The structure of sodium chloride is one of the most classic crystal structures. Under extremely high pressure, sodium chloride can combine with several molecules of chlorine to form $\mathrm { N a C l  _  x  }$. The figure shows a schematic diagram of the $\mathrm{NaCl_7}$ primitive cubic crystal cell, where the coordinates of two $\mathrm{Cl}$ atoms are $(0.5,0,0.1671), (0.5,0,0.8329)$.

<MultiModal>![This figure is the crystal structure diagram of $\mathrm{NaCl_7}$, where the yellow spheres are sodium atoms and the green spheres are chlorine ions. The edges of the cube are black solid lines, the yellow spheres are distributed at the vertices of the cube, there is one green sphere at the body center and two on each face, and the two green spheres on the face are connected by dashed lines, where the coordinates of two $\mathrm{Cl}$ atoms are $(0.5,0,0.1671), (0.5,0,0.8329)$](/media/uploads/f5dc9424-207c-474f-91c8-446821a8164f.png)</MultiModal>

Where the bond length of $\mathrm { C l _ 2 }$ in the crystal is $m { \AA }$.

Another $\mathrm { N a C l } _ { n }$ (substance $\bf{A}$) with a similar crystal structure to the above, where the coordination number of $\mathrm{Na}$ is the same as that of $\mathrm {  N a C l _ { 7 } }$ in $\mathrm{Na}$, and there is only one type of coordination number for $\mathrm {Cl}$.

The following statements are correct:
氯化钠的结构是最经典的晶体结构之一。在极高的压强下，氯化钠可以和数分子氯气化合生成 $\mathrm { N a C l  _  x  }$ 。图为 $\mathrm{NaCl_7}$ 正当立方晶系晶胞示意图，其中两个$\mathrm{Cl}$原子的坐标为$(0.5,0,0.1671)、(0.5,0,0.8329)$。

<MultiModal>![该图为$\mathrm{NaCl_7}$的晶体结构图，其中黄色的球为钠原子，绿色的球为氯离子。正方体的棱为黑色实线，黄色球分布在正方体的顶点，绿色的球在体心有一个、每个面上有两个，面上的两个绿色的球由虚线连接，其中两个$\mathrm{Cl}$原子的坐标为$(0.5,0,0.1671)、(0.5,0,0.8329)$](/media/uploads/f5dc9424-207c-474f-91c8-446821a8164f.png)</MultiModal>

其中 $\mathrm { C l _ 2 }$ 在晶体中的键长为 $m { \AA }$ 。

另一种与上述晶体结构相似的 $\mathrm { N a C l } _ { n }$ （物质$\bf{A}$），其中$\mathrm{Na}$的配位数与 $\mathrm {  N a C l _ { 7 } }$ 中的$\mathrm{Na}$相同，且$\mathrm {Cl}$的配位数只有一种。  

以下说法正确的是：
The coordination numbers of the body-centered $\mathrm {Cl}$ atom and the vertex $\mathrm{Na}$ atom are 6 and 6, respectively.
Body-centered $\mathrm {Cl}$ atom and vertex $\mathrm{Na}$ atom have coordination numbers of 8 and 8, respectively.
The crystal density of $\mathrm {NaCl_7}$ is $\frac{16.8}{m^3} \ g/cm^3$
$\mathrm {NaCl_7}$'s crystal density is $\frac{133}{m^3} \ g/cm^3$
The chemical formula of substance $\bf{A}$ is $\mathrm {NaCl_5}$
The chemical formula of $\bf{A}$ is $\mathrm {NaCl_6}$
The density of substance $\bf{A}$ is less than $\mathrm {NaCl_7}$.
$\mathrm {NaCl_7}$ can be described as the substance $\mathrm {NaCl·3Cl_2}$, then the substance $\bf{A}$ can be described as $\mathrm {NaCl·Cl_2}$
$\mathrm {NaCl_7}$ can be described as the substance $\mathrm {NaCl·3Cl_2}$, then the substance $\bf{A}$ can be described as $\mathrm {NaCl·2Cl_2}$
All of the above options are incorrect.
体心$\mathrm {Cl}$原子和顶点$\mathrm{Na}$原子的配位数分别为6和6
体心$\mathrm {Cl}$原子和顶点$\mathrm{Na}$原子的配位数分别为8和8
$\mathrm {NaCl_7}$的晶体密度为$\frac{16.8}{m^3} \ g/cm^3$
$\mathrm {NaCl_7}$的晶体密度为$\frac{133}{m^3} \ g/cm^3$
物质$\bf{A}$的化学式为$\mathrm {NaCl_5}$
物质$\bf{A}$的化学式为$\mathrm {NaCl_6}$
物质$\bf{A}$的密度比$\mathrm {NaCl_7}$小
$\mathrm {NaCl_7}$可以描述为物质$\mathrm {NaCl·3Cl_2}$，那么物质$\bf{A}$可以描述为$\mathrm {NaCl·Cl_2}$
$\mathrm {NaCl_7}$可以描述为物质$\mathrm {NaCl·3Cl_2}$，那么物质$\bf{A}$可以描述为$\mathrm {NaCl·2Cl_2}$
以上选项均不正确
<MultiModal>![This figure shows the crystal structure of $\mathrm{NaCl_7}$, where the yellow spheres are sodium atoms and the green spheres are chlorine ions. The edges of the cube are black solid lines, the yellow spheres are distributed at the vertices of the cube, and there is one green sphere at the body center and two on each face. The two green spheres on each face are connected by dashed lines, where the coordinates of two $\mathrm{Cl}$ atoms are $(0.5,0,0.1671)$ and $(0.5,0,0.8329)$](/media/uploads/f5dc9424-207c-474f-91c8-446821a8164f.png)</MultiModal>

