A crystal is composed of a metallic element $\mathbf{M}$ and non-metallic elements boron and oxygen, where the mass fraction ratio of $\mathbf{M}$ to B is 6.35, and its anion belongs to the $D_{\mathrm{3h}}$ point group. The crystal can be considered as being composed of stacked single-layer structures. The single-layer structure can be viewed as $\mathbf{M}$ forming a close-packed layer, with anions orderly filling part of the triangular voids, so that the distribution of the centers of the anions is the same as the distribution of carbon atoms in a single layer of graphite. All anions within the layer have the same orientation. It is known that the anions within the layer are equivalent to hexadentate ligands, coordinating to the three adjacent $\mathbf{M}$ atoms. Assume that all $\mathbf{M}$-O bond lengths are $282.3\ \mathrm{pm}$, and all B-O bond lengths are $137.5\ \mathrm{pm}$. Which of the following options is correct:
某晶体由一种金属元素$\mathbf{M}$和非金属元素硼、氧组成，其中$\mathbf{M}$与B的质量分数之比为6.35，其阴离子属于$D_{\mathrm{3h}}$点群。该晶体可看作由单层结构堆叠而成。单层结构可以看作$\mathbf{M}$形成密置层，阴离子有序地填入部分三角形空隙，使得阴离子的中心的分布与石墨单层中碳原子的分布相同。单层内所有的阴离子都具有相同的取向。已知层内阴离子相当于六齿配体，向与之相邻的三个$\mathbf{M}$原子配位。假设所有的$\mathbf{M}$-O键长度均为$282.3\ \mathrm{pm}$，所有B-O键长度均为$137.5\ \mathrm{pm}$。下列选项正确的是：
The occupancy of anions is 1/3, the coordination number of $\mathbf{M}$ within the layer is 6, and the lattice parameter is $a=1233\ \mathrm{pm}$.
The occupancy of anions is 1/3, the coordination number of $\mathbf{M}$ within the layer is 6, and the lattice parameter is $a=1672\ \mathrm{pm}$.
The occupancy of anions is 1/3, the coordination number of $\mathbf{M}$ within the layer is 4, and the lattice parameter is $a=1233\ \mathrm{pm}$.
The occupancy of anions in the interstitial sites is 1/3, the coordination number of $\mathbf{M}$ within the layer is 4, and the unit cell parameter is $a=1672\ \mathrm{pm}$.
The occupancy of anions in interstitial sites is 2/3, the coordination number of $\mathbf{M}$ within the layer is 4, and the lattice parameter is $a=1233\ \mathrm{pm}$.
The occupancy of anions is 2/3, the coordination number of $\mathbf{M}$ within the layer is 4, and the lattice parameter $a=1672\ \mathrm{pm}$
The occupancy of anions is 2/3, the coordination number of $\mathbf{M}$ within the layer is 6, and the lattice parameter is $a=1233\ \mathrm{pm}$.
The occupancy of anions is 2/3, the coordination number of $\mathbf{M}$ within the layer is 6, and the cell parameter is $a=1672\ \mathrm{pm}$.
All other options are incorrect.
阴离子的填隙率为1/3，$\mathbf{M}$在层内的配位数为6，晶胞参数$a=1233\ \mathrm{pm}$
阴离子的填隙率为1/3，$\mathbf{M}$在层内的配位数为6，晶胞参数$a=1672\ \mathrm{pm}$
阴离子的填隙率为1/3，$\mathbf{M}$在层内的配位数为4，晶胞参数$a=1233\ \mathrm{pm}$
阴离子的填隙率为1/3，$\mathbf{M}$在层内的配位数为4，晶胞参数$a=1672\ \mathrm{pm}$
阴离子的填隙率为2/3，$\mathbf{M}$在层内的配位数为4，晶胞参数$a=1233\ \mathrm{pm}$
阴离子的填隙率为2/3，$\mathbf{M}$在层内的配位数为4，晶胞参数$a=1672\ \mathrm{pm}$
阴离子的填隙率为2/3，$\mathbf{M}$在层内的配位数为6，晶胞参数$a=1233\ \mathrm{pm}$
阴离子的填隙率为2/3，$\mathbf{M}$在层内的配位数为6，晶胞参数$a=1672\ \mathrm{pm}$
其他选项均不正确
Starting from the anionic point group, the $D_{\mathrm{3h}}$ point group could be $\mathrm{BO_3^{3-}}$ or $\mathrm{B_3O_6^{3-}}$. Assuming the former, there is no reasonable $\mathbf{M}$ that can simultaneously satisfy the ratio of mass and oxidation state, resulting in no reasonable solution. Therefore, the anion can only be $\mathrm{B_3O_6^{3-}}$, and $\mathbf{M}$ can be determined as Ba, with the simplest crystal formula being $\mathrm{BaB_2O_4}$.

