X-ray diffraction reveals that the anion of the diamagnetic pink potassium salt $\mathrm{K_{2}Mo(SO_{4})_{2}}$ is a binuclear complex ion, in which the sulfate ion acts as a bridging ligand, and the nuclear distance between molybdenum atoms is $211\mathrm{~pm}$. Recrystallization of this double salt yields a potassium salt with the chemical formula $\mathrm{K_{3}Mo_{2}(SO_{4})_{4}\cdot3.5H_{2}O}$, whose anion retains the structure of the $\mathrm{K_{2}Mo(SO_{4})_{2}}$ anion but exhibits a different overall charge. Calculate the product of the $\mathrm{Mo-Mo}$ bond orders for the two anions mentioned above.
X-射线衍射表明，抗磁性粉红色钾盐$\mathrm{K_{2}Mo(SO_{4})_{2}}$的阴离子为双核配离子，其中硫酸根离子为桥连配体，钼原子的核间距为$211\mathrm{~pm}$。若将此复盐进行重结晶，则可制得化学式为$\mathrm{K_{3}Mo_{2}(SO_{4})_{4}\cdot3.5H_{2}O}$的钾盐，该复盐的阴离子仍保持了$\mathrm{K_{2}Mo(SO_{4})_{2}}$阴离子的结构，但整个阴离子的电荷数却有所不同。求上述提到的两种阴离子之间$\mathrm{Mo-Mo}$键级的乘积。
0.5
1.5
3
5
7.5
10.5
14
18
22.5
33
39
12
16
0.5
1.5
3
5
7.5
10.5
14
18
22.5
33
39
12
16
According to the problem statement, the dinuclear anion in $\mathrm{K_{2}Mo(SO_{4})_2}$ is $\mathrm{[Mo_{2}(SO_4)_4]^{4-}}$, which contains 2 $\mathrm{Mo}$ atoms and 4 sulfate groups, with a total charge of -4.

<Checkpoint>[The dinuclear anion is $\mathrm{[Mo_{2}(SO_4)_4]^{4-}}$](1)</Checkpoint>

In this anion, the oxidation state of $\mathrm{Mo}$ is $\frac{1}{2}(-4+2\times4)=+2$, and the number of valence electrons is $6-2=4$.

<Checkpoint>[The oxidation state of $\mathrm{Mo}$ is +2, and the number of valence electrons is 4.](1)</Checkpoint>

Each $\mathrm{Mo}$ atom contributes 4 valence electrons, forming a diamagnetic anion, making it reasonable for a $\mathrm{Mo-Mo}$ quadruple bond to form.

<Checkpoint>[The bond order of $\mathrm{Mo-Mo}$ in the $\mathrm{[Mo_{2}(SO_4)_4]^{4-}}$ anion is 4](1)</Checkpoint>

Based on the above analysis, the structure of $\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$ can be drawn as:

<MultiModal>![[Structure of [Mo_2(SO_4)]^{4-}, containing 2 Mo, 4 S, 16 O, enclosed by square brackets with a 4- superscript at the upper right. A quadruple bond exists between Mo-Mo, with 4 five-membered rings arranged in the Mo-O-S-O-Mo sequence. The Mo-O and S-O bonds in the rings are single bonds. The 8 O atoms bonded to Mo form parallel and orientationally overlapping squares. Each S is connected to two terminal O atoms via S=O double bonds outside the rings.]](/media/uploads/9fe15410-5785-4244-a94e-04a0d3744f9b.png)</MultiModal>

The anion in the crystal $\mathrm{K_{3}Mo_{2}(SO_{4})_{4}\cdot3.5H_{2}O}$ is $\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$, which has one fewer electron compared to $\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$.

<Checkpoint>[The anion in another salt is $\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$](1)</Checkpoint>

Considering the electronic structure of the central metal in $\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$, the 8 coordinating oxygen atoms from the 4 bridging sulfate groups form a square coordination. Each $\mathrm{Mo}$ contributes 4 valence electrons, forming a $\mathrm{Mo-Mo}$ quadruple bond, with the corresponding electronic structure being $\sigma^2\pi^4\delta^2$.

