$\mathrm{ReO_3F}$ forms a disolvate $\mathbf{A}$ in anhydrous $\mathrm{HF}$. $\mathbf{A}$ reacts with $\mathrm{XeF_2}$ at $-30\degree \mathrm{C}$ to yield a 1:2 salt $\mathbf{B}$, where 2/5 of the $\mathrm{XeF_2}$ is reduced during the reaction. $\mathbf{B}$ contains no hydrogen, with mass fractions of $\mathrm{Xe}$ and $\mathrm{Re}$ being $25.17\%$ and $47.60\%$, respectively, and a molecular weight not exceeding 2000. The cation of $\mathbf{B}$ consists of only two elements and excludes fluorine, while the anion contains no $\mathrm{Xe}$ but features a fluorine-bridged structure, with the bridging fluorine atoms positioned trans to oxygen atoms. If the cation and anion are treated as free ions, all $\mathrm{Re}$ atoms exhibit identical chemical environments. $\mathbf{B}$ decomposes above $-20\degree \mathrm{C}$ to produce a $\mathrm{Re}$-containing compound $\mathbf{C}$. This reaction yields no ionic compounds, and no $\mathrm{Xe-O}$ bonds are present in the products. Throughout these processes, the oxidation state of the metal $\mathrm{Re}$ remains unchanged. In this context, if $\mathrm{HF}$ is generated as a reaction product, it is not considered a gas.

Which of the following statements is correct?
$\mathrm{ReO_3F}$ 在无水 $\mathrm{HF}$ 中形成二溶剂合物 $\mathbf{A}$, $\mathbf{A}$ 与 $\mathrm{XeF_2}$ 在$-30\degree \mathrm{C}$ 反应得到 1:2 型盐 $\mathbf{B}$, 反应
中有 2/5 的 $\mathrm{XeF_2}$ 被还原。 $\mathbf{B}$ 不含氢元素, $\mathrm{Xe}$ 和 $\mathrm{Re}$ 的质量分数分别为 $25.17 \%$ 和 $47.60 \%$, 且分子量不超过 2000。$\mathbf{B}$的阳离子仅含两种元素且不含氟元素, 阴离子不含$\mathrm{Xe}$，含有氟桥结构, 桥接氟原子的
对位是氧原子,若将阴阳离子视为自由离子,则所有$\mathrm{Re}$原子的化学环境相同。$\mathbf{B}$在高于$-20\degree \mathrm{C}$ 时分解产生含$\mathrm{Re}$的化合物$\mathbf{C}$，该反应产物中没有离子化合物，且产物中不存在$\mathrm{Xe-O}$键。以上所有过程中，金属$\mathrm{Re}$的价态不发生变化。该题目中，反应若产生$\mathrm{HF}$，不视为气体。

下列说法正确的是：
The relative molecular mass of $\mathbf{A}$ is 292.2
The reaction producing $\mathbf{B}$ and the reaction producing $\mathbf{C}$ have inconsistent gaseous products.
The chemical equation for the formation of $\mathbf{B}$ (with coefficients in the simplest ratio), where the sum of the product coefficients is an odd number.
The chemical equation for the formation of $\mathbf{C}$ (with coefficients in the simplest ratio), where the sum of the product coefficients is an odd number.
The cationic structure of $\mathbf{B}$ is trigonal.
The anionic structure of $\mathbf{B}$ exhibits fourfold rotational symmetry.
In the chemical equation for the formation of $\mathbf{C}$ (with coefficients in the simplest whole-number ratio), the coefficient of $\mathbf{C}$ is 4.
Among the elements that constitute $\mathbf{C}$, there are two elements with the same atomic number.
 $\mathbf{A}$的相对分子质量为292.2
生成$\mathbf{B}$的反应和生成$\mathbf{C}$的反应，气体产物不一致。
生成$\mathbf{B}$的化学方程式（系数为最简系数比），产物系数和为奇数。
生成$\mathbf{C}$的化学方程式（系数为最简系数比），产物系数和为奇数。
$\mathbf{B}$的阳离子结构为三角型
$\mathbf{B}$的阴离子结构具有四重旋转轴对称性
生成$\mathbf{C}$的化学方程式（系数为最简整数比）中，$\mathbf{C}$的系数为4。
构成$\mathbf{C}$的元素中，有两种元素具有相同的原子数。
$\mathrm{ReO_3F}$ forms a disolvate $\mathbf{A}$ in anhydrous $\mathrm{HF}$, so the chemical formula of $\mathbf{A}$ is clearly $\mathrm{ReO_3F(HF)_2}$, with a relative molecular mass of 293.2, making option A incorrect.

