There are the following five processes:

1. In a warm acetonitrile solution, sodium fluoride reacts with sulfur dichloride, and one of the products is a yellow-red liquid;

2. Under high temperature and pressure, $\mathrm{Ni}$ reacts with $\mathrm{PF_3}$ to form a tetrahedral molecule;

3. Passing sufficient carbon dioxide into sodium metaborate produces borax crystals;

4. A small amount of $\mathrm{ClF_3}$ added to a solution of hydrazine reacts violently;

5. Copper reacts with dinitrogen tetroxide in nitromethane, and the product contains $\mathrm{Cu\ \ 22.73\%}$.

   Based on the chemical equations (with the simplest integer coefficients) of the above five processes, which of the following statements is incorrect
有下述五个过程：

1. 在温热的乙腈溶液中氟化钠与二氯化硫反应，产物之一是黄红色液体；

2. 高温高压下$\mathrm{Ni}$和$\mathrm{PF_3}$反应，生成一种四面体型分子；

3. 向偏硼酸钠中通入足量二氧化碳生成硼砂晶体；

4. 少量$\mathrm{ClF_3}$加入肼的溶液当中剧烈反应；

5. 铜与四氧化二氮在硝基甲烷中反应，产物含$\mathrm{Cu\ \ 22.73\%}$。

   根据上述五个过程的化学方程式（系数为最简整数比），下列说法中错误的是
Process 1 The sum of the coefficients in the chemical reaction equation is 13.
Process 2: The sum of the coefficients in the chemical reaction equation is 6.
The sum of the stoichiometric coefficients of the reactants in process 3 is 17.
The sum of the coefficients of the reactants in process 4 is 6.
The product of Process 5 is an important biomolecule
The products of process 4 are all diatomic small molecules.
All of the above options are incorrect.
过程1化学反应方程式系数和为13
过程2化学反应方程式系数和为6
过程3中反应物的系数和为17
过程4中反应物的系数和为6
过程5的产物是一种重要的生物分子
过程4产物均为双原子小分子
以上选项均不正确
This is a chemistry equation problem, mainly examining the deduction of inorganic reaction products.

In process 1, the yellow-red liquid containing $\mathrm{S}$ and $\mathrm{Cl}$ is $\mathrm{S_2Cl_2}$, which can be used to determine the destination of $\mathrm{S}$ and $\mathrm{Cl}$. Since $\mathrm{NaF}$ participates in the reaction, the only destination of $\mathrm{Na}$ is to combine with $\mathrm{Cl}$, and the only destination of $\mathrm{F}$ is to combine with $\mathrm{S}$, forming the common $\mathrm{SF_4}$.

<Checkpoint>[Products are $\mathrm{S_2Cl_2}$, $\mathrm{NaCl}$, and $\mathrm{SF_4}$](1)</Checkpoint>

Therefore, the balanced equation is $\mathrm{4NaF+3SCl_2=S_2Cl_2+4NaCl+SF_4}$, and the sum of the coefficients is 13, so option A is correct.

<Checkpoint>[The sum of the coefficients of the balanced equation in process 1 is 13](0.5)</Checkpoint>

In process 2, the tetrahedral molecule is $\mathrm{Ni(PF_3)_4}$, and the balanced equation is $\mathrm{Ni+4PF_3=Ni(PF_3)_4}$, and the sum of the coefficients is 6, so option B is correct.

<Checkpoint>[The tetrahedral molecule corresponds to four ligands, and the sum of the coefficients of the balanced equation is 6](1)</Checkpoint>

In process 3, borax crystal is $\mathrm{Na_2B_4O_7\cdot 10H_2O}$. In addition, note that $\mathrm{CO_2}$ is in sufficient quantity, and the formed $\mathrm{Na_2CO_3}$ will further react with $\mathrm{CO_2}$, and the product should be $\mathrm{NaHCO_3}$.

