Researchers synthesized and analyzed the chemical reaction of complex $\mathbf{A}$, the structure of complex $\mathbf{A}$ is shown in the following figure:

<MultiModal>![This complex is a 1:1 zwitterionic salt with each ion carrying a single charge. The central metal of the cation is Ni with planar triangular coordination, where two coordination sites are coordinated by 2,2'-bipyridine, and one site is coordinated by N of linear NO; the anion has P as the center, coordinated by 6 F in an octahedral arrangement](/media/uploads/4b857f0f-0fe3-4c43-a9b9-0be349dcac84.png)</MultiModal>

Complex $\mathbf{A}$ can further coordinate with a neutral ligand in another $\mathbf{A}$ to obtain complex $\mathbf{B}$. The following formula ligand $\mathbf{X}$ is added to the latter for reaction:

<MultiModal>![SMILES: CC(CC(C)=O)=O](/media/uploads/1c49de2e-537b-43ef-921d-8e3801800f59.png)</MultiModal>

At this time, a chemical reaction occurs to obtain a gas with a density of $1.97~\mathrm{g/L}$ under standard conditions.

It was found that this is a multi-step reaction: two molecules of complex $\mathbf{B}$ first undergo a ligand substitution reaction to obtain complexes $\mathbf{C}$ and $\mathbf{D}$, and $\mathbf{D}$ contains a $\mathrm{N=N}$ double bond; then, complex $\mathbf{D}$ undergoes a ligand decomposition reaction, and further reacts with the newly added ligand $\mathbf{X}$ and with complex $\mathbf{C}$, finally obtaining a transition metal-containing complex $\mathbf{E}$. The above notations $\mathbf{A\sim E}$ are all complete chemical formulas of the compounds.

The following propositions are given:

1.  The total number of electrons around the transition metal in complex $\mathbf{A}$ is the same as that in complex $\mathbf{D}$.
2.  The symmetry of the complex in $\mathbf{D}$ is higher than that of complex $\mathbf{B}$.
3.  In complex $\mathbf{B}$, considering the most likely situation, the valence state of the transition metal is $0$.
4.  In the chemical formula of one complex $\mathbf{C}$, the sum of the point group orders of all ionic species is $102$.
5.  Among the complexes $\mathbf{A\sim E}$, only one is electrically neutral.
6.  For every $22.4~\mathrm{L}$ of gas at standard temperature and pressure produced in the reaction, approximately $18~\mathrm{mL}$ of small molecule by-products are obtained at room temperature and normal pressure.

Please calculate the ratio of the sum of the serial numbers of the correct propositions to (the sum of the serial numbers of the incorrect propositions + 1), and choose the correct option.
有研究人员合成并分析了配合物 $\mathbf{A}$ 的化学反应，配合物 $\mathbf{A}$ 的结构如下图所示：

<MultiModal>![这个配合物是一个阴阳离子1:1且均带一个电荷的盐，阳离子中心金属为平面三角形配位的Ni，其中两个配位位点由2,2′-联吡啶配位，一个位点由直线型的NO的N配位；阴离子是中心是P，由6个F呈八面体配位](/media/uploads/4b857f0f-0fe3-4c43-a9b9-0be349dcac84.png)</MultiModal>

配合物 $\mathbf{A}$ 可以进一步与另一个 $\mathbf{A}$ 中的电中性配体发生配位，得到配合物 $\mathbf{B}$。在后者中加入以下化学式的配体 $\mathbf{X}$ 进行反应：

<MultiModal>![SMILES: CC(CC(C)=O)=O](/media/uploads/1c49de2e-537b-43ef-921d-8e3801800f59.png)</MultiModal>

这时会发生化学反应，得到一种标准状况下密度为 $1.97~\mathrm{g/L}$ 的气体。

研究发现，这是一个多步反应：两分子的配合物 $\mathbf{B}$ 首先发生配体置换反应，得到配合物 $\mathbf{C}$ 和 $\mathbf{D}$，且 $\mathbf{D}$ 中含有 $\mathrm{N=N}$ 双键；然后，配合物 $\mathbf{D}$ 发生配体分解反应，后进一步与新加入的配体 $\mathbf{X}$ 发生化学反应、与配合物 $\mathbf{C}$ 发生反应，最终得到一种含过渡金属的配合物 $\mathbf{E}$。以上记号 $\mathbf{A\sim E}$ 均为化合物的完整化学式。

