The multidentate ligand **HTPPNOL** (as shown) forms complexes with copper that can be used to simulate the active centers of binuclear copper-containing enzymes.  

<MultiModal>![C1=CC=NC(=C1)CNCC(CN(CC2=NC=CC=C2)CC3=CC=CC=N3)O](/media/uploads/828fe6e9-6a17-4a9c-b78b-6a09306f4978.png)</MultiModal>  

In all the formed complexes, the rings consisting solely of $\mathrm{Cu}^{2+}$ and atoms from **HTPPNOL** are all five-membered rings.  

In a sodium acetate solution, **HTPPNOL** reacts with $\mathrm{Cu}^{2+}$ to form the binuclear complex $\mathrm{[Cu_2(TPPNOL)(CH_3COO)]}^{2+}$. In this complex ion, $\mathrm{Cu}^{2+}$ exhibits two coordination numbers, one of which is five-coordinate (trigonal bipyramidal geometry), and the complex structure includes four six-membered ring.  

Based on the above information, select the correct option.
多齿配体 **HTPPNOL** （如图）与铜形成的配合物可以用于模拟双核含铜酶的活性中心。
 
<MultiModal>![C1=CC=NC(=C1)CNCC(CN(CC2=NC=CC=C2)CC3=CC=CC=N3)O](/media/uploads/828fe6e9-6a17-4a9c-b78b-6a09306f4978.png)</MultiModal>

在形成的所有配合物中，只由 $\mathrm{Cu}^{2+}$ 与来自 **HTPPNOL** 的原子构成的环均为五元环。

在乙酸钠溶液中 **HTPPNOL** 与 $\mathrm{Cu}^{2+}$ 反应生成双核配合物 $\mathrm{[Cu_2(TPPNOL)(CH_3COO)]}^{2+}$ 。这个配离子中，$\mathrm{Cu}^{2+}$ 有两种配位数，其中一个为五配位（三角双锥构型），配合物结构中包含四个六元环。

请根据以上信息，选出正确的选项。
Another coordination number of $\mathrm{Cu}^{2+}$ is $\mathrm5$.
The complex has $\mathrm{5}$ five-membered rings.
In this complex, the two $\mathrm{Cu}^{2+}$ ions have different oxygen coordination numbers.
When the salt of this complex ion is dissolved in water, only one binuclear complexes with identical coordination numbers for two $\mathrm{Cu}^{2+}$ exist in the solution. Moreover, these two complexes can interconvert by adjusting the $\mathrm{pH}$ of the solution. In both complexes, the coordination number of $\mathrm{Cu}^{2+}$ is $\mathrm{4}$.
In the alkaline solution of the $\mathrm{Cu}^{2+}$-$\mathrm{HTPPNOL}$ system without other ligands, there exist binuclear complex ions where the coordination number of $\mathrm{Cu}^{2+}$ is $\mathrm{5}$, and the oxygen coordination numbers of the two $\mathrm{Cu}^{2+}$ ions in this complex are identical.
To measure the crystal structure of the complex ion in option E, you can choose a solution with smaller anions to reduce the solubility of the complex ion in the solution and obtain crystals.
None of the above options is correct.
另一个 $\mathrm{Cu}^{2+}$ 的配位数为 $\mathrm5$ 。
该配合物有 $\mathrm{5}$ 个五元环。
该配合物中两个 $\mathrm{Cu}^{2+}$ 的氧配位数不同。
将该配离子的盐溶于水，溶液中仅存在一种两个 $\mathrm{Cu}^{2+}$ 配位数一致的双核配合物。且这两个配合物之间可以通过调控溶液 $\mathrm{pH}$ 相互转化。那么这两个配合物中 $\mathrm{Cu}^{2+}$ 配位数均为 $\mathrm{4}$ 。
在无其他配体的碱性溶液 $\mathrm{Cu}^{2+}$-$\mathrm{HTPPNOL}$ 体系中存在 $\mathrm{Cu}^{2+}$ 配位数均为 $\mathrm{5}$ 的双核配离子，那么这个配合物中两个 $\mathrm{Cu}^{2+}$ 的氧配位数相同。
若要测量选项 E 中配离子的晶体结构，可选择体积较小阴离子的溶液，减小配离子在溶液中的溶解度以获得晶体。
以上选项均不正确
A new six-membered ring exists in the complex. Based on the information that all rings composed only of $\mathrm{Cu}^{2+}$ and atoms from **HTPPNOL** are five-membered rings, this six-membered ring cannot be composed of **HTPPNOL** and $\mathrm{Cu}^{2+}$. Therefore, it must be formed by bridging acetate ions, namely the $\mathrm{Cu-O-C-O-Cu}$ bridge.

