Deduce and write the equations for the following chemical reactions (requiring the coefficient ratio to be the simplest integer ratio).

1-1 Phosphorus trifluoride decomposes upon heating, producing a molecule with a tetrahedral structure.

1-2 Manganese(II) nitrate hexahydrate reacts with acetic anhydride to produce manganese(III) acetate and release a reddish-brown mixed gas.

1-3 Iodine phosphate ($\mathrm{IPO}_{4}$) decomposes upon contact with water, with the only reduction product being $\mathrm{I}_{2}$, and no gas is generated.

1-4 In dichloromethane, $\mathrm{Br}_{2}$ reacts with $\mathrm{H}_{2} \mathrm{~S}$ in a 1:1 molar ratio, and one of the products is an unstable acid; no element is formed.

1-5 Reactions in non-aqueous solvents can be used to prepare many novel species. Gold(III) fluorosulfate reacts with $\mathrm{CO}$ in fluorosulfonic acid,
resulting in a mononuclear complex molecule with $\mathrm{Au}$ being sp hybridized; the reaction also releases $\mathrm{CO}_{2}$ gas. Note the form of ions in non-aqueous solvents.

There are the following statements:

1: The product of the coefficients of all chemical formulas on both sides of the 1-1 equation is 60.

2: The product of the coefficients of all chemical formulas on both sides of the 1-2 equation is 184320.

3: The product of the coefficients of all chemical formulas on both sides of the 1-3 equation is 320.

4: The number of atoms in each molecule on both sides of the 1-4 equation does not exceed 2.

5: The product of the sum of the number of atoms in each type of molecule on both sides of the 1-5 equation is 6912.

The following options are all correct and have the largest number of correct statements:
推测写出以下化学反应的方程式(要求系数比为最简整数比)。

1-1 三氟化磷受热分解, 产生一种四面体结构的分子。


1-2 六水合硝酸锰(II)与乙酸酐反应, 制得乙酸锰(III)并放出红棕色混合气体。


1-3 磷酸碘 $\left(\mathrm{IPO}_{4}\right)$ 遇水发生分解, 还原产物仅有 $\mathrm{I}_{2}$，无气体生成 。


1-4 在二氯甲烷中, $\mathrm{Br}_{2}$ 与 $\mathrm{H}_{2} \mathrm{~S}$ 按摩尔比 1:1 反应, 产物中有一种不稳定的酸; 无单质生成。


1-5 利用非水溶剂中的反应能制得许多新奇的物种。氟磺酸金(III)与 $\mathrm{CO}$ 在氟磺酸中反应,
得到 $\mathrm{Au}$ 为 $\mathrm{sp}$ 杂化的单核配合物分子; 反应还有 $\mathrm{CO}_{2}$ 气体放出。注意非水溶剂中离子的存在形式。

有以下几种说法：

1：1-1 方程式中左右所有化学式系数乘积为60

2：1-2 方程式中左右所有化学式系数乘积为184320

3：1-3 方程式中左右所有化学式系数乘积为320

4：1-4 方程式左右各分子中原子数均不超过2

5：1-5 方程式左右各类分子内原子数总和的乘积为6912

以下选项说法全部正确且正确数量最多的是：
All other options are incorrect
1
2
3
4
5
1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
1,2,3
1,2,4
1,2,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
3,4,5
1,2,3,4
其他选项均不正确
1
2
3
4
5
1,2
1,3
1,4
1,5
2,3
2,4
2,5
3,4
3,5
4,5
1,2,3
1,2,4
1,2,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
3,4,5
1,2,3,4
1-1 The disproportionation reaction occurs during the decomposition process. The tetrahedral structure molecule is likely $\mathrm{P}_{4}$. The remaining phosphorus element is oxidized to a stable +5 valence state:

$10 \mathrm{PF}_{3} \rightarrow 6 \mathrm{PF}_{5}+\mathrm{P}_{4}$

<Checkpoint>[ $10 \mathrm{PF}_{3} \rightarrow 6 \mathrm{PF}_{5}+\mathrm{P}_{4}$](1)</Checkpoint>

The product of the coefficients of all chemical formulas on both sides of the equation is 60, 1 is correct.

