There are six chemical equations as follows:  
1. Industrially, sodium borohydride is prepared by the reaction of metallic sodium, hydrogen gas, silicon dioxide, and anhydrous borax under high temperature and pressure.  
2. In analytical chemistry, the content of diallyl trisulfide in garlic is analyzed using the potassium permanganate method, with sulfuric acid providing the acidic environment.  
3. Methyl-substituted isocyanic acid reacts with sodium hydroxide, and this reaction can be referenced in organic reaction mechanisms.  
4. Industrially, $\mathrm{K_2[Fe(CN)_5(NO)]\cdot 2H_2O}$ is prepared by the reaction of potassium ferrocyanide and nitric acid.  
5. When iodine is added to concentrated aqueous ammonia, a black precipitate is obtained, in which the mass fraction of iodine is $92.47\%$.  
6. Lithium aluminum hydride reacts with copper(I) iodide to form a new covalent hydride, and this reaction does not involve a redox process.  

Regarding the above six equations, when the coefficients are balanced and simplified to the smallest whole number ratio, the following statements are made:  
1. The sum of the product coefficients in Equation 1 is even.  
2. The sum of the reactant coefficients in Equation 2 is 120.  
3. Equation 3 has only two products, and the sum of their coefficients is 2.  
4. One of the products in Equation 4 is explosive, and the sum of the product coefficients is 4.  
5. The sum of the reactant coefficients in Equation 5 is even.  
6. The product coefficients in Equation 6 include a perfect square number other than 1.  

Which of the following options includes **all** the incorrect statements?
有下列六个化学方程式：
1. 工业上利用金属钠，氢气，二氧化硅和无水硼砂在高温高压下反应制备硼氢化钠。
2. 分析化学中利用高锰酸钾法分析大蒜中二烯丙基三硫醚的含量，酸性环境由硫酸提供。
3. 甲基取代的异氰酸与氢氧化钠反应，该反应可以参考有机反应机理。
4. 工业上利用黄血盐，硝酸反应制备$\mathrm{K_2[Fe(CN)_5(NO)]\cdot 2H_2O}$。
5. 将碘单质加入浓氨水中，得到一种黑色沉淀，该沉淀中碘元素的质量分数为$92.47\%$。
6. 氢化铝锂与碘化亚铜反应，生成一种新的共价型氢化物，该反应不涉及氧化还原过程。

关于上述六个方程式，配平后系数化为最简整数比的情况下，有下列说法：
1. 方程1的产物系数和为偶数。
2. 方程2的反应物系数和为120。
3. 方程3只有两种产物，且产物系数和为2。
4. 方程4的某种产物具有爆炸性，且产物系数和为4。
5. 方程5的反应物系数和为偶数。
6. 方程6的产物系数有除1以外的完全平方数。

以下哪个选项包含**全部**的错误说法？
1,2
3,4
1,4
1,4,6
2,4,5
2,3,6
None of the above options are correct
1,2
3,4
1,4
1,4,6
2,4,5
2,3,6
以上选项均不正确
Equation 1: Anhydrous borax is $\mathrm{Na_2B_4O_7}$. Neither boron nor silicon atoms exhibit strong oxidizing properties, whereas elemental sodium is highly reducing. During the formation of sodium borohydride, four hydride ions are generated, suggesting that hydrogen gas is reduced by elemental sodium to hydride ions, thereby balancing the equation. At high temperatures, silicon dioxide transforms into silicate.  

$\mathrm{Na_2B_4O_7 + 16Na + 7SiO_2 + 8H_2 = 4NaBH_4 + 7Na_2SiO_3}$  

The sum of the product coefficients is 11, an odd number, so Statement 1 is incorrect.  

