Passing an excess of chlorine gas into a hot concentrated potassium iodide solution, cooling it, and precipitating golden needle-shaped crystals $\mathbf{A}$ (the reaction forming $\mathbf{A}$ is a combination reaction).

Heating $\mathbf{A}$ to constant weight yields a white water-soluble substance $\mathbf{B}$, with a weight loss of $75.8\%$. Collecting all volatile substances from the decomposition products in a cold trap yields orange-yellow crystals $\mathbf{C}$, but collecting only the less volatile substances at a relatively high temperature yields a dark red liquid, and further cooling yields dark ruby-colored needle-shaped crystals $\mathbf{D}$.

Hint 1: $\mathbf{A}$ is decomposed by water to give a brown solution, while if $\mathbf{A}$ is directly decomposed with an excess of base, a colorless solution is obtained. Write the equations for the reactions under both conditions, respectively.

Hint 2: The actual form in which $\mathbf{C}$ crystals exist is its dimer.

Select the correct option(s) below.
将过量的氯气通入热浓碘化钾溶液, 冷却后析出金黄色针状晶体 $\mathbf{A}$ (形成 $\mathbf{A}$ 的反应是化合反应)。
将 $\mathbf{A}$ 加热至恒重, 得到白色水溶性物质 $\mathbf{B}$, 失重 $75.8 \%$ 。将分解产物中所有挥发性物质在冷阱中收集, 得到橙黄色晶体$\mathbf{C}$, 但在相对高的温度下只收集其中挥发性较差的物质, 则会得到暗红色液体,进一步冷却后得到暗红宝石色的针状晶体 $\mathbf{D}$。

提示1：$\mathbf{A}$被水分解得到棕色溶液,而若直接用过量的碱分解$\mathbf{A}$,则得到无色溶液。分别写出两种条件下反应的方程式。

提示2：$\mathbf{C}$晶体的实际存在形式是它的二聚体

选择以下正确的选项。
The anion of $\mathbf{A}$ and the hybridization of the central atom of molecule $\mathbf{C}$ are different.
The anion of $\mathbf{A}$ is a tetrahedral structure.
The sum of the reactant coefficients in the ionic equation for the decomposition of $\mathbf{A}$ in water to obtain a brown solution is 14.
The sum of the reactant coefficients in the ionic equation for the hydrolysis of $\mathbf{A}$ to obtain a brown solution is 11.
The sum of the reactant coefficients in the ionic equation for the decomposition of $\mathbf{A}$ by excess alkali to obtain a colorless solution is 18.
The sum of the product coefficients in the ionic equation for the decomposition of $\mathbf{A}$ directly with excess base to obtain a colorless solution is 27.
In the ionic equation for the decomposition of $\mathbf{A}$ directly with excess base to obtain a colorless solution, the sum of the reactant coefficients is 21.
All of the above statements are incorrect.
$\mathbf{A}$的阴离子和$\mathbf{C}$分子的中心原子杂化方式不同。
$\mathbf{A}$的阴离子为四面体结构。
$\mathbf{A}$被水分解得到棕色溶液的离子方程式中，反应物系数之和为14
$\mathbf{A}$被水分解得到棕色溶液的离子方程式中，反应物系数之和为11
直接用过量的碱分解$\mathbf{A}$得到无色溶液的离子方程式中，反应物系数之和为18
直接用过量的碱分解$\mathbf{A}$得到无色溶液的离子方程式中，产物系数之和为27
直接用过量的碱分解$\mathbf{A}$得到无色溶液的离子方程式中，反应物系数之和为21
以上说法均不正确
Since $\mathrm{KI}$ combines with $\mathrm{Cl}_{2}$ to form $\mathbf{A}$, the chemical formula of $\mathbf{A}$ can be assumed to be $\mathrm{KICl}_{x}$

$\mathbf{B}$ is most likely $\mathrm{KI}$ or $\mathrm{KCl}$

If $\mathbf{B}$ is $\mathrm{KI}$, then $\mathrm{Cl}_{x}$ is lost,

$35.45 x /(39.1+35.45 x+126.9)=75.8 \%$, solving for $x=14.7$, which is discarded