<MultiModal>![[O-]B1OB([O-])OB([O-])O1，即阴离子$\mathrm{B_3O_6^{3-}}$结构](/media/uploads/5a83bf1c-0d20-4385-8b05-f8f6362dbd0c.png)</MultiModal>

<Checkpoint>[$\mathbf{M}$为Ba](1)</Checkpoint>

<Checkpoint>[晶体最简式为$\mathrm{BaB_2O_4}$](1)</Checkpoint>

(In the following description, $\mathbf{M}$ is equivalent to Ba.)

$\mathbf{M}$ adopts a close-packed arrangement, and the anions fill the triangular voids formed by $\mathbf{M}$ in a graphite-like manner, as shown in the figure below, which can outline a unit cell.

<MultiModal>![这是一张晶体的二维平面层示意图，由代表Ba的实心黑球和代表阴离子的空心白球组成。其中Ba原子作密堆积排布，阴离子以石墨的方式填充$\mathbf{M}$围成的三角形空隙。图像的下半部分用一个菱形轮廓线标示出了一个最小的二维周期结构（二维晶胞），该菱形的四个顶点都位于Ba原子的三角形空隙中心，菱形的边长为$\sqrt{3}$倍的Ba-Ba间距。在一个二维晶胞中包含有3个Ba原子和2个$\mathrm{B_3O_6^{3-}}$阴离子。](/media/uploads/7930a07c-c7a7-4c7a-9bc1-fffab8eaf09e.png)</MultiModal>

<Checkpoint>[最小的二维周期性结构中包含3个Ba原子和2个$\mathrm{B_3O_6^{3-}}$阴离子](1)</Checkpoint>

In each unit cell, Ba forms 6 triangular voids, while the anions occupy 2, with an anion occupancy rate of 1/3.

<Checkpoint>[在每个晶胞中，Ba围成6个三角形空隙，而阴离子占据2个](1)</Checkpoint>

<Checkpoint>[阴离子填隙率为1/3](1)</Checkpoint>

The anion $\mathrm{B_3O_6^{3-}}$ has 6 oxygen atoms and is a hexadentate ligand. Since each anion is surrounded by 3 Ba, each two adjacent O coordinate to one Ba; and each Ba is surrounded by 2 anions, so the coordination number of Ba within the layer is $2\times 2=4$.

<MultiModal>![[O]12B(O[Ba]2)[O]3B(O[Ba]3)[O]4B1O[Ba]4，仅表示配位连接方式](/media/uploads/f6437a80-8fe0-4ccd-8d73-8cd02db68ea3.png)</MultiModal>

<Checkpoint>[Ba在层内的配位数为4](1)</Checkpoint>

For the unit cell parameters, first calculate the shortest distance between the solid black spheres (Ba) and hollow white spheres (anions) in the figure above:

<MultiModal>![本图中示出了一个阴离子和其最近的一个Ba的几何结构。图中$\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D},\mathbf{E},\mathbf{F}$构成正六边形，其中$\mathbf{A},\mathbf{C},\mathbf{E}$为硼原子，$\mathbf{B},\mathbf{D},\mathbf{F}$为氧原子。每个硼原子分别再向外连接一个氧原子，记为$\mathbf{G},\mathbf{H},\mathbf{I}$；$\mathbf{J}$为钡原子，$\mathbf{K}$为正六边形的中心。钡原子$\mathbf{J}$与氧原子$\mathbf{F},\mathbf{G}$配位。由于$\mathbf{AF=AG}$，$JA$平分$\mathbf{\angle{FAG}}$。求出$\mathbf{JK}$的长度，由二维晶胞示意图可知，$\mathbf{JK}$长度的三倍即为晶胞参数$a$](/media/uploads/ddcaf6a8-6ef5-42b7-ac2d-f1f7fb64a37b.png)</MultiModal>