现有如下命题：

1. 配合物 $\mathbf{A}$ 中过渡金属周围的总电子数与配合物 $\mathbf{D}$ 相同。
2. $\mathbf{D}$ 中配合物的对称性高于配合物 $\mathbf{B}$。
3. 配合物 $\mathbf{B}$ 中，考虑最有可能的情况，过渡金属的价态为 $0$ 价。
4. 一个配合物 $\mathbf{C}$ 的化学式中，所有离子物种的点群阶数之和为 $102$。
5. 配合物 $\mathbf{A\sim E}$ 中，只有一种为电中性。
6. 该反应每生成标准状况下 $22.4~\mathrm{L}$ 的气体，会得到室温常压下约 $18~\mathrm{mL}$ 的小分子副产物。

请计算以上命题中，正确命题的序号和与（错误命题的序号和+1）的比值，选择正确的选项。
0.375
2.143
0.467
1
1.2
0.692
4.5
0
21
0.375
2.143
0.467
1
1.2
0.692
4.5
0
21
First, based on the information given in the question, deduce the chemical formulas and properties of each species.

Complex $\mathbf{A}$:
- Cation: $\mathrm{[Ni(bpy)(NO)]^+}$。 Where, $\mathrm{bpy}$ (2,2'-bipyridine) is a neutral ligand. The $\mathrm{NO}$ ligand is described as linear, which usually means it coordinates as $\mathrm{NO^+}$ (nitrosonium ion), a 2-electron donor. To make the cation have a $+1$ charge, the oxidation state of $\mathrm{Ni}$ must be 0. That is, $\mathrm{[Ni^0(bpy)(NO^+)]^+}$。
- Anion: $\mathrm{[PF_6]^-}$。
- Chemical formula: $\mathrm{[Ni(bpy)(NO)][PF_6]}$。

<Checkpoint>[Complex $\mathbf{A}$ is $\mathrm{[Ni(bpy)(NO)][PF_6]}$](1)</Checkpoint>

Complex $\mathbf{B}$:
- Obtained from the reaction of complex $\mathbf{A}$ with another neutral ligand $\mathrm{bpy}$: $$\mathrm{[Ni(bpy)(NO)]^+ + bpy \rightarrow [Ni(bpy)_2(NO)]^+}$$
- The chemical formula of complex $\mathbf{B}$ is $\mathrm{[Ni(bpy)_2(NO)][PF_6]}$。

<Checkpoint>[Complex $\mathbf{B}$ is $\mathrm{[Ni(bpy)_2(NO)][PF_6]}$](1)</Checkpoint>

Gaseous product of the reaction:

- The reaction produces a gas with a density of $1.97~\mathrm{g/L}$ under standard conditions (STP, $0^\circ\mathrm{C}$, $1~\mathrm{atm}$).
- The molar mass of the gas $M = \text{density} \times V_m = 1.97~\mathrm{g/L} \times 22.4~\mathrm{L/mol} \approx 44.13~\mathrm{g/mol}$.
- This value is very close to the molar mass of nitrous oxide ($\mathrm{N_2O}$) or carbon dioxide ($\mathrm{CO_2}$) (both are $44.02~\mathrm{g/mol}$). Combined with the information of ligand $\mathrm{NO}$ and the presence of $\mathrm{N=N}$ double bond in the complex in the question. Therefore, the gas produced is $\mathrm{N_2O}$.

<Checkpoint>[The gaseous product of the reaction is $\mathrm{N_2O}$](1)</Checkpoint>

Multi-step reaction mechanism:

First step: $\mathrm{2\mathbf{B} \rightarrow \mathbf{C} + \mathbf{D}}$

- This is a ligand redistribution reaction starting from two molecules of cation $\mathrm{[Ni(bpy)_2(NO)]^+}$。
- Complex $\mathbf{D}$ contains $\mathrm{N=N}$ double bond, indicating that two $\mathrm{NO}$ ligands in $\mathbf{D}$ have undergone coupling, the reaction is:
        $$ \mathrm{2[Ni(bpy)_2(NO)]^+ \rightarrow [Ni(bpy)_3]^{2+} + [Ni(bpy)(N_2O_2)]} $$

