The compounds or complex ions $\bf{B,C,D}$ all contain metal $\bf{A}$, and their mass fractions in $\bf{B,C,D}$ are $\rm{26.30\%,31.60\%,26.80\%}$ respectively. All three have high symmetry. $\bf{B}$ is sensitive to oxygen and can be obtained by reacting chloride $\bf{E}$ of $\bf{A}$ with $\bf{F}$ in diethyl ether or benzene under a nitrogen atmosphere. $\bf{F}$ is a sodium salt of an acidic organic compound $\bf{G}$, which is aromatic and contains approximately $\rm{9.15\%}$ hydrogen. $\bf{C}$ is obtained by replacing two adjacent ligands in $\bf{B}$ with chlorine, and $\bf{C}$ can also be obtained by reacting $\bf{E}$ and $\bf{F}$ in dimethoxyethane. Another method for preparing $\bf{C}$ is to react the thallium salt of $\bf{G}$ with $\bf{E}$, followed by reduction with $\rm{SnCl_2}$. $\bf{D}$ is a bimetallic derivative cationic complex. It is known that $\bf{C}$ can react with sodium thiolate to generate $\bf{H}$, and $\bf{H}$ can then react with $\rm{NiCl_2}$ to obtain $\bf{D}$, which contains approximately $\rm{8.465\%}$ nickel.

Which of the following options regarding the unknown species $\bf{A-H}$ is correct:
化合物或配离子$\bf{B,C,D}$中均含有金属$\bf{A}$，其在$\bf{B,C,D}$中的质量分数分别为$\rm{26.30\%,31.60\%,26.80\%}$，三者都具有很高的对称性。$\bf{B}$对氧气敏感,可以由$\bf{A}$的氯化物$\bf{E}$和$\bf{F}$在氮气环境下于乙醚或苯中反应得到。$\bf{F}$是一种具有酸性的有机化合物$\bf{G}$的钠盐,具有芳香性,$\bf{G}$中含有约$\rm{9.15\%}$的氢。$\bf{C}$是$\bf{B}$中两个相邻配体被氯取代后得到的,$\bf{C}$也可以由$\bf{E}$和$\bf{F}$在二甲氧基乙烷中反应得到。$\bf{C}$的另外一种制备方法是$\bf{G}$的铊盐与$\bf{E}$作用,再用$\rm{SnCl_2}$还原。$\bf{D}$是一种双金属衍生物配阳离子。已知$\bf{C}$能通过和硫醇钠发生反应生成$\bf{H}$,$\bf{H}$再与$\rm{NiCl_2}$反应即可得到$\bf{D}$,$\bf{D}$中镍的含量约为$\rm{8.465\%}$。

下列关于未知物种$\bf{A-H}$的选项，正确的是：
All other options are incorrect.
The valence electrons of element $\bf{A}$ possess paired electrons.
The central element of $\bf{B}$ satisfies the EAN rule.
The structure of $\bf{C}$ possesses a threefold rotational axis.
$\bf{D}$ contains a ligand with two chemical environments.
$\bf{G}$'s chemical formula contains three elements
其他选项均不正确
元素$\bf{A}$的价电子具有成对电子
$\bf{B}$的中心元素满足EAN规则
$\bf{C}$的结构具有三重旋转轴
$\bf{D}$中存在一种配体具有两种化学环境
$\bf{G}$的化学式含有三种元素
The entry point for this question is $\bf{G}$, which has acidity, the negative ion has aromaticity, and has approximately $\rm{9.15\%}$ hydrogen.

Consider common acidic compounds containing benzene rings:

The mass fraction of hydrogen in benzoic acid is $\rm{6/(12\times6+16\times2+6)=5.5\%}$, consider reducing the number of heavy atoms;

The mass fraction of hydrogen in phenol is $\rm{6/(12\times6+16+6)=6.4\%}$, the number of heavy atoms also needs to be reduced; therefore, $\bf{G}$ most likely does not contain oxygen.

