120A problem set 1 key

120A problem set 1 key - 1. 2 pts each a. 2...

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Unformatted text preview: 1. 2 pts each a. 2 spherical nodes b. 2 angular nodes c. The angular nodal surface consist of two planes that are coincidental with the z ­axis. One plane diagonally bisects the (x,y) and ( ­x, ­y) quadrants, the other diagonally bisects the (x, ­y) and ( ­x,y) quadrants. d. http://winter.group.shef.ac.uk/orbitron/ 2. 3 pts a. [Kr]5s24d105p3 b. [Kr]5s04d5 c. [Ar]4s03d10 3. 6 pts = 10 ∏e + 1 ∏c = 6 ∏e + 3 ∏c Most favorable configuration is on top 4. 4 pts Ca through V, 4s; Cr, 3d; Mn, 4s. The general increase in the second ionization energies from left to right is due to the increase in Zeff. The decrease between Cr and Mn is a result of Mn1+ having 6 d electrons, requiring the pairing of electrons. These paired electrons repel each other offsetting the greater nuclear charge. 5. 6 pts Bond orders are listed above molecules. As bond order increases the bond length decreases and the reverse is also true. 6. 6 pts The most stable resonance form is bolded. The resonance form furthest to the right is unreasonable because it places three positive charges adjacent to one another. The bolded structure is favored over the middle structure because the negative charge is flanked and consequently stabilized by two positive charges. 7. 8 pts Rnl(r) Pnl(r) 1s 2s 3s The locations of the nodes on the radial wave function of the 3s orbital correspond to the locations at which the radial probability functions of the 3s orbital approach zero. Comparing the radial probability functions of the 1s, 2s, and 3s orbitals, the radial probability function of the 3s orbital approaches zero when the 1s and 2s radial probability functions are near maximal probability, thus it is very unlikely to find a 3s electron in the same vicinity as a 1s or 2s electron. The radial probability function of the 3s orbital shows that the greatest probability for finding an electron occurs after the second node of the radial wave function, therefore the electrons involved bonding will most likely occur at a distance close to this maximum. 8. 3 pts In addition to the increase in the effective nuclear charge, the overall charge increases, making the 2nd and 3rd steps more difficult. The radius degreases slightly with loss of each electron also making the 2nd and 3rd steps more difficult. The large discrepancy between the 2nd and 3rd ionization energies is due to the large energy difference between 3s and 2p orbitals. 9. 4 pts CF3 has a greater attraction for electrons than CH3, so the P in PF2(CF3)3 is more positive that the P in PF2(CH3)3. This draws the F atoms in slightly, so the P ­F bonds are shorter in PF2(CF3)3. 10. 6 pts a. E, C∞, ∞σv b. E, C2(x), C2(y), C2(z), i, σ(xy), σ(xz), σ(yz) c. E, C∞, ∞σv, i, S∞, ∞C2 11. 2 pts each except W(NMe2)6 which is worth 4 Determine the point groups of the following molecules. C4v C3h C2v C∞v C2h Oh D5d D4h D5h D 2d d Cs Ih 11. Continued TOP VIEW SIDE VIEW S8 Th C3v D2h D3 Td 12. 2 pts each 13. 8 pts The irreducible representation is highlighted in red. There are no other orbitals degenerate with the dx2 ­ 2 y in the D4h point group. 14. E = , i = , C2(C2(z)) = , C2’(C2(x)) = C2’(C2(y)) = σV (σX) = i = , C2’’ = , C2’’ = σh (σZ) = , , σV (σY) = , σd = , σd = , , C4(C4(z)) = , S4(S4(z)) = ...
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