Download Chemistry 181H Fall 2006 Bonus Exam and more Exams Chemistry in PDF only on Docsity! Chemistry 181H Fall, 2006 Bonus I Thursday, 26 Oct 2oo6 WRITE your Name Student Number Recitation Time: Monday There are Four questions on this exam. Write your answer, showing all work to each on each page. A list of potentially useful equations is on page 6. A periodic table is contained on page 8. # Score 1 /10 2 /10 3 /10 4 /9 total /39 1 1. In a recent experiment, a gold atom was trapped on a two dimen- sional square surface with an edge length of 50 nm. The surface was flat so that V(x,y)=0. The experiment consisted of observing the emit- ted radiation as the atom slowed down and went into its ground, or lowest energy, state. (a) Write the quantum numbers of the ground state, the first excited state and the second excited state of this system, be sure to indi- cate if each state is degenerate or not. (6 points) (b) What is the energy (in joules) of the photon that will be released when the gold drops from the first excited state to the ground state of this system? (4 points) 2 4. Write the Lewis electron dot diagram for each of the following mole- cules. The central atom in each molecule is listed first. Be sure to indicate the positions of all the atoms and lone pairs of electrons. Also indicate any resonance structures. (3 points each) (a) nitrogen dioxide molecule, NO2 (b) Tellurium tetrachloride, TeCl4 (c) Iodine trichloride, ICl3 5 Potentially Useful Constants & Equations .. 10/23/06 R = 0.08206 ` atm ◦K−1 mole−1 R = 8.314 J ◦K−1 mole−1 kB = R/NA = 1.38066x10 −23J ◦K−1 1. amu = 1.6605 x 10−27 kg h = 6.626 x 10−34 J sec c = 2.99792 x 108 m sec−1 NA = 6.0221 x 10 23 mole−1 ²0 = 8.8542 x 10−12 Coulomb2 J−1 m−1 proton mass = 1.67263 x 10−27 kg electron mass = 9.1094 x 10−31 kg neutron mass = 1.67493 x 10−27 kg electron charge = 1.60218 x 10−19 Coul c1 = 56.705 nWm −2K−4 c2 = 1.44 cm K a0 = ²0h 2/e2πme = 52.92 pm RH = 109, 737 cm −1 Fcoulomb = − q1q2e2/4π²0r2 F = − ddxV(x) E = E0 sin[(2π/λ)(x− ct)] E = hν λ ν = c λdeB = h/p P/A = c1T 4 Tλmax = c2/5 1/λ = RH ( 1/n2low − 1/n2high ) Vcoul = q1 q2 e 2 / 4 π²0 r EK = 1 2 m v2 = p2/2m p = m v E(n) = -2.18 x 10−18 J 1 n2 EB(n,Z) = −Z2e4m 8²20h 2 1 n2 EB(n,Z) = -13.60eV Z2eff n2 ∆ E ∆ t ≥ h/4π ∆ x ∆ p ≥ h/4π rB(n)=²0n2h2/q1q2e2πm ET = EK + EP EK = hν − Φ Zeff = Z − σ −h2 8π2m d2 dx2 Ψ(x) + V(x) Ψ(x) = EΨ(x) [ −h̄2 2m ∇2 + V ] Ψ = ET Ψ P(x) = Ψ∗(x)Ψ(x)dx ∫ Ψ∗(x)Ψ(x)dx = 1 Ψfree(x) = A sin(2πx/λdeB) ET = h 2/2mλ2deB Ψn(x) = √ 2/L sin(nπx/L) E(nx) = n 2 xh 2/8mL2 E(nx,ny,nz) = (n 2 x+n 2 y+n 2 z) h2 8mL2 V(r, θ, φ) = h̄2`(` + 1)/2µr2 − Ze2/4π²0r [ −h̄2 2m ∇2 + V ] R(r)Yl,m(θ, φ) = ET R(r)Yl,m(θ, φ) 6 ∆El pair = NA q1q2e 2/(4π²0 d) ∆El = A NA (q1q2e 2/(4π²0 d)) ( 1− d∗ d ) χA = (FIP + EA)/2 %-IC = 16|χA − χB|+ 3.5|χA − χB|2 |χA − χB| = 0.102 √ ∆ ∆ = BDE(A : B)− √ BDE(A2) ∗ BDE(B2) FC = (#V alence e−)− (#LonePair e−)− 1 2 (#Bonding e−) 7
