The element **X** exhibits stable properties with strong corrosion resistance, capable of withstanding corrosion from hydrochloric acid, sulfuric acid, nitric acid, and aqua regia at room temperature. Fusing the elemental **X** with potassium nitrate and potassium hydroxide yields **A**. After cooling the melt, it is extracted with water and treated with chlorine gas, followed by heating to volatilize **B**. Introducing **B** into a hydrochloric acid solution and heating it, then adding ammonium chloride precipitates element **X** in the form of **C**, which, upon ignition in hydrogen gas, yields spongy elemental **X**. In a hydrochloric acid solution of potassium chloride, reducing **B** with ethanol produces the red crystalline compound **D**, in which the mass fraction of **X** is $26.99\%$. Heating **D** to $200 ℃$ results in a weight loss of $4.81\%$, yielding the brown crystalline compound **E**. If **B** is replaced with **A** as the reactant, a deep-red diamagnetic dinuclear complex **F** is obtained, in which the mass fraction of **X** is $27.73\%$. Dissolving **F** in concentrated hydrochloric acid and adding ammonia water generates a 1:3 ionic compound **G**. Exposing the solution of **G** to air produces a vivid red color, from which the diamagnetic compound **H** can be isolated. **H** is named for its intense color. Oxidizing **H** with ammonium cerium(IV) sulfate yields the paramagnetic brown compound **I**. It is known that both **H** and **I** are trinuclear complexes with identical cationic chemical formulas, where the mass fraction of **X** in the cation is $52.86\%$, while the anion count ratio is 6:7.  

Given that **A**~**I** are all compounds containing **X**, which of the following options are correct?
元素 **X** 的性质稳定，耐腐蚀性强，常温下能经受盐酸、硫酸、硝酸以及王水的腐蚀。将 **X** 的单质与硝酸钾、氢氧化钾共熔可得 **A**。将熔融物冷却后用水萃取，再通入氯气处理，加热使 **B** 挥发。将 **B** 通入盐酸溶液并加热，再加入氯化铵使元素 **X** 以 **C** 的形式沉淀，**C** 在氢气中灼烧得到海绵状单质 **X**。在氯化钾的盐酸溶液中，用乙醇还原 **B** 生成红色晶体 **D**， **D** 中 **X** 的质量分数为 $26.99\%$。加热 **D** 至 $200 ℃$，失重 $4.81\%$得到褐色晶体 **E**。若将反应物 **B** 更换为 **A**，得到深红色的反磁性双核配合物 **F**，**F** 中 **X** 的质量分数为 $27.73\%$。将 **F** 溶于浓盐酸并加入氨水，生成 1： 3 型离子化合物 **G**。 **G** 的溶液暴露于空气中，产生鲜艳的红色，从中能分离出反磁性化合物 **H**。 **H** 因其强烈的颜色而得名。用硫酸铈铵氧化 **H**，得到顺磁性的褐色化合物 **I**。已知 **H**、 **I** 均为三核配合物，其阳离子的化学式完全相同，阳离子中 **X** 的质量分数为$52.86\%$，而阴离子的个数比为6：7。 

已知 **A**~**I** 均为含 **X** 的化合物，下列选项正确的有：
In **A**, the **X** element has a valence of $+4$.
**B** is a regular octahedral structure.
**C** is a 2:1 type ionic compound
**E** has a molecular weight of approximately $392g/mol$
There is only one set of three-center π bonds in <b>F</b>
The ratio of nitrogen atoms to chlorine atoms in **H** is $7:3$.
In **I**, the average oxidation state of **X** is $+3$
None of the above options are correct
**A** 中 **X** 元素为 $+4$ 价
**B** 为正八面体结构
**C** 为 2：1 型离子化合物
**E** 分子量约为 $392.0\ g/mol$
**F** 中只有一组三中心 $\pi$ 键
**H** 中氮原子与氯原子之比为 $7:3$
**I** 中 **X** 平均化合价为 $+3$
以上选项均不正确
Based on the fact that the elemental form of **X** can withstand corrosion by hydrochloric acid, sulfuric acid, nitric acid, and aqua regia at room temperature, and that it can form complexes, it is determined to be a high-period platinum-group metal.  

<Checkpoint>[**X** is a high-period platinum-group metal](1)</Checkpoint>  

The elemental form of **X**, when fused with potassium nitrate and potassium hydroxide, yields **A**, which takes the form of $K_aXO_b$. After treatment with chlorine gas, volatile **B** is obtained, indicating that it is a high-valent chloride or oxide of **X**. When **B** is reduced with ethanol in a hydrochloric acid solution of potassium chloride, **D** is produced. The cation of **D** is $K^+$, while the anion is a chloride or aqua complex of **X**.  

