$\rm [Cu(CH_3CN)_4](PF_6)$ , $\rm Na(S_2CNR_2)$ and $\rm [(^nC_4H_9)_4N](BH_4)$ were mixed in acetonitrile according to stoichiometric ratio, stirred at room temperature for 1 hour under nitrogen protection, the solvent was evaporated, and the residue was recrystallized from chloroform to obtain a yellow solid. The yellow solid is a $\rm Cu(I)$ polynuclear cluster with the chemical formula $\rm [Cu_8H(S_2CNR_2)_x](PF_6)_y$, and the hydride ion is encapsulated in the cluster; elemental analysis shows that the mass fractions of each element in the cluster are: $\rm C, 23. 24\%; H, 4.14\%; N, 5.64\%$

The following statement that is correct is:
1.  $\rm R$ is methyl
2.  In the cation, there are two kinds of environments for $\rm Cu$ (ratio is 1:1), one of which $\rm Cu$ is connected to 6 $\rm Cu$, and the other is connected to 3 $\rm Cu$. Then $\rm Cu$ constitutes a tetracapped tetrahedron.
3.  In the cation, each $\rm Cu$ is coordinated by 3 $\rm S$, then each $\rm S$ is coordinated by 2 $\rm Cu$.
4.  The cluster can also be obtained by mixing $\rm [Cu(CH_3CN)_4](PF_6)$ , $\rm Cu(S_2CNR_2)_2$ and $\rm [(^nC_4H_9)_4N](BH_4)$ in acetonitrile according to stoichiometric ratio. The sum of the stoichiometric numbers of the reaction equation (the coefficient is balanced to the simplest integer ratio, and the species in the equation need to reflect the real structural form) is 41
将$\rm [Cu(CH_3CN)_4](PF_6)$ 、 $\rm Na(S_2CNR_2)$和$\rm [(^nC_4H_9)_4N](BH_4)$按计量比混合溶于乙腈，在氮气保护下室温搅拌反应1小时，蒸干溶剂，用氯仿重结晶得到黄色固体。该黄色固体为$\rm Cu(I)$的多核簇合物，化学式为$\rm [Cu_8H(S_2CNR_2)_x](PF_6)_y$ ，氢负离子被包在簇中；元素分析表明，该簇合物中各元素的质量分数为：$\rm C, 23. 24\%；H, 4.14\%；N, 5.64\%$

下面说法正确的是：
1. $\rm R$是甲基
2. 在阳离子中，$\rm Cu$共有两种环境(比例为1:1），其中一种$\rm Cu$与6个$\rm Cu$连接，另一种与3个$\rm Cu$连接。则$\rm Cu$构成四帽四面体
3. 在阳离子中，每个$\rm Cu$均被3个$\rm S$配位，则每个$\rm S$被2个$\rm Cu$配位
4. $\rm [Cu(CH_3CN)_4](PF_6)$ 、 $\rm Cu(S_2CNR_2)_2$和$\rm [(^nC_4H_9)_4N](BH_4)$在乙腈中按计量比混合反应也可得到该簇合物，该反应方程式（系数配平为最简整数比，方程式中物种需要反映真实结构存在形式）的化学计量数之和为41
All other options are incorrect
2, 3
2, 4
1, 2, 4
1, 2
2, 3
1, 2, 3, 4
2
3, 4
1, 3, 4
2, 3, 4
1, 2, 3
1
4
其他选项均不正确
2, 3
2, 4
1, 2, 4
1, 2
2, 3
1, 2, 3, 4
2
3, 4
1, 3, 4
2, 3, 4
1, 2, 3
1
4
The number of carbon atoms in the ligand is: $14.01/5.64\%×23.24\%/12.01=4.81$, $\rm R$ contains 2 carbons

The number of hydrogen atoms in the ligand is much larger than that outside the ligand, approximately: $14.01/5.64\%×4.14\%/1.008=10.2$, that is, $\rm R$ contains 5 hydrogens
So $\rm R=Et$, statement 1 is incorrect

<Checkpoint>[$\rm R$ is ethyl](2)</Checkpoint>

According to charge balance, we have: $x + y = 7$

According to the mass fraction, we have: $[63.55×8+1.008+148.3x+145.0×y]×5.64%=14.01x$

Solving for: $x=6, y=1$

<Checkpoint>[$x=6, y=1$](2)</Checkpoint>

Each $\rm Cu$ in the $\rm Cu$ tetrahedron is connected to 3 $\rm Cu$, so we can first assemble 4 $\rm Cu$ tetrahedra. To make the 4 $\rm Cu$ in the cation connected to 6 $\rm Cu$, we can combine them into another tetrahedron, which is a capped tetrahedral type, statement 2 is correct

