Yellow Midterm Test Answers
GREEN MIDTERM TEST ANSWERS
1. Face 1 = (2,-1) Face 2 = (-3, -1) Face 3 = (-1,0)
Face 4 = (-1,1) Face 5 = (0, 1) Face 6 = (1,0)
2. A) in hkl, h + k + l = 2n + 1 => I-centered cell
in 0kl, h = 2n + 1 => b-glide perpendicular to a
in h0l, l = 2n + 1 => c-glide perpendicular to b
in hk0, h = 2n + 1 => a-glide perpendicular to c
B) Ibca
C) mmm => orthorhombic
3. C48H48Cl4Cu2N10O16 MW = 1289.84 g/mol
V = abc sinb = (22.3883)(9.5978)(25.4833)sin98.985o = 5408.62 Å3
a) r = ZM/VN Z = rVN/M = (1.599 gcm-3)(540.862)(6.022)/(1289.84 g/mol)
Z = 4 molecules / cell
b) a* = 1/a sinb = 1 / (22.3883 sin98.985o) = 0.0452211 Å-1
b* = 1/b = 1/9.5978 = 0.1041905 Å-1
c* = 1/c sinb = 1 / (25.4833 sin98.985) = 0.0397289 Å-1
a* = g* = 90o ; b* = 180 - b = 180 – 98.985 = 81.015o
c) The angle between a & a* = the angle between c & c*
Therefore, (98.985o – 81.015o) / 2 = 8.985o
d) d*100 = a* = 0.0452211 Å-1
Therefore, 1.54178Å(d*100) / 2 = sin q100 = 0.03486049
q100 = sin-1 (0.03486049) = 1.998o
d*222 = [(2x0.0452211)2+(2x0.1041905) 2+(2x0.0397289) 2+
2(2)(2)( 0.0452211)( 0.0397289)cos81.015]1/2
d*222 = (0.0601613) 1/2 = 0.24528 Å-1
Therefore, 1.54178 Å(d*222) / 2 = sin q222 = 0.189082
q222 = sin-1 (0.190617) = 10.899o
5. A = S fi cos a
A = (47.7 cos -18.23o) + (31.89 cos 154.78o) + (49.7 cos -213.26o)
A = (45.282 - 28.85 -41.559) = -25.127
B = S fi sin a = (47.7 sin -18.23o) + (31.89 sin 154.78o) + (49.7 sin -213.26o)
B = -14.993 + 13.588 + 27.257 = 25.852
│F│ = (A2 + B2) 1/2 = [(-25.127) 2 + (25.852) 2] ½ = 36.051
a = tan-1(B/A) = tan-1(25.852/-25.127) = tan-1(-1.029) = -45.82o
Since A = - and B = +, the resultant sum should be in the 2nd quadrant.
Therefore, a = tan(180o + (-45.82o) = 134.19 o, which is in the 2nd quadrant.
4. a) Cu1 at x,y,z 0.8135 0.08926 0.9610
Cu1A has to be at x, ½-y, z 0.8135 0.41074 0.9610
Therefore, the difference between these is: 0 0.32148 0
b) Bond distance = {[Dy (b)]2}1/2 = (0.32148)(41.247} = 13.26 Å
c) We have to calculate the Cu1-C19 distance:
C19 0.82402 0.2500 1.28101
Cu1 0.8135 0.08926 0.9610_
Delta: 0.01052 0.16074 0.32001
Bond distance = [(Dx a)2 + (Dy b)2 + (Dz c)2+…………..]1/2
= {[(0.01052)(8.5404Å)]2 + [(0.16074)(41.247 Å)]2
+ [(0.32001)(8.5889Å)
+ (2)(8.5404)(0.01052)(8.5889)(0.32001)cos]118.18]}1/2
Bond distance = 7.161 Å
d) Bond angle: let C19 be atom A, Cu1 be atom B, and Cu1A be atom C
Bond angle: Cu1-C19-Cu1A:
Bond angle q = cos-1[(AB)2 + (AC)2 – (BC)2] / 2(AB)(AC)
q = cos-1[(7.161)2 + (7.161) 2 - (13.26)2] / 2(7.161)(7.161)
q = cos-1(-0.714395)
q = 135.59o