Perovskite is one of the classic structures. The perovskite with a common structure ($\mathbf{AB}\mathrm{O_3}$) can be described as follows: the $\mathbf{A}$-site atoms and oxygen atoms form a cubic close-packed structure together, and the $\mathbf{B}$-site atoms fill all the octahedral voids formed only by oxygen atoms. In addition, if the $\mathbf{A}$-site atoms and oxygen atoms form other forms of close-packed structures, perovskite can also form a variety of other structures (hexagonal perovskite). For example, in $5H-\mathrm{BaCrO_3}$ (denoted as $\mathbf{X}$) obtained at high temperature, $\mathrm{Ba}$ atoms and $\mathrm{O}$ atoms together form a …ABABCABABC… type close-packed structure (where the structures of layers A, B, and C are shown in the figure below; the large spheres are $\mathrm{Ba}$ atoms, and the small spheres are $\mathrm{O}$ atoms); $\mathrm{Cr}$ atoms fill all the octahedral voids formed only by $\mathrm{O}$ atoms.

<MultiModal>![The figure shows the structures of layers A, B, and C. In layer A, the coordinates of the large spheres are (2/3, 1/3), and the coordinates of the small spheres are (1/6, 1/3), (1/6, 5/6), and (2/3, 5/6); in layer B, the coordinates of the large spheres are (1/3, 2/3), and the coordinates of the small spheres are (1/3, 1/6), (5/6, 1/6), and (5/6, 2/3); in layer C, the coordinates of the large spheres are (0, 0), and the coordinates of the small spheres are (1/2, 0), (0, 1/2), and (0, 0).](/media/uploads/51e43652-3452-43b1-83dc-e8d8896acdb7.png)</MultiModal>

$\mathbf{X}$ can be reduced by hydrogen at $\rm 350 \degree C$ to obtain $\mathbf{Y}$. Crystal structure determination shows that the main structural difference between $\mathbf{Y}$ and $\mathbf{X}$ is that the number and position of $\mathrm{O}$ atoms in layer C in $\mathbf{Y}$ are different from those in $\mathbf{X}$, which makes the two layers of $\mathrm{Cr}$ atoms originally connected to the $\mathrm{O}$ atoms in layer C in $\mathbf{Y}$ transformed into tetrahedral coordination; except for the $\mathrm{O}$ atoms in layer C, the positions of other atoms have no obvious changes; and the cell parameters of $\mathbf{Y}$ and $\mathbf{X}$ are not significantly different.

The chemical formula of $\mathbf{Y}$ can be determined by thermogravimetric analysis. When $\mathbf{X}$ and $\mathbf{Y}$ are heated to $\rm 720\degree C$ in an oxygen atmosphere, both generate $\mathbf{Z}$; the conversion of $\mathbf{X}$ to $\mathbf{Z}$ increases the weight by $6.74\%$, while the conversion of $\mathbf{Y}$ to $\mathbf{Z}$ increases the weight by $8.20\%$.

Assume that $[\mathrm{CrO_6}]$ in $\mathbf{X}$ is a regular octahedron, and the $\mathrm{Cr-O}$ bond length is 202 pm.

Let $a, b$ be the mass fractions of oxygen in $\mathbf{Y, Z}$ respectively; let $h$ be the height of a stacking layer (unit: pm), and $\rho$ be the density of $\mathbf{X}$ (unit: $g/cm^{3}$).

And there are the following statements about $\mathbf{X}$:

(1) The connection mode between $[\mathrm{CrO_6}]$ octahedra in $\mathbf{X}$ has vertex sharing;

(2) The connection mode between $[\mathrm{CrO_6}]$ octahedra in $\mathbf{X}$ has edge sharing;

(3) The connection mode between $[\mathrm{CrO_6}]$ octahedra in $\mathbf{X}$ has face sharing;

(4) There is no connection between $[\mathrm{CrO_6}]$ octahedra in $\mathbf{X}$;

Let $c$ be the sum of the serial numbers of all the correct statements above.

