At $-78 \degree \mathrm{C}$, $\mathrm{Cl_2}$, $\mathrm{NCl_3}$, and $\mathrm{AsF_5}$ react in liquid $\mathrm{SO_2}$ solvent to produce a salt $\mathbf{A}$. Both the cation and anion of $\mathbf{A}$ exhibit high symmetry, and the molar ratio of $\mathrm{Cl_2}$ to $\mathrm{NCl_3}$ in the reaction is 1:2. Upon heating, $\mathbf{A}$ decomposes without yielding any elemental substances. When $\mathbf{A}$ reacts with $\mathrm{CF_3SCl}$, it forms salt $\mathbf{B}$ and several small-molecule binary compounds, with the original $\mathrm{S-Cl}$ bond remaining intact. The symmetry of the cation in $\mathbf{B}$ is identical to that of $\mathrm{NH_3}$.

Regarding the above process, which of the following statements is incorrect?
$-78 \degree \mathrm{C}$下，$\mathrm{Cl_2}$、$\mathrm{NCl_3}$和$\mathrm{AsF_5}$在液态$\mathrm{SO_2}$溶剂中反应产生一种盐$\mathbf{A}$，$\mathbf{A}$的阴阳离子都具有高的对称性，参与反应的$\mathrm{Cl_2}$和$\mathrm{NCl_3}$的摩尔比为1:2。升温时$\mathbf{A}$发生分解，产物中没有单质。将$\mathbf{A}$与$\mathrm{CF_3SCl}$反应产生盐$\mathbf{B}$和几种小分子的二元化合物，原有$\mathrm{S-Cl}$键未被破坏，$\mathbf{B}$中阳离子对称性与$\mathrm{NH_3}$相同。

对于上述过程，下列说法中错误的是
The anions of $\mathbf{A}$ and $\mathbf{B}$ are identical, and both belong to the $\mathrm{O_h}$ point group.
The cation of $\mathbf{A}$ belongs to the $\mathrm{T_d}$ point group.
During the generation of $\mathbf{A}$, the product of the simplest integer ratio of the balanced chemical equation coefficients is 12.
The gas produced during the decomposition of $\mathbf{A}$ corrodes glass containers in a moist state, so it is necessary to ensure the experiment is strictly anhydrous.
During the generation of $\mathbf{B}$, the cations in $\mathbf{B}$ and one of the products are isoelectronic.
During the generation of $\mathbf{B}$, the sum of the simplest integer ratio of coefficients in the balanced chemical equation is 8.
All of the above statements are correct
All other statements are incorrect
$\mathbf{A}$和$\mathbf{B}$的阴离子相同，且均为$\mathrm{O_h}$点群
$\mathbf{A}$的阳离子为$\mathrm{T_d}$点群
生成$\mathbf{A}$的过程中，配平后化学方程式系数的最简整数比的乘积为12
$\mathbf{A}$分解过程中产生的气体在潮湿状态会腐蚀玻璃容器，因此需要保证实验严格无水
生成$\mathbf{B}$过程中，$\mathbf{B}$中阳离子和产物中的一种互为等电子体
生成$\mathbf{B}$的过程中，配平后化学方程式系数的最简整数比的和为8
以上说法均正确
其他说法均不正确
This is an inorganic compound inference problem. The question states that "both the cation and anion of $\mathbf{A}$ possess high symmetry." Considering that the only reactant capable of forming the anion is $\mathrm{AsF_5}$, the anion should be $\mathrm{AsF_6^{-}}$. Since a reaction must occur, the cation should not contain $\mathrm{As}$. Therefore, the reactant involved in forming the cation is $\mathrm{NCl_3}$, and the resulting cation should be $\mathrm{NCl_4^{+}}$.

<Checkpoint>[Compound $\mathbf{A}$ is $\mathrm{NCl_4AsF_6}$](1)</Checkpoint>

Balancing the equation based on the hint "the molar ratio of $\mathrm{Cl_2}$ to $\mathrm{NCl_3}$ is 1:2," the chemical equation for this process is:

$\mathrm{2NCl_3+Cl_2+3AsF_5=2NCl_4AsF_6+AsF_3}$

<Checkpoint>[The chemical equation for the formation of $\mathbf{A}$ is $\mathrm{2NCl_3+Cl_2+3AsF_5=2NCl_4AsF_6+AsF_3}$](1)</Checkpoint>

According to the question, "when heated, $\mathbf{A}$ decomposes without producing elemental substances," indicating that the decomposition products cannot include $\mathrm{Cl_2}$ or $\mathrm{N_2}$. We can deduce the fate of the elements: during heating, an oxidizing gas must form, and the reduction of $\mathrm{As}$ is impossible, so $\mathrm{As}$ exists as $\mathrm{AsF_5}$ in the products. Since $\mathrm{N_2}$ is absent, there is no stable high oxidation state of $\mathrm{N}$. The oxidation of $\mathrm{N}$ cannot occur during heating, so $\mathrm{N}$ remains as $\mathrm{NCl_3}$. Considering element conservation and the absence of elemental substances in the products, the product should be $\mathrm{ClF}$. The balanced chemical equation is:

