The elemental **X** reacts with oxygen at high temperatures to form a yellow solid **A**. Dissolving **A** in $\rm HClO_4$ yields a perchlorate **B**. Upon dropwise addition of $\rm NaOH$ solution until neutral, a hexanuclear cation **C** with a +6 charge is obtained in the solution. This ion exhibits high symmetry, possessing four threefold axes, with all **X** atoms being tetrahedrally coordinated. Further addition of $\rm NaOH$ produces a white precipitate **D**. Continuing to add $\rm NaOH$ until alkaline and introducing $\rm Cl_2$ yields a yellow precipitate **E**.  

Element **Y** is a transition metal whose elemental form reacts with water under certain conditions to form a white precipitate **F**, which rapidly converts to a brown substance **G** containing approximately $52.35\%$ **Y** by mass upon exposure to air. It is known that dissolving **F** in dilute sulfuric acid and then adding **E** results in a purple solution **H**.  

Deduce the chemical formulas of each substance and determine which of the following statements is correct:
元素 **X** 的单质高温下和氧气反应生成黄色固体 **A**，将 **A** 溶于 $\rm HClO_4$ 中得到高氯酸盐 **B**，滴加 $\rm NaOH$ 溶液至中性，溶液中得到一个带 6 个正电荷的 6 核阳离子 **C**，该离子具有高度对称性，存在四根三重轴，**X** 均为四配位。继续滴加 $\rm NaOH$，得到一白色沉淀 **D**，再滴加 $\rm NaOH$ 至碱性，通入 $\rm Cl_2$，得到一黄色沉淀 **E**。

元素 **Y** 是一种过渡金属元素，其单质在一定条件下能与水反应生成白色的沉淀 **F**，和空气接触会立刻转化为含 **Y** 质量分数约为$52.35\%$的棕色物质 **G**。已知：将 **F** 溶于稀硫酸溶液后，再加入 **E**，可得一紫色溶液 **H**。

推断各物质的化学式，判断下列说法正确的是：
If the electron configuration of element **X** is written, its atomic core should be $\rm[Xe]$.
The molecular point group of the cation **C** is consistent with that of methane.
The white precipitate **D** contains chlorine.
The mass fraction of element **X** in the yellow precipitate **E** is $80.25\%$.
During the process of **F** generating **H**, the oxidation state of the central element increases by 4.
None of the above options are correct.
若写出元素**X**的核外电子排布式，其原子实应当为$\rm[Kr]$。
阳离子**C**的分子点群与甲烷分子一致。
白色沉淀**D**中含有氯元素。
黄色沉淀**E**中**X**元素的质量分数为$80.25\%$。
**F**生成**H**的过程，中心元素的氧化态上升了4。
以上选项均不正确。
First, **X** can react with oxygen when heated in air, indicating its relatively active chemical properties, which rules out the stable noble metals in air.

<Checkpoint>[**X** can react with oxygen when heated in air, so it is not a stable noble metal in air.](1)</Checkpoint>

Its combination with oxygen should produce oxide **A**, which is stable in air and represents its primary oxidation state. **A** dissolves in perchloric acid, with only two possibilities: either no redox reaction occurs, or it is oxidized by perchloric acid to a higher oxidation state. However, perchloric acid is unlikely to perform oxidation reactions under normal conditions that are difficult for oxygen to achieve in dry heating, and perchloric acid also lacks the ability to alter redox potentials through coordination. Therefore, it is highly probable that the reaction from **A** to **B** is a non-redox reaction.

<Checkpoint>[The reaction from **A** to **B** is a non-redox reaction.](1)</Checkpoint>

From **B** to **C**, a hydrolysis reaction occurs, forming a hexanuclear complex with four threefold axes. First, consider the $\rm O_h$ point group, i.e., a regular octahedral shape. Given the possible oxidation states of **X**, it cannot be +1 (as no hydroxide or oxide ions would form), but it could be +2 or +3. Oxidation states of +4 or higher would make it difficult to propose a reasonable structure with hydroxide or oxide ion counts that match the symmetry.<br>

**D** is clearly the precipitate of the fully hydrolyzed hydrate, while the yellow precipitate **E** is more likely the product of a redox reaction with chlorine gas. Therefore, **E** should be a strong oxidizing agent.

