$Mn$ carbonyl compounds and their derivatives are rich and varied. Reacting a $D_{4d}$ point group carbonyl compound $\bf{X}$ of $Mn$ with $Na$ yields a 1:1 ionic compound $\bf{A}$. $\bf{A}$ reacts with allyl chloride to produce $\bf{B}$. Heating $\bf{B}$ results in the loss of one molecule of a toxic gas, yielding $\bf{C}$. Simultaneously, $\bf{B}$ reacts with perchloric acid to produce an ionic compound $\bf{D}$. $\bf{A}$ can react with 1,3-dibromopropane to produce a neutral complex $\bf{E}$, and it can also react with $BrCH_{2}CH_{2}SMe$ to produce $\bf{F}$ (hint: one molecule of a toxic gas is also generated). $\bf{A}$ reacts with phosphoric acid to produce a mononuclear complex $ G $ with a four-fold axis. $ G $ further transforms into a trinuclear complex $\bf{H}$. $\bf{G}$ can react with diazomethane in a 1:1 ratio to produce $\bf{I}$. $\bf{X}$ can react with dimethylamine to produce a 1:2 ionic compound $\bf{J}$, while also generating a common solvent $K$. The mass fraction of the Mn element in $\bf{J}$ is $2 3 . 0 4 \%$ , and $\bf{J}$ and $\bf{A}$ have the same anion.

The following statements are made about $\bf{A}$-$K$:

1.  Both $\bf{A}$ and $\bf{B}$ satisfy the EAN rule.

2.  $\bf{C}$ does not satisfy the EAN rule.

3.  The cation in $\bf{D}$ has a $C_4$ axis.

4.  The molar mass of $\bf{E}$ is approximately 317 g/mol.

5.  $\bf{G}$ and $\bf{I}$ have the same highest order of rotational axis.

6.  $\bf{H}$ belongs to the $D_{3h}$ point group.

7.  If carbon and hydrogen atoms are ignored, the cation of $\bf{J}$ belongs to the $O_h$ point group.

Let the sum of the numbers of the correct statements be a, and the smallest number of the correct statements be b, then a and b are respectively:
$Mn$的羰基化合物及其衍生物丰富多彩。将$Mn$的 $D_{4d}$ 点群的羰基化合物$\bf{X}$与$Na$反应，得到1:1离子化合物$\bf{A}$，$\bf{A}$与烯丙基氯反应得到$\bf{B}$，$\bf{B}$加热失去一分子有毒气体得到$\bf{C}$；同时$\bf{B}$与高氯酸反应得到离子化合物$\bf{D}$。$\bf{A}$可以与1,3-二溴丙烷反应得到电中性配合物$\bf{E}$，还可以与$BrCH_{2}CH_{2}SMe$ 反应得到$\bf{F}$（提示：还生成了一分子有毒气体)。$\bf{A}$与磷酸反应得到具有四次轴的单核配合物 $ G $ , $ G $ 进一步转化得到三核配合物$\bf{H}$，$\bf{G}$可以与重氮甲烷1：1反应得到$\bf{I}$。$\bf{X}$可以与二甲胺反应得到1:2的离子化合物$\bf{J}$，同时产生了某种常见溶剂$K$。$\bf{J}$中Mn元素的质量分数为 $2 3 . 0 4 \%$ ，$\bf{J}$与$\bf{A}$具有相同的阴离子。

有下列关于$\bf{A}$-$K$的说法：

1，$\bf{A}$和$\bf{B}$中的$Mn$都满足EAN规则。

2，$\bf{C}$中的$Mn$不满足EAN规则。

3，$\bf{D}$中的阳离子具有$C_4$轴。

4，$\bf{E}$的摩尔质量约为317 g/mol。

5，$\bf{G}$和$\bf{I}$的最高旋转轴次相同。

6，$\bf{H}$属于$D_{3h}$点群。

7，若忽略碳原子和氢原子，则$\bf{J}$的阳离子属于$O_h$点群。

设其中正确的说法序号之和为a，正确说法的序号的最小值为b，则a和b分别为：
12，1
12，2
12，3
13，1
13，2
13，3
14，1
14，2
14，3
15，1
15，2
15，3
21，1
21，2
21，3
22，1
22，2
22，3
23，1
23，2
23，3
12，1
12，2
12，3
13，1
13，2
13，3
14，1
14，2
14，3
15，1
15，2
15，3
21，1
21，2
21，3
22，1
22，2
22，3
23，1
23，2
23，3
$\bf{X}$ is a carbonyl compound of $Mn$ and the molecular point group is $D_{4d}$, then it is easy to know that $\bf{X}$ is the dinuclear complex of $Mn$, $Mn_2(CO)_{10}$.

<Checkpoint>[$\bf{X}$ is $Mn_2(CO)_{10}$](1)</Checkpoint>

The ionic compound obtained from the reaction of $Mn_2(CO)_{10}$ with $Na$ is $NaMn(CO)_5$.

