To a mixed $\mathrm{CH_2Cl_2}$ solution of $\mathrm{[Au(PPh_3)_2][SbF_6]}$ and $\mathrm{Au}\cdot \mathbf{dpa}$ (the structure of the ligand $\mathbf{dpa^-}$ is shown below), a methanol solution of $\mathrm {CH_3ONa}$ was added. The mixture was stirred at room temperature, and the system turned into a green solution. Under vigorous stirring, a freshly prepared ethanol solution of $\mathrm{NaBH_4}$ was gradually added dropwise to the system, and the solution gradually turned light brown, and finally turned dark brown. The reaction was continued with stirring in the dark for $20\mathrm{h}$. The system was evaporated to dryness, washed with diethyl ether and ethanol to obtain a black solid $\mathbf{A}$. Single crystals of $\mathbf{A}$ were obtained by recrystallization of $\mathbf{A}$ and studied. The following partial information about $\mathbf{A}$ was obtained:

(a) $\mathbf{A}$ is a 1:2 type salt, and its cation is a cluster ion centered on the gold atom cluster $\mathrm {Au_x^{y+}}$; $\mathrm {Au_x^{y+}}$ has a double-layer nested structure in the form of $[ \mathrm { A u } _ { a } @ \mathrm { A u } _ { b } ]$.

(b) The main signal of the cation of $\mathbf{A}$ in time-of-flight mass spectrometry is $m / z = 4 7 1 1 . 02$

(c) $\mathbf{A}$ thermally decomposes at $950 ^ { \circ } \mathrm { C }$ with a weight loss of $36 . 5 \%$ , and the remaining solid residue is $\mathrm{Au}$ element.

(d) Partial elemental content (mass fraction) of $\mathbf{A}$: $\mathrm { C } , 24 . 8\% ; \mathrm { N } , 2 . 5 5\% ; \mathrm { P } , 2 . 5 0 \%$

<MultiModal>![The figure shows the structure of $\mathbf{dpa^-}$, and the smiles is expressed as: C1([N-]C2=NC=CC=C2)=NC=CC=C1](/media/uploads/64d24ca1-2b88-4296-9fd9-18010de4f491.png)</MultiModal>

There are the following statements:

1，$|\frac{x-y}{x+2y}|^{\ln|\frac{x+y}{x-y}|}=0.70$.

2, The molecular weight of $\mathbf{A}$ is 9657.8.

3，$|\frac{a-b}{a+2b}|^{\ln|\frac{a+b}{a-b}|}=0.07$.

4, The phosphorus atoms in the cation of $\mathbf{A}$ are coordinated to the inner gold atoms.

Then the following options contain all the correct options (the error of 1, 3 within 0.01 is considered correct):
向$\mathrm{[Au(PPh_3)_2][SbF_6]}$和$\mathrm{Au}\cdot \mathbf{dpa}$(配体$\mathbf{dpa^-}$的结构如下所示)的混合 $\mathrm{CH_2Cl_2}$ 溶液中，加入$\mathrm {CH_3ONa}$的甲醇溶液，室温搅拌，体系变为绿色溶液。在剧烈搅拌下，向体系中逐渐滴加新制的 $\mathrm{NaBH_4}$的乙醇溶液，溶液逐渐变为淡棕色，最后变为深棕色。继续避光搅拌反应$20\mathrm{h}$，将体系蒸发干燥，经乙醚和乙醇洗涤后，得到一种黑色固体$\mathbf{A}$；对 $\mathbf{A}$进行重结晶可得到$\mathbf{A}$的单晶并进行研究；得出 $\mathbf{A}$的部分信息如下：

(a)$\mathbf{A}$为1:2 型盐，其阳离子为以金原子簇$\mathrm {Au_x^{y+}}$ 为中心的簇离子；$\mathrm {Au_x^{y+}}$具有形如 $[ \mathrm { A u } _ { a } @ \mathrm { A u } _ { b } ]$ 的双层嵌套结构。

(b)$\mathbf{A}$的阳离子在飞行时间质谱中的主要信号的 $m / z = 4 7 1 1 . 0 2$ 

(c)$\mathbf{A}$在 $9 5 0 ^ { \circ } \mathrm { C }$ 下热分解，失重 $3 6 . 5 \%$ ，剩余固体残渣为$\mathrm{Au}$单质。

(d)$\mathbf{A}$中部分元素含量(质量分数)： $\mathrm { C } , 2 4 . 8\% ; \mathrm { N } , 2 . 5 5 \% ; \mathrm { P } , 2 . 5 0 \%$

