Reacting $\mathrm{BiCl}_{3}$ and $\mathrm{AlCl}_{3}$ with stoichiometric $\mathrm{Bi}$ in molten $\mathrm{NaAlCl_4}$, and changing the feeding ratio, ionic compounds $\bf{A}$ and $\bf{B}$ can be prepared separately. If the same amount of substance of $\bf{A}$ and $\bf{B}$ are to be generated, the ratio of $\mathrm{Bi}$投入 [invested] during preparation is $6:11$. It is known that the mass fraction of $\mathrm{Bi}$ in $\bf{A}$ is $67.4\%$, and all atoms in $\bf{A}$ satisfy the octet rule. The cation in the ionic compound $\bf{B}$ has a valence of $+2$, and the anion is the same as in $\bf{A}$.

In addition to the cations in $\bf{A}$ and $\bf{B}$, $\mathrm{Bi}$ can also form a positive ion with the same symmetry as $\mathrm{BF_3}$, which was first discovered in the black subchloride $\bf{C}$ of Bi. It is known that there are two valence states (one of which is $+3$ valence) of $\mathrm{Bi}$ in $\bf{C}$ in a ratio of $3: 1$, and three kinds of $\mathrm{Bi}$-containing ions in a ratio of $2: 4: 1$. In the anion, $\mathrm{Bi}$ is in a square pyramidal coordination, and the mass fraction of $\mathrm{Bi}$ in $\bf{C}$ is $83.48 \%$.

Which of the following statements are correct:

$1$. The ratio of anions to cations in ionic compound $\bf{A}$ is $1:2$.

$2$. The cation in ionic compound $\bf{A}$ has $3$ fewer $\mathrm{Bi}$ atoms than the cation in $\bf{B}$.

$3$. There are $2$ different chemical environments for $\mathrm{Bi}$ atoms in the cation cluster in $\bf{C}$.

$4$. All anions in $\bf{C}$ contain $C_2$ symmetry.
将 $\mathrm{BiCl}_{3}$ 及 $\mathrm{AlCl}_{3}$ 与化学计量的 $\mathrm{Bi}$ 在熔融 $\mathrm{NaAlCl_4}$ 中反应, 改变投料比, 可以分别制得离子化合物 $\bf{A}$ 和 $\bf{B}$, 若要生成相同物质的量的 $\bf{A}$ 和 $\bf{B}$, 则制备时分别投入的 $\mathrm{Bi}$ 的比例为
$6:11$。已知 $\bf{A}$ 中 $\mathrm{Bi}$ 的质量分数为$67.4\%$,且 $\bf{A}$ 中所有原子均满足八隅律。离子化合物 $\bf{B}$ 中阳离子为 $+2$ 价, 阴离子与 $\bf{A}$ 相同。

除了 $\bf{A}$ 和 $\bf{B}$ 中的阳离子外, $\mathrm{Bi}$ 还可以形成一种与 $\mathrm{BF_3}$ 对称性相同的正离子, 其最早
被发现于 Bi 的黑色次氯化物 $\bf{C}$ 中。已知 $\bf{C}$ 中存在着比例为 $3: 1$ 的两种价态(其中一种为$+3$
价)的 $\mathrm{Bi}$, 和三种比例为 $2: 4: 1$ 的含 $\mathrm{Bi}$ 离子, 阴离子中 $\mathrm{Bi}$ 均为四方锥配位, $\bf{C}$ 中 $\mathrm{Bi}$ 的质量分数为 $83.48 \%$ 。

下列说法正确的有：

$1$. 离子化合物$\bf{A}$中阴阳离子之比为$1:2$。

$2$. 离子化合物$\bf{A}$中阳离子比$\bf{B}$中阳离子少$3$个$\mathrm{Bi}$原子。

$3$. $\bf{C}$中阳离子簇中$\mathrm{Bi}$原子有$2$种不同的化学环境。

$4$. $\bf{C}$中阴离子均含有$C_2$对称性。
All other options are incorrect
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
其他选项均不正确
1
2
3
4
1，2
1，3
1，4
2，3
2，4
3，4
1，2，3
1，2，4
1，3，4
2，3，4
1，2，3，4
According to the question, let the chemical formula of $\bf{A}$ be $\mathrm{Bi_m(AlCl_4)_n}$