From this, $\mathbf{C}$ and $\mathbf{D}$ can be determined:
- $\mathbf{C}$: Ionic complex $\mathrm{[Ni(bpy)_3][PF_6]_2}$

<Checkpoint>[Complex $\mathbf{C}$ is $\mathrm{[Ni(bpy)_3][PF_6]_2}$](1)</Checkpoint>

- $\mathbf{D}$: Neutral complex $\mathrm{[Ni(bpy)(N_2O_2)]}$

<Checkpoint>[Complex $\mathbf{D}$ is $\mathrm{[Ni(bpy)(N_2O_2)]}$](1)</Checkpoint>

Second step: $\mathbf{D}$ decomposes and reacts with $\mathbf{X}$

- First, complex $\mathbf{D}$ decomposes. The $\mathrm{N_2O_2}$ ligand therein undergoes a decomposition reaction to generate gas $\mathrm{N_2O}$ and an oxo species:
        $$ \mathrm{[Ni(bpy)(N_2O_2)] \rightarrow [Ni(bpy)(O)] + N_2O} $$
- The ligand $\mathbf{X}$ (i.e., $\mathrm{acacH}$, acetylacetone) does not directly react with the above-generated oxo species $\mathrm{[Ni(bpy)(O)]}$, but is related to the final product. In fact, there are several possible pathways in the reaction system. A more reasonable mechanism is that $\mathrm{[Ni(bpy)(O)]}$ reacts with complex $\mathbf{C}$, and ligand $\mathbf{X}$ participates in the subsequent steps. To simplify and obtain the final product, we consider the overall reaction:
$$ \mathrm{[Ni(bpy)(N_2O_2)] + [Ni(bpy)_3]^{2+} + 2\,acacH \rightarrow 2[Ni(bpy)_2(acac)]^+ + N_2O + H_2O} $$
- Therefore, the final product $\mathbf{E}$ is $\mathrm{[Ni(bpy)_2(acac)][PF_6]}$。

<Checkpoint>[Complex $\mathbf{E}$ is $\mathrm{[Ni(bpy)_2(acac)][PF_6]}$](1)</Checkpoint>

**Proposition Judgment**

**1. The total number of electrons around the transition metal in complex $\mathbf{A}$ is the same as that in complex $\mathbf{D}$.**

- $\mathbf{A}$ (cation): $\mathrm{[Ni^0(bpy)(NO^+)]^+}$。 Total number of electrons = $\mathrm{Ni^0~(d^{10},~10e^-) + bpy~(4e^-) + NO^+~(2e^-) = 16e^-}$。
- $\mathbf{D}$: $\mathrm{[Ni(bpy)(N_2O_2)]}$。 To maintain electroneutrality, if the ligand is $\mathrm{N_2O_2^{2-}}$, then the metal is $\mathrm{Ni^{2+}}$. Total number of electrons = $\mathrm{Ni^{2+}~(d^{8},~8e^-) + bpy~(4e^-) + N_2O_2^{2-}~(4e^-) = 16e^-}$。
- Conclusion: Proposition 1 is correct.

<Checkpoint>[The total number of electrons around the transition metal in complex $\mathbf{A}$ and complex $\mathbf{D}$ are both 16-electron structure](1)</Checkpoint>

