Oxalyl chloride reacts with aniline to obtain the ligand $\mathbf{L}$. The ligand $\mathbf{L}$ reacts with $\mathrm{cis-[Ru(bipy)_2Cl_2]}$ under basic conditions to obtain the corresponding dicationic diruthenium complex.

The absolute configuration of chiral complexes is represented by the symbols $\Delta$ and $\Lambda$. When viewed along the threefold axis of the octahedron, with the thumb parallel to the threefold axis of the complex and pointing out of the paper, $\Delta$ is used for those that conform to a right-handed helix (lower coordinating atoms rotating to the upper coordinating atom position), and $\Lambda$ is used for those that conform to a left-handed helix. The resulting complex has corresponding stereoisomers.

Given the following information, please select all the correct statements:
1. After balancing and simplifying the reaction equation for the formation of the complex, the sum of the coefficients on the left side is greater than the sum of the coefficients on the right side.
2. The hybridization of the central $\mathrm{Ru}$ atom in the complex involves $\mathrm{3}$ types of orbitals (orbitals with the same angular momentum quantum number are considered the same type of orbital).
3. According to the classification by the number of atoms in the ring, there is $\mathrm{1}$ type of ring in the diruthenium complex.
4. According to the above configuration rules for chiral complexes, without considering the chirality of $\mathrm{N}$, the chemical environment of the $\mathrm{Ru}$ element is different in the $\Delta\Delta$ isomer.
5. Considering the chirality of $\mathrm{N}$, there exists the most stable isomer in the absolute configuration of $\Lambda\Delta$. The high stability is due to the structure being centrosymmetric.
草酰氯与苯胺反应得到配体 $\mathbf{L}$ ，配体 $\mathbf{L}$ 在碱性条件下与 $\mathrm{cis-[Ru(bipy)_2Cl_2]}$ 反应可以得到对应的带两个电荷双钌配合物阳离子。

手性配合物的绝对构型用符号 $\Delta$ 和 $\Lambda$ 表示，沿八面体的三重轴方向观察，大拇指方向平行于配合物的三重轴且朝向纸面外时，符合右手螺旋（下层配位原子转到上层配位原子位置）的用 $\Delta$ 表示，符合左手螺旋用 $\Lambda$ 表示。反应得到的配合物有对应的立体异构体。

现有以下信息，请选择所有正确的说法：
1. 生成配合物的反应方程式在配平化简后，左边系数和大于右边系数和。
2. 该配合物中心 $\mathrm{Ru}$ 原子的杂化方式涉及 $\mathrm{3}$ 种轨道（以角量子数相同认定为同种轨道）。
3. 根据环中原子个数分类，双钌配合物中环种类为 $\mathrm{1}$ 种。
4. 根据上述手性配合物的构型规则，不考虑 $\mathrm{N}$ 的手性，绝对构型为  $\Delta\Delta$ 的异构体中 $\mathrm{Ru}$ 元素的化学环境不同。
5. 考虑 $\mathrm{N}$ 的手性，绝对构型为 $\Lambda\Delta$ 中存在最稳定的异构体。稳定性高的原因是结构关于中心对称。
1
2
4
1, 2
2, 4
1, 2, 5
3, 4
2, 3, 4 
4 ,5
3, 4, 5
None of the above options are correct.
1
2
4
1, 2
2, 4
1, 2, 5
3, 4
2, 3, 4 
4, 5
3, 4, 5
以上选项都不正确
The reaction equation for generating the complex is:
$\mathrm{2[Ru(bipy)_2Cl_2]+(CONHPh)_2+2OH^-=[Ru_2(bipy)_4(CONHPh)_2]^{2+}+4Cl^-+H_2O}$

<Checkpoint>[Reaction equation: $\mathrm{2[Ru(bipy)_2Cl_2]+(CONHPh)_2+2OH^-=[Ru_2(bipy)_4(CONHPh)_2]^{2+}+4Cl^-+H_2O}$](2)</Checkpoint>

Therefore, after balancing and simplifying the reaction equation for the formation of the complex, the sum of the coefficients on the left side is less than the sum of the coefficients on the right side, so statement 1 is incorrect.