Then, according to the mass fraction of $\mathrm{Bi}$, we have

 $m M(\mathrm{Bi})=0.674[mM(\mathrm{Bi})+n M(AlCl_4)]$ 

<Checkpoint>[ $m M(\mathrm{Bi})=0.674[mM(Bi)+n M(\mathrm{AlCl_4})]$ ](1)</Checkpoint>

Solving the above Diophantine equation, we obtain the only solution that meets the conditions of the question

That is, $m=5, n=3$, then the chemical formula of $\mathbf{A}$ is $\mathrm{Bi_5(AlCl_4)_3}$ 

<Checkpoint>[$\mathbf{A}$ is $\mathrm{Bi_5(AlCl_4)_3}$ ](1)</Checkpoint>

Its cation is an ion cluster, so the ratio of cation to anion is $1:3$. $1$ is incorrect.

According to the question, let the chemical formula of $\bf{B}$ be $\mathrm{Bi_m(AlCl_4)_2}$

Since the same amount of substance of $\bf{A}$ and $\bf{B}$ are generated, the ratio of $\mathrm{Bi} $ that needs to be invested is $6: 11$
And $\mathrm{Bi_5^{3+}}$ can be regarded as $\mathrm{Bi^{3+}}+4 \mathrm{Bi}$, then $\mathrm{Bi_m^{2+}}$ can be regarded as $\frac{2}{3} \mathrm{Bi^{3+}}+\frac{3 m-2}{3} \mathrm{Bi}$

<Checkpoint>[$\mathrm{Bi_m^{2+}}$ can be regarded as $\frac{2}{3} \mathrm{Bi^{3+}}+\frac{3 m-2}{3} \mathrm{Bi}$](1)</Checkpoint>

Therefore, we can know $m=4$, then the chemical formula of $\bf{B}$ is $\mathrm{Bi_8(AlCl_4)_2}$ 

<Checkpoint>[$\bf{B}$ is $\mathrm{Bi_8(AlCl_4)_2}$ ](1)</Checkpoint>

The cation in the ionic compound $\bf{A}$ has $3$ fewer $\mathrm{Bi}$ atoms than the cation in $\bf{B}$.

<Checkpoint>[2 is correct.](1)</Checkpoint>

According to the question, $\bf{C}$ is only composed of $\mathrm{Bi}$ and $\mathrm{Cl}$

Then $n(\mathrm{Bi}): n(\mathrm{Cl})=\omega(\mathrm{Bi}) M(\mathrm{Cl}) / \omega(\mathrm{Cl}) M(\mathrm{Bi})=6: 7$ in $\bf{C}$
From the valence state, we know that $\bf{C}$ should be $\mathrm{Bi_{24}Cl_{28}}$, and among them, $6$ $\mathrm{Bi}$ have a valence of $+3$, and the other $18$ $\mathrm{Bi}$ have a total valence of $+10$

First of all, it can be inferred that these $18$ $\mathrm{Bi}$ should be positive ions formed by $\mathrm{Bi}$
Secondly, there are two possibilities: the " $1$ " in $2: 4: 1$ is the positive ion $\mathrm{Bi_18}^{10+}$, and the " $2$ " is the positive ion $\mathrm{Bi_9}^{5+}$ 
And since the $\mathrm{Bi}$ in the anion are all square pyramidal coordinated, it can be known that the one that meets the condition is $\mathrm{Bi_9}^{5+}$
And the remaining "$4 $" corresponds to $\mathrm{BiCl_5}^{2-}$, and "$1$ " corresponds to $\mathrm{Bi_2Cl_8}^{2-}$

