The crystal structure of $\mathbf{MN}\mathrm{O}_3$ is similar to a common crystal structure $\mathbf{X}$, and can be regarded as $\mathbf{M}$ and $\mathrm{O}$ jointly forming a cubic close-packed structure, with $\mathbf{N}$ filling the octahedral voids formed only by $\mathrm{O}$.

However, unlike ordinary $\mathbf{X}$, in the close-packed layer formed by $\mathbf{M}$ and $\mathrm{O}$, the position of $\mathbf{M}$ remains unchanged, but each equilateral triangle composed of $\mathrm{O}$ rotates clockwise or counterclockwise by the same and very small angle. It is known that the direction of rotation of every two $[\mathrm{O_3}]$ triangles sharing a vertex is different, and there are two such close-packed layers.

Select the correct option below.
$\mathbf{MN}\mathrm{O}_3$ 的晶体结构与一种常见的晶体结构 $\mathbf{X}$ 类似，可以看作 $\mathbf{M}$ 和 $\mathrm{O}$ 共同作立方最密堆积，$\mathbf{N}$ 填入只由 $\mathrm{O}$ 构成的八面体空隙。

然而与普通的 $\mathbf{X}$ 不同的是，由 $\mathbf{M}$ 和 $\mathrm{O}$ 所形成的密置层中，$\mathbf{M}$ 的位置不变，但每个由 $\mathrm{O}$ 构成的正三角形均顺时针或者逆时针旋转了一个相同且很小的角度。已知共顶点链接的每两个 $[\mathrm{O_3}]$ 三角形旋转的方向均不相同，共有两种这样的密置层。

选择以下正确的选项。
If $A$, $B$, $C$ represent one type of rotated close-packed layer, and $A'$, $B'$, $C'$ represent another type of close-packed layer, then in the $[\mathbf{MN}\mathrm{O}_3]$ structure, one stacking period of the close-packed layer is $\dots A'BCA' \dots$
The proper unit cell of this crystal contains 6 formula units, with one structural motif representing 1 formula unit.
Given that the average bond length of $\mathbf{N}-\mathrm{O}$ in the crystal is $191~\mathrm{pm}$, the volume of the primitive unit cell is $3.86 \times 10^{-22}~\mathrm{cm^3}$.
All $[\mathbf{N}\mathrm{O}_6]$ octahedra are connected by sharing vertices, and each $[\mathbf{N}\mathrm{O}_6]$ octahedron is connected to 3 $[\mathbf{N}\mathrm{O}_6]$ octahedra.
In this crystal, the same atoms have the same chemical environment and the same spatial environment.
If $\mathbf{M}$ and $\mathrm{O}$ are regarded as close-packed layers, then all $\mathbf{N}$ fill into the octahedral voids composed only of $\mathrm{O}$, with a filling fraction of $\dfrac{1}{3}$.
$\mathrm{O}$ exists and exists only in the middle of every two adjacent $\mathbf{N}$ atoms.
All other options are incorrect
如果用 $A$, $B$, $C$ 表示其中一种旋转后的密置层，$A'$, $B'$, $C'$ 表示另一种密置层，$[\mathbf{MN}\mathrm{O}_3]$ 结构中密置层的一个堆积周期为 $\dots A'BCA' \dots$
该晶体的正当晶胞包含6个化学式，一个结构基元代表1个化学式
已知该晶体中 $\mathbf{N}-\mathrm{O}$ 的平均键长为 $191~\mathrm{pm}$，则正当晶胞的体积为 $3.86 \times 10^{-22}~\mathrm{cm^3}$
所有的 $[\mathbf{N}\mathrm{O}_6]$ 八面体均共顶点连接，且每个 $[\mathbf{N}\mathrm{O}_6]$ 八面体均与 3 个 $[\mathbf{N}\mathrm{O}_6]$ 八面体相连。
该晶体中相同原子的化学环境相同，空间环境也相同。
若将 $\mathbf{M}$ 和 $\mathrm{O}$ 视为密置层，则所有 $\mathbf{N}$ 填入只由 $\mathrm{O}$ 构成的八面体空隙中，填隙率为 $\dfrac{1}{3}$。
$\mathrm{O}$ 存在且仅存在于每两个相邻的 $\mathbf{N}$ 原子中间。
其他选项均不正确
According to the problem description, $M$ and $O$ together form a cubic closest packing, and $N$ occupies the octahedral voids composed solely of $O$. This crystal structure is similar to the cubic perovskite structure, where $X$ is the cubic perovskite, with $M$ corresponding to $Ti$ and $N$ corresponding to $Ca$.

The stacking sequence with $N$ at the vertices in the cubic system is: $...ABCABCA...$

In the close-packed layers formed by $M$ and $O$, the positions of $M$ remain unchanged, but each equilateral triangle composed of $O$ is rotated by the same small angle, either clockwise or counterclockwise. It is known that every two adjacent $O_3$ triangles (within the same close-packed layer) sharing a vertex rotate in opposite directions. There are two such close-packed layers. To ensure that the two $O_3$ triangles forming the octahedral voids in adjacent layers rotate in the same direction, the two types of close-packed layers must alternate.

