There is a compound $\mathrm{Bi}$ called $\bf{Y}$:

The cation of $\bf{Y}$ is $\mathrm{Bi^{3+}}$, and $\bf{Y}$ contains only four elements and one anion. The mass fraction of $\mathrm{Bi}$ in $\bf{Y}$ is $\mathrm{27.57}$%, and the mass fraction of $\mathrm{C}$ in $\bf{Y}$ is $\mathrm{28.49}$%.

$\bf{Y}$ can react with the $\mathrm{Ga}$ compound $\bf{X}$ shown in the figure to obtain a compound containing a $-Bi=Bi-$ bonded structure, and another product containing the anion of $\bf{Y}$.

Hint:

1. Generally, the ligands in compounds containing $Bi=Bi$ double bonds are not electrically neutral.
2. The 2,6-diisopropylphenyl (i.e., $Dipp$) group connected to the nitrogen atom in $\bf{X}$ has a very large steric volume.

<MultiModal>![CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga]N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1](/media/uploads/49aa167c-9bee-47eb-8bee-518fb9b2b506.png)</MultiModal>
Which of the following statements are correct:
1. $\bf{Y}$ contains $9$ chlorine atoms
2. All $Ga$ in the products are 4-coordinated
3. The compound containing $Bi=Bi$ double bonds contains $Ga-O$ bonds
4. The byproduct without $Bi$ contains $40$ carbon atoms
5. The stoichiometric ratio of $\bf{Y}$ to $\bf{X}$ is $1:1$
有一种$\mathrm{Bi}$的化合物$\bf{Y}$：

$\bf{Y}$阳离子为$\mathrm{Bi^{3+}}$，$\bf{Y}$仅含有四种元素和一种阴离子。$\bf{Y}$含$\mathrm{Bi}$质量分数为$\mathrm{27.57}$%，含$\mathrm{C}$质量分数为$\mathrm{28.49}$%。

$\bf{Y}$可以与如图所示的$\mathrm{Ga}$化合物$\bf{X}$反应，得到一种含有$-Bi=Bi-$键联结构的化合物，以及另一种含有$\bf{Y}$的阴离子的产物。

提示：

1. 一般含有$Bi=Bi$双键的化合物中配体不是电中性的。
2. $\bf{X}$中连接在氮原子上的2,6-二异丙基苯基(即$Dipp$)基团空间体积十分巨大。

<MultiModal>![CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga]N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1](/media/uploads/49aa167c-9bee-47eb-8bee-518fb9b2b506.png)</MultiModal>
下面说法中哪些是正确的：
1. $\bf{Y}$中含有$9$个氯原子
2. 所有产物中的$Ga$均为4配位
3. 含有$Bi=Bi$双键的化合物中含有$Ga-O$键
4. 不含$Bi$的副产物中含有$40$个碳原子
5. $\bf{Y}$与$\bf{X}$的计量比为$1：1$
All other options are incorrect
1.2.3.
1.3.
2.3.
1.3.4.
3.4.
3.5.
1.5.
2.5.
1.2.4.
其他选项均不正确
1.2.3.
1.3.
2.3.
1.3.4.
3.4.
3.5.
1.5.
2.5.
1.2.4.
According to the question, assuming $\bf{Y}$ contains one $\mathrm{Bi}$, the number of carbons it contains is:
\[
  M_Y = \frac{209}{0.2757} \approx 758 \text{ g/mol}
  \]
\[
  m_C = 758 \times 0.2849 \approx 216 \text{ g/mol}
  \]
 \[
  n_C = \frac{216}{12} = 18
  \]

<Checkpoint>[Assuming $\bf{Y}$ contains one $\mathrm{Bi}$, it contains 18 carbons](1)</Checkpoint>

The remaining molecular weight is: $758-18 \times 12-209=333$

Consider the element to be $Bi$, which may have three identical ligands, each ligand having $6$ $C$s, and a remaining molecular weight of 111, which may be $5$ $F$s and one $O$.

<Checkpoint>[The remaining each ligand has $5$ $F$s and one $O$](1)</Checkpoint>

Therefore, $\bf{Y}$ is $Bi(OC_6F_5)_3$

<Checkpoint>[$\bf{Y}$ is $Bi(OC_6F_5)_3$](2)</Checkpoint>

Therefore, 1 is wrong.

