At room temperature, excess formaldehyde hydrate was added to a dichloromethane solution of $\rm [(n\text{-}C_4H_9)_4N]_2Mo_2O_7$, and after sufficient reaction (solution 1), ethyl ether was added for crystallization; the crude product was purified by recrystallization to obtain compound $\bf X$. Further studies have shown that $\bf X$ is a salt compound, and its crystal contains only $\rm 1$ kind of cation and $\rm 1$ kind of anion; the formula weight of the anion is $\rm 638.8$, and it has a multinuclear cluster structure; among them, all $\rm Mo$ atoms have the same chemical environment. In $\bf X$, the content (mass fraction) of each element is: $\rm C$, $\rm 43.1\%$；$\rm H$, $\rm 8.2\%$；$\rm N$, $\rm 3.1\%$；$\rm O$, $\rm 17.6\%$.

Which of the following statements are correct:
1. The number of $\rm H$ in the chemical formula of $\bf X$ does not exceed 110
2. Solution 1 is neutral, or weakly acidic/basic
3. In one anion of $\bf X$, there are 2 triply coordinated $\rm O$
4. In molybdates formed by molybdenum, the number of terminal oxygen atoms connected to each molybdenum is usually no more than 2
在室温下，向$\rm [(n\text{-}C_4H_9)_4N]_2Mo_2O_7$的二氯甲烷溶液中加入过量的甲醛水合物，充分反应后(溶液1)，加入乙醚析晶；粗产物经重结晶纯化，可得到化合物$\bf X$。进一步研究表明，$\bf X$是盐类化合物，其晶体中仅含 $\rm 1$ 种阳离子和$\rm 1$种阴离子；阴离子式量为$\rm 638.8$，具有多核簇状结构；其中，所有$\rm Mo$原子的化学环境完全相同。在$\bf X$中，各元素的含量（质量分数）为：$\rm C$，$\rm 43.1\%$；$\rm H$，$\rm 8.2\%$；$\rm N$，$\rm 3.1\%$；$\rm O$，$\rm 17.6\%$。

下列说法正确的有：
1. $\bf X$的化学式中$\rm H$的数目不超过110
2. 溶液1为中性，或者是较弱的酸性/碱性
3. 1个$\bf X$的阴离子中，三配位的$\rm O$有2个
4. 在由钼形成的多酸盐中，每个钼所连接的端基氧的数目通常不多于2
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
$\bf X$ contains only $\rm Mo$ in addition to $\rm C$, $\rm H$, $\rm N$, and $\rm O$;

The content of $\rm Mo$ is: $\rm 1 - 43.1~\% - 8.2~\% - 3.1~\% - 17.6~\% = 28.0~\%$
The ratio of the number of $\rm N$ and $\rm Mo$ atoms in $\bf X$ is: $\rm (3.1~\% / 14.01) : (28.0~\% / 95.95) = 0.758 \approx 3 : 4$

<Checkpoint>[$\rm N:Mo=3:4$ in $\bf X$](0.5)</Checkpoint>

The ratio of the number of $\rm N$ and $\rm O$ atoms in $\bf X$ is: $\rm (3.1~\% / 14.01) : (17.6~\% / 16.00) = 0.201 \approx 1 : 5 = 3 : 15$

<Checkpoint>[$\rm N:O=1:5$ in $\bf X$](0.5)</Checkpoint>

The cation of $\bf X$ should be $\rm [(n\text{-}C_4H_9)_4N]^+$; therefore, both $\rm Mo$ and $\rm O$ atoms are in the anion. Given the anion formula weight is $\rm 638.8$,
there should be $\rm 4$ $\rm Mo$ atoms and $\rm 15$ $\rm O$ atoms in the anion;

The formula weight of the remaining part in the $\bf X$ anion is: $\rm 638.8 - 95.95 \times 4 - 16.00 \times 15 = 15.0$

Corresponding to $\rm 1$ $\rm C$ and $\rm 3$ $\rm H$ atoms;
Therefore, the composition of $\bf X$ is $\rm [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{15}CH_3]$

The carbon in the anion of $\bf X$ comes from formaldehyde hydrate $\rm CH_2(OH)_2$, so the chemical formula of $\bf X$ is: $\rm [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{12}(CH_2O_2)(OH)]$

<Checkpoint>[The chemical formula of $\bf X$ is $\rm [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{12}(CH_2O_2)(OH)]$](2)</Checkpoint>

$\bf X$ contains $9 \times 4 \times 3 + 3 = 111$, statement 1 is incorrect.

The reaction of two molecules of $\rm [(n\text{-}C_4H_9)_4N]_2Mo_2O_7$ with one molecule of $\rm CH_2(OH)_2$ will produce a strong base, statement 2 is incorrect: $\rm 2[(n\text{-}C_4H_9)_4N]_2Mo_2O_7 + CH_2(OH)_2 \rightarrow [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{12}(CH_2O_2)(OH)] + [(n\text{-}C_4H_9)_4N]OH$

<Checkpoint>[The reaction will produce $\rm [(n\text{-}C_4H_9)_4N]OH$](1)</Checkpoint>

The chemical environment of $\rm Mo$ is the same, so $\rm OH^-$ must be in the middle of the $\rm Mo_4$ rectangle. There are 12 $\rm O^{2-}$, first consider 4 bridging and 4 terminal, at this time $\rm Mo$ is already four-coordinated, but in multinuclear cluster structures, it is generally six-coordinated. $\rm CH_2O_2^{2-}$ can be placed on the other side of $\rm OH^-$, and coordinated with 4 $\rm Mo$ at the same time, and the remaining 4 individual $\rm O^{2-}$ can be terminally coordinated with $\rm Mo$. There are only 2 three-coordinated $\rm O$, statement 3 is correct:

<MultiModal>![O=[Mo]1([OH]([Mo]2(O3)(O1)(=O)=O)([Mo]345(=O)=O)[Mo]67(O4)(=O)=O)(O6)(O2CO75)=O，带3个负电荷](/media/uploads/cd940c3a-f602-4d8d-a122-62c5c787060a.png)</MultiModal>

<Checkpoint>[The anion structure of $\bf X$ is O=[Mo]1([OH]([Mo]2(O3)(O1)(=O)=O)([Mo]345(=O)=O)[Mo]67(O4)(=O)=O)(O6)(O2CO75)=O(with 3 negative charges)](2)</Checkpoint>

Terminal oxygen $\rm O^{2-}$ has 2 negative charges and is a good $\sigma$ donor and $\pi$ donor; when there are more terminal $\rm O^{2-}$, the $\rm Mo$ center is electron deficient and tends to form low-coordinated monomeric structures. Statement 4 is correct.

<Checkpoint>[Terminal oxygen $\rm O^{2-}$ has 2 negative charges and is a good $\sigma$ donor and $\pi$ donor; when there are more terminal $\rm O^{2-}$, the $\rm Mo$ center is electron deficient and tends to form low-coordinated monomeric structures](1)</Checkpoint>

Statements 3 and 4 are correct, choose K
$\bf X$ 中除 $\rm C$、$\rm H$、$\rm N$、$\rm O$ 外只有 $\rm Mo$；

$\rm Mo$的含量为：$\rm 1 - 43.1~\% - 8.2~\% - 3.1~\% - 17.6~\% = 28.0~\%$
$\bf X$ 中 $\rm N$ 和 $\rm Mo$ 原子的数目比为：$\rm (3.1~\% / 14.01) : (28.0~\% / 95.95) = 0.758 \approx 3 : 4$

<Checkpoint>[$\bf X$中$\rm N:Mo=3:4$](0.5)</Checkpoint>

$\bf X$中$\rm N$和$\rm O$原子的数目比为：$\rm (3.1~\% / 14.01) : (17.6~\% / 16.00) = 0.201 \approx 1 : 5 = 3 : 15$

<Checkpoint>[$\bf X$中$\rm N:O=1:5$](0.5)</Checkpoint>

$\bf X$ 的阳离子应为$\rm [(n\text{-}C_4H_9)_4N]^+$；所以 $\rm Mo$ 和 $\rm O$ 原子均在阴离子中，结合阴离子式量为$\rm 638.8$，
阴离子中应有$\rm 4$个$\rm Mo$原子和$\rm 15$个$\rm O$原子；

$\bf X$ 阴离子中剩余部分的式量为：$\rm 638.8 - 95.95 \times 4 - 16.00 \times 15 = 15.0$

对应$\rm 1$个$\rm C$和$\rm 3$个$\rm H$原子；
所以，$\bf X$的组成为$\rm [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{15}CH_3]$

$\bf X$ 的阴离子中的碳来源于甲醛水合物$\rm CH_2(OH)_2$，因此 $\bf X$ 的化学式为：$\rm [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{12}(CH_2O_2)(OH)]$

<Checkpoint>[$\bf X$的化学式为$\rm [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{12}(CH_2O_2)(OH)]$](2)</Checkpoint>

$\bf X$中含$9 \times 4 \times 3 + 3 = 111$，说法1错误

两分子的$\rm [(n\text{-}C_4H_9)_4N]_2Mo_2O_7$与一分子的$\rm CH_2(OH)_2$反应，会产生强碱，说法2错误：$\rm 2[(n\text{-}C_4H_9)_4N]_2Mo_2O_7 + CH_2(OH)_2 \rightarrow [(n\text{-}C_4H_9)_4N]_3[Mo_4O_{12}(CH_2O_2)(OH)] + [(n\text{-}C_4H_9)_4N]OH$

<Checkpoint>[反应会产生$\rm [(n\text{-}C_4H_9)_4N]OH$](1)</Checkpoint>

$\rm Mo$的化学环境相同，因此$\rm OH^-$必然处于$\rm Mo_4$矩形正中间。有12个$\rm O^{2-}$，先考虑4个桥联，4个端配，此时$\rm Mo$已经四配位了，但在多核簇状结构里一般要六配位。可以把$\rm CH_2O_2^{2-}$放在$\rm OH^-$的另一侧，同时与4个$\rm Mo$配位，剩下的4个单独的$\rm O^{2-}$再与$\rm Mo$端配即可。三配位的$\rm O$仅有2个，说法3正确：

<MultiModal>![O=[Mo]1([OH]([Mo]2(O3)(O1)(=O)=O)([Mo]345(=O)=O)[Mo]67(O4)(=O)=O)(O6)(O2CO75)=O，带3个负电荷](/media/uploads/cd940c3a-f602-4d8d-a122-62c5c787060a.png)</MultiModal>

<Checkpoint>[$\bf X$的阴离子结构为O=[Mo]1([OH]([Mo]2(O3)(O1)(=O)=O)([Mo]345(=O)=O)[Mo]67(O4)(=O)=O)(O6)(O2CO75)=O（带3个负电荷）](2)</Checkpoint>

端基氧$\rm O^{2-}$带有2个负电荷，是好的$\sigma$给体和$\pi$ 给体；端基$\rm O^{2-}$较多时，$\rm Mo$中心缺电子，而倾向于形成低配位的单体结构。说法4正确

<Checkpoint>[端基氧$\rm O^{2-}$带有2个负电荷，是好的$\sigma$给体和$\pi$给体；端基$\rm O^{2-}$较多时，$\rm Mo$中心缺电子，而倾向于形成低配位的单体结构](1)</Checkpoint>

说法3和4正确，选K