Due to the unique properties of metal-metal bonds, the synthesis of metal string complexes has consistently garnered widespread attention from researchers. A common method for synthesizing the ligand $\bf A$ for metal strings is as follows: Under an argon atmosphere, substrate $\bf B$ (1.98 g, 23 mmol) and $\bf C$ (2.16 g, 10 mmol), $\rm t-BuOK$ (2.81 g, 25 mmol), $\rm Pd_2(dba)_3$ (183 mg, 0.20 mmol), $\rm dppp$ (165 mg, 0.40 mmol) were added to a dry flask. The mixture was stirred and refluxed in toluene (100 mL) for 48 hours, and the system was observed to turn dark brown. The solvent was removed under reduced pressure, and then the mixture was poured into water. The precipitate was filtered to obtain a solid, which was purified by silica gel column chromatography using ethyl acetate as the eluent to obtain a bright yellow product $\bf A$. It is known that dppp is a bidentate phosphine ligand, and the structures of $\bf B$ and $\bf C$ are as follows:

<MultiModal>![The SMILES of $\bf B$ is NC1=NC=CC=C1, and the SMILES of $\bf C$ is ClC1=NC2=C(C=C1)C=CC(Cl)=N2](/media/uploads/97623dfa-2721-4207-95cc-b525b514df0a.png)</MultiModal>

A mixture of $\bf A$ (251 mg), naphthalene (25 g), and $\rm DMF$ (5 mL) was placed in a conical flask and heated to 100 $\rm\degree C$. Then $\rm t-BuOK$ (224 mg, 2.0 mmol) was slowly added, and the system immediately turned dark brown. It was heated at approximately 130~140 $\rm\degree C$ for 15 hours, and then refluxed at 200 ℃ for 15 minutes to obtain an orange transparent solution. Anhydrous $\rm CoCl_2$ (182 mg, 1.4 mmol) was added dropwise, and after refluxing for another 30 minutes, $\rm NaSCN$ (32 mg, 0.40 mmol) was slowly added, and the color changed from dark brown to dark green. The mixture was refluxed for another 3 hours, then concentrated to approximately 5 mL, and then cooled to 70 ℃. It was treated with hexane, and the metal complex precipitated. The precipitate was collected by suction filtration and washed with hexane to remove naphthalene and $\rm DMF$. The solid was extracted with $\rm CH_2Cl_2$ and filtered. The $\rm CH_2Cl_2$ solution was treated with excess $\rm KPF_6$ in methanol and stirred overnight, and then the solvent was removed under vacuum. The remaining solid was extracted with $\rm CH_2Cl_2$ and filtered. Pentane vapor was slowly diffused into the filtrate to obtain dark purple 1:1 salt $\bf D$. The mass fractions of each element in $\bf D$ (all theoretical values) are as follows: $\rm Co$ 18.14%, $\rm C$ 46.22%, $\rm H$ 2.59%, $\rm N$ 18.69%. It is known that all coordinating atoms in the complex are the same.

$\bf D$ was treated with $\rm [Cp_2Fe](PF_6)$ in $\rm CH_2Cl_2/MeOH$ in a 1:1 reaction to obtain $\bf E$.

At room temperature, the highest occupied single electron of $\bf D$ occupies $\delta ^*(6)$; the highest occupied two single electrons of $\bf E$ occupy $\delta ^*(6)$ and $\sigma ^*(4)$ with similar energy levels, respectively.