Consider the change in valence: According to the bond structure, $Bi$ is in the +1 valence state and is reduced; in the substrate $\bf{X}$, $Ga$ is in the +1 valence state, so it should act as a reducing agent and is generally oxidized to the stable oxidation state $Ga(III)$. To balance the compound's charge, considering that the product contains the anion of $\bf{Y}$, the oxidation product should be $GaL(OC_6F_5)_2$, containing 41 carbon atoms, so statement 4 is wrong.

<Checkpoint>[$Bi$ is in the +1 valence state and is reduced](1)</Checkpoint>

<Checkpoint>[$Ga$ is in the +1 valence state, so it should act as a reducing agent and is generally oxidized to the stable oxidation state $Ga(III)$](1)</Checkpoint>

<Checkpoint>[The oxidation product should be $GaL(OC_6F_5)_2$](1)</Checkpoint>

Note that the premise "$Dipp$" is a structure with large steric hindrance. If the organic ligand $L$ of $Ga$ in $\bf{X}$ is directly coordinated with the $Bi=Bi$ double bond to form $L-Bi=Bi-L$, then the spatial distance between the two ligands $L$ is too short, so it cannot be formed due to spatial hindrance. Therefore, the ligand $L$ can only continue to coordinate with $Ga$; therefore, the $Bi=Bi$ double bond is likely to be connected to $Ga$. Therefore, the structure of the reduction product is $L(OC_6F_5)Ga-Bi=Bi-Ga(OC_6F_5)L$. This structure contains an $O-Ga$ bond, so statement 3 is correct.

<Checkpoint>[The spatial distance between the two ligands $L$ of $L-Bi=Bi-L$ is too short, so it cannot be formed due to spatial hindrance](1)</Checkpoint>

<Checkpoint>[The ligand $L$ can only continue to coordinate with $Ga$; the $Bi=Bi$ double bond is likely to be connected to $Ga$](1)</Checkpoint>

<Checkpoint>[The structure of the reduction product is $L(OC_6F_5)Ga-Bi=Bi-Ga(OC_6F_5)L$](1)</Checkpoint>

Therefore, the two product structures that can be obtained are:
CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])(OC#CC#CC#C(F)(F)(F)(F)F)N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1

<Checkpoint>[CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])(OC#CC#CC#C(F)(F)(F)(F)F)N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1](1)</Checkpoint>

CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])([Bi]=[Bi][Ga]3(O[C6F5])N(C4=C(C(C)C)C=CC=C4C(C)C)C(C)=CC(C)=[N]3C5=C(C(C)C)C=CC=C5C(C)C)N(C6=C(C(C)C)C=CC=C6C(C)C)C(C)=C1

<Checkpoint>[CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])([Bi]=[Bi][Ga]3(O[C6F5])N(C4=C(C(C)C)C=CC=C4C(C)C)C(C)=CC(C)=[N]3C5=C(C(C)C)C=CC=C5C(C)C)N(C6=C(C(C)C)C=CC=C6C(C)C)C(C)=C1](1)</Checkpoint>

<MultiModal>![CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])(OC#CC#CC#C(F)(F)(F)(F)F)N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1](/media/uploads/bff9cd12-69aa-4422-9dee-c8c62d8a81b1.png)</MultiModal>

<MultiModal>![CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])([Bi]=[Bi][Ga]3(O[C6F5])N(C4=C(C(C)C)C=CC=C4C(C)C)C(C)=CC(C)=[N]3C5=C(C(C)C)C=CC=C5C(C)C)N(C6=C(C(C)C)C=CC=C6C(C)C)C(C)=C1](/media/uploads/9548101b-5590-4a24-bc2a-94310620a5ff.png)</MultiModal>

All products are 4-coordinate $Ga$, so statement 2 is correct; both products contain $Ga$, so the reaction stoichiometric ratio should be 2:3, so statement 5 is wrong.

<Checkpoint>[The stoichiometric ratio of $\bf{Y}$ to $\bf{X}$ is $2:3$](1)</Checkpoint>

Therefore, 2 and 3 are correct, choose D.
由题意可知，假设$\bf{Y}$中含有一个$\mathrm{Bi}$，则含有碳的个数为：
\[
  M_Y = \frac{209}{0.2757} \approx 758 \text{ g/mol}
  \]
\[
  m_C = 758 \times 0.2849 \approx 216 \text{ g/mol}
  \]
 \[
  n_C = \frac{216}{12} = 18
  \]