$\mathrm{NCl_4AsF_6=NCl_3+ClF+AsF_5}$

<Checkpoint>[The decomposition equation of $\mathbf{A}$ is $\mathrm{NCl_4AsF_6=NCl_3+ClF+AsF_5}$](1)</Checkpoint>

$\mathrm{NCl_4AsF_6}$ reacts with $\mathrm{CF_3SCl}$. Based on the question's statement that "the original $\mathrm{S-Cl}$ bond remains intact," it can be inferred that the $\mathrm{C-S}$ bond breaks during the reaction. According to the hint "the symmetry of the cation in $\mathbf{B}$ is the same as that of $\mathrm{NH_3}$," we know $\mathbf{B}$ is also composed of cations and anions. Observing the reactants capable of forming cations, the cation can only be formed from $\mathrm{S}$, $\mathrm{Cl}$, and $\mathrm{CF_3}$. Considering the symmetry matches $\mathrm{NH_3}$, the cation should be $\mathrm{SCl_3^+}$. Examining the reactants capable of forming anions, it is clear that the anion is solely $\mathrm{AsF_6^{-}}$.

<Checkpoint>[Compound $\mathbf{B}$ is $\mathrm{SCl_3AsF_6}$](1)</Checkpoint>

Considering element conservation and the formation of small binary molecular compounds, the products are $\mathrm{CF_4}$ and $\mathrm{NCl_3}$. The balanced equation is:

$\mathrm{2NCl_4AsF_6+CF_3SCl=SCl_3AsF_6+CF_4+2NCl_3+AsF_5}$

<Checkpoint>[The chemical equation for the formation of $\mathbf{B}$ is $\mathrm{2NCl_4AsF_6+CF_3SCl=SCl_3AsF_6+CF_4+2NCl_3+AsF_5}$](1)</Checkpoint>

Analyzing the options:

The anions of both $\mathbf{A}$ and $\mathbf{B}$ are $\mathrm{AsF_6^{-}}$, belonging to the $\mathrm{O_h}$ point group. Option A is correct.

<Checkpoint>[The anions of both $\mathbf{A}$ and $\mathbf{B}$ are $\mathrm{AsF_6^{-}}$, belonging to the $\mathrm{O_h}$ point group. Option A is correct.](1)</Checkpoint>

The cation of $\mathbf{A}$ is $\mathrm{NCl_4^{+}}$, belonging to the $\mathrm{T_d}$ point group. Option B is correct.

<Checkpoint>[The cation of $\mathbf{A}$ is $\mathrm{NCl_4^{+}}$, belonging to the $\mathrm{T_d}$ point group. Option B is correct.](1)</Checkpoint>

In the formation of $\mathbf{A}$, the product of the simplest integer ratio of the balanced equation coefficients is $2\times2\times3=12$. Option C is correct.

<Checkpoint>[In the formation of $\mathbf{A}$, the product of the simplest integer ratio of the balanced equation coefficients is $2\times2\times3=12$. Option C is correct.](0.5)</Checkpoint>

The decomposition of $\mathbf{A}$ produces $\mathrm{ClF}$, which reacts with water to form $\mathrm{HF}$, capable of etching glass. Option D is correct.

<Checkpoint>[The decomposition of $\mathbf{A}$ produces $\mathrm{ClF}$, which reacts with water to form $\mathrm{HF}$, capable of etching glass. Option D is correct.](1)</Checkpoint>

In the formation of $\mathbf{B}$, $\mathrm{SCl_3^+}$ and $\mathrm{NCl_3}$ are isoelectronic. Option E is correct.

<Checkpoint>[In the formation of $\mathbf{B}$, $\mathrm{SCl_3^+}$ and $\mathrm{NCl_3}$ are isoelectronic. Option E is correct.](1)</Checkpoint>

In the formation of $\mathbf{B}$, the sum of the simplest integer ratio of the balanced equation coefficients is $2+1+1+1+2+1=8$. Option F is correct.

<Checkpoint>[In the formation of $\mathbf{B}$, the sum of the simplest integer ratio of the balanced equation coefficients is $2+1+1+1+2+1=8$. Option F is correct.](1)</Checkpoint>

All the above statements are correct, so the answer is G.
这是一道无机物推断。题干中指出，“$\mathbf{A}$的阴阳离子都具有高的对称性”，考虑反应物中能参与阴离子形成的只有$\mathrm{AsF_5}$，阴离子应当为$\mathrm{AsF_6^{-}}$。考虑到需要发生反应，阳离子中不应该还含有$\mathrm{As}$，则反应物中参与阳离子的形成有$\mathrm{NCl_3}$，形成的阳离子应当为$\mathrm{NCl_4^{+}}$。

