Mixing an aqueous solution of sodium acetate with a methanol solution of cobalt nitrate, adding pyridine, and oxidizing with hydrogen peroxide yields a diamagnetic, electrically neutral molecular compound **A**. The mass fraction of cobalt in **A** is $27.66\%$, where four cobalt atoms and four oxygen atoms form a cubane-like core structure, with all acetate groups acting as bridging ligands, and the overall molecular symmetry is identical to that of allene.

**A** reacts with lithium perchlorate in acetonitrile via anodic oxidation to form the perchlorate salt **B** of $\mathbf{A}^{+}$. **B** decomposes in alkaline solution, releasing oxygen gas, during which $\mathbf{A}^{+}$ is reduced back to **A**.

Regarding the reaction mechanism of the above process, the following information is available:

1. When compound **B**, containing no $\mathrm{^{18}O}$, reacts with water and base containing only $\mathrm{^{18}O}$, all the oxygen gas produced is $\mathrm{^{18}O_2}$, and the reduced **A** contains no $\mathrm{^{18}O}$;
2. None of the original ligands fully dissociate during the reaction; when $\mathrm{OH^{-}}$ is in large excess, the reaction rate is proportional to the concentration of $\mathbf{A}^{+}$; when $\mathbf{A}^{+}$ is in large excess, the reaction rate is also proportional to the concentration of $\mathrm{OH^{-}}$;
3. The decomposition of **B** to release oxygen occurs only in alkaline solution and does not proceed in neutral solution;
4. The cation **X** (shown below) has a structure similar to $\mathbf{A}^{+}$, with the same oxidation state of cobalt, and carries more positive charge than $\mathbf{A}^{+}$, making it a theoretically stronger oxidant. However, when reacting with alkaline aqueous solution, only the oxidation of the 2,2'-bipyridine ligand to $\mathrm{CO_2}$ is observed, with no $\mathrm{O_2}$ generated.

<MultiModal>![This is the structure of **X**, with the SMILES string: Cc1o[Co]234([N]5=CC=CC=C5C6=[N]4C=CC=C6)O7[Co]89(o1)([N]%10=CC=CC=C%10C%11=[N]9C=CC=C%11)O2[Co]%12%13([N]%14=C%15C=CC=C%14)([N]%16=CC=CC=C%16%15)O8[Co]7%17(oc(C)o%13)([N]%18=CC=CC=C%18C%19=[N]%17C=CC=C%19)O3%12. In the diagram, N-N represents 2,2'-bipyridine.](/media/uploads/fa44aa0c-79ad-4552-a12d-103d139a25ef.png)</MultiModal>

The decomposition mechanism of **B** in alkaline solution can be described as follows, where **M**, **N**, and **P** are key intermediates:
$$
\mathbf{A}^{+}\rightarrow\mathbf{M}\rightarrow\mathbf{N}\rightarrow\mathbf{P}\rightarrow\mathbf{A}+\mathrm{O_2}
$$

It is known that **M** and **N** are both electrically neutral, **P** carries two negative charges, **M**→**N** is the rate-determining step, and nucleophilic attack on the substrate occurs during both $\mathbf{A}^{+}$→**M** and **M**→**N**. Additionally, **P** contains a peroxide chain structure.

Among the following statements, the incorrect one is:
将乙酸钠的水溶液和硝酸钴的甲醇溶液混合，加入吡啶，用过氧化氢氧化，可得到抗磁性的电中性分子化合物$\mathbf{A}$。$\mathbf{A}$中钴元素的质量分数为$27.66 \%$，四个钴原子和四个氧原子形成类似立方烷的结构核心，乙酸根全部为桥连配体，且整个分子的对称性与丙二烯相同。

$\mathbf{A}$与高氯酸锂在乙腈中经阳极氧化得到$\mathbf{A}^{+}$的高氯酸盐$\mathbf{B}$。$\mathbf{B}$在碱性溶液中分解放出氧气，其中$\mathbf{A}^{+}$被还原为$\mathbf{A}$。

