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CHEMISTRY
introduction
Schroedinger equation why electron stick to nuclei and all stick together
[HYDROGEN(IC) ATOM
NUCLEUS CHARGE Z
1 e-
n = 1,2,3,...
l = 0,1,2,... n-1
s,d,p,f,g,..]\\
[ m = -l,...,l
s = pm 1/2
s shell 2 eigs (count spin)
p shell 6 eigs (count l=1 *spin
d shell 10 eigs
En = - Ry/n^2 Z^2 + small]
What is the relation between hydrogenic atom and the peridoc table? Missing in MS!
Assumes that N electrons will fill progressively these levels, wrong but not so desperately wrong.
[AUFBAU PRINCIPLE
FILL BY SHELL
MADELUNG'S PRINCIPLE
1s
2s 2p
3s 3p 3d
4s 4p 4d 4f
5s 5p ...]
diagonals, 1, 2, 3, 4,
show periodic table with Madelung evidenced
name groups: H-He, Li-Be, B-C-N-O-F-Ne, 3s 3p 4s before 3detc
why 3p before 4d (against n^-2) and why 2p after 2s (same n)?
depends on V deviating from Coulomb
[V(r) != 1/r
discuss the role of all electrons in creening the potential
effective potential
still assume spherical symmetry
all shells have different energy
surprising that hydrogenic works, it does as long as we can approximate the average effect of other electrons as a spherical problem
Show type of bonds: IONIC COVALENT METALLIC MOLECULAR (Van der Waals) HYDROGEN see book
Today ionic and covalent
IONIC BONDS
e- IS TRANSFERRED
FROM ONE ATOM
TO ANOTHER
AND THEN TWO IONS ATTRACT]]
Na + Cl -> Na+ + Cl- -> NaCl
analyse by energy:
E_IONIZTNG = E_(Na+_ + e-) - E_Na>0, required to pull the e out of Na to infinity
E_e-AFFINITY = E(Cl + e-) - E_(Cl-)>0, get back energy when adding the e from infinity to form Cl- (stabler than Cl + e-)
E_(COHESIVE) or _(BONDING) = E_(Na+ + Cl-) - E_(NaCl) >0], get out energy forming NaCl from two ions at infinity wreo
[DE = E_ION - E_AFF - E _COH <0 THEN REACTS.
See MS05Chemistry. Chemist use Gibbs free energy,not energy, because reaction happens at finite T and finite p, and energy must take into account work done (PV) and work lost (TS). We are calculating E at T,p=0.
warning 1 E_COH depends on definitions (some books quote DE as cohesive energy, depends on what you are putting together)
warning 2 E_COH depends whether we put together a molecule or a solid: Na is bonded to one or to say 6 clorines
In the last instance
E_COH (MADELUNG, COULOMB) = 1/2 sum_{i ne j} q_iq_j/4 pi epsi0 |r_ij| ]
q_j = charge on ion j
good if you know the position, but -> -infty fr r -> 0, so how does it work?
draws 1/r and actual potential with repulsive part, showing that diff between Coul and actual is tiny at minimum of actual potential
when putting two ions close their orbital start to overlap and Pauli exclusion comes into play
To determine E_ION and E_AFF one needs to solve the Schroed eq.
Trends easy to keep track of (from chemistry)
Area plot of electron Affinities and Ionization energies,
We search for small EION (ellipse around mostly empty shell, alcaline metals) and large EAFF (ellipse around mostly filled shell, F Cl, etc)
Chemist say that a shell of electron wants to be filled
[Why is that, cartoon picture draw
Na 11th element +11 nucleus 1s 2s 2p 10 e nuclear charge seen by 11th e is +1
E(n) = - Ry 1^2/3^2
nearly right, underestimate because when gettng inside the core it sees that Z>1]
[Draw F 9th element +9 nucleus 2 e in 1s 7 e in 2s 2p jointly as 2 shell
each of them sees a nuclear charge 7+ (large overestimate) E(n)= -Ry 7^2/2^2
explaining why an almost filled shell wants to be filled and an almost empty wants to loose its electron
overestimate but explains why Na sees a small Coulomb attraction and F sees a large one
Issue how well screened each nucleus is.
Great picture for ionic bonding]]
However there is bond also between to equal atoms, eg H2
[[COVALENT BONDS
Chemist say that the two atoms want to share an electron and this binds atoms together
It is particle in a box physics]]
[[ PARTICLE IN A BOX
H + H ----------> H2
draw 2 boxes below each of size L draw wf in each box
hbar^2/2m (pi/L)^2 + hbar^2/2m (pi/L)^2
now draw larger box under H2 size 2L draw larger wf in it
hbar^2/2m (pi/2L)^2 can hold two spin states (draw spin up down)
still 2hbar^2/2m (pi/L)^2 > 2 hbar^2/2m (pi/2L)^2
what is going on? electrons are more delocalized in the larger box, their kinetic energy is smaller
Heisenberg uncertainty: more delta x larger, delta p gets smaller hence smaller kinetic energy
spreading out the wave function lowers the kinetic energy]]
[[Now turn to He cartoon
same drawing but both He boxes have already two e spin up down in each
so under He2 ? draw one wf with two spin up down and then have to go to an excited state (node wf with two spin up down e)
the first two gain energy by delocalization but the second two no energy gain, actually loose because the second level grows as 2^2
Bonding (ground state, lower energy than the two isolated atoms) and antibonding (excited state, higher than isolated atoms)
very roughly why H bonds and He doesn't bond.]
