$\bf M$ is an element of the sixth period. The flue gas obtained by leaching roasted molybdenite with water, the volatile oxide $\bf N$ of $\bf M$ (containing $\bf M$ $\rm 76.9 \%$) is absorbed, and the resulting solution contains acid $\mathbf{A}$; the solution is purified, concentrated, and alkaline gas $\mathbf{B}$ is passed into it, crystallization occurs, and $\mathbf{A}$ is converted into ammonium salt $\mathbf{C}$; $\mathbf{C}$ is reduced by hydrogen at $\rm 1000 \degree C$ to obtain $\mathbf{M}$ element. If $\mathbf{C}$ is directly heated to decompose, then oxide $\mathbf{D}$ is obtained.

Adding $6 \mathrm{~mol/L}$ hydrochloric acid to $\mathbf{A}$ and adding potassium iodide for reduction, a green paramagnetic potassium salt $\mathbf{E}$ can be prepared, its magnetic moment $\mu=3.25 ~\mu_{\mathrm{B}}$; adding $\mathrm{AgNO_3}$ to the saturated solution of $\mathbf{E}$ yields an orange precipitate $\mathbf{F}$; $\mathbf{F}$ decomposes upon heating to give silver chloride and two other chlorides $\bf G,H$; $\bf G,H$ are both chlorine-bridged polymers, where $\bf G$ is a dimer and the metal is six-coordinate; $\bf H$ is a trimer, containing metal-metal bonds, and the ratio of bridging chlorine to terminal chlorine is $1:2$. $\bf G$ can also be prepared by directly heating $\bf M$ element with chlorine gas to $\rm 400 \degree C$; $\bf G$ undergoes a disproportionation reaction with water to obtain a hydrate precipitate of $\bf D$ and a solution of $\bf A$ (the molar ratio of the products is $2:1$).

The following statement is correct:
$\bf M$为第六周期元素。用水淋洗焙烧辉钼矿得到的烟道气, $\bf M$挥发性的氧化物$\bf N$(含$\bf M$ $\rm 76.9 \%$) 被吸收, 得到的溶液中含有酸 $\mathbf{A}$; 该溶液经净化、浓缩, 向其中通入碱性气体$\mathbf{B}$, 析晶, $\mathbf{A}$ 转化为铵盐 $\mathbf{C}$; $\mathbf{C}$在 $\rm 1000 \degree C$下被氢气还原得到 $\mathbf{M}$ 单质。若将 $\mathbf{C}$ 直接加热分解, 则得到氧化物 $\mathbf{D}$ 。

向 $\mathbf{A}$ 中加入 $6 \mathrm{~mol/L}$ 盐酸, 并加入碘化钾还原, 可制得绿色顺磁性钾盐 $\mathbf{E}$, 其磁矩 $\mu=3.25 ~\mu_{\mathrm{B}}$; 向 $\mathbf{E}$的饱和溶液中加入 $\mathrm{AgNO_3}$, 得到橙色沉淀$\mathbf{F}$; $\mathbf{F}$ 受热分解, 得到氯化银和另外两种氯化物$\bf G,H$; $\bf G,H$ 均为含氯桥键的多聚体,其中$\bf G$为二聚体,金属为六配位; $\bf H$为三聚体,含金属-金属键,桥氯与端基氯的比为$1:2$。$\bf G$也可由$\bf M$单质与氯气直接加热至$\rm 400 \degree C$反应制备; $\bf G$遇水发生歧化反应, 得到$\bf D$的水合物沉淀和$\bf A$的溶液(产物的摩尔比为$2:1$)。

下列说法正确的是：
All other options are incorrect
Element $\bf M$ has an even number of unpaired electrons in its ground state.
Element $\bf M$ does not contain common radioactive nuclides.
The mass fraction of $\bf M$ in $\bf D$ is $\rm 79.5~\%$
The oxidation state of $\bf M$ in $\bf E$ is different from that of $\bf D$.
The mass fraction of $\bf M$ in $\bf F$ is $\rm 34.2~\%$
The mass fraction of $\bf M$ in $\bf G$ is greater than $\bf H$
The chemical equation for the reduction of $\bf A$ to obtain $\bf E$, when the coefficients are balanced to the simplest integer ratio, has a sum of 37 for the coefficients of all substances.
其他选项均不正确
元素$\bf M$基态含有偶数个单电子
元素$\bf M$不含有常见的放射性核素
$\bf D$中$\bf M$的质量分数为$\rm 79.5~\%$ 
$\bf E$中$\bf M$的氧化态与$\bf D$不同
$\bf F$中$\bf M$的质量分数为$\rm 34.2~\%$
$\bf G$中$\bf M$的质量分数大于$\bf H$
$\bf A$被还原得到$\bf E$的化学方程式，系数配平为最简整数比时，所有物质的系数和为37
The flue gas obtained from roasting molybdenite can yield the sixth-period $\bf M$, indicating that the position of $\bf M$ on the periodic table is close to $\rm Mo$, and it may be $\rm Ta, W, Re$.

