課程名稱︰化學鍵
課程性質︰選修
課程教師︰彭旭明/王瑜
開課系所︰化學系
考試時間︰2005/04/21
試題：
Give the Character Tables of C3v and D2d
C3v │ E 2C3 3σx │
──┼─────────┼────────────
A1 │ 1 1 1 │ x2+y2 , z2
A2 │ 1 1 -1 │
E │ 2 -1 0 │ (x2-y2,xy) , (xz,yz)
D2d │ E 2S4 C2 2C2' 2σd │(x,y axis coincident with C2')
──┼────────────────┼────────┬───────
A1 │ 1 1 1 1 1 │ │ x2+y2 , z2
A2 │ 1 1 1 -1 -1 │ Rz │
B1 │ 1 -1 1 1 -1 │ │ x2-y2
B2 │ 1 -1 1 -1 1 │ z │ xy
E │ 2 0 -2 0 0 │ (x,y) , (Rx,Ry)│ (xz,yz)
(1) (i) What is the order of C3v point group ?
(ii) What is (are) he possible subgroup(s) of C3v point group ?
(iii) How many classes are in C3v point group ?
What are the orders of those classes ?
(iv) How many irreducible representations are in C3v point group ?
What are the dimensions of those irreducible representations ?
(v)Show that the characters of two different irreducible representations
are orthogonal in C3v point group .
(vi) Reduce the reducible representation with characters : X(E)=7 ,X(C3)=1
, and X(σv)=-3 into irreducible representations in C3v point group .
(vii) Give a real molecule which belongs to C3v point group .(10%)
(2) Answer the same set of problems for D2d point group except (vi) .(10%)
(3) (i) Give a Lewis structure for cyanamine H2NCN which does not place
formal charges on the atoms . What is the spatial geometry for this
structure ?
(ii) Generate another Lewis structure by delocalizing the lone pair on
the amino group . What is the geometry associated with this structure
?
(iii) Experimentally it is found that the amino group is slightly
pyramidal and the inversion barrier very small . Explain this result
.(10%)
(4) Consider all the atoms with Z less than or equal to 20 . In their
electronic ground state .
(i) Which of them are diamagnetic ?
(ii) Which of them have a single unpaired electron ?
(iii) Which of them have two unpaired electrons ?
(iv) Which of the following diatomic molecules are paramagnetic ? Li2, B2
, C2, N2, O2, and F2. (10%)
(5) The calculated MOs for the HHe+ ion are given by the expressions
1σ = 0.977(1s He) + 0.202(1s H)
2σ = 0.798(1s He) - 1.168(1s H)
(i) Justify the relative sizes of the coefficients in 1σ and 2σ.
(ii) Given that the MOs are normalized , calculate the overlap intergral
between the AOs , 1s H and 1s He .
(iii) Verify that 1σ and 2σ are orthogonal .
(iv) The coefficient of 1s H is larger than unity in 2σ . Is this
acceptable ? (10%)
(6) Calculate the atomic orbital coefficients for the MOs of rectangular H4 ,
(i) neglecting overlap between the 1s H orbitals in the normalization
process and (ii) taking overlap into account . Use the following
numerical result :
Ha ●──────● Hd
│ │
│ │
Hb ●──────● Hc
S = ＜ 1sa│1sb ＞ = ＜1sc│1sd ＞ = 0.496
S'= ＜ 1sa│1sd ＞ = ＜1sb│1sc ＞ = 0.231
S"= ＜ 1sa│1sc ＞ = ＜1sb│1sd ＞ = 0.143 (10%)
(7) Trigonal bipyramidal H5
Construct the energy level diagram of trigonal bipyramidal H5 . The atoms
Hd and Hc lie along the z-axis and Ha, Hb and Hc lie at the vertices of a
trigonal plane . Make all of the distances of these hydrogen atoms to the
origin equal and construct the MOs of this system from the two fragments ,
triangular H3 (Ha Hb Hc) and linear H2 (Hd...He) units.
Z
╱l╲
l
·Hd
/l\ ╲
/ l \ ╲·Hc
/ l \ /
/._____l_____.\/
Hb \ ／l / Ha
\／ l /
／ \ l /
／ \l/
└ ·He
Y l
(i) Describe the relative energies of the fragment orbitals , taking into
account the distances between the hydrogen atoms .
(ii) Use the xy plane of symmetry to find one of the MOs of H5 .
(iii) Use the xz plane to determine a second MO .
(iv) Analyze the overlap integrals between the orbital pairs symmetric
with respect to reflection in both of these planes . Hence determine
a third MO of H5 .
(v) Construct he complete orbital interaction faigram , given the fact
that the highest energy orbital is non-degenerate . Give the form of
each MO . (20%)
(8) Analogy between the orbitals of square planar H4 and those of a central A
atom . Decompose a square planar AH4 molecular into the two gragments, A
and square planar H4 .
y↑
Hd │ Ha
╲ │ ╱
╲ │ ╱
╲│╱
Ａ─────→ x
╱ ╲
╱ ╲
Hc╱ ╲Hb
(i) Establish the orbital analogy between the MOs of square planar H4 (φ1
-φ4 ) and the AOs (s,px,py,pz) of the central A atom .
(ii) Do the same but for the case where the four hydrogen atoms lie along
the axes x and y . (10%)
(9) The CH radical
(i) Why is this species called a radical ?
(ii) Give its Lewis structure .
(iii) The AO coefficients in the MOs of CH are given in the Table below ,
the MO's being given in order of increasing energy .
z
C───H────→
─────────────────────
MO 1s H 2s 2px 2py 2pz
─────────────────────
1σ 0.29 0.72 0 0 0.28
2σ -0.29 0.62 0 0 -0.71
π 0 0 1 0 0
π 0 0 0 1 0
3σ 1.56 -0.96 0 0 -1.08
─────────────────────
(a) Draw out the MOs .
(b) Give the electronic configuration of the ground state .
(c) Establish the correspondence with the Lewis structure .
(d) Why do we describe CH as a π-radical ?
(iv) On ionization the bond length of the molecule hardly changes (CH;112pm
,CH+;113pm) . Explain this result. (20%)
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