<Checkpoint>[$\bf{G}$ most likely does not contain oxygen](0.5)</Checkpoint>

Common acidic compounds containing benzene rings are difficult to match, consider five-membered ring aromatic compounds:

The cyclopentadienyl anion has aromaticity, and the mass fraction of hydrogen in cyclopentadiene is $\rm{6/(12\times5+6)=9.1\%}$, which matches.

Other pyrrole/pyran species can also be searched for, but the mass fractions do not match. Therefore, $\bf{G}$ is cyclopentadiene $\rm{C_5H_6}$, and $\bf{F}$ is sodium cyclopentadienide $\rm{C_5H_5Na}$. $\rm{C_5H_6}$ has only two elements, so option F is incorrect.

<Checkpoint>[$\bf{G}$ is cyclopentadiene $\rm{C_5H_6}$](2)</Checkpoint>

<Checkpoint>[$\bf{F}$ is sodium cyclopentadienide $\rm{C_5H_5Na}$](2)</Checkpoint>

$\bf{B}$ is obtained from the reaction of the chloride of element $\bf{A}$ with sodium cyclopentadienide, which can be basically judged as a ligand substitution reaction; the chemical formula of $\bf{B}$ can be written as $\rm{\bf{A}\rm(C_5H_5)_n}$; determine the value of n based on the mass fraction:

<Checkpoint>[The generation of $\bf{B}$ is a ligand substitution reaction, and the chemical formula can be written as $\rm{\bf{A}\rm(C_5H_5)_n}$](1)</Checkpoint>

If n is 1, the relative atomic mass of $\bf{A}$ is $\rm{65/(1-0.2630)-65=23}$, which may be $\rm{Na}$, but since $\bf{F}$ is $\rm{NaC_5H_5}$, this possibility is excluded.

<Checkpoint>[$\bf{F}$ is $\rm{NaC_5H_5}$, excluding the possibility that $\bf{A}$ is $\rm{Na}$](1)</Checkpoint>

If n is 2, the relative atomic mass of $\bf{A}$ is $\rm{130/(1-0.2630)-130=46.4}$, which is not a reasonable divalent metal.

If n is 3, the relative atomic mass of $\bf{A}$ is $\rm{195/(1-0.2630)-195=69.6}$, which is $\rm{Ga}$.

If n is 4, the relative atomic mass of $\bf{A}$ is $\rm{260/(1-0.2630)-260=92.8}$, which is $\rm{Nb}$.

If n is 5 or more, there is no reasonable metal.

Therefore, it can be deduced that the metal $\bf{A}$ may be $\rm{Ga}$ or $\rm{Nb}$, and the corresponding $\bf{B}$ may be $\rm{Ga(C_5H_5)_3}$ and $\rm{Nb(C_5H_5)_4}$.

<Checkpoint>[Metal $\bf{A}$ may be $\rm{Ga}$ or $\rm{Nb}$](2)</Checkpoint>

Then focus on $\bf{C}$, $\bf{C}$ is $\bf{B}$ in which two adjacent ligands are substituted by chlorine, then for these two metals, the chemical formulas of $\bf{C}$ are $\rm{GaCl_2(C_5H_5)}$ and $\rm{NbCl_2(C_5H_5)_2}$ respectively, calculate the mass fraction of the metal respectively:

If $\bf{A}$ is $\rm{Ga}$, calculate $\rm{69.7/(65+35.5\times2+69.7)=33.9\%}$, which is inconsistent with the given $\rm{31.6\%}$.

If $\bf{A}$ is $\rm{Nb}$, calculate $\rm{92.9/(65\times2+35.5\times2+92.9)=31.6\%}$, which is consistent with the given condition.

Therefore, it is finally determined that the metal $\bf{A}$ is $\rm{Nb}$; then $\bf{B}$ is $\rm{Nb(C_5H_5)_4}$, and $\bf{C}$ is $\rm{NbCl_2(C_5H_5)_2}$. $\bf{E}$ is $\rm{NbCl_4}$.

