Molecular Physics. Attraction between the ions causes the chemical bond.

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1 Molecular Physics A molecule is a stable configuration of electron(s) and more than one nucleus. Two types of bonds: covalent and ionic (two extremes of same process) Covalent Bond Electron is in a molecular orbital it doesn t belong to an individual nucleus. Ionic Bond Electron from one atom jumped ship to the other atom, causing the individual atoms to be oppositelycharged ions. Attraction between the ions causes the chemical bond.

2 Trivial Model of a Molecule Consider two protons, both stationary, separated by a distance R. Let there be a single stationary electron, exactly in between the two protons. What is the overall potential energy of the configuration? U =+ ke2 R U =+ ke2 R ke 2 r 1 ke 2 r 2 2ke 2 R 2ke 2 R = 3ke2 R What is optimal distance between protons?

3 Trivial Model of a Molecule What s wrong with the model?

4 JITT Comment The way this book explained Ionic bonding for some reason resonated with me.i always hated the way that Chemistry textbooks worded things. I am still trying to wrap my head around atoms, so molecules are starting off a bit fuzzy overall.

5 Trivial Model of a Molecule What s wrong with the model? Electrons are fuzz balls! (and they aren t stationary) If protons are too close together, electron is less likely to be found between them. Actual Energy diagram for molecular hydrogen:

6 Covalent Bonds Consider again H 2 + (2 protons, 1 electron) 2 2 ke 2 ke 2 2m e r 1 r 2 + ke2 R ( r 1, r 2,R)=E(R) (nuclei assumed to be sta0onary)

7 2 2 ke 2 ke 2 2m e r 1 r 2 + ke2 R ( r 1, r 2,R)=E(R) Complicated! No spherical symmetry. Has to be done numerically (similar to multi-electron atoms) Approximation: Assume R>>Bohr radius. If nucleus #1 has electron: if nucleus #2 has electron: 1 A 1 exp( r 1 /a 0 ) 2 A 2 exp( r 2 /a 0 ) Use theory of superposition for general solution: S = C + ( ) A = C ( 1 2 )

8 Symmetric Wavefunction Antisymmetric (antibonding) Wavefunction

9 For symmetric spatial state (bonding orbital), electron is more likely to be found between nuclei. For anti-bonding orbital, electron is less likely to be found between nuclei. Using atomic 1s orbitals is just an approximation - numerical results also give symmetric and anti-symmetric orbitals

10 JITT Comments The basic concept of the H2 molecule seems to makes sense, but I'm still a little unclear on exactly what the overall wavefunction looks like... I get that the overall function must be antisymmmetric due to the exclusion principle. In the ground state it seems that the symmentric spatial state minimizes KE and is therefore favored, which forces an antisymmetric spin state. It also seems that you must add the wavefunctions for the electrons since there are 2. The most confusing part of the reading was the anti-bonding orbital. I don't understand how the anti-bonding orbital has a positive wave function and a negative wave? What does it mean to have a positive wave function?

11 Schematic Orbital Diagrams: For the superposition of 1s atomic states (sigma bonding orbitals), the orbital diagrams are: the plus/minus refers to the sign of the wave function

12 JITT Comments The rules of QM say that as we bring together two hydrogen atoms we should add the wave functions of the two electrons and then square the result to get the overall probability density. What would change if this procedure were reversed? If We square the wave functions first there would never be a nodal plane because there we would never be a negative wave function. We would not get the cross terms of the two wave functions, which would ignore their interactions with each other. You would get something like overlapping spheres.

13 What is optimum nuclei spacing for H 2 +? trade-off between coulomb repulsion of nuclei and attraction of probability cloud of electron between the nuclei.

14 As nuclei get closer together, probability that electron is between nuclei decreases (see hw). Result is R=0.106 nm, Binding energy B = 2.8 ev (dissociation energy) No stable molecular state using anti-bonding orbital Neutral H 2 : 2nd electron can go into bonding orbital (with S=0). Ignoring repulsion of the electrons, we would expect binding energy to double (5.6 ev). Actual binding energy is B = 4.5 ev How do you expect the optimum separation R to change (relative to H 2+?) 1. Increases 2. Decreases 3. Remains the same

15 Lithium Hydrogen

16 Helium molecules? He 2+ : 2 helium nuclei, 3 electrons: 2 electrons occupy the two available sigma 1s orbital. 1 electron has to go into the sigma* 1s (antibonding) orbital (better than the sigma 2s). Very weakly bound. He 2 : 2 occupied antibonding orbitals required. Not worth it Never observed (as a gas)

