${I}_{2}$ reacts with ozone to form a compound **A** containing a three-fold rotation axis. **A** decomposes at $75^\circ \mathrm{C}$ to yield **B**, releasing 0.75 equivalents of oxygen. **B** further decomposes at $300^{\circ} \mathrm{C}$, releasing 2.5 equivalents of oxygen. The reaction of **B** with $({SO}_3F)_{2}$ yields the ionic compound **C**, in which the mass fraction of iodine is $49.2 \%$. Which of the following options is correct?
${I}_{2}$与臭氧作用可以得到一种含有三重旋转轴的化合物**A**。**A**在$75^\circ \mathrm{C}$ 下分解得到**B**，并释放出 0.75 当量的氧气。**B**在$300^{\circ} \mathrm{C}$ 下进一步分解, 释放出 2.5 当量的氧气。**B**与 $({SO}_3F)_{2}$ 作用得到离子化合物**C**，**C** 中碘元素的质量分数为 $49.2 \%$ 。以下选项正确的是？
All iodine atoms in **A** have the same chemical environment
The oxidation state of iodine in <b>A</b> is higher than that in <b>B</b>.
In the equation for the decomposition reaction of **A** (with coefficients as coprime integers), the sum of the coefficients on the product side is 11.
The cation of **C** carries three positive charges
**C** contains polyiodine cations
The geometric configuration of iodine in **B** is a planar triangle.
**A**中所有碘原子的化学环境相同
**A**中碘的化合价比**B**中碘的更高
**A**的分解反应的方程式中(系数为互质整数)产物侧的系数和为11
**C**的阳离子带3个正电
**C**含有多碘阳离子
**B**中碘的几何构型为平面三角形
From chemical knowledge of elements, it is known that **A** is a high-valent iodine oxide, with possible substances including $I_2O_5$ and $I_4O_9$. Considering **A** has a three-fold rotational axis, it can be deduced that **A** is $I(IO_3)_3$, i.e., $I_4O_9$. **B** is also a high-valent iodine oxide, and the common high-valent iodine oxides are only $I_2O_5$ and $I_4O_9$ (among high-valent iodine compounds, the +5 oxidation state is relatively stable). It is inferred that **B** is $I_2O_5$, which loses $2.5 \times 2 = 5$ oxygen atoms to form elemental iodine, a common substance, making the assumption reasonable.

(However, it must be acknowledged that the condition "**A** decomposes to form **B** and releases $O_2$" might mislead one to think that **A** decomposes into only two products, which is somewhat deceptive.)

<Checkpoint>[A=$I(IO_3)_3$](1)</Checkpoint>

<Checkpoint>[B=$I_2O_5$](1)</Checkpoint>

The decomposition of $I_4O_9$ yields $I_2O_5$ and releases $0.75O_2$. Thus, the decomposition equation for **A** can be written as $4I_4O_9\rightarrow6I_2O_5+3O_2+2I_2$.

<Checkpoint>[The decomposition reaction of A is $4I_4O_9\rightarrow6I_2O_5+3O_2+2I_2$](1)</Checkpoint>

The anion of **C** is $SO_3F^-$. Assuming the chemical formula of **C** is $I_aO_b(SO_3F)_c$, we first test $c=1$. Starting with $b=1$, we check whether an integer or near-integer value of $a$ can be obtained. The combination $a=1$, $b=2$, $c=1$ fits the mass fraction, and iodine in the relatively stable +5 oxidation state also aligns with chemical principles. Other combinations of $(a, b, c)$ do not satisfy the mass fraction. Note that the reactants forming **C** include the strongly oxidizing $(SO_3F)_2$, and polyiodine cations with strong reducing properties cannot exist, which aids in screening the chemical formula of **C**.

<Checkpoint>[C=$IO_2SO_3F$](1)</Checkpoint>

The oxidation states of iodine in **A** are +3 and +5.

<Checkpoint>[The oxidation states of iodine in A are +3 and +5](1)</Checkpoint>

The geometric configuration of iodine in **B** is trigonal pyramidal.

<Checkpoint>[The geometric configuration of iodine in B is trigonal pyramidal](1)</Checkpoint>
由元素化学知识可知**A**为高价碘氧化物，可能的物质有$I_2O_5$和$I_4O_9$等，结合**A**有三重旋转轴，可知**A**为$I(IO_3)_3$即$I_4O_9$。**B**也是高价碘氧化物，而常见高价碘氧化物只有$I_2O_5$和$I_4O_9$（高价碘中较稳定的是+5价碘），推测**B**为$I_2O_5$，其失去$2.5*2=5$个氧原子得到碘单质，为常见物质，猜测合理。

（不过需要承认“**A**分解生成**B**并放出$O_2$”的条件可能让人误以为**A**的分解只生成2种产物，有一定迷惑性）

<Checkpoint>[A=$I(IO_3)_3$](1)</Checkpoint>

<Checkpoint>[B=$I_2O_5$](1)</Checkpoint>

$I_4O_9$分解得到$I_2O_5$并放出$0.75O_2$，可以写出**A**分解的方程式为$4I_4O_9\rightarrow6I_2O_5+3O_2+2I_2$

<Checkpoint>[A的分解反应为$4I_4O_9\rightarrow6I_2O_5+3O_2+2I_2$](1)</Checkpoint>

**C**的阴离子为$SO_3F^-$，设**C**的化学式为$I_aO_b(SO_3F)_c$，先试$c=1$，从$b=1$开始试验能否得到接近整数的$a$，发现$a=1,b=2,c=1$的组合符合质量分数，碘为较稳定的+5价也符合化学规律。其他$(a,b,c)$的组合不能符合质量分数。注意生成**C**的反应物有强氧化性的$(SO_3F)_2$，多碘阳离子还原性强不能存在，这有利于**C**化学式的筛查。

<Checkpoint>[C=$IO_2SO_3F$](1)</Checkpoint>

**A**中碘的化合价为+3和+5

<Checkpoint>[A中碘的化合价为+3和+5](1)</Checkpoint>

**B**中碘的几何构型为三角锥形

<Checkpoint>[B中碘的几何构型为三角锥形](1)</Checkpoint>