The structure of the complex is:

<MultiModal>![The figure shows a diruthenium complex cation with two charges: C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16. Each $\mathrm{Ru}$ is hexacoordinated, coordinated with $\mathrm{4}$ $\mathrm{N}$ from two $\mathrm{bipy}$, and simultaneously coordinated with $\mathrm{1}$ $\mathrm{N}$ and $\mathrm{1}$ $\mathrm{O}$ from $\mathrm{(CONHPh)_2}$. $\mathrm{R = Ph}$ ](/media/uploads/f37cf341-9012-431f-aa4d-90f2e1c4813b.png)</MultiModal>

<Checkpoint>[The structure of the complex is: C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16](2)</Checkpoint>

Therefore, the hybridization mode of the central $\mathrm{Ru}$ atom is $\mathrm{d^2sp^3}$, involving $\mathrm{3}$ kinds of orbitals, so statement 2 is correct.

There are five-membered and six-membered rings in the diruthenium complex, and the ring types are $\mathrm{2}$ kinds. Statement 3 is incorrect.

4. According to the configuration rules of the above chiral complex, without considering the chirality of $\mathrm{N}$, the stereoisomeric structures of the complex are:

<MultiModal>![In the $\Delta\Delta$ isomer of absolute configuration, in the left half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 3}$ coordinated with $\mathrm{Ru}$, located on the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{2, 4}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated below the plane with the left $\mathrm{Ru}$; $\mathrm{O}$ is coordinated above the plane with the left $\mathrm{Ru}$. In the right half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated above the plane with the right $\mathrm{Ru}$; $\mathrm{O}$ is coordinated below the plane with the right $\mathrm{Ru}$.

In the $\Lambda\Lambda$ isomer of absolute configuration, in the left half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated below the plane with the left $\mathrm{Ru}$; $\mathrm{O}$ is coordinated above the plane with the left $\mathrm{Ru}$. In the right half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 3}$ coordinated with $\mathrm{Ru}$, located on the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{2, 4}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated above the plane with the right $\mathrm{Ru}$; $\mathrm{O}$ is coordinated below the plane with the right $\mathrm{Ru}$.

In the $\Lambda\Delta$ isomer of absolute configuration, in the left half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated below the plane with the left $\mathrm{Ru}$; $\mathrm{O}$ is coordinated above the plane with the left $\mathrm{Ru}$. In the right half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated above the plane with the right $\mathrm{Ru}$; $\mathrm{O}$ is coordinated below the plane with the right $\mathrm{Ru}$.
In the figure, the arc indicates the bipyridine ligand; $\mathrm{R = Ph}$ 。](/media/uploads/122acd48-88e0-4fc6-9cf1-c408138627e4.png)</MultiModal>

<Checkpoint>[In the $\Delta\Delta$ isomer of absolute configuration, in the left half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 3}$ coordinated with $\mathrm{Ru}$, located on the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{2, 4}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated below the plane with the left $\mathrm{Ru}$; $\mathrm{O}$ is coordinated above the plane with the left $\mathrm{Ru}$. In the right half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated above the plane with the right $\mathrm{Ru}$; $\mathrm{O}$ is coordinated below the plane with the right $\mathrm{Ru}$.](2)</Checkpoint>

<Checkpoint>[In the $\Lambda\Lambda$ isomer of absolute configuration, in the left half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated below the plane with the left $\mathrm{Ru}$; $\mathrm{O}$ is coordinated above the plane with the left $\mathrm{Ru}$. In the right half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 3}$ coordinated with $\mathrm{Ru}$, located on the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{2, 4}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated above the plane with the right $\mathrm{Ru}$; $\mathrm{O}$ is coordinated below the plane with the right $\mathrm{Ru}$.](2)</Checkpoint>


<Checkpoint>[In the $\Lambda\Delta$ isomer of absolute configuration, in the left half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated below the plane with the left $\mathrm{Ru}$; $\mathrm{O}$ is coordinated above the plane with the left $\mathrm{Ru}$. In the right half, the $\mathrm{N-N}$ in one $\mathrm{bipy}$ is $\mathrm{1, 2}$ coordinated with $\mathrm{Ru}$, located below the plane; the $\mathrm{N-N}$ in the other $\mathrm{bipy}$ is $\mathrm{3, 4}$ coordinated with $\mathrm{Ru}$, located above the plane; the $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ is coordinated above the plane with the right $\mathrm{Ru}$; $\mathrm{O}$ is coordinated below the plane with the right $\mathrm{Ru}$.](2)</Checkpoint>

Therefore, the chemical environment of the $\mathrm{Ru}$ element is different in the $\Delta\Delta$ isomer, so statement 4 is correct.

