$\mathbf{A}$ is an important compound of the element $\mathbf{M}$ with a point group of $D_{2d}$, where the mass fraction of $\mathbf{M}$ is 70.0%. The elemental form of $\mathbf{M}$ can be obtained by the fusion reaction of $\mathbf{A}$ with tin. Heating the elemental $\mathbf{M}$ in air yields $\mathbf{B}$. Refluxing $\mathbf{B}$ with $\mathrm{S_{2}Cl_{2}}$ under chlorine gas produces the oily liquid $\mathbf{C}$. If $\mathbf{B}$ is mixed with Zn and hydrochloric acid, one of the products reacts with silver nitrate solution to form a black precipitate $\mathbf{D}$. Please determine which of the following options is correct.
$\mathbf{A}$为元素$\mathbf{M}$的一种重要化合物，点群为$D_{2d}$，其中$\mathbf{M}$的质量分数为70.0%。$\mathbf{A}$和锡熔融反应可得$\mathbf{M}$的单质。将$\mathbf{M}$单质在空气中加热，可以得到$\mathbf{B}$。在通氯气下将$\mathbf{B}$与$\mathrm{S_{2}Cl_{2}}$加热回流，可以得到油状液体$\mathbf{C}$。若将$\mathbf{B}$与Zn和盐酸混合，产物之一与硝酸银溶液反应，可以观察到黑色沉淀$\mathbf{D}$的生成。请判断以下选项中哪一个是正确的。
Element $\mathbf{M}$ is an element from the fifth period.
There are 5 atoms in compound $\mathbf{A}$
The mass fraction of oxygen in compound $\mathbf{B}$ is 43.66%.
The molar mass of compound $\mathbf{C}$ is approximately 181.27 g/mol
The sum of the molar masses of $\mathbf{M}$ and $\mathbf{D}$ is 304.9 g/mol
In the reaction equation where $\mathbf{B}$ is heated under reflux with $\mathrm{S_{2}Cl_{2}}$ under chlorine gas, there are three products.
元素$\mathbf{M}$是第五周期元素
化合物$\mathbf{A}$中有5个原子
化合物$\mathbf{B}$中氧元素的质量分数为43.66%
化合物$\mathbf{C}$的摩尔质量约为181.27 g/mol
$\mathbf{M}$和$\mathbf{D}$的摩尔质量和为304.9 g/mol
在通氯气下将$\mathbf{B}$与$\mathrm{S_{2}Cl_{2}}$加热回流的反应方程式中，产物有三种
Determine the element $\mathbf{M}$:

1. Heating the simple substance $\mathbf{M}$ in air can yield $\mathbf{B}$, then $\mathbf{B}$ should be the oxide of $\mathbf{M}$.
2. Mixing $\mathbf{B}$ with Zn and hydrochloric acid, one of the products reacts with silver nitrate solution, and the formation of a black precipitate $\mathbf{D}$ can be observed. Zn-HCl is a reducing system, then the oxide $\mathbf{B}$ will be reduced when mixed with Zn and hydrochloric acid, and this product reacts with silver nitrate solution, and the formation of a black precipitate $\mathbf{D}$ can be observed. If no redox reaction occurs, the black precipitate may be $\mathrm{Ag_2S}$ ( $\mathrm{Ag_2O}$ cannot be produced in this way, and $\mathrm{Ag_2Se}$ is a gray crystalline powder); if a redox reaction occurs, the black precipitate may be elemental $\mathrm{Ag}$, and $\mathbf{B}$ may be the reducing gas $\mathrm{PH_3}$, $\mathrm{AsH_3}$, $\mathrm{SbH_3}$.
3. Heating $\mathbf{B}$ with $\mathrm{S_{2}Cl_{2}}$ under chlorine gas reflux can obtain an oily liquid $\mathbf{C}$. This information can rule out the possibility that $\mathbf{M}$ is sulfur, so the black precipitate $\mathbf{D}$ is $\mathrm{Ag}$. Considering that $\mathrm{S_{2}Cl_{2}}$ is generally used as a chlorinating reagent, and the chloride is liquid, it can be locked that $\mathbf{M}$ is one of P, As, and Sb.

