Mix a dilute hydrochloric acid solution of ruthenium trichloride with an ethanol solution of **L** (structure shown below), gently heat at $75^{\circ} C$ for $3 h$, and brick-red powder precipitates;  

<MultiModal>![NC1=NN2C(N=C(C)C=C2C)=N1](/media/uploads/f573bff9-ce1a-44f0-be6c-ffde1dc2b94a.png)</MultiModal>  

The powder is recrystallized with an acetone-water mixed solvent, yielding orange-red crystals **A**. The following characterization results are obtained for **A**:  

(1) Magnetic moment measurements show that the magnetic moment of **A** at $20^{\circ} C$ is $\mu=1.9 \mu_B$.  

(2) Elemental analysis indicates the mass fractions of elements in **A** are: $C, 29.52\%; H, 3.86\%; N, 24.58\%$.  

(3) X-ray diffraction reveals that two types of molecules exist in the crystal of **A**: one is a mononuclear octahedral ruthenium complex molecule, which contains $4$ hydrogen bonds, with **L** coordinating via the nitrogen atoms linked to two carbon atoms on the five-membered ring; the other is a solvent molecule.  

(4) The $^1H$ NMR spectrum of **A** in acetone shows that ligand **L** has only one chemical environment.  

Which of the following statements about crystal **A** are correct?  

1. The electron configuration of $Ru$ in **A** in the octahedral field is $t_{2g}^3e_g^2$.  
2. The chemical formula of crystal **A** is $RuC_{14}H_{22}O_2N_{10}Cl_3$.  
3. **A** contains neither acetone nor ethanol molecules.  
4. The complex contains $N-H…Cl$ hydrogen bonds.  
5. The complex contains $N-H…O$ hydrogen bonds.
将三氯化钌的稀盐酸溶液与**L** （结构如下图所示）的乙醇溶液混合, 在$75^{\circ} C$下温和加热$3 h$, 析出砖红色粉末;

<MultiModal>![NC1=NN2C(N=C(C)C=C2C)=N1](/media/uploads/f573bff9-ce1a-44f0-be6c-ffde1dc2b94a.png)</MultiModal>

 该粉末经丙酮-水混合溶剂重结晶, 析出橙红色晶体**A**。对**A**有如下表征结果:

(1) 磁矩测定实验表明, **A**在$20^{\circ} C$下的磁矩为$\mu=1.9 \mu_B$。

(2) 元素分析表明, **A**中各元素的含量(质量分数)为: $C, 29.52\%; H, 3.86\%; N, 24.58\%$。

(3) X-射线衍射表明,**A**的晶体中存在两种分子:一种是钌的单核八面体配合物分子,分子中存在$4$根氢键, **L**用五元环上与两个碳原子相连的氮原子配位; 另一种是溶剂分子。

(4) **A**的丙酮溶液的核磁共振氢谱表明, 配体**L**仅有一种化学环境。

以下有关**A**晶体的说法中，哪几项是正确的？

1. **A**中$Ru$在八面体场中的电子排布为$t_{2g}^3e_g^2$
2. **A**晶体的化学式为$RuC_{14}H_{22}O_2N_{10}Cl_3$
3. **A**中不含丙酮分子或乙醇分子
4. 配合物中含有$N-H…Cl$氢键
5. 配合物中含有$N-H…O$氢键
1, 2, 3, 4, 5
1, 2, 3, 4
1, 2, 3, 5
1, 2, 4, 5
1,3,4,5
2, 3, 4, 5
1, 2, 3
1, 2, 4
1, 2, 5
1,3,4
1,3,5
1,4,5
2, 3, 4
2,3,5
2,4,5
3,4,5
None of the above
1,2,3,4,5
1,2,3,4
1,2,3,5
1,2,4,5
1,3,4,5
2,3,4,5
1,2,3
1,2,4
1,2,5
1,3,4
1,3,5
1,4,5
2,3,4
2,3,5
2,4,5
3,4,5
以上都不对
From the magnetic moment $\mu = \sqrt{n(n+2)} = 1.9\mu_B$, we can obtain the number of unpaired electrons as $n=1$, so $Ru$ is low spin; the valence electron configuration is $t_{2g}^5 e_g^{0}$, statement 1 is incorrect.

<Checkpoint>[$Ru$ valence electron configuration is $t_{2g}^5 e_g^{0}$, statement 1 is incorrect](1)</Checkpoint>

The ratio of $C, H, N$ in **A** is: $(29.52 / 12.01) : (3.86 / 1.008) : (24.58 / 14.01) \approx 7: 11: 5$. In L, $C:N =7: 5$. Ethanol and acetone contain carbon but not nitrogen, so **A** does not contain acetone or ethanol. The source of the extra hydrogen should be water;
Therefore, **A** contains $Ru$, **L**, $H_2O$, and may contain $Cl$ and $OH$ ligands.

