Solid $\mathbf{X}$ is a crystal with a $\mathrm{ZnS}$ structure. $\mathbf{X}$ can be synthesized through aqueous solution reactions: Oxide $\mathbf{A}$ (containing $25.6\%$ oxygen) is mixed with element $\mathbf{B}$, which has a tetrahedral molecular structure, in dilute $\mathrm{NaOH}$ aqueous solution and heated to react. $\mathbf{A}$ dissolves to obtain $\mathbf{C}$ (Reaction 1), and $\mathbf{B}$ disproportionates to obtain $\mathbf{D}$ (the conversion rate of the reaction is $25\%$ (Reaction 2); then $\mathbf{C}$ reacts with $\mathbf{D}$ to obtain $\mathbf{X}$ (Reaction 3). Adding an appropriate amount of iodine to the system can increase the reaction rate and the yield of $\mathbf{X}$, possibly because: $\mathbf{B}$ and elemental iodine undergo a new reaction in dilute $\mathrm{NaOH}$ aqueous solution in a molar ratio of 1:2, generating products such as $\mathbf{D}$ and $\mathbf{E}$ (the molar ratio of the products is 1:1) (Reaction 4). In addition, $\mathbf{X}$ can also be synthesized by traditional methods at high temperatures. Reacting $\mathbf{A}$ with $\mathbf{D}$ at $950^{\circ} \mathrm{C}$ under a $\mathrm{H_2}$ atmosphere (Reaction 5), or reacting $\mathbf{B}$ with the low-valent chloride $\mathbf{F}$ at $1050^{\circ} \mathrm{C}$ in vacuum (Reaction 6), can both obtain $\mathbf{X}$. $\mathbf{F}$ is a gaseous molecule at high temperature, and it rapidly cools to 77K and transforms into a red solid; upon further heating to above $0^{\circ} \mathrm{C}$, $\mathbf{F}$ undergoes disproportionation decomposition.

The following statements are made:

1, Let the atomic numbers of the elements in $\mathbf{X}$ be a and b (a>b), then $\frac{a-b}{a+b}=0.35$.

2, The mass fractions of hydrogen in $\mathbf{C}$ and $\mathbf{D}$ are 2.508% and 8.896%, respectively.

3, The purpose of evacuating in Reaction 6 is to reduce the boiling point of the reactants.

4, Among Reactions 1-6, Reaction 6 has the smallest difference in the number of molecules between reactants and products.

Then, which of the following options contains all the correct statements (a calculation result error within 0.01 in statement 1 is considered correct):
固体$\mathbf{X}$为$\mathrm{ZnS}$ 结构的晶体。通过水溶液反应可合成$\mathbf{X}$：将氧化物$\mathbf{A}$(含氧 $2 5 . 6 \% )$ 与具有四面体分子结构的单质 $\mathbf{B}$于稀$\mathrm{NaOH}$水溶液中混合加热反应，$\mathbf{A}$溶解得到$\mathbf{C}$(反应1)，$\mathbf{B}$歧化得到$\mathbf{D}$(反应的转化率为 $2 5 \%$ (反应2)；而后$\mathbf{C}$与$\mathbf{D}$反应可得到$\mathbf{X}$(反应3)。向体系中加入适量碘可提高反应的速率和$\mathbf{X}$的产率，原因可能是由于：$\mathbf{B}$与碘单质按摩尔比1:2在稀$\mathrm{NaOH}$水溶液中发生了新的反应，生成$\mathbf{D}$、$\mathbf{E}$等产物(产物的摩尔比为1:1) (反应4)。此外还可通过传统方法高温下合成$\mathbf{X}$。令 $\mathbf{A}$与$\mathbf{D}$于 $9 5 0 ^ { \circ } \mathrm { C }$ 的$\mathrm{H_2}$气氛下反应(反应5)，或令$\mathbf{B}$与低价氯化物$\mathbf{F}$于 $1 0 5 0 ^ { \circ } \mathrm { C }$ 的真空中反应(反应6)，均可得到$\mathbf{X}$；$\mathbf{F}$在高温下为气相分子，迅速冷却至77K转化为红色固体；继续升温至 $0 ^ { \circ } \mathrm { C }$ 以上，$\mathbf{F}$发生歧化分解。

有下列说法：

1，记$\mathbf{X}$中元素的原子序数为a和b(a>b)，则$\frac{a-b}{a+b}=0.35$。

2，$\mathbf{C，D}$中氢元素的质量分数分别为2.508%，8.896%。

3，反应6中抽真空的目的是降低反应物的沸点。

4，反应1-6中，反应物与生成物分子数目相差最小的是反应6。

则下列选项中包含所有正确说法的是（说法1中计算结果误差在0.01以内视为正确）：
All other options are incorrect
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
其他选项均不正确
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
Starting from oxide $\mathbf{A}$, let the chemical formula of $\mathbf{A}$ be $\mathbf{M}_2\mathrm{O}_y$, therefore:

$\frac{16y}{2M+16y}=25.6\%$

Starting from $y=1$, when $y=3$, we solve for $M=69.75~g/mol$. Considering the precision, it can be considered as the element $\mathrm{Ga}$. Therefore, $\mathbf{A}$ is $\mathrm{Ga_2O_3}$.