<MultiModal>![图中展示了$\mathbf{dpa^-}$的结构，smiles表示为：C1([N-]C2=NC=CC=C2)=NC=CC=C1](/media/uploads/64d24ca1-2b88-4296-9fd9-18010de4f491.png)</MultiModal>

有下列说法：

1，$|\frac{x-y}{x+2y}|^{\ln|\frac{x+y}{x-y}|}=0.70$。

2，$\mathbf{A}$的分子量为9657.8。

3，$|\frac{a-b}{a+2b}|^{\ln|\frac{a+b}{a-b}|}=0.07$。

4，$\mathbf{A}$的阳离子中磷原子与内层金原子配位。

则下列选项中包含所有正确选项的是（1，3结果误差在0.01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
First, calculate the ratio of atoms by mass fraction:

$\mathrm{C:N:P}=\frac{24.8\%}{12.01}:\frac{2.55\%}{14.01}:\frac{2.50\%}{30.97}\approx11.35:1:0.4435$

The nitrogen atoms in $\mathbf{A}$ only come from the ligand $\mathbf{dpa^-}$, so it is a multiple of 3. Thus, the above formula is transformed into:

$11.35:1:0.4435\approx34.05:3:1.331\approx102:9:4$

Therefore, $\mathrm{C:N:P}=102:9:4$. The total number of carbon atoms in 4 $\mathrm{PPh_3}$ and 3 $\mathbf{dpa^-}$ is $4\times18+3\times10=102$, which matches the calculation.

<Checkpoint>[Atomic ratio of $\mathbf{A}$ $\mathrm{C:N:P}=102:9:4$](1)</Checkpoint>

<Checkpoint>[The carbon-containing ligands of $\mathbf{A}$ are only $\mathrm{PPh_3}$ and $\mathbf{dpa^-}$ with a ratio of 4:3](2)</Checkpoint>

The total molecular weight of 4 $\mathrm{PPh_3}$ and 3 $\mathbf{dpa^-}$ is $4\times262.3+3\times170.192=1559.8$.

$\mathbf{A}$ is a 1:2 type salt. Here, the anion can only be $\mathrm{SbF_6^-}$, so the cation carries two positive charges.

<Checkpoint>[The cation of $\mathbf{A}$ carries two positive charges](1)</Checkpoint>

Therefore, the mass spectrometry signal m/z = M/2, from which the molecular weight of the cation can be obtained as $2\times4711.02=9422.04$.

If the cation contains 4 $\mathrm{PPh_3}$ and 3 $\mathbf{dpa^-}$, the number of $\mathrm{Au}$ is $\frac{9422.04-1559.8}{197}=39.90$, which means the gold content is too high, inconsistent with the thermogravimetric results.

If the cation contains 8 $\mathrm{PPh_3}$ and 6 $\mathbf{dpa^-}$, the number of $\mathrm{Au}$ is $\frac{9422.04-1559.8\times2}{197}=32.00$, which is consistent with the question. Therefore, the cation is $[\mathrm{Au_{32}(PPh_3)_8(dpa)_6]^{2+}}, x=32, y=8$.

<Checkpoint>[The cation of $\mathbf{A}$ is $[\mathrm{Au_{32}(PPh_3)_8(dpa)_6]^{2+}}$](2)</Checkpoint>

<Checkpoint>[$x=32,y=8$](0.5)</Checkpoint>

Thus, the chemical formula of $\mathbf{A}$ is $[\mathrm{Au_{32}(PPh_3)_8(dpa)_6][SbF_6]_2}$, and the molecular weight is 9893.56.

<Checkpoint>[The molecular weight of $\mathbf{A}$ is 9893.56](1)</Checkpoint>

$\mathrm {Au_{32}^{8+}}$ has a double-layer nested structure in the form of $[ \mathrm { A u } _ { a } @ \mathrm { A u } _ { b } ]$. It is speculated that it is a nesting of two polyhedra. Considering the simplest polyhedron, i.e., a regular polyhedron, then 32 atoms can be exactly divided into 20+12, which is exactly a regular dodecahedron plus a regular icosahedron.

<Checkpoint>[32 atoms form a regular dodecahedron and a regular icosahedron](1)</Checkpoint>

The inside is a regular icosahedron with fewer atoms, and the outside is a regular dodecahedron with more atoms. Therefore, $a=12,b=20$.

<Checkpoint>[$a=12,b=20$](1)</Checkpoint>

Since the space inside the cage is very small, and $\mathrm{PPh_3}$ has a large volume, it can only coordinate on the outside. Therefore, phosphorus atoms coordinate with the outer gold atoms.

