It has been discovered that alkali metal oxides can induce the disproportionation reaction of gold. Dissolving $\mathrm{Rb_2O}$ in molten rubidium metal, adding gold metal, and heating to $200^{\circ}\mathrm{C}$ for 14 days yields an orthorhombic crystal. The structure of this crystal is shown below and can be viewed as a stacking of two types of layers, A and B. In the diagram, all cube vertices represent a rubidium atom, each hollow cube center contains a rubidium atom, while each white and black cube center contains a gold atom; oxygen atoms are located at the interfaces between all white cubes and hollow cubes. The structure is as follows:

<MultiModal>![On the left is a stacking structure composed of hollow cube frameworks, black cubes, and white cubes, and on the right is its layered decomposition, with a black arrow pointing from the right to the left. In the left stacking structure, the left-right direction aligns with the cube edges, while the up-down and front-back directions align with the cube face diagonals. In the right decomposed structure, from top to bottom, it splits into five layers: A, B, A, B, A. Layer A consists of hollow cube frameworks and white cubes, with the front-back direction showing the same type of cubes sharing edges and the left-right direction showing different types of cubes sharing faces. Adjacent A layers are staggered left-right for the same type of cubes. Layer B consists solely of black cubes, with the front-back direction forming a zigzag structure through face-sharing and the left-right direction also face-sharing. In the left stacking structure, adjacent black cube layers are connected via shared edges, while adjacent white-hollow cube layers have no direct connection.](/media/uploads/ee2f93e9-b4e6-4e4e-b482-eabb439a9024.png)</MultiModal>

Calculate the variance of the oxidation states of gold in this crystal.
人们发现碱金属氧化物可以引发金的歧化反应。将$\mathrm{Rb_2O}$溶解在熔融的铷单质中，加入金单质并加热至$200^{\circ}\mathrm{C}$，反应14天，得到一种正交晶系的晶体。该晶体的结构如下图所示，可看做由A、B两种层堆叠而成。图中所有立方体的顶点均代表一个铷原子，每个空心立方体的中心均有一个铷原子，而每个白色立方体和黑色立方体的中心均有一个金原子；氧原子位于所有白色立方体和空心立方体的交界面上。其结构如下：

<MultiModal>![左边为由空心立方体框架、黑色立方体、百色立方体构成的堆积结构，右边为其层状拆分，中间有一黑色箭头从右侧指向左侧，在左侧的堆积结构中，左右方向沿立方体的棱方向，上下和前后方向沿立方体的面对角线方向，在右侧的拆分结构中，从上至下拆分为A,B,A,B,A五层，A层包含空心立方体框架和白色立方体，前后方向为同种立方体共棱连接，左右方向为不同种立方体共面连接，相邻的A层之间同种立方体在左右方向上错开，B层只包含黑色立方体，前后方向为共面连接形成折线形结构，左右方向为共面连接，在左侧的堆积结构中，相邻的黑色立方体层之间通过共棱连接，相邻的白色-空白立方体层之间无直接连接。](/media/uploads/ee2f93e9-b4e6-4e4e-b482-eabb439a9024.png)</MultiModal>

求该晶体中，金的氧化数的方差。
$0$
$\frac{1}{4}$
$\frac{2}{9}$
$\frac{3}{16}$
$\frac{4}{25}$
$1$
$\frac{8}{9}$
$\frac{3}{4}$
$\frac{16}{25}$
$\frac{9}{4}$
$2$
$\frac{27}{16}$
$\frac{36}{25}$
$4$
$\frac{32}{9}$
$3$
$\frac{64}{25}$
$0$
$\frac{1}{4}$
$\frac{2}{9}$
$\frac{3}{16}$
$\frac{4}{25}$
$1$
$\frac{8}{9}$
$\frac{3}{4}$
$\frac{16}{25}$
$\frac{9}{4}$
$2$
$\frac{27}{16}$
$\frac{36}{25}$
$4$
$\frac{32}{9}$
$3$
$\frac{64}{25}$
From the crystal structure:

