Now there is a cubic crystal $\bf{A}$, which contains a cuboctahedral ($4^6 3^8$) polyhedron atomic cluster $\bf{Z}$ formed by boron atoms. The structure of $\bf{A}$ can be regarded as $\bf{Z}$ forming a NaCl-type structure with metal atom $\bf{X}$. It is known that all adjacent $\bf{Z}$ are connected by B-B bonds, and their bond lengths are all 169.2pm; all B-B bonds inside $\bf{Z}$ have bond lengths of 178.8pm. The theoretical density of the crystal is $5.908 \mathrm { g } / \mathrm { c m } ^ { 3 }$. The crystal can also be regarded as the result of $\bf{X}$ atoms being encapsulated by another atomic cluster $\bf{Y}$.

Then the following statements are correct:
今有一立方晶系的晶体$\bf{A}$，其中含有由硼原子形成的立方八面体（$4^6 3^8$）多面体原子簇$\bf{Z}$。  $\bf{A}$的结构可以视作 $\bf{Z}$ 与金属原子$\bf{X}$形成NaCl型结构。已知$\bf{Z}$与邻近的所有 $\bf{Z}$通过B-B键相连，其键长均为169.2pm；$\bf{Z}$内部所有的B-B键键长均为178.8pm。该晶体的理论密度为$5.908 \mathrm { g } / \mathrm { c m } ^ { 3 }$ 。该晶体还可以看作$\bf{X}$原子被另一种原子簇$\bf{Y}$包合的结果。

则下列说法正确的是：
All other options are incorrect
The ratio of two elements in the chemical formula of $\bf{A}$ is 1:8.
The angle formed by the triangular face in $\bf{Z}$ and the (110) crystal plane is unique.
The coordination number of metal $\bf{X}$ in $\bf{Z}$ is 16.
$\bf{A}$ unit cell body diagonal length is greater than 1300 pm.
$\bf{X}$ The mass number of the stable nuclide produced finally by the natural radioactive series to which the most abundant natural isotope belongs is 208.
The structure of $\bf{Y}$ can be represented by the symbol $4^6 6^8$.
其他选项均不正确
$\bf{A}$的化学式中两种元素比例为 1:8.
$\bf{Z}$中的三角形面与(110)晶面形成的角度只有一种。
$\bf{Z}$中金属$\bf{X}$的配位数为16。
$\bf{A}$晶胞体对角线长度大于1300 pm。
$\bf{X}$天然丰度最高的同位素所属天然放射系最终产生的稳定核素的质量数为208.
$\bf{Y}$的结构可以用符号$4^6 6^8$表示。
### Basic Unit $\bf{Z}$:
A cuboctahedral ($4^6 3^8$) atomic cluster composed of boron atoms (B). The cuboctahedron has 12 vertices, 8 equilateral triangular faces, and 6 square faces. Therefore, $\bf{Z}$ is a $B_{12}$ atomic cluster.

<Checkpoint>[The chemical formula of $\bf{Z}$ is $B_{12}$](1)</Checkpoint>

### Crystal Structure:
The structure of $\bf{A}$ can be regarded as $\bf{Z}$ and the metal atom $\bf{X}$ forming a NaCl-type structure. In the NaCl structure, the stoichiometric ratio of cations and anions (or units) is 1:1. This means that the number ratio of $\bf{Z}$ to $\bf{X}$ is 1:1, so the chemical formula of $\bf{A}$ should be $\bold{X}B_{12}$, with a ratio of the two elements being 1:12. Option B is incorrect.

<Checkpoint>[The ratio of the two elements in $\bf{A}$ is 1:12](1)</Checkpoint>

According to the cubic crystal symmetry, the six quadrilateral faces in the atomic cluster $\bf{Z}$ are perpendicular to the a, b, and c axes, respectively. Therefore, it is easy to have two angles for the quadrilateral faces, one perpendicular to the (110) crystal plane and the other non-perpendicular. Option C is incorrect.

<Checkpoint>[The triangular faces in $\bf{Z}$ form 2 types of angles with the [110] crystal plane](1)</Checkpoint>

According to the description, there are 6 $\bf{Z}$ clusters coordinating around $\bf{X}$. Each cluster uses a quadrilateral face for coordination, so the coordination number of $\bf{X}$ is $4*6=24$. Option D is incorrect.

<Checkpoint>[The coordination number of the metal $\bf{X}$ in $\bf{Z}$ is 24](1)</Checkpoint>

The lattice parameter a is determined by $\sqrt2$ times the distance between the centers of two adjacent $B_{12}$ clusters, and the distance between the centers of two adjacent $B_{12}$ clusters is equal to the sum of the B-B bond length within a cluster and the B-B bond length between two clusters. Therefore, the lattice parameter of $\bf{Z}$ is $a=169.2*\sqrt2+178.8*\sqrt2*2=745.0pm$, so the diagonal length is $745.0*\sqrt3=1290.3 pm$. Option E is incorrect.

<Checkpoint>[The body diagonal length of the $\bf{A}$ unit cell is 1290 pm.](1)</Checkpoint>

The body diagonal length of the unit cell is less than 1300 pm.

