The following five chemical processes exist:

1. Liquid $\mathrm{N_2O_4}$ is a common non-aqueous solvent. Sodium azide can undergo a "strong base displacing a weak base" reaction in this solvent, similar to that of sodium ethoxide in water.

2. Determination of formaldehyde content in solution using iodine in sodium hydroxide alkaline conditions.

3. Silver nitrate reacts with a halogen element in potassium chloride solution to produce a mononuclear diamagnetic substance containing 48.39% silver.

4. Sodium reacts with molten sodium hydroxide to obtain a strong base commonly used in organic synthesis.

5. Industrially, $\mathrm{PH_3}$ is prepared by the disproportionation reaction of white phosphorus in alkaline sodium hydroxide solution. In addition, another product is obtained, with a ratio to $\mathrm{PH_3}$ of 3:1.

For the above processes, the sum of the simplest integer ratios of the coefficients of the balanced chemical equations are a, b, c, d, and e, respectively. Which of the following statements is correct for these five numbers?
有下列五个化学过程：

1. 液态$\mathrm{N_2O_4}$是一种常见的非水溶剂。叠氮化钠可以在该溶剂中发生类似于乙醇钠在水中发生的“强碱制弱碱”反应

2. 氢氧化钠碱性条件下用碘单质测定溶液中的甲醛含量

3. 硝酸银与一种卤素单质在氯化钾溶液中生成一单核抗磁性物质，其中含银48.39%

4. 钠与熔融的氢氧化钠反应得到一种有机合成中常用的强碱
5. 工业上利用白磷在氢氧化钠碱性溶液中的歧化反应来制备$\mathrm{PH_3}$。此外还会得到另一种产物，与$\mathrm{PH_3}$比例为3:1

对于上述过程，配平后化学方程式系数的最简整数比之和分别为a、b、c、d、e。对于这五个数下列说法中正确的是
All other options are incorrect
Odd numbers have 3.
Even numbers have 2.
All five numbers are greater than 5.
Prime numbers have 3.
All five numbers are less than 10.
The sum of five numbers is 40.
The smallest number among five numbers is 3.
There exist 2 numbers among 5 numbers that are powers of 2.
其他选项均不正确
奇数有3个
偶数有2个
五个数均大于5
质数有3个
五个数均小于10
五个数的和为40
五个数中最小的数为3
5个数中存在2个数是2的幂指数
For process 1, sodium ethoxide will react with water to produce sodium hydroxide and ethanol. Therefore, $\mathrm{NaNO_3}$ and $\mathrm{[NO] [N_3]}$ will be formed, i.e., the balanced chemical equation is

<Checkpoint>[Formation of $\mathrm{NaNO_3}$ and $\mathrm{[NO] [N_3]}$](1)</Checkpoint>

$\mathrm{NaN_3+N_2O_4=NaNO_3+N_4O}$

<Checkpoint>[$\mathrm{NaN_3+N_2O_4=NaNO_3+N_4O}$](1)</Checkpoint>

The sum of coefficients is 4.

For process 2, according to the knowledge of elements, a redox reaction will occur, and formaldehyde is oxidized to formic acid. The balanced chemical equation is

$\mathrm{I_2+CH_2O+3NaOH=2NaI+HCOONa+2H_2O}$

<Checkpoint>[$\mathrm{I_2+CH_2O+3NaOH=2NaI+HCOONa+2H_2O}$](1)</Checkpoint>

The sum of coefficients is 10.

For process 3, since the electronic configuration of $\mathrm{Ag}$ is $\mathrm{[Kr]4d^{10}5s^1}$, the diamagnetic substance can only be $\mathrm{Ag}$ with a valence of $+1$ or $+3$. Considering that halogen element participates in the reaction, $\mathrm{Ag}$ should be oxidized to $+3$ valence, and the only halogen element with this oxidizing property is $\mathrm{F}$. From the question "containing 48.39% silver", it is assumed that the product is $\mathrm{K_x Ag F_y}$, which should satisfy: $x+3=y$, $\frac{107.9}{107.9+39.1x+19y}=0.4839$, it can be judged that the product is $\mathrm{KAgF_4}$. Similarly, due to the strong oxidizing property of $\mathrm{F_2}$, $\mathrm{Cl^-}$ will be oxidized to $\mathrm{Cl_2}$. The balanced chemical equation is

<Checkpoint>[Halogen element participates in the reaction, $\mathrm{Ag}$ should be oxidized to $+3$ valence, only $\rm{F_2}$](1)</Checkpoint>

<Checkpoint>[The product is $\mathrm{KAgF_4}$](1)</Checkpoint>

<Checkpoint>[Due to the strong oxidizing property of $\mathrm{F_2}$, $\mathrm{Cl^-}$ will be oxidized to $\mathrm{Cl_2}$](1)</Checkpoint>

$\rm{AgNO_3 + 2F_2 + 2KCl = KAgF_4 + Cl_2 + KNO_3}$

<Checkpoint>[$\rm{AgNO_3 + 2F_2 + 2KCl = KAgF_4 + Cl_2 + KNO_3}$](1)</Checkpoint>

The sum of coefficients is 8.