<Checkpoint>[$\mathrm{PPh_3}$ has a large volume, and phosphorus atoms coordinate with the outer gold atoms](1)</Checkpoint>

The following is an analysis of the options:

1，$|\frac{x-y}{x+2y}|^{\ln|\frac{x+y}{x-y}|}=0.70$. Correct.

2, The molecular weight of $\mathbf{A}$ is 9893.56. Incorrect.

3，$|\frac{a-b}{a+2b}|^{\ln|\frac{a+b}{a-b}|}=0.07$. Correct.

4, In the cation of $\mathbf{A}$, phosphorus atoms coordinate with the outer gold atoms. Incorrect.

Choose G
先通过质量分数计算原子数之比：

$\mathrm{C:N:P}=\frac{24.8\%}{12.01}:\frac{2.55\%}{14.01}:\frac{2.50\%}{30.97}\approx11.35:1:0.4435$

$\mathbf{A}$中的氮原子仅来源于配体$\mathbf{dpa^-}$，因此为3的倍数，于是上式化为：

$11.35:1:0.4435\approx34.05:3:1.331\approx102:9:4$

于是$\mathrm{C:N:P}=102:9:4$，4个$\mathrm{PPh_3}$和3个$\mathbf{dpa^-}$的碳原子总数为$4\times18+3\times10=102$,符合计算。

<Checkpoint>[$\mathbf{A}$原子数之比$\mathrm{C:N:P}=102:9:4$](1)</Checkpoint>

<Checkpoint>[$\mathbf{A}$的含碳配体只有$\mathrm{PPh_3}$和$\mathbf{dpa^-}$且比例为4：3](2)</Checkpoint>

4个$\mathrm{PPh_3}$和3个$\mathbf{dpa^-}$的总分子量为$4\times262.3+3\times170.192=1559.8$。

$\mathbf{A}$为1:2 型盐，此处阴离子只可能为$\mathrm{SbF_6^-}$，因此阳离子带两个正电荷。

<Checkpoint>[$\mathbf{A}$的阳离子带两个正电荷](1)</Checkpoint>

因此质谱信号m/z = M/2，由此可得阳离子的分子量为 $2\times4711.02=9422.04$。

若阳离子中为4个$\mathrm{PPh_3}$和3个$\mathbf{dpa^-}$，则$\mathrm{Au}$的数目为$\frac{9422.04-1559.8}{197}=39.90$，含金量过大，不符合热重结果。

若阳离子中为8个$\mathrm{PPh_3}$和6个$\mathbf{dpa^-}$，则$\mathrm{Au}$的数目为$\frac{9422.04-1559.8\times2}{197}=32.00$，符合题意。因此阳离子为$[\mathrm{Au_{32}(PPh_3)_8(dpa)_6]^{2+}}，x=32,y=8$。

<Checkpoint>[$\mathbf{A}$的阳离子为$[\mathrm{Au_{32}(PPh_3)_8(dpa)_6]^{2+}}$](2)</Checkpoint>

<Checkpoint>[$x=32,y=8$](0.5)</Checkpoint>

从而$\mathbf{A}$的化学式为$[\mathrm{Au_{32}(PPh_3)_8(dpa)_6][SbF_6]_2}$，分子量为9893.56。

<Checkpoint>[$\mathbf{A}$的分子量为9893.56](1)</Checkpoint>

$\mathrm {Au_{32}^{8+}}$具有形如 $[ \mathrm { A u } _ { a } @ \mathrm { A u } _ { b } ]$ 的双层嵌套结构，推测其为两个多面体进行嵌套，考虑最简单的多面体即正多面体，则32个原子恰好可以分为20+12个，刚好为正十二面体加上正二十面体，

<Checkpoint>[32个原子组成了正十二面体和正二十面体](1)</Checkpoint>

内部为原子数较少的正二十面体，外部为原子数较多的正十二面体，因此$a=12,b=20$。

<Checkpoint>[$a=12,b=20$](1)</Checkpoint>

由于笼内部的空间非常小，而$\mathrm{PPh_3}$体积较大，因此只能在外部配位，因此磷原子与外层金原子配位。

<Checkpoint>[$\mathrm{PPh_3}$体积较大，磷原子与外层金原子配位](1)</Checkpoint>


下面进行选项分析：

1，$|\frac{x-y}{x+2y}|^{\ln|\frac{x+y}{x-y}|}=0.70$。正确。

2，$\mathbf{A}$的分子量为9893.56。错误。

3，$|\frac{a-b}{a+2b}|^{\ln|\frac{a+b}{a-b}|}=0.07$。正确。

4，$\mathbf{A}$的阳离子中磷原子与外层金原子配位。错误。

选G