2. 南昌航空大学 测试与光电工程学院, 江西 南昌 330063
2. School of Testing and Optoelectronic Engineering, Nanchang Hangkong University, Nanchang 330063, China
晶体Cu1-xHxZr2(PO4)3与NaZr2(PO4)3同为单斜晶体,属于Nasico类型的磷酸盐,该类型晶体具有由一系列PO4四面体相连接而构成的带状三维网络结构,其中四面体的顶点由ZrO6八面体占据[1-4].因具有相对大的表面积,这类晶体在催化、吸附、离子传导、陶瓷等方面具有广阔的应用前景,受到越来越多的重视.晶体Cu1-xHxZr2(PO4)3中Cu2+可处于M(1)和M(2)两种替代位置,其中M(1)位置为六配位三角八面体对称,M(2)与周围近邻8~10个离子构成团簇[1].而通常情况下,含过渡金属离子功能材料的光、电、磁等性能与其中过渡金属离子所处的局域结构密切相关[5-8],因此,探讨晶体Cu1-xHxZr2(PO4)3中Cu2+所处的局域结构非常重要.
顺磁离子的电子顺磁共振(electron paramagnetic resonance,EPR)波谱由EPR参量(如朗德g因子和超精细结构常数A因子)表征,这些参量对顺磁离子的局域结构非常敏感[4].因此,对上述体系EPR参量的实验研究和理论解释不仅能获得顺磁中心离子周围的局部结构,而且能帮助我们了解其化学及物理的特性[8].例如,文献[5]报道了晶体Cu1-xHxZr2(PO4)3中Cu2+中心的谱学实验数据.遗憾的是,到目前为止对这些实验EPR谱及Cu2+所处的局域结构尚欠缺合理、统一的理论解释.文献[9]采用简单二阶微扰公式计算了晶体Cu0.5Zr2(PO4)3中Cu2+的g因子,由于公式中未考虑配体离子的贡献并使用了不太精确的点电荷模型,因此,理论计算与实验数据符合度不太理想.
Cu2+离子属于3d9离子,晶体Cu1-xHxZr2(PO4)3中Cu2+中心为斜方对称,所以本文基于3d9离子EPR参量在斜方对称下的高阶微扰公式计算了晶体Cu1-xHxZr2(PO4)3不同Cu2+浓度的EPR谱,计算公式中所包含的晶场参量由重叠模型结合此类Cu2+中心所处的局域结构给出,并在双旋轨耦合模型的基础上考虑了基态波函数中2A1g(θ)和2A1g(ε)混合对EPR参量的影响[10].
1 模型和公式根据Taoufik等的EPR实验,晶体Cu1-xHxZr2(PO4)3中Cu2+占据三角八面体对称的M(1)位置,并与周围6个最近邻氧离子形成[CuO6]10‒基团[1, 10].由于三角畸变不能消除d9离子基态2Eg的双重轨道简并性,原三角八面体位置上的Cu2+可能由于Jahn-Teller效应由三角对称转变为四角(或斜方)对称,从而消除基态简并.由实验测得的g因子
由于2A1g(θ)和2A1g(ε)属于同一不可约表示,因而有一定程度的混合,基态波函数可以写为[10, 12, 13]:
$\mathit{\Phi} = N[\alpha \left| {{d_{{x^2} - {y^2}}}} \right\rangle + \beta \left| {{d_{3{z^2} - {r^2}}}} \right\rangle ]$ | (1) |
(1) 式中d代表中心离子d轨道,r代表三维空间位置矢径,x,y,z表示三个坐标分量;N为共价因子,体现体系的共价性[10].混合系数α和β满足归一化关系,表征基态的混合程度[10]:
${\alpha ^2} + {\beta ^2} = 1$ | (2) |
由双旋-轨耦合模型和Macfarlance的强场微扰理论,可得3d9离子在斜方伸长八面体对称下EPR参量的高阶微扰公式[10, 13, 14]:
$\begin{align} & {{g}_{x}}={{g}_{s}}+\frac{2k'\varsigma '{{(\alpha +\sqrt{3}\beta )}^{2}}}{{{E}_{4}}}-\frac{2\alpha k'\varsigma \varsigma '(\alpha +\sqrt{3}\beta )}{{{E}_{2}}{{E}_{4}}}+\frac{k'\varsigma \varsigma '({{\alpha }^{2}}-3{{\beta }^{2}})}{{{E}_{2}}{{E}_{4}}}-\frac{2{{\alpha }^{2}}{{g}_{s}}\varsigma {{'}^{2}}}{E_{2}^{2}} \\ & \begin{matrix} {} & {} \\ \end{matrix}-\frac{{{g}_{s}}\varsigma {{'}^{2}}{{(\alpha -\sqrt{3}\beta )}^{2}}}{2E_{3}^{2}}+\frac{2\alpha k\varsigma {{'}^{2}}(\alpha -\sqrt{3}\beta )}{{{E}_{2}}{{E}_{3}}} \\ & {{g}_{y}}={{g}_{s}}+\frac{2k'\varsigma '{{(\alpha -\sqrt{3}\beta )}^{2}}}{{{E}_{3}}}-\frac{2\alpha k\varsigma \varsigma '(\alpha -\sqrt{3}\beta )}{{{E}_{2}}{{E}_{3}}}+\frac{k'\varsigma \varsigma '({{\alpha }^{2}}-3{{\beta }^{2}})}{{{E}_{2}}{{E}_{3}}}-\frac{2{{\alpha }^{2}}{{g}_{s}}\varsigma {{'}^{2}}}{E_{2}^{2}} \\ & \begin{matrix} {} & {} \\ \end{matrix}-\frac{{{g}_{s}}\varsigma {{'}^{2}}{{(\alpha +\sqrt{3}\beta )}^{2}}}{2E_{4}^{2}}+\frac{2\alpha k\varsigma {{'}^{2}}(\alpha +\sqrt{3}\beta )}{{{E}_{2}}{{E}_{4}}} \\ & {{g}_{z}}={{g}_{s}}+\frac{8{{\alpha }^{2}}k'\varsigma '}{{{E}_{2}}}-\frac{2\alpha k'\varsigma \varsigma '(\alpha -\sqrt{3}\beta )}{{{E}_{2}}{{E}_{3}}}-\frac{2\alpha k'\varsigma \varsigma '(\alpha +\sqrt{3}\beta )}{{{E}_{2}}{{E}_{4}}}-\frac{{{g}_{s}}\varsigma {{'}^{2}}{{(\alpha -\sqrt{3}\beta )}^{2}}}{2E_{3}^{2}} \\ & \begin{matrix} {} & {} \\ \end{matrix}-\frac{{{g}_{s}}\varsigma {{'}^{2}}{{(\alpha +\sqrt{3}\beta )}^{2}}}{2E_{4}^{2}}-\frac{k'\varsigma \varsigma '(\alpha -3{{\beta }^{2}})}{{{E}_{3}}{{E}_{4}}} \\ \end{align}$ | (3) |
${{A}_{x}}=P\left[ -\kappa +\frac{2N}{7}({{\alpha }^{2}}-{{\beta }^{2}})+({{g}_{x}}-{{g}_{s}})-\frac{1}{14}(\frac{3\alpha +\sqrt{3}\beta }{\alpha -\sqrt{3}\beta })({{g}_{x}}-{{g}_{s}})+\frac{\sqrt{3}\beta }{14\alpha }({{g}_{z}}-{{g}_{s}})-\frac{4\sqrt{3}k\alpha \beta }{7} \right]$ |
${{A}_{y}}=P\left[ -\kappa +\frac{2N}{7}({{\alpha }^{2}}-{{\beta }^{2}})+({{g}_{y}}-{{g}_{s}})-\frac{1}{14}(\frac{3\alpha +\sqrt{3}\beta }{\alpha -\sqrt{3}\beta })({{g}_{y}}-{{g}_{s}})+\frac{\sqrt{3}\beta }{14\alpha }({{g}_{z}}-{{g}_{s}})-\frac{4\sqrt{3}k\alpha \beta }{7} \right]$ |
${{A}_{z}}=P\left[ -\kappa -\frac{4}{7}({{\alpha }^{2}}-{{\beta }^{2}})+({{g}_{z}}-{{g}_{s}})+\frac{1}{14}(\frac{3\alpha -\sqrt{3}\beta }{\alpha +\sqrt{3}\beta })({{g}_{z}}-{{g}_{s}})+\frac{1}{14}(\frac{3\alpha +\sqrt{3}\beta }{\alpha -\sqrt{3}\beta })({{g}_{z}}-{{g}_{s}}) \right]$ |
(3) 式中,
$\begin{align} & \varsigma ={{N}_{t}}(\varsigma _{d}^{0}+\lambda _{t}^{2}\varsigma _{p}^{0}/2),\text{ }\varsigma '=\sqrt{{{N}_{t}}{{N}_{e}}}(\varsigma _{d}^{0}-{{\lambda }_{t}}{{\lambda }_{e}}\varsigma _{p}^{0}/2) \\ & k={{N}_{t}}(1+\lambda _{t}^{2}/2),\text{ }k'=\sqrt{{{N}_{t}}{{N}_{e}}}[1-{{\lambda }_{t}}({{\lambda }_{e}}+{{\lambda }_{s}}A)/2] \\ \end{align}$ | (4) |
(4) 式中,
$\begin{gathered} {N^2} = N_t^2(1 + \lambda _t^2S_{dpt}^2 - 2{\lambda _t}{S_{dpt}}) \\ {N^2} = N_e^2(1 + \lambda _e^2S_{dpe}^2 + \lambda _s^2S_{ds}^2 - 2{\lambda _e}{S_{dpe}} - 2{\lambda _s}{S_{ds}}) \\ \end{gathered} $ | (5) |
以及归一化条件:
$\begin{gathered} {N_t}(1 - 2{\lambda _t}{S_{dpt}} + \lambda _t^2) = 1 \\ {N_e}(1 - 2{\lambda _e}{S_{dpe}} - 2{\lambda _s}{S_{ds}} + \lambda _e^2 + \lambda _s^2) = 1 \\ \end{gathered} $ | (6) |