### $\bf{X}$ Element Deduction:
Use the density formula to calculate the molar mass of $\bf{A}$. From $\rho=\frac{Z\cdot M}{N_A\cdot V}$, we get $M_{\bold{X}B_{12}}=\frac{\rho\cdot  N_A \cdot V}{Z}=\frac{5.908 *N_A*7.450^{3} * 10^{-24}}{4}=367.8g/\mathrm{mol}$, thus $M_\bold{X}=367.8-12\cdot M_\bold{B}=238.1$, $\bf{X}$ is the element U.

<Checkpoint>[$\bf{X}$ is the element uranium (U)](1)</Checkpoint>

The mass number of the stable nuclide finally produced by the natural radioactive series of its most abundant natural isotope, $\mathrm{^{238}U}$, is 206 (i.e., $^{206}Pb$). Option F is incorrect.

<Checkpoint>[The mass number of the stable nuclide finally produced by $\mathrm{^{238}U}$ is 206](1)</Checkpoint>

### $\bf{Y}$ Structure Deduction
According to the coordination mode, there are 6 quadrilateral faces in $\bf{Y}$. These six faces can be regarded as being on the six vertices of an octahedron. From this, it can be seen that the three-dimensional structure is a truncated octahedron. Therefore, $\bf{Y}$ has 6 quadrilateral faces and 8 hexagonal faces, which can be represented by the symbol $4^6 6^8$. Option G is correct.

<Checkpoint>[The structure of $\bf{Y}$ can be represented by the symbol $4^6 6^8$.](1)</Checkpoint>

## Therefore, the correct option is: G
### 基本单元$\bf{Z}$: 
由硼原子(B)构成的立方八面体($4^6 3^8$
 )原子簇。立方八面体有12个顶点，8个正三角形面，6个正方形面。因此，$\bf{Z}$是一个$B_{12}$ 原子簇。

<Checkpoint>[$\bf{Z}$的化学式为$B_{12}$](1)</Checkpoint>

### 晶体结构: 
$\bf{A}$的结构可以看作$\bf{Z}$与金属原子$\bf{X}$形成NaCl型结构。在NaCl结构中，阴阳离子（或单元）化学计量比为1:1。这意味着$\bf{Z}$与$\bf{X}$的个数比为1:1，因此$\bf{A}$的化学式应该为$\bold{X}B_{12}$，两种元素比例为1:12。选项B错误。

<Checkpoint>[$\bf{A}$中两种元素比例为1:12](1)</Checkpoint>

根据立方晶体对称性可知原子簇$\bf{Z}$中六个四边形面分别垂直与a，b，c轴，因此易有四边形面存在垂直于(110)晶面和非垂直两种角度。选项C错误。

<Checkpoint>[$\bf{Z}$中的三角形面与[110]晶面形成的角度有2种](1)</Checkpoint>

根据描述可知$\bf{X}$周围有6个$\bf{Z}$团簇对其进行配位。每个团簇使用四边形面进行配位，因此$\bf{X}$的配位数为$4*6=24$。选项D错误。


<Checkpoint>[$\bf{Z}$中金属$\bf{X}$的配位数为24](1)</Checkpoint>


晶胞参数 a 是由两个相邻的$B_{12}$簇中心距的$\sqrt2$倍决定的，而两个相邻的$B_{12}$簇中心距等于一个簇内 B-B 键长和两个簇间 B-B 键长之和。因此$\bf{Z}$的晶胞参数为$a=169.2*\sqrt2+178.8*\sqrt2*2=745.0pm$，因此对角线长度为$745.0*\sqrt3=1290.3 pm$。选项E错误。


<Checkpoint>[$\bf{A}$晶胞体对角线长度为1290 pm。](1)</Checkpoint>


晶胞体对角线长度小于1300 pm。

### $\bf{X}$元素推断：
使用密度公式来计算$\bf{A}$的摩尔质量。由$\rho=\frac{Z\cdot M}{N_A\cdot V}$ 得 $M_{\bold{X}B_{12}}=\frac{\rho\cdot  N_A \cdot V}{Z}=\frac{5.908 *N_A*7.450^{3} * 10^{-24}}{4}=367.8g/\mathrm{mol}$，从而$M_\bold{X}=367.8-12\cdot M_\bold{B}=238.1$，$\bf{X}$为元素U。


<Checkpoint>[$\bf{X}$为铀元素（U）](1)</Checkpoint>


其天然丰度最高的同位素为$\mathrm{^{238}U}$所属天然放射系最终产生的稳定核素的质量数为206（即$^{206}Pb$）。选项F错误。

<Checkpoint>[$\mathrm{^{238}U}$最终产生的稳定核素的质量数为206](1)</Checkpoint>

### $\bf{Y}$的结构推断
根据配位方式可知$\bf{Y}$中存在6个四边形面，这六个面可以看作处于一个八面体的六个顶点上，由此可以看出该立体结构为截角八面体，因此$\bf{Y}$中有6个四边形面，8个六边形面，可以用符号$4^6 6^8$表示。选项G正确。


<Checkpoint>[$\bf{Y}$的结构可以用符号$4^6 6^8$表示。](1)</Checkpoint>


## 因此正确选项为：G