<Checkpoint>[**E** should be a strong oxidizing agent.](1)</Checkpoint>

Based on the subsequent text, **F** is oxidized under acidic conditions by **E** to produce a purple-red solution, which immediately suggests the permanganate ion $\rm MnO_4^-$. Thus, it is hypothesized that **Y** should be manganese. Following this logic, **F** is manganese(II) hydroxide $\rm Mn(OH)_2$, and the condition "when exposed to air, it immediately transforms into a brown substance **G** containing **Y** with a mass percentage of approximately $52.35\%$" can be used to verify the hypothesis, calculating **G** as $\rm MnO(OH)_2$. The oxidation state of $\rm Mn^{2+}$ increases from +2 to +7 in $\rm MnO_4^-$, a rise of 5, so option E is incorrect.

<Checkpoint>[**F** is $\rm Mn(OH)_2$.](1)</Checkpoint>

<Checkpoint>[The chemical formula of **G** is $\rm MnO(OH)_2$.](1)</Checkpoint>

<Checkpoint>[$\rm Mn^{2+}$ is oxidized to $\rm MnO_4^-$, with the oxidation state increasing from +2 to +7, a rise of 5, so option E is incorrect.](1)</Checkpoint>

There are few strong oxidizing agents capable of oxidizing $\rm Mn(II)$ to $\rm MnO_4^-$ under acidic conditions. With some thought, the most likely candidate is $\rm NaBiO_3$, as only bismuth can achieve the +3 oxidation state ($\rm Bi(III)$) before being oxidized by chlorine gas, and it forms a characteristic sodium ion precipitate.

<Checkpoint>[Based on oxidizing ability and solubility, **E** is determined to be $\rm NaBiO_3$.](2)</Checkpoint>

Calculations show that the mass percentage of $\rm Bi$ in **E** ($\rm NaBiO_3$) is approximately $74.64\%$, while the given data in the question is $80.25\%$, indicating an error, so option D is incorrect. From this, the central element of the problem is deduced to be bismuth. Bismuth is in the sixth period with an electron core of $\rm [Xe]$, not $\rm [Kr]$, so option A is incorrect.

<Checkpoint>[The mass percentage of $\rm Bi$ in **E** ($\rm NaBiO_3$) is approximately $74.64\%$, so option D is incorrect.](1)</Checkpoint>

<Checkpoint>[The electron core of bismuth is $\rm [Xe]$, so option A is incorrect.](1)</Checkpoint>

Proceeding sequentially, **X** is oxidized in air to form the yellow oxide $$\rm Bi_2O_3$$ (**A**), which then dissolves in perchloric acid in a non-redox reaction to form $$\text{Bi}(\text{ClO}_4)_3$$ (**B**).

<Checkpoint>[**A** is $$\rm Bi_2O_3$$, and **B** is $$\text{Bi}(\text{ClO}_4)_3$$.](1)</Checkpoint>

Next, using oxidation state analysis, the question provides the number and oxidation state of $$\text{Bi}^{3+}$$ (+6), allowing the number of hydroxide ions to be easily determined as 12 based on charge balance. The possibility of oxygen atoms is also considered, but replacing hydroxide ions with oxide ions does not yield a structure with high symmetry.

<Checkpoint>[**C** is $$\text{Bi}(\text{OH})_{12}^{6+}$$, and replacing hydroxide ions with oxide ions does not yield a structure with high symmetry.](1)</Checkpoint>

In reality, this forms an ion resembling a cube, where $$\text{Bi}^{3+}$$ occupies all face centers and $$\text{OH}^{-}$$ occupies all edge centers, consistent with $O_h$ (regular octahedral) symmetry. Methane, however, has $T_d$ (tetrahedral) symmetry, so they are inconsistent, and option B is incorrect.

<Checkpoint>[The point group of $\rm Bi_6(OH)_{12}^{6+}$ is $O_h$ (regular octahedral), which is inconsistent with methane's $T_d$ (tetrahedral), so option B is incorrect.](2)</Checkpoint>

Throughout the process, perchlorate ions do not participate in redox reactions and do not produce chloride ions, so **D** does not contain chloride ions and is $$\text{Bi}(\text{OH})_3$$, making option C incorrect.