<Checkpoint>[$\mathbf{A}$ is $\mathrm{Ga_2O_3}$](1)</Checkpoint>

$\mathrm{Ga_2O_3}$ dissolves in sodium hydroxide solution to form $\mathrm{NaGa(OH)_4}$, so $\mathbf{C}$ is $\mathrm{NaGa(OH)_4}$.

<Checkpoint>[$\mathbf{C}$ is $\mathrm{NaGa(OH)_4}$](1)</Checkpoint>

The tetrahedral molecular element $\mathbf{B}$ is easily obtained as $\mathrm{P_4}$.

<Checkpoint>[$\mathbf{B}$ is $\mathrm{P_4}$](0.5)</Checkpoint>

The disproportionation reaction of $\mathrm{P_4}$ under alkaline conditions is as follows:

$\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$

It can be seen from the equation that the substance with a yield of 25% is $\mathrm{PH_3}$, so $\mathbf{D}$ is $\mathrm{PH_3}$.

<Checkpoint>[$\mathbf{D}$ is $\mathrm{PH_3}$](1)</Checkpoint>

According to the hints given in the question, reaction 4 can be written (the weakly alkaline product should be $\mathrm{NaH_2PO_4}$):

$\mathrm{P_4+2I_2+2H_2O+6NaOH\to 2PH_3+2NaH_2PO_4+4NaI}$

Therefore, $\mathbf{E}$ is $\mathrm{NaH_2PO_4}$.

<Checkpoint>[$\mathbf{E}$ is $\mathrm{NaH_2PO_4}$](1)</Checkpoint>

According to the conservation of elements, it can be inferred that $\mathbf{F}$ contains the element $\mathrm{Ga}$, so it is a low-valent chloride of $\mathrm{Ga}$, namely $\mathrm{GaCl}$. Therefore, $\mathbf{F}$ is $\mathrm{GaCl}$.

<Checkpoint>[$\mathbf{F}$ is $\mathrm{GaCl}$](1)</Checkpoint>

The following is an analysis of the options:

1.  The atomic numbers of the elements in $\mathbf{X}$ are 31 and 15, so $\frac{a-b}{a+b}=0.35$. Correct.

2.  The mass fractions of hydrogen in $\mathbf{C}$ and $\mathbf{D}$ are 2.508% and 8.896%, respectively. Correct.

3.  The reactants in reaction 6 have low boiling points, but $\mathrm{GaCl}$ is easily oxidized, so vacuuming is to protect the reactants. Incorrect.

<Checkpoint>[$\mathrm{GaCl}$ is easily oxidized, and vacuuming can protect the reactants](1)</Checkpoint>

4.  First write out reactions 1-6:

Reaction 1: $\mathrm{Ga_2O_3+2NaOH+3H_2O\to2NaGa(OH)_4}$

<Checkpoint>[Reaction 1: $\mathrm{Ga_2O_3+2NaOH+3H_2O\to2NaGa(OH)_4}$](1)</Checkpoint>

Reaction 2: $\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$

<Checkpoint>[Reaction 2: $\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$](1)</Checkpoint>

Reaction 3: $\mathrm{NaGa(OH)_4+PH_3\to GaP+NaOH+3H_2O}$

<Checkpoint>[Reaction 3: $\mathrm{NaGa(OH)_4+PH_3\to GaP+NaOH+3H_2O}$](1)</Checkpoint>

Reaction 4: $\mathrm{P_4+2I_2+2H_2O+6NaOH\to 2PH_3+2NaH_2PO_4+4NaI}$

<Checkpoint>[Reaction 4: $\mathrm{P_4+2I_2+2H_2O+6NaOH\to 2PH_3+2NaH_2PO_4+4NaI}$](1)</Checkpoint>

Reaction 5: $\mathrm{Ga_2O_3+2PH_3\to2GaP+3H_2O}$

<Checkpoint>[Reaction 5: $\mathrm{Ga_2O_3+2PH_3\to2GaP+3H_2O}$](1)</Checkpoint>

Reaction 6: $6\mathrm{GaCl+P_4\to4GaP+2GaCl_3}$

<Checkpoint>[Reaction 6: $6\mathrm{GaCl+P_4\to4GaP+2GaCl_3}$](1)</Checkpoint>

It can be seen immediately that the reaction with the smallest difference in the number of molecules between reactants and products is reaction 6. Correct.
从氧化物$\mathbf{A}$入手，设$\mathbf{A}$的化学式为$\mathbf{M}_2\mathrm{O}_y$，因此可以有：

