$\mathbf{M}$ can directly react with carbon monoxide under certain conditions to obtain carbonyl compound $\mathbf{A}$. $\mathbf{A}$ reacts with sodium metal in liquid ammonia at low temperature to obtain salt $\mathbf{B}$ containing a tetrahedral mononuclear anion. It is known that the mass fraction of $\mathbf{M}$ in $\mathbf{A}$ is 34.47%. $\mathbf{B}$ can react with carbon tetrachloride in a 3:1 ratio to generate $\mathbf{C}$ with $C_{3v}$ symmetry. Adding the chloride of $\mathbf{M}$ to a concentrated ammonium chloride-ammonia solution, and adding hydrogen peroxide dropwise, can prepare cation $\mathbf{D}$.

Determine the correctness of the following conclusions related to the above content, and select the option corresponding to all the correct conclusions.

Conclusion 1: The mass fraction of $\mathbf{M}$ in $\mathbf{C}$ is 37.11%.

Conclusion 2: The reaction of a metal element with an atomic number 1 greater than that of $\mathbf{M}$ with carbon monoxide to obtain a carbonyl compound can occur under milder conditions.

Conclusion 3: The mass-to-charge ratio of $\mathbf{D}$ is 54.04.

Conclusion 4: There is only one type of ligand in $\mathbf{D}$.

Conclusion 5: The reaction of the iodide of $\mathbf{M}$, copper metal, and carbon monoxide can yield $\mathbf{A}$, while generating a salt with a mass fraction of 20.03% for one of the metals.
$\mathbf{M}$可以在一定条件下与一氧化碳直接反应得到羰基化合物$\mathbf{A}$。$\mathbf{A}$与钠单质在液氨中于低温下反应得到含四面体型单核阴离子的盐$\mathbf{B}$。已知 $\mathbf{A}$ 中 $\mathbf{M}$ 的质量分数为34.47%。$\mathbf{B}$可以与四氯化碳以3:1比例反应生成具有$C_{3v}$对称性的$\mathbf{C}$。将$\mathbf{M}$的氯化物加入浓的氯化铵-氨水溶液中，并滴入过氧化氢，可制得阳离子$\mathbf{D}$。

判断以下与上述内容相关的结论的正确性，选择对应于所有正确结论的选项。

结论1：$\mathbf{C}$中$\mathbf{M}$的质量分数为37.11%。

结论2：原子序数比$\mathbf{M}$大1的金属元素，其与一氧化碳直接作用得到羰基化合物的反应可以在更温和的条件下发生。

结论3：$\mathbf{D}$的质荷比为54.04。

结论4：$\mathbf{D}$中只存在一种配体。

结论5：$\mathbf{M}$的碘化物、铜单质与一氧化碳共同反应可得到 $\mathbf{A}$，同时生成一种金属的质量分数为20.03%的盐。
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 general formula of $\mathbf{A}$ should be $\mathbf{M}(\mathrm{CO})_{x}$, where $x$ can be 4, 4.5, 5, etc. Through trials, it is found that when $x = 4$, $M(\mathbf{M}) = [4 \times 28.01 /(1-0.3447)] \times 0.3447=58.94 (\mathrm{g/mol})$. The molar mass is closest to that of cobalt, followed by nickel.

<Checkpoint>[$M(\mathbf{M}) = [4 \times 28.01 /(1-0.3447)] \times 0.3447=58.94 (\mathrm{g/mol})$](1)</Checkpoint>

If $\mathbf{M}$ is cobalt, then $\mathbf{A}$ is $Co_2(CO)_8$, which can be reduced to $[Co(CO)_4]^-$ by metallic sodium in liquid ammonia, which meets the condition of "tetrahedral mononuclear anion" in the question.

If $\mathbf{M}$ is nickel, then $\mathbf{A}$ is $Ni(CO)_4$, which can be reduced to $[NiH(CO)_3]_2$, $[Ni_4(CO)_9]^{2-}$ and other species by metallic sodium in liquid ammonia, all of which do not meet the condition of "tetrahedral mononuclear anion" in the question.