For process 4, sodium can only act as a reducing agent in this reaction, so it can be inferred that the strong base is $\rm{NaH}$, and the reaction equation is

$\rm{2Na + NaOH = Na_2O + NaH}$

<Checkpoint>[$\rm{2Na + NaOH = Na_2O + NaH}$](1)</Checkpoint>

The sum of reaction coefficients is 5.

For process 5, according to the disproportionation reaction ratio, the oxidation state of phosphorus in the other product should be +1, and the existing form under alkaline conditions should be $\rm{H_2PO_2^-}$. Therefore, the equation can be written as:

$\rm{P_4 + 3NaOH + 3H_2O = PH_3 + 3NaH_2PO_2}$

<Checkpoint>[$\rm{P_4 + 3NaOH + 3H_2O = PH_3 + 3NaH_2PO_2}$](1)</Checkpoint>

The sum of coefficients is 11.

In summary, the five numbers are 4, 10, 8, 5, 11, option I is correct.
对于过程1，乙醇钠会和水反应，生成氢氧化钠和乙醇。因此，会形成$\mathrm{NaNO_3}$和$\mathrm{[NO] [N_3]}$，即配平的化学方程式为

<Checkpoint>[形成$\mathrm{NaNO_3}$和$\mathrm{[NO] [N_3]}$](1)</Checkpoint>

$\mathrm{NaN_3+N_2O_4=NaNO_3+N_4O}$

<Checkpoint>[$\mathrm{NaN_3+N_2O_4=NaNO_3+N_4O}$](1)</Checkpoint>

系数和为4。

对于过程2，根据元素知识，会发生氧化还原反应，甲醛被氧化为甲酸。配平的化学方程式为

$\mathrm{I_2+CH_2O+3NaOH=2NaI+HCOONa+2H_2O}$

<Checkpoint>[$\mathrm{I_2+CH_2O+3NaOH=2NaI+HCOONa+2H_2O}$](1)</Checkpoint>

系数和为10。

对于过程3，由于$\mathrm{Ag}$的电子组态为$\mathrm{[Kr]4d^{10}5s^1}$，抗磁性物质只能为$+1$或$+3$价的$\mathrm{Ag}$。考虑到有卤素单质参与反应，$\mathrm{Ag}$应当被氧化到$+3$价，而有这种氧化性的卤素单质只能为$\mathrm{F}$。由题干“含银48.39%”，不妨假设产物为$\mathrm{K_x Ag F_y}$，应当有：$x+3=y$，$\frac{107.9}{107.9+39.1x+19y}=0.4839$，可以判断产物为$\mathrm{KAgF_4}$。同样，由于$\mathrm{F_2}$的强氧化性，$\mathrm{Cl^-}$会被氧化为$\mathrm{Cl_2}$。配平的化学方程式为

<Checkpoint>[卤素单质参与反应，$\mathrm{Ag}$应当被氧化到$+3$价，只能是$\rm{F_2}$](1)</Checkpoint>

<Checkpoint>[产物为$\mathrm{KAgF_4}$](1)</Checkpoint>

<Checkpoint>[由于$\mathrm{F_2}$的强氧化性，$\mathrm{Cl^-}$会被氧化为$\mathrm{Cl_2}$](1)</Checkpoint>

$\rm{AgNO_3 + 2F_2 + 2KCl = KAgF_4 + Cl_2 + KNO_3}$

<Checkpoint>[$\rm{AgNO_3 + 2F_2 + 2KCl = KAgF_4 + Cl_2 + KNO_3}$](1)</Checkpoint>

系数和为8。

对于过程4，该反应中钠只能作为还原剂，因而可以推测该强碱为$\rm{NaH}$，反应方程式为

$\rm{2Na + NaOH = Na_2O + NaH}$

<Checkpoint>[$\rm{2Na + NaOH = Na_2O + NaH}$](1)</Checkpoint>

反应系数和为5。

对于过程5，根据歧化反应比例，另一种产物中磷的氧化态应当为+1，碱性条件下存在形式应当为$\rm{H_2PO_2^-}$。因此方程式可写为：

$\rm{P_4 + 3NaOH + 3H_2O = PH_3 + 3NaH_2PO_2}$

<Checkpoint>[$\rm{P_4 + 3NaOH + 3H_2O = PH_3 + 3NaH_2PO_2}$](1)</Checkpoint>

系数和为11。

综上，五个数分别为4，10，8，5，11，选项I正确。