1-2 Acetic anhydride $\mathrm{(CH}_{3} \mathrm{CO)}_{2} \mathrm{O}$ acts as a dehydrating agent and reaction medium, removing water of hydration and providing an anhydrous environment, promoting the oxidation of nitrate to $\mathrm{Mn(II)}$ to $\mathrm{Mn(III)}$. Nitrate oxidizes divalent manganese to produce reddish-brown $\mathrm{NO}_{2}$. Excess nitric acid is unstable and decomposes under acetic anhydride dehydration. Another component in the mixed gas may be $\mathrm{O}_{2}$:

$4\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}+30\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \rightarrow 4 \mathrm{Mn}\left(\mathrm{OOCCH}_{3}\right)_{3}+8 \mathrm{NO}_{2}+\mathrm{O}_{2}+48 \mathrm{CH}_{3} \mathrm{COOH}$

The product of the coefficients of all chemical formulas on both sides of the equation is 184320, 2 is correct.

<Checkpoint>[$4\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}+30\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \rightarrow 4 \mathrm{Mn}\left(\mathrm{OOCCH}_{3}\right)_{3}+8 \mathrm{NO}_{2}+\mathrm{O}_{2}+48 \mathrm{CH}_{3} \mathrm{COOH}$](1)</Checkpoint>

1-3 Trivalent iodine is reduced to zero valence. Then, some iodine undergoes disproportionation to generate stable pentavalent iodate. Balancing the equation:

$5 \mathrm{IPO}_{4}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2}+3 \mathrm{HIO}_{3}+5 \mathrm{H}_{3} \mathrm{PO}_{4}$

1-3 The product of the coefficients of all chemical formulas on both sides of the equation is 675, not 320, 3 is incorrect.

<Checkpoint>[ $5 \mathrm{IPO}_{4}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2}+3 \mathrm{HIO}_{3}+5 \mathrm{H}_{3} \mathrm{PO}_{4}$](1)</Checkpoint>

1-4 Under normal circumstances, $\mathrm{Br}_{2}$ will directly oxidize $\mathrm{H_2S}$ to generate elemental sulfur. However, the question indicates a 1:1 reaction, producing an unstable acid and no elemental substance. Therefore, sulfur may be partially oxidized to generate $\mathrm{HSBr}$:

$\mathrm{Br}_{2}+\mathrm{H_2S} \rightarrow \mathrm{HBr}+\mathrm{HSBr}$

The number of atoms in each molecule on the left and right sides of the equation exceeds 2, 4 is incorrect.

<Checkpoint>[$\mathrm{Br}_{2}+\mathrm{H_2S} \rightarrow \mathrm{HBr}+\mathrm{HSBr}$](1)</Checkpoint>

1-5 sp-hybridized gold atoms are two-coordinated. With $\mathrm{CO}_{2}$ released, it indicates that gold is reduced to the most likely +1 valence state. Considering the difficulty of ion solvation in non-aqueous solvents, a fluorosulfate ligand is needed to directly coordinate and balance the charge, then the other is $\mathrm{CO}$. The oxygen atoms in $\mathrm{CO}_{2}$ are provided by the fluorosulfate, forming fluorosulfonic anhydride in a non-aqueous solvent:

$\mathrm{Au}\left(\mathrm{SO}_{3} \mathrm{F}\right)_{3}+2 \mathrm{CO} \rightarrow \mathrm{Au}(\mathrm{CO})\left(\mathrm{SO}_{3} \mathrm{F}\right)+\mathrm{CO}_{2}+\mathrm{S}_{2} \mathrm{O}_{5} \mathrm{F}_{2}$

The product of the sum of the number of atoms in each type of molecule on the left and right sides of the equation is 6912, which is correct.

<Checkpoint>[$\mathrm{Au}\left(\mathrm{SO}_{3} \mathrm{F}\right)_{3}+2 \mathrm{CO} \rightarrow \mathrm{Au}(\mathrm{CO})\left(\mathrm{SO}_{3} \mathrm{F}\right)+\mathrm{CO}_{2}+\mathrm{S}_{2} \mathrm{O}_{5} \mathrm{F}_{2}$](1)</Checkpoint>