<Checkpoint>[**D** does not contain chloride ions and is $$\text{Bi}(\text{OH})_3$$, so option C is incorrect.](1)</Checkpoint>

Thus, the following can be summarized:

**X**: $$\text{Bi}$$  

**A**: $$\rm Bi_2O_3$$  

**B**: $$\text{Bi}(\text{ClO}_4)_3$$  

**C**: $$\text{Bi}(\text{OH})_{12}^{6+}$$  

**D**: $$\text{Bi}(\text{OH})_3$$  

**E**: $$\text{NaBiO}_3$$

**F**: $\rm Mn(OH)_2$

**G**: $\rm MnO(OH)_2$

**H**: $\rm NaMnO_4$

Analysis of the options is as follows:

A: The electron configuration of $\rm Bi$ is $\rm [Xe] 4f^{14}5d^{10}6s^26p^3$, with an electron core of $\rm [Xe]$, not $\rm [Kr]$. Incorrect.

B: The point group of **C** ($\rm Bi_6(OH)_{12}^{6+}$) is $O_h$ (regular octahedral), while methane is $T_d$ (tetrahedral), so they are inconsistent. Incorrect.

C: **D** ($\rm Bi(OH)_3$) does not contain chlorine. Incorrect.

D: The mass percentage of $\rm Bi$ in **E** ($\rm NaBiO_3$) is approximately $74.64\%$, while the given data is $80.25\%$, which may be erroneous. Incorrect.

E: $\rm Mn^{2+}$ is oxidized to $\rm MnO_4^-$, with the oxidation state increasing from +2 to +7, a rise of 5, not 4. Incorrect.

Therefore, options A-E are all incorrect, and option F is correct.
首先**X**可以在空气中加热和氧气反应，说明其化学性质较为活泼，可以排除空气中稳定的贵金属元素。

<Checkpoint>[**X**可以在空气中加热和氧气反应，不是空气中稳定的贵金属元素。](1)</Checkpoint>

其与氧结合应当生成在空气中较为稳定的、应当为主要价态的氧化物**A**，其溶于高氯酸，只有两种可能：不发生氧化还原反应或被高氯酸氧化至更高价。但是高氯酸在一般条件下较难发生干态加热下氧气难以发生的氧化反应，且高氯酸也不具备配位改变氧化还原电势的可能性，故有较大可能**A**到**B**的反应是非氧化还原反应。

<Checkpoint>[**A**到**B**的反应是非氧化还原反应。](1)</Checkpoint>

**B**到**C**发生水解反应，生成6核配合物，其有四根三重轴，首先考虑$\rm O_h$点群即正八面体形。则综合考虑其中**X**的价态，其不可能为+1（则无氢氧根或氧离子），可能为+2或+3，+4以上则难以写出合理的、氢氧根或氧离子个数符合对称性的结构。<br>

**D**显然为彻底水解后的水化物沉淀，而黄色沉淀**E**则更有可能是和氯气发生氧化还原反应。故**E**应当是一种强氧化剂。

<Checkpoint>[**E**应当是一种强氧化剂](1)</Checkpoint>

结合后文，**F**酸性条件下被其氧化得到紫红色溶液，首先想到高锰酸根$\rm MnO_4^-$离子。因而，猜测**Y**应当为锰元素。顺水推舟，**F**为氢氧化锰$\rm Mn(OH)_2$，可以用“和空气接触会立刻转化为含 **Y** 质量分数约为$52.35\%$的棕色物质 **G**”的条件来验证猜想，计算出**G**为$\rm MnO(OH)_2$。$\rm Mn^{2+}$ 被氧化为 $\rm MnO_4^-$，氧化态从 +2 升至 +7，上升了 5 ，选项E错误.

