The researchers reacted $\mathrm{ZrCl_4}$, $\mathrm{Me_3PNSiMe_3}$, and $\mathrm{NaF}$ at $\mathrm{220 \degree C}$, followed by crystallization in dichloromethane, to obtain a $\mathrm{Zr(IV)}$ complex. The chemical formula of the crystal can be written as $\mathrm{Zr_xCl_y(NPMe_3)_z \cdot m CH_2Cl_2}$ (all unknowns are integers). The unit cell parameters are: $a = 1797 \, \mathrm{pm}$, $b = 1183 \, \mathrm{pm}$, $c = 3488 \, \mathrm{pm}$, $\beta = 99.02^{\degree}$; $\rho = 1.685 \, \mathrm{g \cdot cm^{-3}}$; $Z = 4$. Elemental analysis indicates that the mass fractions of $\mathrm{C}$ and $\mathrm{Cl}$ in the crystal are $17.5\%$ and $34.4\%$, respectively.

It is known that the complex in the crystal is actually a 1:1-type salt, with equal numbers of $\mathrm{Cl}$ atoms in the cation and anion, identical ideal point groups (ignoring $\mathrm{PMe_3}$ orientations), and the same coordination number for $\mathrm{Zr}$. Dichloromethane does not participate in coordination.

Deduce the chemical formula of the complex and the structures of the cation and anion, and determine which of the following options are correct:

1. The mass fraction of $\mathrm{H}$ in the crystal is $4.2\%$.
2. After balancing the high-temperature reaction equation and simplifying to the smallest integer ratio, the sum of the reactant coefficients is 21.
3. Both the cation and anion of the complex carry two charges.
4. The point group of the cation and anion of the complex is $\mathrm{D_{3h}}$.
研究人员在 $\mathrm{220 \degree C}$ 下令$ZrCl_4$、$Me_3PNSiMe_3$与$NaF$反应，后在二氯甲烷中结晶，制得了一$Zr(IV)$配合物，其晶体的化学式可写作$Zr_xCl_y(NPMe_3)_z \cdot m CH_2Cl_2$ (未知数均为整数)。晶胞参数为： $a =1797pm$，$b = 1183pm$，$c = 3488pm$，$\beta = 9 9 . 0 2 ^ { \degree } ; \rho = 1 . 6 8 5 g \cdot cm^{-3} ; Z = 4$。元素分析表明，晶体中 $C$ 质量分数为 $1 7 . 5 \%$ ，$Cl$质量分数为 $34 . 4 \%$。

已知晶体中该配合物实为一1:1型盐，阴阳离子中$Cl$原子个数相等，理想点群(忽略$PMe_3$取向)相同，$Zr$的配位数相同。已知二氯甲烷不参与配位。

推测该配合物的化学式以及阴阳离子的结构，判断以下选项中正确的有：

1. 晶体中$H$的质量分数为$4.2\%$
2. 高温反应的方程式配平并化为最简整数比之后，反应物系数之和为21
3. 配合物阴阳离子各含两个电荷
4. 配合物阴阳离子所属点群为$D_{3h}$
1,2
2,3
3,4
1,3
1,4
2,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
None of the above options are correct
1,2
2,3
3,4
1,3
1,4
2,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
以上没有正确选项
$M = \rho V N_A / Z = 1857 ~ g \cdot mol^{-1}$

<Checkpoint>[The relative molecular mass of the crystal is $1857g \cdot mol^{-1}$](0.5)</Checkpoint>

Thus, there are approximately $27$ $C$ atoms ($1857 \times 0.175/12.01 \approx 27$) and $18$ $Cl$ atoms ($1857 \times 0.344/35.45 \approx 18$).

Setting up the equations gives:

$y + 2m = 18$

$3z + m = 27$

Since $z$ must be an integer, $m$ must be a multiple of 3. Additionally, since $y > 0$, $18-2m > 0$, so $m \leq 6$, meaning $m$ is either 3 or 6.

<Checkpoint>[$m$ is 3 or 6](0.5)</Checkpoint>

Given that $Zr$ has a $+4$ oxidation state, $y + z$ must be a multiple of 4.

When $m = 3$, $y = 12$, $z = 8$, and $x = 5$, which is reasonable.

When $m = 6$, $y = 6$, $z = 7$, and $y + z = 17$, which is not a multiple of 4.

In conclusion, the chemical formula of the crystal is $Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2$, with a molecular weight of $1856.95 ~ g \cdot mol^{-1}$.

<Checkpoint>[The chemical formula of the crystal is $Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2$](2)</Checkpoint>

The mass fraction of $H$ is $78 \times 1.008 / 1856.95 = 4.234\%$, so statement 1 is correct.

The balanced high-temperature reaction equation is:

$5ZrCl_4 + 8Me_3PNSiMe_3 + 8NaF = Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2 + 8Me_3SiF + 8NaCl$

<Checkpoint>[The balanced equation is $5ZrCl_4 + 8Me_3PNSiMe_3 + 8NaF = Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2 + 8Me_3SiF + 8NaCl$](1)</Checkpoint>

The sum of the reaction coefficients is $5 + 8 + 8 = 21$, so statement 2 is correct.

Considering the number of $Zr$ in the cation and anion:

If the ratio is 1:4, the anion would be $[ZrCl_6]^{2-}$, but the cation cannot achieve such high symmetry.