对于上述过程的反应机理，有如下信息：

1. 不含 $\mathrm{^{18}O}$的化合物 $\mathbf{B}$ 与只含$\mathrm{^{18}O}$的水和碱反应，生成的氧气全部为 $^ { 1 8 } \mathrm { O _ { 2 } }$ ，且还原得到的$\mathbf{A}$不含 $\mathrm{^ { 1 8 } O}$ ；
2. 反应过程中原有的配体均没有完全解离；当$\mathrm{OH^{-}}$远过量时，反应速率与$\mathbf{A}^{+}$的浓度成正比；$\mathbf{A}^{+}$远过量时，反应速率也与$\mathrm{OH^{-}}$的浓度成正比；
3. $\mathbf{B}$分解放出氧气的反应仅在碱性溶液中进行，在中性溶液中不发生；
4. 阳离子$\mathbf{X}$(如下图)与$\mathbf{A}^{+}$结构相似，钴元素的氧化态相同，且$\mathbf{X}$所带正电荷更多，理应是比$\mathbf{A}^{+}$更强的氧化剂，然而与碱性水溶液反应只能检测到配体2-2‘-联吡啶被氧化成$\mathrm{CO_2}$，没有任何 $\mathrm{O_2}$ 生成。 

<MultiModal>![这是$\mathbf{X}$的结构图，smiles为Cc1o[Co]234([N]5=CC=CC=C5C6=[N]4C=CC=C6)O7[Co]89(o1)([N]%10=CC=CC=C%10C%11=[N]9C=CC=C%11)O2[Co]%12%13([N]%14=C%15C=CC=C%14)([N]%16=CC=CC=C%16%15)O8[Co]7%17(oc(C)o%13)([N]%18=CC=CC=C%18C%19=[N]%17C=CC=C%19)O3%12。图中，N-N为2,2'-联吡啶。](/media/uploads/fa44aa0c-79ad-4552-a12d-103d139a25ef.png)</MultiModal> 

$\mathbf{B}$在碱性溶液中分解的机理可以表述如下，其中$\mathbf{M}$，$\mathbf{N}$，$\mathbf{P}$为关键中间体。
$$
\mathbf{A}^{+}\rightarrow\mathbf{M}\rightarrow\mathbf{N}\rightarrow\mathbf{P}\rightarrow\mathbf{A}+\mathrm{O_2}
$$

已知$\mathbf{M}$与$\mathbf{N}$均为电中性，$\mathbf{P}$带两个负电荷， $\mathbf{M}\rightarrow\mathbf{N}$为反应的决速步， $\mathbf{A}^{+}\rightarrow\mathbf{M}$与  $\mathbf{M}\rightarrow\mathbf{N}$过程中发生了亲核试剂对底物的进攻，$\mathbf{P}$中含有过氧链结构。

下列说法中错误的是
All $\mathrm{Co}$ in $\mathbf{A}$ are 6-coordinated.
All $\mathrm{Co}$ in $\mathbf{A}$ are in the $+3$ valence state.
In the ionic equation of $\mathbf{B}\rightarrow\mathbf{A}$, the sum of the coefficients of the reactants after balancing is greater than the sum of the coefficients of the products.
$\mathbf{B}\rightarrow\mathbf{A}$ is a second-order reaction
The $\mathrm{O}$ in the product $\mathrm{O_2}$ all originates from the reactant $\mathrm{OH}^{-}$.
$\mathbf{X}$ cannot generate $\mathrm{O_2}$ due to the excessively strong coordination ability of 2,2'-bipyridine as a chelating ligand.
The average oxidation state of $\mathrm{Co}$ in $\mathbf{M}$ is $+3.5$.
The average oxidation state of $\mathrm{Co}$ in $\mathbf{N}$ is $+3.5$.
The average oxidation state of $\mathrm{Co}$ in $\mathbf{P}$ is $+3$.
All of the above statements are correct
$\mathbf{A}$中$\mathrm{Co}$均为6配位
$\mathbf{A}$中$\mathrm{Co}$均为$+3$价
$\mathbf{B}\rightarrow\mathbf{A}$的离子方程式中，配平后反应物的系数和大于产物的系数和
$\mathbf{B}\rightarrow\mathbf{A}$是一个二级反应
产物$\mathrm{O_2}$中的$\mathrm{O}$均来自于反应物$\mathrm{OH}^{-}$
$\mathbf{X}$无法反应生成 $\mathrm{O_2}$是由于2-2‘-联吡啶作为螯合配体配位能力过强
$\mathbf{M}$中$\mathrm{Co}$的平均氧化态为$+3.5$价
$\mathbf{N}$中$\mathrm{Co}$的平均氧化态为$+3.5$价
$\mathbf{P}$中$\mathrm{Co}$的平均氧化态为$+3$价
以上说法均正确
Based on the prompt in the question stem, "four cobalt atoms and four oxygen atoms form a cubane-like structural core," it can be inferred that **A** contains four cobalt atoms. Therefore, the relative molecular weight of **A** is $\frac{4\times 58.93}{0.2766}=852.21\ \mathrm{g/mol}$.  

