Complex salt $\bf{A}$ contains 4 elements and exhibits higher conductivity than ordinary salts. It displays one-dimensional antiferromagnetism at low temperatures, and undergoes a phase transition to a paramagnetic structure as the temperature increases.

A chemical formula that reflects the structural characteristics of complex salt $\bf{A}$ is $[\bold{BC}_2]\bold{D}$, where: species $\bf{B}$ is a metallic element; species $\bf{C}$ is an electrically neutral ligand, and all elements within it have only one chemical environment; species $\bf{D}$ contains no metallic elements. Elemental analysis indicates that the mass fractions of $C$ and $H$ in $\bf{A}$ are $43.6\%$ and $7.4\%$, respectively.

$\bf{A}$ is thermally unstable and must be prepared at very low temperatures ($<-40~^\circ\rm{C}$). Vacuum decomposition of $1.54~\rm{g}~\bf{A}$ at low temperature ($<0~^\circ\rm{C}$) yields $0.31~\rm{g}$ of elemental $\bf{B}$. All of this $\bf{B}$ is then reacted with $O_2$ under controlled conditions to obtain oxide $\bf{N}$, with the chemical formula $\bold{B}_yO_z$. All of the above $\bf{N}$ is dissolved in water, and $11.9~\rm{mL}$ of a certain combustible gas is collected under standard conditions. The aforementioned aqueous solution is then titrated with $0.09196~\rm{mol/L}$ $HCl$, consuming $25.32~\rm{mL}$.

Based on calculations, deduce the above unknowns and select the correct answer.
配合物盐$\bf{A}$包含4种元素，具有高于普通盐类的导电性，其在低温下具有一维反铁磁性，而当温度升高会发生相变，转换为顺磁性结构。

一种能体现配合物盐$\bf{A}$结构特点的化学式是$[\bold{BC}_2]\bold{D}$，其中：物种$\bf{B}$是一种金属元素；物种$\bf{C}$是一个电中性配体，且其中所有元素都只有一种化学环境；物种$\bf{D}$无金属元素。元素分析表明，$\bf{A}$中$C$和$H$的质量分数分别为$43.6\%$和$7.4\%$。

$\bf{A}$对热不稳定，需在很低温度（$<-40~^\circ\rm{C}$）下制备。将$1.54~\rm{g}~\bf{A}$在低温（$<0~^\circ\rm{C}$）下真空分解，得到$0.31~\rm{g}~\bf{B}$单质，再将这些$\bf{B}$全部与$O_2$在控制条件下反应，得到氧化物$\bf{N}$，化学式为$\bold{B}_yO_z$。将上述$\bf{N}$全部溶于水，并收集到标况下的某可燃气体$11.9~\rm{mL}$，再用$0.09196~\rm{mol/L}$的$HCl$滴定前述水溶液，耗去$25.32~\rm{mL}$。

根据计算，推断上述未知物，并选择正确答案。
All other options are incorrect
The common valence states of $\bf{B}$ include $+2$.
$1.54\ g\ \bf{A}$ directly dissolved in sufficient sulfuric acid, the resulting gas is approximately $52\ mL$ under standard conditions.
In the structure of $\bf{N}$, the ratio of the number of chemical environments of $\bf{B}$ to the number of chemical environments of $O$ is $2:1$.
$1\ mol\ \bf{C}$ completely combusts, requiring the consumption of $18\ mol\ O_2$.
The mass fraction of $\bf{D}$ in $\bf{A}$ is less than $0.1\%$
其他选项均不正确
$\bf{B}$的常见价态包括$+2$价
$1.54\ g\ \bf{A}$直接溶于足量硫酸，所得气体在标况下约为$52\ mL$
$\bf{N}$的结构中，$\bf{B}$的化学环境数和$O$的化学环境数之比为$2:1$
$1\ mol\ \bf{C}$完全燃烧，需消耗$18\ mol\ O_2$。
$\bf{D}$在$\bf{A}$中的质量占比小于$0.1\%$
We can first determine the elements of $\bf{B}$. The amount of $OH^-$ produced by $0.31~\rm{g}~\bf{B}$ is $n=c_{HCl}V_{HCl}=0.09196~\rm{mol/L} \cdot 25.32~\rm{mL} = 0.00233~\rm{mol}$. Let $+a$ be the valence of $\bf{B}$ during titration, then the formula weight of $\bf{B}$ is $M_{\bf{B}}=0.31~\rm{g}/(0.00233~\rm{mol}/a)=133.0/a~\rm{g/mol}$

<Checkpoint>[Formula weight of $\bf{B}$ is $M_{\bf{B}}=133.0/a$](3)</Checkpoint>

When $a=1$, $\bf{B}$ is $Cs$; when $a=2$, $\bf{B}$ is approximately $Zn$; when $a=3$, $\bf{B}$ is approximately $Sc$. However, the hydroxides of the latter two are insoluble in water, so $\bf{B}$ is $Cs$.

