Ligand $\mathbf{X}$ can form a series of interesting complexes with $\mathrm{Pd(II)}$, and their structures are shown in the figure below:

<MultiModal>![[Se-]C1=CC=NC=C1](/media/uploads/c7ab5177-355a-4233-8a92-473b6cb03185.png)</MultiModal>

The preparation method of this series of complexes is shown below. It is known that $\bf{2}$ is a mononuclear complex, and the molecular weight of $\bf{4}$ is twice that of $\bf{3}$. In $\bf{3}$ and $\bf{4}$, $\mathrm{Pd}$ has only one chemical environment and all $\mathrm{N}$ atoms participate in coordination.

<MultiModal>![The figure shows a multi-step reaction: complex $\bf{1}$ reacts with two equivalents of $\mathrm{Na}\mathbf{X}$ to obtain complex $\bf{2}$; complex $\bf{2}$ then reacts with one equivalent of complex $\bf{1}$ to obtain complex $\bf{3}$ and complex $\bf{4}$. The structure of $\bf{1}$ is: O=S(O[Pd]1(OS(=O)(C(F)(F)F)=O)[P](C2=CC=CC=C2)(C3=CC=CC=C3)CC[P]1(C4=CC=CC=C4)C5=CC=CC=C5)(C(F)(F)F)=O 。](/media/uploads/640dfa3f-00f3-4d27-a32a-5aa3d77e9596.png)</MultiModal>

Which of the following statements are correct:

1.  $\bf{2}$ contains $\mathrm{Pd—N}$ coordination bonds

2.  The cation of $\bf{3}$ has a center of symmetry

3.  The cation of $\bf{3}$ has a mirror plane passing through all $\mathrm{Pd}$ atoms

4.  The ratio of cations to anions in $\bf{4}$ is 6:1

5.  The cation of $\bf{4}$ has the same symmetry as $\mathrm{Re_2Cl_8^{2-}}$

6.  The cation of $\bf{4}$ has the same symmetry as benzene
配体 $\mathbf{X}$ 可以和 $\mathrm{Pd(II)}$ 形成一系列有趣的配合物，其结构如下图所示：

<MultiModal>![[Se-]C1=CC=NC=C1](/media/uploads/c7ab5177-355a-4233-8a92-473b6cb03185.png)</MultiModal>

该系列配合物的制备方法如下所示，已知 $\bf{2}$ 为单核配合物，$\bf{4}$ 的分子量是 $\bf{3}$ 的两倍。$\bf{3}$ 和 $\bf{4}$ 中 $\mathrm{Pd}$ 只具有一种化学环境且所有 $\mathrm{N}$ 原子均参与配位。

<MultiModal>![图中为多步反应：配合物 $\bf{1}$ 与两当量 $\mathrm{Na}\mathbf{X}$ 反应得到配合物 $\bf{2}$；配合物 $\bf{2}$ 再与一当量配合物 $\bf{1}$ 反应可以得到配合物 $\bf{3}$ 和配合物 $\bf{4}$。$\bf{1}$ 的结构为：O=S(O[Pd]1(OS(=O)(C(F)(F)F)=O)[P](C2=CC=CC=C2)(C3=CC=CC=C3)CC[P]1(C4=CC=CC=C4)C5=CC=CC=C5)(C(F)(F)F)=O 。](/media/uploads/640dfa3f-00f3-4d27-a32a-5aa3d77e9596.png)</MultiModal>

下列说法正确的有：

1.$\bf{2}$ 中存在 $\mathrm{Pd—N}$ 配位键

2.$\bf{3}$ 的阳离子具有对称中心

3.$\bf{4}$ 中阴阳离子数量比为6：1

4.$\bf{4}$ 的阳离子对称性与 $\mathrm{Re_2Cl_8^{2-}}$ 相同

5.$\bf{4}$ 的阳离子对称性与苯相同
1,2,3,4
1,2,4
1,2,3
1,2
2,3
2,5
2,3,6
2,3,5
2,4
2
All of the above options are incorrect or the answer is incomplete.
1,2,3,4
1,2,4
1,2,3
1,2
2,3
2,5
2,3,6
2,3,5
2,4
2
以上选项均不正确或答案不完全
Complex $\bf{1}$ undergoes a ligand substitution reaction with two equivalents of $\mathrm{Na}\mathbf{X}$ to generate $\bf{2}$ and two equivalents of $\mathrm{NaOTf}$. According to the hard-soft acid-base theory, $\mathrm{Pd(II)}$ is a cation with a large high-cycle radius, which is relatively soft. At the same time, the negative charge is concentrated on the $\mathrm{Se}$ atom, which is more likely to form a coordination bond with the $\mathrm{Se}$ atom.