According to the preparation method, the ligands in **A** are acetate ions and pyridine (abbreviated as py). Let the number of acetate ions be $x$ and the number of pyridine molecules be $y$. Solving the indeterminate equation $4\times (58.93+16)+59.04x+79.1y=852.21$, where $x$ and $y$ are integers, we find that $x=4$, $y=4$ is a reasonable solution. Thus, the structure of **A** is $\mathrm{Co_4O_4(CH_3COO)_4(py)_4}$.  

<Checkpoint>[The structure of **A** is $\mathrm{Co_4O_4(CH_3COO)_4(py)_4}$](1)</Checkpoint>  

Combining the symmetry information provided in the question stem, the four cobalt atoms and four oxygen atoms form a cubane-like structure, with acetate ions acting as bidentate ligands bridging two $\mathrm{Co}$ atoms, and one $\mathrm{py}$ coordinating with one $\mathrm{Co}$ atom. The $\mathrm{Co}$ atoms are hexacoordinated with a $\mathrm{+3}$ oxidation state, so options A and B are both correct.  

<Checkpoint>[The four cobalt atoms and four oxygen atoms form a cubane-like structure, with acetate ions acting as bidentate ligands bridging two $\mathrm{Co}$ atoms, and one $\mathrm{py}$ coordinating with one $\mathrm{Co}$ atom. Thus, the $\mathrm{Co}$ atoms are hexacoordinated with a $\mathrm{+3}$ oxidation state.](1)</Checkpoint>  

Based on the information in the question stem, the ionic equation for the reaction $\mathbf{B}\rightarrow\mathbf{A}$ is: $\mathrm{4\mathbf{A}^{+}+4OH^{-}=4\mathbf{A}+2H_2O+O_2}$. The sum of the coefficients of the reactants (8) is greater than that of the products (7), so option C is correct.  

<Checkpoint>[$\mathrm{4\mathbf{A}^{+}+4OH^{-}=4\mathbf{A}+2H_2O+O_2}$, where the sum of the reactant coefficients is greater than that of the product coefficients.](1)</Checkpoint>  

Analyzing the reaction mechanism for this process: According to information 1, it can be concluded that all $\mathrm{O}$ atoms in the generated $\mathrm{O_2}$ originate from the reactant $\mathrm{OH}^{-}$.  

<Checkpoint>[All $\mathrm{O}$ atoms in the generated $\mathrm{O_2}$ originate from the reactant $\mathrm{OH}^{-}$](1)</Checkpoint>  

According to information 2, the rate equation for this reaction can be described as $r=k[\mathbf{A}^{+}][\mathrm{OH^{-}}]$, indicating a second-order reaction. Consequently, the rate-determining step must involve the reaction of one $\mathbf{A}^{+}$ molecule with one $\mathrm{OH}^{-}$ molecule.  

<Checkpoint>[The rate-determining step involves the reaction of one $\mathbf{A}^{+}$ molecule with one $\mathrm{OH}^{-}$ molecule](1)</Checkpoint>  

According to information 3, this process must involve a reaction with $\mathrm{OH}^{-}$ and not with $\mathrm{H_2O}$.  

According to information 4, comparing **A** and **X**, it can be observed that although both have hexacoordinated $\mathrm{Co}$ atoms, in **A**, one $\mathrm{Co}$ atom is connected to three $\mathrm{O}$ atoms, two bridging acetate ions, and one pyridine molecule, while in **X**, one $\mathrm{Co}$ atom is connected to three $\mathrm{O}$ atoms, one bridging acetate ion, and one non-bridging bidentate ligand. Therefore, the bridging acetate ions play a critical role in the reaction process. The 2,2'-bipyridine ligand, due to its strong chelating effect, exhibits lower reactivity.  

