Treatment of $\mathrm{NSF_3}$ with $\mathrm{BCl_3}$ yields an ionic compound $\mathbf{A}$ containing N, B, S, and Cl. In $\mathbf{A}$, the ratio of N to B is 1:1. The theoretical mass fraction of N in the cation is 9.40% (relative atomic masses are retained to two decimal places), and the anion is a tetrahedral ion. The decomposition of $\mathbf{A}$ produces two liquids with the same elemental composition but different colors at room temperature. Regarding the decomposition reaction, which of the following statements is correct?
用$\mathrm{BCl_3}$处理$\mathrm{NSF_3}$可以得到一种含N、B、S、Cl四种元素的离子型化合物$\mathbf{A}$，$\mathbf{A}$中N : B = 1 : 1，阳离子中N的理论质量分数为9.40%（相对原子质量保留至小数点后两位），阴离子则是一种四面体型的离子。$\mathbf{A}$发生分解得到的产物中有两种元素组成相同、但颜色不同的液体（室温下）。关于分解反应下列说法正确的是
There are a total of four products; the sum of the stoichiometric coefficients after balancing the reaction equation is 8; elemental substances exist among the products.
There are three products in total; the sum of stoichiometric coefficients after balancing the reaction equation is 8; elemental substances are present among the products.
There are four types of products in total; the sum of stoichiometric coefficients after balancing the reaction equation is 4; no elemental substance exists among the products.
There are three types of products in total; the sum of stoichiometric coefficients after balancing the reaction equation is 4; elemental substances exist among the products.
一共有四种产物；反应方程式配平后化学计量数之和为8；产物里存在单质
一共有三种产物；反应方程式配平后化学计量数之和为8；产物里存在单质
一共有四种产物；反应方程式配平后化学计量数之和为4；产物里不存在单质
一共有三种产物；反应方程式配平后化学计量数之和为4；产物里存在单质
Since there is no F element, the most probable anion is $\mathrm{BCl_4^-}$.

<Checkpoint>[The anion is $\mathrm{BCl_4^-}$.](1)</Checkpoint>

Assuming the presence of one N in the cation, after removing N, the remaining mass is 135.03, which may correspond to S and Cl elements. Through simple enumeration, it can be found that this precisely matches 2S and 2Cl.

<Checkpoint>[Enumeration calculations show that $\mathbf{A}$ with 2S + 2Cl excluding the N atom best fits the mass fraction.](1)</Checkpoint>

Thus, $\mathbf{A}$ is $\mathrm{[N(SCl)_2][BCl_4]}$.

The two liquids with identical elemental compositions correspond to the classic $\mathrm{SCl_2}$ and $\mathrm{S_2Cl_2}$. In a decomposition reaction, it is unlikely for N to form compounds with B and Cl as products; instead, $\mathrm{N_2}$ should be produced, while B can only combine with Cl to yield $\mathrm{BCl_3}$.

<Checkpoint>[The products are $\mathrm{SCl_2}$, $\mathrm{S_2Cl_2}$, $\mathrm{N_2}$, and $\mathrm{BCl_3}$](1)</Checkpoint>

Therefore, the equation is: $\mathrm{2[N(SCl)_2][BCl_4] = N_2 + S_2Cl_2 + 2SCl_2 + 2BCl_3}$, and the answer is A.

<Checkpoint>[$\mathrm{2[N(SCl)_2][BCl_4] = N_2 + S_2Cl_2 + 2SCl_2 + 2BCl_3}$](1)</Checkpoint>
由于不存在F元素，阴离子最可能是$\mathrm{BCl_4^-}$

<Checkpoint>[阴离子为$\mathrm{BCl_4^-}$.](1)</Checkpoint>

假设阳离子中存在一个N，除去N后还剩下135.03，可能对应S和Cl元素，经简单的尝试可以发现刚好对应2S和2Cl

<Checkpoint>[计算枚举，发现$\mathbf{A}$除N原子外2S + 2Cl最符合质量分数。](1)</Checkpoint>

故$\mathbf{A}$为$\mathrm{[N(SCl)_2][BCl_4]}$.

元素组成相同的两种液体对应经典的$\mathrm{SCl_2}$和$\mathrm{S_2Cl_2}$。在一个分解反应中N元素很难再和B、Cl元素形成化合物作为产物，应该产生$\mathrm{N_2}$，而B元素只能与Cl元素结合得到$\mathrm{BCl_3}$

<Checkpoint>[产物为$\mathrm{SCl_2}$、$\mathrm{S_2Cl_2}$、$\mathrm{N_2}$和$\mathrm{BCl_3}$](1)</Checkpoint>

故方程式为：$\mathrm{2[N(SCl)_2][BCl_4] = N_2 + S_2Cl_2 + 2SCl_2 + 2BCl_3}$，选A。

<Checkpoint>[$\mathrm{2[N(SCl)_2][BCl_4] = N_2 + S_2Cl_2 + 2SCl_2 + 2BCl_3}$](1)</Checkpoint>