17 Bonds Involving P Orbitals Atomic p z state: p x and p y states: 2,1,0 (r,, ) = 2pz R(r) cos( ) z 2px 2,1,1 + 2,1, 1 sin( )(e i + e i ) sin cos 2py 2,1,1 2,1, 1 sin( )(e i e i ) sin sin All three states look like dumbbells aligned along each of the three axis. Since the p states are degenerate, we can use these instead of the m=-1,0,1 states. 2s and 2p atomic orbitals

18 PP Bonds sigma (σ) bonds for lobes along nuclei axis pi (π) bonds for lobes perpendicular to axis

19 Clicker Question Which molecular bond is stronger? a) 2P Sigma bond b) 2P Pi bond

20 Sigma binding bonds are stronger than pi bonds (more overlap of wave functions)

21 How many 2p orbital bonding states? 3 spatial bonding (3 spatial antibonding) 2 electrons allowed in each of the three spatial P orbital states. Example: Nitrogen: 1s 2 2s 2 2p 3 Molecular nitrogen has six 2p electrons. 2 electrons in sigma 2p, 4 electrons in the pi 2p (no electrons in antibonding orbital). 2 electrons in 2s bonding and 2 electrons in 2s antibonding orbitals 2 electrons in 1s bonding and 2 ectrons in 1s antibonding orbitals Triple bond (bond order = Difference between the number of bonding electrons and number of antibonding electrons divided by two.) Very stable! (B = 9.8 ev)

22 Use the following atomic numbers for the 2p elements: B C N O F Ne Z = Which one of the following two diatomic molecules requires the larger energy to break the molecular bonds? A) CO B) O 2 22

23 Use the following atomic numbers for the 2p elements: B C N O F Ne Z = Which one of the following two diatomic molecules requires the larger energy to break the molecular bonds? A) N N B) N O 23

24 Sigma bonds involving s and p orbitals: Consider HF (Z=9 for F, so it has 5 2p electrons) H has one 1s electron and F has one 2p hole The one 1s electron can go into a sp molecular orbit. Note: although considered a covalent bond, the molecular orbital isn t symmetric [ for HF electron much more likely to be near Flourine, causing a polar covalent bond (partially ionic)]

25 SP Hybrid Orbitals An atom can spend a slight investment in energy to have one s-electron and several p electrons. The s and p electrons can form hybrid atomic orbitals. These hybrid atomic orbitals can allow for strong molecular bonds with other atoms (stronger than just using p-orbital bonds)

26 SP 3 Hybrid These form 4 degenerate sp 3 orbitals

27 Clicker Question Carbon (Z=6: ground state 1s 2 2s 2 2p 2 ) How many vacancies are there for forming molecular bonds with the SP 3 hybrid orbitals?

28 For methane, each of the four hydrogen atoms forms a sigma bond with the sp 3 states.

29 Diamond Each carbon atom bonds with four other carbon atoms

30 Nitrogen has an atomic number of Z=7. If it forms sp 3 hybrids, how many vacancies will it have in the sp 3 orbital states?

31 Ammonia (NH 3 ). Recall ground state configuration of nitrogen is 1s 2 2s 2 2p 3. There are eight sp 3 hybrid states (including spin). Since there are 5 valence electrons, there are three vacancies, which can form sigma bonds with hydrogen.

32 Water: H 2 0 Oxygen (1s 2 2s 2 2p 4 ) has 6 n=2 electrons which can form sp 3 hybrid orbitals. 2 of the atomic hybrid orbitals have one vacancy each to combine with hydrogen This is a polar covalent bond, where the oxygen side is negatively charged and the hydrogen side is positively charged. Angle between atoms is 104 degrees.

33 SP 2 Hybrid When an s and two p orbitals are mixed to form a set of three sp 2 orbitals, one p orbital remains unchanged and is perpendicular to the plane of the hybrid orbitals. For carbon (four n=2 electrons), three electrons go in the three hybrid orbitals and one electron is in the p z orbital. Four vacancies

34 Ethylene Each carbon has four vacancies for molecular binding orbitals. A double bond occurs between the two carbon atoms due to both a sigma hybrid bond and a pi bond from the p z orbitals.

