Equal molar amounts of pyridine-2-carbohydrazide and gadolinium acetylacetonate dihydrate (denoted as $Gd(acac)_3·2H_2O$) were mixed in a methanol-dichloromethane solution, stirred at room temperature for 2 h, and reacted at $70^\circ \mathrm{C}$ for 48 h. After cooling, yellow crystals precipitated. X-ray diffraction revealed that the crystal is a binuclear gadolinium complex with the chemical formula $[Gd_2(L)_2(acac)_2(CH_3OH)_2]·nCH_3OH$, where the coordination number of gadolinium is 8. Elemental analysis indicated the mass percentages of each element in the crystal as follows: $\mathrm{C~40.21\%, H~4.97\%, N~7.75\%}$. Upon heating to approximately $300^\circ \mathrm{C}$, the crystal exhibited a weight loss of about $6.0\%$, corresponding to the removal of all external methanol molecules. The crystal can be magnetized under an applied magnetic field, leading to the ordered alignment of its magnetic moments.  

Based on calculation and reasoning, determine which of the following statements are correct. Select the option containing **all** correct statements.  

1. In the reaction generating the yellow crystals, dichloromethane is one of the reactants.  

2. A single $L^{2-}$ ligand contains 11 carbon atoms.  

3. The number of coordination bonds a single $L^{2-}$ ligand can form is 4.  

4. The $L^{2-}$ ligand forms a five-membered ring when coordinating with the metal.  

5. The magnetization process of this crystal under an applied magnetic field is exothermic.
将等摩尔的吡啶-2-甲酰肼和二水合乙酰丙酮钆(可记作$Gd(acac)_3·2H_2O)$于甲醇-二氯甲烷溶液中混合,室温搅拌2h,并在$70^\circ \mathrm{C}$下反应48h,降温,析出黄色晶体。晶体衍射表明该晶体为双核钆配合物,化学式为$[Gd_2(L)_2(acac)_2(CH_3OH)_2]·nCH_3OH$,钆的配位数为8;元素分析表明该晶体中各元素的含量分别为:$\mathrm{C~40.21\%,H~4.97\%,N~7.75\%}$(质量分数,%);将该晶体加热至$300^\circ \mathrm{C}$左右,失重约$6.0\%$,对应于失去所有外界甲醇分子。该晶体在外加磁场中可被磁化，使得其磁矩有序排列。

根据计算推理，判断以下说法正确的有哪些，选出包含**所有**正确说法的选项。

1.生成黄色晶体的反应中，反应物有二氯甲烷。

2.单个$L^{2-}$配体含有11个碳原子。

3.单个$L^{2-}$配体能形成的配位键数目是4。

4.$L^{2-}$配体和金属配位时形成了五元环。

5.该晶体在外加磁场中被磁化的过程是放热的。
1, 2, 3, 4, 5
1,2,4,5
2, 3, 4, 5
1,3,4
2,4,5
2,3,5
3,4,5
2,4
1,3
4,5
None of the above options are correct
1,2,3,4,5
1,2,4,5
2,3,4,5
1,3,4
2,4,5
2,3,5
3,4,5
2,4
1,3
4,5
以上选项均不对
Based on the substrate structure, $acac^-$ condenses with at most 1 equivalent of pyridine-2-carbohydrazide, suggesting that $L^{2-}$ contains 3 $N$ atoms.

<Checkpoint>[The ligand $L^{2-}$ contains 3 $N$ atoms](0.5)</Checkpoint>

The molecular weight of the complex is: $14.01 \times 3 \times 2\div 0.0775=1085$.

The weight loss corresponds to a molecular weight of: $1085 \times 0.060=65.1$, which is equivalent to 2 methanol molecules, i.e., $n=2$.

<Checkpoint>[$n=2$](1)</Checkpoint>

Assuming the ligand $L^{2-}$ contains $x$ carbon atoms, then $1085 \times 0.4021 \div 12.01=36.4 \approx 36$. Therefore: $2 x+2 \times 5+2 \times 1+2 \times 1=36, x=11$. If $L^{2-}$ were formed by the reaction of 2 or more equivalents of pyridine-2-carbohydrazide, no reasonable integer solution would be obtained under the above procedure. Thus, $x=11$, and statement 2 is correct. The product of the condensation between $acac^-$ and 1 equivalent of pyridine-2-carbohydrazide contains 11 carbon atoms, so dichloromethane is not a reactant, and statement 1 is incorrect.

