By utilizing cobalt metal ions to capture cyclic ${ \mathrm N _ { 5 } ^- }$ ions in a solution containing ${ \mathrm N _ { 5 } ^- }$ ions, a stable pure nitrogen anion metal salt complex $\mathbf { A }$ was successfully synthesized in air: $\mathrm { [ C o ( N _ { 5 } ) _ { x } ( H _ { 2 } O ) _ { y } ] { \cdot } z H _ { 2 } O }$ (orange). It is known that $\textcircled { 1 }$ the mass fractions are $ \omega _{( \mathrm N )} = 40.82 \%$ and $\omega _{( \mathrm O )}= 37.30 \%$; $\textcircled{2 }$ if the coordinating groups in $\mathbf { A }$ are all regarded as spheres, then there are $z$ $C _ { 2 }$ axes perpendicular to the $C _ { \mathrm { y } }$ axis in the inner sphere of $\mathbf { A }$. Based on the above information, the correct option is:
利用金属钴离子在含${ \mathrm N _ { 5 } ^- }$离子的溶液中捕捉环状的 ${ \mathrm N _ { 5 } ^- }$ 离子，成功制得了空气中稳定的纯氮阴离子金属盐配合物 $\mathbf { A }$ ：$\mathrm { [ C o ( N _ { 5 } ) _ { x } ( H _ { 2 } O ) _ { y } ] { \cdot } z H _ { 2 } O }$ （橙色)。已知$\textcircled { 1 }$质量分数 $ \omega _{( \mathrm N )} = 40.82 \%$ ， $\omega _{( \mathrm O )}= 37.30 \%$ ； $\textcircled{2}$ 若 $\mathbf { A }$ 中配位基团均视为球体，则$\mathbf { A }$的内界存在 $z$ 个垂直于 $C _ { \mathrm { y } }$ 轴的 $C _ { 2 }$ 轴。根据以上信息，下列选项正确的是：
In ${ \mathrm N _ { 5 } ^- }$, there exists a delocalized $\pi$ bond $\Pi_5^5$, which exhibits aromaticity.
$\mathrm{x + y + z} = 8$
The molecular weight of $\mathbf A$ is 307.13
If all the ligand groups in $\mathbf { A }$ are regarded as spheres, then the inner sphere of $\mathbf { A }$ contains an antisymmetric axis $I _ 8$.
If all ligand groups in $\mathbf { A }$ are considered as spheres, then all geometric isomers of the inner sphere of $\mathbf { A }$ are achiral.
If all the ligand groups in $\mathbf { A }$ are regarded as spheres, then the order of the point group to which the inner sphere of $\mathbf { A }$ belongs is 32.
If the ligand groups in $\mathbf { A }$ cannot be considered as spheres, then the highest-order rotation axis that can be contained in the inner sphere of $\mathbf { A }$ is $C _ 4$.
If the coordinated groups in $\mathbf { A }$ cannot be considered as spheres, then at least 6 atoms in the inner sphere of $\mathbf { A }$ are coplanar.
${ \mathrm N _ { 5 } ^- }$中存在离域$\pi$键$\Pi_5^5$，具有芳香性
$\mathrm {x+y+z}=8$
$\mathbf A$的分子量为307.13
若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界含反轴$I _ 8$ 
若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界的所有几何异构体均没有手性
若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界所属点群的阶次为32
若$\mathbf { A }$中配位基团不可视为球体，则$\mathbf { A }$的内界中可能包含的最高次旋转轴为$C _ 4$
若$\mathbf { A }$中配位基团不可视为球体，则$\mathbf { A }$的内界中至少有6个原子共平面
Based on $ \omega _{( \mathrm N )} = 40.82 \%$ and $\omega _{( \mathrm O )}= 37.30 \%$, it can be calculated that $n(\mathrm N)：n(\mathrm O) = 5 : 4$. Thus, it can be confirmed that $n(\mathrm N_5^-)：n(\mathrm {H_2O}) = 1 : 4$.

<Checkpoint>[$n(\mathrm N_5^-)：n(\mathrm {H_2O}) = 1 : 4$](1)</Checkpoint>

First, set the compound as $\mathrm {  C o ( N _ { 5 } ) _ { x } ( H _ { 2 } O ) _ { 4x } }$. When $\mathrm x=2$, the remaining molar mass is $58.98 \mathrm { g } / \mathrm { mol }$, which exactly corresponds to the atomic weight of one $\mathrm {Co}$ atom. Therefore, the chemical formula of $\mathbf { A }$ is $\mathrm { [ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { y } ] { \cdot } (8-y) H _ { 2 } O }$, with Co in the +2 oxidation state.

