Ebook Physical chemistry Part 2

(BQ) Part 2 book Physical chemistry has contents: Free energy and chemical potential, introduction to chemical equilibrium, equilibria in single component systems, equilibria in multiple component systems, electrochemistry and ionic solutions, introduction to quantum mechanics,...and other contents.

WE HAVE SEEN HOW QUANTUM MECHANICS provides tools for

understanding some simple systems, up to and including the gen atom itself An understanding of the H atom is a crucial point because it

hydro-is real, not a model system Quantum mechanics showed that it can describe

the hydrogen atom like Bohr’s theory did It also describes other model tems that have applications in the real world (Recall that all of the modelsystems—particle-in-a-box, 2-D and 3-D rigid rotors, harmonic oscillators—

sys-could be applied to real systems even if the real systems themselves weren’t

exactly ideal.) As such, quantum mechanics is more applicable than Bohr’stheory and can be considered “better.” We will conclude our development ofquantum mechanics by seeing how it applies to more complicated systems:other atoms and even molecules What we will find is that explicit, analyticsolutions to these systems are not possible, but quantum mechanics doessupply the tools for understanding these systems nonetheless

In this chapter, we will consider one more property of the electron, which iscalled spin Spin has dramatic consequences for the structure of matter, con-sequences that could not have been considered by the standards of classicalmechanics We will see that an exact, analytic solution for an atom as simple

as helium is not possible, and so the Schrödinger equation cannot be solvedanalytically for larger atoms or molecules But there are two tools for studyinglarger systems to any degree of accuracy: perturbation theory and variationaltheory Each tool has its advantages, and both of them are used today to studyatoms and molecules and their reactions

Finally, we will consider in a simple way how quantum mechanics siders a molecular system Molecules can get very complicated However, we

con-can apply quantum mechanics to molecules We will finish this chapter with

an introduction to molecular orbitals and how they are defined for a verysimple molecule, H2 Simple as this system is, it paves the way for othermolecules

370

12.1 Synopsis

12.2 Spin

12.3 The Helium Atom

12.4 Spin Orbitals and the

12.8 Linear Variation Theory

12.9 Comparison of Variation and

12.2 Spin

Not long before quantum mechanics was developed, an important tal observation was made In 1922, Otto Stern and W Gerlach attempted tomeasure the magnetic moment of the silver atom They passed vaporized sil-ver atoms through a magnetic field and recorded the pattern that the beam ofatoms made after it passed through the magnetic field Surprisingly, the beamsplit into two parts The experiment is illustrated in Figure 12.1

experimen-Attempts to explain this in terms of the Bohr theory and quantized angularmomentum of electrons in their orbits failed Finally, in 1925, GeorgeUhlenbeck and Samuel Goudsmit proposed that this result could be explained

if it was assumed that the electron had its own angular momentum This

an-gular momentum was an intrinsic property of the electron itself and not a sequence of any motion of the electron In order to explain the experimentalresults, Uhlenbeck and Goudsmit proposed that components of the intrinsic

con-angular momentum, called spin con-angular momentum, had quantized values of

either 1

2  or 1

2  (Recall that h has units of angular momentum.)

Since that proposal, it has become understood that all electrons have an

in-trinsic angular momentum called spin Although commonly compared to the

spinning of a top, the spin angular momentum of an electron is not due to anyrotation about the axis of the particle Indeed, it would be impossible for us todetermine that an electron is actually spinning Spin is a property of a parti-cle’s very existence This property behaves as if it were an angular momentum,

so for all intents and purposes it is considered an angular momentum.Like the angular momentum of an electron in its orbit, there are two mea-surables for spin that can be observed simultaneously: the square of the total

spin and the z component of the spin Because spin is an angular momentum,

there are eigenvalue equations for the spin observables that are the same as for

L

ˆ2and L ˆ, except we use the operators Sˆ z 2and Sˆ to indicate the spin observ-z

ables We also introduce the quantum numbers s and m sto represent the

quan-tized values of the spin of the particles (Do not confuse s, the symbol for the

