$\mathbf{X}$ is a common $\mathbf{M}$-containing compound, which can be obtained by treating an anhydrous ether solution of $\mathbf{M}\mathrm{Cl_3}$ with hydrogen sulfide at low temperature (reaction $\mathbf{1}$). $\mathbf{X}$ is readily soluble in potassium cyanide solution, and the products contain two different sulfur-containing anions (reaction $\mathbf{2}$).  

$\mathbf{Y}$ is a neutral complex containing low-valent $\mathbf{M}$. Refluxing $\mathbf{Z}$ (containing $\mathbf{M}\ 68\ \%$), $\mathrm{Zn(NO_3)_2·6H_2O}$, and $\mathbf{L}$ in anhydrous acetonitrile solvent for 5 min, followed by crystallization of the resulting solution under an inert atmosphere, yields the complex $\mathbf{Y}$ (reaction $\mathbf{3}$). The structure of $\mathbf{L}$ is known as:  

<MultiModal>![C1=CC(=NC=C1)C2=NC=CC=C2, i.e., the structure of [L]](/media/uploads/9a7f4a4c-cf5d-4f90-9a49-8cc286515f36.png)</MultiModal>  

In $\mathbf{Y}$, both $\mathrm{Zn}$ and $\mathbf{M}$ exhibit only one coordination number, and no mirror plane is present. Elemental analysis of $\mathbf{Y}$ yields: $\mathrm{N}\ 12.8; \mathrm{C}\ 32.9; \mathrm{Zn}\ 7.46$ (all in mass fraction, $\%$).  

Which of the following statements are correct?  
1. Reaction $\mathbf{1}$ is a redox reaction, and the oxidation state of $\mathbf{M}$ remains unchanged in reaction $\mathbf{2}$.  
2. In the simplest balanced equation for reaction $\mathbf{1}$, the sum of all coefficients on the reactant side and the product side is 14.  
3. In reaction $\mathbf{2}$, the mass fraction ratio of sulfur in the two anions is 1: 0.55.  
4. In the simplest balanced equation for reaction $\mathbf{2}$, the sum of all coefficients on the reactant side and the product side is 12.  
5. In the simplest balanced equation for reaction $\mathbf{3}$, the sum of all coefficients on the reactant side and the product side is 13.  
6. $\mathbf{Y}$ is chiral.  
7. $\mathbf{Y}$ has a total of 4 reasonable stereoisomers.
$\mathbf{X}$是一种常见的含$\mathbf{M}$化合物，低温下用硫化氢处理$\mathbf{M}\mathrm{Cl_3}$ 的无水乙醚溶液即可得到$\mathbf{X}$(反应$\mathbf{1}$)。$\mathbf{X}$易溶于氰化钾溶液，产物中包含两种不同的含硫阴离子(反应$\mathbf{2}$)。

$\mathbf{Y}$是一种含低价$\mathbf{M}$的电中性配合物。将$\mathbf{Z}$（含$\mathbf{M}\ 68\ \%$）、$\mathrm{Zn(NO_3)_2·6H_2O}$和$\mathbf{L}$在无水乙腈溶剂中回流5 min，所得溶液在惰性气氛下结晶即可得到配合物$\mathbf{Y}$(反应$\mathbf{3}$)。已知$\mathbf{L}$的结构为：

<MultiModal>![C1=CC(=NC=C1)C2=NC=CC=C2，即[L]的结构](/media/uploads/9a7f4a4c-cf5d-4f90-9a49-8cc286515f36.png)</MultiModal>

$\mathbf{Y}$中$\mathrm{Zn}$和$\mathbf{M}$均只有一种配位数，不含镜面；对$\mathbf{Y}$元素分析可得：$\mathrm{N}\ 12.8; \mathrm{C}\ 32.9; \mathrm{Zn}\ 7.46$ (均为质量分数, $\%$).

