Σ=Zr/A0 A0=0.053Nm Θ Is the Polar Angle Down from Z

s=Zr/a0 a0=0.053nm q is the polar angle down from z

FIGURE 12.23: Orbital Energy Levels for H Atom The energies are given by this expression– and only depend on n, not on the other quantum numbers.

2 4 2 $ Z me ' $ Z ' 18 E 2.18*10 − J = −& 2 2 2 ) = −& 2 ) % n 8ε0 h ( % n (

€ l: Angular Momentum ml: The Direction of the Angular Momentum

l ranges from 0 to n-1; ml ranges from –l to l It is possible to prove that there are n2 orbitals for any value of n PollEverywhere Question 51 Statements from Nice Chem websites about nodes. Which is true?

1. For s orbitals, n is the number of angular nodes. 2. Radial nodes are always spherical and angular nodes are planar. 3. Thus, n tells how many radial nodes an orbital will have.

A. T T F B. T F T C. F T T D. T F F E. F F F Nodes in Wavefunctions

• Node: wave function is zero on a plane or a sphere or a cone or another geometric surface, wave function changes sign on crossing this boundary

• Total number of nodes for yn is n-1 Angular vs. Radial Nodes

The shapes of the probability distribution, angular vs. radial nodes, relate to the value of quantized angular momentum for the orbital “l”.

Angular nodes are functions of angles: e.g. planes or cones There are l angular nodes (0 for s, 1 for p, 2 for d etc.)

Radial nodes: function of radius e.g. spheres. There are (n-l-1) radial nodes. From Wikipedia Orbitals are often depicted in terms of the 90% or 95% or sometimes 67% probability surface Angular vs. Radial Nodes

4p vs. 2p at right. Which is which?

http://winter.group.shef.ac.uk/orbitron/AOs/2p/index.html Node for 3dz2

How many nodes does this wave function have? Where does this wavefunction have a node? Radial? Angular?

A 0, radial

B 1, r= 2a0. radial C 1, q=0, angular D 2, q=arccos(+/-sqrt(1/3)), angular E 2, r= 1.9, 7.1, angular Node for 3dz2

How many nodes does this wave function have? Where does this wavefunction have a node? Radial? Angular?

A 0, radial

B 1, r= 2a0. radial C 1, q=0, angular D 2, q=arccos(+/-sqrt(1/3)), angular E 2, r= 1.9, 7.1, angular 3 cos2 q - 1 = 0 q= Arc.cos (+/- sqrt (1/3)) q= 0.9553 rad or 2.1862 rad q = 54.73 deg or 125.248 deg Two cones, at “magic angle” and negative magic angle Node for 3dz2

How many nodes does this wave function have? Where does this wavefunction have a node? ANS: Its found at a specific angle or angles, so they are angular nodes. This is a cone above, and one below.

arccos 1 .995, 2.19rad A 0, θ = (± 3 ) = B 1, r= 2a 0 = 54.735,125.478degrees C 1, q=0 D 2, cos2q=1/3 E 2, r= 1.9, 7.1 STATEMENTS FROM NICE CHEM WEBSITES: WHICH TRUE, WHICH IS FALSE?

1. For s orbitals, n is the number of angular nodes 2. Radial nodes are spherical and angular nodes are always planar 3. Thus, n tells how many radial nodes an orbital will have

A. T T F B. T F T C. F T T D. T F F E. F F F CAVEAT EMPTOR… STATEMENTS FROM “NICE LOOKING” CHEM WEBSITES, CORRECTED

1. For s orbitals, n-1 is the number of angular nodes

2. Radial nodes are spherical and angular nodes are planar, conical sections or other angular dependent surfaces….

3. Thus, n-1- l tells how many radial nodes an orbital will have PollEverywhere Question 52 Ionization Energies The (1st) ionization is the minimum energy needed to release an electron from the atom. The 2nd ionization energy is the minimum energy needed to remove a second electron. Which of the following statements is correct? + • For the transitions n1 ® n2, the frequency is larger for H than for He . • The ionization energy of the H atom is smaller than the second ionization energy of the He atom. • The 1s orbital in He+ is larger (in the sense that the probability density is shifted outward) than the 1s orbital in H.

2 4 2 $ Z me ' $ Z ' 18 E 2.18*10 − J = −& 2 2 2 ) = −& 2 ) % n 8ε0 h ( % n (

€ • b) The ionization energy of the H atom is smaller than the second ionization energy of the He atom.