<Checkpoint>[$\bf{A}$ is $NaMn(CO)_5$](1)</Checkpoint>

$NaMn(CO)_5$ can undergo a substitution reaction with allyl chloride, and the product is $Mn(CO)_5(CH_2CH=CH2)$.

<Checkpoint>[$\bf{B}$ is $Mn(CO)_5(CH_2CH=CH2)$](1)</Checkpoint>

From $\bf{B}$ to $\bf{C}$, the double bond of allyl undergoes ligand substitution reaction with carbonyl, and the product is $Mn(CO)_4(CH_2CHCH2)$.

<Checkpoint>[$\bf{C}$ is $Mn(CO)_4(CH_2CHCH2)$](1)</Checkpoint>

Therefore, $\bf{A,B,C}$ all satisfy the EAN rule. Statement 1 is correct, and 2 is wrong.

$\bf{B}$ reacts with perchloric acid to undergo a protonation reaction on allyl, generating $[Mn(CO)_5(CH_2=CHCH_3)]^+ClO_4^-$.

<Checkpoint>[$\bf{D}$ is $[Mn(CO)_5(CH_2=CHCH_3)]^+ClO_4^-$](1)</Checkpoint>

The propene in $\bf{D}$ destroys the symmetry, so that the cation does not have a $C_4$ axis, statement 3 is wrong.

1,3-dibromopropane and $BrCH_{2}CH_{2}SMe$ can both undergo substitution reactions with $\bf{A}$. Since 1,3-dibromopropane has two leaving groups, and $BrCH_{2}CH_{2}SMe$ has only one leaving group, 1,3-dibromopropane can undergo two substitution reactions, and $BrCH_{2}CH_{2}SMe$ can only undergo one substitution reaction, and the S atom in it can undergo intramolecular ligand substitution reaction after the reaction, so the products are $(CO)_5MnCH_2CH_2CH_2Mn(CO)_5$ and $Mn(CO)_4(CH2CH_2SCH_3)$ respectively.

<Checkpoint>[$\bf{E}$ is $(CO)_5MnCH_2CH_2CH_2Mn(CO)_5$](1)</Checkpoint>

<Checkpoint>[$\bf{F}$ is $Mn(CO)_4(CH2CH_2SCH_3)$](1)</Checkpoint>

The calculated molar mass of $\bf{E}$ is 432 g/mol, statement 4 is wrong.

$\bf{A}$ undergoes a protonation reaction with phosphoric acid, and the product is $Mn(CO)_5H$.

<Checkpoint>[$\bf{G}$ is $Mn(CO)_5H$](1)</Checkpoint>

Three molecules of $\bf{G}$ each lose one carbonyl group and can trimerize to form a trinuclear complex $\bf{H}$, whose chemical formula is $[Mn(CO)_4H]_3$, and its structure is <MultiModal>![This is a color diagram of a chemical structure, where all text and lines are red. At the center of the image is a triangular structure composed of three "Mn" symbols. These three "Mn" symbols are connected to each other by solid red lines, in pairs. Additionally, there are three "H" symbols in the diagram, each located between two "Mn" symbols and connected to these two adjacent "Mn" symbols by a single solid red line. In addition to these connections, each "Mn" symbol is connected to other chemical groups: The "Mn" symbol at the top vertex of the triangle connects to two "OC" groups to its top-left and two "CO" groups to its top-right; The "Mn" symbol on the left side of the triangle connects to four "OC" groups, located at its top-left, left, bottom-left, and directly below it; The "Mn" symbol on the right side of the triangle connects to four "CO" groups, located at its top-right, right, bottom-right, and directly below it.](/media/uploads/c34f90ae-bc59-4f22-8c22-5761f3730838.png)</MultiModal>

<Checkpoint>[$\bf{H}$ is $[Mn(CO)_4H]_3$](2)</Checkpoint>

$\bf{H}$ belongs to the $D_{3h}$ point group, statement 6 is correct.

$\bf{G}$ reacts with diazomethane to lose nitrogen, yielding $Mn(CO)_5CH_3$.

<Checkpoint>[$\bf{I}$ is $Mn(CO)_5CH_3$](1)</Checkpoint>

The highest rotation axis order of $\bf{G}$ is 4, while $\bf{I}$ loses its rotational symmetry due to the presence of methyl, statement 5 is wrong.

$Mn_2(CO)_{10}$ reacts with dimethylamine to obtain a 1:2 type ionic compound. Through mass fraction calculation, $\bf{J}$ is $[Mn(NHMe_2)_6^{2+}][Mn(CO)_5^-]_2$.

<Checkpoint>[$\bf{J}$ is $[Mn(NHMe_2)_6^{2+}][Mn(CO)_5^-]_2$](2)</Checkpoint>

Ignoring carbon and hydrogen atoms, the cation of $\bf{J}$ can be regarded as a regular octahedron formed by $Mn$ and $N$, which is the $O_h$ point group, statement 7 is correct.