<Checkpoint>[$\bf{B}$ is $Cs$](1)</Checkpoint>

The $H_2$ produced by directly dissolving $0.31~{\rm{g}}~Cs$ in water is $V=(0.31~\rm{g}/133~\rm{g/mol})/2 \cdot 22.4~\rm{L/mol} \cdot 1000~\rm{mL/L} \approx 26.1~\rm{mL}$. According to the conservation of electrons in $Cs_yO_z$, we have $(y-2z)/y=11.9/26.1$, and the smallest reasonable integer solution is $y=11,z=3$. Therefore, $\bf{N}$ is $Cs_{11}O_3$. Since there is $Cs$(0) in it, $\bf{N}$ dissolves in water, and the flammable gas produced is the same as $Cs$, which is $H_2$.

<Checkpoint>[$\bf{N}$ is $Cs_{11}O_3$](3)</Checkpoint>

$M_{\bf{A}}=M_{Cs}\cdot 1.54~\rm{g}/0.31~\rm{g} \approx 133~\rm{g/mol} \cdot 4.97 \approx 660.7~\rm{g/mol}$. According to the mass fraction of $C$ and $H$, $n(C)=(660.7~\rm{g/mol} \cdot 0.436)/12.01~\rm{g/mol} \approx 23.98, n(H)=(660.7~\rm{g/mol} \cdot 0.074)/1.008~\rm{g/mol} \approx 48.50$.

Notice that $n(C):n(H)\approx1:2$, and $\bf{C}$ is a ligand for $Cs^+$, and all elements have only one environment. According to coordination chemistry knowledge, it can be thought of as a crown ether ligand, and then $\bf{C}$ is deduced to be 18-crown-6 ($C_{12}H_{24}O_{6}$) based on the number of ligands. At this time, there is still $0.5H$ left, but the remaining molecular weight of $\bf{A}$ after deducting $Cs(C_{12}H_{24}O_{6})_2$ is about 0. Considering the experimental error, combined with the fact that $Cs$ has a $+1$ valence, $\bf{D}$ may be $H^-$ or $e^-$. However, $CsH$ will not decompose into $Cs$ element at $<0~^\circ\rm{C}$. Combining the information in the question, "the complex salt $\bf{A}$ contains 4 elements, has higher conductivity than ordinary salts, and has one-dimensional antiferromagnetism at low temperatures, while a phase transition occurs when the temperature rises, converting to a paramagnetic structure." This highly matches the properties of some electride salts, so $\bf{D}$ should be $e^-$, and $\bf{A}$ is the electride salt of $Cs$.

<Checkpoint>[$\bf{C}$ is 18-crown-6](2)</Checkpoint>

<Checkpoint>[$\bf{D}$ is $e^-$](2)</Checkpoint>

The common valence state of $Cs$ is $+1$ instead of $+2$, so option B is incorrect.

<Checkpoint>[The common valence state of $Cs$ is $+1$](0.5)</Checkpoint>

The amount of $H_2$ released by dissolving $1.54~\rm{g}~\bf{A}$ in sulfuric acid can be regarded as the amount released by all $0.31~\mathrm{g}~Cs$ element contained in it, which is calculated as $26.1~\mathrm{mL}~H_2$ as before, so option C is incorrect.

<Checkpoint>[$1.54~\rm{g}~\bf{A}$ releases $26.1~\rm{mL}~H_2$ upon dissolution in sulfuric acid](1)</Checkpoint>

The structure of $Cs_{11}O_3$ is shown in the figure below:

<MultiModal>![The cluster can be regarded as a triangle arrangement of three Cs₆O octahedra connected by shared faces. According to the different positions of cesium atoms in the cluster, there are three chemical environments: 3-coordinated, 2-coordinated, and 1-coordinated.](/media/uploads/985b3aab-124b-44f7-8eaf-8914ad084349.png)</MultiModal>

Therefore, there are 3 types of $Cs$ and 1 type of $O$ in $Cs_{11}O_3$, and the ratio is $3:1$. Option D is incorrect.

<Checkpoint>[The ratio of the number of chemical environments of $Cs$ to $O$ in $Cs_{11}O_3$ is $3:1$](1.5)</Checkpoint>

Complete combustion of $1~\rm{mol}$ 18-crown-6 ($C_{12}H_{24}O_{6}$) consumes $(12 + 24/4 - 6/2) = 15~{\rm{mol}}~O_2$. Option E is incorrect.

<Checkpoint>[$1~\rm{mol}$ 18-crown-6 completely combusts consumes $15~\rm{mol}~O_2$](1)</Checkpoint>

The mass percentage of $e^-$ in $\bf{A}$ is obviously $<0.1\%$, so option F is correct.

