HW9_key - M 12.17 Use MO theory to describe the bonding in...

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Unformatted text preview: M 12.17 Use MO theory to describe the bonding in N 0+, NO, and NO’. Compare their bond energies and bond lengths. A Since N and O are adjacent in the periodic table and have similar electronegativities, the molecular orbital energy diagram for NO and its ions is essentially the same as that for N2. The three species have the following electron configurations. N0+: KK(023)2 (05;)2 (7:32 (my)2 (r12?)2 with a bond order of 3 (like N2). N0: KK(02_9):z (oils)2 (1132 (1502 (any (1:31 with a bond order of 2.5. it ‘ NO‘:'KK(023)2 (0;)2 (er)2 (75,)2 (02p)2 (31;)1 with a bond order of 2. The bond energies of the three species are expected to increase in the order N 0'" < NO < NO“. The bond lengths are expected to increase in the order NO+ < N0 < N0“. M 12.24 Predict the number of unpaired electrons in the following complex ions: (3) Cr(CN)2— and (b) Cr(H20)§+. (a) The electron configuration of Cr2+ is [Ar13d4. CN‘ is a strong-field ligand, and the four electrons will occupy the three lower orbitals as shown. There will be two unpaired electrons. [Cr(CN)fi]4- E D as _ dz; new cc, as (b) Since H20 is a weak-field ligand. the four 30? electrons of C‘r2+ will be arranged to maximize the number of unpaired electrons as shown below. There will be four unpaired electrons. [CINHJDM2+ E] at: V dz: l ‘l‘ ' ll d... dc 12.26 The absorption maximum for the complex ion C0(NH3)2‘L occurs at 470 nm. (a) Predict the color of the complex, and (b) calculate the crystal field splitting in kJ mol’ ‘. (a) The complex ion is absorbing 470 nm light, which is blue to blue-green. Referring to Table 12.5, this implies that the complex is between yellow and red, or orange. (b) The energy of a 470 nm photon is found using Planck’s relationship and converted to a molar basis. _ he _ A g [6.626 x 1073’4 J 5) (3.00 X 108 ms-l) _ 470 x 10-9 in AB = 4.229 x 10—191 This is the crystal field splitting per molecule. Multiplying by Avogadro’s constant gives the crystal field splitting for a mole of molecules. IR] (4229 x 10—19 J) (6022 x 1023 moi—1) = 255 tamer] r0001 _____—_______-————————-—— 14.15 The J = 3 a 4 transition for a diatomic molecule occurs at 0.50 cm‘ I. What is the wavenum- her for the J = 6 —> 7 transition for this molecule? Assume the molecule is a rigid rotor. Assuming a rigid rotor, the frequency of a rotational transition is given by v : AEM/ h = 28 J’. The wavenumber of the J =2 3 —> 4 transition with J ’ = 4 supplies the value of B. : AEmt/h 2J’ m hCfi/h "" 21' _ Ci) 5? B (3.00 x 103 m s-') (0.50 cm”) (10??) 2(4) = 1.88 x 109 s“1 Then for the J = 6 -> 7 transition with J’ = 7, 1) {5:— C _ AErot/ h _ C _ 231’ — C _ 2(1.88x 1093-1) (7) ( 1m ) 3.00 x 108 m s"1 100 cm = 0.88 cm—1 W 14.19 Draw a vibrational mode of the BF3 molecule that is IR inactive. The symmetric stretching vibration of BF3, where each B—F bond is stretched by the same amount, has no effect on the dipole moment of the trigonal planar molecule, and is IR inactive. l i F/ \F / \ 14.25" Calculate the number of vibrational degrees of freedom of the hemoglobin molecule, which contains 9272 atoms. Hemoglobin is not a linear molecule, and the number of vibrational degrees of freedom is 3N — 6 = 3(9272) — 6 2 27810. 14.34 What is the field strength (in tesla) needed to generate a 1H frequency of 600 MHz? From Equation 14.47 and Table 14.5, 2 271 600 x 1065” Bozflz(—_—):14_1T 7/ 26.75 x 107 T-1 3‘1 —_____.