The stacking period of the close-packed layers in the $MNO_3$ structure is $...AB'CA'BC'A...$, which does not match the description of option A ($...A'BCA'...$), so option A is incorrect.

<Checkpoint>[The stacking period of the close-packed layers in the $MNO_3$ structure is $...AB'CA'BC'A...$, so option A is incorrect](1)</Checkpoint>

Based on the stacking period, it can also be determined that a unit cell contains 6 $N$ atoms, meaning it contains 6 formula units. The spatial environment is identical for every two adjacent layers, exhibiting translational symmetry. A structural motif consists of 2 formula units, which contradicts option B's description of a structural motif representing 1 formula unit. Thus, option B is incorrect.

<Checkpoint>[The proper unit cell of this crystal contains 6 formula units, and a structural motif represents 2 formula units, so option B is incorrect](1)</Checkpoint>

It is known that the average $N-O$ bond length in this crystal is $191pm$.

Let the edge length of the $\mathrm { NO } _ { 6 }$ octahedron be $d$. Then, the height of the octahedron is $h=\frac{\sqrt{6}d}{3}$.

It is straightforward to show that $a=2d$ and $c=6h=2\sqrt{6}d$. Therefore,  
$\frac{c}{a}=\sqrt{6}$

<Checkpoint>[$\frac{c}{a}=\sqrt{6}$](1)</Checkpoint>

$d=\sqrt{2}d(N-O)=270 pm$

<Checkpoint>[$d=\sqrt{2}d(N-O)=270 pm$](1)</Checkpoint>

The volume of the proper unit cell is  
$(270×2)^2 × 2\sqrt{6} × 270 × \frac{\sqrt{3}}{2} × 10^{-30} = 3.34×10^{-22} cm^3$

This contradicts option C's description of the proper unit cell volume as $3.86×10^{-22}cm^3$, so option C is incorrect.

<Checkpoint>[The volume of the proper unit cell is $3.34×10^{-22} cm^3$, so option C is incorrect](1)</Checkpoint>

All $\mathrm { NO } _ { 6 }$ octahedra are vertex-sharing, and each $\mathrm { NO _ { 6 } }$ octahedron is connected to 6 other $\mathrm { NO _ { 6 } }$ octahedra. This contradicts option D's description of each $\mathrm { NO _ { 6 } }$ octahedron being connected to 3 others, so option D is incorrect.

<Checkpoint>[All $\mathrm { NO } _ { 6 }$ octahedra are vertex-sharing, and each $\mathrm { NO _ { 6 } }$ octahedron is connected to 6 other $\mathrm { NO _ { 6 } }$ octahedra, so option D is incorrect](1)</Checkpoint>

From the environment of oxygen atoms in adjacent layers, it can be inferred that atoms of the same type in this crystal have identical chemical environments but different spatial environments. This contradicts option E's description of identical chemical and spatial environments for the same atoms, so option E is incorrect.

<Checkpoint>[In this crystal, atoms of the same type have identical chemical environments but different spatial environments, so option E is incorrect](1)</Checkpoint>

If $M$ and $O$ are treated as close-packed layers, all $N$ atoms occupy the octahedral voids composed solely of $O$. The octahedral voids composed solely of $O$ account for $\frac{1}{4}$ of all octahedral voids, resulting in an occupancy rate of $\frac{1}{4}$. This contradicts option F's description of an occupancy rate of $\frac{1}{3}$, so option F is incorrect.

<Checkpoint>[If $M$ and $O$ are treated as close-packed layers, all $N$ atoms occupy the octahedral voids composed solely of $O$, with an occupancy rate of $\frac{1}{4}$, so option F is incorrect](1)</Checkpoint>.

$N$ occupies the center of the octahedron, and $O$ forms the vertices of the octahedron. Since $N$ only exists in the octahedral voids formed by $O$, it follows that $O$ exists solely between every two adjacent $N$ atoms, meaning option G is correct.

<Checkpoint>[$O$ exists solely between every two adjacent $N$ atoms, so option G is correct](1)</Checkpoint>
根据题目描述， $\mathbf{M}$ 和 $\mathrm{O}$ 共同作立方最密堆积， $\mathbf{N}$ 填入只由 $\mathrm{O}$ 构成的八面体空隙。该晶体结构与立方钙钛矿结构相似，即 $\mathbf{X}$ 为立方钙钛矿，其中 $\mathbf{M}$ 对应 $\mathrm{Ti}$ ，$\mathbf{N}$ 对应 $\mathrm{Ca}$ 。

立方最密堆积中，以 $\mathbf{N}$ 为顶点， $\mathbf{M}$ 和 $\mathrm{O}$ 的密置层堆积方式可写为：$\dots ABCABCA \dots$