Thus, the ratio of $Zr$ counts in the cation and anion is 2:3, meaning the ions consist of 3 $ZrCl_2$ units and 2 $ZrCl_3$ units, respectively. With $Zr$ in octahedral coordination and $NPMe_3$ as a bridge, the structures of the ions are as follows:

Cation $[Zr_3Cl_6(NPMe_3)_5]^+$:

<MultiModal>![Cl[Zr]12(Cl)([N]34[P](C)(C)C)[N]([Zr]3(N5[P](C)(C)C)(Cl)(Cl)N1[P](C)(C)C)([Zr]45(Cl)(Cl)N2[P](C)(C)C)[P](C)(C)C](/media/uploads/72622447-8aea-490b-aa14-30cec69a5410.png)</MultiModal>

<Checkpoint>[The cation is $[Zr_3Cl_6(NPMe_3)_5]^+$](1)</Checkpoint>

Anion $[Zr_2Cl_6(NPMe_3)_3]^-$:

<MultiModal>![Cl[Zr](N1[P](C)(C)C)(N2[P](C)(C)C)(Cl)(Cl)N([P](C)(C)C)[Zr]12(Cl)(Cl)Cl](/media/uploads/2575acf4-3c3c-41ab-bf32-25b910ad645b.png)</MultiModal>

<Checkpoint>[The anion is $[Zr_2Cl_6(NPMe_3)_3]^-$](1)</Checkpoint>

Thus, the cation and anion each carry a single charge, so statement 3 is incorrect. Both ions belong to the $D_{3h}$ point group, so statement 4 is correct.

<Checkpoint>[The point group of the ions is $D_{3h}$](1)</Checkpoint>

Choose 1, 2, 4 (Option H).
$M = \rho V N_A / Z = 1 8 5 7 ~ g \cdot m o l ^ { -1 }$

<Checkpoint>[晶体相对分子质量为$1857g \cdot mol^{-1}$](0.5)</Checkpoint>

故有 $1857 \times 0.175/12.01 \approx 27$个$C$，$1 8 5 7 \times 0 . 3 4 4 / 3 5 . 4 5 \approx 1 8$ 个$Cl$

列方程可得：

$y + 2 m = 1 8$

$3 z + m = 2 7$

由于$z$为整数，故$m$为3的倍数；又$y > 0$，则$18-2m>0$，$m≤6$，$m$为3或6

<Checkpoint>[$m$为3或6](0.5)</Checkpoint>

由于$Zr$为$+4$价，$y+z$应为4的倍数

$m = 3$时$y=12$，$z=8$，$x=5$，合理
  
$m = 6$时$y=6$，$z=7$，$y+z=17$，不为4的倍数

综上，晶体化学式为$Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2$。其分子量为$1856.95 ~ g \cdot m o l ^ { -1 }$

<Checkpoint>[晶体化学式为$Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2$](2)</Checkpoint>

$H$的质量分数为$78 \times 1.008 / 1856.95 = 4.234 \%$，说法1正确。

配平的高温反应方程式为：

$5ZrCl_4+8Me_3PNSiMe_3+8NaF=Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2+8Me_3SiF+8NaCl$

<Checkpoint>[配平的方程式为$5ZrCl_4+8Me_3PNSiMe_3+8NaF=Zr_5Cl_{12}(NPMe_3)_8 \cdot 3 CH_2Cl_2+8Me_3SiF+8NaCl$](1)</Checkpoint>

反应物系数之和为$5+8+8=21$，说法2正确。

考虑阴阳离子中$Zr$的个数：

若为1:4，则阴离子为$[ZrCl_6]^{2-}$，阳离子无法达到这么高的对称性。

因此阴阳离子的$Zr$个数为2:3，则两个离子分别有3个$ZrCl_2$单元和2个$ZrCl_3$单元。而$Zr$的配位是八面体，以$NPMe_3$做桥，可以画出阴阳离子的结构如下：

阳离子$[Zr_3Cl_6(NPMe_3)_5 ]^+$：

<MultiModal>![Cl[Zr]12(Cl)([N]34[P](C)(C)C)[N]([Zr]3(N5[P](C)(C)C)(Cl)(Cl)N1[P](C)(C)C)([Zr]45(Cl)(Cl)N2[P](C)(C)C)[P](C)(C)C](/media/uploads/72622447-8aea-490b-aa14-30cec69a5410.png)</MultiModal>

<Checkpoint>[阳离子为$[Zr_3Cl_6(NPMe_3)_5 ]^+$](1)</Checkpoint>

阴离子$[Zr_2Cl_6(NPMe_3)_3]^-$：

<MultiModal>![Cl[Zr](N1[P](C)(C)C)(N2[P](C)(C)C)(Cl)(Cl)N([P](C)(C)C)[Zr]12(Cl)(Cl)Cl](/media/uploads/2575acf4-3c3c-41ab-bf32-25b910ad645b.png)</MultiModal>

<Checkpoint>[阴离子为$[Zr_2Cl_6(NPMe_3)_3]^-$](1)</Checkpoint>

因此阴阳离子各带一个电荷，说法3错误。阴阳离子均为$D_{3h}$点群，说法4正确。

<Checkpoint>[阴阳离子的点群为$D_{3h}$](1)</Checkpoint>

选择1,2,4（H选项）