For a certain metal $\mathbf{X}$, the oxide $\mathbf{A}$ of $\mathbf{X}$ and a common ternary acid mixed with its $1:1$ ammonium salt are co-heated to $300^\circ C$. After washing with water and drying, compound $\mathbf{Y}$ is obtained, with a formula weight between $240$ and $260$, and contains no water of crystallization. If it is dissolved in dilute sulfuric acid, the solution color does not change significantly, but a precipitate is formed (reaction $1$), while dissolving it in concentrated sulfuric acid yields a clear red solution. Heating compound $\mathbf{Y}$ to $340^\circ C$ in an inert atmosphere results in a weight loss of $3.65\%$, yielding compound $\mathbf{B}$ (reaction $2$); continuing to heat to $460^\circ C$ results in another weight loss of $10.14\%$, yielding compound $\mathbf{C}$ (reaction $3$); subsequently heating to $800^\circ C$ does not result in significant weight loss, and compound $\mathbf{C}$ melts and completely transforms into compound $\mathbf{D}$. The formula weight of $\mathbf{D}$ is 1.5 times that of $\mathbf{C}$.

The chemical equation coefficients for all reactions are the simplest integers. Determine which of the following options is correct through reasoning.
对于某种金属$\mathbf{X}$， 将 $\mathbf{X}$的氧化物  $\mathbf{A}$、 一种常见三元酸与该酸的$1:1$型铵盐混合共热至 $3 0 0 ^ { \circ } C$ ，经水洗、干燥可得到化合物 $\mathbf{Y}$，式量在$240$到$260$之间，且不含结晶水。若将其溶于稀硫酸，溶液颜色无显著变化但有沉淀生成(反应$1$)，而将其溶于浓硫酸则会得到澄清的红色溶液。将化合物 $\mathbf{Y}$于惰性气氛中加热至 $340 ^ { \circ}C$ ，失重 $3.65\% ,$ 得到化合物 $\mathbf{B}$(反应$2$)；继续加热至 $460 ^ { \circ }C$ 又失重 $10.14\%$ ，得到化合物 $\mathbf{C}$(反应$3$)；随后加热至 $800 ^ { \circ } C$ ，未观察到明显失重，化合物 $\mathbf{C}$发生熔化并完全转化为化合物 $\mathbf{D}$。$\mathbf{D}$的式量为$\mathbf{C}$的1.5倍。

所有反应的化学方程式系数为最简整数，通过推理判断下面哪个选项是正确的。
The $+5$ oxidation state is the stable oxidation state of metal $\mathbf{X}$.
$\mathbf{Y}$ has strong oxidizing properties
The number of product species in reaction $1$ is greater than the number of reactant species.
The sum of stoichiometric coefficients of products in reaction $1$ is greater than the sum of stoichiometric coefficients of reactants.
The average relative molecular mass of the weight loss portion $3$ is less than the average relative molecular mass of the weight loss portion $2$.
In all the compounds in this question, the metal elements have a total of $4$ valence states.
None of the above options is correct
$+5$氧化态是金属$\mathbf{X}$的稳定氧化态
$\mathbf{Y}$具有强氧化性
反应$1$中产物的物种数大于反应物的物种数
反应$1$中产物系数和大于反应物系数和
反应$3$失重部分的平均相对分子质量小于反应$2$失重部分 的平均相对分子质量
本题目所有化合物中，金属元素共有$4$种价态
以上选项均不正确
Compound $\mathbf{Y}$ is soluble in concentrated sulfuric acid to obtain a clear red solution, possibly containing $\mathrm{Mn}$ compounds, ferricyanide, iodine, etc. Combined with other information in the question that there are no cyanide ions and iodine, it can be explained that the central metal is $\mathrm{Mn}$; the composition of compound $\mathbf{Y}$ can also be analyzed. The common ternary acid is phosphoric acid, so $\mathbf{Y}$ may be ammonium phosphate or ammonium pyrophosphate. Combining the relative molecular mass range, ammonium pyrophosphate is a reasonable answer. At this time, the metal ion corresponds to the first transition period metal element. Combining the aforementioned chemical properties analysis, it can be determined that $\mathbf{X}$ is $\mathrm{Mn}$. The $+5$ oxidation state is not a stable oxidation state of $\mathrm{Mn}$, so option $A$ is incorrect.

<Checkpoint>[Metal $\mathbf{X}$ is $\mathrm{Mn}$.](1)</Checkpoint>

<Checkpoint>[$+5$ oxidation state is not a stable oxidation state of $\mathrm{Mn}$, so option $A$ is incorrect](0.5)</Checkpoint>

The reaction equation for the formation of compound $\mathbf{Y}$:

$4 \mathrm{MnO}_{2}+4 \mathrm{NH}_{4} \mathrm{H}_{2} \mathrm{PO}_{4}+4 \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 4 \mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$

$\mathrm{Mn}$ in $\mathbf{Y}$ has a $+3$ oxidation state and does not have strong oxidizing properties, so option $B$ is incorrect.

