It is known that the chemical formula of a cluster compound $\mathbf{W}$ containing a p-block non-metallic element can be expressed as $\mathbf{A}_2\mathbf{B}_{14}Cl_{14}$, and its structure contains an electrically neutral fragment of $\mathbf{B}_{10}$ in the middle. The transition metal $\bf{A}$ is coordinated by 6 $\bf{B}$ atoms.

It is known that 5.346 g of the cluster compound $\bf{W}$ can be precisely obtained from 0.768 g of the transition metal element $\bf{A}$, 2.688 g of the element $\bf{B}$, and a certain amount of chloride $\bf{C}$ of $\bf{B}$ through a combination reaction at 250 $\degree \mathrm{C}$. If $\bf{W}$ is heated to 280 $\degree \mathrm{C}$ in a hydrogen atmosphere, then $\bf{W}$ will decompose into the element $\bf{B}$, 2.133 g of $\bf{D}$, and 0.597 g of $\bf{E}$, and 0.4032 L of gas $\bf{F}$ (341.1 $\degree \mathrm{C}$, 1 $\text{atm}$). $\bf{D}$ is a coordination compound of $\bf{A}$, belonging to the $\text{D}_{3d}$ point group. After separating the element $\bf{B}$ from the above decomposition products, the remaining solid residue is further heated to a higher temperature to obtain 1.792 g of a ternary compound $\bf{G}$, and two gases $\bf{C}$ and $\bf{F}$ are produced in a 1:1 ratio. $\bf{C}$ and $\bf{F}$ have the same elemental composition, and the volume of the mixed gas at 409 $\degree \mathrm{C}$ and 1 $\text{atm}$ is 0.224 L. The molar mass of $\bf{E}$ is 0.28 times that of $\bf{D}$. Then the following options are correct.
已知某包含p区非金属元素的簇合物$\mathbf{W}$的化学式可表示为 $\mathbf{A}_2\mathbf{B}_{14}Cl_{14}$，其结构中间包含一个 $\mathbf{B}_{10}$ 的电中性片段，过渡金属 $\bf{A}$ 被 6 个 $\bf{B}$ 原子配位。

已知，5.346 g 簇合物 $\bf{W}$ 恰好可由过渡金属 0.768 g 单质 $\bf{A}$、2.688 g 单质 $\bf{B}$ 和若干 $\bf{B}$ 的氯化物 $\bf{C}$ 在 250 $\degree \mathrm{C}$通过化合反应得到。如果将 $\bf{W}$ 在氢气氛中加热到 280 $\degree \mathrm{C}$，则 $\bf{W}$ 又会分解成单质 $\bf{B}$、2.133 g $\bf{D}$ 和 0.597 g $\bf{E}$，和 0.4032 L 气体 $\bf{F}$(341.1 $\degree \mathrm{C}$, 1 $\text{atm}$)。$\bf{D}$ 是某种 $\bf{A}$ 的配合物，属于 $\text{D}_{3d}$ 点群。将单质 $\bf{B}$ 从上述分解产物中分离后，把剩余的固体残留物继续加热到更高温度，得到 1.792 g 三元化合物 $\bf{G}$，并 1:1 地产生了两种气体 $\bf{C}$ 和 $\bf{F}$，$\bf{C}$ 和 $\bf{F}$ 具有相同的元素组成，该混合气体在 409 $\degree \mathrm{C}$、1 $\text{atm}$ 时的体积为 0.224 L。$\bf{E}$ 的摩尔质量是 $\bf{D}$ 的 0.28 倍。则下列选项正确的是。
$\bf W$ decomposition yields chloride $\bf C$ of $\bf B$ containing 6 chlorines
The molecular weight of $\bf F$ is higher than that of $\bf C$
$\bf D$ contains 6 chlorine atoms.
$\bf G$ can be represented as $\bf A_2B$
5.346 g $\bf W$ decomposes to produce 2.04 g of elemental $\bf B$
The ratio of $\bf G$, $\bf C$, and $\bf F$ generated by heating after removing $\bf B$ is 2:1:1.
The ratio of $\bf B$ and $\bf C$ generated by the $\bf W$ decomposition is 2:1
All of the above options are incorrect.
$\bf W$分解得到的$\bf B$的氯化物$\bf C$中含有6个氯
$\bf F$的分子量比$\bf C$高
$\bf D$中包含6个氯原子
$\bf G$可以表示为$\bf A_2B$
5.346 g $\bf W$分解产生2.04 g的单质$\bf B$
除去$\bf B$后加热生成的$\bf G$、$\bf C$和$\bf F$比例为2：1：1
$\bf W$分解生成的$\bf B$和$\bf C$的比例为2：1
以上选项均不正确
Solution process (Keep the original format, only change the uppercase letters to $\bf{}$ format):

The reaction of synthesizing $\bf{W}$ is a combination reaction. Let the chemical formula of $\bf{C}$ be $\mathbf{B}Cl_x$. We can try to write the equation of the reaction:

$2{\bf A}+(14-14/x){\bf B}+(14/x){\bf B}Cl_x\rightarrow {\bf A_2}{\bf B_{14}}Cl_{14}$

