An oxyacid $\mathbf A$ is an inorganic strong acid with a highly stable corresponding anion. $\mathbf X$ is an extremely unstable, explosive gas that can be prepared using $\mathbf A$ or its salts as substrates. Two methods of preparation are as follows:

(1) Under cooling, a 60% aqueous solution of acid $\mathbf A$ reacts with a pale yellow gas $\mathbf B$ to produce gas $\mathbf X$.

(2) In hydrogen fluoride at $- 78 ^ { \circ } C$, the reaction of the cesium salt of $\mathbf A$ with $\mathrm{NF_4SbF_6}$ yields a precipitate $\mathbf C$ and a readily decomposable $\mathbf D$; $\mathbf D$ slowly decomposes into $\mathbf E$ and $\mathbf X$ at $0 \degree \mathrm { C }$.

Vapor density determination experiments show that at $25 \degree \mathrm C $, the density of $\mathbf X$ is $4 . 85 ~ \mathrm { g / L }$
含氧酸 $\mathbf A$ 是一种无机强酸，其对应的阴离子十分稳定。 $\mathbf X$ 是一种极不稳定，易爆炸的气体，可由 $\mathbf A$ 或其盐为底物制备，其中的两种制备方法如下：

(1)在冷却下，浓度为 $60 \%$ 的酸 $\mathbf A$ 水溶液与淡黄色气体 $\mathbf B$ 反应，生成气体 $\mathbf X$。

(2)在 $- 78 ^ { \circ } C$ 的氟化氢中，使 $\mathbf A$ 的铯盐与 $\mathrm{NF_4SbF_6}$ 反应，得到沉淀 $\mathbf C$ 和易分解 $\mathbf D$；$\mathbf D$ 在 $0 \degree \mathrm { C }$ 缓慢地分解为 $\mathbf E$ 和 $\mathbf X$。

蒸气密度测定实验表明， $25 \degree \mathrm C $ 下， $\mathbf X$ 的密度为 $4 . 85 ~ \mathrm { g / L }$
The configuration of the corresponding anion of substance $\mathbf A$ is trigonal pyramidal.
The substance $\mathbf B$ contains elements from the third period.
The configuration of substance $\mathbf E$ is a trigonal pyramid
The anionic configuration of substance $\mathbf C$ is not octahedral
Only covalent bonds exist in material $\mathbf D$.
Substance $\mathbf X$ contains 3 oxygen atoms
The anions in substance $\mathbf A$ and substance $\mathbf D$ are not identical.
物质 $\mathbf A$ 的对应阴离子的构型是三角锥形
物质 $\mathbf B$ 中含有第三周期元素
物质 $\mathbf E$ 的构型是三角锥
物质 $\mathbf C$ 的阴离子构型不是八面体
物质 $\mathbf D$ 中只存在共价键
物质 $\mathbf X$ 中含有3个氧原子
物质 $\mathbf A$ 和物质 $\mathbf D$ 中的阴离子不相同
According to the ideal gas equation $ PM = \rho RT$, with the default pressure set to 1 atmosphere, the relative molecular mass of substance $\mathbf X$ is calculated to be 118.65.

<Checkpoint>[The relative molecular mass of $\mathbf X$ is 118.65](1)</Checkpoint>

The common pale-yellow gas $\mathbf B$ is likely to be $F_2$. Considering the highly unstable and explosive nature of substance $\mathbf X$, it is inferred to possess strong oxidizing properties, supporting the conclusion that $\mathbf B$ is $F_2$.

<Checkpoint>[$\mathbf B$ is $F_2$](1)</Checkpoint>

Since oxygen-containing acid $\mathbf A$ reacts with $F_2$ to produce $\mathbf X$, it must contain at least $O$ and $F$ elements. Based on the relative molecular mass of $\mathbf X$, a table is constructed for possible combinations of $O$ and $F$. Starting with 1 $F$ atom and $n$ $O$ atoms, it is found that when $n = 4$, the corresponding relative molecular mass is 35.65, which is close to that of $Cl$. Given that $\mathbf A$ is an inorganic strong acid with a highly stable anion, it is deduced that $\mathbf A$ is $HClO_4$ and $\mathbf X$ is $FClO_4$.

