第十六族元素(S、Se、Te)與過渡金屬(Cr、Mn、Fe、Ru)所形成金屬團簇化合物之化性、物性研究

洪昭宇(Chaoyu Hung)

指導教授 ： 謝明惠 博士

國立臺灣師範大學/理學院/化學系/碩士(2002年)

摘要

室溫下將Mn2(CO)10與Cr(CO)6以2：1的莫耳比在4M的KOH甲醇溶液中活化12小時，再加入過量E powder (E = S, Se) 繼續反應，S系統反應1小時，Se系統反應10分鐘後可得啞鈴形化合物 [E10Mn6(CO)18]4- (E = S, Se)，改變反應之E powder、Mn2(CO)10與Cr(CO)6莫耳比及KOH濃度結果顯示，合成此二化合物時須注意主族元素與Mn2(CO)10之比例以及較高濃度之KOH。此外二化合物均符合18電子計算規則，然而經由SQUID磁性測量可發現，於室溫 (300K) 時S系統具有8個未成對電子，Se系統則具有4個未成對電子。
進一步將 [E10Mn6(CO)18]4-與Cr(CO)6反應可得[E2Mn2Cr(CO)9]2- (E = S, Se)；而 [E2Mn2Cr(CO)9]2-與過量E powder反應亦可轉變為 [E10Mn6(CO)18]4- (E = S, Se)。此外 [E10Mn6(CO)18]4-與Mn2(CO)10反應可得 [E2Mn3(CO)9]2- (E = S, Se)；若與Fe(CO)5反應，於S系統得 [SFe3(CO)9]2-，Se系統則會得tbp結構之unknown 4。[S10Mn6(CO)18]4-若與Mn(CO)5Br反應則會被氧化為 [S4Mn3(CO)10]-。
將莫耳數比3：6：1的Se powder、Cr(CO)6、Ru3(CO)12在2.57 M的KOH 甲醇溶液中加熱迴流約24小時，產生八面體結構的 [Se2Ru3Cr(CO)10]2-，此化合物較18電子計算規則缺少4個電子。
以莫耳數比2：1：1的Te powder、Mn2(CO)10、Fe(CO)5在6.33 M的KOH 甲醇溶液中反應7天，或是在1.65 M的KOH甲醇溶液中加熱迴流5天均可得 [Te2Mn2Fe(CO)9]2-。進一步與 [Cu(CH3CN)4][BF4] 反應則會產生鍵的斷裂及重組而得到氧化性產物 [Te2Mn3Fe(CO)12]-。

關鍵字

金屬團簇化合物

並列摘要

To the mixture of Mn2(CO)10 and Cr(CO)6 in the molar ratio of 2 : 1 in 4 M KOH / MeOH solution was added the excess E powder (E = S, Se) which resulted in the formation of the dumbbell compounds [E10Mn6(CO)18]4-. The variation of the molar ratio of E : Mn : Cr or change of the KOH / MeOH concentration showed that the ratio of E : Mn and the higher concentration of KOH / MeOH solution both play important factors in controlling the production of [E10Mn6(CO)18]4- (E = S, Se). [E10Mn6(CO)18]4- (E = S, Se) are both electron precise species, however, the SQUID analysis shows that [S10Mn6(CO)18]4- has 8 unpaired electrons and [Se10Mn6(CO)18]4- has 4 unpaired electrons at room temperature (300 K).
The reactions of [E10Mn6(CO)18]4- (E = S, Se) with Cr(CO)6 formed the trigonal bipyramidal clusters [E2Mn2Cr(CO)9]2-, conversely, the reactions of [E2Mn2Cr(CO)9]2- with the excess E powder converted to [E10Mn6(CO)18]4-. Further reactions of [E10Mn6(CO)18]4- (E = S, Se) with Mn2(CO)10 generated the trigonal bipyramidal clusters [E2Mn3(CO)9]2-. The metallation reaction of [S10Mn6(CO)18]4- with Fe(CO)5 produced the known cluster [SFe3(CO)9]2- while the reaction in the Se case yielded the trigonal bipyramidal unknown cluster 4. If [S10Mn6(CO)18]4- was treated with Mn(CO)5Br, the oxidized product [S4Mn3(CO)10]- was obtained.
Refluxing Se powder, Cr(CO)6 and Ru3(CO)12 in the molar ratio of 3 : 6 : 1 in 2.57 M KOH / MeOH solution for 24h led to the formation of an octahedral cluster [Se2Ru3Cr(CO)10]2- which is a 4 electron deficient species based on the EAN rule.
The reaction of Te powder, Mn2(CO)10, and Fe(CO)5 in the molar ratio of 2 : 2 : 1 in KOH / MeOH solution formed the square pyramidal cluster [Te2Mn2Fe(CO)9]2-. Further reaction with [Cu(CH3CN)4][BF4] afforded the oxidized product [Te2Mn3Fe(CO)12]- resulted from bond breaking and formation processes.

並列關鍵字

cluster

參考文獻

1. Suzuki, H. Eur. J. Inorg. Chem. 2002, 1009.

3. Fachinetti, G.; Fochi, C.; Funaioli, T. Inorg. Chem. 1994, 33, 1719.

4. Adams, R. D.; Falloon, S. B. Chem. Rev. 1995, 95, 2587.

6. Reynolds, M. A.; Guzei, L. A.; Angelici, R. J. J. Am. Chem. Soc. 2002, 124, 1689.

7. Bianchini, C.; Meli, A. Acc. Chem. Res. 1998, 31, 109.