At room temperature, 0.15 g of the compound $\mathrm{H}_{3} \mathbf{L}$ (structure shown in Figure 1) was dissolved in anhydrous ethanol, followed by the addition of 1 mL of pyridine solution and 0.09 g of nickel acetate. The mixture was stirred at 50°C for 30 min to obtain a solution. Slow evaporation in air yielded block-shaped single crystals $\mathbf{X}$ with a 48% yield. Elemental analysis of $\mathbf{X}$ revealed:  $\mathrm{C}$ 52.94; $\mathrm{H}$ 3.33; $\mathrm{N}$ 12.87; $\mathrm{O}$
14.69 (all in mass percent, %).  

<MultiModal>![Fig. 1. The SMILES representation of this molecule is: O=C(NNC(C1=CC=C([N+]([O-])=O)C=C1)=O)C2=CC=CC=C2O](/media/uploads/f248b3c0-c33f-4a33-83b4-497b2cc8caf5.png)</MultiModal>  

The chemical formula of $\mathbf{X}$ was deduced through calculations.  

Infrared spectroscopy was performed on both compound $\mathrm{H}_{3} \mathbf{L}$ and the resulting complex $\mathbf{X}$, yielding the following results.  

Key IR absorption peaks $\left(\mathrm{KBr}, \mathrm{cm}^{-1}\right): \mathrm{H}_{3} \mathbf{L}: 1692 ; 1590 ; 1543 ; 1520 ; \mathbf{X}: 1602 ; 1563 ; 1518$.  

By analyzing the differences in their IR spectra, the coordination structure of  $\mathbf{L}$  was deduced.  

Which of the following options is correct?
在室温下将 $0.15 \mathrm{~g}$ 化合物 $\mathrm{H}_{3} \mathbf{L}$ (结构如图1所示)溶解于无水乙醇中, 加入 $1 \mathrm{~mL}$ 吡啶溶液
和 $0.09 \mathrm{~g}$ 乙酸镍, 将混合物在 $50^{\circ} \mathrm{C}$ 下搅拌 $30 \mathrm{~min}$, 得到溶液。在空气中缓慢蒸发,
以 $48 \%$ 的产率得到了块状单晶 $\mathbf{X}$ 。对 $\mathbf{X}$ 元素分析可知: $\mathrm{C}$ 52.94; $\mathrm{H}$ 3.33; $\mathrm{N}$ 12.87; $\mathrm{O}$
14.69 (均为质量分数,\%)。

<MultiModal>![Fig. 1，该分子以SMILES表示为：O=C(NNC(C1=CC=C([N+]([O-])=O)C=C1)=O)C2=CC=CC=C2O](/media/uploads/f248b3c0-c33f-4a33-83b4-497b2cc8caf5.png)</MultiModal>

通过计算, 推出 $\mathbf{X}$ 的化学式。

对化合物 $\mathrm{H}_{3} \mathbf{L}$ 以及所得的配合物 $\mathbf{X}$ 分别进行红外光谱测定, 得到如下所示的结果。

主要的 IR 吸收峰 $\left(\mathrm{KBr}, \mathrm{cm}^{-1}\right): \mathrm{H}_{3} \mathbf{L}: 1692 ; 1590 ; 1543 ; 1520 ; \mathbf{X}: 1602 ; 1563 ; 1518$ 。

