University Chemistry ›› 2022, Vol. 37 ›› Issue (11): 2112078.doi: 10.3866/PKU.DXHX202112078
• Survey of Chemistry • Previous Articles Next Articles
Xiaoming Wang1, Huan Jiao1,*(
), Xiping Jing2,*(
)
Received:2021-12-27
Accepted:2022-01-20
Published:2022-06-08
Contact:
Huan Jiao,Xiping Jing
E-mail:jiaohuan@snnu.edu.cn;xpjing@pku.edu.cn
Xiaoming Wang, Huan Jiao, Xiping Jing. What Is Nonstoichiometric Compound?[J].University Chemistry, 2022, 37(11): 2112078.
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| 序号 | 类别 | 组成 | 形成机理 | 缺陷 | 说明 | 文献 |
| 1 | 非本征填隙 原子缺陷类 | PdHx (0 ≤ x ≤ 0.7) γ-FeCx (0 ≤ x ≤ 0.1) | H2 = 2Hi× C = Ci× | Hi× Ci× | Pd为密堆积结构,H进入Pd的八面体空隙中;吸氢材料 C进入Fe构筑的填隙位置;1100 ℃下,x ≈ 0.1 | [ |
| 2 | 原子取代 缺陷类 | Fe1?xZnx (0 ≤ x ≤ 0.42) Hf1?xScx (0 ≤ x ≤ 1.0) | Zn = ZnFe× 富Hf固溶体:Sc = ScHf× 富Sc固溶体:Hf = HfSc× | ZnFe× ScHf× HfSc× | Zn (r = 0.133 nm)和Fe (r = 0.124 nm)原子半径接近。Zn占据Fe的格位,782 ℃下溶解度达x = 0.42 Sc (r = 0.165 nm)和Hf (r = 0.160 nm)原子半径接近;两者可以互相取代 | [ |
| 3 | 本征填隙阳离子缺陷 | Li2+4xTi1?xO3 (0 ≤ x ≤ 0.08) | 2Li2O = LiTi[O]"' + 3Lii[T]? + 2OO× | Li"'Ti[O],Lii[T]? | Li+ (r = 0.076 nm)和Ti4+ (r = 0.0605 nm)价态差别较大,但半径接近,这里,4个Li+可以取代1个Ti4+,1个占据八面体空隙[O],其余3个占据四面体空隙[T] | [ |
| 4 | 本征填隙阴离子缺陷类 | La2CuO4+x (0 ≤ x ≤ 0.09) | CuCu× +1/2O2 = 2CuCu? + Oi" | Oi",CuCu? | Cu3+ (CuCu?)含量2x。形成Oi"需要铜的变价缺陷CuCu?进行电荷补偿。含有Oi"的材料是超导体,超导转变温度35 K | [ |
| 5 | 本征阳离子 空位缺陷类 | Mn1?xO (0 ≤ x ≤ 0.18) Fe1?xO (0.045 ≤ x ≤ 0.20) | 2MM× + 1/2O2 = 2Mi[T]??? + 3VM[O]" + OO× (M = Mn, Fe) | Mi[T]???,VM[O]" (M = Mn, Fe) | O2将M2+氧化为M3+,O2?进入晶格形成八面体格位的阳离子空位VM[O]",M3+进入四面体格位形成填隙Mi[T]??? (含量为2x)。Mn1?xO中固溶体组成可以从x = 0开始;而Fe1-xO中x ≠ 0 | [ |
| 6 | 本征阴离子 空位缺陷类 | Y2Ti2O7?x (0 ≤ x → 1.0) | OO× + TiTi× = VO? + TiTi' +1/2O2 | VO?,TiTi' | 低氧分压或还原性气氛下,O2?会溢出,在晶格中形成VO×。VO×部分电离成VO?,将Ti4+还原成Ti3+ (TiTi')。合适的还原条件可使x值接近1。存在氧空位的物相颜色变深,电导率增加 | [ |