Obviously, the mass of $\bf{C}$ is $5.346 - 0.768 - 2.688 = 1.890(g)$

So we have:

$\frac{2.688}{M(\bf{B})} \cdot \frac{14}{x} =\frac{1.890}{M({\bf{B}}) + 35.45x} \times (14 - 14/x)$

Simplifying, we get:

$M({\bf{B}}) = \frac{35.45x}{0.703(x - 1) - 1}$

<Checkpoint>[$M({\bf{B}}) = \frac{35.45x}{0.703(x - 1) - 1}$](1)</Checkpoint>

Obviously $x > 2$. A simple discussion shows that when $x = 4$, $M({\bf{B}}) = 127.9$, corresponding to the $Te$ element.

So the molecular formula of $\bf{C}$ is $TeCl_4$.

<Checkpoint>[$\bf{B}$ is $Te$, $\bf{C}$ is $TeCl_4$](1)</Checkpoint>

Therefore, option $\bf{A}$ is incorrect.

At the same time, it is obvious that

$\frac{0.768}{2M(\bf{A})} = \frac{5.346}{2M({\bf{A}}) + 35.45 \times 14 + 127.9 \times 4}$

Solving for $M({\bf{A}}) = 191.8$, corresponding to the $Ir$ element

So the molecular formula of $\bf{W}$ is $Ir_2Te_{14}Cl_{14}$

<Checkpoint>[$\bf{A}$ is $Ir$, $\bf{W}$ is $Ir_2Te_{14}Cl_{14}$](1)</Checkpoint>

According to the thermal decomposition experiment of the mixture $\bf{D} + \bf{E}$, the total mass of $1:1$ $\bf{C}$ and $\bf{F}$ gases is $m = 2.133 + 0.597 - 1.792 = 0.938(g)$

The average molar mass of the mixed gas of $\bf{C}$ and $\bf{F}$ is $M = mRT/pV = 234.5 \ g \ mol^{-1}$

It is known that $\bf{C}$ is $TeCl_4$, $M({\bf{C}}) = 269.7 \ g \ mol^{-1}$. Obviously, $M({\bf{F}}) = 199.3 \ g \ mol^{-1}$, and $\bf{F}$ is composed of the same elements as $\bf{C}$. It is easy to know that $\bf{F}$ is $TeCl_2$

<Checkpoint>[$\bf{F}$ is $TeCl_2$](1)</Checkpoint>

Therefore, option $\bf{B}$ is incorrect.

In the decomposition process of $\bf{W}$, the mass of $\bf{F}$ produced is $m({\bf{F}}) =  pVM/RT = 1.597 \ g$

So $m({\bf{B}}) = 5.346 - 2.133 - 0.597 - 1.597 = 1.019(g)$

<Checkpoint>[Mass of the simple substance produced during decomposition $m({\bf{B}})=1.02\, g$ ](0.5)</Checkpoint>

Therefore, option $\bf{E}$ is incorrect.

Comparing, we can find that $n({\bf{B}}) = n({\bf{F}}) = 4n(\bf{W})$

Therefore, option $\bf{G}$ is incorrect.

Now try to write the decomposition equation of $\bf{W}$. First, assume that $1$ molecule of $\bf{D}$ and $\bf{E}$ are obtained:

$Ir_2Te_{14}Cl_{14} \rightarrow 4Te + 4TeCl_2 + \bf{D} + \bf{E}$

$\bf{D}$ and $\bf{E}$ add up to $Ir_2Te_6Cl_3$. According to the question, $M({\bf{E}}) = 0.28M({\bf{D}})$

We can get $M({\bf{E}}) = 298.6 \ g \ mol^{-1}$, corresponding to $\bf{E}$ being $IrCl_3$, so $\bf{D}$ is $IrTe_6Cl_3$, which also meets the symmetry of $\bf{D}$ in the question, that is, $[IrCl_3]$ is at the center of $[Te_6]$.

<Checkpoint>[$\bf{E}$ is $IrCl_3$, $\bf{D}$ is $IrTe_6Cl_3$](1)</Checkpoint>

Therefore, option $\bf{C}$ is incorrect.

The equation for the thermal decomposition experiment of the $\bf{D}+\bf{E}$ mixture:

$IrCl_3 + IrTe_6Cl_3 \rightarrow {\bf{G}} + TeCl_2 + TeCl_4$

$\bf{G}$ is $IrTe_2$, and the reaction equation is:

$IrCl_3 + IrTe_6Cl_3 =2IrTe_2 + TeCl_2 + TeCl_4$

<Checkpoint>[The composition of $\bf{G}$ is $IrTe_2$](1)</Checkpoint>

<Checkpoint>[$IrCl_3 + IrTe_6Cl_3 =2IrTe_2 + TeCl_2 + TeCl_4$](1)</Checkpoint>

Therefore, options $D$, $G$, and $H$ are incorrect, and $F$ is correct.
### 解答过程（保持原格式，仅将大写字母改为$\bf{}$格式）：