<Checkpoint>[$\mathbf{Y}$ is $\mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}$, in which $\mathrm{Mn}$ has a $+3$ oxidation state and does not have strong oxidizing properties, so option $B$ is incorrect.](1)</Checkpoint>

Reaction $1$: $2 \mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+\mathrm{MnSO}_{4}+\mathrm{MnO}_{2}+4 \mathrm{H}_{3} \mathrm{PO}_{4}$

The number of species of products in reaction 1 is $4$, the sum of the coefficients of products is $7$, the number of species of reactants is $3$, and the sum of the coefficients of reactants is $8$, so option $C$ is correct and option $D$ is incorrect.

<Checkpoint>[The number of species of products in reaction 1 is $4$, and the number of species of reactants is $3$, the former is greater than the latter, so option $C$ is correct.](1)</Checkpoint>

<Checkpoint>[The sum of the coefficients of products in reaction 1 is $7$, and the sum of the coefficients of reactants is $8$, the former is less than the latter, so option $D$ is incorrect.](1)</Checkpoint>

Inert atmosphere heating weight loss process:

First step weight loss: $\mathrm{M}_{1}=w_{1} \times \mathrm{H}=3.65 \times 10^{-2} \times 246.922 \mathrm{~g} \cdot \mathrm{mol}^{-1}=9.01 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, corresponding to 0.5 molecules of $\mathrm{H}_{2} \mathrm{O}$;

<Checkpoint>[First step weight loss $9.01 \ \ g/mol$, corresponding to 0.5 $H_2O$](1)</Checkpoint>

Second step weight loss: $\mathrm{H}_{2}=w_{2} \times \mathrm{H}=10.14 \times 10^{-2} \times 246.922 \mathrm{~g} \cdot \mathrm{mol}^{-1}=25.04 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, corresponding to 0.5 molecules of $\mathrm{NH}_{3}$ and 0.5 molecules of $\mathrm{NH}_{2} \mathrm{OH}$ (further decomposed into $\mathrm{N}_{2} 、 \mathrm{H}_{2} \mathrm{O} 、 \mathrm{NH}_{3}$ );

<Checkpoint>[Second step weight loss $25.04 \ \ g/mol$](1)</Checkpoint>

Reaction $2$: $2 \mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7} \rightarrow \mathrm{Mn}_{2}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{P}_{4} \mathrm{O}_{13}+\mathrm{H}_{2} \mathrm{O}$

Reaction $3$: $3 \mathrm{Mn}_{2}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{P}_{4} \mathrm{O}_{13} \rightarrow 3 \mathrm{Mn}_{2} \mathrm{P}_{4} \mathrm{O}_{12}+4 \mathrm{NH}_{3}+\mathrm{N}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

From the above, the average relative molecular mass of the weight loss portion of reaction $3$:

$m=(4\times17.034+28.02+3\times18.016)/8=18.78$

The average relative molecular mass of the weight loss portion of reaction $3$ is $18.78$, and the average relative molecular mass of the weight loss portion of reaction $2$ is $18.02$, so option $E$ is incorrect.

In the compounds involved in the entire question, there are three oxidation states of $\mathrm{Mn}$: $+2$ oxidation state ($\mathrm{Mn}_{2} \mathrm{P}_{4} \mathrm{O}_{12}$), $+3$ oxidation state ($\mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}$), and $+4$ oxidation state ($\mathrm{MnO}_{2}$), so option $F$ is incorrect.

<Checkpoint>[The average relative molecular mass of the weight loss portion of reaction $3$ is $18.78$, and the average relative molecular mass of the weight loss portion of reaction $2$ is $18.02$, so option $E$ is incorrect.](0.5)</Checkpoint>

<Checkpoint>[In the compounds involved in the entire question, there are three oxidation states of $\mathrm{Mn}$: $+2$ oxidation state ($\mathrm{Mn}_{2} \mathrm{P}_{4} \mathrm{O}_{12}$), $+3$ oxidation state ($\mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}$), and $+4$ oxidation state ($\mathrm{MnO}_{2}$), so option $F$ is incorrect.](1)</Checkpoint>
化合物$\mathbf{Y}$可溶于浓硫酸得到澄清红色溶液，可能是含$\mathrm{Mn}$化合物、铁氰化物、碘单质等，结合其它题干不存在氰根离子和碘，可说明中心金属是$\mathrm{Mn}$；还可分析化合物$\mathbf{Y}$组成，常见三元酸为磷酸，则$\mathbf{Y}$可能是磷酸铵盐、焦磷酸铵盐，结合式量区间焦磷酸铵盐为合理答案，此时金属离子对应第一过渡周期金属元素，结合前述化学性质分析可以确定$\mathbf{X}$为$\mathrm{Mn}$。$+5$氧化态不是$\mathrm{Mn}$的稳定氧化态，选项$A$错误。