The most stable isomeric structure exists in $\Lambda\Delta$:

<MultiModal>![The complex structure is: C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16. Each $\mathrm{Ru}$ is hexacoordinated, coordinated with $\mathrm{4}$ $\mathrm{N}$ in two $\mathrm{bipy}$ at $\mathrm{1, 2}$ below the plane and $\mathrm{3, 4}$ above the plane, respectively. The left $\mathrm{Ru}$ is coordinated with $\mathrm{1}$ $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ below the plane, and with $\mathrm{1}$ $\mathrm{O}$ above the plane, where the benzene ring connected to $\mathrm{N}$ is located above the plane; the right $\mathrm{Ru}$ is coordinated with $\mathrm{1}$ $\mathrm{N}$ in $\mathrm{(CONHPh)_2}$ above the plane, and with $\mathrm{1}$ $\mathrm{O}$ below the plane, where the benzene ring connected to $\mathrm{N}$ is located below the plane.](/media/uploads/4d99c3d4-cc65-46b6-9257-9dbbb0002027.png)</MultiModal>

<Checkpoint>[The most stable isomeric structure exists in $\Lambda\Delta$: C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16。](2)</Checkpoint>

The reason for the high stability is that the structure has $\mathrm{π-π}$ stacking, so statement 5 is incorrect.

<Checkpoint>[The reason for the high stability is that the structure has $\mathrm{π-π}$ stacking](2)</Checkpoint>

Therefore, statements 2 and 4 are correct, so choose option E.
生成配合物的反应方程式为：
$\mathrm{2[Ru(bipy)_2Cl_2]+(CONHPh)_2+2OH^-=[Ru_2(bipy)_4(CONPh)_2]^{2+}+4Cl^-+2H_2O}$

<Checkpoint>[反应方程式为： $\mathrm{2[Ru(bipy)_2Cl_2]+(CONHPh)_2+2OH^-=[Ru_2(bipy)_4(CONPh)_2]^{2+}+4Cl^-+2H_2O}$](2)</Checkpoint>

因此，生成配合物的反应方程式在配平化简后，左边系数和小于右边系数和，说法 1 错误。

配合物的结构为：

<MultiModal>![图中为带两个电荷双钌配合物阳离子：C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16。每个 $\mathrm{Ru}$ 为六配位，与两个 $\mathrm{bipy}$ 中的 $\mathrm{4}$ 个 $\mathrm{N}$ 配位，同时与 $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{1}$ 个 $\mathrm{N}$ 和 $\mathrm{1}$ 个 $\mathrm{O}$ 配位。 $\mathrm{R = Ph}$ ](/media/uploads/f37cf341-9012-431f-aa4d-90f2e1c4813b.png)</MultiModal>

<Checkpoint>[配合物的结构为：C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16](2)</Checkpoint>

因此，中心 $\mathrm{Ru}$ 原子的杂化方式为 $\mathrm{d^2sp^3}$ ，涉及 $\mathrm{3}$ 种轨道，说法 2 正确。

双钌配合物中有五元环和六元环，环种类为 $\mathrm{2}$ 种。说法 3 错误。

根据上述手性配合物的构型规则，不考虑 $\mathrm{N}$ 的手性，配合物的立体异构体结构有：

<MultiModal>![绝对构型为 $\Delta\Delta$ 的异构体中，左半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 3}$ 配位，位于面上；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{2, 4}$ 配位，位于面下； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与左边 $\mathrm{Ru}$ 面下配位；$\mathrm{O}$  与左边 $\mathrm{Ru}$ 面上配位。右半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与右边 $\mathrm{Ru}$ 面上配位；$\mathrm{O}$  与右边 $\mathrm{Ru}$ 面下配位。

绝对构型为 $\Lambda\Lambda$ 的异构体中，左半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与左边 $\mathrm{Ru}$ 面下配位；$\mathrm{O}$  与左边 $\mathrm{Ru}$ 面上配位。右半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 3}$ 配位，位于面上；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{2, 4}$ 配位，位于面下； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与右边 $\mathrm{Ru}$ 面上配位；$\mathrm{O}$  与右边 $\mathrm{Ru}$ 面下配位。