<Checkpoint>[Hydrogen gas is reduced by elemental sodium to hydride ions](0.5)</Checkpoint>  

<Checkpoint>[At high temperatures, silicon dioxide transforms into silicate](0.5)</Checkpoint>  

<Checkpoint>[Equation 1 $\mathrm{Na_2B_4O_7 + 16Na + 7SiO_2 + 8H_2 = 4NaBH_4 + 7Na_2SiO_3}$](1)</Checkpoint>  

<Checkpoint>[The sum of the product coefficients is 11, an odd number, so Statement 1 is incorrect.](0.5)</Checkpoint>  

Equation 2: Under acidic conditions, potassium permanganate oxidizes to form divalent manganese sulfate, while diallyl trisulfide, as an organic compound, is completely oxidized to produce carbon dioxide and sulfate ions. This allows the equation to be balanced:  

$\mathrm{5(CH_2CHCH_2)_2S_3 + 52 KMnO_4 + 63H_2SO_4 = 26K_2SO_4 + 52 MnSO_4 + 30 CO_2 + 88H_2O}$  

The sum of the reactant coefficients is 120, so Statement 2 is correct.  

<Checkpoint>[Under acidic conditions, potassium permanganate oxidizes to form divalent manganese sulfate](0.5)</Checkpoint>  

<Checkpoint>[Diallyl trisulfide is completely oxidized to produce carbon dioxide and sulfate ions](0.5)</Checkpoint>  

<Checkpoint>[$\mathrm{5(CH_2CHCH_2)_2S_3 + 52 KMnO_4 + 63H_2SO_4 = 26K_2SO_4 + 52 MnSO_4 + 30 CO_2 + 88H_2O}$](1)</Checkpoint>  

<Checkpoint>[The sum of the reactant coefficients is 120, so Statement 2 is correct.](0.5)</Checkpoint>  

Equation 3: Considering the organic reaction mechanism, hydroxide ions act as strong nucleophiles, and the carbon atom in the isocyanate group is electrophilic, leading to nucleophilic attack and the formation of an amide-like intermediate. Under alkaline conditions, the amide-like intermediate undergoes hydrolysis, producing carbon dioxide and a primary amine, with carbon dioxide further converting to carbonate in the alkaline environment. Thus, the equation is:  

$\mathrm{CH_3NCO + 2NaOH = CH_3NH_2 + Na_2CO_3}$  

There are only two products, and the sum of their coefficients is 2, so Statement 3 is correct.  

<Checkpoint>[Hydroxide ions nucleophilically attack the carbon atom in the isocyanate group to form an amide-like intermediate](1)</Checkpoint>  

<Checkpoint>[$\mathrm{CH_3NCO + 2NaOH = CH_3NH_2 + Na_2CO_3}$](1)</Checkpoint>  

Equation 4: Since the product contains only five cyanide groups, if the $\mathrm{NO}$ in the complex is assumed to be neutral, the iron would be oxidized to the +3 oxidation state. Nitric acid is a strong oxidizer, suggesting that the -3 nitrogen in the cyanide group is oxidized to +2 in $\mathrm{NO}$, while the nitrate ion itself is reduced to -3 as ammonium. The +2 carbon is also oxidized to +4, appearing as carbon dioxide. Electron calculations show that one nitrate ion can oxidize exactly one molecule of $\mathrm{K_4Fe(CN)_6}$. Thus, the balanced equation is:  

$\mathrm{K_4Fe(CN)_6 + 4HNO_3 = K_2[Fe(CN)_5(NO)]\cdot 2H_2O + CO_2 + NH_4NO_3 + 2KNO_3}$  

<Checkpoint>[One nitrate ion can oxidize exactly one molecule of $\mathrm{K_4Fe(CN)_6}$](1)</Checkpoint>  

<Checkpoint>[$\mathrm{K_4Fe(CN)_6 + 4HNO_3 = K_2[Fe(CN)_5(NO)]\cdot 2H_2O + CO_2 + NH_4NO_3 + 2KNO_3}$](1)</Checkpoint>  

Ammonium nitrate in the products is explosive, and the sum of the product coefficients is 5, so Statement 4 is incorrect.  