$\frac{16y}{2M+16y}=25.6\%$

从$y=1$开始尝试，当$y=3$时解得$M=69.75~g/mol$，考虑到精度问题，可以认为是$\mathrm{Ga}$元素。因此$\mathbf{A}$为$\mathrm{Ga_2O_3}$。

<Checkpoint>[$\mathbf{A}$为$\mathrm{Ga_2O_3}$](1)</Checkpoint>

$\mathrm{Ga_2O_3}$溶解在氢氧化钠溶液中生成$\mathrm{NaGa(OH)_4}$，因此$\mathbf{C}$为$\mathrm{NaGa(OH)_4}$。

<Checkpoint>[$\mathbf{C}$为$\mathrm{NaGa(OH)_4}$](1)</Checkpoint>

四面体分子单质$\mathbf{B}$容易得到为$\mathrm{P_4}$。

<Checkpoint>[$\mathbf{B}$为$\mathrm{P_4}$](0.5)</Checkpoint>

$\mathrm{P_4}$在碱性条件下发生的歧化反应如下：

$\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$

从方程式可以看到产率为25%的物质为$\mathrm{PH_3}$，因此$\mathbf{D}$为$\mathrm{PH_3}$。

<Checkpoint>[$\mathbf{D}$为$\mathrm{PH_3}$](1)</Checkpoint>

根据题目所给提示可以写出反应4(弱碱性产物应为$\mathrm{NaH_2PO_4}$)：

$\mathrm{P_4+2I_2+2H_2O+6NaOH\to 2PH_3+2NaH_2PO_4+4NaI}$

因此$\mathbf{E}$为$\mathrm{NaH_2PO_4}$。

<Checkpoint>[$\mathbf{E}$为$\mathrm{NaH_2PO_4}$](1)</Checkpoint>

根据元素守恒可以推测$\mathbf{F}$中含有$\mathrm{Ga}$元素，因此为$\mathrm{Ga}$的低价氯化物，即$\mathrm{GaCl}$。因此$\mathbf{F}$为$\mathrm{GaCl}$。

<Checkpoint>[$\mathbf{F}$为$\mathrm{GaCl}$](1)</Checkpoint>

下面进行选项分析：

1，$\mathbf{X}$中元素的原子序数为31和15，因此$\frac{a-b}{a+b}=0.35$。正确

2，$\mathbf{C，D}$中氢元素的质量分数分别为2.508%，8.896%。正确。

3，反应6中反应物沸点都较低，但是$\mathrm{GaCl}$很容易被氧化，因此抽真空是为了保护反应物。错误。

<Checkpoint>[$\mathrm{GaCl}$易被氧化，抽真空可以保护反应物](1)</Checkpoint>

4，先写出反应1-6：

反应1：$\mathrm{Ga_2O_3+2NaOH+3H_2O\to2NaGa(OH)_4}$

<Checkpoint>[反应1：$\mathrm{Ga_2O_3+2NaOH+3H_2O\to2NaGa(OH)_4}$](1)</Checkpoint>

反应2：$\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$

<Checkpoint>[ 反应2：$\mathrm{P_4+3NaOH+3H_2O\to PH_3+3NaH_2PO_2}$](1)</Checkpoint>

反应3：$\mathrm{NaGa(OH)_4+PH_3\to GaP+NaOH+3H_2O}$

<Checkpoint>[ 反应3：$\mathrm{NaGa(OH)_4+PH_3\to GaP+NaOH+3H_2O}$](1)</Checkpoint>

反应4：$\mathrm{P_4+2I_2+2H_2O+6NaOH\to 2PH_3+2NaH_2PO_4+4NaI}$

<Checkpoint>[反应4：$\mathrm{P_4+2I_2+2H_2O+6NaOH\to 2PH_3+2NaH_2PO_4+4NaI}$](1)</Checkpoint>

反应5：$\mathrm{Ga_2O_3+2PH_3\to2GaP+3H_2O}$

<Checkpoint>[反应5：$\mathrm{Ga_2O_3+2PH_3\to2GaP+3H_2O}$](1)</Checkpoint>

反应6：$6\mathrm{GaCl+P_4\to4GaP+2GaCl_3}$

<Checkpoint>[反应6：$6\mathrm{GaCl+P_4\to4GaP+2GaCl_3}$](1)</Checkpoint>

立即看出反应物与生成物分子数目相差最小的是反应6。正确。