Then the chemical formula of $\bf{C}$ is $\mathrm{(Bi_9^{5+})_{2}(BiCl_5^{2-})_4(Bi_2 Cl_8^{2-})}$

<Checkpoint>[$\bf{C}$ is $\mathrm{(Bi_9^{5+})_{2}(BiCl_5^{2-})_4(Bi_2 Cl_8^{2-})}$](1)</Checkpoint>

Its cation is a tricapped trigonal prism, which belongs to the $D_{3h}$ point group, as shown in the figure:

<MultiModal>![The $\mathrm{Bi_9^{5+}}$ cation is a tricapped trigonal prism structure. Each rectangular face of the trigonal prism extends a vertex to form a tricapped trigonal prism. Each $\mathrm{Bi}$ atom is located at the vertex of the structure, which belongs to the $D_{3h}$ point group.](/media/uploads/68895c98-262d-4c4d-a55d-9d32ec81ad5d.png)</MultiModal>

There are $2$ different chemical environments for $\mathrm{Bi}$ atoms in the cation cluster in $\bf{C}$.

<Checkpoint>[$3$ is correct.](1)</Checkpoint>

Since the anion given in the question is square pyramidal coordinated, combined with chloride ions, it can be bridged, and it is inferred that it is a mononuclear and binuclear anion, as shown in the figure:

<MultiModal>![$\mathrm{Bi_2Cl_8}^{2-}$: This structure is composed of two square pyramids sharing one side of the base square, and the vertices of the two square pyramids are on both sides of the formed common base. The vertex of the square pyramid is $\mathrm{Cl}$, and the center of the base is $\mathrm{Bi}$. $\mathrm{Bi}$ is five-coordinated, and two $\mathrm{BiCl_5}$ are bridged by 2 $\mathrm{Cl}$. $\mathrm{BiCl}_{5}^{2-}$: Square pyramid structure, $\mathrm{Cl}$ is at the vertex and four corners of the base, and $\mathrm{Bi}$ is five-coordinated at the center of the base square.](/media/uploads/7065114e-a610-4d4d-b6fc-25441c87be8d.png)</MultiModal>

<Checkpoint>[Both contain a $C_2$ axis, $4$ is correct.](1)</Checkpoint>

Therefore, choose O.
由题意可设 $\bf{A}$ 的化学式为 $\mathrm{Bi_m(AlCl_4)_n}$

则根据 $\mathrm{Bi}$ 的质量分数可知

 $m M(\mathrm{Bi})=0.674[mM(\mathrm{Bi})+n M(AlCl_4)]$ 

<Checkpoint>[ $m M(\mathrm{Bi})=0.674[mM(Bi)+n M(\mathrm{AlCl_4})]$ ](1)</Checkpoint>

解上述不定方程, 求得唯一符合题目条件的解

即 $m=5, n=3$, 则 $\mathbf{A}$ 的化学式为 $\mathrm{Bi_5(AlCl_4)_3}$ 

<Checkpoint>[$\mathbf{A}$ 为 $\mathrm{Bi_5(AlCl_4)_3}$ ](1)</Checkpoint>

其阳离子为离子簇，故阴阳离子之比为$1:3$。$1$错误。

由题意可设 $\bf{B}$ 的化学式为 $\mathrm{Bi_m(AlCl_4)_2}$

由于生成相同物质的量的 $\bf{A}$和 $\bf{B}$, 需分别投入的$\mathrm{Bi} $的比例为 $6: 11$
而 $\mathrm{Bi_5^{3+}}$ 可看做 $\mathrm{Bi^{3+}}+4 \mathrm{Bi}$, 则 $\mathrm{Bi_m^{2+}}$ 则可看做 $\frac{2}{3} \mathrm{Bi^{3+}}+\frac{3 m-2}{3} \mathrm{Bi}$

<Checkpoint>[$\mathrm{Bi_m^{2+}}$ 则可看做 $\frac{2}{3} \mathrm{Bi^{3+}}+\frac{3 m-2}{3} \mathrm{Bi}$](1)</Checkpoint>