**2. The symmetry of the complex in $\mathbf{D}$ is higher than that of the complex in $\mathbf{B}$.**

- First, analyze the valence state of the metal in the cation in complex $\mathbf{B}$: $\mathrm{[Ni(bpy)_2(NO)]^+}$。 If the metal is in the $0$ valence state, then it is $\mathrm{Ni^0~(d^{10},~10e^-)}$, and the ligand is $\mathrm{NO^+}$, the total number of electrons is $\mathrm{10e^- + 2 \times bpy~(8e^-) + NO^+~(2e^-) = 20e^-}$. This does not conform to the common 18-electron rule. If the metal is in the $+2$ valence state, then it is $\mathrm{Ni^{2+}~(d^8,~8e^-)}$, and the ligand is $\mathrm{NO^-}$ (bent, 2e- donor), the total number of electrons is $\mathrm{8e^- + 8e^- + 2e^- = 18e^-}$. The 18-electron structure is more stable, so the metal valence state is more likely to be $+2$ valence.
- Complex $\mathbf{D}$ is $\mathrm{[Ni(bpy)(N_2O_2)]}$, in which the $\mathrm{N_2O_2}$ ligand forms a five-membered ring with the metal, and the arrangement of the four coordination sites is planar quadrilateral, and the whole molecule has $C_{2v}$ symmetry.
- Complex $\mathbf{B}$ has a bent coordinated $\mathrm{NO}^-$ in the cation. Considering the possibility that its five coordination sites are arranged in a square pyramidal or trigonal bipyramidal arrangement, it is impossible to achieve a complex cation structure with a symmetry higher than or equal to $C_{2v}$.
- Conclusion: Proposition 2 is correct.

<Checkpoint>[The point group of the complex in $\mathbf{D}$ is $C_{2v}$, which is higher than complex $\mathbf{B}$](1)</Checkpoint>

**3. In complex $\mathbf{B}$, considering the most likely situation, the valence state of the transition metal is $0$ valence.**

- According to the previous analysis, the metal is $\mathrm{Ni^{2+}~(d^8,~8e^-)}$ with a valence of $+2$, and the total number of electrons is $\mathrm{8e^- + 8e^- + 2e^- = 18e^-}$. The 18-electron structure is more stable, so the metal valence state is more likely to be $+2$ valence.
- Conclusion: Proposition 3 is incorrect.

<Checkpoint>[In complex $\mathbf{B}$, considering the most likely situation, the valence state of the transition metal is $+2$ valence](1)</Checkpoint>

**4. In the chemical formula of one complex $\mathbf{C}$, the sum of the orders of the point groups of all ionic species is $102$.**
- The chemical formula of complex $\mathbf{C}$ is $\mathrm{[Ni(bpy)_3][PF_6]_2}$, where the cation $\mathrm{[Ni(bpy)_3]^{2+}}$ has $D_3$ symmetry (group of order 6) and the anion $\mathrm{[PF_6]^-}$ has $O_h$ symmetry (group of order 48).
- Since there are two anions in the chemical formula, the sum of the orders of the point groups of all ionic species $= 6 + (48 \times 2) = 102$.
- Conclusion: Proposition 4 is correct.

<Checkpoint>[In the chemical formula of one complex $\mathbf{C}$, the sum of the orders of the point groups of all ionic species is $102$](1)</Checkpoint>

**5. Among complexes $\mathbf{A\sim E}$, only one is neutral.**

- $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{E}$ are all ionic salts containing $\mathrm{[PF_6]^-}$ counterions. Complex $\mathbf{D}$, i.e., $\mathrm{[Ni(bpy)(N_2O_2)]}$, is an uncharged inner complex.
- Conclusion: Proposition 5 is correct.

<Checkpoint>[Among complexes $\mathbf{A\sim E}$, only $\mathbf{D}$ is a neutral complex](1)</Checkpoint>

**6. For every $22.4~\mathrm{L}$ of gas at standard temperature and pressure produced in the reaction, approximately $18~\mathrm{mL}$ of small molecule by-product is obtained under ambient conditions.**

- According to the mechanism derivation, for every $1~\mathrm{mol}$ of gas ($\mathrm{N_2O}$) produced, $2~\mathrm{mol}$ of $\mathrm{acacH}$ will be consumed in the subsequent steps and $1~\mathrm{mol}$ of small molecule by-product ($\mathrm{H_2O}$) will be generated.
- $1~\mathrm{mol}$ of $\mathrm{N_2O}$ has a volume of $22.4~\mathrm{L}$ at STP.
- $1~\mathrm{mol}$ of $\mathrm{H_2O}$ has a mass of $18~\mathrm{g}$, which is liquid under ambient conditions with a density of approximately $1~\mathrm{g/mL}$ and a volume of approximately $18~\mathrm{mL}$.
- Conclusion: Proposition 6 is correct.

