Under anhydrous conditions, $PPh_3$ and $CCl_4$ are reacted in a certain proportion to obtain an ionic compound $\bf{A}$. The mass fraction of $Cl$ in $\bf{A}$ is $11.7\%$. $\bf{A}$ is treated with $P(NMe_2)_3$ to obtain compound $\bf{B}$. $\bf{B}$ does not contain nitrogen and the mass fraction of $P$ is $11.6\%$. In the structure of $\bf{B}$, all bond angles centered on carbon atoms are close to $120^\circ$. $\bf{B}$ reacts with an equimolar amount of $AuCl$ to obtain complex $\bf{C}$, and $\bf{C}$ reacts further with an equimolar amount of $AuCl$ to obtain $\bf{D}$. The mass fraction of $Au$ in $\bf{D}$ is $39.3\%$. $\bf{B}$ is treated with 2 equivalents of n-butyllithium at low temperature and then reacted with $BeCl_2$ to obtain $\bf{E}$, which contains 2 rings containing $Be$. In addition, $\bf{B}$ can also be used as a catalyst for certain organic reactions.

Regarding the above content, there are the following statements:
1. All carbon atoms in $\bf{A}$ have the same hybridization.
2. Theoretically, there are 4 sets of peaks in the $^{13}C NMR$ spectrum of $\bf{B}$.
3. $\bf{B}$ can be stored normally in air.
4. $\bf{C}$ does not contain chlorine.
5. The coordination number of $Au$ in $\bf{D}$ is 3.
6. When $\bf{B}$ is used as a catalyst, its function is more likely to be a Lewis acid than a Lewis base.
7. The coordination number of $Be$ in $\bf{E}$ is 4.
8. The theoretical molar ratio of $PPh_3$ to $CCl_4$ when preparing $\bf{A}$ is $2:1$.

Which of the following options corresponds to a set of completely incorrect statements?
无水条件下，将$PPh_3$和$CCl_4$按一定比例反应，得到离子化合物$\bf{A}$，$\bf{A}$中$Cl$的质量分数为$11.7\%$。将$\bf{A}$用$P(NMe_2)_3$处理，得到化合物$\bf{B}$，$\bf{B}$不含氮元素，且$P$的质量分数为$11.6\%$。$\bf{B}$的结构中，所有以碳原子为中心的键角均接近$120^\circ$。将$\bf{B}$与等当量$AuCl$反应可得配合物$\bf{C}$，$\bf{C}$再与等当量的$AuCl$反应得到$\bf{D}$。$\bf{D}$中$Au$的质量分数是$39.3\%$。低温下将$\bf{B}$用2当量的正丁基锂处理，再与$BeCl_2$反应，得到$\bf{E}$，$\bf{E}$中有2个含$Be$的环。此外，$\bf{B}$也可以用作某些有机反应的催化剂。

关于以上内容，有以下几个陈述：
1. $\bf{A}$中所有碳原子的杂化形式相同。
2. $\bf{B}$的$^{13}C NMR$谱中理论上有4组峰。
3. $\bf{B}$可以正常保存在空气中。
4. $\bf{C}$中不含氯元素。
5. $\bf{D}$中$Au$的配位数为3。
6. $\bf{B}$在作为催化剂时，其功能更可能是Lewis酸而非Lewis碱。
7. $\bf{E}$中$Be$的配位数为4。
8. 制备$\bf{A}$时$PPh_3$和$CCl_4$的理论摩尔比是$2:1$。

以下各选项中对应的陈述，哪一组是完全错误的？
All other options are incorrect
1, 3, 5, 7
1, 3, 6, 8
3, 4, 5, 6
1, 2, 7, 8
2, 3, 5, 8
2, 4, 6, 8
2, 3, 4, 5
其他选项均不正确
1，3，5，7
1，3，6，8
3，4，5，6
1，2，7，8
2，3，5，8
2，4，6，8
2，3，4，5
According to organic chemistry knowledge, $PPh_3$ and $CCl_4$ easily react to yield $Ph_3P=CCl_2$ and $Ph_3PCl_2$. The latter is obviously not consistent, and the mass fraction of $Cl$ in the former is too high. Considering its further addition with $PPh_3$, the chemical formula of $\bf{A}$ can be obtained as $(Ph_3P)_2CCl_2$, and the mass fraction is also consistent with the question. Since it is an ionic compound, considering the electron-donating property of $PPh_3$, its actual structure is $[(Ph_3P)_2CCl]^+Cl^-$.

<Checkpoint>[The structure of $\bf{A}$ is $[(Ph_3P)_2CCl]^+Cl^-$](2)</Checkpoint>

$P(NMe_2)_3$ is a reducing agent and should reduce the central carbon atom of $\bf{A}$. Therefore, $\bf{B}$ is $Ph_3P=C=PPh_3$, which also matches the mass fraction in the question.

<Checkpoint>[$\bf{B}$ is $Ph_3P=C=PPh_3$](2)</Checkpoint>

$\bf{C}$ is $(Ph_3P)_2CAuCl$. According to the mass fraction, $\bf{D}$ is $(Ph_3P)_2C(AuCl)_2$.

