The chemical formula of salt $\mathbf{A}$ can be represented as $\mathbf{M}(\mathbf{XY})_2$, where the anion is highly toxic. The mass fraction of $\mathbf{M}$ in $\mathbf{A}$ is 53.10%, and both $\mathbf{X}$ and $\mathbf{Y}$ are second-period elements.

Dissolving $\mathbf{A}$ in a potassium salt solution with the same anion forms a red solution. Boiling it in air causes an oxidation reaction to produce a solution of yellow salt $\mathbf{B}$. Reacting $\mathbf{B}$ with hydrochloric acid and diethyl ether yields colorless crystals $\mathbf{C}$. $\mathbf{C}$ reacts with nitromethane in methanol to obtain $\mathbf{D}$. Neglecting water of crystallization, the molar mass of $\mathbf{D}$ is 42.08 g/mol greater than that of $\mathbf{C}$.

Determine the correctness of the conclusions regarding the above content and select the option corresponding to all the correct conclusions.

Conclusion 1: The element $\mathbf{M}$ can react directly with carbon monoxide to form a mononuclear carbonyl complex.

Conclusion 2: Theoretical prediction is that $\mathbf{B}$ should be diamagnetic.

Conclusion 3: Compound $\mathbf{D}$ contains two ligands in a 1:1 ratio.

Conclusion 4: There is only one type of atom coordinating to $\mathbf{M}$ in compound $\mathbf{D}$.

Conclusion 5: $\mathbf{C}$ exhibits a weak acidity comparable to that of hydrocyanic acid in aqueous solution.
盐$\mathbf{A}$的化学式可以表示为$\mathbf{M}(\mathbf{XY})_2$的形式，其中阴离子有剧毒。$\mathbf{A}$中$\mathbf{M}$的质量分数为53.10%，$\mathbf{X}$和$\mathbf{Y}$均为第二周期元素。

将$\mathbf{A}$溶解在相同阴离子的钾盐溶液中，形成红色溶液，在空气中煮沸，发生氧化反应生成黄色盐$\mathbf{B}$的溶液。用盐酸和乙醚作用于$\mathbf{B}$可制得无色晶体$\mathbf{C}$，$\mathbf{C}$在甲醇中与硝基甲烷反应得到$\mathbf{D}$。若不考虑结晶水，则$\mathbf{D}$的摩尔质量比$\mathbf{C}$多42.08 g/mol。

判断关于上述内容的结论的正确性，并选择对应于所有正确结论的选项。

结论1：单质$\mathbf{M}$可以与一氧化碳直接反应生成单核羰基配合物。

结论2：理论预测$\mathbf{B}$应该是抗磁性的。

结论3：化合物$\mathbf{D}$中存在两种比例为1:1的配体。

结论4：化合物$\mathbf{D}$中对$\mathbf{M}$配位的原子只有一种。

结论5：$\mathbf{C}$在水溶液中体现与氢氰酸相当的弱酸性。
All other options are incorrect
Conclusion 1, Conclusion 2, Conclusion 3
Conclusion 1, Conclusion 2, Conclusion 4
Conclusion 1, Conclusion 2, Conclusion 5
Conclusion 1, Conclusion 3, Conclusion 4
Conclusion 1, Conclusion 3, Conclusion 5
Conclusion 1, Conclusion 4, Conclusion 5
Conclusion 2, Conclusion 3, Conclusion 4
Conclusion 2, Conclusion 3, Conclusion 5
Conclusion 2, Conclusion 4, Conclusion 5
Conclusion 3, Conclusion 4, Conclusion 5
其他选项均不正确
结论1、结论2、结论3
结论1、结论2、结论4
结论1、结论2、结论5
结论1、结论3、结论4
结论1、结论3、结论5
结论1、结论4、结论5
结论2、结论3、结论4
结论2、结论3、结论5
结论2、结论4、结论5
结论3、结论4、结论5
The highly toxic ion composed of only second-period elements in a 1:1 ratio is $CN^-$, so we can write the equation:

$$
\frac{M(\mathbf M)}{M(\mathbf M)+2\times(12.01+14.01)} = 53.10\%
$$

Solving for $M(\mathbf M) = 58.9 \mathrm{(g/mol)}$, corresponding to cobalt or nickel. However, nickel cyanide complexes cannot be oxidized by air, which does not meet the requirements of the question. Therefore, $\mathbf{M}$ is cobalt, and $\mathbf{A}$ is $Co(CN)_2$.

