10 Z=12 0.4 0.6 0.8 1.0 r(bohr) Si 2p Z=14 0.0 0.5 1.5 r(bohr) 16
16 0.0 0.2 0.4 0.6 0.8 1.0 -10 0 10 20 30 40 Si 2s r (bohr) Radial Function R(r) Z=14 Z=12 0.0 0.5 1.0 1.5 0 2 4 6 8 Si 2p r (bohr) Radial Function R(r) Z=12 Z=14
Si 3s 15 12 6 Z=4 Z=14 r(bohr) Si 3p Z=14 Z=4 1 2 3 r(bohr)
17 0 1 2 3 -30369 12 15 Si 3s r (bohr) Radial Function R(r) Z=14 Z=4 0 1 2 3 4 5 -2 -1012345 Si 3p r (bohr) Radial Function R(r) Z=14 Z=4 8
i. In ammonia, the only "core"orbital is the n ls and this becomes an aj orbital in C3 symmetry. The n 2s orbitals and 3 H ls orbitals become 2 ar and an e set of orbitals The remaining n 2p orbitals also become 1 al and a set of e orbitals. The total valence orbitals in C3y symmetry are 3a1 and 2e orbitals ii. In water, the only core orbital is theo 1s and this becomes an a] orbital in C2v symmetry. Placing the molecule in the yz plane allows us to further analyze the remaining valence orbitals as: 0 2pz=a1,0 2py as b2, and O 2px as b. The(H Is+ H 1s)combination is an al whereas the(h Is-H ls)combination is a b iii. Placing the oxygens of H202 in the yz plane(z bisecting the oxygens)and the(cis) hydrogens distorted slightly in +x and -x directions allows us to analyze the orbitals as follows. The core is+o is combination is an a orbital whereas theois-o Is combination is a b orbital. The valence orbitals are: 0 2S+o 2s=a, 0 2s-o2s=b, o 2px+O2px=b,02px-02px=a,02py +O 2py=a,O 2p -O2p=b,02pz +O2pz b,O2pz-02pz=a, H ls+H Is=a, and finally the H Is-H ls-b iv For the next two problems we will use the convention of choosing the z axis principal axis for the Dooh, D2h, and C2v point groups and the xy plane as the horizontal reflection plane in Cs syr Dooh D2h
18 i. In ammonia, the only "core" orbital is the N 1s and this becomes an a1 orbital in C3v symmetry. The N 2s orbitals and 3 H 1s orbitals become 2 a1 and an e set of orbitals. The remaining N 2p orbitals also become 1 a1 and a set of e orbitals. The total valence orbitals in C3v symmetry are 3a1 and 2e orbitals. ii. In water, the only core orbital is the O 1s and this becomes an a1 orbital in C2v symmetry. Placing the molecule in the yz plane allows us to further analyze the remaining valence orbitals as: O 2pz = a1, O 2py as b2, and O 2px as b1. The (H 1s + H 1s) combination is an a1 whereas the (H 1s - H 1s) combination is a b2. iii. Placing the oxygens of H2O2 in the yz plane (z bisecting the oxygens) and the (cis) hydrogens distorted slightly in +x and -x directions allows us to analyze the orbitals as follows. The core O 1s + O 1s combination is an a orbital whereas the O 1s - O 1s combination is a b orbital. The valence orbitals are: O 2s + O 2s = a, O 2s - O 2s = b, O 2px + O 2px = b, O 2px - O 2px = a, O 2py + O 2py = a, O 2py - O 2py = b, O 2pz + O 2pz = b, O 2pz - O 2pz = a, H 1s + H 1s = a, and finally the H 1s - H 1s = b. iv. For the next two problems we will use the convention of choosing the z axis as principal axis for the D¥h, D2h, and C2v point groups and the xy plane as the horizontal reflection plane in Cs symmetry. D¥h D2h C2v Cs N 1s sg ag a1 a' N 2s sg ag a1 a' N 2px pxu b3u b1 a
N2p 2u N2pz lu nTX Pn(x) Pn(x)dx=In(x)dx The probability that the particle lies in the interval 0<x s4 is given by nTX Pn=JPn(x)d This integral can be integrated to give: nπxlnπx P P P A Sin20+ 2nπ nr-4Sin 4+(2)(4
19 N 2py pyu b2u b2 a' N 2pz su b1u a1 a'' 9. a. Yn(x) = è ç æ ø ÷ 2ö L 1 2 Sin npx L Pn(x)dx = |Y | n 2 (x) dx The probability that the particle lies in the interval 0 £ x £ L 4 is given by: Pn = õó 0 L 4 Pn(x)dx = è ç æ ø ÷ 2ö L õ ô ó 0 L 4 Sin2 è ç æ ø ÷ npxö L dx This integral can be integrated to give : Pn = è ç æ ø ÷ L ö np è ç æ ø ÷ 2ö L õ ô ó 0 np 4 Sin2 è ç æ ø ÷ npxö L dè ç æ ø ÷ npxö L Pn = è ç æ ø ÷ L ö np è ç æ ø ÷ 2ö L õó 0 np 4 Sin2qdq Pn = 2 npè ç ç æ ø ÷ ÷ ö - 1 4 Sin2q + q 2 ï ï ï ïnp 4 0 = 2 npè ç æ ø ÷ ö - 1 4 Sin 2np 4 + np (2)(4)
ah Sin( b If n is even, Sin( 2=0 and Pn=4 f'n is odd and n=15913,…sn、2)=1 nd P If n is odd and n =3.7.11.15 and Pn=4+2n The higher Pn is when n=3. Then Pn=4+2T3 Pn= 0.303 IEnt c P(t)=e t ay +bYm =aYne i +bye h IEmt Hp=apnEne h +bmE h I(En-Emt <YHY>=lal?En+|bEm+a"be <pnlHIYm'> i( Em-Ent +b h PmIN, Since <InIHYm> and <pmlH'Yn> are zero, <YHY>=la2En +lb/2Em(note the time independence)
20 = 1 4 - 1 2pn Sinè ç æ ø ÷ npö 2 b. If n is even, Sinè ç æ ø ÷ npö 2 = 0 and Pn = 1 4 . If n is odd and n = 1,5,9,13, ... Sinè ç æ ø ÷ npö 2 = 1 and Pn = 1 4 - 1 2pn If n is odd and n = 3,7,11,15, ... Sinè ç æ ø ÷ npö 2 = -1 and Pn = 1 4 + 1 2pn The higher Pn is when n = 3. Then Pn = 1 4 + 1 2p3 Pn = 1 4 + 1 6p = 0.303 c. Y(t) = e -iHt -h [ ] aYn + bYm = aYne -iEnt -h + bYme -iEmt -h HY = aYnEne -iEnt -h + bYmEme iEmt -h <Y|H|Y> = |a|2En + |b|2Em + a*be i(En-Em)t -h <Yn|H|Ym> + b*ae -i(Em-En)t -h <Ym|H|Yn> Since <Yn|H|Ym> and <Ym|H|Yn> are zero, <Y|H|Y> = |a|2En + |b|2Em (note the time independence)