<Checkpoint>[The six-membered ring is formed by bridging acetate ions, namely the $\mathrm{Cu-O-C-O-Cu}$ bridge.](1)</Checkpoint>

According to the structure of **HTPPNOL** and the clue that all rings formed by $\mathrm{Cu}^{2+}$ and atoms from **HTPPNOL** are five-membered rings, it can be known that $\mathrm{N/N}$ or $\mathrm{N/O}$ every two carbons in **HTPPNOL** can form a five-membered ring with $\mathrm{Cu}^{2+}$. A total of $\mathrm{5}$ five-membered rings can be generated.

<Checkpoint>[$\mathrm{N/N}$ or $\mathrm{N/O}$ every two carbons in **HTPPNOL** can form a five-membered ring with $\mathrm{Cu}^{2+}$, and a total of $\mathrm{5}$ five-membered rings can be generated.](1)</Checkpoint>

Based on the distance between $\mathrm{Cu}^{2+}$ and the coordinating atoms on **HTPPNOL**, the approximate structure of the dinuclear complex formed by **HTPPNOL** and $\mathrm{Cu}^{2+}$ can be inferred. For five-coordinated $\mathrm{Cu}^{2+}$, it is coordinated with two pyridine nitrogens , one amine nitrogen connected to the two methylpyridines, one hydroxyl oxygen in **HTPPNOL**, and one carboxyl oxygen in $\mathrm{CH_3COO^-}$.

<Checkpoint>[The five-coordinated $\mathrm{Cu}^{2+}$ is coordinated with two pyridine nitrogens, one amine nitrogen connected to the two methylpyridines, one hydroxyl oxygen in **HTPPNOL**, and one carboxyl oxygen in $\mathrm{CH_3COO^-}$.](2)</Checkpoint>

The other $\mathrm{Cu}^{2+}$ is coordinated with the remaining one pyridine nitrogen, one amine nitrogen connected to the methylpyridine, one hydroxyl oxygen in **HTPPNOL**, and one carboxyl oxygen in $\mathrm{CH_3COO^-}$. Its coordination number is $\mathrm{4}$.

<Checkpoint>[The other $\mathrm{Cu}^{2+}$ is coordinated with one pyridine nitrogen, one amine nitrogen connected to the methylpyridine, one hydroxyl oxygen in **HTPPNOL**, and one carboxyl oxygen in $\mathrm{CH_3COO^-}$, and the coordination number is $\mathrm{4}$.](2)</Checkpoint>

Option A: The coordination number of the other $\mathrm{Cu}^{2+}$ is $\mathrm{4}$, A is incorrect.

Option B: The complex has $\mathrm{5}$ five-membered rings, B is correct.

Option C: The $\mathrm{O}$ coordination numbers of $\mathrm{Cu}^{2+}$ in the complex are the same, both are $\mathrm{2}$, C is incorrect.

Option D: Dissolving the salt of the complex ion in water yields two dinuclear complexes with the same $\mathrm{Cu}^{2+}$ coordination number. It is possible that $\mathrm{H_2O}$ coordinates with $\mathrm{Cu}^{2+}$ with a coordination number of $\mathrm{4}$, resulting in complexes with $\mathrm{Cu}^{2+}$ coordination numbers of $\mathrm{5}$.

<Checkpoint>[$\mathrm{H_2O}$ coordinates with $\mathrm{Cu}^{2+}$ with a coordination number of $\mathrm{4}$, and the $\mathrm{Cu}^{2+}$ coordination numbers of the complex are all $\mathrm{5}$.](1)</Checkpoint>

Among them, the protonation/deprotonation of $\mathrm{H_2O}$/$\mathrm{OH^-}$ can be achieved by adjusting the $\mathrm{pH}$, forming these two dinuclear complexes.

<Checkpoint>[The protonation/deprotonation of $\mathrm{H_2O}$/$\mathrm{OH^-}$ can be achieved by adjusting the $\mathrm{pH}$ to complete the conversion of the two complexes](1)</Checkpoint>

D is incorrect.