<Checkpoint>[$L^{2-}$ ligand contains 11 carbon atoms, statement 2 is correct](1)</Checkpoint>

<Checkpoint>[Dichloromethane is not a reactant, statement 1 is incorrect](0.5)</Checkpoint>

The structure of $L^{2-}$ is as follows: <MultiModal>![[O-]/C(C1=NC=CC=C1)=N\N=C(C)\C=C([O-])\C](/media/uploads/539012ab-9782-4d11-95c0-db3cb2e782ad.png)</MultiModal>

Analyzing its coordination mode: $Gd$ is 8-coordinated, requiring a total of 16 coordination bonds. $2acac^-$ and $2CH_3OH$ provide $2\times 2=4$ and $2\times 1=2$ coordination bonds, respectively, so each $L^{2-}$ provides $(16-4-2)\div 2=5$ coordination bonds. Combining with the configuration of $L^{2-}$, it can be concluded that the pyridine nitrogen, two oxygen atoms, and the nitrogen marked in blue in the structure serve as coordination atoms, forming a total of 5 coordination bonds. The pyridine nitrogen and the oxygen atom from pyridine-2-carbohydrazide are in one chelating five-membered ring, the nitrogen marked in blue and the oxygen atom from pyridine-2-carbohydrazide are in another chelating five-membered ring, and the nitrogen marked in blue and the oxygen atom from $acac^-$ are in a chelating six-membered ring. Therefore, statement 3 is incorrect, and statement 4 is correct.

(Note: Here, $acac^-$ is unlikely to act as a bidentate ligand while also coordinating to another metal center to form 3 coordination bonds. This point is not required to be written.)

<Checkpoint>[Each $L^{2-}$ provides 5 coordination bonds, statement 3 is incorrect](1)</Checkpoint>

<Checkpoint>[The complex contains two chelating five-membered rings, statement 4 is correct](1)</Checkpoint>

When the crystal is magnetized, the magnetic moments align in an ordered manner, $\Delta S<0$, $-T\Delta S>0$, and since $\Delta G=\Delta H-T\Delta S<0$, it follows that $\Delta H<0$. Thus, the magnetization process is exothermic, and statement 5 is correct.

<Checkpoint>[$\Delta S<0$ during magnetization](0.5)</Checkpoint>

<Checkpoint>[The magnetization process is exothermic, statement 5 is correct](0.5)</Checkpoint>
根据底物结构，$acac^-$最多和1当量吡啶-2-甲酰肼缩合，推测$L^{2-}$含有3个$N$。

<Checkpoint>[配体$L^{2-}$中有3个$N$](0.5)</Checkpoint>

配合物式量为: $14.01 \times 3 \times 2\div 0.0775=1085$ 。

失重所对应的式量为: $1085 \times 0.060=65.1$, 为 2 个甲醇分子, 即$n=2$。

<Checkpoint>[$n=2$](1)</Checkpoint>

设$L^{2-}$配体含碳原子数为$x$，而$1085 \times 0.4021 \div 12.01=36.4 \approx 36$。因此: $2 x+2 \times 5+2 \times 1+2 \times 1=36, x=11$ 。若$L^{2-}$由2个及以上吡啶-2-甲酰肼反应得到,按上述流程无合理整数解。因此, $x=11$，说法2正确。$acac^-$和1当量吡啶-2-甲酰肼缩合的产物碳原子数为11，故二氯甲烷不是反应物，说法1错误。

<Checkpoint>[$L^{2-}$配体含碳原子数为11，说法2正确](1)</Checkpoint>

<Checkpoint>[二氯甲烷不是反应物，说法1错误](0.5)</Checkpoint>

$L^{2-}$的结构如下：<MultiModal>![[O-]/C(C1=NC=CC=C1)=N\N=C(C)\C=C([O-])\C](/media/uploads/539012ab-9782-4d11-95c0-db3cb2e782ad.png)</MultiModal>

分析其配位模式：$Gd$为8配位，则一共需要形成16个配位键。$2acac^-$和$2CH_3OH$分别提供$2\times 2=4$和$2\times 1=2$ 个配位键，所以每个$L^{2-}$提供$(16-4-2)\div 2=5$ 个配位键。结合$L^{2-}$的构型，可知：吡啶氮、两个氧原子和结构式中标蓝的氮作为配位原子，共形成5个配位键；吡啶氮和来自吡啶-2-甲酰肼的氧原子在一个螯合五元环中，标蓝的氮和来自吡啶-2-甲酰肼的氧原子在另一个螯合五元环中，标蓝的氮和来自$acac^-$的氧原子在一个螯合六元环中。所以说法3错误，说法4正确。

（注：$acac^-$在这里很难在作为二齿配体时再配位给另一个金属中心，不能形成3个配位键。这一点不要求写出）

<Checkpoint>[每个$L^{2-}$提供5个配位键，说法3错误](1)</Checkpoint>

<Checkpoint>[配合物中存在两个螯合五元环，说法4正确](1)</Checkpoint>

该晶体被磁化时磁矩有序排列，$\Delta S<0，-T\Delta S>0$，而$\Delta G=\Delta H-T\Delta S<0$，故$\Delta H<0$
，磁化过程放热，说法5正确。

<Checkpoint>[磁化过程$\Delta S<0$](0.5)</Checkpoint>

<Checkpoint>[磁化过程放热，说法5正确](0.5)</Checkpoint>