<Checkpoint>[The electronic structure of $\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$ is $\sigma^2\pi^4\delta^2$.](1)</Checkpoint>

After losing one electron, $\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$ is obtained, with the lost electron originating from the highest-energy δ bonding orbital. The electronic structure becomes $\sigma^2\pi^4\delta^1$, and the bond order is 3.5.

<Checkpoint>[The electronic structure of $\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$ is $\sigma^2\pi^4\delta^1$, and the bond order is 3.5.](1)</Checkpoint>

Thus, the product of the $\mathrm{Mo-Mo}$ bond orders for the two anions is $4\times3.5=14$.

<Checkpoint>[The product of the $\mathrm{Mo-Mo}$ bond orders for the two anions is $4\times3.5=14$.](1)</Checkpoint>
由题意，$\mathrm{K_{2}Mo(SO_{4})_2}$中的双核阴离子为$\mathrm{[Mo_{2}(SO_4)_4]^{4-}}$，其中含有2个$\mathrm{Mo}$原子和4个硫酸根，总电荷数为-4。

<Checkpoint>[双核阴离子为$\mathrm{[Mo_{2}(SO_4)_4]^{4-}}$](1)</Checkpoint>

在该阴离子中，$\mathrm{Mo}$的氧化数为$\frac{1}{2}(-4+2\times4)=+2$，价电子数为$6-2=4$。

<Checkpoint>[$\mathrm{Mo}$的氧化数为+2，价电子数为4。](1)</Checkpoint>

每个$\mathrm{Mo}$原子贡献4个价电子，形成抗磁性阴离子，形成$\mathrm{Mo-Mo}$四重键是合理的。

<Checkpoint>[$\mathrm{[Mo_{2}(SO_4)_4]^{4-}}$阴离子中$\mathrm{Mo-Mo}$的键级为4](1)</Checkpoint>


由上述分析，可画出$\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$的结构为：

<MultiModal>![[Mo_2(SO_4)]^{4-}的结构，包含2个Mo，4个S，16个O，整体由中括号包围，中括号右上角标有4-，Mo-Mo之间为四重键，有4个按Mo-O-S-O-Mo顺序构成的五元环，五元环上的Mo-O和S-O均为单键，与Mo相连的8个O构成相互平行且取向重叠的正方形，每个S在环外以S=O双键连接两个端基O。](/media/uploads/9fe15410-5785-4244-a94e-04a0d3744f9b.png)</MultiModal>

$\mathrm{K_{3}Mo_{2}(SO_{4})_{4}\cdot3.5H_{2}O}$晶体的阴离子为$\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$，与$\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$相比，减少了一个电子。

<Checkpoint>[另一种盐的阴离子为$\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$](1)</Checkpoint>

考虑$\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$的中心金属的电子结构，4个桥联硫酸根的8个配位氧以四方形配位，每个$\mathrm{Mo}$各自贡献4个价电子，形成$\mathrm{Mo-Mo}$四重键，相应的电子结构为$\sigma^2\pi^4\delta^2$。

<Checkpoint>[$\mathrm{[Mo_{2}(SO_{4})_{4}]^{4-}}$的电子结构为$\sigma^2\pi^4\delta^2$.](1)</Checkpoint>

失去一个电子后，得到$\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$，失去的电子位于能量最高的δ成键轨道，电子结构变为$\sigma^2\pi^4\delta^1$，键级为3.5。

<Checkpoint>[$\mathrm{[Mo_{2}(SO_{4})_{4}]^{3-}}$的电子结构为$\sigma^2\pi^4\delta^1$，键级为3.5。](1)</Checkpoint>

故两种阴离子的$\mathrm{Mo-Mo}$键级乘积为$4\times3.5=14$。

<Checkpoint>[两种阴离子的$\mathrm{Mo-Mo}$键级乘积为$4\times3.5=14$。](1)</Checkpoint>