<Checkpoint>[$\mathbf{M}$ is cobalt, and $\mathbf{A}$ is $Co(CN)_2$](2)</Checkpoint>

Cobalt reacts directly with carbon monoxide to only produce $Co_2(CO)_8$, which is not a mononuclear carbonyl complex. Conclusion 1 is incorrect.

<Checkpoint>[Cobalt reacts directly with carbon monoxide to only produce $Co_2(CO)_8$, which is not a mononuclear carbonyl complex.](1)</Checkpoint>

When $Co(CN)_2$ is dissolved in potassium cyanide solution, $Co$ is coordinated to form the deep red $[Co(CN)_6]^{4-}$. Boiling the solution in air, $[Co(CN)_6]^{4-}$ is oxidized to the yellow $[Co(CN)_6]^{3-}$. Therefore, $\mathbf{B}$ is $[Co(CN)_6]^{3-}$.

<Checkpoint>[$\mathbf{B}$ is $[Co(CN)_6]^{3-}$](1)</Checkpoint>

$CN^-$ is a strong field ligand, and $Co^{3+}$ has a high valence, resulting in a large splitting energy. Therefore, $[Co(CN)_6]^{3-}$ is a low-spin complex, where $Co$ has a $d^6$ electron configuration, which in an octahedral field is a $(t_{2g})^6(e_g)^0$ electron configuration. All electrons are paired, so the complex is predicted to be diamagnetic. Conclusion 2 is correct.

<Checkpoint>[$[Co(CN)_6]^{3-}$ has a $(t_{2g})^6(e_g)^0$ electron configuration in an octahedral field, all electrons are paired, and it is diamagnetic](1)</Checkpoint>

$\mathbf{B}$ is treated with hydrochloric acid and diethyl ether to obtain $\mathbf{C}$. Since complexes formed by +3 valent cobalt have kinetic stability, it can be inferred that $\mathbf{C}$ is likely the acid or acid hydrate corresponding to the anion, where the complex anion remains unchanged, still being $[Co(CN)_6]^{3-}$. Therefore, it is speculated that $\mathbf{C}$ is $H_3[Co(CN)_6]·nH_2O$, where $n$ represents the number of crystal water molecules, which cannot be deduced from the existing conditions.

<Checkpoint>[$\mathbf{C}$ is $H_3[Co(CN)_6]·nH_2O$](1)</Checkpoint>

Ignoring crystal water, the molar mass of compound $\mathbf{D}$ is 42.08 g/mol more than that of $\mathbf{C}$, which corresponds precisely to 3 methylene groups. The possible reactions of nitromethane include oxidation and substitution reactions. Here, $Co$ has already been oxidized to the stable +3 valence, so a substitution reaction should have occurred. Note that the $N$ atom in $[Co(CN)_6]^{3-}$ has a lone pair of electrons and is nucleophilic. Therefore, it is speculated that the $N$ atom undergoes an $S_N2$ reaction by nucleophilic attack on nitromethane. Based on the molar mass, it can be deduced that $\mathbf{D}$ is $[Co(CN)_3(CNCH_3)_3]$.

<Checkpoint>[$\mathbf{D}$ is $[Co(CN)_3(CNCH_3)_3]$](1)</Checkpoint>

$\mathbf{D}$ contains cyanide and methyl isocyanide ligands in a 1:1 ratio. Conclusion 3 is correct.

Since complexes formed by +3 valent cobalt have kinetic stability, it can be inferred that the coordination of the $C$ atom to $Co$ will not change in the above nucleophilic substitution reaction. In $\mathbf{D}$, only the $C$ atom is still coordinated to cobalt. Conclusion 4 is correct.

<Checkpoint>[Only the $C$ atom is still coordinated to cobalt in $\mathbf{D}$.](1)</Checkpoint>

Since the basicity of the cyanide coordinated to cobalt is greatly reduced, its ability to bind to protons is reduced. Therefore, the acidity of $\mathbf{C}$ is much stronger than that of hydrocyanic acid, and it is a strong acid in aqueous solution. Conclusion 5 is incorrect.