<Checkpoint>[$\bf{C}$ is $(Ph_3P)_2CAuCl$](1)</Checkpoint>

<Checkpoint>[$\bf{D}$ is $(Ph_3P)_2C(AuCl)_2$](1)</Checkpoint>

n-Butyllithium abstracts one hydrogen each from the $PPh_3$ on both sides of $\bf{B}$. Also note that the central carbon atom of $\bf{B}$ has a strong coordination ability. Therefore, $\bf{E}$ is

<MultiModal>![C1([Be-]2C(C=CC=C3)=C3[P+](C4=CC=CC=C4)1C5=CC=CC=C5)=P(C6=C2C=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8](/media/uploads/e2687f6a-8da7-4816-95a4-6c095e39fcbe.png)</MultiModal>

<Checkpoint>[$\bf{E}$ is <MultiModal>![C1([Be-]2C(C=CC=C3)=C3[P+](C4=CC=CC=C4)1C5=CC=CC=C5)=P(C6=C2C=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8](/media/uploads/e2687f6a-8da7-4816-95a4-6c095e39fcbe.png)</MultiModal>](3)</Checkpoint>

All carbons in $\bf{A}$ are $sp^2$ hybridized. Statement 1 is correct.

There are 5 types of carbon in $\bf{B}$. $\bf{B}$ is a strong Lewis base and is obviously sensitive to water and oxygen. Statements 2, 3, and 6 are incorrect.

$\bf{C}$ contains chlorine element. Statement 4 is incorrect.

In $\bf{D}$, due to the presence of $Au-Au$ interaction, $Au$ is 3-coordinate. Statement 5 is correct.

In $\bf{E}$, $Be$ is 3-coordinate. Statement 7 is incorrect.

The theoretical molar ratio of $PPh_3$ and $CCl_4$ in the preparation of $\bf{A}$ is $3:1$. Statement 8 is incorrect.

<Checkpoint>[Statements 1 and 5 are correct, the rest are incorrect](2)</Checkpoint>

Only the statement in option G is completely incorrect. Choose G.
根据有机化学知识，$PPh_3$和$CCl_4$容易反应得到$Ph_3P=CCl_2$和$Ph_3PCl_2$，后者显然不符合，前者中$Cl$的质量分数过高，考虑其与$PPh_3$的进一步加成，可得$\bf{A}$的化学式是$(Ph_3P)_2CCl_2$，且质量分数也与题目一致。由于是离子化合物，考虑到$PPh_3$的给电子性，所以其实际结构是$[(Ph_3P)_2CCl]^+Cl^-$。

<Checkpoint>[$\bf{A}$的结构是$[(Ph_3P)_2CCl]^+Cl^-$](2)</Checkpoint>

$P(NMe_2)_3$是还原剂，应当还原$\bf{A}$中心的碳原子，因而$\bf{B}$是$Ph_3P=C=PPh_3$，也符合题干的质量分数。

<Checkpoint>[$\bf{B}$是$Ph_3P=C=PPh_3$](2)</Checkpoint>

$\bf{C}$是$(Ph_3P)_2CAuCl$，根据质量分数可知$\bf{D}$是$(Ph_3P)_2C(AuCl)_2$。

<Checkpoint>[$\bf{C}$是$(Ph_3P)_2CAuCl$](1)</Checkpoint>

<Checkpoint>[$\bf{D}$是$(Ph_3P)_2C(AuCl)_2$](1)</Checkpoint>

正丁基锂在$\bf{B}$两边的$PPh_3$上各攫取一个氢，同时注意到$\bf{B}$的中心碳原子具有很强的配位能力，因此$\bf{E}$是

<MultiModal>![C1([Be-]2C(C=CC=C3)=C3[P+](C4=CC=CC=C4)1C5=CC=CC=C5)=P(C6=C2C=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8](/media/uploads/e2687f6a-8da7-4816-95a4-6c095e39fcbe.png)</MultiModal>

<Checkpoint>[$\bf{E}$是<MultiModal>![C1([Be-]2C(C=CC=C3)=C3[P+](C4=CC=CC=C4)1C5=CC=CC=C5)=P(C6=C2C=CC=C6)(C7=CC=CC=C7)C8=CC=CC=C8](/media/uploads/e2687f6a-8da7-4816-95a4-6c095e39fcbe.png)</MultiModal>](3)</Checkpoint>

$\bf{A}$中所有碳都是$sp^2$杂化，陈述1正确。

$\bf{B}$中有5种碳，$\bf{B}$是强Lewis碱，显然对水氧敏感，陈述2，3，6错误。

$\bf{C}$中含氯元素，陈述4错误。

$\bf{D}$中因含$Au-Au$作用，$Au$为3配位，陈述5正确。

$\bf{E}$中$Be$为3配位，陈述7错误。

制备$\bf{A}$时$PPh_3$和$CCl_4$的理论摩尔比是$3:1$，陈述8错误。

<Checkpoint>[陈述1，5正确，其余错误](2)</Checkpoint>

只有选项G中的陈述完全错误。选G。