As can be seen from the figure, the atoms closest to the sodium atom and the body-centered chlorine atom are both chlorine atoms belonging to chlorine gas. For the sodium atom at the vertex $(0,0,0)$, the closest one is a chlorine atom slightly above the edge center $(0,0.5,0.1671)$. There are 12 faces in total, with one on each face, so the coordination number of the sodium atom is 12;

<Checkpoint>[The coordination number of the sodium atom is 12](1)</Checkpoint>

For the body-centered chlorine atom, there are 2 atoms belonging to chlorine gas closest to each face, with a total of 6 faces, so the coordination number of the body-centered chlorine atom is 12.

<Checkpoint>[The coordination number of the body-centered chlorine atom is 12](1)</Checkpoint>

The bond length of $\mathrm { C l } _ { 2 }$ in the crystal is $m \ \mathrm { \AA }$. It should be noted that the two atoms constituting the chlorine gas are not the two atoms connected by dashed lines (the distance between them is $0.8329-0.1671=0.6658$), but rather the two chlorine atoms separated by an edge (the distance between them is $0.1671-（-0.1671）=0.3342$). Obviously, these two chlorine atoms are closer.

<Checkpoint>[Two chlorine atoms separated by an edge form chlorine gas](1)</Checkpoint>

Calculate the crystal density of $\mathrm {NaCl_7}$:

$$0.1671a=m/2$$

$$a=m/0.3242 {\AA} $$

$$D=\frac{ZM_r}{N_AV}=\frac{（22.99+35.45×7)g/mol}{6.022×10^{23}\ mol^{-1} ×(m/0.3242 {\AA})^3}=\frac{16.8}{m^3}\ g/cm^3$$

<Checkpoint>[Crystal density of $\mathrm {NaCl_7}$ is $$D=\frac{16.8}{m^3}\ g/cm^3$$](1)</Checkpoint>

Another $\mathrm { N a C l } _ { n }$ (substance $\bf{A}$) with a similar crystal structure to the above, where the coordination number of $\mathrm {Na}$ is the same as that of $\mathrm{Na}$ in $\mathrm { N a C l _ { 7 } }$, and the coordination number of $\mathrm {Cl}$ is only one type. The fact that the coordination number of $\mathrm {Cl}$ is only one type indicates that there is only one type of chlorine atom environment. In $\mathrm { N a C l } _ { 7 }$, there are chlorine atoms in chlorine gas and body-centered chlorine atoms. After replacing the body-centered chlorine atom with a sodium atom, the coordination number of the sodium atom remains at 12 (6 faces, with two chlorine atoms on each face), and the atoms adjacent to each chlorine atom are the other atom that makes up chlorine gas, with a coordination number of 1. The resulting substance $A$ has $\mathrm{Na}=8×1/8+1=2$ and $\mathrm{Cl}=12×1/2=6$ per unit cell, so the chemical formula of $A$ is $\mathrm { N a_2 C l_6 } $, which simplifies to $\mathrm { N a C l_3 } $

<Checkpoint>[The structure of substance $\bf{A}$ can be regarded as replacing the body-centered chlorine atom of $\mathrm { N a C l _ { 7 } }$ with a sodium atom](1)</Checkpoint>

<Checkpoint>[The chemical formula of $\bf{A}$ is $\mathrm { N a C l_3 } $](1)</Checkpoint>

Because the body-centered chlorine atom is replaced with a smaller sodium atom, the unit cell volume becomes smaller, and the density of $\bf{A}$ should be greater than that of $\mathrm { N a C l_7 } $

<Checkpoint>[The unit cell volume becomes smaller, and the density of $\bf{A}$ should be greater than that of $\mathrm { N a C l_7 } $](1)</Checkpoint>

$\mathrm { N a C l_7 } $ can be described as $\mathrm { N a C l·3Cl_2 } $, because there are indeed three chlorine gases in the unit cell, and the rest is $\mathrm { N a C l } $, but substance $A$($\mathrm { N a C l_3 } $) cannot be described by $\mathrm { N a C l·nCl_2 } $, because all chlorine atoms in the unit cell have the same chemical environment, and there is only one type of $\mathrm{Cl-Cl}$ bond length