则下列说法正确的有几个？
1. 反应$\mathbf{1}$是氧化还原反应，反应$\mathbf{2}$中$\mathbf{M}$的价态不改变
2. 反应$\mathbf{1}$的最简整系数方程式中反应物一侧与产物一侧的所有系数之和为14
3. 反应$\mathbf{2}$中两种阴离子的硫元素质量分数之比为1: 0.55
4. 反应$\mathbf{2}$的最简整系数方程式中反应物一侧与产物一侧的所有系数之和为12
5. 反应$\mathbf{3}$的最简整系数方程式中反应物一侧与产物一侧的所有系数之和为13
6. $\mathbf{Y}$具有手性
7. $\mathbf{Y}$总计有4种合理的立体异构体
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
Analysis of the mass fractions of $\mathbf{Y}$:

$n(\mathrm{N}):n(\mathrm{Zn})=\frac{12.8}{14.01}:\frac{7.46}{65.41}=8:1$

$n(\mathrm{C}):n(\mathrm{Zn})=\frac{32.9}{12.01}:\frac{7.46}{65.41}=24:1$

<Checkpoint>[In $\mathbf{M}$, $n(\mathrm{C}):n(\mathrm{N}):n(\mathrm{Zn})=24:8:1$](1)</Checkpoint>

The chemical formula of $\mathbf{L}$ is $\mathrm{C_{10}H_8N_2}$, containing 10 $\mathrm{C}$ and 2 $\mathrm{N}$. Assuming only one $\mathrm{Zn}$, there should be 2 $\mathbf{L}$ molecules, leaving 4 $\mathrm{C}$ and 4 $\mathrm{N}$. Considering $\mathrm{CN^-}$, the molecular weight of $\mathbf{Y}$ can be calculated:

$Mr(\mathbf{Y})=\frac{65.41}{7.46\ \%}=876.8$

After subtracting one $\mathrm{Zn}$ and two $\mathbf{L}$ ($Mr=156.184$), 499.022 remains. Assuming the remainder is 4 $\mathrm{CN^-}$, 394.942 is left, corresponding to two $\mathrm{Au}$. Examining the oxidation states: $\mathrm{Zn}$ is +2, so $\mathrm{Au}$ is +1, which aligns with chemical principles. Therefore, $\mathbf{M}$ is $\mathrm{Au}$, and the chemical formula of $\mathbf{Y}$ is $\mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}$.

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

<Checkpoint>[The chemical formula of $\mathbf{Y}$ is $\mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}$](1)</Checkpoint>

Knowing $\mathbf{M}$ is $\mathrm{Au}$ and $\mathbf{Y}$ contains $\mathrm{CN^-}$, the chemical formula of $\mathbf{Z}$ (containing $\mathbf{M}\ 68\ \%$) can be deduced as $\mathrm{KAu(CN)_2}$.

<Checkpoint>[The chemical formula of $\mathbf{Z}$ is $\mathrm{KAu(CN)_2}$](1)</Checkpoint>

Thus, reaction $\mathbf{3}$ is:

$\mathrm{2K[Au(CN)_2]+Zn(NO_3)_2·6H_2O}+2\mathbf{L} \to \mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}+\mathrm{2KNO_3+6H_2O}$

<Checkpoint>[Reaction $\mathbf{3}$: $\mathrm{2K[Au(CN)_2]+Zn(NO_3)_2·6H_2O}+2\mathbf{L} \to \mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}+\mathrm{2KNO_3+6H_2O}$](1)</Checkpoint>

The total coefficient sum is 14, so option 5 is incorrect.

For reaction $\mathbf{1}$:

$\mathrm{2AuCl_3+3H_2S \to Au_2S_3+6HCl}$

<Checkpoint>[Reaction $\mathbf{1}$: $\mathrm{2AuCl_3+3H_2S \to Au_2S_3+6HCl}$](1)</Checkpoint>

Thus, $\mathbf{X}$ is $\mathrm{Au_2S_3}$. The sum of all coefficients on the reactant and product sides in the simplest integer equation for reaction $\mathbf{1}$ is 12, so option 2 is incorrect.