<Checkpoint>[Borax is $\mathrm{Na_2B_4O_7\cdot 10H_2O}$, and the reaction product is $\mathrm{NaHCO_3}$](1)</Checkpoint>

Therefore, the balanced equation is $\mathrm{4NaBO_2+2CO_2+11H_2O=Na_2B_4O_7\cdot 10H_2O+2NaHCO_3}$, and the sum of the reactant coefficients is 17, so option C is correct.

<Checkpoint>[The sum of the reactant coefficients of the balanced equation in process 2 is 17](0.5)</Checkpoint>

In process 4, the question mentions "violent reaction", indicating that a violent redox reaction has occurred, and $\mathrm{N_2}$ is produced. $\mathrm{Cl(III)}$ has strong oxidizing properties and will be reduced to a valence of $\mathrm{-1}$. It should be pointed out that since 0-valent $\mathrm{Cl}$ (i.e., $\mathrm{Cl_2}$) also has strong oxidizing properties, it will continue to react with $\mathrm{N_2H_4}$, and $\mathrm{Cl_2}$ should not appear in the products.

<Checkpoint>[In the products of process 4, Cl has a valence of $\mathrm{-1}$](1)</Checkpoint>

The question indicates that the reaction environment is a hydrazine solution, and attention should be paid to the existence form of the products. The generated $\mathrm{HCl}$ and $\mathrm{HF}$ will combine with hydrazine to form $\mathrm{N_2H_5Cl}$ and $\mathrm{N_2H_5F}$.

<Checkpoint>[Reaction products should combine with hydrazine to form $\mathrm{N_2H_5Cl}$ and $\mathrm{N_2H_5F}$](1)</Checkpoint>

Therefore, the balanced chemical equation is $\mathrm{ClF_3+5N_2H_4=N_2+N_2H_5Cl+3N_2H_5F}$. The sum of the coefficients of the reactants is 6, so option D is correct; the products are not only diatomic molecules, so option F is incorrect.

<Checkpoint>[The sum of the reactant coefficients in process 4 is 6](0.5)</Checkpoint>

In process 5, let's assume the product is $\mathrm{Cu N_x O_y}$, solve the indeterminate equation $\mathrm{\frac{63.55}{63.55+14.01x+16y}=0.2273}$, where x and y are both integers. One set of solutions is $\mathrm{x=4,y=10}$, i.e., $\mathrm{Cu N_4 O_{10}}$. Considering element conservation, 3 $\mathrm{N_2O_4}$ leaving the product $\mathrm{Cu N_4 O_{10}}$ still have two $\mathrm{N}$ and two $\mathrm{O}$, i.e., two $\mathrm{NO}$ molecules.

<Checkpoint>[The reaction product of process 5 is $\mathrm{Cu N_4 O_{10}}$](1)</Checkpoint>

Therefore, the balanced chemical equation is $\mathrm{Cu+3N_2O_4=CuN_4O_{10}+2NO}$, in fact, it is $\mathrm{Cu(NO_3)_2\cdot N_2O_4}$. Since $\mathrm{NO}$ is an important biological molecule, option E is correct.

<Checkpoint>[Another product of process 5 is NO](1)</Checkpoint>

<Checkpoint>[Option D is incorrect, A, B, C, E, and F are all correct](1)</Checkpoint>
这是一道化学方程式题目，主要考察对于无机反应产物的推断。

过程1中，包含$\mathrm{S}$和$\mathrm{Cl}$的黄红色液体为$\mathrm{S_2Cl_2}$，可以据此判断$\mathrm{S}$和$\mathrm{Cl}$的去向。由于$\mathrm{NaF}$参与了反应，$\mathrm{Na}$的去向就仅为和$\mathrm{Cl}$结合，$\mathrm{F}$的去向也仅为和$\mathrm{S}$结合，形成常见的$\mathrm{SF_4}$。

<Checkpoint>[产物为$\mathrm{S_2Cl_2}$、$\mathrm{NaCl}$和$\mathrm{SF_4}$](1)</Checkpoint>