<Checkpoint>[Determine that the metal $\bf{A}$ is $\rm{Nb}$ through the mass fraction of $\bf{C}$](2)</Checkpoint>

<Checkpoint>[$\bf{B}$ is $\rm{Nb(C_5H_5)_4}$](0.5)</Checkpoint>

<Checkpoint>[$\bf{C}$ is $\rm{NbCl_2(C_5H_5)_2}$](0.5)</Checkpoint>

The ground state valence electron configuration of $\rm{Nb}$ is $\rm{4d^45s^1}$, there are no paired electrons, so option B is incorrect; in $\rm{Nb(C_5H_5)_4}$, $\rm{Nb}$ is +4 valent, there is a single electron, and the EAN rule cannot be achieved no matter how the four cyclopentadienyl groups are coordinated, so option C is incorrect; $\bf{C}$ is $\rm{NbCl_2(C_5H_5)_2}$, the structure is similar to dichloromethane, and there is no threefold axis, so option D is incorrect.

<Checkpoint>[The ground state valence electron configuration of $\rm{Nb}$ is $\rm{4d^45s^1}$, there are no paired electrons](0.5)</Checkpoint>

<Checkpoint>[$\rm{Nb(C_5H_5)_4}$ cannot reach the EAN rule](0.5)</Checkpoint>

<Checkpoint>[The structure of $\rm{NbCl_2(C_5H_5)_2}$ is similar to dichloromethane, and there is no threefold axis](0.5)</Checkpoint>

$\rm{NbCl_2(C_5H_5)_2}$ can react with sodium thiomethoxide to generate $\bf{H}$, obviously a ligand substitution reaction occurs, the thiomethoxide anion substitutes the chloride ion, and the generated $\bf{H}$ is $\rm{Nb(CH_3S)_2(C_5H_5)_2}$.

<Checkpoint>[The thiomethoxide anion substitutes the chloride ion, and the generated $\bf{H}$ is $\rm{Nb(CH_3S)_2(C_5H_5)_2}$](1)</Checkpoint>

$\rm{Nb(CH_3S)_2(C_5H_5)_2}$ reacts with $\rm{NiCl_2}$ to obtain the bimetallic derivative cationic complex $\bf{D}$, it can also be guessed that it is a ligand substitution reaction, and the product should not contain chlorine;

<Checkpoint>[The reaction to generate $\bf{D}$ can also be guessed as a ligand substitution reaction, and the product should not contain chlorine](1)</Checkpoint>

The content of nickel in $\bf{D}$ is about $\rm{8.465\%}$, assuming it contains one nickel atom, the remaining relative molecular mass is exactly two molecules of $\rm{Nb(CH_3S)_2(C_5H_5)_2}$, so it can be calculated that the chemical formula of $\bf{D}$ is $\rm{[NiNb_2(CH_3S)_4(C_5H_5)_4]^{2+}}$.

<Checkpoint>[It can be calculated that the chemical formula of $\bf{D}$ is $\rm{[NiNb_2(CH_3S)_4(C_5H_5)_4]^{2+}}$](2)</Checkpoint>

In the structure of this cationic complex, the cyclopentadienyl group is coordinated with $\rm{Nb}$, and according to symmetry, there is only one chemical environment; $\rm{Ni}$ is obviously tetracoordinate, coordinated with four thiomethoxide groups, the thiomethoxide group is a bridging ligand coordinated with $\rm{Ni}$ and $\rm{Nb}$, and there is also only one chemical environment, so option E is incorrect.

<Checkpoint>[The cyclopentadienyl group is coordinated with $\rm{Nb}$, and according to symmetry, there is only one chemical environment](1)</Checkpoint>

<Checkpoint>[The thiomethoxide group is a bridging ligand coordinated with $\rm{Ni}$ and $\rm{Nb}$, and there is also only one chemical environment](1)</Checkpoint>

The structure of the cationic complex is as follows:

<MultiModal>![$\bf{D}$结构为：[A][Nb]1(S[Ni]2(S1)S[Nb]([A])(S2)[A])[A]；其中A代表环戊二烯基[cH-]1cccc1。](/media/uploads/02746e30-4e71-42ee-8c4e-6dde7ba8c1c9.png)</MultiModal>

<Checkpoint>[$\bf{D}$ structure is: [A][Nb]1(S[Ni]2(S1)S[Nb]([A])(S2)[A])[A]; where A represents the cyclopentadienyl group [cH-]1cccc1](1)</Checkpoint>