绝对构型为 $\Lambda\Delta$ 的异构体中，左半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与左边 $\mathrm{Ru}$ 面下配位；$\mathrm{O}$  与左边 $\mathrm{Ru}$ 面上配位。右半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与右边 $\mathrm{Ru}$ 面上配位；$\mathrm{O}$  与右边 $\mathrm{Ru}$ 面下配位。
图中用弧线示意联吡啶配体；$\mathrm{R = Ph}$ 。](/media/uploads/122acd48-88e0-4fc6-9cf1-c408138627e4.png)</MultiModal>

<Checkpoint>[绝对构型为 $\Delta\Delta$ 的异构体中，左半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 3}$ 配位，位于面上；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{2, 4}$ 配位，位于面下； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与左边 $\mathrm{Ru}$ 面下配位；$\mathrm{O}$  与左边 $\mathrm{Ru}$ 面上配位。右半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与右边 $\mathrm{Ru}$ 面上配位；$\mathrm{O}$  与右边 $\mathrm{Ru}$ 面下配位。](2)</Checkpoint>

<Checkpoint>[绝对构型为 $\Lambda\Lambda$ 的异构体中，左半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与左边 $\mathrm{Ru}$ 面下配位；$\mathrm{O}$  与左边 $\mathrm{Ru}$ 面上配位。右半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 3}$ 配位，位于面上；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{2, 4}$ 配位，位于面下； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与右边 $\mathrm{Ru}$ 面上配位；$\mathrm{O}$  与右边 $\mathrm{Ru}$ 面下配位。](2)</Checkpoint>


<Checkpoint>[绝对构型为 $\Lambda\Delta$ 的异构体中，左半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与左边 $\mathrm{Ru}$ 面下配位；$\mathrm{O}$  与左边 $\mathrm{Ru}$ 面上配位。右半边一个 $\mathrm{bipy}$ 中的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{1, 2}$ 配位，位于面下；另一个 $\mathrm{bipy}$ 中 的 $\mathrm{N-N}$ 与 $\mathrm{Ru}$ 为 $\mathrm{3, 4}$ 配位，位于面上； $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{N}$ 与右边 $\mathrm{Ru}$ 面上配位；$\mathrm{O}$  与右边 $\mathrm{Ru}$ 面下配位。](2)</Checkpoint>

因此，绝对构型为 $\Delta\Delta$ 的异构体中 $\mathrm{Ru}$ 元素的化学环境不同，说法 4 正确。

$\Lambda\Delta$ 中存在最稳定的异构体结构为：

<MultiModal>![配合物结构为：C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16。其中每个 $\mathrm{Ru}$ 为六配位，分别与两个 $\mathrm{bipy}$ 中的 $\mathrm{4}$ 个 $\mathrm{N}$ 呈 $\mathrm{1, 2}$ 面下和 $\mathrm{3, 4}$ 面上配位。左边 $\mathrm{Ru}$ 与 $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{1}$ 个 $\mathrm{N}$ 为面下配位，与 $\mathrm{1}$ 个 $\mathrm{O}$ 为面上配位，其中 $\mathrm{N}$ 相连的苯环位于面上；右边 $\mathrm{Ru}$ 与 $\mathrm{(CONHPh)_2}$ 中的 $\mathrm{1}$ 个 $\mathrm{N}$ 为面上配位， $\mathrm{1}$ 个 $\mathrm{O}$ 为面下配位，其中 $\mathrm{N}$ 相连的苯环位于面下。](/media/uploads/4d99c3d4-cc65-46b6-9257-9dbbb0002027.png)</MultiModal>

<Checkpoint>[$\Lambda\Delta$ 中存在最稳定的异构体结构为：C1(C=CC=C2)=[N]2[Ru+]34([N]5=C(C6=CC=CC=[N]63)C=CC=C5)(O=C7C(N4C8=CC=CC=C8)=O[Ru+]9%10([N]%11=CC=CC=C%11C%12=CC=CC=[N]%12%10)(N7C%13=CC=CC=C%13)[N]%14=C(C%15=CC=CC=[N]%159)C=CC=C%14)[N]%16=CC=CC=C1%16。](2)</Checkpoint>

稳定性高的原因是结构具有 $\mathrm{π-π}$ 堆积，因此说法 5 错误。

<Checkpoint>[稳定性高的原因是结构具有 $\mathrm{π-π}$ 堆积](2)</Checkpoint>

因此，说法 2, 4 正确，选择选项 E 。