<Checkpoint>[**F**为$\rm Mn(OH)_2$](1)</Checkpoint>

<Checkpoint>[**G**的化学式为$\rm MnO(OH)_2$](1)</Checkpoint>

<Checkpoint>[$\rm Mn^{2+}$ 被氧化为 $\rm MnO_4^-$，氧化态从 +2 升至 +7，上升了 5 ，选项E错误](1)</Checkpoint>

可以在酸性下把$\rm Mn(II)$氧化至$\rm MnO_4^-$的强氧化剂并不多。稍加联想，最有可能的结果是$\rm NaBiO_3$，因为只有其在被氯气氧化前可以达到+3的氧化态即$\rm Bi(III)$，且是钠离子较为特征性的沉淀。

<Checkpoint>[通过氧化性和可溶性判断**E**为$\rm NaBiO_3$](2)</Checkpoint>

计算可知**E**（$\rm NaBiO_3$） 中$\rm Bi$ 的质量分数约为 $74.64\%$，题目数据 $80.25\%$ 有误，选项D错误。由此推出本题的中心元素为铋元素。铋元素位于第六周期，原子实为$\rm [Xe]$，不是$\rm [Kr]$，选项A错误。

<Checkpoint>[**E**（$\rm NaBiO_3$） 中$\rm Bi$ 的质量分数约为 $74.64\%$，选项D错误](1)</Checkpoint>

<Checkpoint>[铋元素的原子实为$\rm [Xe]$，选项A错误](1)</Checkpoint>

向后依次整理，其在空气中被氧化形成黄色的氧化物$$\rm Bi_2O_3$$（**A**），随后溶于高氯酸进行非氧化还原反应生成$$\text{Bi}(\text{ClO}_4)_3$$（**B**）。

<Checkpoint>[**A**是$$\rm Bi_2O_3$$，**B**是$$\text{Bi}(\text{ClO}_4)_3$$  ](1)</Checkpoint>

随后利用价态分析，题目给出了 $$\text{Bi}^{3+}$$的个数和价态（+6），能够轻易通过电荷得到氢氧根的数量为12。此处可能还涉及到氧原子的个数，但是将氢氧根替换为氧离子得不到符合高对称性的结构。

<Checkpoint>[**C**是$$\text{Bi}(\text{OH})_{12}^{6+}$$，将氢氧根替换为氧离子得不到符合高对称性的结构  ](1)</Checkpoint>

实际上，这里形成了类似正方体的离子，  $$\text{Bi}^{3+}$$占据所有面心，$$\text{OH}^{-}$$占据所有棱心，符合 $O_h$（正八面体）对称性。而甲烷是 $T_d$（四面体），不一致，B错误。

<Checkpoint>[$\rm Bi_6(OH)_{12}^{6+}$ 的点群是 $O_h$（正八面体），与甲烷 $T_d$（四面体）不一致，选项B错误](2)</Checkpoint>

整个过程中高氯酸根没有氧化还原过程，不会产生氯离子，故**D**不含氯离子，为$$\text{Bi}(\text{OH})_3$$ ，选项C错误。

<Checkpoint>[**D**不含氯离子，为$$\text{Bi}(\text{OH})_3$$ ，选项C错误](1)</Checkpoint>

于是可以整理得出：

**X**: $$\text{Bi}$$  

**A**: $$\rm Bi_2O_3$$  

**B**: $$\text{Bi}(\text{ClO}_4)_3$$  

**C**: $$\text{Bi}(\text{OH})_{12}^{6+}$$  

**D**: $$\text{Bi}(\text{OH})_3$$  

**E**: $$\text{NaBiO}_3$$

**F**：$\rm Mn(OH)_2$

**G**：$\rm MnO(OH)_2$

**H**：$\rm NaMnO_4$

选项分析整理如下：

A：$\rm Bi$ 的核外电子排布式为 $\rm [Xe] 4f^{14}5d^{10}6s^26p^3$，原子实为$\rm [Xe]$，不是$\rm [Kr]$。错误。

B：**C**（$\rm Bi_6(OH)_{12}^{6+}$） 的点群是 $O_h$（正八面体），甲烷是 $T_d$（四面体），不一致。错误。

C：**D**（$\rm Bi(OH)_3$） 不含氯元素。错误。

D：**E**（$\rm NaBiO_3$） 中$\rm Bi$ 的质量分数约为 $74.64\%$，题目数据 $80.25\%$ 可能有误。错误。

E：$\rm Mn^{2+}$ 被氧化为 $\rm MnO_4^-$，氧化态从 +2 升至 +7，上升了 5 而非4。错误。

因此，A-E选项均错误，选项F正确。