<Checkpoint>[The chemical formula of $\mathbf { A }$ is $\mathrm { [ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { y } ] { \cdot } (8-y) H _ { 2 } O }$, with Co in the +2 oxidation state](2)</Checkpoint>

According to condition $\textcircled{2}$: If the coordinating groups in $\mathbf { A }$ are treated as spheres, the inner sphere of $\mathbf { A }$ contains $z$ $C _ { 2 }$ axes perpendicular to the $C _ { \mathrm { y } }$ axis. $\mathrm { y }$ could be 2 or 4. However, when $\mathrm y=2$, it is impossible to have 6 $C _ { 2 }$ axes perpendicular to the $C _ {2}$ axis, so this case can be excluded. When $\mathrm y=4$ and $\mathrm z=4$, the trans-configuration of $\mathrm {  C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { 4 } }$ has 4 $C _ { 2 }$ axes perpendicular to the $C _ {4}$ axis, satisfying the condition. Therefore, the complex $\mathbf { A }$ is $\mathrm { trans-[ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { 4 } ] { \cdot } 4H _ { 2 } O }$.

<Checkpoint>[$\mathrm y=4$, $\mathrm z=4$](1)</Checkpoint>

<Checkpoint>[$\mathbf { A }$ is $\mathrm { trans-[ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { 4 } ] { \cdot } 4H _ { 2 } O }$](2)</Checkpoint>

Now, evaluate each option one by one:

Option A: The ${ \mathrm N _ { 5 } ^- }$ ion contains a delocalized $\pi$ bond $\Pi_5^6$ and exhibits aromaticity. Each $\mathrm N$ atom contributes 1 electron, and the additional electron from the negative charge gives a total of 6 electrons, forming $\Pi_5^6$. This satisfies Hückel's rule ($4 \times 2 + 2 = 6$), confirming aromaticity. Thus, the statement is incorrect.

<Checkpoint>[Hückel's rule $4 \times 2 + 2 = 6$, indicating aromaticity](1)</Checkpoint>

Option B: $\mathrm {x+y+z}=8$. Based on the above derivation, $\mathrm {x+y+z}=10$, so the statement is incorrect.

<Checkpoint>[$\mathrm {x+y+z}=10$](1)</Checkpoint>

Option C: The molecular weight of $\mathbf A$ is 307.13. According to the chemical formula, the calculated molecular weight is $M_\mathbf { A }=343.16$, so the statement is incorrect.

<Checkpoint>[Molecular weight $M_\mathbf { A }=343.16$](1)</Checkpoint>

Option D: If the coordinating groups in $\mathbf { A }$ are treated as spheres, the inner sphere of $\mathbf { A }$ contains an improper rotation axis $I _ 8$. If the coordinating groups are treated as spheres, the point group of the inner sphere is $D_{4\mathrm h}$, which does not include an $I _ 8$ axis. Thus, the statement is incorrect.

<Checkpoint>[If the coordinating groups in $\mathbf { A }$ are treated as spheres, the inner sphere's point group is $D_{4\mathrm h}$, which lacks an $I _ 8$ axis](1)</Checkpoint>

Option E: If the coordinating groups in $\mathbf { A }$ are treated as spheres, all geometric isomers of the inner sphere of $\mathbf { A }$ are achiral. The cis-isomer has a $C_{2\mathrm v}$ point group, and the trans-isomer has a $D_{4\mathrm h}$ point group. Both contain mirror planes and are achiral, so the statement is correct.

<Checkpoint>[$C_{2\mathrm v}$ and $D_{4\mathrm h}$ point groups are both achiral](1)</Checkpoint>

Option F: If the coordinating groups in $\mathbf { A }$ are treated as spheres, the order of the point group of the inner sphere is 32. If the coordinating groups are treated as spheres, the inner sphere's point group is $D_{4\mathrm h}$, which has an order of 16. Thus, the statement is incorrect.

<Checkpoint>[$D_{4\mathrm h}$ has an order of 16](1)</Checkpoint>

Option G: If the coordinating groups in $\mathbf { A }$ are not treated as spheres, the highest-order rotation axis in the inner sphere of $\mathbf { A }$ may be $C _ 2$. Due to the structure of $\mathrm N_5^-$, the $C _ 4$ axis is necessarily disrupted. If the water molecules are arranged with specific orientations, the original $C _ 4$ axis degenerates to $C _ 2$, meaning the highest-order rotation axis possible is $C _ 2$. Thus, the statement is incorrect.

<Checkpoint>[$C _ 4$ degenerates to $C _ 2$](1)</Checkpoint>

Option H: If the coordinating groups in $\mathbf { A }$ are not treated as spheres, at least 7 atoms in the inner sphere of $\mathbf { A }$ are coplanar. One $\mathrm N_5^-$ group has 5 coplanar atoms, plus the central $\mathrm {Co}$ atom and one coordinating $\mathrm N$ atom from the other $\mathrm N_5^-$ group, totaling 7 coplanar atoms (1 $\mathrm {Co}$ and 6 $\mathrm N$ atoms). Thus, the statement is incorrect.