Then the value of $\log_{\rho}\frac{\lfloor abh\rfloor}{c}$ is (where $\lfloor~\rfloor$ is the floor function, and the final result with an error within 0.01 is considered correct):
钙钛矿是经典结构之一，常见结构的钙钛矿（$\mathbf{AB}\mathrm{O_3}$）可描述为：$\mathbf{A}$位原子和氧原子共同作立方最密堆积，$\mathbf{B}$位原子填入所有仅有氧原子形成的八面体空隙中。除此之外，若$\mathbf{A}$位原子和氧原子做其他形式的最密堆积，则钙钛矿还可形成多种多样的其他结构(六方钙钛矿)。例如，在高温制得的 $5H-\mathrm{BaCrO_3}($记为$\mathbf{X}$)中，$\mathrm{Ba}$原子与$\mathrm{O}$原子共同作…ABABCABABC…型最密堆积（其中A、B、C层的结构均如下图所示；大球为$\mathrm{Ba}$原子，小球为$\mathrm{O}$原子)；$\mathrm{Cr}$原子填入所有仅由$\mathrm{O}$原子形成的八面体空隙中。

<MultiModal>![图中展示了A、B、C层的结构，A层中大球的坐标为（2/3，1/3），小球的坐标为（1/6，1/3），（1/6，5/6）和（2/3，5/6）；B层中大球的坐标为（1/3，2/3），小球的坐标为（1/3，1/6），（5/6，1/6）和（5/6，2/3）；C层中大球的坐标为（0，0），小球的坐标为（1/2，0），（0，1/2）和（0，0）。](/media/uploads/51e43652-3452-43b1-83dc-e8d8896acdb7.png)</MultiModal>

$\mathbf{X}$在 $\rm 3 5 0 \degree C$ 下经氢气还原可得到$\mathbf{Y}$，晶体结构测定表明：$\mathbf{Y}$与$\mathbf{X}$的主要结构差别在于，$\mathbf{Y}$中位于C层中的$\mathrm{O}$原子的数目和位置与$\mathbf{X}$中不同，使得$\mathbf{Y}$中原本与C层$\mathrm{O}$原子相连的两层$\mathrm{Cr}$原子转变为四面体配位；除C层的$\mathrm{O}$原子之外，其他原子的位置无明显变化；且$\mathbf{Y}$与$\mathbf{X}$的晶胞参数差别并不显著。

通过热重分析，可以确定$\mathbf{Y}$的化学式。$\mathbf{X}$和$\mathbf{Y}$在氧气气氛中加热至 $ \rm 720\degree C$ ，均生成$\mathbf{Z}$；$\mathbf{X}$转变为$\mathbf{Z}$增重 $6 . 7 4 \%$ ，而$\mathbf{Y}$转变为$\mathbf{Z}$则增重 $8 . 2 0 \%$ 。

假设$\mathbf{X}$中的$[\mathrm{CrO_6}]$为正八面体，设$\mathrm{Cr-O}$键长均为202 pm。

记$\mathbf{Y，Z}$中氧元素的质量分数分别为$a,b$；记一层堆积层的高度为$h$（单位：pm），$\mathbf{X}$的密度为$\rho$（单位：$g/cm^{3}$）。

以及有下列关于$\mathbf{X}$的说法：

（1）$\mathbf{X}$中$[\mathrm{CrO_6}]$八面体之间的连接方式存在共顶点连接；

（2）$\mathbf{X}$中$[\mathrm{CrO_6}]$八面体之间的连接方式存在共棱连接；

（3）$\mathbf{X}$中$[\mathrm{CrO_6}]$八面体之间的连接方式存在共面连接；

（4）$\mathbf{X}$中$[\mathrm{CrO_6}]$八面体之间无任何连接；

记上述说法中所有正确的说法的序号之和为$c$。

则$\log_{\rho}\frac{\lfloor abh\rfloor}{c}$的值为（$\lfloor~\rfloor$为向下取整函数，最终结果误差在0.01之内视为正确）：
All other options are incorrect
-0.89
-0.73
-0.52
-0.38
-0.11
0.14
0.39
0.57
0.66
0.89
1.05
其他选项均不正确
-0.89
-0.73
-0.52
-0.38
-0.11
0.14
0.39
0.57
0.66
0.89
1.05
The molecular weight of $\mathbf{X}$ is $M(\mathbf{X})=137.3+52+3\times16=237.3$

Therefore, the molecular weight of $\mathbf{Z}$ is $M(\mathbf{Z})=237.3\times(1+6.74\%)=253.3$, which is 16 more than $\mathbf{X}$, which is one oxygen atom, so $\mathbf{Z}$ is $\mathrm{BaCrO_4}$

<Checkpoint>[The molecular weight of $\mathbf{Z}$ is $253.3$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{Z}$ is $\mathrm{BaCrO_4}$](1)</Checkpoint>

The molecular weight of $\mathbf{Y}$ is $M(\mathbf{Y})=253.3/(1+8.20\%)=234.1$, so $\mathbf{Y}$ is $\mathrm{BaCrO_{2.8}}$

<Checkpoint>[The molecular weight of $\mathbf{Y}$ is $234.1$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$ is $\mathrm{BaCrO_{2.8}}$](2)</Checkpoint>

Therefore, $a=19.14\%,b=25.26\%$.

