It is known that $\mathrm{[XeF][AsF_{6}]}$ reacts with $\mathrm{H_2O}$ in anhydrous $\mathrm{HF}$ according to the following equilibrium:
$$
[\mathrm{XeF}]\left[\mathrm{AsF}_{6}\right]+\mathrm{H}_{2} \mathrm{O}+\mathrm{HF} \rightleftharpoons \mathrm{XeF}_{2}+\left[\mathrm{H}_{3} \mathrm{O}\right]\left[\mathrm{AsF}_{6}\right]
$$
When equimolar amounts of $\mathrm{[XeF][AsF_{6}]}$ and $\mathrm{H_2O}$ react in anhydrous $\mathrm{HF}$ at −78 °C, the adduct $\mathrm{[H_3O][AsF_6]·2XeF_2}$ is initially formed. Subsequently, $\mathrm{[XeF][AsF_{6}]}$ reacts with free $\mathrm{XeF_2}$ in a 1:1 ratio to yield compound $\mathbf{A}$. This adduct further reacts with free $\mathrm{XeF_2}$ to form a 1:1 salt $\mathbf{B}$. The anion of $\mathbf{B}$ is identical to that of $\mathbf{A}$, while its cation contains oxygen with a coordination number of 2. The mass fraction of $\mathrm{Xe}$ in $\mathbf{B}$ is approximately 60%. The cation of $\mathbf{B}$ is obtained from the cation of $\mathbf{A}$ by reacting with the electrically neutral species $\mathbf{C}$, a process involving a single proton transfer.

Which of the following statements is correct:
已知$\mathrm{[XeF][AsF_{6}]}$ 与 $\mathrm{H_2O}$ 在无水 $\mathrm{HF}$ 中有下列平衡:
$$
[\mathrm{XeF}]\left[\mathrm{AsF}_{6}\right]+\mathrm{H}_{2} \mathrm{O}+\mathrm{HF} \rightleftharpoons \mathrm{XeF}_{2}+\left[\mathrm{H}_{3} \mathrm{O}\right]\left[\mathrm{AsF}_{6}\right]
$$
无水 $\mathrm{HF}$ 中等摩尔的 $\mathrm{[XeF][AsF_{6}]}$ 与 $\mathrm{H_2O}$  在 -78 °C 时反应,先生成加合物 $\mathrm{[H_3O][AsF_6]·2XeF_2}$，随后 $\mathrm{[XeF][AsF_{6}]}$ 与游离的 $\mathrm{XeF_2}$ 按 1:1 反应生成化合物 $\mathbf{A}$，该加合物与游离的 $\mathrm{XeF_2}$ 反应生成
1:1 型盐 $\mathbf{B}$。$\mathbf{B}$ 的阴离子与 $\mathbf{A}$ 相同,阳离子含氧且氧的配位数为 2, $\mathbf{B}$ 中 $\mathrm{Xe}$ 的质量分数约为 60%。$\mathbf{B}$的阳离子是由$\mathbf{A}$的阳离子与电中性物种$\mathbf{C}$反应得到，该反应经过一步质子转移。

下列说法中正确的是：
The cation of $\mathbf{A}$ is linear in shape.
The relative molecular mass of $\mathbf{A}$ is smaller than that of $\mathbf{C}$.
The cations of $\mathbf{B}$ are triangular in shape.
The chemical reaction equation for the formation of $\mathbf{B}$ (with coefficients in the simplest whole-number ratio), where the sum of the coefficients on the right side of the equation is 3.
$\mathbf{C}$ contains 3 types of elements.
The number of atoms contained in $\mathbf{C}$ is 4
$\mathbf{A}$的阳离子呈直线型
$\mathbf{A}$的相对分子质量比$\mathbf{C}$更小
$\mathbf{B}$的阳离子呈三角型
生成$\mathbf{B}$的化学反应方程式（系数为最简整数比），方程式右边的系数和为3
$\mathbf{C}$所含的元素种类为3
$\mathbf{C}$所含的原子个数为4
The formation of $\mathbf{A}$ is straightforward. Since it is a 1:1 reaction, $\mathrm{XeF_2}$ can only insert into $\mathrm{XeF^+}$ to yield $\mathrm{Xe_2F_3^+}$, which is the cation of $\mathbf{A}$. According to VSEPR theory, one $\mathrm{F}$ atom bridges the two $\mathrm{Xe}$ atoms and adopts $\mathrm{sp^3}$ hybridization, resulting in a trigonal shape rather than a linear geometry. Therefore, option A is incorrect.

<Checkpoint>[The cation of $\mathbf{A}$ is $\mathrm{Xe_2F_3^+}$ with a trigonal shape](2)</Checkpoint>

The anion of $\mathbf{B}$ is the same as that of $\mathbf{A}$, both being hexafluoroarsenate anions. The cation contains oxygen with a coordination number of 2, suggesting it is composed of $\mathrm{Xe, O,}$ and $\mathrm{F}$. Combining the mass fraction, the chemical formula of $\mathbf{B}$'s cation is determined to be $\mathrm{Xe_3OF_3^+}$.