If $\mathbf{B}$ is $\mathrm{KCl}$, then $\mathrm{ICl}_{x-1}$ is lost,

$[126.9+35.45(x-1)] /(39.1+35.45 x+126.9)=75.8 \%$
Solving for $x=4.0$, which meets the requirements

Therefore, $\mathbf{A}$ is $\mathrm{KICl}_{4}$, $\mathbf{B}$ is $\mathrm{KCl}$

<Checkpoint>[$\mathbf{A}$ is $\mathrm{KICl}_{4}$, $\mathbf{B}$ is $\mathrm{KCl}$](2)</Checkpoint>

Wherein, in $\mathbf{A}$, $\mathrm{I}$ is $\mathrm{sp^3d^2}$ hybridized, and the ion configuration is planar.

<Checkpoint>[In $\mathbf{A}$, $\mathrm{I}$ is $\mathrm{sp^3d^2}$ hybridized, and the ion configuration is planar](0.5)</Checkpoint>

The weight loss is $\mathrm{ICl}_{3}$, so $\mathbf{C}$ is $\mathrm{ICl}_{3}$ or written as $\mathrm{I_2Cl_6}$. According to the question, $\mathrm{ICl}_{3}$ will further decompose when the temperature rises, so $\mathbf{D}$ is $\mathrm{ICl}$.

<Checkpoint>[$\mathrm{C}$ is $\mathrm{ICl}_{3}$ or written as $\mathrm{I_2Cl_6}$, $\mathbf{D}$ is $\mathrm{ICl}$](2)</Checkpoint>

Wherein, in $\mathrm{I_2Cl_6}$, $\mathrm{I}$ is $\mathrm{sp^3d^2}$ hybridized, and two $\mathrm{Cl}$ act as bridging ligands, and the molecular configuration is planar.

<Checkpoint>[In $\mathrm{I_2Cl_6}$, $\mathrm{I}$ is $\mathrm{sp^3d^2}$ hybridized, and two $\mathrm{Cl}$ act as bridging ligands, and the molecular configuration is planar.](0.5)</Checkpoint>

In prompt 1, $\mathbf{A}$ is decomposed by water to obtain a brown solution. According to the conditions of the question, it is determined that the cause of the brown color is the disproportionation of $\mathbf{A}$ to generate $\mathrm{I_2}$. The ion equation of the reaction is as follows:

$ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} +3 \mathrm{IO}_{3}^{-}+20  \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $

Or
$ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} \mathrm{Cl}^{-}+3 \mathrm{IO}_{3}^{-}+19 \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $

<Checkpoint>[$ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} +3 \mathrm{IO}_{3}^{-}+20  \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $  Or $ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} \mathrm{Cl}^{-}+3 \mathrm{IO}_{3}^{-}+19 \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $](1)</Checkpoint>

If excess alkali is directly used to decompose $\mathbf{A}$ to obtain a colorless solution, it directly disproportionates to generate $\mathrm{I}^{-}$ and $\mathrm{IO}_{3}^{-}$. The ion equation of the reaction is as follows:

$ 3 \mathrm{ICl}_{4}^{-}+12 \mathrm{OH}^{-} \rightarrow \mathrm{I}^{-}+2 \mathrm{IO}_{3}^{-}+12 \mathrm{Cl}^{-}+6 \mathrm{H}_{2} \mathrm{O}$

<Checkpoint>[$ 3 \mathrm{ICl}_{4}^{-}+12 \mathrm{OH}^{-} \rightarrow \mathrm{I}^{-}+2 \mathrm{IO}_{3}^{-}+12 \mathrm{Cl}^{-}+6 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

In summary, option $\mathrm{C}$ is correct.
由于 $\mathrm{KI}$ 与 $\mathrm{Cl}_{2}$ 化合形成 $\mathbf{A}$, 故可设 $\mathbf{A}$ 的化学式为 $\mathrm{KICl}_{x}$

