Rhodium(III) chloride trihydrate $( \mathrm{RhCl_3{\cdot}3H_2O})$ is a red solid, but its composition is relatively complex; column chromatographic separation and spectroscopic studies of its aqueous solution reveal the presence of two hexacoordinate mononuclear complex molecules, $\mathbf{A_1,A_2}$; in $\mathbf{A_1}$, the chemical environment of the same kind of atom is the same. The reaction of $\mathrm{RhCl_3\cdot 3H_2O}$ with $\mathrm{CO}$ gas flow yields $\mathbf{B}$; $\mathbf{B}$ is a dinuclear complex molecule, $\mathrm{Rh}$ is 4-coordinate, $+1$ oxidation state, and the ligands are only $\mathrm{Cl^-}$ and $\mathrm{CO}$. $\mathbf{B}$ reacts with tetramethylammonium iodide to obtain $\mathbf{C}$; $\mathbf{C}$ is a 1:1 type salt, $\mathrm{Rh}$ is in the $+1$ oxidation state, and the ligands are only $\mathrm{I^-}$ and $\mathrm{CO}$. $\mathbf{B}$ is a diamagnetic species.

Using the anion of $\mathbf{C}$ as a catalyst and iodomethane as a co-catalyst, the preparation of acetic acid can be achieved: $\mathrm{CH_3OH+CO\to CH_3COOH}$ . In the reaction, the anion of $\mathbf{C}$ undergoes an oxidative addition reaction with iodomethane to obtain $\mathbf{D}$; $\mathbf{D}$ undergoes methyl migration to obtain $\mathbf{E}$; $\mathbf{E}$ coordinates with $\mathrm{CO}$ to obtain $\mathbf{F}$; $\mathbf{F}$ undergoes reductive elimination, releasing the small molecule $\mathbf{G}$ and regenerating $\mathbf{C}$; $\mathbf{G}$ reacts with water to produce acetic acid and $\mathrm{HI}$; $\mathrm{HI}$ reacts with methanol to produce iodomethane and water.

There are the following statements:

1, Ignoring hydrogen atoms, the point groups to which $\mathbf{A_1,A_2,B}$ belong are $C_{2v},C_{3v},D_{2h}$, respectively.

2, The theoretical magnetic moments of $\mathbf{A_1,C}$ are 0 and 2.83 B.M., respectively.

3, The oxidation states of $\mathrm{Rh}$ in $\mathbf{D,E,F}$ are not all the same.

4, The molecular weight of $\mathbf{G}$ is 297.9.

Then the option that contains all the correct statements is:
三水合三氯化铑$( \mathrm{RhCl_3{\cdot}3H_2O})$ 是一种红色固体，但其组成比较复杂；对其水溶液进行柱色谱分离和光谱研究，发现其中包含$\mathbf{A_1,A_2}$两种六配位单核配合物分子；$\mathbf{A_1}$中同种原子化学环境相同。$\mathrm{RhCl_3\cdot 3H_2O}$与$\mathrm{CO}$气流反应，得到$\mathbf{B}$；$\mathbf{B}$是双核配合物分子，$\mathrm{Rh}$为4配位，$+1$氧化态，配体只有$\mathrm{Cl^-}$和$\mathrm{CO}$。$\mathbf{B}$与四甲基碘化铵反应得到$\mathbf{C}$；$\mathbf{C}$为1:1型盐，$\mathrm{Rh}$为 $+1$ 氧化态，配体只有$\mathrm{I^-}$和$\mathrm{CO}$。$\mathbf{B}$为抗磁性物种。

以$\mathbf{C}$的阴离子为催化剂，以碘甲烷为助催化剂，可以实现醋酸的制备：$\mathrm{CH_3OH+CO\to CH_3COOH}$ 。反应中，$\mathbf{C}$的阴离子与碘甲烷发生氧化加成反应得到$\mathbf{D}$；$\mathbf{D}$发生甲基迁移得到$\mathbf{E}$；$\mathbf{E}$与$\mathrm{CO}$配位得到$\mathbf{F}$；$\mathbf{F}$发生还原消除，脱去小分子$\mathbf{G}$并得回$\mathbf{C}$；$\mathbf{G}$遇水生成醋酸和$\mathrm{HI}$；$\mathrm{HI}$与甲醇反应生成碘甲烷和水。

有以下几个说法：

1，忽略氢原子，则$\mathbf{A_1,A_2,B}$所属的点群分别为$C_{2v},C_{3v},D_{2h}$。

2，$\mathbf{A_1,C}$的理论磁矩分别为0，2.83 B.M. 。

3，$\mathbf{D,E,F}$中 $\mathrm{Rh}$的化合价不完全相同。

4，$\mathbf{G}$的分子量为297.9。

则下列选项中包含所有正确说法的选项为：
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
$\mathrm{RhCl_3{\cdot}3H_2O}$ exists in both mer and fac isomers, where the fac isomer has the same chemical environment for the same type of atoms. Therefore, $\mathbf{A_1}$ is the fac isomer, and $\mathbf{A_2}$ is the mer isomer.

