ch10-2web-Revised

Chem 5 Chapter 10 The Periodic Table and Some Atomic Properties Part 2 November 4, 2002

Artificial Elements?

?

Probability Distribution

Galton board http://probability.ca/jeff/java/utday/

Define Probability Density, P(x) such that

Probability = P(x) ∆x

Probability is - unitless - proportional to ∆x - equal to unity when integrated over the whole range or space. ∞

∫ P(x)dx = 1

−∞

Probability density is - a continuous function of x - has units of per unit distance (for 1D distribution), per unit area (for 2D distribution) or per unit volume (for 3D distribution )

P(x)

x

∆x

Neither probability nor probability density is negative.

1-D Standing Wave Tie the ends down two nodes

+ yo + yo -

Not allowed one node

no node

Traveling wave

+ yo x Waves with positive and negative amplitudes, amplitude variation along x, and nodes

Ends not tied down

Probability Distribution of the Particle

Particle in a Box

n quantum No. x

n – 1 nodes 0

0

x

p322 Solutions of

Schrödinger Eq

ψ2 are the probability densities for finding the particle, not probabilities!

1-D Schrödinger Equation h 2 d 2ψ − 2 + V(x)ψ = Eψ 2 8π m dx Total Energy E = V + Ek, m particle mass

p323

Potential Energy V(x) = 0,

h 2 d 2ψ − 2 = Eψ 2 8π m dx

Verify the solutions

ψ n ( x) =

2  nπx  sin   L  L 

n = 1, 2, 3 … n2h2 with En = 8mL2

h2 d2ψ h 2 nπ Left side − 2 =− 2 8π m dx 2 8π m L

2 d   nπ x    cos  L dx   L   h 2 nπ nπ 2  nπ x  = 2 sin   8π m L L L  L 

h2 n2 = 8m L2 = Enψ n

2  n πx  sin   L  L  right side

Energy levels for a particle in a box • The energy is quantized. n=4

• There are n-1 nodes in the nth wave functions– same as in a H atom. The more nodes, the higher the energy.

n=3

• Zero point energy. E=Ek > 0 for the lowest energy, n=1, which is a consequence of the uncertainty principle.

n=2 n=1 n2h2 En = 8mL2

• The energy spacing is inversely proportional to L2.

Work by Prof. Moungi Bowendi at MIT

Quantum dots - nanometer crystals of CdSe The different colors of emission under ultraviolet illumination are due to different diameters of the nanocrystals in different vials. The size dependence of the emission frequency can be explained by quantum mechanics of the particle (electron) in a box.

Radial probability distribution Fig. 9-32 of text, the vertical axis should be R2r2, not 4πR2r2

1-D Probability Density R2r2

2 2 2 Ψ = R ( r ) Y Ψ = R ( r ) Y ( θ , φ ) Wave Function nlm 3-D Prob. Density nlm lm (θ , φ ) lm nl nl

Probability in dV Ψnlm dV = R nl ( r )Y lm (θ , φ )dV = R nl ( r )Y lm (θ , φ ) r sin θdrdθdφ 2

Integrating over θ and φ

2

2

2

2

1-D Radial Probability Density

2

R2(r)r2

R2Y2(θ,φ) 3D Prob. density

R Radial wave function

R2r2 1D Prob. density

0

Potential Energy Potential Energy, V Kinetic Energy, Ek Total Energy, E

p222

E = Ek + V high V

Gravitational Potential V linear with respect to height

low V

Columbic Potential

V

r→∞

V=0

er +Z

r Z2 En = − RH 2 n

Potential Energy Ze 2 V (r ) = − r

r Columbic Attraction Force

Ze2 dV ( r ) F= 2 = r dr

e - electron charge

p323

V

Total orbital En = Ek + V = 1 V 2 energy Z2 En = − RH 2 n

RH Ryberg const

Screening (or Shielding) in Multi-electron Atoms Screening reduces the apparent nuclear charge. e+Z

+Z

eZeff = Z - S

Effective nuclear charge

Zeff approaches +1 V (r ) = −

Zeff approaches Z

Z eff (r )e 2 r

Orbital Energy

En = − RH

Z eff2 n

2

Eq. 10.4

n

En

Zeff

En

RH Ryberg Const

Atom or Ion Radius n 2 a0 rnl =< r >= Z eff

 1  l (l + 1)   1 + 1 −  2  n 2   

n



Zeff



a0 Bohr radius Eq. 10.5

An example F-, Na+,Mg2+ are isoelectronic

1s22s22p6

What are relative sizes? r ∝

n2 Z eff

Mg2+ is 66 pm

Same n

Na+ is 95 pm

But different Z

F- is 133 pm

Z eff Mg2+ > Z effNa+ > Z effF-

Another Example of Screening and Atomic Radii

r ∝

n

•

2

Z eff

H- (1s2)

• Z=1

•

H

e-

e• Zeff = 1- 0.3 = 0.7

• Z=1 •

e-

Z=2

•

He(1s2)

• e-

Zeff~2-0=2 Z=2 •

• -

e-

•

e Zeff = 2 – 0.2 =1.8

e- • Zeff~2-1=1

He(1s12p1)

Penetration

- to circumvent screening

-The ability of electrons in s (compared to p) orbitals or p (compared to d) orbitals to get close to nucleus

Compare E2s and E2p Higher probability to be close, larger Zeff

R2r2

Z eff ( r )e 2

Potential energy

V (r) = −

Total energy

2 2 Z eff 1 1 Z eff ( r )e = − RH 2 E = V (r) = − 2 2 r n

Zeff2s > Zeff2p

r

E2s < E2p

Compare E3s, E3p, E3d

Higher probability to be close, large Zeff

R2r2

Zeff3s > Zeff3p> Zeff3d

E3s < E3p < E3d

Degenerate

Energy splitting

Energy cross over

Energy crossover is dependent on the atomic No. (different Zeff)

K

Mn

Mn2+

E4s < E3d

E3d < E4s

4f105d16s2

4f36s2

5d first Exceptions

4f46s2

4f56s2

Normal filling

4f66s2

4f76s2

Ionization Energy Mg(g) → Mg+(g) + e-

I1=738kJ/mol

The energy needed to ionize a gaseous atom. 2 2 2  Z eff  Z eff Z eff I = ∆E = E f − Ei = RH  2 − 2  = RH 2 n ∞   n

I = En = RH

Z eff2 n2

I = RH n=2

n=3

n

2

Increasing across period Noble gases high I.E.

n=3 Zeff

Z eff2

Alkali metals low I.E.