$\mathbf{B}$ 最可能是 $\mathrm{KI}$ 或者 $\mathrm{KCl}$ 

如果 $\mathbf{B}$ 是 $\mathrm{KI}$ , 则失去的是 $\mathrm{Cl}_{x}$,

有 $35.45 x /(39.1+35.45 x+126.9)=75.8 \%$, 解得 $x=14.7$, 舍去

如果 $\mathbf{B}$ 是 $\mathrm{KCl}$, 则失去的是 $\mathrm{ICl}_{x-1}$, 

有 $[126.9+35.45(x-1)] /(39.1+35.45 x+126.9)=75.8 \%$
解得 $x=4.0$, 符合题意

故 $\mathbf{A}$ 是$\mathrm{KICl}_{4}$,  $\mathbf{B}$ 是$\mathrm{KCl}$ 

<Checkpoint>[$\mathbf{A}$ 是$\mathrm{KICl}_{4}$,  $\mathbf{B}$ 是$\mathrm{KCl}$ ](2)</Checkpoint>

其中，$\mathbf{A}$ 中 $\mathrm{I}$ 为 $\mathrm{sp^3d^2}$杂化，离子构型为平面型。

<Checkpoint>[$\mathbf{A}$ 中 $\mathrm{I}$ 为 $\mathrm{sp^3d^2}$杂化，离子构型为平面型](0.5)</Checkpoint>

失重的是 $\mathrm{ICl}_{3}$, 故 $\mathbf{C}$ 为 $\mathrm{ICl}_{3}$或写为 $\mathrm{I_2Cl_6}$, 根据题意, 温度升高时 $\mathrm{ICl}_{3}$ 会进一步分解, 故 $\mathbf{D}$ 是 $\mathrm{ICl}$ 。

<Checkpoint>[$\mathrm{C}$ 是 $\mathrm{ICl}_{3}$或写为 $\mathrm{I_2Cl_6}$, $\mathbf{D}$ 是 $\mathrm{ICl}$ ](2)</Checkpoint>

其中$\mathrm{I_2Cl_6}$中$\mathrm{I}$ 为 $\mathrm{sp^3d^2}$杂化，两个$\mathrm{Cl}$做桥联配体，分子构型为平面型。

<Checkpoint>[$\mathrm{I_2Cl_6}$中$\mathrm{I}$ 为 $\mathrm{sp^3d^2}$杂化，两个$\mathrm{Cl}$做桥联配体，分子构型为平面型。](0.5)</Checkpoint>

提示1中$\mathbf{A}$被水分解得到棕色溶液，根据题目条件，判断棕色的成因是$\mathbf{A}$歧化生成了$\mathrm{I_2}$，反应的离子方程式如下：

$ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} +3 \mathrm{IO}_{3}^{-}+20  \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $

或
$ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} \mathrm{Cl}^{-}+3 \mathrm{IO}_{3}^{-}+19 \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $

<Checkpoint>[$ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} +3 \mathrm{IO}_{3}^{-}+20  \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $  或 $ 5 \mathrm{ICl}_{4}^{-}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2} \mathrm{Cl}^{-}+3 \mathrm{IO}_{3}^{-}+19 \mathrm{Cl}^{-}+18 \mathrm{H}^{+} $](1)</Checkpoint>

若直接用过量的碱分解$\mathbf{A}$得到无色溶液，直接歧化生成$\mathrm{I}^{-}$和$\mathrm{IO}_{3}^{-}$，反应的离子方程式如下：

$ 3 \mathrm{ICl}_{4}^{-}+12 \mathrm{OH}^{-} \rightarrow \mathrm{I}^{-}+2 \mathrm{IO}_{3}^{-}+12 \mathrm{Cl}^{-}+6 \mathrm{H}_{2} \mathrm{O}$

<Checkpoint>[$ 3 \mathrm{ICl}_{4}^{-}+12 \mathrm{OH}^{-} \rightarrow \mathrm{I}^{-}+2 \mathrm{IO}_{3}^{-}+12 \mathrm{Cl}^{-}+6 \mathrm{H}_{2} \mathrm{O}$](1)</Checkpoint>

综上所述，选项$\mathrm{C}$正确。