考虑二者 IR 光谱的区别的来源, 推算 $\mathbf{L}$配位时的结构。

以下选项正确的是？
None of the other options are correct
In a $\mathbf{X}$ molecule, the product of the total quantities of various atoms is 172800.
In a molecule of $\mathbf{X}$, the product of the total number of various atoms is 345600.
In a $\mathbf{X}$ molecule, the product of the total numbers of various atoms is 10800.
The UV-visible absorption peak of the coordinated $\mathbf{L}$ molecular structure (excluding the absorption of the central $\mathrm{Ni}$) may exhibit a significant red shift.
The variation in IR absorption peaks is caused by the formation of coordinate bonds when $\mathbf{L}$ molecules coordinate, leading to the weakening of the original infrared-active four chemical bonds.
There is no change in the position of the double bond in the $\mathbf{L}$ molecule before and after coordination.
其他选项均不正确
 一个$\mathbf{X}$分子中，各种原子总数量的乘积为172800
一个$\mathbf{X}$分子中，各种原子总数量的乘积为345600
一个$\mathbf{X}$分子中，各种原子总数量的乘积为10800
配位后的$\mathbf{L}$分子结构（不考虑中心$\mathrm{Ni}$的吸收）紫外-可见吸收峰可能发生明显红移
IR 吸收峰的变化是$\mathbf{L}$分子配位时形成配位键，导致原来有红外吸收的四根化学键强度变弱引起的
$\mathbf{L}$分子配位前后未发生双键位置的变化
From the elemental analysis results, we obtain $n(\mathrm{C}): n(\mathrm{O})=4.80$, $n(\mathrm{C}): n(\mathrm{H})=1.33$, and $n(\mathrm{N}): n(\mathrm{O})=1.00$.

The nitrogen-containing ligands in $\mathbf{X}$ can only be $\mathbf{L}$ and pyridine, with the chemical formulas $\mathrm{C}_{14} \mathrm{H}_{8} \mathrm{N}_{3} \mathrm{O}_{5}$ and $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}$, respectively.

From $n(\mathrm{N}): n(\mathrm{O})=1.00$, it can be deduced that the ratio of $\mathbf{L}$ to pyridine is $1: 2$.

<Checkpoint>[The ratio of $\mathbf{L}$ to pyridine in the ligands is $1: 2$](1)</Checkpoint>

Assuming $\mathbf{X}$ contains 1 $\mathbf{L}$ and 2 pyridines, the number of $\mathrm{C}$ atoms would be 24, $\mathrm{N}$ and $\mathrm{O}$ atoms would each be 5, and $\mathrm{H}$ atoms would be 18, which precisely matches the ratios obtained from the elemental analysis. Therefore, $\mathbf{X}$ contains only $\mathbf{L}$ and pyridine as ligands. If the number of $\mathbf{L}$ in $\mathbf{X}$ is $x=2$, then the number of pyridines is $2x$.

The central metal is clearly $\mathrm{Ni}$. When $x=2$, the remaining mass $M=176.1 \mathrm{~g} / \mathrm{mol}$ corresponds to 3 $\mathrm{Ni}$ atoms. Thus, the chemical formula of $\mathbf{X}$ is $\mathrm{Ni}_{3} \mathbf{L}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{4}$, written as $\mathrm{Ni_3C_{48}H_{36}O_{10}N_{10}}$, with a coefficient product of 518400. Options A, B, C, and D are incorrect.

<Checkpoint>[The chemical formula of $\mathbf{X}$ is $\mathrm{Ni}_{3} \mathbf{L}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{4}$, with a coefficient product of 518400](1)</Checkpoint>

When the compound $\mathrm{H}_{3} \mathbf{L}$ exists alone, it exhibits the equilibrium shown in Figure 2:

<MultiModal>![Fig. 2, the figure depicts a reversible reaction: O=C(C1=CC=CC=C1O)NNC(C2=CC=C(C=C2)[N+]([O-])=O)=O>>O=[N+]([O-])C(C=C3)=CC=C3/C(O)=N/N=C(O)/C4=CC=CC=C4O](/media/uploads/42faa357-ca57-4cac-ad13-17240ee01363.png)</MultiModal>

Thus, $\mathrm{H_3} \mathbf{L}$ displays four major infrared absorption peaks. The difference between $\mathbf{X}$ and $\mathrm{H_3}\mathbf{L}$ lies in the absence of the carbonyl absorption peak near $1700 \mathrm{~cm}^{-1}$, making option F incorrect. When $\mathbf{L}$ coordinates with the central metal, it undergoes keto-enol tautomerization, with the structure shown in Figure 3, making option G incorrect.

<MultiModal>![Fig. 3, the molecular structure is represented in SMILES as: [O-]C(C(C=C1)=CC=C1[N+]([O-])=O)=NN=C(C2=CC=CC=C2[O-])[O-]](/media/uploads/8c5d6d1d-e516-4a9f-8e14-637ee1481ec9.png)</MultiModal>

When coordinated, $\mathbf{L}$ forms a large conjugated system, causing the absorption peak to redshift, making option E correct.