<Checkpoint>[The position of $\bf M$ on the periodic table is close to $\rm Mo$](1)</Checkpoint>

After roasting, elements are generally oxidized to their highest valence state. The volatile oxide $\bf N$ of $\bf M$ may be $\rm Ta_2O_5, WO_3, Re_2O_7$; based on the mass fraction of $\rm 76.9\%$, $\bf N$ can be calculated to be $\rm Re_2O_7$. Therefore, element $\bf M$ is $\rm Re$.

<Checkpoint>[$\bf N$ is $\rm Re_2O_7$, element $\bf M$ is $\rm Re$](1)</Checkpoint>

The valence electron configuration of $\rm Re$ is $\rm 5d^56s^2$, containing five unpaired electrons, so option B is incorrect. $\rm ^{187}Re$ is a radioactive nuclide, so option C is incorrect.

<Checkpoint>[$\rm ^{187}Re$ is a radioactive nuclide](1)</Checkpoint>

$\rm Re_2O_7$ reacts with water to produce acid $\bf A$, so $\bf A$ is $\rm HReO_4$; its reaction with alkaline gas produces ammonium salt, so the ammonium salt $\bf C$ is obviously $\rm NH_4ReO_4$.

<Checkpoint>[Ammonium salt $\bf C$ is obviously $\rm NH_4ReO_4$](1)</Checkpoint>

$\rm NH_4ReO_4$ decomposes upon direct heating. $\rm Re(VII)$ in the molecule has strong oxidizing properties, which can oxidize the ammonium ion to nitrogen gas, while it is reduced to the stable valence oxide $\rm ReO_2$; that is, $\bf D$ is $\rm ReO_2$.

<Checkpoint>[$\rm Re(VII)$ has strong oxidizing properties, which can oxidize the ammonium ion to nitrogen gas, while it is reduced to the stable valence oxide $\rm ReO_2$](1)</Checkpoint>

<Checkpoint>[$\bf D$ is $\rm ReO_2$](1)</Checkpoint>

$\rm HReO_4$ is reduced by hydrochloric acid and potassium iodide to obtain potassium salt $\mathbf{E}$. According to the magnetic moment $\mu=3.25 ~\mu_{\mathrm{B}}=\sqrt{n(n+2)}~\mu_{\mathrm{B}}$, it can be calculated that $\mathbf{E}$ contains 3 unpaired electrons. (The theoretical magnetic moment should be $\mu=3.87 ~\mu_{\mathrm{B}}$, but the sixth-period elements have a large deviation due to relativistic effects.)

<Checkpoint>[According to the magnetic moment $\mu=3.25 ~\mu_{\mathrm{B}}=\sqrt{n(n+2)}~\mu_{\mathrm{B}}$, it can be calculated that $\mathbf{E}$ contains 3 unpaired electrons](1)</Checkpoint>

Therefore, $\mathbf{E}$ contains $\rm{Re(IV)}$; considering the stable octahedral hexacoordination, the chemical formula of $\mathbf{E}$ is $\rm K_2ReCl_6$.

<Checkpoint>[$\mathbf{E}$ contains $\rm{Re(IV)}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{E}$ chemical formula is $\rm K_2ReCl_6$](1)</Checkpoint>

Adding nitrate to $\rm K_2ReCl_6$ yields a precipitate. According to the principle of like dissolves like, the precipitate $\bf F$ is $\rm Ag_2ReCl_6$.

<Checkpoint>[According to the principle of like dissolves like, the precipitate $\bf F$ is $\rm Ag_2ReCl_6$](1)</Checkpoint>

According to the chemical formulas, options D, E, and F are all incorrect.

$\bf G$ contains chlorine bridges and is dimeric, which is obviously $\rm Re_2Cl_{10}$ with two octahedra connected by edge-sharing. $\bf H$ is trinuclear and contains metal-metal bonds, and should be a triangular configuration; there should be 3 bridging chlorines as the sides of the triangle, and the terminal chlorines are 6 according to the ratio, so $\bf H$ is $\rm Re_3Cl_{9}$. From the chemical formula, option G is incorrect.