<Checkpoint>[For every $22.4~\mathrm{L}$ of $\mathrm{N_2O}$  at standard temperature and pressure produced in the reaction, approximately $18~\mathrm{mL}$ of $\mathrm{H_2O}$ is obtained under ambient conditions](1)</Checkpoint>

Calculate the final result:
- Correct proposition numbers: 1, 2, 4, 5, 6. The sum is $1+2+4+5+6 = 18$.
- Incorrect proposition number: 3. The sum is 3.
- Calculate the ratio: $$ \dfrac{\text{Sum of correct proposition numbers}}{\text{Sum of incorrect proposition numbers} + 1} = \dfrac{18}{3+1} = \dfrac{18}{4} = 4.5$$

Therefore, choose option **G**.

<Checkpoint>[The final calculated result is $4.5$](1)</Checkpoint>
首先，根据题目中给出的信息，推导各物种的化学式与性质。

配合物 $\mathbf{A}$:
- 阳离子: $\mathrm{[Ni(bpy)(NO)]^+}$。其中，$\mathrm{bpy}$ (2,2'-联吡啶) 是电中性配体。$\mathrm{NO}$ 配体被描述为直线型，这通常意味着它作为 $\mathrm{NO^+}$ (亚硝𬭩离子) 配位，是2电子供体。为使阳离子带 $+1$ 电荷，$\mathrm{Ni}$ 的氧化态必须为 0。即 $\mathrm{[Ni^0(bpy)(NO^+)]^+}$。
- 阴离子: $\mathrm{[PF_6]^-}$。
- 化学式: $\mathrm{[Ni(bpy)(NO)][PF_6]}$。

<Checkpoint>[配合物 $\mathbf{A}$ 是 $\mathrm{[Ni(bpy)(NO)][PF_6]}$](1)</Checkpoint>

配合物 $\mathbf{B}$:
- 由配合物 $\mathbf{A}$ 与另一个电中性配体 $\mathrm{bpy}$ 反应得到：$$\mathrm{[Ni(bpy)(NO)]^+ + bpy \rightarrow [Ni(bpy)_2(NO)]^+}$$
- 配合物 $\mathbf{B}$ 化学式为 $\mathrm{[Ni(bpy)_2(NO)][PF_6]}$。

<Checkpoint>[配合物 $\mathbf{B}$ 是 $\mathrm{[Ni(bpy)_2(NO)][PF_6]}$](1)</Checkpoint>

反应的气体产物:

- 反应生成一种气体，在标准状况 (STP, $0^\circ\mathrm{C}$, $1~\mathrm{atm}$) 下密度为 $1.97~\mathrm{g/L}$。
- 气体的摩尔质量 $M = \text{密度} \times V_m = 1.97~\mathrm{g/L} \times 22.4~\mathrm{L/mol} \approx 44.13~\mathrm{g/mol}$。
- 这个值非常接近一氧化二氮 ($\mathrm{N_2O}$) 或二氧化碳 ($\mathrm{CO_2}$) 的摩尔质量 (两者均为$44.02~\mathrm{g/mol}$)。结合题目中配体 $\mathrm{NO}$、配合物中含有$\mathrm{N=N}$ 双键等信息。因此，生成的气体为 $\mathrm{N_2O}$。

<Checkpoint>[反应的气体产物是 $\mathrm{N_2O}$](1)</Checkpoint>

多步反应机理:

第一步: $\mathrm{2\mathbf{B} \rightarrow \mathbf{C} + \mathbf{D}}$

- 这是一个起始于两分子阳离子 $\mathrm{[Ni(bpy)_2(NO)]^+}$ 的配体重新分配反应。
- 配合物 $\mathbf{D}$ 中含有 $\mathrm{N=N}$ 双键，说明 $\mathbf{D}$ 中有两个 $\mathrm{NO}$ 配体产生了偶联，反应为：
        $$ \mathrm{2[Ni(bpy)_2(NO)]^+ \rightarrow [Ni(bpy)_3]^{2+} + [Ni(bpy)(N_2O_2)]} $$

由此可确定 $\mathbf{C}$ 和$\mathbf{D}$：
- $\mathbf{C}$: 离子配合物 $\mathrm{[Ni(bpy)_3][PF_6]_2}$