——_———m—II_-—fl—-II—— 14.37 For each of the following molecules, state how many proton NMR peaks occur and whether each peak is a singlet, doublet, triplet, etc. (a) CH30CH3, (b) C2H50C2H5, (c) C2H6, (d) CH3F, (e) CH3COOC2H5. (a) One singlet, (b) two peaks, one triplet (12:1) and one quartet (113:3:1), (c) one singlet, ((1) one doublet (1:1) due to the coupling of the protons to the fluorine nucleus which has I = 1 / 2, (c) three peaks, one singlet, one triplet (1 :21), one quartet (1:3:3: l). 14.40 The methyl radical has a planar geometry. How many lines would you observe in the ESR spectrum of -CH3? Of -CD3? Normal methyl radical, -CH3, has three equivalent H atoms, each with I = 1 / 2. Thus, there are 2n] + l : 2(3)(l/2) + l = 4 lines in its ESR spectrum with intensity ratio 1:32321. The deuterated methyl radical, -CD3, has three equivalent D atoms, each with I = 1, giving 2(3)(]) + l = 7 lines in its ESR spectrum. The intensity pattern does not follow the binomial distribution, which applies only to nuclei with I = l / 2. 15.5 In the photochemical decomposition of a certain compound, light intensity of 5.4 x 10"6 einstein s‘1 was employed. Assuming the most favorable conditions, estimate the time needed to decompose 1 mole of the compound. The rate of a photochemical reaction is Rater-Wit»? = 1 (all incident light is absorbed), f = 1 (all absorbed The most favorable conditions are: F ‘ oduct formation is 1). Under light produces the reactive state), and (D p = 1 (quantum yield of pr these conditions, the rate is Rate = 5.4 x 10'6 mol S_1 Thus, to decompose 1 mole of the compound, the time needed is “"01 = 1.85 x 105 s=2.1 days _,____....._——-—4—-—— 5.4 x 10‘6 mols"1 ____________________________————— _______________——-——————— 15.8 Why does one have to irradiate a sample for hours or even days to achieve acceptable yields in some photochemical reaction even though the lifetimes of excited electronic states are on the order of micro— or nanoseconds? Assume that the rate of light absorption is 2.0 x 1019 photons s4. Even if every photon led directly to product, which would correspond to F = f 2 (DP 2 l in Equation 15.4, the rate of product formation would still be limited by the rate of light absorption. At 2.0 X 1019 photons 5’1 = 3.3 X 10—5 einstein 5’1, the rate of product formation is only 3.3 X 10—5 mol 3". Atthis rate, itwould take 1 mol/3.3 X 10*5 mol s‘1 : 3.0 X 104 s = 8.4 hr to produce a mole of product. ——__—————-_——————,——_——_ 15.19 A light source of power 2 x 10‘ 16 W is sufficient to be detected by the human eye. Assuming the wavelength of light is at 550 nm, calculate the number of photons that must be absorbed by rhodopsin per second. (Hint: Vision persists for only 1/30 of a second.) The energy of 1 photon at 550 am is 6.626 x 10-34 Is 3.00 x 108 m s-1 E 2,11) = E = (————————)—(——————) =3_614 x10‘19J A 550 x 10-9 m The light source produces 2 x 10‘16 J s“. Therefore, the number of photons that must be absorbed by rhodopsin each second is 2 x 10*“ ______ = 5.53 x 102 z 5.5 x 102 3.614 x 10—191 Since vision persists for only 1/30 of a second, the number of photons detected is (5.53 x 102) :18 W ...
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This note was uploaded on 05/29/2011 for the course CHM 3400 taught by Professor Seabra during the Spring '08 term at University of Florida.

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HW9_key - M 12.17 Use MO theory to describe the bonding in...

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