Option E:
In an alkaline solution without other ligands, only $\mathrm{OH}^-$ coordinates with $\mathrm{Cu}^{2+}$ besides **HTPPNOL**.

<Checkpoint>[$\mathrm{OH}^-$ coordinates with $\mathrm{Cu}^{2+}$.](1)</Checkpoint>

It is known that a dinuclear complex with $\mathrm{Cu}^{2+}$ coordination numbers of $\mathrm{5}$ is formed in the system, indicating that one $\mathrm{OH}^-$ replaces the position of $\mathrm{CH_3COO^-}$ in the original complex and coordinates with two $\mathrm{Cu}^{2+}$ respectively.

<Checkpoint>[One $\mathrm{OH}^-$ coordinates with two $\mathrm{Cu}^{2+}$ respectively.](1)</Checkpoint>

Another $\mathrm{OH}^-$ coordinates with the original $\mathrm{Cu}^{2+}$ with a coordination number of $\mathrm{4}$ to form the complex.

<Checkpoint>[Another $\mathrm{OH}^-$ coordinates with the original $\mathrm{Cu}^{2+}$ with a coordination number of $\mathrm{4}$](1)</Checkpoint>

Adding the hydroxyl oxygen in **HTPPNOL** that coordinates with the two $\mathrm{Cu}^{2+}$, the oxygen coordination numbers of $\mathrm{Cu}^{2+}$ in the complex are $\mathrm{2}$ and $\mathrm{3}$ respectively.

<Checkpoint>[The oxygen coordination numbers of $\mathrm{Cu}^{2+}$ in the complex are $\mathrm{2}$ and $\mathrm{3}$ respectively.](1)</Checkpoint>

E is incorrect.

Option F: According to the Bartolho rule, crystals composed of large anions and large cations have lower solubility. A solution with larger volume anions should be selected to reduce the solubility of the complex ion in the solution to obtain crystals.

<Checkpoint>[According to the Bartolho rule, a solution with larger volume anions should be selected](1)</Checkpoint>

F is incorrect.

Option G: G is incorrect.
配合物中存在四个六元环，除了三个吡啶环外。根据只由 $\mathrm{Cu}^{2+}$ 与来自 **HTPPNOL** 的原子构成的环均为五元环这个信息，这个六元环不可能是由 **HTPPNOL** 和 $\mathrm{Cu}^{2+}$ 构成的。因此，它必定是由桥联的醋酸根离子形成的，即 $\mathrm{Cu-O-C-O-Cu}$ 桥。

<Checkpoint>[六元环由桥联的醋酸根离子形成的，即 $\mathrm{Cu-O-C-O-Cu}$ 桥。](1)</Checkpoint>

根据 **HTPPNOL** 的结构以及 $\mathrm{Cu}^{2+}$ 与来自 **HTPPNOL** 的原子构成的环均为五元环这两个线索，可知 **HTPPNOL** 中每隔两个碳的 $\mathrm{N/N}$ 或 $\mathrm{N/O}$ 可与 $\mathrm{Cu}^{2+}$ 形成五元环。一共可能生成 $\mathrm{5}$ 个五元环。

<Checkpoint>[HTPPNOL中每隔两个碳的 $\mathrm{N/N}$ 或 $\mathrm{N/O}$ 可与 $\mathrm{Cu}^{2+}$ 形成五元环，共可生成 $\mathrm{5}$ 个五元环。](1)</Checkpoint>

根据 $\mathrm{Cu}^{2+}$ 与 **HTPPNOL** 上配位原子的距离，可以推测 **HTPPNOL** 与 $\mathrm{Cu}^{2+}$ 形成的双核配合物的大致结构。对于五配位的 $\mathrm{Cu}^{2+}$而言，它分别与 **HTPPNOL** 中的两个吡啶氮、一个与该两个甲基吡啶相连的胺基氮、一个羟基氧和 $\mathrm{CH_3COO^-}$ 中一个羧基氧配位。

<Checkpoint>[五配位的 $\mathrm{Cu}^{2+}$ 分别与 **HTPPNOL** 中的两个吡啶氮、一个与该两个甲基吡啶相连的胺基氮、一个羟基氧和 $\mathrm{CH_3COO^-}$ 中一个羧基氧配位。](2)</Checkpoint>

