Element **A** is generally found only in associated minerals, and independent deposits are rare. When a mineral containing **A** is mixed with potassium carbonate and burned in the presence of air, a potassium salt **B** containing **A** is obtained. Upon acidification and evaporative crystallization of **B**, colorless crystals **C** are produced. **C** readily dehydrates to form a yellow solid **D**. When **D** reacts with $\rm SF_4$ at temperatures above $120℃$, only a gas **E** containing **A** with a mass fraction of $62.67\%$ and another gas **F** are obtained. **F** reacts with an aqueous solution of **C** to precipitate black crystals **G**. When **G** reacts with $\rm SbF_5$ in $\rm SO_2$ at $-30℃$ in a stoichiometric ratio of $8:5$, an ionic compound **H** and $\rm SbF_3$ are produced in a $1:1$ product ratio. The cation of compound **H** has a symmetric and cyclic structure.  

Please analyze the substances described above through reasonable logic or conjecture, determine what each substance is (providing molecular formulas rather than simplest formulas), and answer whether the following options are correct.
某元素 **A** 一般仅见于伴生矿产，独立矿床罕见。在含 **A** 的矿物中加入碳酸钾并通入空气灼烧，得到含 **A** 的钾盐 **B**，**B** 酸化后蒸发结晶，得到无色晶体 **C**，**C** 易脱水得到黄色固体 **D**。**D** 与 $\rm SF_4$在 $120℃$以上反应，仅得到含 **A** 质量分数为 $62.67\%$ 的气体 **E** 和另一种气体 **F**。**F** 与 **C** 的水溶液反应可沉淀出黑色晶体 **G**。**G** 与 $\rm SbF_5$在 $-30℃$ 下于 $\rm SO_2$中按照 $8:5$ 的计量比反应，生成一种离子化合物 **H** 和 $\rm SbF_3$，产物比例为 $1:1$，化合物 **H** 的阳离子为对称并环状结构。

请你理解上述题目，经过合理的逻辑或猜想，分析以上各物质各是什么（注意给出分子式而非最简式），并回答以下选项是否正确。
In an infinitely layered **D**, let the other element besides **A** be denoted as **Y**. Then, the ratio of terminal **Y** to bridging **Y** is $2:1$.
The gas **E** has a square planar structure.
The element in gas **F** with the highest oxidation state has a structure similar to that of element **G**.
**H** is a $1:1$ type ionic compound
There is no **A**-**A** bond in the cation of **H**.
It is known that one of the conformations of the **H** cation has two mirror planes, thus its molecular point group is the same as that of the water molecule.
None of the other options are correct.
G has a crystal structure with $\mathbf{A}_8$ as the structural motif.
无限层状的**D**中，记除**A**以外的另一种元素为**Y**，则端基**Y**和桥连**Y**的比例为$2:1$。
气体**E**是平面四方形结构
气体**F**中最高氧化态的元素的单质有和**G**类似的结构
**H**是一种$1:1$型的离子化合物
**H**的阳离子中不存在**A**-**A**键
已知**H**的阳离子的其中一种构象具有两个镜面，则其所属分子点群和水分子一致。
其他选项均不正确。
G具有以$\mathbf{A}_8$为结构基元的晶体结构
At first glance, this question is difficult to answer directly. Therefore, start with **E**, which has a definite mass fraction, and first determine which other elements **E** contains.
From the question, $\rm SF_4$ acts as a fluorinating agent here, and undergoes a metathesis-like reaction with the oxide **D** formed after acid dehydration. **F** should be $\rm SO_2$, then **E** is a fluoride.<br>

<Checkpoint>[$\rm SF_4$ acts as a fluorinating agent here, **F** is $\rm SO_2$, **E** is a fluoride.](1)</Checkpoint><br>

Substituting the mass fraction, the relative atomic mass of **A** can be calculated. The mass fraction of **A** in $$\text{AF}_4$$ is $$\frac{x}{x+4×19}≈45.23\%$$ , solving for $x=127.6(\rm g/mol)$, corresponding to the element tellurium.

