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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 strongﬁeld 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 weakﬁeld 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 bluegreen. 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 msl)
_ 470 x 109 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 10931) (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 T1 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 RaterWit»? = 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 1034 Is 3.00 x 108 m s1
E 2,11) = E = (————————)—(——————) =3_614 x10‘19J
A 550 x 109 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.
 Spring '08
 SEABRA
 Physical chemistry, pH

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