In summary, option A is correct.
本题入手点在$\bf{G}$，其具有酸性，负离子具有芳香性，且有约$\rm{9.15\%}$的氢。

考虑含苯环的常见酸性化合物：

苯甲酸中氢质量分数为$\rm{6/(12\times6+16\times2+6)=5.5\%}$，考虑降低重原子数目；

苯酚为$\rm{6/(12\times6+16+6)=6.4\%}$，还需要降低重原子数目；因此$\bf{G}$大概率不含氧。；

<Checkpoint>[$\bf{G}$大概率不含氧](0.5)</Checkpoint>

常见的含苯环酸性化合物难以匹配，考虑五元环芳香化合物：

环戊二烯负离子具有芳香性，本体环戊二烯中氢质量分数为$\rm{6/(12\times5+6)=9.1\%}$，符合。

还可以寻找其他吡咯/吡喃物种，但均质量分数不符。因此，$\bf{G}$为环戊二烯$\rm{C_5H_6}$，$\bf{F}$为环戊二烯钠$\rm{C_5H_5Na}$。$\rm{C_5H_6}$只有两种元素，选项F错误。

<Checkpoint>[$\bf{G}$为环戊二烯$\rm{C_5H_6}$](2)</Checkpoint>

<Checkpoint>[$\bf{F}$为环戊二烯钠$\rm{C_5H_5Na}$](2)</Checkpoint>

$\bf{B}$由元素$\bf{A}$的氯化物与环戊二烯钠反应得到，可以基本判断是配体取代反应；$\bf{B}$的化学式可以写为$\rm{\bf{A}\rm(C_5H_5)_n}$；根据质量分数判断n的取值：

<Checkpoint>[生成$\bf{B}$为配体取代反应，化学式可以写为$\rm{\bf{A}\rm(C_5H_5)_n}$](1)</Checkpoint>

若n取1，$\bf{A}$的相对原子质量为$\rm{65/(1-0.2630)-65=23}$，可能为$\rm{Na}$，但因为$\bf{F}$就是$\rm{NaC_5H_5}$，所以排除这种可能性。

<Checkpoint>[$\bf{F}$就是$\rm{NaC_5H_5}$，排除$\bf{A}$为$\rm{Na}$的可能性](1)</Checkpoint>

若n取2，$\bf{A}$的相对原子质量为$\rm{130/(1-0.2630)-130=46.4}$，不为合理的二价金属。

若n取3，$\bf{A}$的相对原子质量为$\rm{195/(1-0.2630)-195=69.6}$，为$\rm{Ga}$。

若n取4，$\bf{A}$的相对原子质量为$\rm{260/(1-0.2630)-260=92.8}$，为$\rm{Nb}$。

若n取5及以上，没有合理金属。

从而推出金属$\bf{A}$可能是$\rm{Ga}$或者$\rm{Nb}$,对应的$\bf{B}$可能是$\rm{Ga(C_5H_5)_3}$和$\rm{Nb(C_5H_5)_4}$。

<Checkpoint>[金属$\bf{A}$可能是$\rm{Ga}$或者$\rm{Nb}$](2)</Checkpoint>

那么关注$\bf{C}$，$\bf{C}$是$\bf{B}$中两个相邻配体被氯取代，那么对于这两种金属，$\bf{C}$的化学式分别为$\rm{GaCl_2(C_5H_5)}$和$\rm{NbCl_2(C_5H_5)_2}$，分别计算金属的质量分数：

若$\bf{A}$为$\rm{Ga}$，计算$\rm{69.7/(65+35.5\times2+69.7)=33.9\%}$，与题设$\rm{31.6\%}$不符。

若$\bf{A}$为$\rm{Nb}$，计算$\rm{92.9/(65\times2+35.5\times2+92.9)=31.6\%}$，符合题设。

因此最终确定金属$\bf{A}$为$\rm{Nb}$；则$\bf{B}$为$\rm{Nb(C_5H_5)_4}$，$\bf{C}$为$\rm{NbCl_2(C_5H_5)_2}$。$\bf{E}$为$\rm{NbCl_4}$。

