Two identical samples of a common compound **A** (relative molecular mass not exceeding $80$) found in daily life are boiled until completely evaporated, leaving no solid residue. The gas obtained from the first sample is passed sequentially through anhydrous calcium chloride and soda lime, with the latter increasing in weight by $33.0\textup{ mg}$; the gas obtained from the second sample is passed sequentially through soda lime and anhydrous calcium chloride, with the latter increasing in weight by $25.5\textup{ mg}$. Under an $\mathrm{Ar}$ atmosphere, $5.0\textup{ mL}$ of a $2\textup{ M}$ $\mathrm{MCl}_3$ solution is mixed with a large excess of an ethanol solution of **A**, immediately producing a precipitate. After filtration and drying, a purple fluffy powder **B** $5.15\textup{ g}$ is obtained. **B** is readily soluble in water, stable in air, and has a non-zero magnetic moment measured at room temperature. Please select the correct option from the following choices.
将相同的两份生活中常见的化合物**A**（相对分子质量不超过$80$）的水溶液煮沸至完全蒸干，无固体残留物，第一份所得气体依次通过无水氯化钙和碱石灰，后者增重$33.0\textup{ mg}$；第二份所得气体依次通过碱石灰和无水氯化钙，后者增重$25.5\textup{ mg}$。在$\mathrm{Ar}$气氛下，将$5.0\textup{ mL}$ $2\textup{ M}$的$\mathrm{MCl}_3$溶液与大量**A**的乙醇溶液混合，立即产生沉淀物质，过滤干燥得到紫色蓬松粉末**B** $5.15\textup{ g}$。**B**极易溶于水，对空气稳定，常温下测得磁矩不为零。请从下列选项中选出正确的一项。
The relative molecular mass of A is $79$
+4 is also a common valence of $\mathrm{M}$
Aqueous solutions of $\mathrm{MCl}_3$ are typically colorless
The metal cations in **B** may be in a tetrahedral coordination environment
A的相对分子质量为$79$
$+4$也是$\mathrm{M}$的常见化合价
$\mathrm{MCl}_3$的溶液通常没有颜色
**B**中的金属阳离子可能处于正四面体的配位环境中
Anhydrous calcium chloride can absorb gases including $\mathrm{H}_2\mathrm{O}$ and $\mathrm{NH}_3$, while soda lime can absorb acidic gases such as $\mathrm{H}_2\mathrm{O}$ and $\mathrm{CO}_2$. From this, it can be inferred that during the experiment with the second sample, anhydrous calcium chloride only absorbed $\mathrm{NH}_3$. By testing various acidic gases, it can be observed that during the experiment with the first sample, if the gas absorbed by soda lime is $\mathrm{CO}_2$, the molar ratio of the two gases is an integer ratio:

$$
(33.0÷44.01)/(25.5÷17.034)=0.501\approx1/2
$$

<Checkpoint>[$n(\mathrm{CO}_2):n(\mathrm{NH}_3)=1:2$](1)</Checkpoint>

This implies that **A** can decompose into $\mathrm{CO}_2$ and $\mathrm{NH}_3$ in a $1:2$ ratio, so **A** could be ammonium carbonate $(\mathrm{NH}_4)_2\mathrm{CO}_3$ ($96\textup{ g/mol}$), ammonium carbamate $\mathrm{H}_2\mathrm{NCOO}\mathrm{NH}_4$ ($78 \textup{ g/mol}$), or urea $\mathrm{CO}(\mathrm{NH}_2)_2$ ($60\textup{ g/mol}$). Considering that **A** is a common compound in daily life with a relative molecular mass not exceeding $80$, ammonium carbamate and ammonium carbonate can be ruled out. Therefore, **A** is urea, with the chemical formula $\mathrm{CO}(\mathrm{NH}_2)_2$.

<Checkpoint>[The chemical formula of **A** is $\mathrm{CO}(\mathrm{NH}_2)_2$](1)</Checkpoint>

The molar mass of **B** is:

$$
M_\mathrm{B}=\frac{5.15}{2\times5.0\times10^{-3}}=515\textup{ g/mol}
$$

<Checkpoint>[$M_\mathrm{B}=515$ g/mol](1)</Checkpoint>

In an Ar atmosphere and ethanol solution, urea most likely only undergoes a coordination reaction with $\mathrm{MCl}_3$. Assuming the molecular formula of **B** can be expressed as $\mathrm{MCl}_3\cdot n\mathrm{CO}(\mathrm{NH}_2)_2\cdot m\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$, substituting the relative molecular masses of urea and ethanol and testing multiple values reveals that only when $n=6$ and $m=0$ does **B** have a plausible chemical formula: $\mathrm{Ti}[\mathrm{CO}(\mathrm{NH}_2)_2]_6\mathrm{Cl}_3$. This is also consistent with details such as $\mathrm{Ti(III)}$ being easily oxidized (hence the need for an Ar atmosphere), the product appearing purple, and having a non-zero magnetic moment.