由 $\mathbf{M}$ 和 $\mathrm{O}$ 所形成的密置层中， $\mathbf{M}$ 的位置不变，但每个由 $\mathrm{O}$ 构成的正三角形均顺时针或者逆时针旋转了一个相同且很小的角度。已知共顶点链接的每两个 $[\mathrm{O_3}]$ 三角形（同一密置层内）旋转的方向均不相同。共有两种这样的密置层。为了满足上下相邻两个密置层中形成的八面体空隙的两个 $[\mathrm{O_3}]$ 三角形旋转的方向相同，则两种密置层交替分布。

$\mathbf{MN}\mathrm{O}_3$ 结构中密置层的一个堆积周期为 $\dots AB'CA'BC'A \dots$，与选项A描述的 $\dots A'BCA' \dots$ 不符，选项A错误。

<Checkpoint>[$\mathbf{MN}\mathrm{O}_3$ 结构中密置层的一个堆积周期为$\dots AB'CA'BC'A\dots $，所以选项A错误](1)</Checkpoint>

根据堆积周期也可以判断一个晶胞中包含6个 $\mathbf{N}$ ，即包含6个化学式，以上下相邻两层为一个单位的空间环境相同，具有平移对称性，一个结构基元包含2个化学式，与选项B描述的一个结构基元代表1个化学式不符，选项B错误。

<Checkpoint>[该晶体的正当晶胞包含6个化学式，一个结构基元代表2个化学式，所以选项B错误](1)</Checkpoint>

已知该晶体中 $\mathbf{N}$ 一 $\mathrm{O}$ 的平均键长为 $191~\mathrm{pm}$ ，

设 $\mathbf { N } \mathrm{O} _ { 6 }$ 八面体的边长为 $d$，则八面体的高 $h=\frac{\sqrt{6}d}{3}$

易知 $a=2d$，$c=6h=2\sqrt{6}d$ 故
$\frac{c}{a}=\sqrt{6}$

<Checkpoint>[$\frac{c}{a}=\sqrt{6}$](1)</Checkpoint>

$d=\sqrt{2}d(N-O)=270 ~\mathrm{pm}$

<Checkpoint>[$d=\sqrt{2}d(N-O)=270 ~\mathrm{pm}$](1)</Checkpoint>

正当晶胞的体积为
$(270×2)^2 × 2\sqrt{6} × 270 × \frac{\sqrt{3}}{2} × 10^{-30} = 3.34×10^{-22} ~\mathrm{cm^3}$

与选项C描述的正当晶胞的体积为$3.86×10^{-22} ~\mathrm{cm^3}$不符，选项C错误。

<Checkpoint>[正当晶胞的体积为$3.34×10^{-22} ~\mathrm{cm^3}$，所以选项C错误](1)</Checkpoint>

所有的 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体均共顶点连接，且每个 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体均与6个 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体相连，与选项D描述的每个 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体均与3个 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体相连不符，选项D错误。

<Checkpoint>[所有的 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体均共顶点连接，且每个 $[\mathbf{N}\mathrm{O}_{6}]$ 八面体均与 6 个 $[\mathbf{N}\mathrm{O}_{6}]$八面体相连，所以选项D错误](1)</Checkpoint>

从相邻两层的氧原子的环境可以得到，该晶体中相同原子的化学环境相同，空间环境不同，与选项E描述的该晶体中相同原子的化学环境相同，空间环境也相同不符，选项E错误。

<Checkpoint>[该晶体中相同原子的化学环境相同，空间环境不同，所以选项E错误](1)</Checkpoint>

若将 $\mathbf{M}$ 和 $\mathrm{O}$ 视为密置层，则所有 $\mathbf{N}$ 填入只由 $\mathrm{O}$ 构成的八面体空隙中，只由 $\mathrm{O}$ 构成的八面体空隙占所有八面体空隙的 $\frac{1}{4}$，填隙率为 $\frac{1}{4}$，与选项F描述的填隙率为 $\frac{1}{3}$ 不符，选项F错误。

<Checkpoint>[若将 $\mathbf{M}$ 和 $\mathrm{O}$ 视为密置层，则所有 $\mathbf{N}$ 填入只由 $\mathrm{O}$ 构成的八面体空隙中，填隙率为 $\frac{1}{4}$，所以选项F错误](1)</Checkpoint>


$\mathbf{N}$ 占据八面体中心， $\mathrm{O}$ 是八面体的顶点，且 $\mathbf{N}$ 仅存在于 $\mathrm{O}$ 形成的八面体空隙中，因此 $\mathrm{O}$ 仅存在于每两个相邻 $\mathbf{N}$ 原子之间是正确的，即选项G正确。

<Checkpoint>[ $\mathrm{O}$ 存在且仅存在于每两个相邻的 $\mathbf{N}$ 原子中间，所以选项G正确](1)</Checkpoint>