其中,
(3) 式中,能级差
$\begin{gathered} {E_2} \approx 10{D_q} \\ {E_3} \approx 10{D_q} + 3{D_s} - 5{D_t} - 3{D_\xi } + 4{D_\eta } \\ {E_4} \approx 10{D_q} + 3{D_s} - 5{D_t} + 3{D_\xi } - 4{D_\eta } \\ \end{gathered} $ | (7) |
根据已有[CuO6]10-基团的光谱,立方场参量取为
基于重叠模型[25]并联系晶体Cu1-xHxZr2(PO4)3中Cu2+中心所处的局域结构,斜方场参量可表示为:
$\begin{gathered} {D_s} = \frac{2}{7}{\overline A _2}(R)\left[ {{{(\frac{R}{{R_{{\rm{||}}}^{}}})}^{{t_2}}} - {{(\frac{R}{{R_ \bot ^{}}})}^{{t_2}}}} \right] \\ {D_t} = \frac{4}{7}{\overline A _4}(R)\left[ {(7\cos 2\tau + 3){{(\frac{R}{{R_ \bot ^{}}})}^{{t_4}}} + 4{{(\frac{R}{{R_{||}^{}}})}^{{t_4}}}} \right] \\ {D_\xi } = \frac{4}{{21}}{\overline A _2}(R){(\frac{R}{{{R_ \bot }}})^{{t_2}}}{\rm{cos}}\tau \\ {D_\eta } = \frac{{20}}{{21}}{\overline A _4}(R){(\frac{R}{{{R_ \bot }}})^{{t_4}}}{\rm{cos}}\tau \\ \end{gathered} $ | (8) |
(8) 式中,指数律系数
晶体Cu1-xHxZr2(PO4)3中Cu2+占据三角八面体对称的M(1)位置,Jahn-Teller效应使得中心金属[CuO6]10-基团由三角向斜方转变.此时,轴向键角相对于立方情形(
(8) 式中基团[CuO6]10-中平行和垂直与z轴的Cu-O键长
${R_{{\rm{||}}}} \approx \frac{{3\overline R }}{{1 + 2\tan ({{45}^ \circ } - \Delta \varphi )}},{\rm{ }}{R_ \bot } \approx \frac{{3\overline R }}{{2 + \cot ({{45}^ \circ } - \Delta \varphi )}}$ | (9) |
由此,晶体Cu1-xHxZr2(PO4)3中Cu2+局部结构参数,包括
以上所有公式中仅有三个未知参量(轴向畸变角
$\Delta \varphi =3.2,\text{ }\tau ={{80.1}^{{}^\circ }}$ | (10) |
$\mathit{\Phi} = 0.87[0.995\left| {{d_{{x^2} - {y^2}}}} \right\rangle + 0.9{\rm{99}}\left| {{d_{3{z^2} - {r^2}}}} \right\rangle ]$ | (11) |
对应的自旋哈密顿参量计算结果(Cal.c)列于表 2.同时,考虑配体贡献但忽略基态混合[即(1)式中的基态波函数混合系数
从表 2可以看出,本工作基于上述局部结构和基态混合所得的EPR参量计算结果(Cal. c)与实验值符合度很好.可见,本文所采用的计算公式以及通过拟合EPR参量所得的晶体Cu1-xHxZr2(PO4)3中Cu2+的基态波函数及其局域结构是合理的.
(1)在斜方对称下,3d9离子立方对称时的轨道简并被完全解除,原基态轨道双重态(2Eg)分裂为两个有一定程度混合的轨道单重态2A1g(θ)和2A1g(ε),这种混合对体系的EPR谱有一定影响[10].同时,旋-轨耦合也会对EPR参量产生一定的贡献[10].从表 2可以看出,与文献[10]类似,考虑2A1g(θ)和2A1g(ε)混合以及配体轨道及自旋-轨道耦合作用贡献的结果(Cal.c)与实验值符合很好,然而忽略基态混合的计算值(Cal.a)和忽略配体及旋-轨耦合作用的结果(Cal.b)都与实验值符合较差.而且,通过拟合Cu1-xHxZr2(PO4)3中Cu2+的EPR参量实验值得到的基态波函数混合系数α(≈0.995)与Zn(C3H3O4)2(H2O)2中类似斜方Cu2+中心的混合系数α≈0.995[13]一致,进一步说明本工作的可靠性.
(2)本文通过计算EPR参量获得Cu1-xHxZr2(PO4)3中[CuO6]10-基团的Cu-O键长为:
(3)以上计算公式中的平均共价因子N和芯区极化常数κ可由半经验公式
基于离子簇模型,考虑基态波函数混合及配体轨道和旋轨耦合EPR参量高阶微扰公式,并计算了Cu1-xHxZr2(PO4)3中Cu2+的自旋哈密顿参量,与实验吻合很好.研究表明,Cu1-xHxZr2(PO4)3中处于三角八面体对称M(1)位置的Cu2+,由于Jahn-Teller效应导致中心金属对称性由三角八面体对称转变为斜方伸长八面体对称,[CuO6]10-基团的Cu-O键长为:
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