Adamantane frameworks are not only found in organic compounds, but also commonly seen in inorganic clusters, of which $\mathrm { { P _ { 4 } O _ { 10 } } }$ is a typical representative.

1.  Dissolve $\mathrm { G e B r _ { 4 } }$ in $\mathrm { C S } _ { 2 }$ containing $\mathrm {  A l _ { 2 } B r _ { \mathrm { 6 } } }$, and then introduce $\mathrm { H } _ { 2 } \mathrm { S }$ gas into it to obtain a similar compound $\mathbf A$, containing $\omega _ { \mathrm { G e } } = 0 . 3620$.

2.  The structure of $\mathrm { P } _ { 4 } \mathrm { O } _ { 10 }$ can be described as $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$, that is, a set of interpenetrating $\mathrm { { P _ { 4 } } }$ polyhedra and ${ { \mathrm { O } } _ { 6 } }$ polyhedra encapsulated in $\mathrm { O } _ { 4 }$ polyhedra. Try to describe the structure of adamantane (substance $\mathbf B$) in this way, and give the corresponding polyhedra for each layer.

3.  Adding more atoms to the periphery of the adamantane framework can result in more complex cluster structures. There is a highly symmetrical anion $[ \mathrm { M o } _ { m } \mathrm { S } _ { x } \mathrm { C u } _ { n } \mathrm { C l } _ { y } ] ^ { 4 - }$ (substance $\mathbf C$), containing $\omega _ { \mathrm { C u } } = 0 . 4945$, $\omega _ { \mathrm { M o } } = 0 . 0747$, and each element has a common valence state. Where $\mathrm{Cl}$ is 2-coordinate and has only one chemical environment, and S is 4-coordinate.

Which of the following statements is correct?
金刚烷骨架不仅存在于有机化合物中，在无机簇合物里也常见到这样的骨架，其中$\mathrm { { P _ { 4 } O _ { 1 0 } } }$ 是此类化合物的一个典型代表。

1.将 $\mathrm { G e B r _ { 4 } }$ 溶于含 $\mathrm {  A l _ { 2 } B r _ { \mathrm { 6 } } }$ 的 $\mathrm { C S } _ { 2 }$ ，再向其中通入 $\mathrm { H } _ { 2 } \mathrm { S }$ 气体，得到了一种相似的化合物$\mathbf A$，其中含 $\omega _ { \mathrm { G e } } = 0 . 3 6 2 0$ 。

2.$\mathrm { P } _ { 4 } \mathrm { O } _ { 1 0 }$ 的结构可描述为 $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$ ，即一组互相穿插的 $\mathrm { { P _ { 4 } } }$ 多面体和 ${ { \mathrm { O } } _ { 6 } }$ 多面体包裹于 $\mathrm { O } _ { 4 }$ 多面体中。试以此方法描述金刚烷（物质$\mathbf B$）的结构，给出每一层对应的多面体。

3.向金刚烷骨架外围增加更多原子，可以得到更加复杂的簇结构。有一种高对称性的阴离子 $[ \mathrm { M o } _ { m } \mathrm { S } _ { x } \mathrm { C u } _ { n } \mathrm { C l } _ { y } ] ^ { 4 - }$（物质$\mathbf C$） ，含 $\omega _ { \mathrm { C u } } = 0 . 4 9 4 5$ ， $\omega _ { \mathrm { M o } } = 0 . 0 7 4 7$ ，各元素均为常见价态。其中$\mathrm{Cl}$均为2配位且只有一种化学环境，S为4配位。