• Section 12.4, The Bohr Model

• In both cases, only one electron is present. For helium, Z = 2; for hydrogen, Z = 1. Ionization energy is related to Z2, so the second ionization energy for helium is 4 times as great as the ionization energy for the hydrogen atom. PollEverywhere Question 53 Transition Energies and Light Frequency

A Li2+ atom undergoes the n=3 to n=1 transition. What would the energy of the emitted light be?

-17 2 4 2 A) 3.3 x 10 J $ Z me ' $ Z ' −18 E = − 2 = −& )2.18 *10 J B) 1.7 x 10 17 J & 2 2 ) 2 % n 8ε0 h ( % n ( C) 9.0 x 10-17 J D) 1.7 x 10-17 J E) 1.6 x 10-18 J € Can we use the provided equation for Li2+? (Is Li2+ a one electron atom?)

2 4 2 $ Z me ' $ Z ' 18 E 2.18*10 − J = −& 2 2 2 ) = −& 2 ) % n 8ε0 h ( % n (

€ Emission. Final energy is lower, photon is given off.

DE= Ei - Ef is positive when f >i, higher energy minus lower energy. 2 2 2 2 2 2 2 Ep = ( [-(Z /ni )R ] - [-(Z /nf )R]) = Z (1/nf - 1/ni )R

2 2 2 -18 Ep= Z (1/nf - 1/ni )2.18*10 J • Photon energy is positive and is equal to energy lost in molecule.

Absorption. Final energy is higher. Photon is absorbed. 2 2 2 -18 Ep= Z (1/ni - 1/nf )2.18*10 J. • Photon energy is positive and is equal to energy gain in molecule. A Li2+ atom undergoes the n=3 to n=1 transition.

2 2 2 -18 Ep = Z (1/nf - 1/ni )2.18*10 J

2 2 2 -18 -17 Ep = 3 (1/1 - 1/3 )R = 8 x 2.18 x 10 J = 1.7 x 10 J = Ep. PollEverywhere Question 53 Transition Energies and Light Frequency

A Li2+ atom undergoes the n=3 to n=1 transition. What would the energy of the emitted light be? 2 2 2 -18 -17 Ep= Z (1/nf - 1/ni )2.18*10 J = 1.7 x 10 J

A) 3.3 x 10-17 J B) 1.7 x 10 17 J C) 9.0 x 10-17 J D) 1.7 x 10-17 J E) 1.6 x 10-18 J PollEverywhere Question 54 Transition Energies and Light Frequency A Li2+ atom undergoes the n=3 to n=1 transition. What would the -17 wavelength of the emitted light be? Recall Ep= DE = 1.7 x 10 J

8 -34 2 -1 -34 Ep=hn; ln= c; c = 3x10 M/s; h = 6.6310 kg M s = .6310 Js

A) 114 nm or 1140 Angstroms is in the UV range B) 11.4 nm or 114 Angstroms is in the soft X-ray range C) 1.1 nm or 11.4 A is in the X-ray range D) 1.1 A or 114 pm is in the X-ray range E) 0.11A or 11.4 pm l=hc/DE=(6.63 x 10-34Js)(3x108 M/s)/17.4 x 10-18 J =1.14 x 10-8 M

−34 2 8 ⎡6.6x10 Kgm / s•3x10 m / s⎤ −8 λ = hc / E = ⎢ ⎥ =1.1x10 m ⎣ 1.7x10−17 Kgm2s−2 ⎦

11.4 nM or 114 Angstroms is in the soft X-ray range PollEverywhere Question 54 Transition Energies and Light Frequency A Li2+ atom undergoes the n=3 to n=1 transition. What would the -17 wavelength of the emitted light be? Recall 1.7 x 10 J = DE=Ep; 8 -34 2 -1 -34 Ep=hn; ln= c; c = 3x10 M/s; h = 6.6310 kg M s = .6310 Js

A) 114 nm or 1140 Angstroms is in the UV range B) 11.4 nm or 114 Angstroms is in the soft X-ray range C) 1.1 nm or 11.4 A is in the X-ray range D) 1.1 A or 114 pm is in the X-ray range E) 0.11A or 11.4 pm PollEverywhere Question 55 Transition Energies and Light Frequency