<Checkpoint>[Substance $\bf{A}$($\mathrm { N a C l_3 } $) cannot be described by $\mathrm { N a C l·nCl_2 } $, because all chlorine atoms in the unit cell have the same chemical environment](1)</Checkpoint>

Therefore, option C is correct.
<MultiModal>![该图为$\mathrm{NaCl_7}$的晶体结构图，其中黄色的球为钠原子，绿色的球为氯离子。正方体的棱为黑色实线，黄色球分布在正方体的顶点，绿色的球在体心有一个、每个面上有两个，面上的两个绿色的球由虚线连接，其中两个$\mathrm{Cl}$原子的坐标为$(0.5,0,0.1671)、(0.5,0,0.8329)$](/media/uploads/f5dc9424-207c-474f-91c8-446821a8164f.png)</MultiModal>

从图中可以看出，距离钠原子和体心氯原子最近的原子都为属于氯气的氯原子。对于顶点处的钠原子$（0,0,0）$来说，距离最近的是处于棱心略靠上的一个氯原子$(0,0.5,0.1671)$，一共有12个面，每面上有一个，因此钠原子的配位数为12；

<Checkpoint>[钠原子的配位数为12](1)</Checkpoint>

对于体心氯原子来说，每个面上都有2个属于氯气的原子离得最近，一共6个面，体心氯原子的配位数为12。


<Checkpoint>[体心氯原子的配位数为12](1)</Checkpoint>

$\mathrm { C l } _ { 2 }$ 在晶体中的键长为 $m \ \mathrm { \AA }$ ，需要注意的是构成氯气的两个原子不是虚线相连的两个原子（两者之间的距离为$0.8329-0.1671=0.6658$），而应该是隔着一条棱的两个氯原子（两者之间的距离为$0.1671-（-0.1671）=0.3342$），显然这两个氯原子离得更近。

<Checkpoint>[隔着一条棱的两个氯原子组成氯气](1)</Checkpoint>

计算$\mathrm {NaCl_7}$的晶体密度：

$$0.1671a=m/2$$

$$a=m/0.3242 {\AA} $$

$$D=\frac{ZM_r}{N_AV}=\frac{（22.99+35.45×7)g/mol}{6.022×10^{23}\ mol^{-1} ×(m/0.3242 {\AA})^3}=\frac{16.8}{m^3}\ g/cm^3$$

<Checkpoint>[$\mathrm {NaCl_7}$的晶体密度$$D=\frac{16.8}{m^3}\ g/cm^3$$](1)</Checkpoint>

另一种与上述晶体结构相似的 $\mathrm { N a C l } _ { n }$ （物质$\bf{A}$），其中$\mathrm {Na}$的配位数与 $\mathrm {  N a C l _ { 7 } }$ 中的$\mathrm{Na}$相同，且$\mathrm {Cl}$的配位数只有一种。$\mathrm {Cl}$的配位数只有一种说明只有一种环境的氯原子，在$\mathrm { N a C l } _ { 7 }$中有氯气的氯原子和体心氯原子，将体心氯原子替换为钠原子后，仍旧保持了钠原子配位数12（6个面，每个面两个氯原子），而每个氯原子相邻的原子都是组成氯气的另一个原子，配位数为1。得到的物质$A$每个晶胞中$\mathrm{Na}=8×1/8+1=2$，$\mathrm{Cl}=12×1/2=6$，因此$A$的化学式为 $\mathrm { N a_2 C l_6 } $，化简后为$\mathrm { N a C l_3 } $

<Checkpoint>[物质$\bf{A}$的结构可视为$\mathrm {  N a C l _ { 7 } }$将体心氯原子替换为钠原子](1)</Checkpoint>

<Checkpoint>[$\bf{A}$的化学式为$\mathrm { N a C l_3 } $](1)</Checkpoint>

因为将体心氯原子换成半径更小的钠原子，因此晶胞体积变小，$\bf{A}$的密度应该比$\mathrm { N a C l_7 } $大

<Checkpoint>[晶胞体积变小，$\bf{A}$的密度应该比$\mathrm { N a C l_7 } $大](1)</Checkpoint>

$\mathrm { N a C l_7 } $可以描述为$\mathrm { N a C l·3Cl_2 } $，因为晶胞中确实有三个氯气，剩下的为$\mathrm { N a C l } $，但是物质$A$($\mathrm { N a C l_3 } $)不能用$\mathrm { N a C l·nCl_2 } $描述，因为晶胞中所有氯原子的化学环境相同，仅存在一种$\mathrm{Cl-Cl}$键长

<Checkpoint>[物质$\bf{A}$($\mathrm { N a C l_3 } $)不能用$\mathrm { N a C l·nCl_2 } $描述，因为晶胞中所有氯原子的化学环境相同](1)</Checkpoint>

因此选项C正确。