以下说法正确的是
The structure of $\mathbf A$ is $\mathrm {Ge_4S_4Br_6}$
$\mathbf A$ contains the element $\mathrm {Al}$
In $\mathbf A$, a $\mathrm {Ge}$ atom connects to two $\mathrm {S}$ atoms and a $\mathrm {Br}$ atom.
Following the notation of $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$, substance $\mathbf B$ can be written as $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 8 }  \mathrm { H } _ { 8 }$, and the polyhedra from left to right are a regular tetrahedron, a regular octahedron, a regular hexahedron, and a regular hexahedron.
Following the notation of $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$, the substance $\mathbf B$ can be written as $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 12 }  \mathrm { H } _ { 4 }$, the polyhedra from left to right are regular tetrahedron, regular octahedron, regular icosahedron, regular tetrahedron.
In substance $\mathbf C$, $ m=1,n=10,x=2,y=14$
Substance $\mathbf C$ with $ m=1,n=10,x=3,y=13$
In substance $\mathbf C$, $m=1,n=10,x=6,y=10$.
Each $\mathrm{Cu}$ atom in substance $\mathbf C$ is connected to $\mathrm{S}$.
Substance $\mathbf C$ is divided into core and periphery, the core being $\mathrm{Mo@S_4{\cdot}Cu_6}$, and the periphery being $\mathrm { Cu } _ { 4 } { \cdot } \mathrm { Cl } _ { 12 }$
All of the above options are incorrect.
$\mathbf A$的结构为$\mathrm {Ge_4S_4Br_6}$
$\mathbf A$中含有$\mathrm {Al}$元素
$\mathbf A$中一个$\mathrm {Ge}$原子连接两个$\mathrm {S}$原子和$\mathrm {Br}$原子
按照 $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$ 的写法，物质$\mathbf B$可以写为 $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 8 }  \mathrm { H } _ { 8 }$，多面体从左至右为正四面体、正八面体、正六方体、正六方体
按照 $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$ 的写法，物质$\mathbf B$可以写为 $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 12 }  \mathrm { H } _ { 4 }$，多面体从左至右为正四面体、正八面体、正二十面体、正四面体
物质$\mathbf C$中$ m=1,n=10,x=2,y=14$
物质$\mathbf C$中$ m=1,n=10,x=3,y=13$
物质$\mathbf C$中$ m=1,n=10,x=6,y=10$
物质$\mathbf C$中每个$\mathrm{Cu}$原子都和$\mathrm{S}$相连
物质$\mathbf C$分为核心和外围，核心为$\mathrm{Mo@S_4{\cdot}Cu_6}$，外围为$\mathrm { Cu } _ { 4 } { \cdot } \mathrm { Cl } _ { 12 }$
以上选项均不正确
First question

<MultiModal>![Structure diagram of substance A, composed of Ge4S6Br4, where Ge is connected to 3 S and 1 Br](/media/uploads/85774f10-8194-43bc-99e6-d0bb6a7e63c8.png)</MultiModal>

According to the question, the structure of substance $\mathbf A$ is consistent with that of $\mathrm{P_4O_{10}}$. The possible constituent elements of this substance are $\mathrm{Ge、S、Br}$, where $\mathrm{Ge}$ is most likely to replace the position of $\mathrm{P}$, $\mathrm S$ should be 2-coordinated and located on the bridge, and $\mathrm {Br}$ is more suitable as a terminal group. The resulting substance has the composition $\mathrm{Ge_4S_6Br_4}$.

Verification of $\omega _ { \mathrm { G e } } $: The relative molecular mass of the chemical formula is $4 × 72.63 + 6 × 32.07 + 4 × 79.90 = 290.52 + 192.42 + 319.60 = 802.54$

Then $\omega _ { \mathrm { G e } } = (4 × 72.63) / 802.54 = 290.52 / 802.54 ≈ 0.3620$, which meets the requirements of the question.

Therefore, the composition of substance $\mathbf A$ is $\mathrm{Ge_4S_6Br_4}$, where $\mathrm{Ge}$ is connected to 3 $\mathrm S$ and 1 $\mathrm{Br}$, ABC are incorrect.