[Better job by
MOLECULAR ORBITAL THEORY aka
TIGHT BINDING aka
LCAO = LINEAR COMBINATION OF ATOMIC ORBITALS
see easier than we do it properly for H2+
[[R1 and R2 two nuclei and one electron
BORN OPPENHEIMER APPROXIMATION
fix R1 and R2 slow electrons can rearrange and find their energy and then move nuclei, do that again, find where e+nuclei have their lowest energy. Write the Hamiltonian:
H = p^2/2m + V(r-R1) + V(r-R2) = Kappa + V1 + V2
with V(r-Ri) = e^2/4 pi espi0 |r-Ri|
simplify by throwing away 1 nucleus
(Kappa+V1)|1> = E0 |1> and this is the Hydrogen problem
(Kappa+V2)|2> = E0 |2> again]
[draw two H atoms and \1> over the left, \2> over the right: electron on nucleus 1 as if nucleus 2 were not ther at all and ...
This is not so bad if the two nuclei are very far apart draw two levels E0 with two possibility, either the electron sits on 1 or on 2, when drawing them together there will be a lower energy state when te two electrons stay between the atoms (BONDING) and an excited state (ANTIBONDING). This is a MO diagram]
[variational trial wf |psi> = phi1 |1> + phi2 2>
and vary to find lowest energy, name LCAO justified, also TB because the two items in the trial wf are for e tightly bound to either 1 or 2
very good because you can improve adding terms on the RHS: eg excites states on 1 and 2
the more terms you add the closer you get to the exact wf on the LHS
solution do exactly in the exercise session
ASSUME <i|j> = delta_ij i.e. <1|2> = 0
(BAD assumption, it is good only for far apart 1 2, wrong for 1 2 close, but the correction is trivial (although lengthy, keep algebra simple to understand better the point))
answer looks like a Schroed eq
sum_j H_ij phi_j = E phi_j
where H_ij = <i|H|j> and it is a 2x2 matrix
just H projected onto our two state Hilbert space
projected Schroed eq
write matrix elements
[
<1|H|1> = <1|Kappa+V1|1> + <1|V2|1> = E0 + V_cross
[sketch V_cross physically: interaction (dashed line) between electron 1 on nucleus 1 and nucleus 2 important below
is also called DIRECT integral, or J, but in condensed matter J is a somewhat different concept]
<2|H|2> = <2|Kappa+V2|2> + <2|V1|2> = E0 + V_cross
more interesting
<1|H|2> = <1|Kappa+V2|2> + <1|V1|2> (keep for end of lecture)]
[[= <1|E0|2> + <1|V1|2> = -t (HOPPING)
also called EXCHANGE (K)
exchange is different in cond mat
called hopping because in time dependent Schroed eq. it gives the probab o an electron on |2> to go on |1>
matrix( E0 + V_cross -t
-t* E0 + V_cross)]]
diagonalize it
[[E = E0+V_cross +- |t| eigenvalues
left out n-n Coulomb interaction, V12, go to V_cross phys sketch, it turns out that V12 = -V_cross
and this is obvious from electrostatics:
total Coulomb interaction energy of atom 1 with nucleus 2 is zero (two eq opposite densities charge seen from outside)
so V_cross + V12 = 0 and, go back to eigenvalues
E +- = E0 +- |t|
gain energy by how much you allow the e to hop, the more it hops the more it delocalizes, the more it lowers its energy
but hopping is large only if tow nuclei are close to each other.
eigenvectors
|psi+> = 1/sqrt 2 (|1> - |2>) ANTIBONDING (higher energy)
|psi-> = 1/sqrt 2 (|1> + |2>) BONDING (lower energy)]
very similar to particles in box sketch
[ draw large box with BONDing wf and ANTIBONDing, node in the middle
and below draw 1 + 2 + with two wf symmetric BONDING
and angain 1 + 2 + with two wf opposite sign node in the middle ANTIBONDING
like in the particle in the box, node raises the energy]
[ Plot E vs distance between 1 and 2 |R1-R2|
antibonding goes up
bonding goes down]
[ replot more accurate
E vs |R1-R2|
Antibondig same
Bonding goes neg then climbs positive again, when nuclei are very close becomes repulsive.
Reason is that the approximations break down,
the orthogonaity approx breaks down
also V12= -V_cross breaks down because for 2 inside the e wf you cannot cancel any more the nucleus by electron 1 ]
[do the same for different atoms
Na (a high level, wants the electron less) and F (a low level, wants the electron more)
Now the 2x2 Schroed eq is the same but the two energies on the diagonal, E0, become ENa and EF
and the eigenvalues are two different levels again, a BONDING (little lower than F) and ANTIBONDING (little higher than NA)
but coefficients of b and ab wf will not be the same
in ANTIB |psi+> = beta |Na> - alpha |F> with beta > alpha, higher prob to find e on Na
in BOND |psi-> = alpha |NA> + \beta |F> higher prob to find e on F
This fits the IONIC picture again: you on from sharing equally the e among the two nuclei to a more localized, ionic picture.
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