<Checkpoint>[通过$\bf{C}$的质量分数确定金属$\bf{A}$为$\rm{Nb}$](2)</Checkpoint>

<Checkpoint>[$\bf{B}$为$\rm{Nb(C_5H_5)_4}$](0.5)</Checkpoint>

<Checkpoint>[$\bf{C}$为$\rm{NbCl_2(C_5H_5)_2}$](0.5)</Checkpoint>

$\rm{Nb}$基态价电子组态为$\rm{4d^45s^1}$，没有成对电子，选项B错误；$\rm{Nb(C_5H_5)_4}$中$\rm{Nb}$为+4价，存在单电子，无论四个环戊二烯基如何配位均达不到EAN规则，选项C错误；$\bf{C}$为$\rm{NbCl_2(C_5H_5)_2}$，结构与二氯甲烷类似，不存在三重轴，选项D错误。

<Checkpoint>[$\rm{Nb}$基态价电子组态为$\rm{4d^45s^1}$，没有成对电子](0.5)</Checkpoint>

<Checkpoint>[$\rm{Nb(C_5H_5)_4}$达不到EAN规则](0.5)</Checkpoint>

<Checkpoint>[$\rm{NbCl_2(C_5H_5)_2}$结构与二氯甲烷类似，不存在三重轴](0.5)</Checkpoint>

$\rm{NbCl_2(C_5H_5)_2}$能通过和甲硫醇钠发生反应生成$\bf{H}$，明显发生配体取代反应，甲硫醇负离子取代氯离子，生成的$\bf{H}$为$\rm{Nb(CH_3S)_2(C_5H_5)_2}$。

<Checkpoint>[甲硫醇负离子取代氯离子，生成的$\bf{H}$为$\rm{Nb(CH_3S)_2(C_5H_5)_2}$](1)</Checkpoint>

$\rm{Nb(CH_3S)_2(C_5H_5)_2}$与$\rm{NiCl_2}$得到双金属衍生物配阳离子$\bf{D}$，同样可以猜测是配体取代反应，产物应当不含氯；

<Checkpoint>[生成$\bf{D}$的反应同样可以猜测是配体取代反应，产物应当不含氯](1)</Checkpoint>

$\bf{D}$中镍的含量约为$\rm{8.465\%}$，假设含有一个镍原子，剩余的相对分子质量刚好是两分子$\rm{Nb(CH_3S)_2(C_5H_5)_2}$，从而计算可得$\bf{D}$的化学式为$\rm{[NiNb_2(CH_3S)_4(C_5H_5)_4]^{2+}}$。

<Checkpoint>[计算可得$\bf{D}$的化学式为$\rm{[NiNb_2(CH_3S)_4(C_5H_5)_4]^{2+}}$](2)</Checkpoint>

该配阳离子结构中，环戊二烯基与$\rm{Nb}$配合，根据对称性只有一种化学环境；$\rm{Ni}$很明显为四配位，与四个甲硫醇基配位，甲硫醇基为桥连配体与$\rm{Ni}$和$\rm{Nb}$配位，化学环境同样只有一种，选项E错误。

<Checkpoint>[环戊二烯基与$\rm{Nb}$配合，根据对称性只有一种化学环境](1)</Checkpoint>

<Checkpoint>[甲硫醇基为桥连配体与$\rm{Ni}$和$\rm{Nb}$配位，化学环境同样只有一种](1)</Checkpoint>

配阳离子结构如下图：

<MultiModal>![$\bf{D}$结构为：[A][Nb]1(S[Ni]2(S1)S[Nb]([A])(S2)[A])[A]；其中A代表环戊二烯基[cH-]1cccc1。](/media/uploads/02746e30-4e71-42ee-8c4e-6dde7ba8c1c9.png)</MultiModal>

<Checkpoint>[$\bf{D}$结构为：[A][Nb]1(S[Ni]2(S1)S[Nb]([A])(S2)[A])[A]；其中A代表环戊二烯基[cH-]1cccc1](1)</Checkpoint>

综上，选项A正确。