Therefore, choose S.
1-1 分解过程发生歧化反应，四面体结构分子很可能是$\mathrm{P}_{4}$，剩余磷元素氧化至稳定的+5价：

$10 \mathrm{PF}_{3} \rightarrow 6 \mathrm{PF}_{5}+\mathrm{P}_{4}$

<Checkpoint>[ $10 \mathrm{PF}_{3} \rightarrow 6 \mathrm{PF}_{5}+\mathrm{P}_{4}$](1)</Checkpoint>

方程式中左右所有化学式系数乘积为60，1正确

1-2 乙酸酐$\mathrm{(CH}_{3} \mathrm{CO)}_{2} \mathrm{O}$作为脱水剂和反应介质，去除水合水并提供无水环境，促进硝酸根氧化 $\mathrm{Mn(II)}$ 到 $\mathrm{Mn(III)}$。硝酸根氧化二价锰，生成红棕色$\mathrm{NO}_{2}$。多余的硝酸在乙酸酐脱水下不稳定分解，混合气体中的另一组分可能是$\mathrm{O}_{2}$：

$4\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}+30\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \rightarrow 4 \mathrm{Mn}\left(\mathrm{OOCCH}_{3}\right)_{3}+8 \mathrm{NO}_{2}+\mathrm{O}_{2}+48 \mathrm{CH}_{3} \mathrm{COOH}$

方程式中左右所有化学式系数乘积为184320，2正确

<Checkpoint>[$4\left[\mathrm{Mn}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]\left(\mathrm{NO}_{3}\right)_{2}+30\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O} \rightarrow 4 \mathrm{Mn}\left(\mathrm{OOCCH}_{3}\right)_{3}+8 \mathrm{NO}_{2}+\mathrm{O}_{2}+48 \mathrm{CH}_{3} \mathrm{COOH}$](1)</Checkpoint>

1-3 三价碘还原为零价，那么有一部分碘发生岐化生成稳定的五价碘酸，配平即可：

$5 \mathrm{IPO}_{4}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2}+3 \mathrm{HIO}_{3}+5 \mathrm{H}_{3} \mathrm{PO}_{4}$

1-3 方程式中左右所有化学式系数乘积为675，不是320，3错误

<Checkpoint>[ $5 \mathrm{IPO}_{4}+9 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{I}_{2}+3 \mathrm{HIO}_{3}+5 \mathrm{H}_{3} \mathrm{PO}_{4}$](1)</Checkpoint>

1-4 通常情况下$\mathrm{Br}_{2}$会直接氧化$\mathrm{H_2S}$生成单质硫。但题中指出1:1反应，产物有一种不稳定酸且无单质生成，因此硫可能被部分氧化生成$\mathrm{HSBr}$：

$\mathrm{Br}_{2}+\mathrm{H_2S} \rightarrow \mathrm{HBr}+\mathrm{HSBr}$

 方程式左右各分子中原子数有超过2的，4错误

<Checkpoint>[$\mathrm{Br}_{2}+\mathrm{H_2S} \rightarrow \mathrm{HBr}+\mathrm{HSBr}$](1)</Checkpoint>

1-5 sp杂化的金原子为二配位，有$\mathrm{CO}_{2}$放出，说明金被还原到最可能为+1价。考虑到非水溶剂离子溶剂化困难，需要一个氟磺酸根配体直接配位平衡电荷，那么另一个为$\mathrm{CO}$。$\mathrm{CO}_{2}$中的氧原子由氟磺酸根提供，非水溶剂下形成氟磺酸酐：

$\mathrm{Au}\left(\mathrm{SO}_{3} \mathrm{F}\right)_{3}+2 \mathrm{CO} \rightarrow \mathrm{Au}(\mathrm{CO})\left(\mathrm{SO}_{3} \mathrm{F}\right)+\mathrm{CO}_{2}+\mathrm{S}_{2} \mathrm{O}_{5} \mathrm{F}_{2}$

 方程式左右各类分子内原子数总和的乘积为6912，正确

<Checkpoint>[$\mathrm{Au}\left(\mathrm{SO}_{3} \mathrm{F}\right)_{3}+2 \mathrm{CO} \rightarrow \mathrm{Au}(\mathrm{CO})\left(\mathrm{SO}_{3} \mathrm{F}\right)+\mathrm{CO}_{2}+\mathrm{S}_{2} \mathrm{O}_{5} \mathrm{F}_{2}$](1)</Checkpoint>

因此，选择S。