<Checkpoint>[配合物 $\mathbf{C}$ 是 $\mathrm{[Ni(bpy)_3][PF_6]_2}$](1)</Checkpoint>

- $\mathbf{D}$: 电中性配合物 $\mathrm{[Ni(bpy)(N_2O_2)]}$

<Checkpoint>[配合物 $\mathbf{D}$ 是 $\mathrm{[Ni(bpy)(N_2O_2)]}$](1)</Checkpoint>

第二步: $\mathbf{D}$ 分解并与 $\mathbf{X}$ 反应

- 首先，配合物 $\mathbf{D}$ 分解。其中的 $\mathrm{N_2O_2}$ 配体发生分解反应，生成气体 $\mathrm{N_2O}$ 和一个氧代物种：
        $$ \mathrm{[Ni(bpy)(N_2O_2)] \rightarrow [Ni(bpy)(O)] + N_2O} $$
- 配体 $\mathbf{X}$ (即 $\mathrm{acacH}$, 乙酰丙酮) 与上述生成的氧代物种 $\mathrm{[Ni(bpy)(O)]}$ 并不直接反应，而是与最终产物有关。实际上，反应体系中存在多种可能的路径。一个更合理的机理是 $\mathrm{[Ni(bpy)(O)]}$ 与配合物 $\mathbf{C}$ 反应，而配体 $\mathbf{X}$ 参与后续步骤。为简化并得到最终产物，我们考虑总反应：
$$ \mathrm{[Ni(bpy)(N_2O_2)] + [Ni(bpy)_3]^{2+} + 2\,acacH \rightarrow 2[Ni(bpy)_2(acac)]^+ + N_2O + H_2O} $$
- 因此，最终产物 $\mathbf{E}$ 是 $\mathrm{[Ni(bpy)_2(acac)][PF_6]}$。

<Checkpoint>[配合物 $\mathbf{E}$ 是 $\mathrm{[Ni(bpy)_2(acac)][PF_6]}$](1)</Checkpoint>

**命题判断**

**1. 配合物 $\mathbf{A}$ 中过渡金属周围的总电子数与配合物 $\mathbf{D}$ 相同。**

- $\mathbf{A}$ (阳离子): $\mathrm{[Ni^0(bpy)(NO^+)]^+}$。总电子数 = $\mathrm{Ni^0~(d^{10},~10e^-) + bpy~(4e^-) + NO^+~(2e^-) = 16e^-}$。
- $\mathbf{D}$: $\mathrm{[Ni(bpy)(N_2O_2)]}$。为保持电中性，若配体为 $\mathrm{N_2O_2^{2-}}$，则金属为 $\mathrm{Ni^{2+}}$。总电子数 = $\mathrm{Ni^{2+}~(d^{8},~8e^-) + bpy~(4e^-) + N_2O_2^{2-}~(4e^-) = 16e^-}$。
- 结论: 命题 1 正确。

<Checkpoint>[配合物 $\mathbf{A}$ 中过渡金属周围的总电子数与配合物 $\mathbf{D}$ 均为 16 电子结构](1)</Checkpoint>

**2. $\mathbf{D}$ 中配合物的对称性高于配合物 $\mathbf{B}$。**

- 先分析配合物 $\mathbf{B}$ 中阳离子中金属的价态: $\mathrm{[Ni(bpy)_2(NO)]^+}$。若金属为 $0$ 价，则为 $\mathrm{Ni^0~(d^{10},~10e^-)}$，配体为 $\mathrm{NO^+}$，总电子数为 $\mathrm{10e^- + 2 \times bpy~(8e^-) + NO^+~(2e^-) = 20e^-}$。这不符合常见的18电子规则。若金属为 $+2$ 价，则为 $\mathrm{Ni^{2+}~(d^8,~8e^-)}$，配体为 $\mathrm{NO^-}$ (弯曲型, 2e- 供体)，总电子数为 $\mathrm{8e^- + 8e^- + 2e^- = 18e^-}$。18电子结构更稳定，因此金属价态更可能为 $+2$ 价。
- 配合物 $\mathbf{D}$ 为 $\mathrm{[Ni(bpy)(N_2O_2)]}$，其中的 $\mathrm{N_2O_2}$ 配体与金属形成五元环，四个配位位点排布方式为平面四边形，整个分子具有 $C_{2v}$ 对称性。
- 配合物 $\mathbf{B}$ 阳离子中存在弯曲配位的 $\mathrm{NO}^-$，考虑其五个配位位点呈四方锥型排布或三角双锥型排布的可能性，均无法实现高于或等于 $C_{2v}$ 对称性的配合物阳离子结构。
- 结论: 命题 2 正确。