<Checkpoint>[$a=19.14\%,b=25.26\%$](1)</Checkpoint>

First, calculate the edge length of the $[\mathrm{CrO_6}]$ octahedron: $d=202~pm\times\sqrt{2}=285.7~pm$

The height of a stacking layer is the distance between opposite faces of the octahedron: $h=202~pm\times\frac{2}{3}\sqrt{3}=233.3~pm$

<Checkpoint>[The height of a stacking layer $h=233.3~pm$](1)</Checkpoint>

Therefore, the unit cell parameters $\mathbf{a}=2d=571~pm,\mathbf{c}=5h=1166~pm$

Therefore, the density $\rho=\frac{ZM}{N_AV}=5.98~g/cm^{3}$

<Checkpoint>[Density $\rho=5.98~g/cm^{3}$](2)</Checkpoint>

According to the given layered structure, it can be seen that in ABA stacking, the octahedra formed by AB and BA are connected by sharing faces, while in BCA, the octahedra formed by BC and CA are connected by sharing vertices, so (1) and (3) are correct.

<Checkpoint>[The connection mode between $[\mathrm{CrO_6}]$ octahedra in $\mathbf{X}$ includes vertex-sharing and face-sharing](1)</Checkpoint>

Therefore, $c=4$

<Checkpoint>[$c=4$](1)</Checkpoint>

In summary, $a=19.14\%,b=25.26\%,c=4,h=233.3,\rho=5.98$

$\log_{\rho}\frac{\lfloor abh\rfloor}{c}=0.57$

<Checkpoint>[$\log_{\rho}\frac{\lfloor abh\rfloor}{c}=0.57$](1)</Checkpoint>
$\mathbf{X}$的分子量为$M(\mathbf{X})=137.3+52+3\times16=237.3$

因此$\mathbf{Z}$的分子量为$M(\mathbf{Z})=237.3\times(1+6.74\%)=253.3$，相对于$\mathbf{X}$多出16，为一个氧原子，因此$\mathbf{Z}$为$\mathrm{BaCrO_4}$

<Checkpoint>[$\mathbf{Z}$的分子量为$253.3$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{Z}$为$\mathrm{BaCrO_4}$](1)</Checkpoint>

$\mathbf{Y}$的分子量为$M(\mathbf{Y})=253.3/(1+8.20\%)=234.1$，因此$\mathbf{Y}$为$\mathrm{BaCrO_{2.8}}$

<Checkpoint>[$\mathbf{Y}$的分子量为$234.1$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$为$\mathrm{BaCrO_{2.8}}$](2)</Checkpoint>

因此$a=19.14\%,b=25.26\%$。

<Checkpoint>[$a=19.14\%,b=25.26\%$](1)</Checkpoint>

首先计算$[\mathrm{CrO_6}]$八面体的棱长：$d=202~pm\times\sqrt{2}=285.7~pm$

一层堆积层的高为八面体相对面的距离：$h=202~pm\times\frac{2}{3}\sqrt{3}=233.3~pm$

<Checkpoint>[一层堆积层的高$h=233.3~pm$](1)</Checkpoint>

因此晶胞参数$\mathbf{a}=2d=571~pm,\mathbf{c}=5h=1166~pm$

因此密度$\rho=\frac{ZM}{N_AV}=5.98~g/cm^{3}$

<Checkpoint>[密度$\rho=5.98~g/cm^{3}$](2)</Checkpoint>

根据所给的层状结构可以看出在ABA堆积中AB和BA形成的八面体为共面连接，而BCA中BC和CA形成的八面体为共顶点连接，因此（1）（3）为正确的。

<Checkpoint>[$\mathbf{X}$中$[\mathrm{CrO_6}]$八面体之间的连接方式存在共顶点连接和共面连接](1)</Checkpoint>

因此$c=4$

<Checkpoint>[$c=4$](1)</Checkpoint>

综上，$a=19.14\%,b=25.26\%,c=4,h=233.3,\rho=5.98$

$\log_{\rho}\frac{\lfloor abh\rfloor}{c}=0.57$

<Checkpoint>[$\log_{\rho}\frac{\lfloor abh\rfloor}{c}=0.57$](1)</Checkpoint>