<Checkpoint>[假设$\bf{Y}$中含有一个$\mathrm{Bi}$，则含有碳18个$](1)</Checkpoint>

余下的分子量为：$758-18 \times 12-209=333$

考虑该元素为$Bi$，可能具有三个相同的配体，每个配体上有$6$个$C$,余下111的分子量，可能为$5$个$F$和一个$O$

<Checkpoint>[余下每个配体有$5$个$F$和一个$O$](1)</Checkpoint>

因此，$\bf{Y}$为$Bi(OC_6F_5)_3$

<Checkpoint>[$\bf{Y}$为$Bi(OC_6F_5)_3$](2)</Checkpoint>

因此，1错误。

考虑化合价的变化：根据键联结构可知$Bi$为+1价态，被还原；底物$\bf{X}$中$Ga$为+1价态，从而应当作为还原剂，一般氧化到稳定氧化态$Ga(III)$，为配平化合物电荷，考虑到产物含有$\bf{Y}$的阴离子，氧化产物应当为$GaL(OC_6F_5)_2$，含有41个碳原子，说法4错误。

<Checkpoint>[$Bi$为+1价态，被还原](1)</Checkpoint>

<Checkpoint>[$Ga$为+1价态，从而应当作为还原剂，一般氧化到稳定氧化态$Ga(III)$](1)</Checkpoint>

<Checkpoint>[氧化产物应当为$GaL(OC_6F_5)_2$](1)</Checkpoint>

注意到题设“$Dipp$”是空间位阻很大的结构，如果将$\bf{X}$中$Ga$的有机配体$L$直接与$Bi=Bi$双键配合形成$L-Bi=Bi-L$，此时两个配体$L$之间的空间距离太短，因为空间阻碍无法形成。因此，配体$L$只能与$Ga$继续配位；从而$Bi=Bi$双键大概率与$Ga$相连。因此还原产物结构为$L(OC_6F_5)Ga-Bi=Bi-Ga(OC_6F_5)L$。该结构存在$O-Ga$键，说法3正确。

<Checkpoint>[$L-Bi=Bi-L$的两个配体$L$之间的空间距离太短，因为空间阻碍无法形成](1)</Checkpoint>

<Checkpoint>[配体$L$只能与$Ga$继续配位；$Bi=Bi$双键大概率与$Ga$相连](1)</Checkpoint>

<Checkpoint>[还原产物结构为$L(OC_6F_5)Ga-Bi=Bi-Ga(OC_6F_5)L$](1)</Checkpoint>

因此，可得到的两个产物结构分别为：
CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])(OC#CC#CC#C(F)(F)(F)(F)F)N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1

<Checkpoint>[CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])(OC#CC#CC#C(F)(F)(F)(F)F)N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1](1)</Checkpoint>

CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])([Bi]=[Bi][Ga]3(O[C6F5])N(C4=C(C(C)C)C=CC=C4C(C)C)C(C)=CC(C)=[N]3C5=C(C(C)C)C=CC=C5C(C)C)N(C6=C(C(C)C)C=CC=C6C(C)C)C(C)=C1

<Checkpoint>[CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])([Bi]=[Bi][Ga]3(O[C6F5])N(C4=C(C(C)C)C=CC=C4C(C)C)C(C)=CC(C)=[N]3C5=C(C(C)C)C=CC=C5C(C)C)N(C6=C(C(C)C)C=CC=C6C(C)C)C(C)=C1](1)</Checkpoint>

<MultiModal>![CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])(OC#CC#CC#C(F)(F)(F)(F)F)N(C3=C(C(C)C)C=CC=C3C(C)C)C(C)=C1](/media/uploads/bff9cd12-69aa-4422-9dee-c8c62d8a81b1.png)</MultiModal>

<MultiModal>![CC1=[N](C2=C(C(C)C)C=CC=C2C(C)C)[Ga](O[C6F5])([Bi]=[Bi][Ga]3(O[C6F5])N(C4=C(C(C)C)C=CC=C4C(C)C)C(C)=CC(C)=[N]3C5=C(C(C)C)C=CC=C5C(C)C)N(C6=C(C(C)C)C=CC=C6C(C)C)C(C)=C1](/media/uploads/9548101b-5590-4a24-bc2a-94310620a5ff.png)</MultiModal>

所有产物均为4配位的$Ga$，说法2正确；两种产物均含有$Ga$，反应计量比应为2：3，说法5错误。

<Checkpoint>[$\bf{Y}$与$\bf{X}$的计量比为$2：3$](1)</Checkpoint>

因此，2.3正确，选择D。