<Checkpoint>[The composition of substance $\mathbf A$ is $\mathrm{Ge_4S_6Br_4}$, where $\mathrm{Ge}$ is connected to 3 $\mathrm S$ and 1 $\mathrm{Br}$, ABC are incorrect](1)</Checkpoint>

Second question

<MultiModal>![Structure diagram of adamantane, which can be described as C4·C6@H12·H4, from left to right are regular tetrahedron, regular octahedron, truncated regular tetrahedron, regular tetrahedron](/media/uploads/0306e901-ccda-4bda-9c3b-4b9c61011ed4.png)</MultiModal>

According to the writing method of $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$, substance $\mathbf B$ can be written as $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 12 }  \mathrm { H } _ { 4 }$. The polyhedra from left to right are regular tetrahedron, regular octahedron, truncated regular tetrahedron, regular tetrahedron, DE are incorrect.

<Checkpoint>[According to the writing method of $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$, substance $\mathbf B$ can be written as $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 12 }  \mathrm { H } _ { 4 }$. The polyhedra from left to right are regular tetrahedron, regular octahedron, truncated regular tetrahedron, regular tetrahedron, DE are incorrect](1)</Checkpoint>

Third question

$\mathrm {Cu : Mo= (0.4945/63.55) : (0.0747/95.96)= 10:1} $

<Checkpoint>[$\mathrm {Cu : Mo= 10:1} $](1)</Checkpoint>

Assuming that it contains only 1 $\mathrm{Mo}$, i.e., $m=1$, then the sum of the molar masses of $\mathrm S$ and $\mathrm {Cl}$ in this ion is:

$10×63.55×(1 - 0.4945)/0.4945 - 95.96= 553.7$

<Checkpoint>[The sum of the molar masses of $\mathrm S$ and $\mathrm {Cl}$ is 553.7](1)</Checkpoint>

$32.07x+35.45y=553.7$

The integer solution can be obtained by listing: $x = 4 , y = 1 2 $
At this time, $\mathrm {Cl}$ is $-1$ valence, $\mathrm {S}$ is $-2$ valence, $\mathrm {Cu}$ is $+ 1$ valence, and $\mathrm {Mo}$ is $+ 6$ valence, which are in line with common valence states.
If $m>1$, then $0$ valence $\mathrm {Cu}$ or non-$+6$ valence $\mathrm {Mo}$ will appear, so it is discarded.
Therefore, $m=1$, $n = 1 0$, $x=4$, $y = 1 2$, FGH are incorrect.

<Checkpoint>[ $m=1$ ，$n = 1 0$ ，$x=4$， $y = 1 2$，FGH are incorrect ](1)</Checkpoint>

<MultiModal>![Structure diagram of MoCu10S4Cl12, where the core is Mo@S4·Cu6, and the periphery is Cu4·Cl12](/media/uploads/1540f2ef-5834-448a-b876-33eedc2b8865.png)</MultiModal>

Substance $\mathbf C$ is $\mathrm {MoCu_{10}S_4Cl_{12}}$, where the core is $\mathrm{Mo@S_4{\cdot}Cu_6}$, and the periphery is $\mathrm{Cu_4{\cdot}Cl_{12}}$. The peripheral $\mathrm {Cu}$ is only connected to $\mathrm {Cl}$, I is incorrect, and J is correct.

<Checkpoint>[Substance $\mathbf C$ is $\mathrm {MoCu_{10}S_4Cl_{12}}$, where the core is $\mathrm{Mo@S_4{\cdot}Cu_6}$, and the periphery is $\mathrm{Cu_4{\cdot}Cl_{12}}$, I is incorrect, and J is correct](1)</Checkpoint>
第一题

<MultiModal>![物质A的结构图，组成为Ge4S6Br4，其中Ge和3个S和1个Br相连](/media/uploads/85774f10-8194-43bc-99e6-d0bb6a7e63c8.png)</MultiModal>

按照题目所说，物质$\mathbf A$的结构和$\mathrm{P_4O_{10}}$一致，该物质可能的组成元素为$\mathrm{Ge、S、Br}$，其中$\mathrm{Ge}$最可能替代$\mathrm{P}$的位置，$\mathrm S$应该2配位，处于桥上，$\mathrm {Br}$更适合作为端基，这样得到的物质组成为$\mathrm{Ge_4S_6Br_4}$

验算$\omega _ { \mathrm { G e } } $ :化学式相对分子质量为$4 × 72.63 + 6 × 32.07 + 4 × 79.90 = 290.52 + 192.42 + 319.60 = 802.54$ 