| 7 | 非本征填隙 阳离子缺陷类 | NaxWO3 (0 ≤ x ≤ 1.0) LixTiS2 (0 ≤ x ≤ 1.0) | Na + Ww× = Nai? + Ww' Li + TiTi× = Lii? + TiTi' | Nai?,Ww' Lii?,TiTi' | WO3被Na适度地还原后,颜色由浅黄变蓝色,电导率提高。W5+ (Ww')含量为x TiS2是层状结构,控制反应速度可以使Li+进入层间,这一过程是锂离子电池负极放电过程。Ti3+ (TiTi')含量为x | [ |
| 8 | 阳离子取代 缺陷类 | 等价取代: (Ca1?xSrx)5(PO4)3Cl (0 ≤ x ≤ 1.0) | 富Ca固溶体: SrO = SrCa× + OO× 富Sr固溶体: CaO = CaSr× + OO× | SrCa× CaSr× | M5(PO4)3Cl [M = Ca2+ (r = 0.10 nm),Sr2+ (r = 0.118 nm),Ba2+ (r = 1.35 nm)]具有磷灰石结构。Ca相和Sr相及Sr相和Ba相形成全范围固溶体,Ca相和Ba相只形成部分固溶体。在该体系中掺入Eu2+可以获得蓝色或蓝绿色荧光,分别用于三基色或高显色荧光灯中 | [ |
| 高价取代: Ba6-3xNd8+2xTi18O54 (0.25 ≤ x ≤ 0.75) | Nd2O3 = 2NdBa? + VBa" + 3OO× | NdBa?,VBa" | 阳离子空位补偿:2个Nd3+取代3个Ba2+,生成1个VBa"补偿价态不平衡。固溶体范围不能伸展到x = 0。重要的电介质材料 | [ | ||
| Ca1?xYxF2+x (0 ≤ x ≤ 0.38) | YF3 = YCa? + Fi' | YCa?,Fi' | 填隙阴离子补偿:Y3+取代Ca2+,增加Fi'补偿价态不平衡 | [ | ||
| Sr1?xNdxCuO2 (0 ≤ x ≮ 0.16) | 2Nd2O3 + 4CuO = 4NdSr? + 4CuCu' + 8OO× + O2 | NdSr?,CuCu' | 阳离子变价(降价)补偿:Nd3+取代Sr2+,Cu2+降价为Cu+ (CuCu',含量约为x)补偿价态不平衡。还原气氛处理。含有CuCu'的物相是超导材料,转变温度约为40 K | [ | ||
| 低价取代: Zr1?xYxO2?x/2 (0 ≤ x ≤ 0.70) | Y2O3 = 2YZr' + VO?? + 3OO× | YZr',VO?? | 阴离子空位补偿:Y3+占据Zr4+格位,产生1个VO??补偿价态不平衡。掺Y量x ≈ 0.05时,正交晶系,增韧陶瓷;掺杂量x > 0.15,氧离子导体 | [ | ||
| Li1+xAl1+xSi2?xO8 (0 ≤ x ≤ 0.14) | LiAlO2 = Lii? + AlSi' + 2OO× | AlSi',Lii? | 本征填隙阳离子补偿:Al3+取代Si4+,占据其格位,引入Lii?补偿价态不平衡 | [ | ||
| La2?xCaxCuO4 (0 ≤ x ≮ 0.30) | CaO + CuO + 1/4O2 ? CaLa′ + CuCu? + 2.5OO× | CaLa′,CuCu? | 阳离子变价(升价)补偿:Ca2+取代La3+,在氧化气氛中处理,Cu2+升价为Cu3+ (CuCu?,含量为x)。超导转变温度~20 K | [ | ||
| 9 | 阴离子取代 缺陷类 | 等价取代: BaFCl1?xBrx (0.0 ≤ x ≤ 1.0) 不等价取代: Y4Al2?2xSi2xO9?2xN2x (0.0 ≤ x ≤ 0.4) | 富Cl固溶体: BaBr2 = BaBa× + 2BrCl× 富Br固溶体: BaCl2 = BaBa× + 2ClBr× Si2ON2 = 2NO' + 2SiAl? + OO× | BrCl× ClBr× NO',OO× | BaFX [X = Cl (r = 0.181 nm), Br (r = 0.196 nm), Ba (r = 0.220 nm)]具有氟氯铅矿结构。Cl相和Br相及Br相和I相形成全范围固溶体,Cl相和I相形成很窄的部分固溶体。在该体系中掺入Eu2+发射近紫外到紫色荧光,用于X射线增感屏或存储屏中 在N3?取代O2?时,Si4+同时取代Al3+进行电荷补偿 | [ [ |
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