<MultiModal>![On the left is a packing structure composed of a hollow cubic framework, black cubes, and white cubes. On the right is its layered decomposition, with a black arrow in the middle pointing from the right to the left. In the packing structure on the left, the left-right direction aligns with the cube edges, while the up-down and front-back directions align with the cube face diagonals. In the decomposed structure on the right, it is split into five layers from top to bottom: A, B, A, B, A. Layer A consists of the hollow cubic framework and white cubes, with the front-back direction showing edge-sharing connections of the same type of cubes and the left-right direction showing face-sharing connections of different types of cubes. Adjacent A layers have the same type of cubes staggered in the left-right direction. Layer B consists only of black cubes, with the front-back direction showing face-sharing connections forming a zigzag structure and the left-right direction showing face-sharing connections. In the packing structure on the left, adjacent black cube layers are connected via edge-sharing, while adjacent white-hollow cube layers have no direct connections.](/media/uploads/ee2f93e9-b4e6-4e4e-b482-eabb439a9024.png)</MultiModal>

It can be seen that the ratio of the three types of cubes is hollow:white:black = $1:1:4$.

<Checkpoint>[The ratio of the three types of cubes is hollow:white:black = $1:1:4$](1)</Checkpoint>

According to the problem description, the hollow cube contains 2 $\mathrm{Rb}$ atoms at the vertices and body center, and 1 $\mathrm{O}$ atom on the faces, giving the chemical formula $\mathrm{Rb_2O}$.

<Checkpoint>[The hollow cube contains $\mathrm{Rb_2O}$](1)</Checkpoint>

The white cube contains $\mathrm{Rb}$ at the vertices, $\mathrm{Au}$ at the body center, and 1 $\mathrm{O}$ atom on the faces, giving the chemical formula $\mathrm{RbAuO}$.

<Checkpoint>[The white cube contains $\mathrm{RbAuO}$](1)</Checkpoint>

The black cube contains $\mathrm{Rb}$ at the vertices and $\mathrm{Au}$ at the body center, giving the chemical formula $\mathrm{RbAu}$.

<Checkpoint>[The black cube contains $\mathrm{RbAu}$](1)</Checkpoint>

Therefore, the overall chemical formula of the crystal is 1×hollow + 1×white + 4×black = $\mathrm{Rb_7Au_5O_2}$.

<Checkpoint>[The overall chemical formula of the crystal is $\mathrm{Rb_7Au_5O_2}$](1)</Checkpoint>

From the crystal structure, it can be observed that there are no clusters with unconventional oxidation states of $\mathrm{Rb}$, nor structures like peroxide or superoxide. Thus, it can be concluded that the oxidation number of $\mathrm{Rb}$ in the crystal is $+1$, and that of $\mathrm{O}$ is $-2$.

<Checkpoint>[In the crystal, the oxidation number of $\mathrm{Rb}$ is $+1$, and that of $\mathrm{O}$ is $-2$](1)</Checkpoint>

Therefore, the average oxidation number of $\mathrm{Au}$ is $\frac{-1\times7+2\times2}{5}=-\frac{3}{5}$.

<Checkpoint>[The average oxidation number of $\mathrm{Au}$ is $-\frac{3}{5}$](1)</Checkpoint>

Considering the charge balance of each $\mathrm{Au}$ with its surrounding environment, the oxidation number of $\mathrm{Au}$ at the center of the white cube is $+1$, and that at the center of the black cube is $\mathrm{-1}$.

<Checkpoint>[The oxidation number of $\mathrm{Au}$ at the center of the white cube is $+1$, and that at the center of the black cube is $\mathrm{-1}$](1)</Checkpoint>

Using the formula for variance, we have $\sigma^2=\frac{1}{1+4}\times\left[+1-\left(-\frac{3}{5}\right)\right]^2+\frac{4}{1+4}\times\left[-1-\left(-\frac{3}{5}\right)\right]^2=\frac{16}{25}$. The correct answer is $\mathbf{I}$.