When **D** is heated at $200 ℃$, it loses $4.81\%$ of its weight, which is attributed to the loss of water of crystallization. Assuming *y* water molecules are lost, when *y* = 1, the relative molecular mass of **D** can be deduced. Subsequently, based on the mass fraction of **X** in **D** being $26.99\%$, the relative atomic mass of **X** is calculated.  

When *y* = 1, the relative atomic mass of **X** is calculated to be 101.1, corresponding to $Ru$, which fits the context.  

For higher values of *y*, no matching element is found.  

Thus, it is concluded that **X** is $Ru$, **A** is $K_2RuO_4$, **B** is $RuO_4$, **D** is $K_2[RuCl_5(H_2O)]$, and **E** is $K_2RuCl_5$.  

**A** is $K_2RuO_4$, where the oxidation state of $Ru$ is $+6$, making option A incorrect. **B** is $RuO_4$, which has a tetrahedral structure, making option B incorrect. **E** is $K_2RuCl_5$, so its molecular weight is approximately $356.6\ g/mol$, making option D incorrect.  

<Checkpoint>[**X** is $Ru$](1)</Checkpoint>  

<Checkpoint>[**A** is $K_2RuO_4$, where $Ru$ is in the $+6$ oxidation state; option A is incorrect](1)</Checkpoint>  

<Checkpoint>[**B** is $RuO_4$, with a tetrahedral structure; option B is incorrect](1)</Checkpoint>  

<Checkpoint>[**D** is $K_2[RuCl_5(H_2O)]$](1)</Checkpoint>  

<Checkpoint>[**E** is $K_2RuCl_5$, with a molecular weight of approximately $356.6\ g/mol$; option D is incorrect](1)</Checkpoint>  

When **B** is passed into a hydrochloric acid solution and heated, followed by the addition of ammonium chloride, **X** precipitates as **C**, which is the complex $(NH_4)_3RuCl_6$, a 3:1 ionic compound, making option C incorrect.  

<Checkpoint>[**C** is the complex $(NH_4)_3RuCl_6$, a 3:1 ionic compound; option C is incorrect](1)</Checkpoint>  

**F** is a deep-red, diamagnetic binuclear complex with a **X** mass fraction of $27.73\%$. Calculations show that it contains an oxo bridge, and **F** is identified as $K_4[Ru_2OCl_{10}]$. It has two sets of three-center $\pi$ bonds perpendicular to the $Ru-O-Ru$ direction, making option E incorrect.  

<Checkpoint>[**F** is $K_4[Ru_2OCl_{10}]$, with two sets of three-center $\pi$ bonds perpendicular to the $Ru-O-Ru$ direction; option E is incorrect](1)</Checkpoint>  

When **F** is dissolved in concentrated hydrochloric acid and treated with ammonia, a 1:3 ionic compound **G** is formed, with $Cl^-$ as the anion. The cation is a trivalent ammonia complex, so **G** is $[Ru(NH_3)_6]Cl_3$.  

<Checkpoint>[**G** is $[Ru(NH_3)_6]Cl_3$](1)</Checkpoint>  

When the solution of **G** is exposed to air, a diamagnetic compound **H** can be isolated. Oxidizing **H** with ammonium cerium(IV) sulfate yields a paramagnetic brown compound **I**. Both **H** and **I** are trinuclear complexes with identical cationic formulas. The mass fraction of **X** in the cation is $52.86\%$, allowing the molar ratio of the cation to be deduced. Apart from $Ru$, the cation may contain $Cl^-$, $NH_3$, or $O^{2-}$. Ultimately, the cation is determined to be $[Ru_3O_2(NH_3)_{14}]^+$, while the ratio of $Cl^-$ anions is 6:7 to maintain reasonable oxidation states for **Ru**. Thus, **H** is $[Ru_3O_2(NH_3)_{14}]Cl_6$, and **I** is $[Ru_3O_2(NH_3)_{14}]Cl_7$. In **H**, the ratio of nitrogen to chlorine atoms is $7:3$, making option F correct. In **I**, the average oxidation state of $Ru$ is $+\frac{11}{3}$, making option G incorrect.  