<Checkpoint>[$\mathbf A$ is perchloric acid](1)</Checkpoint>

<Checkpoint>[$\mathbf X$ is $FClO_4$](1)</Checkpoint>

The cesium salt of $\mathbf A$ reacts with $NF_4SbF_6$ to form precipitates $\mathbf C$ and $\mathbf D$, suggesting a double displacement reaction. Therefore, $\mathbf C$ is $CsSbF_6$, and $\mathbf D$ is $NF_4ClO_4$. The decomposition of $\mathbf D$ yields $\mathbf X$, with the remaining molecular formula being $NF_3$, which is $\mathbf E$.

<Checkpoint>[$\mathbf C$ is $CsSbF_6$](1)</Checkpoint>

<Checkpoint>[$\mathbf D$ is $NF_4ClO_4$](1)</Checkpoint>

The anion of $\mathbf A$ has a tetrahedral geometry, $\mathbf E$ has a trigonal pyramidal geometry, and the anion of $\mathbf C$ has an octahedral geometry. $\mathbf D$ contains both covalent and ionic bonds, so option C is correct.

<Checkpoint>[The anion of $\mathbf A$ has a tetrahedral geometry](1)</Checkpoint>

<Checkpoint>[$\mathbf E$ has a trigonal pyramidal geometry](1)</Checkpoint>

<Checkpoint>[The anion of $\mathbf C$ has an octahedral geometry](1)</Checkpoint>
根据理想气体方程 $ PM = \rho RT$，默认压力为1个大气压，计算得出物质 $\mathbf X$ 的相对分子质量为118.65

<Checkpoint>[$\mathbf X$ 的相对分子质量为118.65](1)</Checkpoint>

常见的淡黄色气体$\mathbf B$可能为$F_2$，结合物质$\mathbf X$极不稳定易爆炸的性质，推测其具有强氧化性，佐证$\mathbf B$即为$F_2$。

<Checkpoint>[$\mathbf B$为$F_2$](1)</Checkpoint>

由于含氧酸$\mathbf A$和$F_2$反应生成$\mathbf X$，则其中至少应该含有$O$元素和$F$元素。由此根据$\mathbf X$的相对分子质量，对$O$和$F$进行打表，首先尝试1个$F$原子和n个$O$原子，发现在$n = 4$时对应相对分子质量为35.65，与$Cl$接近。结合含氧酸$\mathbf A$是无机强酸且阴离子十分稳定的性质，推出$\mathbf A$为$HClO_4$，$\mathbf X$为$FClO_4$

<Checkpoint>[$\mathbf A$为高氯酸](1)</Checkpoint>

<Checkpoint>[$\mathbf X$为$FClO_4$](1)</Checkpoint>

$\mathbf A$的铯盐与$NF_4SbF_6$反应生成沉淀$\mathbf C$和$\mathbf D$，推测发生复分解反应，因此$\mathbf C$为$CsSbF_6$，$\mathbf D$为$NF_4ClO_4$。$\mathbf D$分解产生$\mathbf X$，剩余分子式为$NF_3$，即为$\mathbf E$。

<Checkpoint>[$\mathbf C$为$CsSbF_6$](1)</Checkpoint>

<Checkpoint>[$\mathbf D$为$NF_4ClO_4$](1)</Checkpoint>

$\mathbf A$的阴离子为正四面体型，$\mathbf E$ 的构型是三角锥型，$\mathbf C$ 的阴离子构型为八面体型，$\mathbf D$中含共价键和离子键，故选项C正确

<Checkpoint>[$\mathbf A$的阴离子为正四面体型](1)</Checkpoint>

<Checkpoint>[$\mathbf E$ 的构型是三角锥型](1)</Checkpoint>

<Checkpoint>[$\mathbf C$ 的阴离子构型为八面体型](1)</Checkpoint>