那么$\omega _ { \mathrm { G e } } = (4 × 72.63) / 802.54 = 290.52 / 802.54 ≈ 0.3620$，符合题目要求

所以物质$\mathbf A$的组成为$\mathrm{Ge_4S_6Br_4}$，其中$\mathrm{Ge}$和3个$\mathrm S$和1个$\mathrm{Br}$相连，ABC错误

<Checkpoint>[物质$\mathbf A$的组成为$\mathrm{Ge_4S_6Br_4}$，其中$\mathrm{Ge}$和3个$\mathrm S$和1个$\mathrm{Br}$相连，ABC错误](1)</Checkpoint>

第二题

<MultiModal>![金刚烷的结构图，可以描述为C4·C6@H12·H4，从左至右分别为正四面体、正八面体、截角正四面体、正四面体](/media/uploads/0306e901-ccda-4bda-9c3b-4b9c61011ed4.png)</MultiModal>

按照 $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$ 的写法，物质$\mathbf B$可以写为 $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 12 }  \mathrm { H } _ { 4 }$，多面体从左至右为正四面体、正八面体、截角正四面体、正四面体，DE错误

<Checkpoint>[按照 $\mathrm { P } _ { 4 } { \cdot } \mathrm { O } _ { 6 } @ \mathrm { O } _ { 4 }$ 的写法，物质$\mathbf B$可以写为 $\mathrm { C } _ { 4 }{ \cdot } \mathrm{C}_{6}@\mathrm { H } _ { 12 }  \mathrm { H } _ { 4 }$，多面体从左至右为正四面体、正八面体、截角正四面体、正四面体，DE错误](1)</Checkpoint>

第三题

$\mathrm {Cu : Mo= (0.4945/63.55) : (0.0747/95.96)= 10:1} $

<Checkpoint>[$\mathrm {Cu : Mo= 10:1} $](1)</Checkpoint>

假设只含1个$\mathrm{Mo}$，即 $m=1$ ,则该离子中$\mathrm S$与$\mathrm {Cl}$摩尔质量和为：  

$10×63.55×(1 - 0.4945)/0.4945 - 95.96= 553.7$

<Checkpoint>[$\mathrm S$与$\mathrm {Cl}$摩尔质量和为553.7](1)</Checkpoint>

$32.07x+35.45y=553.7$

列表可解得整数解： $x = 4 , y = 1 2 $   
此时$\mathrm {Cl}$为$-1$价，$\mathrm {S}$为$-2$价，$\mathrm {Cu}$为 $+ 1$ 价，$\mathrm {Mo}$为 $+ 6$ 价，符合常见价态。  
若 $m>1$ ，则会出现$0$价 $\mathrm {Cu}$或非 $+6$价$\mathrm {Mo}$，故舍去。  
因此， $m=1$ ，$n = 1 0$ ，$x=4$， $y = 1 2$ ,FGH错误

<Checkpoint>[ $m=1$ ，$n = 1 0$ ，$x=4$， $y = 1 2$，FGH错误 ](1)</Checkpoint>

<MultiModal>![MoCu10S4Cl12的结构图，其中核心为Mo@S4·Cu6，外围为Cu4·Cl12](/media/uploads/1540f2ef-5834-448a-b876-33eedc2b8865.png)</MultiModal>

物质$\mathbf C$为$\mathrm {MoCu_{10}S_4Cl_{12}}$,其中核心为$\mathrm{Mo@S_4{\cdot}Cu_6}$，外围为$\mathrm{Cu_4{\cdot}Cl_{12}}$，外围$\mathrm {Cu}$只和$\mathrm {Cl}$相连，I错误，J正确

<Checkpoint>[物质$\mathbf C$为$\mathrm {MoCu_{10}S_4Cl_{12}}$,其中核心为$\mathrm{Mo@S_4{\cdot}Cu_6}$，外围为$\mathrm{Cu_4{\cdot}Cl_{12}}$，I错误，J正确](1)</Checkpoint>