<Checkpoint>[The relative atomic mass of **A** is approximately 127.6, corresponding to the element tellurium.](1)</Checkpoint><br>

The answers from **A** to **F** in this question are obvious.<br>
Under the action of alkali, tellurium can be oxidized by oxygen, but it is difficult to be directly oxidized to the highest valence +6, it can only be oxidized to the +4 valence, to obtain potassium tellurite **B**.<br>

<Checkpoint>[In **B**, $\rm Te$ has a +4 valence, i.e., it is $\rm K_2TeO_3$](1)</Checkpoint>

Acidification of potassium tellurite yields tellurous acid **C**, dehydration yields tellurium dioxide **D**, and the logic up to this point is very smooth.<br>

<Checkpoint>[**C** is tellurous acid $\rm H_2TeO_3$, **D** is tellurium dioxide $\rm TeO_2$.](1)</Checkpoint>

Based on this, answer options A and B: Analyzing the atomic ratio, in layered $$\text{TeO}_2$$, the ratio of tellurium to bridging oxygen is obviously $1:4$, each tellurium corresponds to 4 bridging oxygens, and there are no terminal oxygen atoms, A is incorrect;<br>

<Checkpoint>[Tellurium dioxide has no terminal oxygens](1)</Checkpoint>

Using VSEPR to judge, in $$\text{TeF}_4$$, $\rm Te$ has 6 electrons, forms four bonds, and has one lone pair of electrons remaining, which is a distorted tetrahedral structure, B is incorrect.<br>

<Checkpoint>[Tellurium tetrafluoride has a distorted tetrahedral structure](1)</Checkpoint>

$$\text{SO}_2$$ is a reducing agent, and undergoes a redox reaction with the oxidizing agent $$\text{H}_2\text{TeO}_3$$, the product should be elemental tellurium (acid-base reaction may be considered here, but the product cannot obtain a precipitate). $\mathrm{Te}$ does not have a $\rm Te_8$ cyclic structure, but a spiral crystal, H is incorrect<br>

<Checkpoint>[$\rm SO_2$ acts as a reducing agent here](1)</Checkpoint><br>

$$\text{SbF}_5$$ can act as both an oxidizing agent and a coordinating agent. The production of $$\text{SbF}_3$$ indicates that it does act as an oxidizing agent.
Trying it out, it can be found that $$\text{SbF}_5$$ cannot avoid acting as a Lewis acid. Discuss the common anions such as $$\rm [Sb_2F_{11}]^-$$ or $$\rm [SbF_{6}]^-$$ respectively, try to balance the equation, and finally it is concluded that it is the former, and the chemical formula of **H** can be obtained as $$\text{Te}_8[\text{Sb}_2\text{F}_{11}]_2$$ .<br>

<Checkpoint>[Analyze that **G** is elemental tellurium, and give the information that **H** contains $\rm Te_8^{2+}$](2)</Checkpoint><br>

Based on this, it can be known that the element in the highest oxidation state in the gas **F** is $\rm S_8$, which has a different structure from the spiral crystal structure of Te, C is incorrect.<br>

 **H** is obviously a $1:2$ type ionic compound, D is incorrect.<br>

The $$\text{Te}_8^{2+}$$ ion combined with the question prompt (symmetric and cyclic structure) should be a fused five-membered ring structure. Combining VSEPR knowledge, $\rm Te$ and its homolog $\rm O$ form two bonds when uncharged; when forming three bonds, it should carry a formal positive charge to satisfy the octet rule, each $\rm Te$ bridging two $\rm Te$ carries one positive charge, and the ion as a whole carries two positive charges. Therefore, there is bonding interaction between $\rm Te$ and $\rm Te$, E is incorrect.<br>

<Checkpoint>[$\rm Te_8^{2+}$ ion contains $\rm Te$-$\rm Te$ bonds](2)</Checkpoint><br>

Combining organic chemistry knowledge, there are two structures for the fused five-membered ring, i.e., cis-trans isomers, where the cis structure has one two-fold axis and two mirror planes, $C_{2v}$ point group, which is consistent with the water molecule and meets the requirements of the question, F is correct.<br>

<Checkpoint>[There exists a certain (cis) $\rm Te_8^{2+}$ structure that is consistent with the molecular point group of the water molecule](1)</Checkpoint>