<Checkpoint>[The ratio of $C, H, N$ in **A** is: $7: 11: 5$, from which it is determined that **A** does not contain acetone or ethanol, and contains water molecules](1)</Checkpoint>

Let the chemical formula of **A** have $n$ ligands **L**, then the molar mass of **A** is:
$M(A)=12.01 \mathrm{~g} \cdot \mathrm{mol}^{-1} \times 7 n / 29.52 \%=284.8 n \mathrm{~g} \cdot \mathrm{mol}^{-1}$

When $n = 1$, the molar mass of $1$ ligand **L** and $1$ $ Ru$ has already reached $264.3 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, which is unreasonable;

When $n = 2$, the molar mass $M = 569.6 \mathrm{~g} \cdot \mathrm{mol}^{-1}$; since $Ru$ is in the $+3$ oxidation state, there are $3$ $OH$ and $Cl$ ligands in total; since $Ru$ is $6$-coordinate and **L** is a monodentate ligand, there should be $1$ water molecule within the coordination sphere, so the chemical formula of **A** can be expressed as: $RuL_2Cl_{3-x}(OH)_x(H_2O)_n$

In **L**, $C : H=7 : 11$, so $x+2n=4$; substitute $x=2, n=1$ and $x=0, n=2$ into the chemical formula in turn, and calculate the molar mass: when $x=0, n=2$, the molar mass is $569.8 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, which is consistent with the question, i.e., **A** is $RuL_2Cl_3(H_2O)_2$; then statements 2 and 3 are correct.

<Checkpoint>[**A** is $RuL_2Cl_3(H_2O)_2$](3)</Checkpoint>

The actual structure should be $[RuL_2Cl_3(H_2O)] \cdot H_2O$. The structure of the complex is
SMILES: 

Cl[Ru-3]([N+]1=C2N(C(C)=CC(C)=N2)N=C1N([H])[H])([O+]([H])[H])(Cl)(Cl)[N+]3=C(N=C(C)C=C4C)N4N=C3N([H])[H]

According to the description of the NMR test in the question, the amino group in **L** can form $N-H…Cl$ hydrogen bonds with chlorine atoms, and the water molecule as a ligand can form $O-H…N$ hydrogen bonds with the six-membered ring nitrogen atom in **L**. Therefore, there are $N-H…Cl$ and $O-H…N$ hydrogen bonds in the complex, statement 4 is correct, and 5 is incorrect.

<Checkpoint>[There are $N-H…Cl$ hydrogen bonds, statement 4 is correct](1)</Checkpoint>

<Checkpoint>[There are $O-H…N$ hydrogen bonds, statement 5 is incorrect](1)</Checkpoint>

Therefore, choose option M.
由磁矩 $\mu = \sqrt{n(n+2)} = 1.9\mu_B$可得单电子数为$n=1$，则$Ru$为低自旋; 价电子排布为 $t_{2g}^5 e_g^{0}$, 说法1错误.

<Checkpoint>[$Ru$价电子排布为 $t_{2g}^5 e_g^{0}$, 说法1错误](1)</Checkpoint>

**A** 中 $C, H, N$ 数目比为: $(29.52 / 12.01) : (3.86 / 1.008) : (24.58 / 14.01) \approx 7: 11: 5$ L 中的 $C:N =7: 5$, 乙醇和丙酮中含碳而不含氮, 因此 **A** 中无丙酮或乙醇, 多余的氢的来源应为水;
因此 **A** 中含有 $Ru$, **L**, $H_2O$, 可能含有 $Cl$ 和 $OH$ 配体. 

<Checkpoint>[**A** 中 $C, H, N$ 数目比为: $7: 11: 5$，由此判断**A** 中无丙酮或乙醇, 且含有水分子](1)</Checkpoint>

设 **A** 的化学式中有 $n$ 个配体 **L**, 则 **A** 的摩尔质量为:
$M(A)=12.01 \mathrm{~g} \cdot \mathrm{mol}^{-1} \times 7 n / 29.52 \%=284.8 n \mathrm{~g} \cdot \mathrm{mol}^{-1}$ 

当 $n = 1$ 时, $1$ 个配体 **L** 和 $1$ 个$ Ru$ 的摩尔质量就已达到 $264.3 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, 不合理; 

当 $n = 2$ 时, 摩尔质量 $M = 569.6 \mathrm{~g} \cdot \mathrm{mol}^{-1}$;由于 $Ru$ 为$+3$ 氧化态, 因此 $OH$ 配体和 $Cl$ 配体共有 $3$ 个; 又因为 $Ru$ 为 $6$ 配位, **L** 为单齿配体, 因此配合物内界应有 $1$ 个水分子, 故 **A** 的化学式可表示为: $RuL_2Cl_{3-x}(OH)_x(H_2O)_n$

**L** 中的 $C : H=7 : 11$, 得 $x+2n=4$; 依次将 $x=2, n=1$ 和 $x=0, n=2$ 代入化学式中, 计算摩尔质量得: $x=0, n=2$ 时摩尔质量为 $569.8 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, 符合题意, 即 **A** 为 $RuL_2Cl_3(H_2O)_2$; 则说法2 3正确. 

<Checkpoint>[**A** 为 $RuL_2Cl_3(H_2O)_2$](3)</Checkpoint>

实际结构应为 $[RuL_2Cl_3(H_2O)] \cdot H_2O$. 配合物的结构为
SMILES: 

Cl[Ru-3]([N+]1=C2N(C(C)=CC(C)=N2)N=C1N([H])[H])([O+]([H])[H])(Cl)(Cl)[N+]3=C(N=C(C)C=C4C)N4N=C3N([H])[H]

根据题目NMR测试的描述，**L**中氨基可以与氯原子形成$N-H…Cl$氢键，而作为配体的水分子可以与**L**中的六元环氮原子形成$O-H…N$氢键. 因此配合物中存在$N-H…Cl$氢键与$O-H…N$氢键, 说法4正确, 5错误. 

<Checkpoint>[存在$N-H…Cl$氢键, 说法4正确](1)</Checkpoint>

<Checkpoint>[存在$O-H…N$氢键, 说法5错误](1)</Checkpoint>

因此选择选项M.