<Checkpoint>[The chemical formula of $\mathbf{A}$ is $\mathrm{ReO_3F(HF)_2}$, with a relative molecular mass of 293.2](1)</Checkpoint>

$\mathrm{ReO_3F(HF)_2}$ reacts with $\mathrm{XeF_2}$ to form a 1:2 salt $\mathbf{B}$, with 2/5 of the $\mathrm{XeF_2}$ being reduced. The reduction product is clearly $\mathrm{Xe}$.

<Checkpoint>[The reduction product is $\mathrm{Xe}$](0.5)</Checkpoint>

Assuming the reaction requires 5 molecules of $\mathrm{XeF_2}$, then $\mathbf{B}$ contains 3 $\mathrm{Xe}$ atoms. Based on the molecular weight not exceeding 2000 and the mass fraction information, it can be concluded that $\mathbf{B}$ indeed contains 3 $\mathrm{Xe}$ atoms.

<Checkpoint>[$\mathbf{B}$ contains 3 $\mathrm{Xe}$ atoms](1)</Checkpoint>

According to the mass fraction information, the atomic ratio of $\mathrm{Xe}$ to $\mathrm{Re}$ in $\mathbf{B}$ is $(0.2517/131.3 : 0.4760/186.2) = 3 : 4$, so $\mathbf{B}$ contains 4 $\mathrm{Re}$ atoms.

<Checkpoint>[The atomic ratio of $\mathrm{Xe}$ to $\mathrm{Re}$ in $\mathbf{B}$ is $3 : 4$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{B}$ contains 4 $\mathrm{Re}$ atoms](1)</Checkpoint>

Thus, the molecular weight of $\mathbf{B}$ can be calculated as $3 \times 131.2 / 0.2517 = 1565$.

<Checkpoint>[The molecular weight of $\mathbf{B}$ is $1565$](1)</Checkpoint>

$\mathbf{B}$ contains no hydrogen, and based on the reactants in the formation reaction of $\mathbf{B}$, it can be inferred that it also contains oxygen and fluorine. The remaining molecular weight of $\mathbf{B}$ after accounting for $\mathrm{Xe}$ and $\mathrm{Re}$ is 426, which corresponds to exactly 10 oxygen atoms and 14 fluorine atoms. Therefore, the chemical formula of $\mathbf{B}$ is $\mathrm{Xe_3Re_4O_{10}F_{14}}$.

<Checkpoint>[The chemical formula of $\mathbf{B}$ is $\mathrm{Xe_3Re_4O_{10}F_{14}}$](2)</Checkpoint>

In the reaction forming $\mathbf{B}$, since the $\mathrm{Re}$ in $\mathbf{A}$ is already in its highest oxidation state, only the oxygen is oxidized by $\mathrm{Xe}$, and the oxidation product is clearly oxygen. Thus, the equation for the formation of $\mathbf{B}$ can be written as:

<Checkpoint>[The oxidation product is clearly oxygen](0.5)</Checkpoint>

$\mathrm{5XeF_2 + 4ReO_3F(HF)_2 = 2Xe + O_2 + Xe_3Re_4O_{10}F_{14} + 8HF}$

<Checkpoint>[ $\mathrm{5XeF_2 + 4ReO_3F(HF)_2 = 2Xe + O_2 + Xe_3Re_4O_{10}F_{14} + 8HF}$](1)</Checkpoint>

The sum of the coefficients of the products in the equation is 12, which is not odd, so option C is incorrect.

The anion of $\mathbf{B}$ does not contain $\mathrm{Xe}$, so the cation contains 3 $\mathrm{Xe}$ atoms and no fluorine. Since $\mathbf{B}$ is a 1:2 ionic compound, the cation must carry a +2 charge. The 3 $\mathrm{Xe}$ atoms in the cation are not reduced and are in the +2 oxidation state, requiring an additional 4 negative charges, which can be balanced by two oxygen atoms. Therefore, the chemical formula of the cation in $\mathbf{B}$ is $\mathrm{Xe_3O_2^{2+}}$, and its structure is bent, with oxygen atoms serving as bridging atoms. Thus, option E is incorrect.

<Checkpoint>[The chemical formula of the cation in $\mathbf{B}$ is $\mathrm{Xe_3O_2^{2+}}$, and its structure is bent](2)</Checkpoint>

Therefore, the anion of $\mathbf{B}$ carries a single negative charge and there are two of them, with the chemical formula $\mathrm{Re_2O_4F_7^{-}}$.

<Checkpoint>[The chemical formula of the anion in $\mathbf{B}$ is $\mathrm{Re_2O_4F_7^{-}}$](1)</Checkpoint>

The anion of $\mathbf{B}$ contains fluorine bridges. $\mathrm{Re}$ in the +7 oxidation state commonly forms octahedral coordination with six ligands, so each $\mathrm{Re}$ is connected to two terminal oxygen atoms, three terminal fluorine atoms, and one bridging fluorine atom, forming two octahedra sharing a vertex.