$\mathbf{X}$ readily dissolves in potassium cyanide solution, producing two different sulfur-containing anions (reaction $\mathbf{2}$): $\mathrm{S^{2-}}$ and $\mathrm{SCN^-}$. The mass fraction of sulfur in $\mathrm{S^{2-}}$ is 100 %, while in $\mathrm{SCN^-}$ it is 55 %, so option 3 is correct.

<Checkpoint>[The two anions produced in reaction $\mathbf{2}$ are $\mathrm{S^{2-}}$ and $\mathrm{SCN^-}$](1)</Checkpoint>

Since the oxidation state of sulfur changes, the oxidation state of $\mathbf{M}$ must also change in reaction $\mathbf{2}$, whereas it remains unchanged in reaction $\mathbf{1}$. Thus, option 1 is incorrect.

<Checkpoint>[The oxidation state of $\mathbf{M}$ does not change in reaction $\mathbf{1}$ but changes in reaction $\mathbf{2}$](1)</Checkpoint>

The equation for reaction $\mathbf{2}$ can be written as:

$\mathrm{Au_2S_3+6KCN \to 2K[Au(CN)_2]+K_2S+2KSCN}$

The sum of all coefficients on the reactant and product sides in the simplest integer equation for reaction $\mathbf{2}$ is 12, so option 4 is correct.

From $\mathbf{Y}$, where $\mathrm{Zn}$ and $\mathbf{M}$ each have only one coordination number and no mirror plane, the structure of $\mathbf{Y}$ is:

<MultiModal>![N#C[Au]C#[N][Zn]1([N]2=C3C=CC=C2)([N]#C[Au]C#N)([N]4=CC=CC=C43)[N]5=CC=CC=C5C6=[N]1C=CC=C6](/media/uploads/6da10dcc-4318-4157-8a5f-8a4cee808aa4.png)</MultiModal>

<Checkpoint>[The structure of $\mathbf{Y}$ is N#C[Au]C#[N][Zn]1([N]2=C3C=CC=C2)([N]#C[Au]C#N)([N]4=CC=CC=C43)[N]5=CC=CC=C5C6=[N]1C=CC=C6](2)</Checkpoint>

Clearly, $\mathbf{Y}$ is chiral and has 3 reasonable stereoisomers. Option 6 is correct, and option 7 is incorrect.

<Checkpoint>[$\mathbf{Y}$ is chiral](1)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$ has 3 reasonable stereoisomers](1)</Checkpoint>

In summary, 3 options are correct (3, 4, 6), so the answer is D.
对$\mathbf{Y}$的质量分数进行分析：

$n(\mathrm{N}):n(\mathrm{Zn})=\frac{12.8}{14.01}:\frac{7.46}{65.41}=8:1$

$n(\mathrm{C}):n(\mathrm{Zn})=\frac{32.9}{12.01}:\frac{7.46}{65.41}=24:1$

<Checkpoint>[$\mathbf{M}$中$n(\mathrm{C}):n(\mathrm{N}):n(\mathrm{Zn})=24:8:1$](1)</Checkpoint>

$\mathbf{L}$化学式为$\mathrm{C_{10}H_8N_2}$中有10个$\mathrm{C}$和2个$\mathrm{N}$，假设只有一个$\mathrm{Zn}$，则应该有2个$\mathbf{L}$，除此之外还剩4个$\mathrm{C}$和4个$\mathrm{N}$，考虑$\mathrm{CN^-}$，且可由此求得$\mathbf{Y}$的分子量：

$Mr(\mathbf{Y})=\frac{65.41}{7.46\ \%}=876.8$

除去一个$\mathrm{Zn}$和两个$\mathbf{L}（Mr=156.184）$后还剩499.022，假设剩下是4个$\mathrm{CN^-}$，则剩余394.942，正好对应于两个$\mathrm{Au}$，考察一下价态，$\mathrm{Zn}$为+2价，则$\mathrm{Au}$为+1价，符合化学常识。因此$\mathbf{M}$为$\mathrm{Au}$，$\mathbf{Y}$的化学式为$\mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}$.