35 SP Hybrid When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at 180 degrees results. Example: Carbon Dioxide: Carbon: 2 electrons in sp2 hybrid, 2 electrons in other p orbitals. Oxygen: 5 electrons in sp2 hybrid (one vacancy) and one electron in P state (one vacancy). Thus double bond with carbon. O = C = O

36 IONIC BONDS Extreme case of an asymmetric covalent bond is where electrons spends all of its time around one atom, essentially in an atomic orbital. Easier to model as atomic ions attracted to each other. Electron affinity: energy decrease of an atom in accepting an electron. Ionic bonds form if the electron affinity plus attractive ionic bond is greater than ionization energy of other atom(s). Tends to occur when both resulting ions have full shells. (example: NaCl next slide)

37 Sodium Chloride: Na: Z=11 (3s 1 valence electron) Cl: Z= 17 (3p 5 valence electrons)

38 Vibrational and Rotational Energy Molecules can vibrate and rotate. There are energy levels associated with these motions. Vibrational States of a diatomic molecule Equilibrium inter-atomic disstance = R 0. U(R) acts like a spring for small deviations from equilibrium (x=r-r 0 ). Therefore we can use quantum results for the harmonic oscillator : 2 2µ 2 (x)+ 1 2 kx2 (x) =E (x) µ = m 1m 2 m 1 + m 2

39 What are the energy states? E n = (n +1/2) = k (n +1/2) µ Spring constant determined from spectrum, or from numerical modeling of U(R). Example: Hydrogen: what is reduced mass?

40 µ = m p /2 k = d2 U = J/m2 dr p 2! = k/µ = s 1 E n =0.547 (n +1/2) ev Selection Rule for Dipole Emission: n = ±1 How many emission/absorption lines are there?

41 µ = m p /2 k = d2 U = J/m2 dr p 2! = k/µ = s 1 E n =0.547 (n +1/2) ev n = ±1 = hc/ E =2.3 µm

42

43 Infrared Spectra of Molecules and Characteris0c Vibra0onal Frequencies of Common Bonds A typical infrared spectrum plots intensity versus vibra0onal frequency (wavenumber, cm - 1 ) Each bond type has a characteris0c stretching frequency, which is approximately the same in different molecules. Figure 14.61: The infrared spectrum of CH 2 Cl 2.

44 Rotational Kinetic Energy Recall that the rotating kinetic energy of a rigid body is given by E rot = K rot = 1 2 I 2 = 1 2 L 2 Rotational States of a Diatomic Molecule Treat molecule as dumbbell, rotating about its center of mass. I I = m 1 r m 2 r 2 2 I = µr 2

45 Rotational Kinetic Energy Rotational States of a Diatomic Molecule The allowed values of the angular momentum squared come from the eigenvalues of the L 2 operator: L 2 l = 2 l(l + 1) l E rot = 1 2 l(l + 1) 2 µr 2

46 Example: Molecular Hydrogen µ = m p /2 R 0 =0.074 nm 2 /(2µR0)= ev E rot = l(l + 1) ev Notice that this energy is much lower than the vibrational energy states.

47 Rotational State Transitions E rot = 1 2 l(l + 1) 2 µr 2 What is the selection rule for transitions from one rotational state to another? l = ±1 Consider a transition in which Δl = -1. Is there more than one resulting value of ΔE? 1. yes there are three possible values 2. yes there are many possible values 3. no like vibrational transitions, all transitions give rise to the same change in energy.

48 The energy levels are not evenly spaced (like those of the harmonic oscillator). Typically we use the letter J or R to represent rotational angular momentum. E rot = BJ(J + 1) For transitions where ΔJ= -1 (P branch) E rot (J) =E rot (J 1) E rot (J) = 2BJ For transitions where ΔJ= +1 (R branch) E rot (J) =E rot (J + 1) E rot (J) =+2B(J + 1) E rot (J + 1) E rot (J) =2B

49 Ro-Vibrational Energy Levels

50 Vibration-Rotational Spectrum of HCl Simultaneous transitions of vibrational and rotational state. Lines are centered on ΔE vib = ħω

51 As l increases, the real energy of the rotational states differs slightly from the theoretical prediction: How would you expect the actual energy to differ from the predicted energy? A) The actual energy is a little larger than the predicted energy. B) The actual energy is a little smaller than the predicted energy. (think about our diatomic molecule model). E l = l l + 1 ( ) 2 2I h 51

52

53 Electronic Spectra Now combine electronic orbital states with rotational and vibrational states.

54 E = E e +(E 0 E 00 )+(E 0 rot E 00 rot) For transitions involving a change in the electron s orbital energy state, =0, ±1, ±2,... This can give rise to a band structure in the spectra.

55

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