故可知 $m=4$, 则 $\bf{B}$ 的化学式为 $\mathrm{Bi_8(AlCl_4)_2}$ 

<Checkpoint>[$\bf{B}$ 为 $\mathrm{Bi_8(AlCl_4)_2}$ ](1)</Checkpoint>

离子化合物$\bf{A}$中阳离子比$\bf{B}$中阳离子少$3$个$\mathrm{Bi}$原子。

<Checkpoint>[2正确。](1)</Checkpoint>

由题意可知 $\bf{C}$ 只由 $\mathrm{Bi}$ 和 $\mathrm{Cl}$ 构成

则 $\bf{C}$ 中 $n(\mathrm{Bi}): n(\mathrm{Cl})=\omega(\mathrm{Bi}) M(\mathrm{Cl}) / \omega(\mathrm{Cl}) M(\mathrm{Bi})=6: 7$ 
由价态可知 $\bf{C}$ 应为 $\mathrm{Bi_{24}Cl_{28}}$, 且其中由 $6 $个 $\mathrm{Bi}$ 为$+3$ 价, 其他 $18$ 个 $\mathrm{Bi}$ 一共呈现$+10$

首先可以推断这 $18$ 个 $\mathrm{Bi}$ 应是 $\mathrm{Bi}$ 形成的正离子
其次存在两种可能 $2: 4: 1$ 中的 “ $1$ ”为该正离子 $\mathrm{Bi_18}^{10+}$, “ $2$ ”为该正离子 $\mathrm{Bi_9}^{5+}$ 
而由阴离子中 $\mathrm{Bi}$ 均为四方锥配位可知符合条件的是 $\mathrm{Bi_9}^{5+}$
且剩下的 “ $4 $”对应着 $\mathrm{BiCl_5}^{2-}$, “ $1$ ”对应着 $\mathrm{Bi_2Cl_8}^{2-}$

则 $\bf{C}$ 的化学式为 $\mathrm{(Bi_9^{5+})_{2}(BiCl_5^{2-})_4(Bi_2 Cl_8^{2-})}$

<Checkpoint>[$\bf{C}$ 为 $\mathrm{(Bi_9^{5+})_{2}(BiCl_5^{2-})_4(Bi_2 Cl_8^{2-})}$](1)</Checkpoint>

其阳离子为三加帽三棱柱，为$D_{3h}$点群，如图所示：

<MultiModal>![$\mathrm{Bi_9^{5+}}$阳离子为三加帽三棱柱结构，三棱柱的每一个矩形面延伸出一个顶点形成三加帽三棱柱，每个$\mathrm{Bi}$原子位于该结构的顶点处，为$D_{3h}$点群，](/media/uploads/68895c98-262d-4c4d-a55d-9d32ec81ad5d.png)</MultiModal>

 $\bf{C}$中阳离子簇中$\mathrm{Bi}$原子有$2$种不同的化学环境。

<Checkpoint>[$3$正确。](1)</Checkpoint>

由于题中所给阴离子为四方锥配位，结合氯离子可进行桥连，推知其为单核和双核阴离子，如图所示：

<MultiModal>![$\mathrm{Bi_2Cl_8}^{2-}$：该结构为两个四方锥共用底面正方形的一条边组成，且两个四方锥的顶点处于形成的共同底面的两侧，四方锥的顶点为$\mathrm{Cl}$，底面中心为$\mathrm{Bi}$，$\mathrm{Bi}$为五配位，两个$\mathrm{BiCl_5}$以2个$\mathrm{Cl}$桥连。$\mathrm{BiCl}_{5}^{2-}$：四方锥结构，$\mathrm{Cl}$处于顶点和底面四个角上，$\mathrm{Bi}$为五配位处于底面正方形的中心。](/media/uploads/7065114e-a610-4d4d-b6fc-25441c87be8d.png)</MultiModal>

<Checkpoint>[二者均含有$C_2$轴，$4$正确。](1)</Checkpoint>

故选择O。