Since the Ba-O distances are the same, $\mathbf{JA}$ bisects $\mathbf{\angle{FAG}}$, and the angle $\mathbf{\angle{FAJ}}$ of O-B-Ba is $180\degree-120\degree=60\degree$. From the law of sines, the angle $\mathbf{\angle{AJF}}$ formed by B-Ba-O is $\arcsin{\frac{d_{\mathrm{B-O}}}{d_{\mathrm{M-O}}}\times \sin{60\degree}}=24.95\degree$

<Checkpoint>[$\mathbf{\angle{AJF}=\angle{AJG}}=24.95\degree$](1)</Checkpoint>

Therefore, the angle $\mathbf{\angle{JFA}=\angle{JGA}}$ of Ba-O-B is $180\degree-60\degree-24.95\degree=95.05\degree$

<Checkpoint>[$\mathbf{\angle{JFA}=\angle{JGA}}=95.05\degree$](1)</Checkpoint>

Obviously, $\mathbf{G},\mathbf{A},\mathbf{K}$ are collinear. In the triangle $\mathbf{JGK}$, from the law of cosines, the shortest distance $\mathbf{JK}$ (denoted as $d$) between the solid black sphere (Ba) and the hollow white sphere (anion center) is:

$d=\sqrt{d_{\mathrm{M-O}}^2+(2d_{\mathrm{B-O}})^2-2\cdot d_{\mathrm{M-O}}\cdot(2d_{\mathrm{B-O}})\cdot \cos{95.05\degree}}=411\ \mathrm{pm}$

<Checkpoint>[$d=411\ \mathrm{pm}$](1)</Checkpoint>

Therefore, the unit cell parameter $a$ is three times the shortest distance between the solid black sphere (Ba) and the hollow white sphere (anion), i.e., $a=3d=1233\ \mathrm{pm}$

<Checkpoint>[$a=1233\ \mathrm{pm}$](2)</Checkpoint>

In summary, option C should be selected for this question.
从阴离子点群入手，$D_{\mathrm{3h}}$点群可能是$\mathrm{BO_3^{3-}}$或$\mathrm{B_3O_6^{3-}}$，假设为前者，则没有合理的$\mathbf{M}$能够同时满足质量之比与氧化态，无合理解。因此阴离子只可能是$\mathrm{B_3O_6^{3-}}$，可求得$\mathbf{M}$为Ba，晶体最简式为$\mathrm{BaB_2O_4}$.

<MultiModal>![[O-]B1OB([O-])OB([O-])O1，即阴离子$\mathrm{B_3O_6^{3-}}$结构](/media/uploads/5a83bf1c-0d20-4385-8b05-f8f6362dbd0c.png)</MultiModal>

<Checkpoint>[$\mathbf{M}$为Ba](1)</Checkpoint>

<Checkpoint>[晶体最简式为$\mathrm{BaB_2O_4}$](1)</Checkpoint>

（在以下的描述中$\mathbf{M}$与Ba等价。）

$\mathbf{M}$作密堆积排布，阴离子以石墨的方式填充$\mathbf{M}$围成的三角形空隙，如下图所示，可以框出一个晶胞。

<MultiModal>![这是一张晶体的二维平面层示意图，由代表Ba的实心黑球和代表阴离子的空心白球组成。其中Ba原子作密堆积排布，阴离子以石墨的方式填充$\mathbf{M}$围成的三角形空隙。图像的下半部分用一个菱形轮廓线标示出了一个最小的二维周期结构（二维晶胞），该菱形的四个顶点都位于Ba原子的三角形空隙中心，菱形的边长为$\sqrt{3}$倍的Ba-Ba间距。在一个二维晶胞中包含有3个Ba原子和2个$\mathrm{B_3O_6^{3-}}$阴离子。](/media/uploads/7930a07c-c7a7-4c7a-9bc1-fffab8eaf09e.png)</MultiModal>

<Checkpoint>[最小的二维周期性结构中包含3个Ba原子和2个$\mathrm{B_3O_6^{3-}}$阴离子](1)</Checkpoint>

在每个晶胞中，Ba围成6个三角形空隙，而阴离子占据2个，阴离子填隙率为1/3.