Which of the following statements are correct:
1.  Given that the average bond order of the $\rm Co-Co$ bond in $\bf D$ is 0.5, then $\rm Co$ all satisfy the 18-electron rule.
2.  There are 38 atoms involved in coordination in all anions and cations of $\bf E$, and there are 2 types of coordinating atoms.
3.  In the process of preparing $\bf D$, a substance must have been oxidized in the reaction system.
4.  Considering the distance between the farthest $\rm Co$ atoms, $\bf D$>$\bf E$
5.  The magnetic moments of $\bf D$ and $\bf E$ vary with temperature as shown in the figure below:
<MultiModal>![This image is a two-dimensional scatter plot with lines against a white background. The vertical Y-axis is labeled "$\rm μ_{eff}$ (B.M.)", with a scale starting from 0 and incrementing upwards to 1, 2, and 3. The horizontal X-axis is labeled "$\rm T$ (K)", with a scale starting from 0 and incrementing to the right with values 50, 100, 150, 200, 250, and 300. The plot shows two curves formed by discrete data points, as indicated in the legend: black squares represent compound $\bf 1$, and white circles represent compound $\bf 2$. The first curve, represented by solid black squares, has a $\rm μ_{eff}$ value of approximately 1.9 in the very low temperature region near 0 K, which then remains essentially constant, forming a flat region up to about T = 75 K. After 75 K, the $\rm μ_eff$ value of this curve increases very slowly with increasing temperature, reaching about 2.1 at 300 K. The second curve, represented by hollow white circles, has a $\rm μ_{eff}$ value below 1.0, around 0.8, near 0 K. As the temperature T increases, the $\rm μ_{eff}$ value of this curve continues to rise significantly, with a higher rate of increase in the low-temperature region, which gradually slows down as the temperature increases. At approximately T = 80 K, this hollow circle curve intersects with the solid square curve, where the $\rm μ_{eff}$ value is slightly below 2.0. At T = 300 K, the hollow circle curve reaches its highest point in the graph, with a $\rm μ_{eff}$ value of approximately 2.8.](/media/uploads/12faa5ef-ddf2-4ec6-a909-5bbf99e93106.png)</MultiModal>

Then the black squares represent $\bf E$, and the white circles represent $\bf D$
由于金属-金属键的独特性质，金属串络合物的合成一直以来受到研究人员的广泛关注。一种金属串常用的配体$\bf A$合成方法如下：在氩气气氛下，将底物$\bf B$(1.98 g, 23 mmol)及$\bf C$(2.16 g, 10 mmol)、$\rm t-BuOK$(2.81 g, 25 mmol)、$\rm Pd_2(dba)_3$(183 mg, 0.20 mmol)、$\rm dppp$(165 mg, 0.40 mmol)加入干燥的烧瓶中，搅拌混合物并在甲苯(100 mL)中回流48小时，观察到体系变为深棕色。在减压下除去溶剂，而后将混合物倒入水中，过滤沉淀物得到固体，用乙酸乙酯作为洗脱剂，通过硅胶柱色谱法得到亮黄色产物$\bf A$。已知dppp为一种双齿膦配体，$\bf B$和$\bf C$的结构如下：

<MultiModal>![$\bf B$的SMILES为NC1=NC=CC=C1，$\bf C$的SMILES为ClC1=NC2=C(C=C1)C=CC(Cl)=N2](/media/uploads/97623dfa-2721-4207-95cc-b525b514df0a.png)</MultiModal>

将$\bf A$(251 mg)、萘(25 g)及$\rm DMF$(5 mL)的混合物置入锥形瓶中，加热至100 $\rm\degree C$。然后缓慢加入$\rm t-BuOK$(224 mg, 2.0 mmol)，体系立刻变深棕色。在约130~140 $\rm\degree C$下加热15小时，而后再200 ℃下回流15分钟，得到橙色透明溶液。逐滴加入无水$\rm CoCl_2$(182 mg, 1.4 mmol)，再回流30分钟后，缓慢加入$\rm NaSCN$(32 mg, 0.40 mmol)，颜色由深棕色变为深绿色。混合物再回流3小时，再浓缩至约5 mL，然后冷却至70 ℃。用己烷处理，沉淀出金属络合物。抽滤收集沉淀物，用己烷洗涤以除去萘及$\rm DMF$。固体用$\rm CH_2Cl_2$萃取并过滤，将$\rm CH_2Cl_2$溶液用过量的$\rm KPF_6$于甲醇中处理并搅拌过夜后，于真空下除去溶剂。残余固体用$\rm CH_2Cl_2$萃取并过滤，正戊烷蒸汽缓慢扩散到滤液中以获得深紫色的1:1型盐$\bf D$。$\bf D$中各元素质量分数(均为理论值)如下：$\rm Co$ 18.14%，$\rm C$ 46.22%，$\rm H$ 2.59%，$\rm N$ 18.69%。已知配合物中所有配位原子相同。