Magnet

Glass plate

Figure 12.1 A diagram of the Stern-Gerlach experiment A beam of silver atoms passed through a magnetic field splits into two separate beams This finding was used to propose the ex- istence of spin on the electron.

spin angular momentum, with s, an orbital that has   0.) The eigenvalueequations are therefore

out, is a characteristic of a type of subatomic particle, and all electrons have

the same value for their s quantum number For the possible values of the z

component of the spin, there is a similar relationship to the possible values of

mand : msgoes from s to s in integral steps, so mscan equal 12or 12

Thus, there is only one possible value of s for electrons, and two possible ues for m s

val-Spin also has no classical counterpart Nothing in classical mechanics dicts or explains the existence of a property we call spin Even quantum me-chanics, at first, did not provide any justification for spin It wasn’t until 1928when Paul A M Dirac incorporated relativity theory into the Schrödingerequation that spin appeared as a natural theoretical prediction of quantummechanics The incorporation of relativity into quantum mechanics was one

pre-of the final major advances in the development pre-of the theory pre-of quantum chanics Among other things, it led to the prediction of antimatter, whose ex-istence was verified experimentally by Carl Anderson (with the discovery of thepositron) in 1932

me-Example 12.1

What is the value, in Js, of the spin of an electron? Compare this to the value

of the angular momentum for an electron in s and p orbitals of an H-like

atom

Solution

The value of the spin angular momentum of an electron is determined by ing equation 12.1 We must recognize that the operator is the square of thetotal spin, and to find the value for spin we will have to take a square root

The angular momentum of an electron in an s orbital is zero, since   0 for

an electron in an s orbital In a p orbital,  1, so the angular momentum is

(+ 1)  1  2 6.626

2

1034Js

  1.491 1034J  swhich is almost twice as great as the spin The magnitude of the spin angu-lar momentum is not much smaller than the angular momentum of an elec-tron in its orbit Its effects, therefore, cannot be ignored

The existence of an intrinsic angular momentum requires some additionalspecificity when referring to angular momenta of electrons One must now

differentiate between orbital angular momentum and spin angular

momen-tum Both observables are angular momenta, but they arise from differentproperties of the electron: one from its motion about a nucleus, the other fromits very existence

The spin angular momentum of an electron can have only certain specific

values Spin is quantized Like the z component of orbital angular momentum,

m s has 2s  1 possible values In the case of the electron, s 12, so the only

possible values of m sare 12 and 12 The specification of an electron’s spintherefore represents two other quantum numbers that can be used to label the

state of that electron In practice, however, it is convenient to not specify s,

since it is always 12for electrons This gives us a total of four quantum

num-bers: the principal quantum number n, the orbital angular momentum

quan-tum number , the orbital angular momenquan-tum z component m, and the spin

angular momentum (z component) m s These are the only four quantum bers needed to specify the complete state of an electron

num-Example 12.2

List all possible combinations of all four quantum numbers for an electron

in the 2p orbital of a hydrogen atom.

Solution

In tabular form, the possible combinations are

a hydrogen atom changes its spin, there is a concurrent energy change that isequivalent to light having a frequency of 1420.40575 MHz, or a wavenumber

of about 21 cm1, as shown in Figure 12.2 Because of the pervasiveness of drogen in space, this “21-cm1radiation” is important for radio astronomerswho are studying the structure of the universe

hy-Finally, since spin is part of the properties of an electron, its observable ues should be determined from the electron’s wavefunction That is, thereshould be a spin wavefunction part of the overall  A discussion of the exactform of the spin part of a wavefunction is beyond our scope here However,

val-since there is only one possible observable value of the total spin (s 1

2 ) and

only two possible values of the z component of the spin (m s 1

2 or 1

2 ), it istypical to represent the spin part of the wavefunction by the Greek letters and

, depending on whether the quantum number m sis 12or 12, respectively

Figure 12.2 A very high resolution spectrum

of the hydrogen atom shows a tiny splitting due

to the spin on the electron This splitting is caused

by the electron spin interacting with the nuclear

spin of the hydrogen atom’s nucleus (a proton).