<Checkpoint>[The bridging acetate ions play a key role in the reaction, while 2,2'-bipyridine, due to its strong chelating effect, hinders the reaction.](1)</Checkpoint>  

According to the prompt in the question stem, "the processes $\mathbf{A}^{+}\rightarrow\mathbf{M}$ and $\mathbf{M}\rightarrow\mathbf{N}$ involve nucleophilic attack on the substrate by the nucleophile," it can be inferred that both processes involve the attack of $\mathrm{OH}^{-}$ on $\mathrm{Co}$, followed by the dissociation of two bridging acetate ions.  

<Checkpoint>[Both $\mathbf{A}^{+}\rightarrow\mathbf{M}$ and $\mathbf{M}\rightarrow\mathbf{N}$ involve nucleophilic attack by $\mathrm{OH^{-}}$ on $\mathrm{Co}$, converting two bidentate acetate ligands into monodentate ligands.](1)</Checkpoint>  

Thus, **M** can be represented as $\mathrm{Co_4O_4(CH_3COO)_4(py)_4(OH)}$, with an average $\mathrm{Co}$ oxidation state of $+3.25$. **N** can be represented as $\mathrm{Co_4O_4(CH_3COO)_4(py)_4(OH)_2}$, with an average $\mathrm{Co}$ oxidation state of $+3.5$. Option G is incorrect, while H is correct.  

<Checkpoint>[The average $\mathrm{Co}$ oxidation state in **M** is $+3.25$, while in **N** it is $+3.5$.](1)</Checkpoint>  

Subsequently, combining the information that "**P** carries two negative charges" and "**P** contains a peroxide chain structure," it can be concluded that the $\mathbf{N}\rightarrow\mathbf{P}$ process involves the removal of two $\mathrm{H}$ atoms, i.e., the formation of a peroxide bond from two $\mathrm{OH}$ groups on one $\mathrm{Co}$ atom. Since **P** carries two negative charges, it can be expressed as $\mathrm{Co_4O_4(CH_3COO)_4(py)_4(O_2)^{2-}}$, with an average $\mathrm{Co}$ oxidation state of $+3$.  

<Checkpoint>[**P** can be expressed as $\mathrm{Co_4O_4(CH_3COO)_4(py)_4(O_2)^{2-}}$, with an average $\mathrm{Co}$ oxidation state of $+3$.](1)</Checkpoint>
根据题干提示"四个钴原子和四个氧原子形成类似立方烷的结构核心"，不妨猜测$\mathbf{A}$中含有4个钴原子。因此，$\mathbf{A}$的相对分子量为$\frac{4\times 58.93}{0.2766}=852.21\ \mathrm{g/mol}$。根据制备方法，$\mathbf{A}$中的配体为乙酸根和吡啶(记作py)。设乙酸根数目为$x$，吡啶数目为$y$，解不定方程$4\times (58.93+16)+59.04x+79.1y=852.21$，其中，$x$和$y$为整数。可以发现，$x=4,y=4$是一组较合理的解。因此$\mathbf{A}$的结构为$\mathrm{Co_4O_4(CH_3COO)_4(py)_4}$。

<Checkpoint>[$\mathbf{A}$的结构为$\mathrm{Co_4O_4(CH_3COO)_4(py)_4}$](1)</Checkpoint>

结合题干中给出的对称性信息，四个钴原子和四个氧原子形成类似立方烷结构，乙酸根作为双齿配体桥连2个$\mathrm{Co}$，1个$\mathrm{py}$与一个$\mathrm{Co}$配合。$\mathrm{Co}$为6配位，$\mathrm{+3}$价，A、B均正确。

<Checkpoint>[四个钴原子和四个氧原子形成类似立方烷结构，乙酸根作为双齿配体桥连2个$\mathrm{Co}$，1个$\mathrm{py}$与一个$\mathrm{Co}$配合。因此$\mathrm{Co}$为6配位，$\mathrm{+3}$价](1)</Checkpoint>