<Checkpoint>[The chemical formula of **B** is $\mathrm{Ti}[\mathrm{CO}(\mathrm{NH}_2)_2]_6\mathrm{Cl}_3$](1)</Checkpoint>

From this, it can be concluded that option A "The relative molecular mass of **A** is $79$" is incorrect, option B "$+4$ is also a common oxidation state of $\mathrm{M}$" is correct, option C "Solutions of $\mathrm{MCl}_3$ are usually colorless" is incorrect, and option D "The metal cation in **B** may be in a tetrahedral coordination environment" is incorrect.

<Checkpoint>[Option B is correct](1)</Checkpoint>
无水氯化钙可吸收的气体包括$\mathrm{H}_2\mathrm{O}$和$\mathrm{NH}_3$，而碱石灰可吸收$\mathrm{H}_2\mathrm{O}$和$\mathrm{CO}_2$等酸性气体，由此可知第二份样品进行实验时无水氯化钙仅吸收$\mathrm{NH}_3$。尝试各种酸性气体，可发现当第一份样品进行实验时，如果碱石灰吸收的气体为$\mathrm{CO}_2$，两种气体的物质的量之比为整数比：

$$
(33.0÷44.01)/(25.5÷17.034)=0.501\approx1/2
$$

<Checkpoint>[$n(\mathrm{CO}_2):n(\mathrm{NH}_3)=1:2$](1)</Checkpoint>

也即**A**可以分解为$1:2$的$\mathrm{CO}_2$和$\mathrm{NH}_3$，则A可能为碳酸铵$(\mathrm{NH}_4)_2\mathrm{CO}_3$（$96\textup{ g/mol}$）、氨基甲酸铵$\mathrm{H}_2\mathrm{NCOO}\mathrm{NH}_4$（$78 \textup{ g/mol}$）、尿素$\mathrm{CO}(\mathrm{NH}_2)_2$（$60\textup{ g/mol}$）。结合**A**为生活中常见的化合物且相对分子量不超过$80$，可以排除氨基甲酸铵和碳酸铵，故**A**为尿素，化学式为$\mathrm{CO}(\mathrm{NH}_2)_2$。

<Checkpoint>[**A**的化学式为$\mathrm{CO}(\mathrm{NH}_2)_2$](1)</Checkpoint>

**B**的摩尔质量为：

$$
M_\mathrm{B}=\frac{5.15}{2\times5.0\times10^{-3}}=515\textup{ g/mol}
$$

<Checkpoint>[$M_\mathrm{B}=515$ g/mol](1)</Checkpoint>

在Ar氛围和乙醇溶液中，尿素大概率只与$\mathrm{MCl}_3$发生配位反应，假设**B**的分子式可以表示为$\mathrm{MCl}_3\cdot n\mathrm{CO}(\mathrm{NH}_2)_2\cdot m\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$，代入尿素和乙醇的相对分子质量多次尝试可得，仅当$n=6$, $m=0$时，**B**有较为可靠的化学式：$\mathrm{Ti}[\mathrm{CO}(\mathrm{NH}_2)_2]_6\mathrm{Cl}_3$。这与$\mathrm{Ti(III)}$容易被氧化所以要在Ar氛围中进行实验、产物呈现紫色且磁矩不为零等细节也是一致的。

<Checkpoint>[**B**的化学式为$\mathrm{Ti}[\mathrm{CO}(\mathrm{NH}_2)_2]_6\mathrm{Cl}_3$](1)</Checkpoint>

此时可知，选项A“A的相对分子质量为$79$”是错误的，选项B“$+4$也是$\mathrm{M}$的常见化合价”是正确的，选项C“$\mathrm{MCl}_3$的溶液通常没有颜色”是错误的，选项D“**B**中的金属阳离子可能处于正四面体的配位环境中”是错误的。

<Checkpoint>[B选项正确](1)</Checkpoint>