用$\rm [Cp_2Fe](PF_6)$在$\rm CH_2Cl_2/MeOH$中处理$\bf D$,按1:1反应可以得到$\bf E$。

室温下$\bf D$能量最高的单电子占据$\delta ^*(6)$；$\bf E$能量最高的两个单电子分别占据能差不大的$\delta ^*(6)$和$\sigma ^*(4)$。

下列说法正确的有：
1. 已知$\bf D$中$\rm Co-Co$键平均键级为0.5，则$\rm Co$均满足18电子规则
2. $\bf E$的所有阴阳离子中共有38个原子参与配位，且配位原子种类为2
3. 制备$\bf D$的过程中，反应体系必然有物质被氧化了
4. 考虑相距最远的$\rm Co$原子之间的距离，$\bf D$>$\bf E$
5. $\bf D$和$\bf E$的磁矩随温度变化如下图所示：
<MultiModal>![这张图片是一幅背景为白色的二维散点线图。垂直的Y轴标签为“$\rm μ_{eff}$ (B.M.)”，刻度从0开始，向上依次标有1、2、3。水平的X轴标签为“$\rm T$ (K)”，刻度从0开始，向右依次标有50、100、150、200、250和300。图中展示了两条由离散数据点构成的曲线，图例中指出黑色方块代表化合物$\bf 1$，白色圆圈代表化合物$\bf 2$。第一条曲线由实心黑色方块表示，它在接近0 K的极低温度区，$\rm μ_{eff}$值约为1.9，随后该值基本保持恒定，形成一个平坦的区域，直到T约为75 K。在75 K之后，这条曲线的$\rm μ_eff$值随着温度的升高而非常缓慢地上升，在300 K时达到约2.1。第二条曲线由空心白色圆圈表示，它在接近0 K的温度下，$\rm μ_{eff}$值低于1.0，大约在0.8左右。随着温度T的升高，这条曲线的$\rm μ_{eff}$值持续显著上升，其上升的速率在低温区较大，并随着温度的增高而逐渐减缓。在T约等于80 K时，这条空心圆圈曲线与实心方块曲线相交，此时$\rm μ_{eff}$值略低于2.0。在T等于300 K时，空心圆圈曲线达到其在图中的最高点，$\rm μ_{eff}$值约为2.8。](/media/uploads/12faa5ef-ddf2-4ec6-a909-5bbf99e93106.png)</MultiModal>

则黑色方块代表$\bf E$，白色圆圈代表$\bf D$
All other options are incorrect
1,3,4
2,3,5
3,4
3,4,5
1,3
2,4
4,5
1,3,4,5
2,5
3,4,5
2
1,3,5
2,3,4,5
4
2,3,4
其他选项均不正确
1,3,4
2,3,5
3,4
3,4,5
1,3
2,4
4,5
1,3,4,5
2,5
3,4,5
2
1,3,5
2,3,4,5
4
2,3,4
$\bf B$ and $\bf C$ react, undergoing two couplings to obtain $\bf A$: <MultiModal>![SMILES为C1(C=CC(NC2=CC=CC=N2)=N3)=C3N=C(NC4=NC=CC=C4)C=C1](/media/uploads/3f779255-723c-4ed5-bbf4-9ee764ba9556.png)</MultiModal>

<Checkpoint>[$\bf A$ is C1(C=CC(NC2=CC=CC=N2)=N3)=C3N=C(NC4=NC=CC=C4)C=C](1)</Checkpoint>

The anion of $\bf A$ corresponds to the ligand $\rm L^{2-}$. From the mass fraction, $\rm n(C)/n(Co)=12.50, n(H)/n(Co)=8.35, n(N)/n(Co)=4.33$, and combined with the structure of the ligand $\rm L^{2-}$, there should be 6 $\rm Co$ atoms on a metal string ($\bf A$).