In summary, the substances are:
- **A**: $$\text{Te}$$  
  (Tellurium, verified by mass fraction calculation)
- **B**: $$\text{K}_2\text{TeO}_3$$  
  (Potassium carbonate + mineral calcined in air, oxidized to potassium tellurite)
- **C**: $$\text{H}_2\text{TeO}_3$$  
  (B acidified to obtain tellurous acid)
- **D**: $$\text{TeO}_2$$  
  (Tellurous acid dehydrated to obtain tellurium dioxide, yellow solid)
- **E**: $$\text{TeF}_4$$  
  ($$\text{TeO}_2 + \text{SF}_4 → \text{TeF}_4 + \text{SO}_2$$, distorted tetrahedral structure)
- **F**: $$\text{SO}_2$$  
  (Byproduct)
- **G**: $$\text{Te}$$  
  ($$\text{SO}_2$$ reacts with $$\text{H}_2\text{TeO}_3$$ to precipitate fused cyclic $$\text{Te}_8$$, black crystal)
- **H**: $$\text{Te}_8[\text{Sb}_2\text{F}_{11}]_2$$  
  ($$8\text{Te} + 5\text{SbF}_5 → \text{Te}_8^{2+}·2[\text{Sb}_2\text{F}_{11}]^- + \text{SbF}_3)$$) <br>

The following is a summary of the answers for each option:<br>

A. In the long-chain $$\text{TeO}_2$$, the ratio of tellurium to bridging oxygen is obviously $1:1$, and each tellurium has one terminal oxygen, so the required ratio is $1:1$, which is incorrect.<br>

B. $$\text{TeF}_4$$ is a distorted tetrahedral structure, which is incorrect.<br>

C. The element in the highest oxidation state in the gas **F** is $\rm S_8$, which has a similar structure to $\rm Te_8$, but is yellow in color, the colors are different, which is incorrect.<br>

D. **H** is a $1:2$ type ionic compound, which is incorrect.<br>

E. $\rm Te_8^{2+}$ has a fused five-membered ring structure, and the two bridging $Te$ have bonding interactions, which is incorrect.<br>

F. There are two types of fused five-membered ring structures, cis and trans. The cis structure has one two-fold axis and two mirror planes, $C_{2v}$ point group, which is consistent with the water molecule and meets the requirements of the question, which is correct. Therefore, G is incorrect.

The answer to this question is F.
乍看之下本题较难直接推出答案，因此从有明确质量分数的 **E** 入手，先判断 **E** 还含有哪些元素。
从题中判断，$\rm SF_4$ 在此处做氟化剂，与酸脱水后生成的氧化物 **D** 做类复分解反应，**F** 应该是 $\rm SO_2$，则 **E** 是氟化物。<br>

<Checkpoint>[$\rm SF_4$ 在此处做氟化剂，**F** 是 $\rm SO_2$， **E** 是氟化物。](1)</Checkpoint><br>

带入质量分数可算出 **A** 的相对原子质量。$$\text{AF}_4$$中**A**质量分数为$$\frac{x}{x+4×19}≈45.23\%$$，解得$x=127.6(\rm g/mol)$，对应元素碲。

<Checkpoint>[**A** 的相对原子质量大约为127.6，对应元素碲。](1)</Checkpoint><br>

此题从 **A** 到 **F** 答案跃然纸上。<br>
在碱的作用下，碲可以被氧气氧化，但难以被直接氧化到最高价+6，只能氧化到+4价，得到亚碲酸钾**B**。<br>

<Checkpoint>[**B**中$\rm Te$为+4价，即其为$\rm K_2TeO_3$](1)</Checkpoint>

亚碲酸钾酸化得到亚碲酸**C**，脱水得到二氧化碲**D**，至此的逻辑都是很顺畅的。<br>

<Checkpoint>[**C**为亚碲酸$\rm H_2TeO_3$，**D**为二氧化碲$\rm TeO_2$。](1)</Checkpoint>

据此回答选项A、B：分析原子比例，层状$$\text{TeO}_2$$ 中显然碲和桥连氧$1:4$，每个碲对应4个桥连氧，没有端基氧原子，A错误；<br>

<Checkpoint>[二氧化碲没有端基氧](1)</Checkpoint>

利用VSEPR判断，$$\text{TeF}_4$$ 中$\rm Te$有6个电子，成四根键，余一个孤对电子，是变形四面体结构，B错误。<br>

<Checkpoint>[四氟化碲为变形四面体结构](1)</Checkpoint>

$$\text{SO}_2$$是还原剂，与氧化剂$$\text{H}_2\text{TeO}_3$$ 发生氧化还原反应，生成的应该是单质碲（此处有可能考虑酸碱反应，但产物不可能得到沉淀）。$\mathrm{Te}$没有$\rm Te_8$环状结构，为螺旋状晶体，H错误<br>