Therefore, $\mathbf{M}$ is cobalt, $\mathbf{A}$ is $Co_2(CO)_8$, and $\mathbf{B}$ is $Na[Co(CO)_4]$.

<Checkpoint>[$\mathbf{M}$ is cobalt, $\mathbf{A}$ is $Co_2(CO)_8$, and $\mathbf{B}$ is $Na[Co(CO)_4]$.](2)</Checkpoint>

$Na[Co(CO)_4]$ can react with carbon tetrachloride in a 3:1 ratio to generate $ClC[Co(CO)_3]_3$, so $\mathbf{C}$ is $ClC[Co(CO)_3]_3$.

<Checkpoint>[$\mathbf{C}$ is $ClC[Co(CO)_3]_3$](1)</Checkpoint>

Calculate the mass fraction of $Co$ in it:

$$
w(Co) = \frac{3\times58.93}{35.45+12.01+3\times(58.93+3\times(12.01+16.00))}\times 100\%  = 37.11\%
$$

<Checkpoint>[$w(Co) = \frac{3\times58.93}{35.45+12.01+3\times(58.93+3\times(12.01+16.00))}\times 100\%  = 37.11\%$](1)</Checkpoint>

Therefore, conclusion 1 is correct.

The direct reaction of cobalt with carbon monoxide generally takes place at 100 atm and 200 degrees Celsius. The metal element with an atomic number 1 greater than $Co$ is $Ni$, and the typical conditions for the reaction of $Ni$ with carbon monoxide to obtain a carbonyl compound are 1 atm and 50 degrees Celsius, and the reaction conditions are milder. Therefore, conclusion 2 is correct.

<Checkpoint>[The direct reaction of cobalt with carbon monoxide generally takes place at 100 atm and 200 degrees Celsius. The metal element with an atomic number 1 greater than $Co$ is $Ni$, and the typical conditions for the reaction of $Ni$ with carbon monoxide to obtain a carbonyl compound are 1 atm and 50 degrees Celsius, and the reaction conditions are milder. ](1)</Checkpoint>

By adding cobalt chloride to a concentrated ammonium chloride-ammonia solution and dropping in hydrogen peroxide, the cation $[Co(NH_3)_5(H_2O)]^{3+}$ can be prepared. Calculate its mass-to-charge ratio:

<Checkpoint>[the cation $\mathbf{D}$ is $[Co(NH_3)_5(H_2O)]^{3+}$](1)</Checkpoint>

$$
m/z = (58.93+5\times(14.01+3\times1.008)+(16.00+2\times1.008))\div3 = 54.04
$$

<Checkpoint>[$ m/z = (58.93+5\times(14.01+3\times1.008)+(16.00+2\times1.008))\div3 = 54.04 $](1)</Checkpoint>

Therefore, conclusion 3 is correct.

$[Co(NH_3)_5(H_2O)]^{3+}$ contains two types of ligands, ammonia molecules and water molecules, so conclusion 4 is incorrect.

<Checkpoint>[$[Co(NH_3)_5(H_2O)]^{3+}$ contains two types of ligands, ammonia molecules and water molecules](1)</Checkpoint>

The equation for the reaction of $Co$ iodide, elemental copper, and carbon monoxide is

$$
2CoI_2+4Cu+8CO \rightarrow Co_2(CO)_8+4CuI
$$

The only salt produced is $CuI$, and the only metal element in it is copper. The mass fraction of copper is

$$
w(Cu) = \frac{63.55}{63.55+126.9} \times 100\% = 33.37\%
$$

which is not equal to 20.03%, so conclusion 5 is incorrect.