<Checkpoint>[There should be 6 $\rm Co$ atoms on a metal string](1)</Checkpoint>

From the mass fraction, we know $\rm n(C)=75$, $\rm n(H)=50$, $\rm n(N)=26$

<Checkpoint>[$\rm n(C)=75$, $\rm n(H)=50$, $\rm n(N)=26$](1)</Checkpoint>

From the number of $\rm N$, it can be seen that $\bf D$ contains 4 $\rm L^{2-}$ and 2 $\rm SCN^-$, containing a total of 74 $\rm C$ and 48 $\rm H$, with an excess of 1 $\rm C$ atom and 2 $\rm H$ atoms.

The anion of the compound should be $\rm PF_6^-$. Assuming there is only 1 anion, the remainder is: $\rm 75M(C)/46.22\% - 6M(Co) - 4M(L^{2-}) - 2M(SCN^-) - M(PF_6^-) = 84.76~g/mol$

<Checkpoint>[Assuming there is only 1 anion, the remainder of $\bf D$ is 84.76 g/mol](2)</Checkpoint>

This corresponds exactly to $\rm CH_2Cl_2$. Combined with the excess of 1 $\rm C$ atom and 2 $\rm H$ atoms, the complex has one solvated molecule of $\rm CH_2Cl_2$.

All coordinating atoms are the same, so it can be seen that $\rm SCN^-$ should coordinate with the $\rm N$ atom.

Therefore, the chemical formula of $\bf D$ is $\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)·CH_2Cl_2$

<Checkpoint>[The chemical formula of $\bf D$ is $\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)·CH_2Cl_2$](1)</Checkpoint>

The metal core in $\bf D$ is $\rm Co_6^{11+}$. $\rm Co^{2+}$ is reduced, so a substance must be oxidized in the reaction system. Statement 3 is correct. Counting the electrons provided by the 26 $\rm N$ atoms and the 5 $\rm Co-Co$ bonds with a bond order of 0.5, a total of $6 \times 9-11+26 \times 2+5 \times 0.5 \times 2=100<6 \times 18$. Statement 1 is incorrect.

<Checkpoint>[In $\bf D$, $\rm 6Co$ has a total of 100 electrons, and it is impossible to satisfy the 18-electron rule for all of them](1)</Checkpoint>

<Checkpoint>[A substance must be oxidized in the reaction system for the synthesis of $\bf D$](0.5)</Checkpoint>

$\bf E$ is $\bf D$ oxidized by a single electron, i.e., $\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)_2·2CH_2Cl_2$ (the coordinating solvent is not required and does not affect the judgment of the options). There are 26 $\rm N$ in the cation and 12 $\rm F$ in the anion participating in coordination. Statement 2 is correct. The average oxidation state of $\rm Co$ in $\bf E$ is higher, so the $\rm Co-Co$ bond order is larger to alleviate the electron deficiency, and $\rm Co-Co$ is shorter. Statement 4 is correct.

<Checkpoint>[$\bf E$ is $\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)_2·2CH_2Cl_2$](0.5)</Checkpoint>

<Checkpoint>[Compared to $\bf D$, the average oxidation state of $\rm Co$ in $\bf E$ is higher, and $\rm Co-Co$ is shorter](1)</Checkpoint>

The magnetic moment of $\bf D$ with 1 unpaired electron is basically unaffected by temperature changes. While the electrons of $\bf E$ can be excited as the temperature rises, transitioning between the singlet/triplet state. Therefore, the correct answer is: the black square corresponds to $\bf D$, and the white circle corresponds to $\bf E$. Statement 5 is incorrect.