For which of the following electron transitions in a hydrogen atom does the emitted light have the longest wavelength? 2 2 2 -18 Ep= Z (1/nf - 1/ni )2.18*10 J

a) n = 4 to n = 3 b) n = 4 to n = 2 c) n = 4 to n = 1 d) n = 3 to n = 2 e) n = 2 to n = 1 ⎛ 1 1 ⎞ E Z 2 2.18*10−18 J Δ = ⎜ 2 − 2 ⎟ ⎝ ni n f ⎠ • a) n = 4 to n = 3

• Section 12.4, The Bohr Model

• Energy is inversely related to wavelength: the smaller the energy change, the longer the wavelength. The transition n = 4 to n = 3 has the smallest energy change of all of the options. Question 56 The Hydrogen Lamp A strong line is predicted for H n=1 to n=2 at 121.57 nm and is observed experimentally. Was this visible to you?

A red line was observed at 653 nm (also 486 nm which is blue violet).

Explain this observation. • 121.57 nm corresponds to this photon energy:

−34 2 8 ⎡6.63x10 Kgm / s•3x10 m / s⎤ −17 2 −2 E = hυ = hc / λ = ⎢ ⎥ =1.636x10 Kgm s ⎣ 1.216x10−8 m ⎦

• This matches the Bohr Energy difference Z=1, n=2 to n=1

2 2 2 -18 Ep= Z (1/nf - 1/ni )*2.18*10 J = (3/4)* 2.18*10-18J =1.635*10-18J • 653.3 nm the photon energy, and energy change in the molecule is

−34 2 8 ⎡6.63x10 Kgm / s•3x10 m / s⎤ −19 2 −2 E = hυ = hc / λ = ⎢ ⎥ = 3.03x10 Kgm s ⎣ 6.53.x10−7 m ⎦

-18 • All of the transitions that start in n=1 have Ep > 1.6 x 10 J • What can you conclude? Both initial and final states are excited states! Try/guess n=3 to n=2

⎛ 1 1 ⎞ E Z 2 2.18*10−18 J Δ = ⎜ 2 − 2 ⎟ ⎝ ni n f ⎠ ⎛ 1 1 ⎞ E 12 2.18*10−18 J Δ = ⎜ 2 − 2 ⎟ ⎝ 2i 3 f ⎠ ⎛ 9 4 ⎞ ΔE = ⎜ − ⎟2.18*10−18 J ⎝36 36 ⎠ = (5 / 36)*2.18*10−18 J = 0.3027*10−18 J Hydrogen Lamp: Summary

• Strong lines are observed at 121.57 nm and 102.57 in the UV – not visible. • Weaker visible lines at 653.3 nm and 483.6 nm involving initial and final states being excited states • These lines are seen in the solar spectrum providing evidence of the presence of hydrogen in the sun (and analogously elsewhere in the universe). How many electrons per orbital?

Postulate: every electron has to be in a different state.

• The n=1 l=0 (1s) orbital has two electrons in it!!!

• In fact, orbitals generally can have up to 2 electrons.

• Why? Are the two electrons in the 1s orbital in different states? Return to this question next week when we discuss: intrinsic spin Aufbau, periodicity. Stern-Gerlach

Otto Stern and Walther Gerlach in 1922 sent a neutral gaseous silver atom spray through an inhomogeneous magnet. An uncharged particle should not be deflected unless is has a magnetic moment, or is magnetic. Is a neutral atom magnetic? Two beams came out, both deflected about a mm, (which they could see on the collection paper thanks to their cigar smoke)…. So not only the neutral atom was magnetic, it had apparently two quantized magnetic states! Why are the beams deflected?

What could cause the neutral atom to have 2 quantized magnetic states?

Not the orbital angular momentum! That would be three states if the orbital were l = 1 (p). That would be five states if l = 2 (d). Stern-Gerlach How many electrons does a neutral Ag atom have?

A 25 B 46 C 47 D 48 E none of the above Stern-Gerlach Ag has an odd number of electrons (47). This turned out to be the key feature. With an odd number of electrons, the beam split into two beams in a magnetic field.

Generally electrons in atoms are paired. If there are an odd number of total electrons, then they cannot all be paired up in the same wavefucntion (level), at least one is unpaired. Stern-Gerlach They postulated that an isolated electron, has “spin” (spins on its axis) and therefore has a magnetic moment.