<Checkpoint>[Therefore, $Pd$ should bond to $Se$ rather than $N$](1)</Checkpoint>

Therefore, there is no $\mathrm{Pd—N}$ coordination bond in $\bf{2}$. The structure of $\bf{2}$ is shown in the figure below:

<MultiModal>![C1([Se][Pd]2([P](CC[P]2(C3=CC=CC=C3)C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)[Se]C7=CC=NC=C7)=CC=NC=C1](/media/uploads/507c5f4a-4319-4000-945b-d535643d1869.png)</MultiModal>


Since $\mathrm{Pd}$ is in a planar square coordination, the bond angle is 90 degrees, and it can form dimers and tetramers, but it is difficult to form trimers, hexamers, etc. with bond angles of 60 degrees or 120 degrees. $\bf{3}$ is obtained by reacting $\bf{2}$ with one equivalent of complex $\bf{1}$. Since the molecular weight of $\bf{4}$ is twice that of $\bf{3}$, it is speculated that $\bf{3}$ is a dimer and $\bf{4}$ is a tetramer. Since $\mathrm{Pd}$ in $\bf{3}$ and $\bf{4}$ has only one chemical environment and all $\mathrm{N}$ atoms are involved in coordination,

<Checkpoint>[All ligands $\mathbf{X}$ act as bidentate ligands](1)</Checkpoint>

, forming a cyclic structure. Therefore, the structure of $\bf{3}$ is as follows:

<MultiModal>![阳离子带两个正电荷，结构为：C1([P]2(CC[P](C3=CC=CC=C3)([Pd]24[N]5=CC=C([Se][Pd]6([P](C7=CC=CC=C7)(CC[P]6(C8=CC=CC=C8)C9=CC=CC=C9)C%10=CC=CC=C%10)[N]%11=CC=C([Se]4)C=C%11)C=C5)C%12=CC=CC=C%12)C%13=CC=CC=C%13)=CC=CC=C1；阴离子为两个 $\mathrm{TfO^-}$](/media/uploads/b15d9bb2-de9e-45ee-9077-76d14abe18b4.png)</MultiModal>

So the cation of $\bf{3}$ has a center of symmetry.


The structure of tetramer $\bf{4}$ is as follows:

<MultiModal>![阳离子带四个正电荷，结构为：[N]1([Pd]2([Se]3)[P](CC[P](C4=CC=CC=C4)2C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)=CC=C([Se][Pd]8([N](C=C9)=CC=C9[Se][Pd]%10([N]%11=CC=C([Se][Pd]%12([N]%13=CC=C3C=C%13)[P](CC[P](C%14=CC=CC=C%14)%12C%15=CC=CC=C%15)(C%16=CC=CC=C%16)C%17=CC=CC=C%17)C=C%11)[P](CC[P](C%18=CC=CC=C%18)%10C%19=CC=CC=C%19)(C%20=CC=CC=C%20)C%21=CC=CC=C%21)[P](C%22=CC=CC=C%22)(C%23=CC=CC=C%23)CC[P](C%24=CC=CC=C%24)8C%25=CC=CC=C%25)C=C1；阴离子为四个 $\mathrm{TfO^-}$](/media/uploads/2a6efc60-2a4d-4224-a241-19e4fef2129f.png)</MultiModal>


<Checkpoint>[The cation-anion ratio in $\bf{4}$ is 1:4](1)</Checkpoint>

The point group of $\mathrm{Re_2Cl_8^{2-}}$ is $\mathrm{D_{4h}}$, but as shown in the figure, due to the symmetry restriction of ligand $\mathbf{X}$, there is no $\mathrm{C_2}$ axis that bisects the secondary axes perpendicularly to the principal axis.

<Checkpoint>[The symmetry of the cation of $\bf{4}$ is different from that of $\mathrm{Re_2Cl_8^{2-}}$](1)</Checkpoint>

The cation of $\bf{4}$ obviously does not have a six-fold rotation axis.