<Checkpoint>[The acidity of $\mathbf{C}$ is much stronger than that of hydrocyanic acid](1)</Checkpoint>

In summary, choose option H.
仅由第二周期元素1:1组成的剧毒离子只有$CN^-$，则可以列出方程：

$$
\frac{M(\mathbf M)}{M(\mathbf M)+2\times(12.01+14.01)} = 53.10\%
$$

解得$M(\mathbf M) = 58.9 \mathrm{(g/mol)}$，对应于钴或镍。而镍的氰配合物无法被空气氧化，不符合题意，因此$\mathbf{M}$是钴，$\mathbf{A}$是$Co(CN)_2$。

<Checkpoint>[$\mathbf{M}$是钴，$\mathbf{A}$是$Co(CN)_2$](2)</Checkpoint>

钴与一氧化碳直接反应只得到$Co_2(CO)_8$，不是单核羰基配合物。结论1错误。

<Checkpoint>[钴与一氧化碳直接反应只得到$Co_2(CO)_8$，不是单核羰基配合物。](1)</Checkpoint>

$Co(CN)_2$溶于氰化钾溶液时，$Co$被配位生成深红色的$[Co(CN)_6]^{4-}$。在空气中煮沸溶液，$[Co(CN)_6]^{4-}$被氧化为黄色的$[Co(CN)_6]^{3-}$。故$\mathbf{B}$是$[Co(CN)_6]^{3-}$。

<Checkpoint>[$\mathbf{B}$是$[Co(CN)_6]^{3-}$](1)</Checkpoint>

$CN^-$为强场配体，$Co^{3+}$价态高，分裂能大，因此$[Co(CN)_6]^{3-}$是低自旋配合物，其中$Co$为$d^6$电子组态，在八面体场中为$(t_{2g})^6(e_g)^0$电子组态，电子全部成对，可以预测配合物为抗磁性。结论2正确。

<Checkpoint>[$[Co(CN)_6]^{3-}$在八面体场中为$(t_{2g})^6(e_g)^0$电子组态，电子全部成对，为抗磁性](1)</Checkpoint>

$\mathbf{B}$用盐酸和乙醚处理得到$\mathbf{C}$。由于+3价钴形成的配合物具有动力学稳定性，因此可以推测$\mathbf{C}$很可能是阴离子对应的酸或酸的水合物，其中配合物阴离子并不发生变化，仍然是$[Co(CN)_6]^{3-}$。故推测$\mathbf{C}$为$H_3[Co(CN)_6]·nH_2O$，$n$表示结晶水数目不能根据已有条件推出。

<Checkpoint>[$\mathbf{C}$为$H_3[Co(CN)_6]·nH_2O$](1)</Checkpoint>

不考虑结晶水，化合物$\mathbf{D}$的摩尔质量比$\mathbf{C}$多42.08 g/mol，恰对应于3个亚甲基。而硝基甲烷可能发生的反应包括氧化反应和取代反应。此处$Co$已经被氧化到稳定的+3价，故应该发生了取代反应。注意到$[Co(CN)_6]^{3-}$中$N$原子由孤对电子，具有亲核性，因此推测由$N$原子亲核进攻硝基甲烷发生$S_N2$反应。根据摩尔质量可以推出$\mathbf{D}$为$[Co(CN)_3(CNCH_3)_3]$。

<Checkpoint>[$\mathbf{D}$为$[Co(CN)_3(CNCH_3)_3]$](1)</Checkpoint>

$\mathbf{D}$中存在氰根和甲基异腈两种配体，比例为1:1。结论3正确。

由于+3价钴形成的配合物具有动力学稳定性，可以推断$C$原子对$Co$的配位不会在上述亲核取代反应中变化，$\mathbf{D}$中仍然只有$C$原子对钴配位。结论4正确。

<Checkpoint>[$\mathbf{D}$中仍然只有$C$原子对钴配位。](1)</Checkpoint>

由于氰根对钴配位，其碱性大大减弱，对质子的结合能力降低，故$\mathbf{C}$的酸性远强于氢氰酸，在水溶液中是强酸。结论5错误。

<Checkpoint>[$\mathbf{C}$的酸性远强于氢氰酸](1)</Checkpoint>

综上所述，选择H选项