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Unformatted text preview: Problem 5% 1a ® In this problem, we consider the heteronuclear diatomic molecule CO. The ionization energies
' of an electron from the valence atomic orbitals on the carbon atom and the oxygen atom are listed
below.
Atom Valence orbital Ionization energy /MJ ~mol‘l 0 2s 3.116 2p 1.524 C 25 1.872 2p 1 .023 Use these data to construct a molecularorbital energylevel diagram for CO. What are the symmetry
designations of the molecular orbitals of CO? What is the electron conﬁguration of the ground state
of CO? What is the bond order of CO? Is CO paramagnetic or diamagnetic? O CO C
orbitals orbitals orbitals
I \ \ \ 2p
2::
2p ,, '
—ooo—L'=:._ >1 ‘ ‘$’ 25
S I c If
Lu
2: (I I!
——o—l.
\~_O___,’ Since C and 0 have similar orbital energies, the bonding will be similar to that in a homonuclear
diatomic molecule. In order of increasing energy, the symmetry designations of the molecular ,
orbitals of C0 are (ignoring the is orbitals) 025, 0‘25, 7Tpr and n2p5,, and ﬁnally asz. Note that
we do not use the subscripts g and u because CO does not have an inversion center. The electron
conﬁguration of CO is (from Example 9—6) KK(0‘25)2(o‘*25)2(7T2px)2(7r2py)2(0‘2pz)2, so the
bond order is 3 and CO is diamagnetic because there are no unpaired electrons. The following lines were observed in the microwave absorption spectrum of Hml and D1271 between 60 cm’I and 90 cm". 17/cm‘l I 
HmI 64.275 77.130 89.985
D1271 65.070 71.577 78.084 84.591 Use the rigid—rotator approximation to determine the values of B, I , and Rg for each molecule.
Do your results for the bond length agree with what you would expect based upon the Born
Oppenheimer approximation? Take the mass of 127I to be 126.904 amu and the mass of D to be 2.014 amu. In the rigidrotator approximation, the spacing between the lines in a microwave absorption
spectrum is 23 (Equation 13.19). The spacing between the lines given for H1271 is 12.855 cm‘I
and the spacing between the lines given for DmI is 6.507 cm". Therefore, 9 B“...l = §(12.855 cm") = 6.428 cm" and BDml = CID—l) = CITI—l We now use Equation 13.9 to ﬁnd I for both molecules: h  8712c]? I 6.626 x 1034 Js
1HI = 2 10 —1 —1
871 (2.998x 10 cms )(6.428cm ) I _ 6.626 x 10‘34 Js
1" ‘ 8212(2998 >< 10'0cms‘)(3.254 em") = 4.355 x 10~47 kgm2 = 8.604 x 10“7 kgmz Now we use the fact that I = ,uRZ for a diatomic molecule to ﬁnd Re: I=MRZ I l/Z
Rio
11 4.355 x 1047 kgmz "2
Realll : T“ 1.008>< 126.904 [27.912 amu) (1.661 x 10‘27 kgamu"‘)
= 1.619 ><10"O m =161.9pm 8.604 x 1047 kgmz "2
Reth = (—‘ 2.014x126.904 128.917 am”)(1661 X 10—27 kgamu“)
= 1.617 x 10'10 m =161.7 pm These values differ by approximately 0.1%. In the BornOppenheimer approximation, the bond
length is independent of the isotope of theatoms, in agreement with the above calculations. Calculate the H'Lickel n—electron energies of cyclobutadiene. What do Hund’s rules say about the ground state of cyclobutadiene? Compare the stability of cyclobutadiene with that of two
isolated ethylene molecules. The structure of cyclobutadiene is Letting x = (a — E)/ﬁ, the H'Lickcl determinantal equation is given by xlOl‘
lxlO
01x1 #0
101x Expanding the determinant gives 1
x=0 10 x
l l
x 01 x
xlx
O 0
l
I x l l l
— 0 x — 0
l l l
and expanding the above determinants gives x(x3——2x)——(x2—l—l—l)—(l—l—xl—l)=0
x4—4x2=0 x = 2, 0, 0, —2
Because x = (01 — E)/i3, the four nelectron energies of cyclobutadiene are
E=ot—2ﬁ E=Ol E=ot E=ot+2ﬂ
There are four 7r electrons, and so the two lowest energy levels will be occupied and
En =2(oz+2ﬁ)+2oz=4oz+4ﬁ The second energy level (E z or) is doubly degenerate. We need to place two electrons in
these orbitals, and according to Hund’s rules, each orbital will contain one electron and these
electrons will have the same spin. Therefore, the ground state of cyclobutadiene should be
a triplet state. We showed in the text that the energy of the n' orbital in ethene is 2a + 213,
so E = E"(cyelobutadiene) — 2E”(ethene) = 0 deloc Cyclobutadiene has the same stability as two isolated ethene molecules. Explain why the doubly degenerate In“ orbitals for a linear XY2 molecule do not remain
degenerate when the molecule is bent. Take the z—axis to lie along the bonds of the linear molecule and let the molecule bend in the
xzp'lane. Then one of the 171“ orbitals lies in the plane of the molecule and the other lies in a plane
that is perpendicular to the plane of the molecule. For a linear molecule, the overlap between the
2px atomic orbitals (one of the In“ orbitals) and the 2pv atomic orbitals (the other If“ orbital)
are identical. Hence these molecular orbitals are degenerate. As the molecule bends, the overlap
between the 2 pl orbitals on the three atoms is affected differently than the overlap between the 2 pv
atomic orbitals. Therefore, the energy associated with the two 171“ molecular orbitals depends on
the bond angle. Explain why the 30“ molecular orbital of a linear XY2 molecule increases in energy as the
molecule bends. (Hint: The 30” molecular orbital is a linear combination of the 2pz orbitals from
each atom.) The 30“ orbital is a bonding orbital formed from a linear combination of the 2pZ orbitals from each
atom. (Recall that we have taken the z—axis to lie along the molecular bond.) When the molecule
bends, the overlap between the 2pZ orbitals decreases and the energy associated with this linear combination of atomic orbitals increases. — a ‘g /¥ 4 , Z
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have (using Equations 8.49, 8.50, and 8.52) m, 111‘ ML Ms M] 1 +§ 1 +§ +§ 1 —r 1 —; +r 0. +% 0 +§ +g o —% o a —% —1 +§ —1 +§ —§ r —1 ~r —1 —r —% The M and M s values given here correspond to a 2P state, and the values of M 1 correspond to a value
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2P3/2 and 2PW. The ground state is determined by using Hund’s rules; by Rule 3, the most stable
state (and therefore the ground state) is 2Pl a. An rip5 conﬁguration can be thought of as an np’ conﬁguration because two of the rip orbitals
are ﬁlled and so Ms and ML are determined by the remaining halfﬁlled p—orbital. Therefore,
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conﬁguration. From Equation 8.53 we also see that the number of sets of mi, and m” remains the
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symbol of N; is 22:, because the molecular wave function does not change when reﬂected through a plane containing the two nuclei. N2 (lag)2(lau)2(20g)2(20“)2(1IT")Z(17r”)2(30g)2 corresponds to ML = 0 and M5 = 0, or
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This note was uploaded on 01/25/2011 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at Berkeley.
 Spring '07
 Whaley

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