因此，配平的方程式为$\mathrm{4NaF+3SCl_2=S_2Cl_2+4NaCl+SF_4}$，系数和为13，A选项正确。

<Checkpoint>[过程1配平后方程系数和为13](0.5)</Checkpoint>


过程2中，四面体型分子为$\mathrm{Ni(PF_3)_4}$，配平方程式为$\mathrm{Ni+4PF_3=Ni(PF_3)_4}$，系数和为6，B选项正确。

<Checkpoint>[四面体型分子对应四个配体，配平后方程式系数和为6](1)</Checkpoint>

过程3中，硼砂晶体为$\mathrm{Na_2B_4O_7\cdot 10H_2O}$。此外，注意到$\mathrm{CO_2}$足量，形成的$\mathrm{Na_2CO_3}$会进一步和$\mathrm{CO_2}$反应，产物应为$\mathrm{NaHCO_3}$。

<Checkpoint>[硼砂为$\mathrm{Na_2B_4O_7\cdot 10H_2O}$，且反应产物为$\mathrm{NaHCO_3}$](1)</Checkpoint>

因此，配平的方程式为$\mathrm{4NaBO_2+2CO_2+11H_2O=Na_2B_4O_7\cdot 10H_2O+2NaHCO_3}$，反应物系数和为17，C选项正确。

<Checkpoint>[过程2配平后方程式反应物系数和为17](0.5)</Checkpoint>

过程4中，题干提出"剧烈反应"，说明发生了剧烈的氧化还原反应，有$\mathrm{N_2}$生成。$\mathrm{Cl(III)}$具有强氧化性，会被还原至$\mathrm{-1}$价。需要指出，由于0价$\mathrm{Cl}$ (即$\mathrm{Cl_2}$) 也具有强氧化性，会和$\mathrm{N_2H_4}$继续反应，产物中不应出现$\mathrm{Cl_2}$。

<Checkpoint>[过程4反应产物中Cl为$\mathrm{-1}$价](1)</Checkpoint>

题干中指出反应环境为肼的溶液，应当注意产物的存在形式。生成的$\mathrm{HCl}$和$\mathrm{HF}$会和肼结合，形成$\mathrm{N_2H_5Cl}$和$\mathrm{N_2H_5F}$。

<Checkpoint>[反应产物应当与肼结合形成$\mathrm{N_2H_5Cl}$和$\mathrm{N_2H_5F}$](1)</Checkpoint>

因此，配平的化学方程式为$\mathrm{ClF_3+5N_2H_4=N_2+N_2H_5Cl+3N_2H_5F}$。反应物的系数和为6，选项D正确；产物中不仅只有双原子分子，选项F错误。

<Checkpoint>[过程4反应物系数和为6](0.5)</Checkpoint>

过程5中，不妨假设产物为$\mathrm{Cu N_x O_y}$，解不定方程$\mathrm{\frac{63.55}{63.55+14.01x+16y}=0.2273}$，其中，x和y均为整数。其一组解为$\mathrm{x=4,y=10}$，即$\mathrm{Cu N_4 O_{10}}$。考虑元素守恒，3个$\mathrm{N_2O_4}$除去产物$\mathrm{Cu N_4 O_{10}}$还剩两个$\mathrm{N}$和两个$\mathrm{O}$，即两个$\mathrm{NO}$分子。

<Checkpoint>[过程5反应产物为$\mathrm{Cu N_4 O_{10}}$](1)</Checkpoint>

因此，配平的化学方程式为$\mathrm{Cu+3N_2O_4=CuN_4O_{10}+2NO}$，事实上，即$\mathrm{Cu(NO_3)_2\cdot N_2O_4}$。由于$\mathrm{NO}$是重要的生物分子，选项E正确。

<Checkpoint>[过程5另一个产物为NO](1)</Checkpoint>

<Checkpoint>[选项D错误，A,B,C,E,F均正确](1)</Checkpoint>