$\mathrm{9.34 ~g}$ of elemental **A** was heated in concentrated $\mathrm{HNO_3}$, and after complete reaction, $\mathrm{12.48~g}$ of solid **B** (an oxide of **A**) was obtained, with the release of reddish-brown gas (Reaction 1). **A** can also react with molten $\mathrm{KOH}$ in the presence of $\mathrm{O_2}$ to yield solid **C** (Reaction 2), where the oxidation state of element **A** in **C** is lower than that in **B**. **B** is a commonly used oxidizing agent in organic chemistry. Treating the $\mathrm{KOH}$ solution of **B** with ammonia yields a light-yellow complex **D**.  

Choose the correct option from the following.
$\mathrm{9.34 ~g}$元素**A**的单质在浓$\mathrm{HNO_3}$中加热，完全反应后得到$\mathrm{12.48~g}$ 固体**B**(**A**的氧化物)，并放出红棕色气体(反应1)。**A**也可以在 $\mathrm{O_2}$ 存在下与熔融的$\mathrm{KOH}$反应，得到固体**C**(反应2)，**C**中元素**A**的价态低于**B**。**B**是有机化学中的常用氧化剂，将**B**的$\mathrm{KOH}$溶液用氨处理，可以制得淡黄色配合物**D**。

选择以下选项中正确的一项。
The ratio of the sum of the stoichiometric coefficients of reactants to products in Reaction 1 is 7:11 (the coefficients are in their simplest integer ratio).
The oxygen content of **B** is 20.15%.
Reaction 2 produces two products in a 2:1 ratio.
The anion in D belongs to the $T_d$ point group.
The potassium content in C is 23.52%.
All of the above options are incorrect.
反应1的反应物与产物的系数和之比为7:11（系数为最简整数比）
**B**的含氧量为20.15%
反应2产生两种生成物且比例为2:1
**D**中阴离子属于$T_d$点群
**C**中钾元素的含量为23.52%
以上选项均不正确
The composition of **B** is $\mathrm{AO_x}$, and the relative atomic mass of **A** is $\mathrm{M_A}$. Then, $\mathrm{M_ A/( M_ A+16x) = 9. 34/12.48}$. Simplifying, we obtain $\mathrm{M_A= 47.59x}$. Enumerating the values in a table, when x=4, $\mathrm{M_A = 190.36}$. Combining the fact that **B** is a commonly used oxidizing agent in organic chemistry, **A** is determined to be **Os**.

<Checkpoint>[**A** is **Os**](1)</Checkpoint>

Since **B** is a commonly used oxidizing agent in organic chemistry, **B** is $\mathrm{OsO_4}$, with an oxygen content of 25.17%.

<Checkpoint>[**B** is $\mathrm{OsO_4}$, with an oxygen content of 25.17%, option B is incorrect](1)</Checkpoint>

Reaction 1 simultaneously produces a reddish-brown gas, which is nitrogen dioxide. The reaction equation is: $\mathrm{Os + 8HNO_3 \xrightarrow\ OsO_4 + 8NO_2 + 4H_2O}$. The ratio of the sum of the coefficients of reactants to products in the simplest whole numbers is 9:13.

<Checkpoint>[$\mathrm{Os + 8HNO_3 \xrightarrow\ OsO_4 + 8NO_2 + 4H_2O}$](1)</Checkpoint>

<Checkpoint>[The ratio of the sum of the coefficients of reactants to products in the simplest whole numbers is 9:13, option A is incorrect](1)</Checkpoint>

The oxidation state of Os in **C** is lower than that in **B**, which should be +6. The equation for reaction 2 is: $\mathrm{2Os + 4KOH + 3O_2 \xrightarrow\ 2K_2OsO_4 + 2H_2O}$. There are two products with a ratio of 1:1, so option C is incorrect.

<Checkpoint>[$\mathrm{2Os + 4KOH + 3O_2 \xrightarrow\ 2K_2OsO_4 + 2H_2O}$](1)</Checkpoint>

<Checkpoint>[Reaction 2 has two products with a ratio of 1:1, option C is incorrect](1)</Checkpoint>

From reaction 2, the chemical formula of **C** is determined to be $\mathrm{K_2OsO_4}$, with a potassium content of 23.52%.

<Checkpoint>[The chemical formula of **C** is $\mathrm{K_2OsO_4}$, with a potassium content of 23.52%, option E is correct](1)</Checkpoint>

Treating the $\mathrm{KOH}$ solution of $\mathrm{OsO_4}$ with ammonia yields a yellow complex **D**. Based on elemental knowledge, the oxidation state of Os does not change in this process, resulting in $\mathrm{KOsNO_3}$. The anion has a tetrahedral geometry, but due to the unequal distribution of N and O, it belongs to the $C_{3v}$ point group.

<Checkpoint>[The anion in **D** is a distorted tetrahedron, belonging to the $C_{3v}$ point group, option D is incorrect](1)</Checkpoint>
设**B**的组成为$\mathrm{AO_x}$，**A**的相对原子质量为$\mathrm{M_A}$，则$\mathrm{M_ A/( M_ A+16x) = 9. 34/12.48}$。化简得 $\mathrm{M_A= 47.59x}$ ，列表枚举得当x=4时，有 $\mathrm{M_A = 190.36}$。结合**B**是有机化学中的常用氧化剂，确定**A**为**Os**。

<Checkpoint>[**A**为**Os**](1)</Checkpoint>

由于**B**是有机化学中常用的氧化剂，因此**B**是$\mathrm{OsO_4}$，含氧量为25.17%。

<Checkpoint>[**B**是$\mathrm{OsO_4}$，含氧量为25.17%，选项B错误](1)</Checkpoint>

反应1同时产生红棕色气体，为二氧化氮，反应方程式为：$\mathrm{Os + 8HNO_3 \xrightarrow\ OsO_4 + 8NO_2 + 4H_2O}$，反应物与产物系数和最简整数之比为9:13。

<Checkpoint>[$\mathrm{Os + 8HNO_3 \xrightarrow\ OsO_4 + 8NO_2 + 4H_2O}$](1)</Checkpoint>

<Checkpoint>[反应物与产物系数和最简整数之比为9:13，选项A错误](1)</Checkpoint>

**C**中Os的价态比**B**的低，应为+6价，反应2的方程式为：$\mathrm{2Os + 4KOH + 3O_2 \xrightarrow\ 2K_2OsO_4 + 2H_2O}$，有两种生成物且比例为1:1，选项C错误。

<Checkpoint>[$\mathrm{2Os + 4KOH + 3O_2 \xrightarrow\ 2K_2OsO_4 + 2H_2O}$](1)</Checkpoint>

<Checkpoint>[反应2有两种生成物且比例为1:1，选项C错误](1)</Checkpoint>

由反应2得出**C**的化学式为$\mathrm{K_2OsO_4}$，钾元素含量为23.52%。

<Checkpoint>[**C**的化学式为$\mathrm{K_2OsO_4}$，钾元素含量为23.52%，选项E正确](1)</Checkpoint>

用氨处理$\mathrm{OsO_4}$的$\mathrm{KOH}$溶液，得到黄色的配合物**D**，结合元素知识这一过程Os的价态不发生改变，得到$\mathrm{KOsNO_3}$，阴离子为四面体构型，但是由于N与O不均等，因此是$C_{3v}$点群。

<Checkpoint>[**D**中阴离子为一畸变四面体，属于$C_{3v}$点群，选项D错误](1)</Checkpoint>