<Checkpoint>[The structure of the anion in $\mathbf{B}$ consists of two octahedra sharing a vertex](1)</Checkpoint>

However, it can be observed that each $\mathrm{Re}$ is bonded to two oxygen atoms and four fluorine atoms. Since one fluorine atom is bridging, the resulting octahedron cannot have four atoms coplanar and thus lacks fourfold rotational symmetry, making option F incorrect.

<Checkpoint>[The resulting octahedron cannot have four atoms coplanar](1)</Checkpoint>

The decomposition of $\mathbf{B}$ yields a $\mathrm{Re}$-containing compound $\mathbf{C}$. The reaction products contain no ionic compounds, and no $\mathrm{Xe-O}$ bonds are present, so the $\mathrm{Xe}$ products are in the form of elemental $\mathrm{Xe}$ and $\mathrm{XeF_2}$. Since the oxidation state of $\mathrm{Re}$ does not change, the reaction can be interpreted as $\mathrm{Xe}$ oxidizing oxygen to produce $\mathrm{O_2}$, and the chemical formula of $\mathbf{C}$ can be written as $\mathrm{ReO_xF_y}$, satisfying $2x + y = 7$. Given that $\mathrm{Re}$ in $\mathbf{B}$ is bonded to two oxygen atoms, assuming $x = 2$, the equation for the formation of $\mathbf{C}$ is:

$\mathrm{Xe_3Re_4O_{10}F_{14} = 4ReO_2F_3 + XeF_2 + 2Xe + O_2}$

<Checkpoint>[ $\mathrm{Xe_3Re_4O_{10}F_{14} = 4ReO_2F_3 + XeF_2 + 2Xe + O_2}$](2)</Checkpoint>

Thus, the chemical formula of $\mathbf{C}$ is $\mathrm{ReO_2F_3}$, making option H incorrect.

<Checkpoint>[The chemical formula of $\mathbf{C}$ is $\mathrm{ReO_2F_3}$](1)</Checkpoint>

The sum of the coefficients of the products in the formation reaction of $\mathbf{C}$ is 8, which is not odd, so option D is incorrect. The gaseous products of this reaction are the same as those in the formation reaction of $\mathbf{B}$, making option B incorrect. The coefficient of $\mathbf{C}$ is 4, so option G is correct.

In conclusion, option G is correct.

<Checkpoint>[Option G is correct](1)</Checkpoint>
$\mathrm{ReO_3F}$ 在无水 $\mathrm{HF}$ 中形成二溶剂合物 $\mathbf{A}$，从而$\mathbf{A}$的化学式很明显为$\mathrm{ReO_3F(HF)_2}$，相对分子质量为293.2，A选项错误。 

<Checkpoint>[$\mathbf{A}$的化学式为$\mathrm{ReO_3F(HF)_2}$，相对分子质量为293.2](1)</Checkpoint>

$\mathrm{ReO_3F(HF)_2}$ 与 $\mathrm{XeF_2}$  反应得到 1:2 型盐 $\mathbf{B}$, 反应
中有 2/5 的 $\mathrm{XeF_2}$ 被还原，还原产物很明显为$\mathrm{Xe}$。

<Checkpoint>[还原产物为$\mathrm{Xe}$](0.5)</Checkpoint>

那么先假设反应需要5个
$\mathrm{XeF_2}$，则$\mathbf{B}$中有3个$\mathrm{Xe}$原子；根据分子量不大于2000和质量分数信息，可以判断$\mathbf{B}$中有3个$\mathrm{Xe}$原子是正确的。

<Checkpoint>[$\mathbf{B}$中有3个$\mathrm{Xe}$原子](1)</Checkpoint>

根据质量分数信息，$\mathbf{B}$中$\mathrm{Xe}$和$\mathrm{Re}$的原子数量比为$(0.2517/131.3 : 0.4760/186.2) = 3 : 4$，从而$\mathbf{B}$中有4个$\mathrm{Re}$原子；

<Checkpoint>[$\mathbf{B}$中$\mathrm{Xe}$和$\mathrm{Re}$的原子数量比为$3 : 4$](0.5)</Checkpoint>

<Checkpoint>[$\mathbf{B}$中有4个$\mathrm{Re}$原子；](1)</Checkpoint>

从而根据质量分数可求出$\mathbf{B}$的分子量为$3\times 131.2/0.2517 = 1565$

<Checkpoint>[$\mathbf{B}$的分子量为$1565$](1)</Checkpoint>

$\mathbf{B}$不含氢元素，根据生成
$\mathbf{B}$的反应的反应物可判断还有氧元素和氟元素。
$\mathbf{B}$的分子量除去
$\mathrm{Xe,Re}$之后剩余426，刚好为10个氧原子和14个氟原子。从而
$\mathbf{B}$的化学式为$\mathrm{Xe_3Re_4O_{10}F_{14}}$.