而另一个 $\mathrm{Cu}^{2+}$ 则与 **HTPPNOL** 中剩下的一个吡啶氮、一个与该甲基吡啶相连的胺基氮、一个羟基氧和 $\mathrm{CH_3COO^-}$ 中一个羧基氧配位。它的配位数为 $\mathrm{4}$ 。

<Checkpoint>[另一个 $\mathrm{Cu}^{2+}$ 与 **HTPPNOL** 中的一个吡啶氮、一个与该甲基吡啶相连的胺基氮、一个羟基氧和 $\mathrm{CH_3COO^-}$ 中一个羧基氧配位，配位数为 $\mathrm{4}$ 。](2)</Checkpoint>

选项 A ：另一个 $\mathrm{Cu}^{2+}$ 的配位数为 $\mathrm{4}$，A 错误。

选项 B ：该配合物有 $\mathrm{5}$ 个五元环，B 正确。

选项 C ：该配合物中 $\mathrm{Cu}^{2+}$ 的 $\mathrm{O}$ 配位数相同，均为 $\mathrm{2}$ ，C 错误。

选项 D ：将该配离子的盐溶于水，得到了 $\mathrm{Cu}^{2+}$ 配位数相同的两个双核配合物。可能是 $\mathrm{H_2O}$ 与配位数为 $\mathrm{4}$ 的 $\mathrm{Cu}^{2+}$ 配位，得到了$\mathrm{Cu}^{2+}$ 配位数均为 $\mathrm{5}$ 的配合物。

<Checkpoint>[$\mathrm{H_2O}$ 与配位数为 $\mathrm{4}$ 的 $\mathrm{Cu}^{2+}$ 配位，配合物的 $\mathrm{Cu}^{2+}$ 配位数均为 $\mathrm{5}$ 。](1)</Checkpoint>

其中  可以通过调节 $\mathrm{pH}$ 实现 $\mathrm{H_2O}$/$\mathrm{OH^-}$ 质子/去质子化，形成这两种双核配合物。

<Checkpoint>[可以通过调节 $\mathrm{pH}$ 实现 $\mathrm{H_2O}$/$\mathrm{OH^-}$ 质子/去质子化,完成两种配合物的转化](1)</Checkpoint>

D 错误。

选项 E ：
在无其他配体的碱性溶液中，除了 **HTPPNOL** 只有 $\mathrm{OH}^-$ 与 $\mathrm{Cu}^{2+}$ 配位。

<Checkpoint>[$\mathrm{OH}^-$ 与 $\mathrm{Cu}^{2+}$ 配位。](1)</Checkpoint>

已知体系中形成 $\mathrm{Cu}^{2+}$ 配位数均为 $\mathrm{5}$ 的双核配离子，可知一个 $\mathrm{OH}^-$ 取代了原配合物中 $\mathrm{CH_3COO^-}$ 的位置，分别与两个 $\mathrm{Cu}^{2+}$ 配位。

<Checkpoint>[一个 $\mathrm{OH}^-$ 分别与两个 $\mathrm{Cu}^{2+}$ 配位。](1)</Checkpoint>

另一个 $\mathrm{OH}^-$ 与原配位数为 $\mathrm{4}$ 的 $\mathrm{Cu}^{2+}$ 配位，形成该配合物。

<Checkpoint>[另一个 $\mathrm{OH}^-$ 与原配位数为 $\mathrm{4}$ 的 $\mathrm{Cu}^{2+}$ 配位](1)</Checkpoint>

加上 **HTPPNOL** 中与两个 $\mathrm{Cu}^{2+}$ 配位的羟基氧，配合物中 $\mathrm{Cu}^{2+}$ 的氧配位数分别 $\mathrm{2}$ 和 $\mathrm{3}$ 。

<Checkpoint>[配合物中 $\mathrm{Cu}^{2+}$ 的氧配位数分别 $\mathrm{2}$ 和 $\mathrm{3}$ 。](1)</Checkpoint>

E 错误。

选项 F ：根据巴索洛规则，大阴离子与大阳离子组成的晶体溶解度更小。应选择体积较大阴离子的溶液，以达到减小配离子在溶液中的溶解度来获得晶体的效果。

<Checkpoint>[根据巴索洛规则，应选择体积较大阴离子的溶液](1)</Checkpoint>

F 错误。

选项 G ：G 错误。