Spin is unaffected by any other property or observable of the electron, and thespin component of a one-electron wavefunction is separable from the spatialpart of the wavefunction Like the three parts of the hydrogen atom’s electronicwavefunction, the spin function multiplies the rest of So for example, thecomplete wavefunctions for an electron in a hydrogen atom are

  R n, ,m m

for an electron having m sof12 A similar wavefunction, in terms of , can be

written for an electron having m s  1

2 

In the previous chapter, it was shown how quantum mechanics provides an act, analytic solution to the Schrödinger equation when applied to the hydro-gen atom Even the existence of spin, discussed in the last section, does not al-ter this solution (it only adds a little more complexity to the solution, acomplexity we will not consider further here) The next largest atom is the he-lium atom, He It has a nuclear charge of 2, and it has two electrons aboutthe nucleus The helium atom is illustrated in Figure 12.3, along with some ofthe coordinates used to describe the positions of the subatomic particles.Implicit in the following discussion is the idea that both electrons of heliumwill occupy the lowest possible energy state

ex-In order to properly write the complete form of the Schrödinger equationfor helium, it is important to understand the sources of the kinetic and po-tential energy in the atom Assuming only electronic motion with respect to amotionless nucleus, kinetic energy comes from the motion of the two elec-trons It is assumed that the kinetic energy part of the Hamiltonian operator

is the same for the two electrons and that the total kinetic energy is the sum

of the two individual parts To simplify the Hamiltonian, we will use the bol 2, called del-squared, to indicate the three-dimensional second derivativeoperator:

This definition makes the Schrödinger equation look less complicated.2is

also called the Laplacian operator It is important to remember, however, that

del-squared represents a sum of three separate derivatives The kinetic energypart of the Hamiltonian can be written as

between electron 2 and the nucleus, and a repulsion between electron 1 and

electron 2 (since they are both negatively charged) Each part depends on the

distance between the particles involved; the distances are labeled r1, r2, and r12

as illustrated in Figure 12.3 Respectively, the potential energy part of theHamiltonian is thus

Figure 12.3 Definitions of the radial

coordi-nates for the helium atom.

where the other variables have been defined in the previous chapter The 2 inthe numerator of each of the first two terms is due to the 2 charge on thehelium nucleus The first two terms are negative, indicating an attraction, andthe final term is positive, indicating a repulsion The complete Hamiltonianoperator for the helium atom is

where Etotrepresents the total electronic energy of a helium atom

The Hamiltonian (and thus the Schrödinger equation) can be rearranged bygrouping together the two terms (one kinetic, one potential) that deal withelectron 1 only and also grouping together the two terms that deal with elec-tron 2 only:

“separation of electrons” approach will allow us to solve the Schrödinger tion for helium

equa-The problem is with the last term: e2/4 0r12 It contains a term, r12, that

depends on the positions of both of the electrons It does not belong only with

the terms for just electron 1, nor does it belong only with the terms for justelectron 2 Because this last term cannot be separated into parts involving only

one electron at a time, the complete Hamiltonian operator is not separable and

it cannot be solved by separation into smaller, one-electron pieces In order forthe Schrödinger equation for the helium atom to be solved analytically, iteither must be solved completely or not at all

To date, there is no known analytic solution to the second-order tial Schrödinger equation for the helium atom This does not mean that there

differen-is no solution, or that wavefunctions do not exdifferen-ist It simply means that we

know of no mathematical function that satisfies the differential equation In

fact, for atoms and molecules that have more than one electron, the lack of

separability leads directly to the fact that there are no known analytical solutions

to any atom larger than hydrogen Again, this does not mean that the

wave-functions do not exist It simply means that we must use other methods to derstand the behavior of the electrons in such systems (It has been provenmathematically that there is no analytic solution to the so-called three-bodyproblem, as the He atom can be described Therefore, we must approach multi-electron systems differently.)

un-Nor should this lack be taken as a failure of quantum mechanics In this text,

we can only scratch the surface of the tools that quantum mechanics provides.Quantum mechanics does provide tools to understand such systems Atomsand molecules having more than one electron can be studied and understood

by applying such tools to more and more exacting detail The level of detail pends on the time, resources, and patience of the person applying the tools Intheory, one can determine energies and momenta and other observables to the

same level that one can know such observables for the hydrogen atom—if onehas the tools.