<Checkpoint>[$\mathbf{A_1}$ is the <MultiModal>![fac structure of $\mathrm{RhCl_3{\cdot}3H_2O}$](/media/uploads/5bbc2b51-a462-4ecd-8369-c780d834fb9e.png)</MultiModal>](1)</Checkpoint>

<Checkpoint>[$\mathbf{A_2}$ is the <MultiModal>![mer structure of $\mathrm{RhCl_3{\cdot}3H_2O}$](/media/uploads/65645dec-d3e5-4d95-bdea-0d392adc0fae.png)</MultiModal>](1)</Checkpoint>

The reaction of $\mathrm{RhCl_3\cdot 3H_2O}$ with $\mathrm{CO}$ gas flow yields $\mathbf{B}$, which is a dinuclear complex. $\mathrm{Rh}$ is 4-coordinated, in a $+1$ oxidation state, and $\mathbf{B}$ is a diamagnetic species; therefore, the coordination geometry is square planar. Thus, the structure of $\mathbf{B}$ can be deduced as

<MultiModal>![O=C[Rh]1(Cl[Rh](C=O)(Cl1)C=O)C=O](/media/uploads/3dae8ddf-e07f-4439-884c-f18ae1271b24.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$ is <MultiModal>![O=C[Rh]1(Cl[Rh](C=O)(Cl1)C=O)C=O](/media/uploads/3dae8ddf-e07f-4439-884c-f18ae1271b24.png)</MultiModal>](2)</Checkpoint>

Therefore, the point groups to which $\mathbf{A_1,A_2,B}$ belong are $C_{3v},C_{2v},D_{2h}$ respectively. Statement 1 is incorrect.

$\mathbf{B}$ reacts with tetramethylammonium iodide to yield $\mathbf{C}$; $\mathbf{C}$ is a 1:1 type salt, $\mathrm{Rh}$ is in the $+1$ oxidation state, and the ligands are only $\mathrm{I^-}$ and $\mathrm{CO}$. Therefore, there are two $\mathrm{I^-}$ and two $\mathrm{CO}$ ligands. Based on the positional relationship of $\mathrm{CO}$ in $\mathbf{B}$, the structure of $\mathbf{C}$ can be obtained as

<MultiModal>![I[Rh-](I)([C]=O)[C]=O，the two CO groups are in cis position](/media/uploads/5ba248d7-e2ef-4fc4-9c40-35e1649583af.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$ does not have any unpaired electrons](1)</Checkpoint>

The theoretical magnetic moments of $\mathbf{A_1,C}$ are both 0. Statement 2 is incorrect.

$\mathbf{C}$ undergoes oxidative addition with iodomethane, readily yielding $\mathbf{D}$ with the chemical formula $\mathrm{[Rh(CO)_2I_3(CH_3)]^-}$. $\mathbf{D}$ undergoes methyl migration to yield $\mathrm{[Rh(CO)I_3(CH_3CO)]^-}$; $\mathrm{[Rh(CO)I_3(CH_3CO)]^-}$ coordinates with $\mathrm{CO}$ to yield $\mathrm{[Rh(CO)_2I_3(CH_3CO)]^-}$; $\mathrm{[Rh(CO)_2I_3(CH_3CO)]^-}$ undergoes reductive elimination, eliminating the small molecule $\mathrm{CH_3COI}$ and regenerating $\mathbf{C}$.

<Checkpoint>[$\mathbf{D}$ is $\mathrm{[Rh(CO)_2I_3(CH_3)]^-}$, $\mathbf{E}$ is $\mathrm{[Rh(CO)I_3(CH_3CO)]^-}$, $\mathbf{F}$ is $\mathrm{[Rh(CO)_2I_3(CH_3CO)]^-}$](3)</Checkpoint>

<Checkpoint>[$\mathbf{G}$ is $\mathrm{CH_3COI}$](1)</Checkpoint>

The oxidation state of $\mathrm{Rh}$ in $\mathbf{D,E,F}$ is all +3, and the molecular weight of $\mathbf{G}$ is 169.9. Statements 3 and 4 are incorrect.