<Checkpoint>[$\mathbf{M}$ is one of P, As, Sb](1)</Checkpoint>

4. The point group $D_{2d}$ is common in **tetrahedral twisted structures**. Let $\mathbf{A = MX_n}$, the molar mass of $\mathbf{M}$ is $M(\mathbf{M})$, and the molar mass of $\mathbf{X}$ is $M(\mathbf{X})$, then the mass fraction expression is: $w=\frac{M(\mathbf{M})}{M(\mathbf{M})+nM(\mathbf{X})}$
, try to substitute common combinations to find a compound with a mass fraction of about 70%. A reasonable result is $\mathbf{M}=\mathrm{As}$, $\mathbf{X}=\mathrm{S}$, $n=1$, then $\mathbf{A}$ is $\mathrm{As_4S_4}$.

<Checkpoint>[$\mathbf{M} = \mathrm{As}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}=\mathrm{As_4S_4}$](1)</Checkpoint>

Verify subsequent compounds:

- $\mathbf{B}$: As is heated in air to obtain an oxide. Since pentavalent As has strong oxidizing properties, it can only be $\mathrm{As_2O_3}$

<Checkpoint>[$\mathbf{B}= \mathrm{As_2O_3}$](1)</Checkpoint>

- $\mathbf{C}$: $\mathrm{As_2O_3}$ and $\mathrm{S_{2}Cl_{2}}$ are heated under reflux in chlorine gas to form an oily liquid $\mathbf{C}$.
  - Reaction: $\mathrm{2As_2O_3 + 3S_{2}Cl_{2} + 2Cl_2 \rightarrow 2AsCl_3 + 6SO_2}$ 
  - $\mathbf{C}$ = $\mathrm{AsCl_3}$ , oily liquid, molar mass = 181.28 g/mol

<Checkpoint>[$\mathbf{C}$ = $\mathrm{AsCl_3}$, molar mass = 181.28 g/mol](1)</Checkpoint>

- $\mathbf{D}$: $\mathrm{As_2O_3}$ is mixed with Zn and hydrochloric acid
  - Reaction: $\mathrm{As_2O_3 + 6Zn + 12HCl \rightarrow 2AsH_3\uparrow + 6ZnCl_2 + 3H_2O}$ 
  - $\mathrm{AsH_3}$ passed into silver nitrate will produce silver element!

 <Checkpoint>[$\mathbf{D} = \mathrm{Ag}$](1)</Checkpoint>

Analyze options:

**A**: As is a fourth-period element, incorrect

**B**: $As_4S_4$ has 8 atoms, incorrect
 
**C**: Mass fraction of oxygen in $As_2O_3$ = 24.26%, incorrect

**D**: The molar mass of $AsCl_3$ is 181.27 g/mol, correct

**E**: The sum of the molar masses of As and Ag is 182.79 g/mol, incorrect

**F**: $2As_2O_3 + 3S_{2}Cl_{2} + 2Cl_2 \rightarrow 2AsCl_3 + 6SO_2$ , there are two products, incorrect

Therefore, the correct answer is option **D**.
确定元素$\mathbf{M}$：

1. 将$\mathbf{M}$单质在空气中加热，可以得到$\mathbf{B}$，则$\mathbf{B}$应为$\mathbf{M}$的氧化物。
2. 将$\mathbf{B}$与Zn和盐酸混合，产物之一与硝酸银溶液反应，可以观察到黑色沉淀$\mathbf{D}$的生成。Zn-HCl为还原性体系，则氧化物$\mathbf{B}$与Zn和盐酸混合会被还原，且这种产物与硝酸银溶液反应，可以观察到黑色沉淀$\mathbf{D}$的生成。若未发生氧化还原反应，则黑色沉淀可能为$\mathrm{Ag_2S}$（$\mathrm{Ag_2O}$无法通过这种方式产生，而$\mathrm{Ag_2Se}$为灰色结晶性粉末）；若发生氧化还原反应，则黑色沉淀可能为$\mathrm{Ag}$单质，$\mathbf{B}$可能为还原型气体$\mathrm{PH_3}$、$\mathrm{AsH_3}$、$\mathrm{SbH_3}$。
3. 在通氯气下将$\mathbf{B}$与$\mathrm{S_{2}Cl_{2}}$加热回流，可以得到油状液体$\mathbf{C}$。通过这一信息可以排除$\mathbf{M}$为硫的可能，因此产生的黑色沉淀$\mathbf{D}$为$\mathrm{Ag}$。考虑到$\mathrm{S_{2}Cl_{2}}$一般作为氯化试剂，氯化物为液态，因此可以锁定$\mathbf{M}$为P、As、Sb中的一个。

<Checkpoint>[$\mathbf{M}$为P、As、Sb中的一个](1)</Checkpoint>

4. 点群 $D_{2d}$ 常见于**四面体扭曲结构**，设 $\mathbf{A = MX_n}$，$\mathbf{M}$ 的摩尔质量为 $M(\mathbf{M})$，$\mathbf{X}$ 的摩尔质量为 $M(\mathbf{X})$，则质量分数表达式为：$w=\frac{M(\mathbf{M})}{M(\mathbf{M})+nM(\mathbf{X})}$
，尝试代入常见组合，找到质量分数约为 70% 的化合物。合理结果为$\mathbf{M}=\mathrm{As}$，$\mathbf{X}=\mathrm{S}$，$n=1$，则$\mathbf{A}$为$\mathrm{As_4S_4}$.

<Checkpoint>[$\mathbf{M} = \mathrm{As}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}=\mathrm{As_4S_4}$](1)</Checkpoint>

验证后续化合物：

- $\mathbf{B}$：As在空气中加热得到氧化物，由于5价As氧化性较强，只能是$\mathrm{As_2O_3}$

<Checkpoint>[$\mathbf{B}= \mathrm{As_2O_3}$](1)</Checkpoint>

- $\mathbf{C}$：$\mathrm{As_2O_3}$与$\mathrm{S_{2}Cl_{2}}$在氯气下加热回流生成油状液体$\mathbf{C}$。
  - 反应：$\mathrm{2As_2O_3 + 3S_{2}Cl_{2} + 2Cl_2 \rightarrow 2AsCl_3 + 6SO_2}$ 
  - $\mathbf{C}$ = $\mathrm{AsCl_3}$ ，油状液体，摩尔质量 = 181.28 g/mol

<Checkpoint>[$\mathbf{C}$ = $\mathrm{AsCl_3}$，摩尔质量 = 181.28 g/mol](1)</Checkpoint>

- $\mathbf{D}$： $\mathrm{As_2O_3}$与Zn和盐酸混合
  - 反应：$\mathrm{As_2O_3 + 6Zn + 12HCl \rightarrow 2AsH_3\uparrow + 6ZnCl_2 + 3H_2O}$ 
  - $\mathrm{AsH_3}$通入硝酸银会生成银单质！

 <Checkpoint>[$\mathbf{D} = \mathrm{Ag}$](1)</Checkpoint>

分析选项：

**A**：As是第四周期元素，错误

**B**：$As_4S_4$中有8个原子，错误
 
**C**：$As_2O_3$氧元素的质量分数 = 24.26%，错误

**D**：$AsCl_3$的摩尔质量为181.27 g/mol，正确

**E**：As和Ag的摩尔质量和为182.79 g/mol，错误

**F**：$2As_2O_3 + 3S_{2}Cl_{2} + 2Cl_2 \rightarrow 2AsCl_3 + 6SO_2$ ，产物为两个，错误

因此，正确答案为选项**D**。