合成 $\bf{W}$ 的反应是化合反应，设 $\bf{C}$ 的化学式为 $\mathbf{B}Cl_x$，我们可以尝试写出反应的方程式：

$2{\bf A}+(14-14/x){\bf B}+(14/x){\bf B}Cl_x\rightarrow {\bf A_2}{\bf B_{14}}Cl_{14}$

显然 $\bf{C}$ 的质量为 $5.346 - 0.768 - 2.688 = 1.890(g)$

于是有：

$\frac{2.688}{M(\bf{B})} \cdot \frac{14}{x} =\frac{1.890}{M({\bf{B}}) + 35.45x} \times (14 - 14/x)$

整理可得：

$M({\bf{B}}) = \frac{35.45x}{0.703(x - 1) - 1}$

<Checkpoint>[$M({\bf{B}}) = \frac{35.45x}{0.703(x - 1) - 1}$](1)</Checkpoint>

显然 $x > 2$，简单讨论可知当 $x = 4$ 时，$M({\bf{B}}) = 127.9$，对应 $Te$ 元素。

于是 $\bf{C}$ 的分子式为 $TeCl_4$。

<Checkpoint>[$\bf{B}$为$Te$，$\bf{C}$为$TeCl_4$](1)</Checkpoint>

故选项$\bf{A}$错误。

同时，显然有

$\frac{0.768}{2M(\bf{A})} = \frac{5.346}{2M({\bf{A}}) + 35.45 \times 14 + 127.9 \times 4}$

解得 $M({\bf{A}}) = 191.8$，对应 $Ir$ 元素

于是 $\bf{W}$ 的分子式为 $Ir_2Te_{14}Cl_{14}$

<Checkpoint>[$\bf{A}$为$Ir$，$\bf{W}$为$Ir_2Te_{14}Cl_{14}$](1)</Checkpoint>

根据混合物 $\bf{D} + \bf{E}$ 的热分解实验，得到 $1:1$ 的 $\bf{C}$ 和 $\bf{F}$ 气体的总质量 $m = 2.133 + 0.597 - 1.792 = 0.938(g)$

$\bf{C}$ 和 $\bf{F}$ 混合气体的平均摩尔质量 $M = mRT/pV = 234.5 \ g \ mol^{-1}$

已知 $\bf{C}$ 是 $TeCl_4$，$M({\bf{C}}) = 269.7 \ g \ mol^{-1}$，显然 $M({\bf{F}}) = 199.3 \ g \ mol^{-1}$，而 $\bf{F}$ 又与 $\bf{C}$ 由相同元素组成，易知 $\bf{F}$ 是 $TeCl_2$

<Checkpoint>[$\bf{F}$为$TeCl_2$](1)</Checkpoint>

故选项$\bf{B}$错误。

在 $\bf{W}$ 分解过程中，产生的 $\bf{F}$ 的质量 $m({\bf{F}}) =  pVM/RT = 1.597 \ g$

于是 $m({\bf{B}}) = 5.346 - 2.133 - 0.597 - 1.597 = 1.019(g)$

<Checkpoint>[分解时产生的单质$m({\bf{B}})=1.02\, g$ ](0.5)</Checkpoint>

故选项$\bf{E}$错误。

对比可发现 $n({\bf{B}}) = n({\bf{F}}) = 4n(\bf{W})$

故选项$\bf{G}$错误。

现在尝试写出 $\bf{W}$ 的分解方程式，先假设都得到 $1$ 分子的 $\bf{D}$ 和 $\bf{E}$：

$Ir_2Te_{14}Cl_{14} \rightarrow 4Te + 4TeCl_2 + \bf{D} + \bf{E}$

$\bf{D}$ 与 $\bf{E}$ 加起来为 $Ir_2Te_6Cl_3$。由题意，$M({\bf{E}}) = 0.28M({\bf{D}})$

可以得到 $M({\bf{E}}) = 298.6 \ g \ mol^{-1}$，对应 $\bf{E}$ 为 $IrCl_3$，于是 $\bf{D}$ 为 $IrTe_6Cl_3$，也符合题干中 $\bf{D}$ 的对称性，即$[IrCl_3]$于$[Te_6]$中心。

<Checkpoint>[$\bf{E}$为$IrCl_3$，$\bf{D}$为$IrTe_6Cl_3$](1)</Checkpoint>

故选项$\bf{C}$错误。

$\bf{D}+\bf{E}$ 混合物热分解实验的方程式：

$IrCl_3 + IrTe_6Cl_3 \rightarrow {\bf{G}} + TeCl_2 + TeCl_4$

$\bf{G}$ 为 $IrTe_2$，反应方程式为：

$IrCl_3 + IrTe_6Cl_3 =2IrTe_2 + TeCl_2 + TeCl_4$

<Checkpoint>[$\bf{G}$的组成为$IrTe_2$](1)</Checkpoint>

<Checkpoint>[$IrCl_3 + IrTe_6Cl_3 =2IrTe_2 + TeCl_2 + TeCl_4$](1)</Checkpoint>

故选项$D$、$G$、$H$错误，$F$正确。