<Checkpoint>[The difference between $\mathbf{X}$ and $\mathrm{H_3}\mathbf{L}$ is the absence of the carbonyl absorption peak near $1700 \mathrm{~cm}^{-1}$, and $\mathbf{L}$ forms an enol isomer upon coordination.](2)</Checkpoint>
由元素分析结果可得 $n(\mathrm{C}): n(\mathrm{O})=4.80, n(\mathrm{C}): n(\mathrm{H})=1.33, n(\mathrm{N}): n(\mathrm{O})=1.00$

$\mathbf{X}$中含氮配体只可能为 $\mathbf{L}$ 和吡啶, 化学式分别为 $\mathrm{C}_{14} \mathrm{H}_{8} \mathrm{N}_{3} \mathrm{O}_{5}$ 和 $\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}$

由 $n(\mathrm{N}): n(\mathrm{O})=1.00$ 可知, $\mathbf{L}$ 和吡啶的个数比为 $1: 2$ 

<Checkpoint>[配体中$\mathbf{L}$ 和吡啶的个数比为 $1: 2$ ](1)</Checkpoint>

假设$\mathbf{X}$中$\mathbf{L}$为1个,吡啶为2个,则$\mathrm{C}$的个数为24,$\mathrm{N}$和$\mathrm{O}$的个数均为5,$\mathrm{H}$的个数为18,恰好符合上述元素分析所得的比例,故$\mathbf{X}$中只含有$\mathbf{L}$和吡啶两种配体。设$\mathbf{X}$中$\mathbf{L}$的个数为$x=2$,则吡啶的个数为$2x$

中心金属显然为 $\mathrm{Ni}$, 故当 $x=2$ 时, 剩余 $M=176.1 \mathrm{~g} / \mathrm{mol}$, 对应于 3 个 $\mathrm{Ni}$，故 $\mathbf{X}$ 的化学式为 $\mathrm{Ni}_{3} \mathbf{L}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{4}$ ，写作 $\mathrm{Ni_3C_{48}H_{36}O_{10}N_{10}}$，各系数乘积为518400，选项A、B、C、D错误。

<Checkpoint>[$\mathbf{X}$的化学式为 $\mathrm{Ni}_{3} \mathbf{L}_{2}\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)_{4}$，各系数乘积为518400](1)</Checkpoint>

化合物 $\mathrm{H}_{3} \mathbf{L}$ 单独存在时,存在图2所示的平衡:

<MultiModal>![Fig. 2，图中为一可逆反应：O=C(C1=CC=CC=C1O)NNC(C2=CC=C(C=C2)[N+]([O-])=O)=O>>O=[N+]([O-])C(C=C3)=CC=C3/C(O)=N/N=C(O)/C4=CC=CC=C4O](/media/uploads/42faa357-ca57-4cac-ad13-17240ee01363.png)</MultiModal>

故 $\mathrm{H_3} \mathbf{L}$ 显示了 4 个主要的红外吸收峰, $\mathbf{X}$ 与$\mathrm{H_3}\mathbf{L}$ 的区别在于 $1700 \mathrm{~cm}^{-1}$ 附近的羰基吸收峰没有显示，选项F错误。$\mathbf{L}$与中心金属配位时发生烯醇式异构变化，结构如图3所示，选项G错误。

<MultiModal>![Fig. 3，图中分子以SMILES表示为：[O-]C(C(C=C1)=CC=C1[N+]([O-])=O)=NN=C(C2=CC=CC=C2[O-])[O-]](/media/uploads/8c5d6d1d-e516-4a9f-8e14-637ee1481ec9.png)</MultiModal>

配位时的$\mathbf{L}$整体形成较大的共轭体系，吸收峰发生红移，选项E正确。

<Checkpoint>[$\mathbf{X}$ 与$\mathrm{H_3}\mathbf{L}$ 的区别在于 $1700 \mathrm{~cm}^{-1}$ 附近的羰基吸收峰没有显示，$\mathbf{L}$配位时形成烯醇异构体。](2)</Checkpoint>