Therefore, choose option G
$\bf{X}$为$Mn$的羰基化合物且分子点群为$D_{4d}$，则易有$\bf{X}$为$Mn$的双核配合物$Mn_2(CO)_{10}$。

<Checkpoint>[$\bf{X}$为$Mn_2(CO)_{10}$](1)</Checkpoint>

$Mn_2(CO)_{10}$与$Na$反应得到的离子化合物为$NaMn(CO)_5$。

<Checkpoint>[$\bf{A}$为$NaMn(CO)_5$](1)</Checkpoint>

$NaMn(CO)_5$与烯丙基氯可以发生取代反应，产物为$Mn(CO)_5(CH_2CH=CH2)$。

<Checkpoint>[$\bf{B}$为$Mn(CO)_5(CH_2CH=CH2)$](1)</Checkpoint>

$\bf{B}$到$\bf{C}$发生烯丙基的双键对羰基的配体取代反应，产物为$Mn(CO)_4(CH_2CHCH2)$。

<Checkpoint>[$\bf{C}$为$Mn(CO)_4(CH_2CHCH2)$](1)</Checkpoint>

因此$\bf{A,B,C}$均满足EAN规则。说法1正确，2错误。

$\bf{B}$与高氯酸反应可以在烯丙基上发生质子化反应，生成$[Mn(CO)_5(CH_2=CHCH_3)]^+ClO_4^-$。

<Checkpoint>[$\bf{D}$为$[Mn(CO)_5(CH_2=CHCH_3)]^+ClO_4^-$](1)</Checkpoint>

$\bf{D}$中的丙烯破坏了对称性，使得阳离子没有$C_4$轴，说法3错误。

1,3-二溴丙烷和$BrCH_{2}CH_{2}SMe$均可以与$\bf{A}$ 发生取代反应，由于1,3-二溴丙烷上有两个离去基团，$BrCH_{2}CH_{2}SMe$上只有一个离去基团，因此1,3-二溴丙烷可以发生两次取代反应，$BrCH_{2}CH_{2}SMe$只能发生一次取代反应，且其中的S原子在反应后可以在分子内发生配体取代反应，因此产物分别为$(CO)_5MnCH_2CH_2CH_2Mn(CO)_5$和$Mn(CO)_4(CH2CH_2SCH_3)$。

<Checkpoint>[$\bf{E}$为$(CO)_5MnCH_2CH_2CH_2Mn(CO)_5$](1)</Checkpoint>

<Checkpoint>[$\bf{F}$为$Mn(CO)_4(CH2CH_2SCH_3)$](1)</Checkpoint>

计算得到$\bf{E}$的摩尔质量为432 g/mol，说法4错误。

$\bf{A}$与磷酸发生质子化反应，产物为$Mn(CO)_5H$。

<Checkpoint>[$\bf{G}$为$Mn(CO)_5H$](1)</Checkpoint>

三分子$\bf{G}$各脱去一个羰基后可以三聚从而形成三核配合物$\bf{H}$，其化学式为$[Mn(CO)_4H]_3$，其结构为<MultiModal>![这是一张彩色的化学结构示意图，其中所有文字和线条均为红色。图像的中心是由三个"Mn"符号组成的三角形结构，这三个"Mn"符号之间由红色实线两两相连。此外，图中还有三个"H"符号，每一个"H"符号都位于两个"Mn"符号之间，并分别通过一条红色实线与这两个相邻的"Mn"符号相连接。除了这些连接外，每个"Mn"符号还连接了其他的化学基团：位于三角形上顶点的"Mn"符号，向其左上方连接两个"OC"基团，向其右上方连接两个"CO"基团；位于三角形左侧的"Mn"符号，连接了四个"OC"基团，这些基团分别位于其左上方、左方、左下方和正下方；位于三角形右侧的"Mn"符号，连接了四个"CO"基团，这些基团分别位于其右上方、右方、右下方和正下方。](/media/uploads/c34f90ae-bc59-4f22-8c22-5761f3730838.png)</MultiModal>

<Checkpoint>[$\bf{H}$为$[Mn(CO)_4H]_3$](2)</Checkpoint>

$\bf{H}$属于$D_{3h}$点群，说法6正确。

$\bf{G}$与重氮甲烷反应脱去氮气，得到$Mn(CO)_5CH_3$。

<Checkpoint>[$\bf{I}$为$Mn(CO)_5CH_3$](1)</Checkpoint>

$\bf{G}$的最高旋转轴次为4，而$\bf{I}$由于甲基的存在使其失去旋转对称性，说法5错误。

$Mn_2(CO)_{10}$与二甲胺反应得到1:2型离子化合物，通过质量分数计算得到$\bf{J}$为$[Mn(NHMe_2)_6^{2+}][Mn(CO)_5^-]_2$。

<Checkpoint>[$\bf{J}$为$[Mn(NHMe_2)_6^{2+}][Mn(CO)_5^-]_2$](2)</Checkpoint>

忽略碳原子与氢原子，则$\bf{J}$的阳离子可以看成$Mn$和$N$形成的正八面体，为$O_h$点群，说法7正确。

因此选择选项G