根据题干信息，$\mathbf{B}\rightarrow\mathbf{A}$的离子方程式为：$\mathrm{4\mathbf{A}^{+}+4OH^{-}=4\mathbf{A}+2H_2O+O_2}$反应物系数和8大于产物系数和7，C正确。

<Checkpoint>[$\mathrm{4\mathbf{A}^{+}+4OH^{-}=4\mathbf{A}+2H_2O+O_2}$，反应物系数和大于产物系数和](1)</Checkpoint>

对于该过程反应机理的信息进行分析：根据信息1，可以判断生成$\mathrm{O_2}$中的$\mathrm{O}$均来自于反应物$\mathrm{OH}^{-}$；

<Checkpoint>[生成$\mathrm{O_2}$中的$\mathrm{O}$均来自于反应物$\mathrm{OH}^{-}$](1)</Checkpoint>

根据信息2，该反应的速率方程可以描述为$r=k[\mathbf{A}^{+}][\mathrm{OH^{-}}]$，即二级反应。也因此，反应的决速步只能是一分子$\mathbf{A}^{+}$和一分子$\mathrm{OH}^{-}$发生反应；

<Checkpoint>[反应的决速步只能是一分子$\mathbf{A}^{+}$和一分子$\mathrm{OH}^{-}$发生反应](1)</Checkpoint>

根据信息3，该过程必须和$\mathrm{OH}^{-}$反应而不能和$\mathrm{H_2O}$反应；

根据信息4，比较$\mathbf{A}$和$\mathbf{X}$，可以发现虽然二者中$\mathrm{Co}$均为6配位，$\mathbf{A}$中一个$\mathrm{Co}$和3个$\mathrm{O}$、2个桥接乙酸根和1个吡啶相连，而$\mathbf{X}$中一个$\mathrm{Co}$和3个$\mathrm{O}$、1个桥接乙酸根和1个非桥连双齿配体相连。因此，桥连的乙酸根在反应过程中起到了重要的作用。2,2'-联吡啶由于螯合作用配位过强，反应性较低。

<Checkpoint>[桥连的乙酸根在反应中起到关键作用，2,2'-联吡啶由于螯合作用配位过强从而不利于反应](1)</Checkpoint>

根据题干中的提示，“$\mathbf{A}^{+}\rightarrow\mathbf{M}$与  $\mathbf{M}\rightarrow\mathbf{N}$过程中发生了亲核试剂对底物的进攻”，可以判断出，这两个过程均为$\mathrm{OH}^{-}$对于$\mathrm{Co}$的进攻，然后发生两个桥连的乙酸根的解配位。

<Checkpoint>[$\mathbf{A}^{+}\rightarrow\mathbf{M}$与  $\mathbf{M}\rightarrow\mathbf{N}$均为$\mathrm{OH^{-}}$对$\mathrm{Co}$的亲核进攻，使两个双齿配体乙酸根变为单齿](1)</Checkpoint>

因此，$\mathbf{M}$可以表示为$\mathrm{Co_4O_4(CH_3COO)_4(py)_4(OH)}$，$\mathrm{Co}$平均价态$+3.25$。$\mathbf{N}$可以表示为$\mathrm{Co_4O_4(CH_3COO)_4(py)_4(OH)_2}$，$\mathrm{Co}$平均价态$+3.5$。G错误，H正确。

<Checkpoint>[$\mathbf{M}$中$\mathrm{Co}$平均价态$+3.25$。$\mathbf{N}$中$\mathrm{Co}$平均价态$+3.5$。](1)</Checkpoint>

随后，结合“$\mathbf{P}$带两个负电荷”和“$\mathbf{P}$中含有过氧链结构”，可以判断$\mathbf{N}\rightarrow\mathbf{P}$的过程脱去两个$\mathrm{H}$，即一个$\mathrm{Co}$上的两个$\mathrm{OH}$形成一个过氧键。由于$\mathbf{P}$带两个负电，可以表述为$\mathrm{Co_4O_4(CH_3COO)_4(py)_4(O_2)^{2-}}$，$\mathrm{Co}$平均价态$+3$。

<Checkpoint>[$\mathbf{P}$可以表述为$\mathrm{Co_4O_4(CH_3COO)_4(py)_4(O_2)^{2-}}$，$\mathrm{Co}$平均价态$+3$。](1)</Checkpoint>