<Checkpoint>[$\bf G$ is $\rm Re_2Cl_{10}$ with two octahedra connected by edge-sharing](1)</Checkpoint>

<Checkpoint>[$\bf H$ is $\rm Re_3Cl_{9}$](1)</Checkpoint>

$\rm HReO_4$ is reduced by hydrochloric acid and potassium iodide to obtain potassium salt $\rm K_2ReCl_6$, and iodine is oxidized to elemental form. Therefore, the balanced equation is:

$\rm 2HReO_4 + 14HCl + 6KI = 2K_2ReCl_6 + 3I_2 + 2KCl + 8H_2O$

<Checkpoint>[$\rm 2HReO_4 + 14HCl + 6KI = 2K_2ReCl_6 + 3I_2 + 2KCl + 8H_2O$](1)</Checkpoint>

The sum of the coefficients is 37, so option H is correct.
焙烧辉钼矿得到的烟道气可得到第六周期的$\bf M$，说明$\bf M$在元素周期表上的位置与$\rm Mo$接近，可能为$\rm Ta,W,Re$。

<Checkpoint>[$\bf M$在元素周期表上的位置与$\rm Mo$接近](1)</Checkpoint>

焙烧后元素一般被氧化到最高价，$\bf M$的挥发性氧化物$\bf N$可能为$\rm Ta_2O_5,WO_3,Re_2O_7$；根据质量分数$\rm 76.9 \%$计算可得$\bf N$为$\rm Re_2O_7$。故元素$\bf M$为$\rm Re$。

<Checkpoint>[$\bf N$为$\rm Re_2O_7$，元素$\bf M$为$\rm Re$](1)</Checkpoint>

$\rm Re$价电子组态为$\rm 5d^56s^2$，含有五个单电子，选项B错误。$\rm ^{187}Re$为放射性核素，选项C错误。

<Checkpoint>[$\rm ^{187}Re$为放射性核素](1)</Checkpoint>

$\rm Re_2O_7$与水产生酸$\bf A$，则$\bf A$为$\rm HReO_4$；其与碱性气体反应产生铵盐，则铵盐$\bf C$明显为$\rm NH_4ReO_4$。

<Checkpoint>[铵盐$\bf C$明显为$\rm NH_4ReO_4$](1)</Checkpoint>

$\rm NH_4ReO_4$直接加热分解，分子中的$\rm Re(VII)$具有强氧化性，可以氧化铵根至氮气，自身被还原为稳定价态氧化物$\rm ReO_2$；即$\bf D$为$\rm ReO_2$。

<Checkpoint>[$\rm Re(VII)$具有强氧化性，可以氧化铵根至氮气，自身被还原为稳定价态氧化物$\rm ReO_2$](1)</Checkpoint>

<Checkpoint>[$\bf D$为$\rm ReO_2$](1)</Checkpoint>

$\rm HReO_4$加入盐酸和碘化钾还原得到钾盐$\mathbf{E}$，根据磁矩$\mu=3.25 ~\mu_{\mathrm{B}}=\sqrt{n(n+2)}~\mu_{\mathrm{B}}$可计算得到$\mathbf{E}$含有3个单电子。（理论磁矩应当为$\mu=3.87 ~\mu_{\mathrm{B}}$，但第六周期元素由于相对论效应有较大偏差。）

<Checkpoint>[根据磁矩$\mu=3.25 ~\mu_{\mathrm{B}}=\sqrt{n(n+2)}~\mu_{\mathrm{B}}$可计算得到$\mathbf{E}$含有3个单电子](1)</Checkpoint>

因此$\mathbf{E}$含有$\rm{Re(IV)}$；考虑到八面体稳定六配位，$\mathbf{E}$化学式为$\rm K_2ReCl_6$。

<Checkpoint>[$\mathbf{E}$含有$\rm{Re(IV)}$](1)</Checkpoint>

<Checkpoint>[$\mathbf{E}$化学式为$\rm K_2ReCl_6$](1)</Checkpoint>

$\rm K_2ReCl_6$加入硝酸盐得到沉淀，根据相似相溶原理可知沉淀$\bf F$为$\rm Ag_2ReCl_6$。

<Checkpoint>[根据相似相溶原理可知沉淀$\bf F$为$\rm Ag_2ReCl_6$](1)</Checkpoint>

根据化学式可知选项D,E,F均错误。

$\bf G$含氯桥且二聚，很明显为两个八面体共边连接的$\rm Re_2Cl_{10}$。$\bf H$为三核且含有金属-金属键，应当为三角形构型；桥氯应当有3个作为三角形的边，端基氯根据比例为6个，则$\bf H$为$\rm Re_3Cl_{9}$。由化学式可知选项G错误。

<Checkpoint>[$\bf G$为两个八面体共边连接的$\rm Re_2Cl_{10}$](1)</Checkpoint>

<Checkpoint>[$\bf H$为$\rm Re_3Cl_{9}$](1)</Checkpoint>

$\rm HReO_4$加入盐酸和碘化钾还原得到钾盐$\rm K_2ReCl_6$，碘被氧化为单质。故配平方程式：

$\rm 2HReO_4 + 14HCl + 6KI = 2K_2ReCl_6 + 3I_2 + 2KCl + 8H_2O$

<Checkpoint>[$\rm 2HReO_4 + 14HCl + 6KI = 2K_2ReCl_6 + 3I_2 + 2KCl + 8H_2O$](1)</Checkpoint>

系数和为37，选项H正确。