<Checkpoint>[Ammonium nitrate in the products is explosive, and the sum of the product coefficients is 5, so Statement 4 is incorrect.](0.5)</Checkpoint>  

Equation 5: The mass fraction of iodine in the precipitate is $92.47\%$. Testing possible numbers of iodine atoms, when the count is 3, the remaining molecular weight is 31, corresponding to two nitrogen atoms and three hydrogen atoms. Thus, the precipitate's chemical formula is $\mathrm{NI_3\cdot NH_3}$. Clearly, iodine undergoes disproportionation, forming ammonium iodide. The equation is:  

<Checkpoint>[The precipitate's chemical formula is $\mathrm{NI_3\cdot NH_3}$. Iodine undergoes disproportionation](1)</Checkpoint>  

$\mathrm{5NH_3 + 3I_2 = NI_3\cdot NH_3 + 3NH_4I}$  

The sum of the reactant coefficients is 8, an even number, so Statement 5 is correct.  

<Checkpoint>[$\mathrm{5NH_3 + 3I_2 = NI_3\cdot NH_3 + 3NH_4I}$](1)</Checkpoint>  

<Checkpoint>[The sum of the reactant coefficients is 8, an even number, so Statement 5 is correct.](0.5)</Checkpoint>  

Equation 6: Since no redox occurs, the newly formed covalent hydride is determined to be copper hydride, leading to the equation:  

<Checkpoint>[The newly formed covalent hydride is copper hydride](0.5)</Checkpoint>  

$\mathrm{LiAlH_4 + 4CuI = 4CuH + LiI + AlI_3}$  

<Checkpoint>[$\mathrm{LiAlH_4 + 4CuI = 4CuH + LiI + AlI_3}$](1)</Checkpoint>  

The product coefficients include 4, a perfect square, so Statement 6 is correct.  

<Checkpoint>[The product coefficients include 4, a perfect square, so Statement 6 is correct.](0.5)</Checkpoint>  

In conclusion, statement 1 and 4 are incorrect, so option C is selected.
方程1，无水硼砂为$\mathrm{Na_2B_4O_7}$，硼原子和硅原子均无强氧化性，而钠单质有强还原性；生成硼氢化钠时有四个负氢原子生成，推测氢气被钠单质还原为负氢，从而配平方程。高温下，二氧化硅会转化为硅酸盐。

$\mathrm{Na_2B_4O_7 + 16Na + 7SiO_2 + 8H_2 = 4NaBH_4 + 7Na_2SiO_3}$

产物系数和为11，奇数，说法1错误。

<Checkpoint>[氢气被钠单质还原为负氢](0.5)</Checkpoint>

<Checkpoint>[高温下，二氧化硅会转化为硅酸盐](0.5)</Checkpoint>

<Checkpoint>[方程1$\mathrm{Na_2B_4O_7 + 16Na + 7SiO_2 + 8H_2 = 4NaBH_4 + 7Na_2SiO_3}$](1)</Checkpoint>

<Checkpoint>[产物系数和为11，奇数，说法1错误。](0.5)</Checkpoint>

方程2，高锰酸钾在酸性条件下氧化产物为二价的硫酸锰，二烯丙基三硫醚作为有机物酸性条件下被彻底氧化，生成二氧化碳和硫酸根。从而可以配平方程：

$\mathrm{5(CH_2CHCH_2)_2S_3 + 52 KMnO_4 + 63H_2SO_4 = 26K_2SO_4 + 52 MnSO_4 + 30 CO_2 + 88H_2O}$

反应物系数和为120，说法2正确。

<Checkpoint>[高锰酸钾在酸性条件下氧化产物为二价的硫酸锰](0.5)</Checkpoint>

<Checkpoint>[二烯丙基三硫醚被彻底氧化，生成二氧化碳和硫酸根](0.5)</Checkpoint>

<Checkpoint>[$\mathrm{5(CH_2CHCH_2)_2S_3 + 52 KMnO_4 + 63H_2SO_4 = 26K_2SO_4 + 52 MnSO_4 + 30 CO_2 + 88H_2O}$](1)</Checkpoint>