<Checkpoint>[1 $\mathrm {Co}$ atom and 6 $\mathrm N$ atoms, totaling 7 coplanar atoms](1)</Checkpoint>

In conclusion, only option E is correct.
根据$ \omega _{( \mathrm N )} = 40.82 \%$ ， $\omega _{( \mathrm O )}= 37.30 \%$ ，可以求出$n(\mathrm N)：n(\mathrm O) = 5 : 4$ ，可以确认，$n(\mathrm N_5^-)：n(\mathrm {H_2O}) = 1 : 4$

<Checkpoint>[$n(\mathrm N_5^-)：n(\mathrm {H_2O}) = 1 : 4$](1)</Checkpoint>

先把化合物设为$\mathrm {  C o ( N _ { 5 } ) _ { x } ( H _ { 2 } O ) _ { 4x } }$。当$\mathrm x=2$，剩余的摩尔质量为 $58.98 \mathrm { g } / \mathrm { mol }$ ，恰好为1个$\mathrm {Co}$ 的原子量，故$\mathbf { A }$的化学式为$\mathrm { [ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { y } ] { \cdot } (8-y) H _ { 2 } O }$，Co为+2价。 

<Checkpoint>[$\mathbf { A }$的化学式为$\mathrm { [ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { y } ] { \cdot } (8-y) H _ { 2 } O }$, Co为+2价](2)</Checkpoint>

根据条件$\textcircled{2}$ ：若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界存在 $z$ 个垂直于 $C _ { \mathrm { y } }$ 轴的 $C _ { 2 }$ 轴，$\mathrm { y } $可能为2或4，但在y为2时，不可能存在6个垂直于$C _ {2}$ 轴的 $C _ { 2 }$ 轴，可排除；而当$\mathrm y=4$、$\mathrm z=4$时，反式构型的$\mathrm {  C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { 4 } }$具有4个垂直于$C _ {4}$ 轴的 $C _ { 2 }$ 轴，满足条件。故配合物$\mathbf { A }$为$\mathrm { trans-[ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { 4 } ] { \cdot } 4H _ { 2 } O }$

<Checkpoint>[$\mathrm y=4$，$\mathrm z=4$](1)</Checkpoint>

<Checkpoint>[$\mathbf { A }$为$\mathrm { trans-[ C o ( N _ { 5 } ) _ { 2 } ( H _ { 2 } O ) _ { 4 } ] { \cdot } 4H _ { 2 } O }$](2)</Checkpoint>

下面逐个判断正误：

选项A：${ \mathrm N _ { 5 } ^- }$中存在离域$\pi$键$\Pi_5^6$，具有芳香性。每个$\mathrm N$原子提供1个电子，加深阴离子负电荷的1个电子，共6个电子，因此为$\Pi_5^6$，符合休克尔规则$4\times 2+2=6$，因此具有芳香性，说法错误。

<Checkpoint>[休克尔规则$4\times 2+2=6，$具有芳香性](1)</Checkpoint>

选项B：  $\mathrm {x+y+z}=8$，根据上述推导可知$\mathrm {x+y+z}=10$，说法错误。

<Checkpoint>[$\mathrm {x+y+z}=10$](1)</Checkpoint>

选项C： $\mathbf A$的分子量为307.13。根据化学式计算分子量$M_\mathbf { A }=343.16$，说法错误。

<Checkpoint>[分子量$M_\mathbf { A }=343.16$](1)</Checkpoint>

选项D： 若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界含反轴$I _ 8$ 。若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界点群为$D_{4\mathrm h}$，不含反轴$I _ 8$，说法错误。

<Checkpoint>[若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界点群为$D_{4\mathrm h}$，不含反轴$I _ 8$](1)</Checkpoint>

选项E： 若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界的所有几何异构体均没有手性。顺式异构体点群为$C_{2\mathrm v}$，反式异构体点群为$D_{4\mathrm h}$，二者均含对称面，无手性，说法正确。

<Checkpoint>[$C_{2\mathrm v}$与$D_{4\mathrm h}$点群均无手性](1)</Checkpoint>

选项F： 若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界所属点群的阶次为32。

若$\mathbf { A }$中配位基团均视为球体，则$\mathbf { A }$的内界点群为$D_{4\mathrm h}$，$D_{4\mathrm h}$阶次为16，说法错误。

<Checkpoint>[$D_{4\mathrm h}$阶次为16](1)</Checkpoint>

选项G：若$\mathbf { A }$中配位基团不可视为球体，则$\mathbf { A }$的内界中可能包含的最高次旋转轴为$C _ 2$。

由于$\mathrm N_5^-$的结构，$C _ 4$一定会被破坏，若水分子按一定取向排列，则原本的$C _ 4$退化为$C _ 2$，即可能包含的最高次旋转轴为$C _ 2$，说法错误。

<Checkpoint>[$C _ 4$退化为$C _ 2$](1)</Checkpoint>

选项H： 若$\mathbf { A }$中配位基团不可视为球体，则$\mathbf { A }$的内界中至少有7个原子共平面。

一个$\mathrm N_5^-$中的5个原子必定共面，加上中心$\mathrm {Co}$原子与另一个$\mathrm N_5^-$中配位的$\mathrm N$原子，1个$\mathrm {Co}$原子与6个$\mathrm N$原子共计7个原子共平面，说法错误。

<Checkpoint>[1个$\mathrm {Co}$原子与6个$\mathrm N$原子共计7个原子共平面](1)</Checkpoint>

综上，选项E正确。