<Checkpoint>[The chemical formula of $\mathbf{B}$'s cation is $\mathrm{Xe_3OF_3^+}$](2)</Checkpoint>

Considering its structure, oxygen cannot be a terminal atom and must act as a bridging atom. Based on the structure of $\mathrm{Xe_2F_3^+}$, $\mathrm{Xe_3OF_3^+}$ can be derived by replacing a terminal fluorine atom in $\mathrm{Xe_2F_3^+}$ with an oxygen atom and then attaching a $\mathrm{XeF}$ unit, described as $\mathrm{F-Xe-F-Xe-O-Xe-F}$. According to hybridization theory, it adopts a bent shape, making option C incorrect.

<Checkpoint>[$\mathrm{Xe_3OF_3^+}$ has a bent shape](1)</Checkpoint>

Thus, the reaction for forming $\mathbf{B}$ can be written as:

$$\mathrm{3XeF_2 + [H_3O][AsF_6] = [Xe_3OF_3^+][AsF_6] + 3HF}$$

The sum of the coefficients on the right side of the equation is 4, so option D is incorrect.

$\mathrm{Xe_3OF_3^+}$ is generated from $\mathrm{Xe_2F_3^+}$ and $\mathbf{C}$ via a proton transfer process. Given that $\mathrm{Xe_3OF_3^+}$ is equivalent to replacing a terminal fluorine atom in $\mathrm{Xe_2F_3^+}$ with an oxygen atom and attaching a $\mathrm{XeF}$ unit, the displaced fluorine atom likely combines with the proton. Therefore, the structure of $\mathbf{C}$ is clearly $\mathrm{HOXeF}$, containing four elements and four atoms. Thus, option F is correct, while options E and C are incorrect.

<Checkpoint>[The structure of $\mathbf{C}$ is $\mathrm{HOXeF}$](1)</Checkpoint>
生成$\mathbf{A}$的反应很简单，由于是1：1反应，那么$\mathrm{XeF_2}$只能插入$\mathrm{XeF^+}$得到$\mathrm{Xe_2F_3^+}$，这就是$\mathbf{A}$的阳离子，根据VSEPR理论可以判断有一个$\mathrm{F}$原子桥连两个$\mathrm{Xe}$且为$\mathrm{sp^3}$杂化，其形状为三角而非直线型，故选项A错误。

<Checkpoint>[$\mathbf{A}$的阳离子为$\mathrm{Xe_2F_3^+}$，形状为三角型](2)</Checkpoint>

$\mathbf{B}$ 的阴离子与 $\mathbf{A}$ 相同，则均为六氟化砷阴离子，阳离子含氧且氧的配位数为 2, 可以判断阳离子由$\mathrm{Xe, O, F}$组成，结合质量分数可以得到$\mathbf{B}$的阳离子化学式为$\mathrm{Xe_3OF_3^+}$。

<Checkpoint>[$\mathbf{B}$的阳离子化学式为$\mathrm{Xe_3OF_3^+}$](2)</Checkpoint>

考虑其结构，氧元素不能做端基原子，则氧原子必为桥连原子，结合$\mathrm{Xe_2F_3^+}$的结构，可以得出$\mathrm{Xe_3OF_3^+}$相当于将$\mathrm{Xe_2F_3^+}$的一个端基氟原子换为氧原子后，再连接一个$\mathrm{XeF}$，可描述为$\mathrm{F-Xe-F-Xe-O-Xe-F}$，根据杂化理论判断其为折线形，选项C错误。

<Checkpoint>[$\mathrm{Xe_3OF_3^+}$为折线形](1)</Checkpoint>

由此可以写出生成$\mathbf{B}$的反应：

$$\mathrm{3XeF_2 + [H_3O][AsF_6] = [Xe_3OF_3^+][AsF_6^-] + 3HF}$$

方程式右边系数和为4，D选项错误。

$\mathrm{Xe_3OF_3^+}$由$\mathrm{Xe_2F_3^+}$与$\mathbf{C}$生成，反应过程经过质子转移，考虑到$\mathrm{Xe_3OF_3^+}$相当于将$\mathrm{Xe_2F_3^+}$的一个端基氟原子换为氧原子后，再连接一个$\mathrm{XeF}$，那么被换的氟原子很可能就与这个质子结合，从而$\mathbf{C}$的结构就很明显为$\mathrm{HOXeF}$，含四种元素和四个原子，故F选项正确，E，C选项错误。

<Checkpoint>[$\mathbf{C}$的结构为$\mathrm{HOXeF}$](1)</Checkpoint>