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Synthesis and Structural Characterization of ... - Jacobs University

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Chapter 3<br />

Unpublished Results<br />

Table 3.5. Crystal Data <strong>and</strong> Structure Refinement for Na-15 <strong>and</strong> Na-16.<br />

Code Na-15 Na-16<br />

Emp. formula Na 10[Yb(H 2O) 3Sb 2W 21O 72(OH)]·41H 2O Na 10[Lu(H 2O) 3Sb 2W 21O 72(OH)]·42H 2O<br />

M r 6469.0 6489.0<br />

Crystal system triclinic triclinic<br />

Space group (no.) P1̅ (2) P1̅ (2)<br />

a / Å 13.3196(8) 13.3368(18)<br />

b / Å 18.1807(13) 18.198(2)<br />

c/Å 23.8781(17) 23.895(3)<br />

α / deg 107.974(3) 108.071(5)<br />

ß/ deg 90.071(3) 90.048(5)<br />

γ/ deg 108.460(3) 108.357(4)<br />

V / Å 3 5185.6(6) 5201.5(12)<br />

Z 2 2<br />

T / ºC -100 -100<br />

λ / Å 0.71073 0.71073<br />

D / g cm -3 4.14 4.14<br />

μ / mm -1 24.8 24.7<br />

R [I ≥ 2σ(I)] a 0.0436 0.0510<br />

R w (all data) b 0.1545 0.1518<br />

__________________________<br />

a<br />

R = ∑|F o| – |F c| |∑| |F o|; b R w = {∑[w(F o 2 – F c 2 ) 2 ] / ∑[w(F o 2 ) 2 ]} 1/2 .<br />

Table 3.6. Crystal Data <strong>and</strong> Structure Refinement for Na-18 – Na-20.<br />

Code Na-18 Na-19 Na-20<br />

Emp. formula Na 8[Yb 2(H 2O) 6Sb 2W 20O 70]·37H 2O Na 8[Lu 2(H 2O) 6Sb 2W 20O 70]·37H 2O Na 8[Yb 2(H 2O) 6Sb 2W 20O 70]·37H 2O<br />

M r 6344.9 6367.1 6176.9<br />

Crystal system triclinic triclinic triclinic<br />

Space group (no.) P1̅ (2) P1̅ (2) P1̅ (2)<br />

a / Å 13.510(3) 13.5753(7) 13.6116(7)<br />

b / Å 13.516(3) 13.6389(8) 13.6742(8)<br />

c/Å 14.290(2) 14.2889(7) 14.3433(7)<br />

α / deg 106.174(11) 93.268(3) 93.180(3)<br />

ß/ deg 93.469(12) 106.357(2) 106.647(3)<br />

γ/ deg 106.313(13) 106.282(3) 106.078(4)<br />

V / Å 3 2377.6(7) 2410.0(2) 2431.3(2)<br />

Z 1 1 1<br />

T / ºC -100 -100 -100<br />

λ / Å 0.71073 0.71073 0.71073<br />

D / g cm -3 4.16 4.39 4.22<br />

μ / mm -1 26.7 26.5 25.4<br />

R [I ≥ 2σ(I)] a 0.066 0.056 0.059<br />

R w (all data) b 0.1874 0.1782 0.1781<br />

__________________________<br />

a<br />

R = ∑|F o| – |F c| |∑| |F o|; b R w = {∑[w(F o 2 – F c 2 ) 2 ] / ∑[w(F o 2 ) 2 ]} 1/2 .<br />

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