<Checkpoint>[**H** is $[Ru_3O_2(NH_3)_{14}]Cl_6$, with a nitrogen-to-chlorine ratio of $7:3$; option F is correct](1)</Checkpoint>  

<Checkpoint>[**I** is $[Ru_3O_2(NH_3)_{14}]Cl_7$, with an average $Ru$ oxidation state of $+\frac{11}{3}$; option G is incorrect](1)</Checkpoint>  

In conclusion, the correct answer for this question is F.
根据单质 **X** 在常温下能经受盐酸、硫酸、硝酸以及王水的腐蚀，及其能够形成配合物，判断其为高周期铂族金属。

<Checkpoint>[**X**是高周期铂族金属](1)</Checkpoint>

**X** 的单质与硝酸钾、
氢氧化钾共熔可得 **A**，则 **A** 为 $K_aXO_b$ 的形式，氯气处理后得到 **B** 具有挥发性，说明其为 **X** 的高价氯化物或氧化物。在氯化钾的盐酸溶液中，用乙醇还原 **B** 生成 **D**，则 **D** 的阳离子为 $K^+$，阴离子为 **X** 的氯离子或水的配合物。

**D**在$200 ℃$，失重 $4.81\%$，考虑为失去结晶水的反应；假设失去y个结晶水，当y取1时可以反推出**D**的相对分子质量，之后根据 **D** 中 **X** 的质量分数为 $26.99\%$，推出元素**X**的相对原子质量。

当y取1时，**X**相对原子质量计算为101.1，为$Ru$，符合题意。

y取更高时均无符合元素。

从而推出 **X** 是$Ru$，**A** 是$K_2RuO_4$，**B** 是 $RuO_4$，**D** 是$K_2[RuCl_5(H_2O)]$，**E** 是 $K_2RuCl_5$。

**A** 是 $K_2RuO_4$，$Ru$ 元素为$+6$价，选项A错误。**B** 是 $RuO_4$，所以 **B** 为正四面体结构，选项B错误。**E** 是 $K_2RuCl_5$，所以 **E** 分子量约为 $356.6\ g/mol$，选项D错误

<Checkpoint>[**X** 是 $Ru$](1)</Checkpoint>

<Checkpoint>[**A** 是 $K_2RuO_4$，$Ru$ 元素为$+6$价，选项A错误](1)</Checkpoint>

<Checkpoint>[**B** 是 $RuO_4$，正四面体结构，选项B错误](1)</Checkpoint>

<Checkpoint>[**D** 是 $K_2[RuCl_5(H_2O)]$](1)</Checkpoint>

<Checkpoint>[ **E** 是 $K_2RuCl_5$，分子量约为 $356.6\ g/mol$，选项D错误](1)</Checkpoint>

将 **B** 通入盐酸溶液并加热，再加入氯化铵使元素 **X** 以 **C** 的形式沉淀，则 **C** 是配合物$(NH_4)_3RuCl_6$，为 3：1 型离子化合物，选项C错误。

<Checkpoint>[ **C** 是 配合物$(NH_4)_3RuCl_6$，为 3：1 型离子化合物，选项C错误](1)</Checkpoint>

**F** 为深红色的反磁性双核配合物，其中 **X** 的质量分数为 $27.73\%$，经计算得其中含有氧桥，**F** 是 $K_4[Ru_2OCl_{10}]$。其在垂直于 $Ru-O-Ru$ 方向上有两组三中心 $\pi$ 键，选项E错误

<Checkpoint>[**F** 是 $K_4[Ru_2OCl_{10}]$，在垂直于 $Ru-O-Ru$ 方向上有两组三中心 $\pi$ 键，选项E错误](1)</Checkpoint>

 **F** 溶于浓盐酸并加入氨水，生成 1： 3 型离子化合物 **G**，阴离子为$Cl^-$，则阳离子为正三价氨配合物，**G** 是 $[Ru(NH_3)_6]Cl_3$。

<Checkpoint>[**G** 是 $[Ru(NH_3)_6]Cl_3$](1)</Checkpoint>

**G** 的溶液暴露于空气中，能分离出反磁性化合物**H**。用硫酸铈铵氧化 **H**，得到顺磁性的褐色化合物 **I**。根据**H**、 **I** 均为三核配合物，阳离子的化学式完全相同，阳离子中 **X** 的质量分数为$52.86\%$，可推出其阳离子摩尔量，除$Ru$之外还可能含有$Cl^-$、$NH_3$、$O^{2-}$，最终得到阳离子为$[Ru_3O_2(NH_3)_{14}]^+$，而阴离子$Cl^-$的比例为 6： 7，为满足 **Ru** 价态合理所以分别为6个和7个，因此 **H** 是 $[Ru_3O_2(NH_3)_{14}]Cl_6$，**I** 是 $[Ru_3O_2(NH_3)_{14}]Cl_7$。**H** 中氮原子与氯原子之比为 $7:3$，选项F正确，**I** 中 $Ru$ 平均化合价为 $+\frac{11}{3}$，选项G错误。

<Checkpoint>[**H** 是 $[Ru_3O_2(NH_3)_{14}]Cl_6$，氮原子与氯原子之比为 $7:3$，选项F正确](1)</Checkpoint>

<Checkpoint>[**I** 是 $[Ru_3O_2(NH_3)_{14}]Cl_7$， $Ru$ 平均化合价为 $+\frac{11}{3}$，选项G错误](1)</Checkpoint>

综上所述，此题答案为F。