Extracting metal elements from lithium battery cathode materials is one of the important ways to obtain elemental manganese. A more traditional method for obtaining elemental manganese is to dissolve $\rm LiMn_2O_4$ solid with a sulfuric acid solution of hydrogen peroxide, where manganese is reduced to $\rm Mn(II)$ in the process (Reaction 1), and then elemental manganese can be obtained by electrochemical means. This traditional method requires the use of high concentrations of inorganic acids, and the conditions are relatively harsh. Therefore, another method for extracting manganese is to treat $\rm LiMn_2O_4$ solid with an aqueous solution of citric acid (3-carboxy-3-hydroxypentanedioic acid) at $120~^\degree\rm C$. As time changes, the reaction system changes from an initial black turbid liquid to a colorless transparent solution (Reaction 2), and then to a white turbid liquid after a period of time (Reaction 3), and a white solid $\bf X$ is obtained after filtration.

XRD shows that $\bf X$ is a mononuclear complex of $\rm Mn$, and the mass fraction of the $\rm Mn$ element in $\bf X$ is 20.89%. It is known that $\rm Mn$ in $\bf X$ is tetracoordinate and contains a five-membered ring.

$\rm LiMn_2O_4$ can also be used as a raw material to prepare manganese dioxide with specific crystal forms. Mixing $\rm LiMn_2O_4$ solid with $0.5\rm~mol/L$ $\rm H_2SO_4$ and magnetically stirring at room temperature for $3\rm~h$ can obtain $\lambda$ crystal form $\rm MnO_2$ (Reaction 4).

The following statements are made:

1. In the ionic equation of Reaction 1, the sum of the coefficients on the reactant side and the product side are both odd numbers.

2. The coefficient of $\rm CO_2$ in the ionic equation of Reaction 2 is an even number. (If there is no $\rm CO_2$, this statement is considered correct)

3. The coefficient of $\rm H_2O$ in the ionic equation of Reaction 4 is an odd number. (If there is no $\rm H_2O$, this statement is considered correct)

4. There is a seven-membered ring in the structure of $\bf X$.

Select the sum of the numbers of the correct statements.
从锂电池阴极材料中提取金属元素是获取锰单质的重要途径之一。比较传统的获得锰单质的方法是将 $\rm LiMn_2O_4$ 固体用过氧化氢的硫酸溶液溶解，锰在该过程中被还原为 $\rm Mn(II)$（反应1），之后通过电化学的手段即可获得锰单质。这种传统的方法需要使用高浓度的无机酸，条件相对来说比较剧烈。因此，另一种提取锰元素的方法是在 $120~^\degree\rm C$ 下用柠檬酸(3-羧基-3-羟基戊二酸)水溶液处理 $\rm LiMn_2O_4$ 固体，随着时间的变化，反应体系从一开始的黑色浑浊液转化为无色透明溶液（反应2），一段时间后又变为白色浑浊液（反应3），过滤后得到白色固体 $\bf X$。

XRD 表明 $\bf X$ 为 $\rm Mn$ 的单核配合物，且 $\bf X$ 中 $\rm Mn$ 元素的质量分数为 20.89%。已知 $\bf X$ 中 $\rm Mn$ 为四配位，含有一个五元环。

以 $\rm LiMn_2O_4$ 为原料还可以制备特定晶型的二氧化锰。将 $\rm LiMn_2O_4$ 固体与 $0.5\rm~mol/L$ $\rm H_2SO_4$ 混合，室温下磁搅拌 $3\rm~h$ 即可得到 $\lambda$ 晶型的 $\rm MnO_2$（反应4）。

有如下说法：

1.反应1离子方程式中反应物侧、产物侧分别的系数和均为奇数。

2.反应2离子方程式中 $\rm CO_2$ 的系数为偶数。（若无 $\rm CO_2$，该说法视为正确）

3.反应4离子方程式中 $\rm H_2O$ 的系数为奇数。（若无 $\rm H_2O$，该说法视为正确）

4.$\bf X$的结构中存在七元环。

选出正确说法的编号之和。
All other options are incorrect
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其他选项均不正确
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The molecular weight of $\bf X$ is $54.94\div 20.89\%=263.0$, and after removing manganese, $208.1$ remains. Assuming it contains 1 molecule of citric acid, and citric acid has reducing properties, it can be reasonably inferred that Mn is divalent. Therefore, we first assume that citric acid is a hydrogen citrate ion. After subtracting the hydrogen citrate ion, the remaining molecular weight is $208.1-190.1=18.0$, which is likely 1 molecule of water. Therefore, it can be known that $\bf X$ is $\rm Mn(C_6H_6O_7)H_2O$.