<Checkpoint>[$\mathbf{D}$ 中配合物的点群为 $C_{2v}$，高于配合物 $\mathbf{B}$](1)</Checkpoint>

**3. 配合物 $\mathbf{B}$ 中，考虑最有可能的情况，过渡金属的价态为 $0$ 价**

- 根据之前的分析，金属为 $+2$ 价的 $\mathrm{Ni^{2+}~(d^8,~8e^-)}$，总电子数为 $\mathrm{8e^- + 8e^- + 2e^- = 18e^-}$。18电子结构更稳定，因此金属价态更可能为 $+2$ 价。
- 结论: 命题 3 错误。

<Checkpoint>[配合物 $\mathbf{B}$ 中，考虑最有可能的情况，过渡金属的价态为 $+2$ 价](1)</Checkpoint>

**4. 一个配合物 $\mathbf{C}$ 的化学式中，所有离子物种的点群阶数之和为 $102$。**
- 配合物 $\mathbf{C}$ 的化学式为 $\mathrm{[Ni(bpy)_3][PF_6]_2}$，其中，阳离子 $\mathrm{[Ni(bpy)_3]^{2+}}$ 具有 $D_3$ 对称性（6阶群），阴离子 $\mathrm{[PF_6]^-}$ 具有 $O_h$ 对称性（48阶群）。
- 由于化学式中有两个阴离子，故所有离子物种的点群阶数之和 $= 6 + (48 \times 2) = 102$。
- 结论: 命题 4 正确。

<Checkpoint>[一个配合物 $\mathbf{C}$ 的化学式中，所有离子物种的点群阶数之和为 $102$](1)</Checkpoint>

**5. 配合物 $\mathbf{A\sim E}$ 中，只有一种为电中性。**

- $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $\mathbf{E}$ 均为离子盐，含有 $\mathrm{[PF_6]^-}$ 抗衡离子。配合物 $\mathbf{D}$，即 $\mathrm{[Ni(bpy)(N_2O_2)]}$，是一个不带电荷的内配合物。
- 结论: 命题 5 正确。

<Checkpoint>[配合物 $\mathbf{A\sim E}$ 中只有 $\mathbf{D}$ 是电中性配合物](1)</Checkpoint>

**6. 该反应每生成标准状况下 $22.4~\mathrm{L}$ 的气体，会得到室温常压下约 $18~\mathrm{mL}$ 的小分子副产物。**

- 根据机理推导，每生成 $1~\mathrm{mol}$ 的气体 ($\mathrm{N_2O}$)，就会在后续步骤中消耗 $2~\mathrm{mol}$ 的 $\mathrm{acacH}$ 并生成 $1~\mathrm{mol}$ 的小分子副产物 ($\mathrm{H_2O}$)。
- $1~\mathrm{mol}$ 的 $\mathrm{N_2O}$ 在 STP 下体积为 $22.4~\mathrm{L}$。
- $1~\mathrm{mol}$ 的 $\mathrm{H_2O}$ 质量为 $18~\mathrm{g}$，其在室温常压下为液体，密度约为 $1~\mathrm{g/mL}$，体积约为 $18~\mathrm{mL}$。
- 结论: 命题 6 正确。

<Checkpoint>[该反应每生成标准状况下 $22.4~\mathrm{L}$ 的 $\mathrm{N_2O}$，会得到室温常压下约 $18~\mathrm{mL}$ 的 $\mathrm{H_2O}$](1)</Checkpoint>

计算最终结果：
- 正确命题序号: 1, 2, 4, 5, 6。 和为 $1+2+4+5+6 = 18$。
- 错误命题序号: 3。 和为 3。
- 计算比值：$$ \dfrac{\text{正确命题的序号和}}{\text{错误命题的序号和} + 1} = \dfrac{18}{3+1} = \dfrac{18}{4} = 4.5$$

因此，选择选项 **G**。

<Checkpoint>[最终计算结果为 $4.5$](1)</Checkpoint>