An “unpaired” electron could be randomly pointed up or down in the field: the energies are not identical but they are very similar. The experimental result says that it is decisively one or the other: the z component of the magnetic moment is quantized!

Note the analogy to orbital angular momentum. The total is l and the projection along an axis is quantized from –l to +l George Uhlenbeck and Samuel Goudsmit

Independently (later) proposed the existence of intrinsic spin. They were graduate students in 1925 in Leiden Isidor Rabi

• CU Faculty • Demonstrated how to force transitions between the up and down state of the electron spin. • Director of Radiation Lab at MIT for development of Radar and of the Atomic Bomb • Our Rabi program is named for him. Pauli Paired Two electrons can be in the 1s orbital but are actually not in the same state! There are two distinct states for the 1s orbital. They differ in terms of having opposing spin direction or intrinsic angular momentum (spinning on their own axis), therefore different magnetic moments, but are otherwise the same. Wolfgang Pauli (1900-1958) predicted that the 2 states should pair into the same orbital. Once they are paired the magnetism cancels and apparently “disappears” . They are called “Pauli paired” and this is called the Pauli principle. States of the Electron in an Atom 4 quantum numbers • n : principal quantum number is an integer greater than or equal to 1 • l : angular momentum quantum number is an integer from 0 to n-1

• ml : magnetic quantum number is an integer between -l and +l

• ms : spin angular momentum (orientation of intrinsic electron spin) either +1/2 or -1/2 States of the Electron in an Atom

Examples:

Ground state of H is 1s (n=1, l=0, ml=0) but can be ms= +1/2 or ms= -1/2 2 ground states of equal energy

st 1 excited states of H are 2s (n=2, l=0, ml=0), or 2p (x, y, or z) (n=2, l=1, mll = +1,0,-1)

and each can be ms= +1/2 or ms= -1/2 8 excited states of equal energy Practical consequences of intrinsic spin

• The magnetic moments of both protons and electrons can be used for analytic chemistry: Magnetic Resonance Spectroscopy and Imaging are very sensitive informative tools for chemists, in contexts ranging from land mine detection, security, forensic, medicine, materials science, to biochemistry, and medicine. • Only some nuclei have a magnetic moment (depends on the specific isotope or nuclide). Uses of Unpaired Electron Detection by Magnetic Resonance

• Radiation dosimetry: Radiation damage, Detection of irradiated foods, Archaeological dating, etc. • Free Radicals in Biology and Medicine Most organic (non metal) compounds have all paired spins, and if a compound has an unpaired spin it is likely to be very reactive and an important metabolism indicator, and even a cancer indicator, therefore interesting to detect diagnostically. • Spectroscopy used to determine structures of molecules Magnetic Resonance Imaging

• Detecting the proton spin

• Used for medical diagnostics especially soft tissue analysis

• Used for materials analysis NY Times, Oct 7, 2003

• American and Briton win Medicine Nobel for Using Chemists Test for MRI • unlike X-rays, which use radiation, MRU scans the body using only radio waves and magnets The hydrogen orbitals provide approximations for the states in a Aufbau Building Up Atoms multi-electron system, because the spatial overlap of these orbitals is minimal. Treat multi-electron state as a combination of the single electron states.

2s 2px 2py 2pz

1s For example, the two electron Aufbau Building Up Atoms: atom, He, H-, or Li+. The Two Electrons – Lowest State lowest state is if they are both in n=1.

n ℓ mℓ ms 1s 2s 2px 2py 2pz •1st el 1 0 0 ½ •2nd el 1 0 0 -½

1s 2s 2px 2py 2pz

1s 2s 2px 2py 2pz Aufbau Building Up Atoms: • They cannot be both in the Two Electrons – Impermissible State exact same state!

1s 2s 2px 2py 2pz

n ℓ mℓ ms •1st el 1 0 0 -½ 1s 2s 2px 2py 2pz •2nd el 1 0 0 -½ Aufbau Building Up Atoms: • The NEXT lowest state has Two Electrons –Excited State one in n=1 and the other is in n=2. Now the electron intrinsic spins orientations can be in any combination.

1s 2px 2py 2pz n ℓ mℓ ms 2s •1st el 2 0 0 -½ •2nd el 1 0 0 -½

n ℓ mℓ ms •1st el 2 0 0 ½ 1s 2s 2px 2py 2pz •2nd el 1 0 0 -½

n ℓ mℓ ms •1st el 2 0 0 ½ 1s 2s 2px 2py 2pz •2nd el 1 0 0 ½ Interactions in the in the multielectron Aufbau” Building Up Atoms atom also lead to differences -- the Energies orbitals and their energies are not the same as for the one electron atom. The 1-electron wavefunctions are only an approximation or a basis set.