<Checkpoint>[在每个晶胞中，Ba围成6个三角形空隙，而阴离子占据2个](1)</Checkpoint>

<Checkpoint>[阴离子填隙率为1/3](1)</Checkpoint>

阴离子$\mathrm{B_3O_6^{3-}}$中有6个氧原子，且为6齿配体，由每个阴离子周围有3个Ba可知，每两个相邻的O对一个Ba进行配位；而每个Ba周围有2个阴离子，因此Ba在层内的配位数为$2\times 2=4$.

<MultiModal>![[O]12B(O[Ba]2)[O]3B(O[Ba]3)[O]4B1O[Ba]4，仅表示配位连接方式](/media/uploads/f6437a80-8fe0-4ccd-8d73-8cd02db68ea3.png)</MultiModal>

<Checkpoint>[Ba在层内的配位数为4](1)</Checkpoint>

对于晶胞参数，先计算上图中实心黑球（Ba）与空心白球（阴离子）的最短距离：

<MultiModal>![本图中示出了一个阴离子和其最近的一个Ba的几何结构。图中$\mathbf{A},\mathbf{B},\mathbf{C},\mathbf{D},\mathbf{E},\mathbf{F}$构成正六边形，其中$\mathbf{A},\mathbf{C},\mathbf{E}$为硼原子，$\mathbf{B},\mathbf{D},\mathbf{F}$为氧原子。每个硼原子分别再向外连接一个氧原子，记为$\mathbf{G},\mathbf{H},\mathbf{I}$；$\mathbf{J}$为钡原子，$\mathbf{K}$为正六边形的中心。钡原子$\mathbf{J}$与氧原子$\mathbf{F},\mathbf{G}$配位。由于$\mathbf{AF=AG}$，$JA$平分$\mathbf{\angle{FAG}}$。求出$\mathbf{JK}$的长度，由二维晶胞示意图可知，$\mathbf{JK}$长度的三倍即为晶胞参数$a$](/media/uploads/ddcaf6a8-6ef5-42b7-ac2d-f1f7fb64a37b.png)</MultiModal>

由于Ba-O距离相同，因此$\mathbf{JA}$平分$\mathbf{\angle{FAG}}$，O-B-Ba的夹角$\mathbf{\angle{FAJ}}$为$180\degree-120\degree=60\degree$，由正弦定理可得B-Ba-O所形成的夹角$\mathbf{\angle{AJF}}$为$\arcsin{\frac{d_{\mathrm{B-O}}}{d_{\mathrm{M-O}}}\times \sin{60\degree}}=24.95\degree$

<Checkpoint>[$\mathbf{\angle{AJF}=\angle{AJG}}=24.95\degree$](1)</Checkpoint>

因此Ba-O-B的夹角$\mathbf{\angle{JFA}=\angle{JGA}}$为$180\degree-60\degree-24.95\degree=95.05\degree$

<Checkpoint>[$\mathbf{\angle{JFA}=\angle{JGA}}=95.05\degree$](1)</Checkpoint>

显然$\mathbf{G},\mathbf{A},\mathbf{K}$共线。在三角形$\mathbf{JGK}$中，由余弦定理，实心黑球（Ba）与空心白球（阴离子中心）的最短距离$\mathbf{JK}$（记为$d$）：

$d=\sqrt{d_{\mathrm{M-O}}^2+(2d_{\mathrm{B-O}})^2-2\cdot d_{\mathrm{M-O}}\cdot(2d_{\mathrm{B-O}})\cdot \cos{95.05\degree}}=411\ \mathrm{pm}$

<Checkpoint>[$d=411\ \mathrm{pm}$](1)</Checkpoint>

因此晶胞参数$a$为三倍实心黑球（Ba）与空心白球（阴离子）的最短距离，即$a=3d=1233\ \mathrm{pm}$

<Checkpoint>[$a=1233\ \mathrm{pm}$](2)</Checkpoint>

综上所述，本题应选择C选项。