<Checkpoint>[The black square corresponds to $\bf D$, and the white circle corresponds to $\bf E$](1)</Checkpoint>

Statements 2, 3, and 4 are correct, choose P
$\bf B$和$\bf C$反应，发生两次偶联得到$\bf A$：<MultiModal>![SMILES为C1(C=CC(NC2=CC=CC=N2)=N3)=C3N=C(NC4=NC=CC=C4)C=C1](/media/uploads/3f779255-723c-4ed5-bbf4-9ee764ba9556.png)</MultiModal>

<Checkpoint>[$\bf A$为C1(C=CC(NC2=CC=CC=N2)=N3)=C3N=C(NC4=NC=CC=C4)C=C](1)</Checkpoint>

$\bf A$的阴离子对应配体$\rm L^{2-}$。由质量分数，$\rm n(C)/n(Co)=12.50, n(H)/n(Co)=8.35, n(N)/n(Co)=4.33$, 再结合配体$\rm L^{2-}$的结构可知，一个金属串上($\bf A$)应有6个$\rm Co$原子。

<Checkpoint>[一个金属串上应有6个$\rm Co$原子](1)</Checkpoint>

由质量分数可知$\rm n(C)=75$，$\rm n(H)=50$，$\rm n(N)=26$

<Checkpoint>[$\rm n(C)=75$，$\rm n(H)=50$，$\rm n(N)=26$](1)</Checkpoint>

由$\rm N$的数量可知，$\bf D$中含有4个$\rm L^{2-}$和2个$\rm SCN^-$，共含74个$\rm C$和48个$\rm H$，其中有多余的1个$\rm C$原子和2个$\rm H$原子。

化合物的阴离子应为$\rm PF_6^-$，假设只有1个阴离子，则剩余：$\rm 75M(C)/46.22\% - 6M(Co) - 4M(L^{2-}) - 2M(SCN^-) - M(PF_6^-) = 84.76~g/mol$

<Checkpoint>[假设只有1个阴离子，则$\bf D$剩余84.76 g/mol](2)</Checkpoint>

这正好对应$\rm CH_2Cl_2$，结合多余的1个$\rm C$原子和2个$\rm H$原子可知，配合物有一分子$\rm CH_2Cl_2$的溶剂合。

所有配位原子相同，由此可知$\rm SCN^-$应以$\rm N$原子配位。

故$\bf D$的化学式为$\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)·CH_2Cl_2$

<Checkpoint>[$\bf D$的化学式为$\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)·CH_2Cl_2$](1)</Checkpoint>

$\bf D$中金属主体为$\rm Co_6^{11+}$，$\rm Co^{2+}$被还原了，反应体系必然有物质被氧化，说法3正确；算上26个$\rm N$原子提供的电子以及5根0.5键级的$\rm Co-Co$键，共$6 \times 9-11+26 \times 2+5 \times 0.5 \times 2=100<6 \times 18$，说法1错误

<Checkpoint>[$\bf D$中$\rm 6Co$共100个电子，不可能均满足18电子规则](1)</Checkpoint>

<Checkpoint>[合成$\bf D$的反应体系必然有物质被氧化](0.5)</Checkpoint>

$\bf E$为$\bf D$被单电子氧化，即$\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)_2·2CH_2Cl_2$（配位溶剂不作要求，不影响选项的判断），阳离子中有26个$\rm N$、阴离子中有12个$\rm F$参与配位，说法2正确；$\bf E$中$\rm Co$的平均氧化态更高，因此$\rm Co-Co$键键级更大来缓解缺电子性，$\rm Co-Co$也就更短，说法4正确

<Checkpoint>[$\bf E$为$\rm [Co_6(\mu_6-L)_4(NCS)_2](PF_6)_2·2CH_2Cl_2$](0.5)</Checkpoint>

<Checkpoint>[相比$\bf D$，$\bf E$中$\rm Co$的平均氧化态更高，$\rm Co-Co$也就更短](1)</Checkpoint>

有1个单电子的$\bf D$温度变化基本上不影响磁矩；而温度升高$\bf E$的电子可以被激发，在单线态/三线态之间转变。故正确答案为：黑色方块对应$\bf D$，白色圆圈对应$\bf E$，说法5错误

<Checkpoint>[黑色方块对应$\bf D$，白色圆圈对应$\bf E$](1)</Checkpoint>

说法2,3,4正确，选P