In dichloroethane at $- 7 8 ^ { \circ } \mathrm { C }$, $[ \mathrm { C _ { 6 } F _ { 5 } X e } ] [ \mathrm { A s F _ { 6 } } ]$ reacts with trimethylchlorosilane in a molar ratio of 1:3 to yield $\mathrm{n}$ products ($\mathrm{n}$ > 3), among which three products are a certain 1:1-type salt, $\mathrm { A s C l _ { 3 } }$, and $\mathrm { C l } _ { 2 }$, in a molar ratio of 1:1:1. What is the correct statement about this reaction?
在二氯乙烷中， $- 7 8 ^ { \circ } \mathrm { C }$ 下， $[ \mathrm { C _ { 6 } F _ { 5 } X e } ] [ \mathrm { A s F _ { 6 } } ]$ 与三甲基氯硅烷按摩尔比1:3反应，得到$\mathrm{n}$种产物（$\mathrm{n}$＞3），其中三种产物为某个1:1型盐、 $\mathrm {  A s C l _ { 3 } }$ 和 $\mathrm { C l } _ { 2 }$ ，三者摩尔比为1:1:1。关于该反应说法正确的是什么？
Chemical equation (with coefficients in their simplest integer ratios) where the sum of the coefficients on the product side is odd and the sum of the coefficients on the reactant side is even
The xenon element is oxidized
There are 5 types of products.
In xenon-containing products, the xenon is in the $sp^3$ hybridization state.
If more than one of A~D is correct, select this option
If all options A~D are incorrect, please select this option
反应方程式（系数为最简整数）产物一侧系数和为奇数且反应物一侧系数和为偶数
氙元素被氧化
$\mathrm{n}$的值为5
含氙产物中氙为$sp^3$杂化
若A~D中不止1项正确请选此项
若A~D全部错误请选此项
The 1:1 salt of the reaction product is likely to contain $\mathrm{Xe}$, and $\mathrm{[(C_6F_5)Xe]^+}$ still appears as an intact unit. Considering the large volume of the cation in the 1:1 salt, the anion should also be a bulky complex anion, most likely $\mathrm{[AsF_6]^-}$.  

<Checkpoint>[The cation of the 1:1 salt contains $\mathrm{[(C_6F_5)Xe]^+}$, and the anion is $\mathrm{[AsF_6]^-}$](1)</Checkpoint>  

Assuming the coefficients of $\mathrm{Cl_2}$ and $\mathrm{AsCl_3}$ in the product are 1, the product side contains 2 $\mathrm{As}$ atoms. Setting the coefficient of the reactant $\mathrm{[C_6F_5Xe][AsF_6]}$ to 2, the coefficient of $\mathrm{SiCl(CH_3)_3}$ becomes 6.  

It is observed that one $\mathrm{As}$ on the product side is no longer bonded to fluorine atoms. Combined with stoichiometric considerations, it can be inferred that the product includes a substance with a $\mathrm{Si-F}$ bond and a coefficient of 6, namely $\mathrm{SiF(CH_3)_3}$.  

<Checkpoint>[The product contains $\mathrm{SiF(CH_3)_3}$](1)</Checkpoint>  

Combining the above information, it is clear that the cation of the 1:1 salt is $\mathrm{[(C_6F_5Xe)_2Cl]^+}$, with a bonding arrangement of $\mathrm{C_6F_5-Xe-Cl-Xe-C_6F_5}$. According to VSEPR theory, $\mathrm{Xe}$ has $(8+2)/2=5$ electron pairs, with 3 lone pairs, resulting in a linear geometry and $sp^3d$ hybridization. The $\mathrm{[(C_6F_5)Xe]^+}$ unit remains unchanged, and the oxidation state of xenon is unaffected.  

<Checkpoint>[$\mathrm{Xe}$ adopts $sp^3d$ hybridization](0.5)</Checkpoint>  

<Checkpoint>[$\mathrm{Xe}$ undergoes no redox changes](0.5)</Checkpoint>  

The reaction equation is:  
$\mathrm{2[(C_6F_5)Xe][AsF_6]+6SiCl(CH_3)_3 \rightarrow AsCl_3+Cl_2+[(C_6F_5Xe)_2Cl][AsF_6]+6SiF(CH_3)_3}$  

<Checkpoint>[$\mathrm{2[(C_6F_5)Xe][AsF_6]+6SiCl(CH_3)_3 \rightarrow AsCl_3+Cl_2+[(C_6F_5Xe)_2Cl][AsF_6]+6SiF(CH_3)_3}$](2)</Checkpoint>  

Only option A is correct. The answer is **A**.
反应产物的1:1型盐很可能含$\mathrm{Xe}$，且$\mathrm{[(C_6F_5)Xe]^+}$仍作为整体出现。考虑到1:1盐中阳离子体积较大，阴离子也应为大体积的复杂阴离子，很可能为$\mathrm{[AsF_6]^-}$。

<Checkpoint>[1:1型盐阳离子含有$\mathrm{[(C_6F_5)Xe]^+}$，阴离子为$\mathrm{[AsF_6]^-}$](1)</Checkpoint>

设产物中$\mathrm{Cl_2}$和$\mathrm{AsCl_3}$的系数为1，则产物一侧含有2个$\mathrm{As}$，设反应物$\mathrm{[C_6F_5Xe][AsF_6]}$的系数为2，则$\mathrm{SiCl(CH_3)_3}$的系数为6。

观察到产物侧的1个$\mathrm{As}$不再键连氟原子，再结合计量数可以推测产物有系数为6的含$\mathrm{Si-F}$键的物质，也就是$\mathrm{SiF(CH_3)_3}$。

<Checkpoint>[产物有$\mathrm{SiF(CH_3)_3}$](1)</Checkpoint>

结合上述已知信息，可知1:1型盐的阳离子为$\mathrm{[(C_6F_5Xe)_2Cl]^+}$，键连方式是$\mathrm{C_6F_5-Xe-Cl-Xe-C_6F_5}$。根据VSEPR，$\mathrm{Xe}$周围有$(8+2)/2=5$ 对电子，3对为孤对电子，几何构型为直线型，杂化方式为$sp^3d$。$\mathrm{[(C_6F_5)Xe]^+}$整体没有变化，氙元素没有发生氧化还原。

<Checkpoint>[$\mathrm{Xe}$的杂化方式为$sp^3d$杂化](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{Xe}$没有发生氧化还原](0.5)</Checkpoint>

反应方程式为$\mathrm{2[(C_6F_5)Xe][AsF_6]+6SiCl(CH_3)_3 \rightarrow AsCl_3+Cl_2+[(C_6F_5Xe)_2Cl][AsF_6]+6SiF(CH_3)_3}$

<Checkpoint>[$\mathrm{2[(C_6F_5)Xe][AsF_6]+6SiCl(CH_3)_3 \rightarrow AsCl_3+Cl_2+[(C_6F_5Xe)_2Cl][AsF_6]+6SiF(CH_3)_3}$](2)</Checkpoint>

仅选项A正确，答案为A