<Checkpoint>[ $\mathbf{B}$的化学式为$\mathrm{Xe_3Re_4O_{10}F_{14}}$](2)</Checkpoint>

生成$\mathbf{B}$的反应，由于$\mathbf{A}$的$\mathrm{Re}$已经为最高价态，从而只有氧元素被$\mathrm{Xe}$氧化，氧化产物很明显为氧气。由此可以写出生成$\mathbf{B}$的方程式：

<Checkpoint>[氧化产物很明显为氧气](0.5)</Checkpoint>

$\mathrm{5XeF_2 + 4ReO_3F(HF)_2 = 2Xe + O_2 + Xe_3Re_4O_{10}F_{14} + 8HF}$

<Checkpoint>[ $\mathrm{5XeF_2 + 4ReO_3F(HF)_2 = 2Xe + O_2 + Xe_3Re_4O_{10}F_{14} + 8HF}$](1)</Checkpoint>

方程式产物系数和为12，不为奇数，选项C错误。

$\mathbf{B}$的阴离子不含$\mathrm{Xe}$，则阳离子含3个$\mathrm{Xe}$且不含氟元素，且根据$\mathbf{B}$为1：2型离子化合物可知阳离子带两个正电荷。阳离子的3个$\mathrm{Xe}$未被还原，为正二价，故还需4个负电荷，可以用两个氧配平。故$\mathbf{B}$的阳离子化学式为$\mathrm{Xe_3O_2^{2+}}$，其结构为折线形，氧原子为桥连原子。故E选项错误。

<Checkpoint>[$\mathbf{B}$的阳离子化学式为$\mathrm{Xe_3O_2^{2+}}$，其结构为折线形](2)</Checkpoint>

因此，$\mathbf{B}$的阴离子含一个负电荷且有两个，化学式为$\mathrm{Re_2O_4F_7^{-}}$

<Checkpoint>[$\mathbf{B}$的阴离子化学式为$\mathrm{Re_2O_4F_7^{-}}$](1)</Checkpoint>

$\mathbf{B}$的阴离子含氟桥，+7价的$\mathrm{Re}$常见六配位八面体，故每个$\mathrm{Re}$连两个端基氧，三个端基氟和一个桥连的氟原子，形成两个共顶点相连的八面体。

<Checkpoint>[$\mathbf{B}$的阴离子结构为两个共顶点相连的八面体](1)</Checkpoint>

但可以发现，每个$\mathbf{Re}$与两个氧和四个氟相连，由于有一个氟为桥连原子，形成的八面体必不可能有同样的四个原子共面从而没有四重旋转对称性，选项F错误。

<Checkpoint>[形成的八面体必不可能有同样的四个原子共面](1)</Checkpoint>

$\mathbf{B}$分解产生含$\mathrm{Re}$的化合物$\mathbf{C}$，该反应产物中没有离子化合物，且产物中不存在$\mathrm{Xe-O}$键，那么$\mathrm{Xe}$的产物存在形式为单质和$\mathrm{XeF_2}$。由于$\mathrm{Re}$价态不发生变化，该反应可以判断为$\mathrm{Xe}$氧化氧元素产生氧气，且$\mathbf{C}$的化学式可写为$\mathrm{ReO_xF_y}$，满足$2x+y=7$；由于$\mathbf{B}$的结构中$\mathrm{Re}$有两个氧相连，考虑$x=2$，写出产生$\mathbf{C}$的方程：

$\mathrm{Xe_3Re_4O_{10}F_{14} = 4ReO_2F_3 + XeF_2 + 2Xe + O_2}$

<Checkpoint>[ $\mathrm{Xe_3Re_4O_{10}F_{14} = 4ReO_2F_3 + XeF_2 + 2Xe + O_2}$](2)</Checkpoint>

故$\mathbf{C}$化学式为$\mathrm{ReO_2F_3}$，选项H错误。

<Checkpoint>[$\mathbf{C}$化学式为$\mathrm{ReO_2F_3}$](1)</Checkpoint>

生成$\mathbf{C}$反应产物系数和为8，不为奇数，选项D错误，该反应与生成$\mathbf{B}$的反应气体产物一致，选项B错误。$\mathbf{C}$的系数为4，选项G正确。

综上，选项G正确。

<Checkpoint>[选项G正确](1)</Checkpoint>