<Checkpoint>[$\rm Cu$ in the cation forms a capped tetrahedron](2)</Checkpoint>

The cation contains a total of $\rm 8Cu+12S$, $\rm 8×3/12=2$, so each $\rm S$ is coordinated by 2 $\rm Cu$, statement 3 is correct

<Checkpoint>[Each $\rm S$ is coordinated by 2 $\rm Cu$](1)</Checkpoint>

Let the coefficient of $\rm Cu(S_2CNEt_2)_2$ be $\rm 3a$, then the coefficient of $\rm [Cu(CH_3CN)_4](PF_6)$ is $\rm 5a$, $\rm [(^nC_4H_9)_4N]^+$ needs to form a salt with $\rm (PF_6)^-$, so the coefficient of $\rm [(^nC_4H_9)_4N](BH_4)$ is $\rm 4a$. $\rm Cu$ is reduced and $\rm H_2$ is released, and the remaining boron element should exist in the form of $\rm B_2H_6$.

<Checkpoint>[The remaining boron element should exist in the form of $\rm B_2H_6$](1)</Checkpoint>

$\rm 5[Cu(CH_3CN)_4](PF_6)+3Cu(S_2CNEt_2)_2+4[(^nC_4H_9)_4N](BH_4)=[Cu_8H(S_2CNEt_2)_6](PF_6)+4[(^nC_4H_9)_4N](PF_6)+2B_2H_6+2H_2+20CH_3CN$

<Checkpoint>[$\rm 5[Cu(CH_3CN)_4](PF_6)+3Cu(S_2CNEt_2)_2+4[(^nC_4H_9)_4N](BH_4)=[Cu_8H(S_2CNEt_2)_6](PF_6)+4[(^nC_4H_9)_4N](PF_6)+2B_2H_6+2H_2+20CH_3CN$](2)</Checkpoint>

The sum of the coefficients is 41, statement 4 is correct

Select K
配体含碳数目为：$14.01/5.64\%×23.24\%/12.01=4.81$，$\rm R$含2个碳

配体中含氢远大于配体外的氢数，约为：$14.01/5.64\%×4.14\%/1.008=10.2$ ，即$\rm R$含5个氢
所以$\rm R=Et$，说法1错误

<Checkpoint>[$\rm R$为乙基](2)</Checkpoint>

根据电荷平衡，有：$x + y = 7$ 

根据质量分数，有：$[63.55×8+1.008+148.3x+145.0×y]×5.64%=14.01x$

解得：$x=6, y=1$

<Checkpoint>[$x=6, y=1$](2)</Checkpoint>

$\rm Cu$四面体满足每个$\rm Cu$与3个$\rm Cu$相连，因此可以先拼出来4个$\rm Cu$四面体。若要使阳离子中4个$\rm Cu$满足与6个$\rm Cu$相连，则将它们再组成一个四面体即可，即四加帽四面体型，说法2正确

<Checkpoint>[阳离子中$\rm Cu$形成四加帽四面体](2)</Checkpoint>

阳离子共有$\rm 8Cu+12S$，$\rm 8×3/12=2$，故每个$\rm S$被2个$\rm Cu$配位，说法3正确

<Checkpoint>[每个$\rm S$被2个$\rm Cu$配位](1)</Checkpoint>

令$\rm Cu(S_2CNEt_2)_2$系数为$\rm 3a$，则$\rm [Cu(CH_3CN)_4](PF_6)$系数为$\rm 5a$，$\rm [(^nC_4H_9)_4N]^+$要与$\rm (PF_6)^-$组成盐，故$\rm [(^nC_4H_9)_4N](BH_4)$系数为$\rm 4a$。有$\rm Cu$被还原，要放出$\rm H_2$，剩下的硼元素应该以$\rm B_2H_6$形式存在。

<Checkpoint>[剩下的硼元素应该以$\rm B_2H_6$形式存在](1)</Checkpoint>

$\rm 5[Cu(CH_3CN)_4](PF_6)+3Cu(S_2CNEt_2)_2+4[(^nC_4H_9)_4N](BH_4)=[Cu_8H(S_2CNEt_2)_6](PF_6)+4[(^nC_4H_9)_4N](PF_6)+2B_2H_6+2H_2+20CH_3CN$

<Checkpoint>[$\rm 5[Cu(CH_3CN)_4](PF_6)+3Cu(S_2CNEt_2)_2+4[(^nC_4H_9)_4N](BH_4)=[Cu_8H(S_2CNEt_2)_6](PF_6)+4[(^nC_4H_9)_4N](PF_6)+2B_2H_6+2H_2+20CH_3CN$](2)</Checkpoint>

系数之和为41，说法4正确

选K