Note the use of the red arrow to denote

1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz

Excited States 1s2 2p

1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz Lowest States 1s2 2s For example Li or Be+ : filled 1s Aufbau Building Up Atoms shell (two electrons), and third Energies electron in the n=2 level.

The states with 2s are lower energy than the ones with 2p.

The electrons’ intrinsic spin must be opposite for the first two electrons, and the third can be in either direction.

1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz

Excited States 1s2 2p

1s 2s 2px 2py 2pz 1s 2s 2px 2py 2pz Lowest States 1s2 2s Aufbau Building Up Multielectron Atoms Energies

1312 kJ /mole: ionization of hydrogen atoms, observed and predicted

2372 kJ/mol: ionization of helium atoms.

What would Helium’s ionization erergy be predicted to be, based on the Bohr or Schroedinger ?

2 4 2 2 " Z me % " Z % −18 " Z % E = −$ 2 2 2 ' = −$ 2 '2.18*10 J = −$ 2 'R∞ # n 8ε0 h & # n & # n & Aufbau Building Up Multielectron Atoms Energies

1312 kJ /mole: ionization of hydrogen atoms n=1 Z=1 (predicted and observed) 2 4 2 2 " Z me % " Z % −18 " Z % E = −$ 2 2 2 ' = −$ 2 '2.18*10 J = −$ 2 'R∞ # n 8ε0 h & # n & # n &

4* 1312 = 5248 kJ /mole: ionization of the one electron He+ ion with n=1 Z=2 (predicted and observed)

2372 kJ/mol: ionization of helium atoms with two electrons. The binding energy for the second electron is much less than the single electron in He+. The electrons are less strongly bound because they repel each other or “screen” the nucleus. Repulsion of the Electrons in the Multielectron Atom

1 electron in 1s orbital. 1312 kJ /mole: ionization of hydrogen atoms. Helium’s second ionization energy is 5250 kJ/mol (Bohr value, 4 times the value for hydrogen). Lithium’s third ionization energy is 11,815.0 kJ / mol (Bohr value, 9 times the value for hydrogen). 1 electron system in 1s orbital.

2 electrons in 1s orbital. Helium’s first ionization is only 2372 kJ/mol; Lithium’s second ionization energy is only 7300 kJ/mol. Hydrogen’s electron affinity is much less than the IE.

These large differences between removing the first vs. the second 1s electrons related to the repulsion between the two 1s electrons and the screening of one electron by the other; the resulting electron-electron repulsion subtracts from the force holding the electron to the nucleus, substantially reducing the local binding of each. Filling Order

The electrons fill in a somewhat predictable order that is different from the Bohr atom energy order because multi-electron states are not really a sum of the single electron states: electron-electron interactions are important. The order of filling derives from the energy. Were discussing the rules for the multielectron, isolated, gas phase neutral atom .

• 2s<2p

• 4s<3d<4p 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p Na Z=11 1s2 2s2 2p6 3s Mg Z=12 1s2 2s2 2p6 3s2

Ionization energies support the idea of shells ( but not the Born energies). Aufbau: Hund’s Rule

When there is a group of orbitals that are the same energy, and they are not completely filled, the electrons are placed in separate orbitals to the extent possible.

Although 2s and 2p are not the same energy, for an isolated atom with 5 electrons the various 2p orbitals are all the same– similar properties just pointing in different directions.

2px 2py 2pz

Lowest States 1s2 2s2 2p2 are: 1s2 2s2 2px 2py or 1s2 2s2 2py 2pz

1s Aufbau: Hund’s Rule

Putting the electrons together in the same orbital is higher energy because the electrons would overlap in space more– more repulsion.

2px 2py 2pz

2s Excited States 1s2 2s2 2p2 are 1s2 2s2 2px2 or 1s2 2s2 2py2 or 1s2 2s2 2pz2

1s Aufbau: Hund’s Rule

This can lead to unpaired electrons in the lowest (most populated state), or magnetic atoms or molecules. Nevertheless magnetism is rare in organic materials (materials made mainly of Carbon and other light elements). We’ll see why soon!

2px 2py 2pz