Choose A
$\mathrm{RhCl_3{\cdot}3H_2O}$存在经式和面式两种异构体，其中面式中同种原子化学环境相同。因此$\mathbf{A_1}$为面式，$\mathbf{A_2}$为经式。

<Checkpoint>[$\mathbf{A_1}$为<MultiModal>![$\mathrm{RhCl_3{\cdot}3H_2O}$的面式结构](/media/uploads/5bbc2b51-a462-4ecd-8369-c780d834fb9e.png)</MultiModal>](1)</Checkpoint>

<Checkpoint>[$\mathbf{A_2}$为<MultiModal>![$\mathrm{RhCl_3{\cdot}3H_2O}$的经式结构](/media/uploads/65645dec-d3e5-4d95-bdea-0d392adc0fae.png)</MultiModal>](1)</Checkpoint>

$\mathrm{RhCl_3\cdot 3H_2O}$与$\mathrm{CO}$气流反应得到$\mathbf{B}$是双核配合物，$\mathrm{Rh}$为4配位，$+1$氧化态且$\mathbf{B}$为抗磁性物种，因此配位构型为平面四边形。从而可以得出$\mathbf{B}$的结构为

<MultiModal>![O=C[Rh]1(Cl[Rh](C=O)(Cl1)C=O)C=O](/media/uploads/3dae8ddf-e07f-4439-884c-f18ae1271b24.png)</MultiModal>

<Checkpoint>[$\mathbf{B}$为<MultiModal>![O=C[Rh]1(Cl[Rh](C=O)(Cl1)C=O)C=O](/media/uploads/3dae8ddf-e07f-4439-884c-f18ae1271b24.png)</MultiModal>](2)</Checkpoint>

因此$\mathbf{A_1,A_2,B}$所属的点群分别为$C_{3v},C_{2v},D_{2h}$，说法1错误。

$\mathbf{B}$与四甲基碘化铵反应得到$\mathbf{C}$；$\mathbf{C}$为1:1型盐，$\mathrm{Rh}$为 $+1$ 氧化态，配体只有$\mathrm{I^-}$和$\mathrm{CO}$，因此配体中有两个$\mathrm{I^-}$和两个$\mathrm{CO}$，根据$\mathbf{B}$中$\mathrm{CO}$的位置关系就可以得到$\mathbf{C}$的结构为

<MultiModal>![I[Rh-](I)([C]=O)[C]=O，两个$CO$处于顺式](/media/uploads/5ba248d7-e2ef-4fc4-9c40-35e1649583af.png)</MultiModal>

<Checkpoint>[$\mathbf{C}$中没有单电子](1)</Checkpoint>

$\mathbf{A_1,C}$的理论磁矩均为0，说法2错误。

$\mathbf{C}$与碘甲烷发生氧化加成，容易得到$\mathbf{D}$化学式为$\mathrm{[Rh(CO)_2I_3(CH_3)]^-}$,$\mathbf{D}$发生甲基迁移得到$\mathrm{[Rh(CO)I_3(CH_3CO)]^-}$；$\mathrm{[Rh(CO)I_3(CH_3CO)]^-}$与$\mathrm{CO}$配位得到$\mathrm{[Rh(CO)_2I_3(CH_3CO)]^-}$；$\mathrm{[Rh(CO)_2I_3(CH_3CO)]^-}$发生还原消除，脱去小分子$\mathrm{CH_3COI}$并得回$\mathbf{C}$。

<Checkpoint>[$\mathbf{D}$为$\mathrm{[Rh(CO)_2I_3(CH_3)]^-}$，$\mathbf{E}$为$\mathrm{[Rh(CO)I_3(CH_3CO)]^-}$，$\mathbf{F}$为$\mathrm{[Rh(CO)_2I_3(CH_3CO)]^-}$](3)</Checkpoint>

<Checkpoint>[$\mathbf{G}$为$\mathrm{CH_3COI}$](1)</Checkpoint>

$\mathbf{D,E,F}$中 $\mathrm{Rh}$的化合价均为+3，$\mathbf{G}$的分子量为169.9。说法3和说法4错误

选A