<Checkpoint>[反应物系数和为120，说法2正确。](0.5)</Checkpoint>

方程3，考虑有机反应机理，氢氧根作为强亲核试剂，异氰酸根的碳原子具有亲电性，从而被亲核后生成酰胺类似物；碱性条件下酰胺类似物发生水解过程，产生二氧化碳和一级胺，二氧化碳碱性条件下变为碳酸盐。因此方程式为

$\mathrm{CH_3NCO + 2NaOH = CH_3NH_2 + Na_2CO_3}$

只有两种产物且产物系数和为2，说法3正确。

<Checkpoint>[氢氧根亲核异氰酸根的碳原子生成酰胺类似物](1)</Checkpoint>

<Checkpoint>[$\mathrm{CH_3NCO + 2NaOH = CH_3NH_2 + Na_2CO_3}$](1)</Checkpoint>

方程4，由于产物只含有五个氰根，若设配合物的$\mathrm{NO}$为中性，则铁元素会被氧化为+3价。硝酸具有强氧化性，推测氰根中的-3价的氮原子被氧化为$\mathrm{NO}$的+2价，同时硝酸根自身被还原为-3价的铵根，+2价的碳原子也会被氧化为+4价，以二氧化碳形式存在。计算电子，发现一个硝酸根刚好可以氧化一分子的$\mathrm{K_4Fe(CN)_6}$。因此配平方程：

$\mathrm{K_4Fe(CN)_6 + 4HNO_3 = K_2[Fe(CN)_5(NO)]\cdot 2H_2O + CO_2 + NH_4NO_3 + 2KNO_3}$

<Checkpoint>[一个硝酸根刚好可以氧化一分子的$\mathrm{K_4Fe(CN)_6}$](1)</Checkpoint>

<Checkpoint>[$\mathrm{K_4Fe(CN)_6 + 4HNO_3 = K_2[Fe(CN)_5(NO)]\cdot 2H_2O + CO_2 + NH_4NO_3 + 2KNO_3}$](1)</Checkpoint>

产物中硝酸铵具有爆炸性，产物系数和为5，说法4错误。

<Checkpoint>[产物中硝酸铵具有爆炸性，产物系数和为5，说法4错误。](0.5)</Checkpoint>

方程5，沉淀中碘元素的质量分数为$92.47\%$，可以逐一尝试碘元素的原子个数，当原子个数为3时，剩余的相对分子质量为31，为两个氮原子和三个氢原子的和，故沉淀化学式为$\mathrm{NI_3\cdot NH_3}$。明显碘单质发生了歧化反应，生成碘化铵。方程为

<Checkpoint>[沉淀化学式为$\mathrm{NI_3\cdot NH_3}$。碘单质发生了歧化反应](1)</Checkpoint>

$\mathrm{5NH_3 + 3I_2 = NI_3\cdot NH_3 + 3NH_4I}$

反应物系数和为8，为偶数，说法5正确。

<Checkpoint>[$\mathrm{5NH_3 + 3I_2 = NI_3\cdot NH_3 + 3NH_4I}$](1)</Checkpoint>

<Checkpoint>[反应物系数和为8，为偶数，说法5正确。](0.5)</Checkpoint>

方程6，由于无氧化还原，可以判断新生成的共价氢化物为氢化亚铜，从而方程式为

<Checkpoint>[新生成的共价氢化物为氢化亚铜](0.5)</Checkpoint>

$\mathrm{LiAlH_4 + 4CuI = 4CuH + LiI + AlI_3}$

<Checkpoint>[$\mathrm{LiAlH_4 + 4CuI = 4CuH + LiI + AlI_3}$](1)</Checkpoint>

产物系数含有4，为完全平方数，说法6正确。

<Checkpoint>[产物系数含有4，为完全平方数，说法6正确。](0.5)</Checkpoint>

综上，错误说法为1,4，选择C选项。