In summary, choose option B.
$\mathbf{A}$的通式应为$\mathbf{M}(\mathrm{CO})_{x}$, 其中$x$可以为4、4.5、5 等，通过尝试发现当$x = 4$时，有$M(\mathbf{M}) = [4 \times 28.01 /(1-0.3447)] \times 0.3447=58.94 (\mathrm{g/mol})$，摩尔质量最接近于钴元素，其次是镍元素。

<Checkpoint>[$M(\mathbf{M}) = [4 \times 28.01 /(1-0.3447)] \times 0.3447=58.94 (\mathrm{g/mol})$](1)</Checkpoint>

若$\mathbf{M}$为钴，则$\mathbf{A}$为$Co_2(CO)_8$，可在液氨中被金属钠还原为$[Co(CO)_4]^-$，符合题目中“四面体型单核阴离子”的条件。

若$\mathbf{M}$为镍，则$\mathbf{A}$为$Ni(CO)_4$，可在液氨中被金属钠还原为$[NiH(CO)_3]_2$、$[Ni_4(CO)_9]^{2-}$等，均不符合题目中“四面体型单核阴离子”的条件。

因此$\mathbf{M}$为钴，$\mathbf{A}$为$Co_2(CO)_8$，$\mathbf{B}$为$Na[Co(CO)_4]$。

<Checkpoint>[$\mathbf{M}$为钴，$\mathbf{A}$为$Co_2(CO)_8$，$\mathbf{B}$为$Na[Co(CO)_4]$。](2)</Checkpoint>

$Na[Co(CO)_4]$可以与四氯化碳以3:1比例反应生成$ClC[Co(CO)_3]_3$，因此$\mathbf{C}$为$ClC[Co(CO)_3]_3$。

<Checkpoint>[$\mathbf{C}$为$ClC[Co(CO)_3]_3$](1)</Checkpoint>

计算其中$Co$的质量分数为：

$$
w(Co) = \frac{3\times58.93}{35.45+12.01+3\times(58.93+3\times(12.01+16.00))}\times 100\%  = 37.11\%
$$

<Checkpoint>[$w(Co) = \frac{3\times58.93}{35.45+12.01+3\times(58.93+3\times(12.01+16.00))}\times 100\%  = 37.11\%$](1)</Checkpoint>

故结论1正确。

钴与一氧化碳直接反应一般在100个大气压和200摄氏度下进行。原子序数比$Co$大1的金属元素是$Ni$，其与一氧化碳直接作用得到羰基化合物的反应的典型条件是一个大气压和50摄氏度，反应条件更温和。故结论2正确。

<Checkpoint>[钴与一氧化碳直接反应一般在100个大气压和200摄氏度下进行。原子序数比$Co$大1的金属元素是$Ni$，其与一氧化碳直接作用得到羰基化合物的反应的典型条件是一个大气压和50摄氏度，反应条件更温和。](1)</Checkpoint>

将$Co$的氯化物加入浓的氯化铵-氨水溶液中，并滴入过氧化氢，可制得阳离子$[Co(NH_3)_5(H_2O)]^{3+}$，计算其质荷比为：

<Checkpoint>[阳离子$\mathbf{D}$为$[Co(NH_3)_5(H_2O)]^{3+}$](1)</Checkpoint>

$$
m/z = (58.93+5\times(14.01+3\times1.008)+(16.00+2\times1.008))\div3 = 54.04
$$

<Checkpoint>[$ m/z = (58.93+5\times(14.01+3\times1.008)+(16.00+2\times1.008))\div3 = 54.04 $](1)</Checkpoint>

故结论3正确。

$[Co(NH_3)_5(H_2O)]^{3+}$中存在氨分子和水分子两种配体，因此结论4错误。

<Checkpoint>[$[Co(NH_3)_5(H_2O)]^{3+}$中存在氨分子和水分子两种配体](1)</Checkpoint>

$Co$的碘化物、铜单质与一氧化碳共同反应的方程式为

$$
2CoI_2+4Cu+8CO \rightarrow Co_2(CO)_8+4CuI
$$

<Checkpoint>[$ 2CoI_2+4Cu+8CO \rightarrow Co_2(CO)_8+4CuI $](1)</Checkpoint>

生成的盐只有$CuI$，其中金属元素只有铜，铜的质量分数为

$$
w(Cu) = \frac{63.55}{63.55+126.9} \times 100\% = 33.37\%
$$

<Checkpoint>[$ w(Cu) = \frac{63.55}{63.55+126.9} \times 100\% = 33.37\% $](1)</Checkpoint>

不等于20.03%，故结论5错误。

综上所述，选择B选项。