<Checkpoint>[The variance of the oxidation numbers of gold is $\sigma^2=\frac{16}{25}$](1)</Checkpoint>
由晶体结构：

<MultiModal>![左边为由空心立方体框架、黑色立方体、百色立方体构成的堆积结构，右边为其层状拆分，中间有一黑色箭头从右侧指向左侧，在左侧的堆积结构中，左右方向沿立方体的棱方向，上下和前后方向沿立方体的面对角线方向，在右侧的拆分结构中，从上至下拆分为A,B,A,B,A五层，A层包含空心立方体框架和白色立方体，前后方向为同种立方体共棱连接，左右方向为不同种立方体共面连接，相邻的A层之间同种立方体在左右方向上错开，B层只包含黑色立方体，前后方向为共面连接形成折线形结构，左右方向为共面连接，在左侧的堆积结构中，相邻的黑色立方体层之间通过共棱连接，相邻的白色-空白立方体层之间无直接连接。](/media/uploads/ee2f93e9-b4e6-4e4e-b482-eabb439a9024.png)</MultiModal>

可以看出三种立方体的数目比为空：白：黑=$1:1:4$，

<Checkpoint>[三种立方体的数目比为空：白：黑=$1:1:4$](1)</Checkpoint>

由题目描述，空立方体中含有顶点和体心2个$\mathrm{Rb}$，面上的1个$\mathrm{O}$，化学式为$\mathrm{Rb_2O}$。

<Checkpoint>[空立方体的内容为$\mathrm{Rb_2O}$](1)</Checkpoint>

白立方体中含有顶点$\mathrm{Rb}$，体心$\mathrm{Au}$，面上的1个$\mathrm{O}$，化学式为$\mathrm{RbAuO}$。

<Checkpoint>[白立方体的内容为$\mathrm{RbAuO}$](1)</Checkpoint>

黑立方体中含有顶点$\mathrm{Rb}$，体心$\mathrm{Au}$，化学式为$\mathrm{RbAu}$。

<Checkpoint>[黑立方体的内容为$\mathrm{RbAu}$](1)</Checkpoint>

故晶体的总化学式为1×空+1×白+4×黑=$\mathrm{Rb_7Au_5O_2}$。

<Checkpoint>[晶体总化学式为$\mathrm{Rb_7Au_5O_2}$](1)</Checkpoint>

由晶体结构可看出，不存在非常规氧化态$\mathrm{Rb}$构成的簇，也不存在过氧根、超氧根等结构，故可认为晶体中$\mathrm{Rb}$的氧化数为$+1$，$\mathrm{O}$的氧化数为$-2$。

<Checkpoint>[晶体中$\mathrm{Rb}$的氧化数为$+1$，$\mathrm{O}$的氧化数为$-2$](1)</Checkpoint>

故$\mathrm{Au}$的平均氧化数为$\frac{-1\times7+2\times2}{5}=-\frac{3}{5}$。

<Checkpoint>[$\mathrm{Au}$的平均氧化数为$-\frac{3}{5}$](1)</Checkpoint>

考虑每个$\mathrm{Au}$与周边环境的电荷平衡，有白立方体中心的$\mathrm{Au}$的氧化数为$+1$，黑立方体中心的$\mathrm{Au}$的氧化数为$\mathrm{-1}$。

<Checkpoint>[白立方体中心的$\mathrm{Au}$的氧化数为$+1$，黑立方体中心的$\mathrm{Au}$的氧化数为$\mathrm{-1}$](1)</Checkpoint>

由方差的计算公式，有$\sigma^2=\frac{1}{1+4}\times\left[+1-\left(-\frac{3}{5}\right)\right]^2+\frac{4}{1+4}\times\left[-1-\left(-\frac{3}{5}\right)\right]^2=\frac{16}{25}$，正确答案为$\mathbf{I}$。

<Checkpoint>[金的氧化数的方差为$\sigma^2=\frac{16}{25}$](1)</Checkpoint>