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

<Checkpoint>[$\mathbf{Y}$的化学式为$\mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}$](1)</Checkpoint>

知道$\mathbf{M}$为$\mathrm{Au}$，且$\mathbf{Y}$中含有$\mathrm{CN^-}$后便可以推得$\mathbf{Z}$（含$\mathbf{M}\ 68\ \%$）的化学式为$\mathrm{KAu(CN)_2}$.

<Checkpoint>[$\mathbf{Z}$的化学式为$\mathrm{KAu(CN)_2}$](1)</Checkpoint>

则反应$\mathbf{3}$为：

$\mathrm{2K[Au(CN)_2]+Zn(NO_3)_2·6H_2O}+2\mathbf{L} \to \mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}+\mathrm{2KNO_3+6H_2O}$

<Checkpoint>[反应$\mathbf{3}$：$\mathrm{2K[Au(CN)_2]+Zn(NO_3)_2·6H_2O}+2\mathbf{L} \to \mathrm{Zn}\mathbf{L}_2\mathrm{[Au(CN)_2]_2}+\mathrm{2KNO_3+6H_2O}$](1)</Checkpoint>

总系数和为14，选项5错误。

由此，反应$\mathbf{1}$：

$\mathrm{2AuCl_3+3H_2S \to Au_2S_3+6HCl}$

<Checkpoint>[反应$\mathbf{1}$：$\mathrm{2AuCl_3+3H_2S \to Au_2S_3+6HCl}$](1)</Checkpoint>

因此$\mathbf{X}$为$\mathrm{Au_2S_3}$，反应$\mathbf{1}$的最简整系数方程式中反应物一侧与产物一侧的所有系数之和为12，选项2错误。

$\mathbf{X}$易溶于氰化钾溶液，产物中包含两种不同的含硫阴离子(反应$\mathbf{2}$)，只能为$\mathrm{S^{2-}}$和$\mathrm{SCN^-}$，$\mathrm{S^{2-}}$硫的质量分数为100 %，$\mathrm{SCN^-}$为55 %，选项3正确。

<Checkpoint>[反应$\mathbf{2}$中产生的两种阴离子为$\mathrm{S^{2-}}$和$\mathrm{SCN^-}$](1)</Checkpoint>

由于硫的价态发生改变，因此反应$\mathbf{2}$中$\mathbf{M}$的价态一定会改变，而反应$\mathbf{1}$中$\mathbf{M}$的价态不改变，选项1错误。

<Checkpoint>[反应$\mathbf{1}$中$\mathbf{M}$的价态不改变，而反应$\mathbf{2}$中$\mathbf{M}$的价态改变](1)</Checkpoint>

可以写出反应$\mathbf{2}$的方程式：

$\mathrm{Au_2S_3+6KCN \to 2K[Au(CN)_2]+K_2S+2KSCN}$

反应$\mathbf{2}$最简整系数方程式中反应物一侧与产物一侧的所有系数之和为12，选项4正确。

由$\mathbf{Y}$中$\mathrm{Zn}$和$\mathbf{M}$均只有一种配位数，不含镜面，可以得到$\mathbf{Y}$：

<MultiModal>![N#C[Au]C#[N][Zn]1([N]2=C3C=CC=C2)([N]#C[Au]C#N)([N]4=CC=CC=C43)[N]5=CC=CC=C5C6=[N]1C=CC=C6](/media/uploads/6da10dcc-4318-4157-8a5f-8a4cee808aa4.png)</MultiModal>

<Checkpoint>[$\mathbf{Y}$的结构为N#C[Au]C#[N][Zn]1([N]2=C3C=CC=C2)([N]#C[Au]C#N)([N]4=CC=CC=C43)[N]5=CC=CC=C5C6=[N]1C=CC=C6](2)</Checkpoint>

显然$\mathbf{Y}$具有手性且拥有3种合理的立体异构体。6正确，7错误。

<Checkpoint>[$\mathbf{Y}$具有手性](1)</Checkpoint>

<Checkpoint>[$\mathbf{Y}$拥有3种合理的立体异构体](1)</Checkpoint>

综上，共3个选项正确（3，4，6），答案选D.