<Checkpoint>[物质$\mathbf{A}$为$\mathrm{NCl_4AsF_6}$](1)</Checkpoint>

根据题目提示“参与反应的$\mathrm{Cl_2}$和$\mathrm{NCl_3}$的摩尔比为1:2”配平方程式，则该过程的化学方程式为

$\mathrm{2NCl_3+Cl_2+3AsF_5=2NCl_4AsF_6+AsF_3}$

<Checkpoint>[形成$\mathbf{A}$的化学方程式为$\mathrm{2NCl_3+Cl_2+3AsF_5=2NCl_4AsF_6+AsF_3}$](1)</Checkpoint>

根据题干中信息，"升温时$\mathbf{A}$发生分解，产物中没有单质"，说明升温分解过程中，产物中不可能含有$\mathrm{Cl_2}$和$\mathrm{N_2}$。不妨推断元素去向：升温过程中，一定会生成氧化性气体，不可能发生$\mathrm{As}$的还原过程，产物中$\mathrm{As}$以$\mathrm{AsF_5}$存在。由于产物中不含$\mathrm{N_2}$，不存在高氧化态稳定的$\mathrm{N}$。升温过程中，不可能发生$\mathrm{N}$的氧化过程，产物中$\mathrm{N}$以$\mathrm{NCl_3}$存在。考虑元素守恒且产物中不含单质，产物应当为$\mathrm{ClF}$。配平的化学方程式为

$\mathrm{NCl_4AsF_6=NCl_3+ClF+AsF_5}$

<Checkpoint>[$\mathbf{A}$的分解方程式为$\mathrm{NCl_4AsF_6=NCl_3+ClF+AsF_5}$](1)</Checkpoint>

$\mathrm{NCl_4AsF_6}$与$\mathrm{CF_3SCl}$发生反应，根据题干"原有$\mathrm{S-Cl}$键未被破坏"可以判断，反应过程中$\mathrm{C-S}$键发生断裂。根据题干“$\mathbf{B}$中阳离子对称性与$\mathrm{NH_3}$相同”，可以知道$\mathbf{B}$也由阴阳离子构成。观察反应物中能够形成阳离子的物质，阳离子仅能为$\mathrm{S}$、$\mathrm{Cl}$和$\mathrm{CF_3}$形成。考虑到对称性与$\mathrm{NH_3}$相同，阳离子应当为$\mathrm{SCl_3^+}$。考虑反应物中能形成阴离子的物质，不难发现阴离子仅为$\mathrm{AsF_6^{-}}$。

<Checkpoint>[物质$\mathbf{B}$为$\mathrm{SCl_3AsF_6}$](1)</Checkpoint>

考虑元素守恒，结合产物为小分子二元化合物，可以发现产物为$\mathrm{CF_4}$和$\mathrm{NCl_3}$，配平的方程式为：

$\mathrm{2NCl_4AsF_6+CF_3SCl=SCl_3AsF_6+CF_4+2NCl_3+AsF_5}$

<Checkpoint>[形成$\mathbf{B}$的化学方程式为$\mathrm{2NCl_4AsF_6+CF_3SCl=SCl_3AsF_6+CF_4+2NCl_3+AsF_5}$](1)</Checkpoint>

下对选项进行分析：

$\mathbf{A}$和$\mathbf{B}$的阴离子均为$\mathrm{AsF_6^{-}}$，为$\mathrm{O_h}$点群，A正确。

<Checkpoint>[$\mathbf{A}$和$\mathbf{B}$的阴离子均为$\mathrm{AsF_6^{-}}$，为$\mathrm{O_h}$点群，A正确](1)</Checkpoint>

$\mathbf{A}$的阳离子为$\mathrm{NCl_4^{+}}$，为$\mathrm{T_d}$点群，B正确。

<Checkpoint>[$\mathbf{A}$的阳离子为$\mathrm{NCl_4^{+}}$，为$\mathrm{T_d}$点群，B正确。](1)</Checkpoint>

生成$\mathbf{A}$的过程中，配平后化学方程式系数的最简整数比的乘积为$2\times2\times3=12$，C正确。

<Checkpoint>[生成$\mathbf{A}$的过程中，配平后化学方程式系数的最简整数比的乘积为$2\times2\times3=12$，C正确。](0.5)</Checkpoint>

$\mathbf{A}$分解会形成$\mathrm{ClF}$，遇水会形成$\mathrm{HF}$，会刻蚀玻璃，D正确。

<Checkpoint>[$\mathbf{A}$分解会形成$\mathrm{ClF}$，遇水会形成$\mathrm{HF}$，会刻蚀玻璃，D正确。](1)</Checkpoint>

生成$\mathbf{B}$过程中，$\mathrm{SCl_3^+}$和$\mathrm{NCl_3}$是等电子体，E正确。

<Checkpoint>[生成$\mathbf{B}$过程中，$\mathrm{SCl_3^+}$和$\mathrm{NCl_3}$是等电子体，E正确。](1)</Checkpoint>

生成$\mathbf{B}$的过程中，配平后化学方程式系数的最简整数比的和为$2+1+1+1+2+1=8$，F正确。

<Checkpoint>[生成$\mathbf{B}$的过程中，配平后化学方程式系数的最简整数比的和为$2+1+1+1+2+1=8$，F正确。](1)</Checkpoint>

以上说法均正确，选择G。