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Unformatted text preview: l. a) What is the radial distribution function?
b) To what part of the Schrodinger equation is it related to? 1. c) Sketch the radial distribution function of the ls, ZS, 3s, 2p, and 3d orbitals and
explain the main features of the distributions ﬁring rarer Ragga} See Martel ‘lttlleltitt The radial distribution function is the probability of ﬁnding an electron in a volume (3R ﬁlm i. element at a given distance ‘r’ from the nucleus. It is related to the radial part of the 511357, lit wavefunction solution of the Schrodinger equation. 8 Q The Schrodinger equation is HUIEEK where the H is the Hamiltonian operator (which
contains a kinetic~energy term and a potential~energy term), ‘l” is the wavefunction, and E
is the energy. This equation is used to describe the behavior of electrons in atoms and
molecules. The wavefunction solution of this equation has three components: W=R<r ) (D (t) (9(9) R(r) is the radial part of the wavefunction and CD(¢)®(9) is the angular part of the
equation. The square of the radial part of the wavefunction (R(r )2) gives us the
probability of finding the electron at a given distance r from the nucleus in one given
direction. The radial distribution function (47hr2 (R(r))2) gives us the probability of ﬁnding
the electron in a volume element at a distance r from the nucleus. W The radial distributions functions for the ls, 2p and 3d orbitals don’t have any nodes. ln
the case of the ls orbital, the electron spends most of its time very close to the nucleus
(the peak of the distribution occurs at 0.52 Angstrom). For the 2p and 3d orbitals, the
electron can be, further from the nucleus. The 25 orbital has 1 node, and the 3s orbital has 4","... Ann 1 +, v ' ‘ ' ' '
LWU “Ouco. 1f the election is in these orbitals, it Will . . in the regions 'of space where the nodes are. The size of the orbital increases with i 7‘ reasing principal quantum number. Whig (it? i ’l S I Q s 'w . Hﬁﬁw‘l /\ H“\\\l\‘\lll _.._........,____.\"'r_t_.__ m k mm/K . W h
V tr , _ 2. a) Explain the chemical bond in a N2 molecule using valence—bond theory and
molecularorbital theory. 2. b) What is the criterion used in each of the theories to justify thtat the N2 bond is
stronger than, for example, the F2 bond? Valence—bond theory indicates that bonds are formed by the overlap of atomic
orbitals. The electrons of the molecule “live” in the atomic orbitals that are
overlapping. A bond will be stronger is many overlaps of atomic orbitals occur
between the two atoms involved in the bond.
Molecularorbital theory indicates that when two atoms interact, their atomic
orbitals combine to give molecular orbitals, which belong to the molecule formed. K
Some of the molecular orbitals will be bonding, and some antibonding. If there affix? Writ/r @élﬂlﬁjé
atomic orbitals that are not involved in bonding, they will be called ' ' .” The
strength of a bond is given by the bond order, which is related to the relative
occupation of bonding and antibonding orbitals. 2a) To draw a valencebond theory picture, we must ﬁrst think about the electronic
conﬁguration and hybridization of the atomic orbitals. N is a 1522522p3 atom. If we
assume that N is not hybridized when forming bonds with another N atom, we will
have that three overlaps are possible between the two atoms, because there are three
singlyoccupied orbitals in each atom available for bonding. One overlap will be
sigma, (say between the 2px singlyoccupied orbitals), and the other two will be pi
(between the 2py’s and 2pz’s). This is a triple bond. The molecularorbital theory diagram indicates that the bond order is 3. In other
words, the bond between the two N atoms is a trlple bond. £0 Vtttettt’éaeav o THQEQ M a F is a 1522322p5 atom, and only one of the 2p orbitals (say the 2pz) is singly occupied an
available for bonding. This singlyoccupied orbital will overlap with the counterpart
singly—occupied 2pz orbital of another F atom to give a sigma bond. According to
r , .\ Molecular—orbital theor , the bond in N is stron er than in F because the bond order in  "' .
V‘g WW” N2 is 3 but that in F2 islbnly 1 2 g 2 z .2 s F MD l lg?!
a I  (t. n a, 2 .. 2; ' m m .— Draw the Lewis structures and molecularmorbital diagrams of CO and 02 (Assume that A e ordering of the energies of the molecular orbitals in CO is identical to that of N2.
Are the Lewis structures reasonable for both molecules? Although the CO and N2 molecules are isoelectronic, their molecular=orbital diagram is
slightly different. What are the differences and why are they different? (10 minutes) , ‘ w
/' rm " W KY
nu ., ,k/ U, w... (A337 E ‘ , ‘ if
‘ I, 7% Rx '
L“ .4, Q, , s
W” Va} \ \ 447» (7‘: mutt .. P
it 9.25, 93'
La 9.3?
ﬂ (Nix AH: 0‘23
<5 Whereas for CO the Lewis structure seems reasonable, the 02 Lewis structure is not
reasonable. If we look at the molecular orbital diagram of 02 we can see that 02 will be a
paramagnetic species, with two uncoupled electronic spins. However, the Lewis structure
indicates that no electrons will be uncoupled. Indeed, the 02 Lewis structure that follows
the octet rule corresponds to an excited state of 02 in which the two electrons of the two antibonding 713* orbitals are paired in one of these two orbitals (which violates Hund’s
rule). The molecular orbital diagram of CO is slightly different from that of N2 because the
electronegativities of O and C are different. First, the energies of the atomic orbitals of C
and O are different, with the energies of O orbitals being lower than those of C. The
reason for this is that in oxygen, there are more protons in the nucleus, which attract more strongly the electrons in the 2p orbitals, lowering their energy. Second, 0 is more electronegative, and tends to attract more strongly the electrons in the
bonding orbitals. This means that whereas the bonding molecular orbitals of C0 are closer in energy to the 0 atomic orbitals, the antibonding molecular orbitals are closer in
energy to the C atomic orbitals. 4. a) We can promote the transition of a pi electron from the 3rd to the 4th state in an
unknown polyene molecule by shinning electromagnetic radiation of 354
wavelength. Calculate the effective length of the polyene. 4. b) What polyene is this? If you don’t know the name, draw it. (Assume that the pi electrons in the polyene behave as a particle in a one—dimensional box) y/v/V’ Hemmwé 5. The % ionic character of HP is 43.2%, and its dipole moment is 1.91 D. What is
its internuclear distance? (Charge of 1 e'=1.6xiO'19C, 1Debye= 3.3X10'30 Cm). SEE, Mme m n
i’ \(CH F) t. 0.1% ’i ...
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This note was uploaded on 11/03/2009 for the course CHEM 4616 taught by Professor Troya,d during the Spring '08 term at Virginia Tech.
 Spring '08
 Troya,D
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