Example 12.3

Assume that the helium wavefunction is a product of two hydrogen-likewavefunctions (that is, neglect the term for the repulsion between the elec-

trons) in the n 1 principal quantum shell Determine the electronic energy

of the helium atom and compare it to the experimentally determined energy

of1.265 1017J (Total energies are determined experimentally by suring how much energy it takes to remove all of the electrons from an atom.)

e h

4 2



n2

  8

2



2 2 0

e h

4 2

4 2

EHe 1.743 1017J

which is low by 37.8% compared to experiment Ignoring the repulsionbetween the electrons leads to a significant error in the total energy of the

system, so a good model of the He atom should not ignore electron-electron

repulsion

The example above shows that assuming that the electrons in helium—andany other multielectron atom—are simple combinations of hydrogen-like elec-trons is naive assumption, and predicts quantized energies that are far from theexperimentally measured values We need other ways to better estimate theenergies of such systems

12.4 Spin Orbitals and the Pauli Principle

Example 12.3 for the helium atom assumed that both electrons have a pal quantum number of 1 If the hydrogen-like wavefunction analogy were

princi-taken further, we might say that both electrons are in the s subshell of the first shell—that they are in 1s orbitals Indeed, there is experimental evidence

(mostly spectra) for this assumption What about the next element, Li? It has

a third electron Would this third electron also go into an approximate 1s

hydrogen-like orbital? Experimental evidence (spectra) shows that it doesn’t

Instead, it occupies what is approximately the s subshell of the second pal quantum shell: it is considered a 2s electron Why doesn’t it occupy the 1s

princi-shell?

We begin with the assumption that the electrons in a multielectron atom

can in fact be assigned to approximate hydrogen-like orbitals, and that the wavefunction of the complete atom is the product of the wavefunctions of each occupied orbital These orbitals can be labeled with the n quantum number labels: 1s, 2s, 2p, 3s, 3p, and so on Each s, p, d, f, subshell can also be labeled

by an mquantum number, where mranges from  to  (2  1 possible

values) But it can also be labeled with a spin quantum number m s, either 1

2

or 1 The spin part of the wavefunction is labeled with either or ,

de-pending on the value of m sfor each electron Therefore, there are several ple possibilities for the approximate wavefunction for, say, the lowest-energystate (the ground state) of the helium atom:

sim-He (1s1)(1s2)

He (1s1)(1s2 )

He (1s1 )(1s2)

He (1s1 )(1s2 )

where the subscript on 1s refers to the individual electron We will assume that

each individual Heis normalized Because each He is a combination of aspin wavefunction and an orbital wavefunction,He’s are more properly called

spin orbitals.

Because spin is a vector and because vectors can add and subtract from

each other, one can easily determine a total spin for each possible helium spin orbital (It is actually a total z component of the spin.) For the first spin or-

bital equation above, both spins are , so the total spin is (1

2 )  (1

2 )  1.Similarly, for the last spin orbital, the total spin is (12)  (12)  1 For

the middle two spin orbitals, the total (z-component) spin is exactly zero To

summarize:

12.4 Spin Orbitals and the Pauli Principle 377

Approximate wavefunction Total z-component spin

an overall angular momentum Since spin is a form of angular momentum, itshould not be surprising that magnetic fields can be used to determine theoverall spin in an atom Experiments show that ground-state helium atoms

have zero z-component spin This means that of the four approximate

wave-functions listed above, the first and last are not acceptable because they do not

agree with experimentally determined facts Only the middle two, (1s1)(1s2 )

and (1s1 ... 1

2

27

2 2

2 2

2 2

1

m

2< /small>h a

2 2

... 1s2< /sup>2s2< /sup>2p y12p z1or 1s2< /sup>2s2< /sup>2p x12p z1... as 1s2< /sup>2s2< /sup>2p2< /sup> An unacceptable ground-state electron

config-uration might be 1s2< /sup>2s2< /sup>2p x2< /sup>,