<Checkpoint>[$\bf X$ is $\rm Mn(C_6H_6O_7)H_2O$](1)</Checkpoint>

In $\bf X$, two carboxyl groups of citric acid are deprotonated. Since there is a five-membered ring in the structure, it can be seen that the hydroxyl group and the carboxylate radical at the same position of the hydroxyl group are coordinated. Since it is tetracoordinate, the carboxylate radical at the end of the chain is also coordinated, forming the following structure. It can be seen that there are five-, six-, and seven-membered rings in the structure, so statement 4 is correct.
<MultiModal>![O=C(CC1(CC(O)=O)[OH]2)O[Mn]2([OH2])OC1=O](/media/uploads/8cf7e9de-d67e-4129-9ba3-5dd32cffcba3.png)</MultiModal>

<Checkpoint>[There is a seven-membered ring in the structure of $\bf X$](1)</Checkpoint>

Reaction 1:
$\rm 2LiMn_2O_4 + 3H_2O_2 + 10H^+ = 4Mn^{2+} + 2Li^+ + 8H_2O + 3O_2$

Note that the ratio of the coefficients of $\rm LiMn_2O_4$ and $\rm H_2O_2$ must be $2:3$. Other balanced equations are the coupling of this reaction and the decomposition of hydrogen peroxide. It is easy to know that statement 1 is correct.

<Checkpoint>[$\rm 2LiMn_2O_4 + 3H_2O_2 + 10H^+ = 4Mn^{2+} + 2Li^+ + 8H_2O + 3O_2$](1)</Checkpoint>

Reaction 2:
$\rm 6LiMn_2O_4 + 30H^+ + C_6H_8O_7 = 6Li^+ + 12Mn^{2+} + 19H_2O + 6CO_2$

It is easy to know that statement 2 is correct.

<Checkpoint>[$\rm 6LiMn_2O_4 + 30H^+ + C_6H_8O_7 = 6Li^+ + 12Mn^{2+} + 19H_2O + 6CO_2$](1)</Checkpoint>

Reaction 3:
$\rm Mn^{2+} + H_2O + C_6H_6O_7^{2-} = Mn(C_6H_6O_7)H_2O$

Reaction 4:
$\rm 2LiMn_2O_4 + 4H^+ = Mn^{2+} + 3MnO_2 + 2Li^+ + 2H_2O$

It is easy to know that statement 3 is wrong.

<Checkpoint>[$\rm 2LiMn_2O_4 + 4H^+ = Mn^{2+} + 3MnO_2 + 2Li^+ + 2H_2O$](1)</Checkpoint>

In summary, the correct statements are 1, 2, and 4, and the sum is 7.
$\bf X$的分子量为 $54.94\div 20.89\%=263.0$，除去锰还剩 $208.1$。假设含有1分子柠檬酸，柠檬酸有还原性，可以合理推测Mn为二价，故首先假设柠檬酸为柠檬酸氢根，扣除柠檬酸氢根后分子量剩余 $208.1-190.1=18.0$，很可能是1分子水。故可知$\bf X$为 $\rm Mn(C_6H_6O_7)H_2O$。

<Checkpoint>[$\bf X$为 $\rm Mn(C_6H_6O_7)H_2O$](1)</Checkpoint>

$\bf X$中柠檬酸的两个羧基被去质子，由于结构存在五元环，可知羟基和羟基同位的羧酸根进行配位。由于为四配位，链末端的羧酸根也进行配位，形成如下结构。可以看出结构中有五、六、七元环，说法4正确。
<MultiModal>![O=C(CC1(CC(O)=O)[OH]2)O[Mn]2([OH2])OC1=O](/media/uploads/8cf7e9de-d67e-4129-9ba3-5dd32cffcba3.png)</MultiModal>

<Checkpoint>[$\bf X$的结构中存在七元环](1)</Checkpoint>

反应1：
$\rm 2LiMn_2O_4 + 3H_2O_2 + 10H^+ = 4Mn^{2+} + 2Li^+ + 8H_2O + 3O_2$

注意 $\rm LiMn_2O_4$ 和 $\rm H_2O_2$的系数之比必为 $2:3$，其他能配平的方程式都是此反应和过氧化氢分解的耦合。易知说法1正确。

<Checkpoint>[$\rm 2LiMn_2O_4 + 3H_2O_2 + 10H^+ = 4Mn^{2+} + 2Li^+ + 8H_2O + 3O_2$](1)</Checkpoint>

反应2：
$\rm 6LiMn_2O_4 + 30H^+ + C_6H_8O_7 = 6Li^+ + 12Mn^{2+} + 19H_2O + 6CO_2$

易知说法2正确。

<Checkpoint>[$\rm 6LiMn_2O_4 + 30H^+ + C_6H_8O_7 = 6Li^+ + 12Mn^{2+} + 19H_2O + 6CO_2$](1)</Checkpoint>

反应3：
$\rm Mn^{2+} + H_2O + C_6H_6O_7^{2-} = Mn(C_6H_6O_7)H_2O$

反应4：
$\rm 2LiMn_2O_4 + 4H^+ = Mn^{2+} + 3MnO_2 + 2Li^+ + 2H_2O$

易知说法3错误。

<Checkpoint>[$\rm 2LiMn_2O_4 + 4H^+ = Mn^{2+} + 3MnO_2 + 2Li^+ + 2H_2O$](1)</Checkpoint>

综上，正确的说法为1,2,4，和为7。