<Checkpoint>[$\rm SO_2$在这里充当还原剂](1)</Checkpoint><br>

$$\text{SbF}_5$$既可以做氧化剂又可以做配位剂，产生$$\text{SbF}_3$$说明确实作氧化剂。
尝试可以发现$$\text{SbF}_5$$不能不作 Lewis 酸。分别讨论阴离子是$$\rm [Sb_2F_{11}]^-$$ 或 $$\rm [SbF_{6}]^-$$等常见阴离子，试进行配平，最后得出是前者，可以得到 **H** 的化学式为$$\text{Te}_8[\text{Sb}_2\text{F}_{11}]_2$$ 。<br>

<Checkpoint>[分析出**G**为单质碲，并给出**H**含$\rm Te_8^{2+}$的信息](2)</Checkpoint><br>

据此可知，气体**F**中最高氧化态的元素的单质即$\rm S_8$，与Te的螺旋晶体结构不同，C错误。<br>

 **H**显然是一种$1:2$型的离子化合物，D错误。<br>

其中的$$\text{Te}_8^{2+}$$离子结合题目提示（对称并环状结构）应为五并五的结构。结合VSEPR知识，$\rm Te$与同族的$\rm O$在不带电荷时成两根键；成三根键时为满足八隅律应带形式正电荷，桥接两个$\rm Te$各带一个正电荷，离子整体带两个正电荷。因此，$\rm Te$与$\rm Te$之间存在键连作用，E错误。<br>

<Checkpoint>[$\rm Te_8^{2+}$离子中含$\rm Te$-$\rm Te$键](2)</Checkpoint><br>

结合有机化学知识，五并五环有两种结构即顺反异构，其中的顺式结构具有一根二重轴和两个镜面，$C_{2v}$点群，与水分子一致，符合题目要求，F正确。<br>

<Checkpoint>[存在某种（顺式）$\rm Te_8^{2+}$的结构与水分子的分子点群一致](1)</Checkpoint>

综上所述，各物质为：
- **A**: $$\text{Te}$$  
  （碲，质量分数计算验证）
- **B**: $$\text{K}_2\text{TeO}_3$$  
  （碳酸钾+空气灼烧矿物，氧化成亚碲酸钾）
- **C**: $$\text{H}_2\text{TeO}_3$$  
  （B酸化得亚碲酸）
- **D**: $$\text{TeO}_2$$  
  （亚碲酸脱水得二氧化碲，黄色固体）
- **E**: $$\text{TeF}_4$$  
  （$$\text{TeO}_2 + \text{SF}_4 → \text{TeF}_4 + \text{SO}_2$$，变形四面体结构）
- **F**: $$\text{SO}_2$$  
  （副产物）
- **G**: $$\text{Te}$$  
  （$$\text{SO}_2$$与$$\text{H}_2\text{TeO}_3$$反应析出并环状$$\text{Te}_8$$，黑色晶体）
- **H**: $$\text{Te}_8[\text{Sb}_2\text{F}_{11}]_2$$  
  （$$8\text{Te} + 5\text{SbF}_5 → \text{Te}_8^{2+}·2[\text{Sb}_2\text{F}_{11}]^- + \text{SbF}_3)$$ <br>

下面整理各选项回答：<br>

A. 长链$$\text{TeO}_2$$  中显然碲和桥连氧$1:1$，每个碲有一个端基氧，故所求比例为$1:1$，错误。<br>

B. $$\text{TeF}_4$$  是变形四面体结构，错误。<br>

C. 气体**F**中最高氧化态的元素的单质即$\rm S_8$，结构与$\rm Te_8$相似，但为黄色，颜色不同，错误。<br>

D. **H**是一种$1:2$型的离子化合物，错误。<br>

E. $\rm Te_8^{2+}$为五并五环状结构，桥接的两个$Te$有键联作用，错误。<br>

F. 五并五环状结构有顺反两种，顺式结构具有一根二重轴和两个镜面，$C_{2v}$点群，与水分子一致，符合题目要求，正确。因此，G错误。

本题选F。