<Checkpoint>[The mass percentage of $e^-$ in $\bf{A}$ is $<0.1\%$](1)</Checkpoint>
本题首先可以判断$\bf{B}$的元素种类。$0.31~\rm{g}~\bf{B}$产生的$OH^-$数量$n=c_{HCl}V_{HCl}=0.09196~\rm{mol/L} \cdot 25.32~\rm{mL} = 0.00233~\rm{mol}$。设$\bf{B}$在滴定时为$+a$价，则$\bf{B}$的式量$M_{\bf{B}}=0.31~\rm{g}/(0.00233~\rm{mol}/a)=133.0/a~\rm{g/mol}$

<Checkpoint>[$\bf{B}$的式量$M_{\bf{B}}=133.0/a$](3)</Checkpoint>

$a=1$时$\bf{B}$为$Cs$，$a=2$时$\bf{B}$近似为$Zn$，$a=3$时$\bf{B}$近似为$Sc$，但后两者的氢氧化物均不溶于水，因此$\bf{B}$为$Cs$。

<Checkpoint>[$\bf{B}$为$Cs$](1)</Checkpoint>

$0.31~{\rm{g}}~Cs$直接溶于水产生的$H_2$为$V=(0.31~\rm{g}/133~\rm{g/mol})/2 \cdot 22.4~\rm{L/mol} \cdot 1000~\rm{mL/L} \approx 26.1~\rm{mL}$。根据$Cs_yO_z$的电子数守恒，有$(y-2z)/y=11.9/26.1$，其最小的合理整数解是$y=11,z=3$。因此$\bf{N}$为$Cs_{11}O_3$。由于其中存在$Cs$(0)，因此$\bf{N}$溶于水，生成的可燃气体与$Cs$相同，为$H_2$。

<Checkpoint>[$\bf{N}$为$Cs_{11}O_3$](3)</Checkpoint>

$M_{\bf{A}}=M_{Cs}\cdot 1.54~\rm{g}/0.31~\rm{g} \approx 133~\rm{g/mol} \cdot 4.97 \approx 660.7~\rm{g/mol}$，根据$C$和$H$的质量分数，$n(C)=(660.7~\rm{g/mol} \cdot 0.436)/12.01~\rm{g/mol} \approx 23.98, n(H)=(660.7~\rm{g/mol} \cdot 0.074)/1.008~\rm{g/mol} \approx 48.50$。

注意到$n(C):n(H)\approx1:2$，且$\bf{C}$是对$Cs^+$的配体，所有元素只有一种环境，根据配位化学知识可以想到是冠醚类配体，进而根据配体数量推出$\bf{C}$是18-冠-6（$C_{12}H_{24}O_{6}$）。此时还剩$0.5H$，但$\bf{A}$在扣除$Cs(C_{12}H_{24}O_{6})_2$后剩余分子量约为0，考虑实验误差，结合$Cs$为$+1$价，$\bf{D}$可能是$H^-$或$e^-$。而$CsH$在$<0~^\circ\rm{C}$下不会分解为$Cs$单质，结合题目中的信息“配合物盐$\bf{A}$包含4种元素，具有高于普通盐类的导电性，其在低温下具有一维反铁磁性，而当温度升高会发生相变，转换为顺磁性结构。”这与一些电子盐的性质高度吻合，因此$\bf{D}$应该为$e^-$，$\bf{A}$为$Cs$的电子盐。

<Checkpoint>[$\bf{C}$是18-冠-6](2)</Checkpoint>

<Checkpoint>[$\bf{D}$为$e^-$](2)</Checkpoint>

$Cs$常见价态为$+1$而不是$+2$，选项B错误。

<Checkpoint>[$Cs$的常见价态为$+1$](0.5)</Checkpoint>

$1.54~\rm{g}~\bf{A}$溶于硫酸释放的$H_2$量可视同其所含全部$0.31~\mathrm{g}~Cs$单质的释放量，如前计算为$26.1~\mathrm{mL}~H_2$，选项C错误。

<Checkpoint>[$1.54~\rm{g}~\bf{A}$溶于硫酸释放$26.1~\rm{mL}~H_2$](1)</Checkpoint>

$Cs_{11}O_3$ 的结构如下图所示：

<MultiModal>![该团簇可以看作是由三个共用面连接的Cs₆O八面体成三角形排列而成。根据铯原子在团簇中的不同位置，有三种化学环境，分别为3配位、2配位、1配位](/media/uploads/985b3aab-124b-44f7-8eaf-8914ad084349.png)</MultiModal>

因此，$Cs_{11}O_3$中有3种$Cs$和1种$O$，比例为$3:1$。选项D错误。

<Checkpoint>[$Cs_{11}O_3$中$Cs$与$O$的化学环境数之比为$3:1$](1.5)</Checkpoint>

$1~\rm{mol}$ 18-冠-6（$C_{12}H_{24}O_{6}$）完全燃烧消耗$(12 + 24/4 - 6/2) = 15~{\rm{mol}}~O_2$。选项E错误。

<Checkpoint>[$1~\rm{mol}$  18-冠-6完全燃烧消耗$15~\rm{mol}~O_2$](1)</Checkpoint>

$e^-$在$\bf{A}$中的质量占比显然$<0.1\%$，选项F正确。

<Checkpoint>[$e^-$在$\bf{A}$中的质量占比$<0.1\%$](1)</Checkpoint>