<Checkpoint>[The symmetry of the cation of $\bf{4}$ is different from that of benzene](1)</Checkpoint>

So the correct answer is J.
配合物 $\bf{1}$ 与两当量 $\mathrm{Na}\mathbf{X}$ 发生配体取代反应，生成 $\bf{2}$ 和两当量 $\mathrm{NaOTf}$。根据软硬酸碱理论，$\mathrm{Pd(II)}$ 为高周期半径大的阳离子，比较软，同时负电荷集中在 $\mathrm{Se}$ 原子上，更易与 $\mathrm{Se}$ 原子形成配位键，

<Checkpoint>[因此$Pd$应与$Se$而非$N$成键](1)</Checkpoint>

因此 $\bf{2}$ 中不存在 $\mathrm{Pd—N}$ 配位键。$\bf{2}$ 的结构如下图所示：

<MultiModal>![C1([Se][Pd]2([P](CC[P]2(C3=CC=CC=C3)C4=CC=CC=C4)(C5=CC=CC=C5)C6=CC=CC=C6)[Se]C7=CC=NC=C7)=CC=NC=C1](/media/uploads/507c5f4a-4319-4000-945b-d535643d1869.png)</MultiModal>


由于 $\mathrm{Pd}$ 为平面四方配位，键角为90度，可以形成二聚物与四聚物，而难以形成键角为60度或120度的三聚物、六聚物等。而 $\bf{3}$ 是由 $\bf{2}$ 再与一当量配合物 $\bf{1}$ 反应，由于 $\bf{4}$ 的分子量是 $\bf{3}$ 的两倍，猜测 $\bf{3}$ 为二聚物，$\bf{4}$ 为四聚物。由于 $\bf{3}$ 和 $\bf{4}$ 中 $\mathrm{Pd}$ 只具有一种化学环境且所有 $\mathrm{N}$ 原子均参与配位，

<Checkpoint>[所有配体 $\mathbf{X}$ 充当双齿配体](1)</Checkpoint>

，形成环状结构。因此 $\bf{3}$ 结构如下：

<MultiModal>![阳离子带两个正电荷，结构为：C1([P]2(CC[P](C3=CC=CC=C3)([Pd]24[N]5=CC=C([Se][Pd]6([P](C7=CC=CC=C7)(CC[P]6(C8=CC=CC=C8)C9=CC=CC=C9)C%10=CC=CC=C%10)[N]%11=CC=C([Se]4)C=C%11)C=C5)C%12=CC=CC=C%12)C%13=CC=CC=C%13)=CC=CC=C1；阴离子为两个 $\mathrm{TfO^-}$](/media/uploads/b15d9bb2-de9e-45ee-9077-76d14abe18b4.png)</MultiModal>

所以 $\bf{3}$ 的阳离子具有对称中心.


四聚体$\bf{4}$ 结构如下：

<MultiModal>![阳离子带四个正电荷，结构为：[N]1([Pd]2([Se]3)[P](CC[P](C4=CC=CC=C4)2C5=CC=CC=C5)(C6=CC=CC=C6)C7=CC=CC=C7)=CC=C([Se][Pd]8([N](C=C9)=CC=C9[Se][Pd]%10([N]%11=CC=C([Se][Pd]%12([N]%13=CC=C3C=C%13)[P](CC[P](C%14=CC=CC=C%14)%12C%15=CC=CC=C%15)(C%16=CC=CC=C%16)C%17=CC=CC=C%17)C=C%11)[P](CC[P](C%18=CC=CC=C%18)%10C%19=CC=CC=C%19)(C%20=CC=CC=C%20)C%21=CC=CC=C%21)[P](C%22=CC=CC=C%22)(C%23=CC=CC=C%23)CC[P](C%24=CC=CC=C%24)8C%25=CC=CC=C%25)C=C1；阴离子为四个 $\mathrm{TfO^-}$](/media/uploads/2a6efc60-2a4d-4224-a241-19e4fef2129f.png)</MultiModal>


<Checkpoint>[$\bf{4}$ 中阴阳离子数量比为4：1](1)</Checkpoint>

$\mathrm{Re_2Cl_8^{2-}}$ 的点群为 $\mathrm{D_{4h}}$，而如图由于配体 $\mathbf{X}$ 的对称性限制，不存在垂直主轴平分副轴的 $\mathrm{C_2}$ 轴，

<Checkpoint>[$\bf{4}$ 的阳离子对称性与 $\mathrm{Re_2Cl_8^{2-}}$ 不同](1)</Checkpoint>

$\bf{4}$ 的阳离子显然没有六次旋转轴

<Checkpoint>[$\bf{4}$ 的阳离子对称性与苯不同](1)</Checkpoint>

所以正确答案为J。