrons in the 1s orbital have the same spin, say a. This changes the determinant as follows

Now two rows (the first and second) are identical

This causes the determinant to vanish (for all orders)

If the Slater determinant vanishes, the wavefunction it represents also vanishes

Thus a state in which two electrons have identical sets of space and spin variables does not exist

[see Module 9 for shorthand ways of writing SDs]

Now examine the wavefunction for the Li atom (Z=3) using SDs.

Firstly is the 1s3 configuration possible? If so

The top row is the principal diagonal

Two of the rows (first and third) are identical, therefore the determinant vanishes and the 1s3 configuration is not possible.

So the next to try is the configuration 1s22s

The 2s electron has been given a spin, but this is purely arbitrary

Thus the ground state of Li exists as a pair of degenerate states, it is a doublet of states, or a doublet state.

No two rows are identical and interchanging the labels (1) and (2) switches the first and second columns

The determinant changes sign, indicating the wavefunction is antisymmetric, (Pauli requirement)

Slater Orbitals

In Module 8 we applied Variation Theory to the He ground state in order to evaluate its energy.

In the calculation we used the atomic number, Z, as a variational parameter and the trial function was taken as

Minimization of the energy with respect to variation of Z

whereas the experimental result is –2.9033 au

One variational parameter brings us within 2% of the actual value.

At first sight this appears to be quite good agreement

However, the ionization energy (IE) of He is given by the energy difference between He and He+

The first term can be calculated exactly because He+ is a hydrogenic ion with Z = 2, and we can use the variation theory value of EHe1s as the second term

The experimental value of IE is 2372 kJ mol-1.

Thus our calculated value is ca 150 kJ mol-1 in error, about the strength of some chemical bonds

We need to do better

One way to improve is to change our trial function.

In the He1s calculation we chose the hydrogenic 1s wavefunction, but almost any function, or combination could be used.

In 1930 Slater introduced a new set of orbitals, now called Slater orbitals which are of the form

Y is the appropriate spherical harmonic

z =Z/n for hydrogenic orbitals, but in the Slater-type orbitals (STOs) it becomes an arbitrary parameter.

The radial parts of STOs have no nodes, unlike the hydrogenic counterparts for n > 1.

is a normalizing factor

In the calculation of He1s that we did in Module 8 we effectively used the S100.

Thus the trial wavefunction, following the procedure in Module 8 and remembering that l = 0 is

giving our earlier values of z = 1.6875 and Emin = -2.8477 au

which gave a value of IP that is about 150 kJ mol-1 in error.

A slight improvement can be gained by also letting n vary (in the normalizing factor and in the exponent of the r term).

Chemists find the concept of orbitals very satisfactory, especially when we think about chemical bonds

Within the orbital concept, improvements to Emin can be achieved by using trial functions of even more flexibility

Such as by defining the spatial function as a product of one electron orbitals that are completely general in nature

This leads us to a limiting value for Emin -- the Hartree-Fock limit.

This is the best that can be done with retention of the orbital concept.

For He1s the Hartree-Fock limit is Emin = -2.8617 au.

Removing the restriction that the trial function be a product of one-electron orbitals and expressing it as a series of variational parameters can lead to improvements in Emin

Pekeris (1959) used 1078 parameters to obtain

Emin = -2.903724375 au

in exact agreement with the true value.

In this approach, the emphasis is on obtaining the exact energy by crunching numbers and the idea of atomic orbitals has been abandoned.

In modern times, the procedure is usually to find the H-F orbitals (that generate the H-F limit) and then correct the energy by applying a Perturbation Theory treatment (see later).

It is instructive for us to see how the H-F procedure works in a simple case, viz., the He atom.

Any procedure that is going to be successful clearly must account for the interaction between electrons in some way.

In the HF approach the electron-electron repulsion is treated in an average way; each electron is considered to be moving within the field of the nucleon and the average field of the remainder of the electrons.

As we shall see this is not detailed enough, but it will get us started.

We start by writing the two-electron wavefunction as the product of one-electron orbitals, as outlined above

The functions f(rj) are identical since we allow both electrons in He 1s2 to be in the same orbital under the Pauli restriction.

The probability distribution of electron 2 (its “charge-cloud”) is given by f*(r2)f(r2)dr2 and thinking classically, we can interpret this as a charge density.

The average potential energy that is experienced by electron 1 at some value of r1 because of the presence of the charge-cloud of electron 2 is (in au) given by

So we can define a one-electron hamiltonian that includes this averaged potential energy term (Coulomb operator)

There is a similar equation for the other electron but since the wavefunctions are identical we need only consider one equation.

This is the H-F equation for He 1s2 and solving this equation will generate the optimum wavefunction for this atom within the orbital concept and the H-F approximation.

But there appears to be a problem

The operator in the Schrödinger equation depends on Vav,1 which depends in turn on f(r2).

Thus we need to know the wavefunction solution to the Schrödinger equation before we can write down its operator.

The way forward uses the self-consistent field method (Hartree)

First we guess a form for f(r2) and use it to evaluate Vav,1(r1) using

then we solve for f(r1) using

After the first cycle the output f(r) will differ from the input f(r), unless we had guessed the correct function in the first place.

Use the calculated f(r)to estimate a new Vav,1(r1) and thence another hamiltonian and subsequently another f(r1).

(Recall that the two orbital functions are identical).

Then the cycle is repeated until the output and input wavefunctions are sufficiently close, or are self-consistent.

At this point we have the Hartree-Fock orbitals.

As implied above single functional orbitals do not yield the best energy and in practice linear combinations of STO’s are employed.

The optimum result (five STO’s with different coefficients) for the He ground state, as stated earlier, was found to give an energy that was 109 kJ mol-1 too high,

It is the best that can be achieved within the orbital approximation.

The way the Coulomb operator was calculated in the above procedure treats each electron as though it were moving in the time-averaged field of the others in the system.

However, this is only partly true because electrons are mutually repulsive and they tend to avoid each other. In other words their motions are correlated.

The HF procedure is not constructed to take this into account and it generates EHF > Eexpt.

The difference between the HF limit and the true energy is the correlation energy.

In the He case this was 109 kJ mol-1, only ca 1.5% of the total electronic energy, but still a large absolute error.

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