<Checkpoint>[金属$\mathbf{X}$为$\mathrm{Mn}$。](1)</Checkpoint>

<Checkpoint>[$+5$氧化态不是$\mathrm{Mn}$的稳定氧化态，选项$A$错误](0.5)</Checkpoint>

生成化合物$\mathbf{Y}$的反应方程式：

$4 \mathrm{MnO}_{2}+4 \mathrm{NH}_{4} \mathrm{H}_{2} \mathrm{PO}_{4}+4 \mathrm{H}_{3} \mathrm{PO}_{4} \rightarrow 4 \mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}+10 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}$

$\mathbf{Y}$中$\mathrm{Mn}$为$+3$氧化态，不具有强氧化性，选项$B$错误。

<Checkpoint>[$\mathbf{Y}$为$\mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}$，其中$\mathrm{Mn}$为$+3$氧化态，不具有强氧化性，选项$B$错误。](1)</Checkpoint>

反应$1$：$2 \mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+4 \mathrm{H}_{2} \mathrm{O} \rightarrow\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}+\mathrm{MnSO}_{4}+\mathrm{MnO}_{2}+4 \mathrm{H}_{3} \mathrm{PO}_{4}$

反应1产物的物种数为$4$，产物系数和为$7$，反应物的物种数为$3$，反应物系数和为$8$，选项$C$正确，选项$D$错误。

<Checkpoint>[反应1产物的物种数为$4$，反应物的物种数为$3$，前者大于后者，选项$C$正确。](1)</Checkpoint>

<Checkpoint>[反应1产物系数和为$7$，反应物系数和为$8$，前者小于后者，选项$D$错误。](1)</Checkpoint>

惰性气氛加热失重过程：

第一步失重: $\mathrm{M}_{1}=w_{1} \times \mathrm{H}=3.65 \times 10^{-2} \times 246.922 \mathrm{~g} \cdot \mathrm{mol}^{-1}=9.01 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, 对应 0.5 个
$\mathrm{H}_{2} \mathrm{O}$ 分子;

<Checkpoint>[第一步失重$9.01 \ \ g/mol$ ，对应0.5个$H_2O$](1)</Checkpoint>

第二步失重: $\mathrm{H}_{2}=w_{2} \times \mathrm{H}=10.14 \times 10^{-2} \times 246.922 \mathrm{~g} \cdot \mathrm{mol}^{-1}=25.04 \mathrm{~g} \cdot \mathrm{mol}^{-1}$, 对应 0.5
个 $\mathrm{NH}_{3}$ 分子和 0.5 个 $\mathrm{NH}_{2} \mathrm{OH}$ 分子 (进一步分解为 $\mathrm{N}_{2} 、 \mathrm{H}_{2} \mathrm{O} 、 \mathrm{NH}_{3}$ );

<Checkpoint>[第二步失重$25.04 \ \ g/mol$](1)</Checkpoint>

反应$2$：$2 \mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7} \rightarrow \mathrm{Mn}_{2}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{P}_{4} \mathrm{O}_{13}+\mathrm{H}_{2} \mathrm{O}$

反应$3$：$3 \mathrm{Mn}_{2}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{P}_{4} \mathrm{O}_{13} \rightarrow 3 \mathrm{Mn}_{2} \mathrm{P}_{4} \mathrm{O}_{12}+4 \mathrm{NH}_{3}+\mathrm{N}_{2}+3 \mathrm{H}_{2} \mathrm{O}$

由上，反应$3$失重部分的平均相对分子量：

$m=(4\times17.034+28.02+3\times18.016)/8=18.78$

反应$3$失重部分平均相对分子量为$18.78$，反应$2$失重部分的平均相对分子量为$18.02$，选项$E$错误。

整个题目涉及化合物中，共涉及到$+2$氧化态($\mathrm{Mn}_{2} \mathrm{P}_{4} \mathrm{O}_{12}$)、$+3$氧化态($\mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}$)、$+4$氧化态($\mathrm{MnO}_{2}$)三种氧化态的$\mathrm{Mn}$，选项$F$错误。。

<Checkpoint>[反应$3$失重部分平均相对分子量为$18.78$，反应$2$失重部分的平均相对分子量为$18.02$，选项$E$错误。](0.5)</Checkpoint>

<Checkpoint>[整个题目涉及化合物中，涉及到$+2$氧化态($\mathrm{Mn}_{2} \mathrm{P}_{4} \mathrm{O}_{12}$)、$+3$氧化态($\mathrm{NH}_{4} \mathrm{MnP}_{2} \mathrm{O}_{7}$)、$+4$氧化态($\mathrm{MnO}_{2}$)三种氧化态的$\mathrm{Mn}$，选项$F$错误。](1)</Checkpoint>