$\bf A$ is a metallic element, and its powder is grayish-black. The element $\bf A$ reacts with fluorine gas upon heating to generate a fluoride molecule $\bf B$ with an octahedral configuration. $\bf B$ disproportionates in water to produce $\bf C$ and $\bf D$. $\bf C$ is a dark brown solid with a rutile structure; $\bf D$ is a diamagnetic light yellow solid with a low melting point and is easily volatile. $\bf C$ can also be obtained by heating in air. In potassium hydroxide solution, $\bf D$ reacts with an equimolar amount of potassium nitrite to produce a 2:1 type potassium salt $\bf E$; $\bf E$ can be heated in nitric acid to obtain $\bf D$. $\bf B$ reacts with an equimolar amount of $\bf D$ at $\rm 150\degree C$ to produce a compound $\bf F$ (containing $\bf A$ $68.1\%$). Structural determination shows that $\bf F$ is a polymer with two chemical environments of $\bf A$ atoms, $\bf A1$ and $\bf A2$ (both octahedrally coordinated), and the bond lengths of the chemical bonds they form are as follows:

$\bf A1$: $170.7 \mathrm{pm}, 182.0 \mathrm{pm}, 182.8 \mathrm{pm}, 185.5 \mathrm{pm}, 202.9 \mathrm{pm}, 204.4 \mathrm{pm}$

$\bf A2$: $170.6 \mathrm{pm}, 171.1 \mathrm{pm}, 171.1 \mathrm{pm}, 183.4 \mathrm{pm}, 214.0 \mathrm{pm}, 214.2 \mathrm{pm}$

Which of the following statements is correct:
$\bf A$为金属元素,其粉末为灰黑色。$\bf A$单质与氟气加热反应生成八面体构型的氟化物分子$\bf B$。$\bf B$遇水歧化生成$\bf C$和$\bf D$。$\bf C$为深褐色固体,具有金红石结构; $\bf D$为抗磁性浅黄色固体,熔点低,易挥发。$\bf C$在空气中加热亦可得到$\bf D$。在氢氧化钾溶液中,$\bf D$与等摩尔的亚硝酸钾反应,可制得2:1型钾盐$\bf E$; $\bf E$在硝酸中加热又可得到$\bf D$。$\bf B$与等摩尔的$\bf D$在$\rm 150\degree C$下反应, 可制得化合物$\bf F$(含$\bf A$ $68.1\%$)。结构测定表明,$\bf F$为聚合物,具有$\bf A1$和$\bf A2$两种化学环境$\bf A$的原子(均为八面体配位),它们形成的各化学键键长如下:

$\bf A1$: $170.7 \mathrm{pm}, 182.0 \mathrm{pm}, 182.8 \mathrm{pm}, 185.5 \mathrm{pm}, 202.9 \mathrm{pm}, 204.4 \mathrm{pm}$

$\bf A2$: $170.6 \mathrm{pm}, 171.1 \mathrm{pm}, 171.1 \mathrm{pm}, 183.4 \mathrm{pm}, 214.0 \mathrm{pm}, 214.2 \mathrm{pm}$

下列说法正确的是：
All other options are incorrect
The atomic ground state of element $\bf A$ has four unpaired electrons.
$\bf D$ is non-toxic
The chemical equation that generates $\bf E$, when the coefficients are balanced to the simplest integer ratio, has a sum of coefficients of all species greater than 10.
The chemical equation for $\bf E$ heated in nitric acid, when the coefficients are balanced to the simplest integer ratio, the sum of the reactant coefficients and the sum of the product coefficients are both not prime numbers.
$\bf B$ molecule has an octahedral configuration.
According to the information in the question, the magnetic moment of the $\bf F$ molecule can be predicted to be almost 0.
其他选项均不正确
元素$\bf A$的原子基态具有四个单电子
$\bf D$不具有毒性
生成$\bf E$的化学方程式，系数配平为最简整数比时，所有物种的系数和大于10
$\bf E$在硝酸中加热的化学方程式，系数配平为最简整数比时，反应物系数和和产物系数和均不为质数
$\bf B$分子为正八面体构型
根据题目信息，可预测$\bf F$分子的磁矩几乎为0
According to the information in the question, it is obvious that $\bf B$ is a hexafluoride, $\bf C$ is a dioxide, and $\bf D$ is a highest-valence oxide.

<Checkpoint>[$\bf B$ is a hexafluoride, $\bf C$ is a dioxide, and $\bf D$ is a highest-valence oxide](1)</Checkpoint>

$\bf B, D$ react in equimolar amounts to obtain $\bf F$, indicating that $\bf F$ contains $\rm F, O$ elements. Let the chemical formula of $\bf F$ be $\rm \bf A\rm F_xO_y$, and try different values for x and y:

<Checkpoint>[$\bf F$ contains $\rm F, O$ elements](1)</Checkpoint>

x=y=1, $\rm Mr(\bf A\rm) = 74.7$, no reasonable metal.

x=1,y=2, $\rm Mr(\bf A\rm) = 108.9$, no reasonable metal.

x=y=2, $\rm Mr(\bf A\rm) = 149.4$, no reasonable metal.

x=2,y=1, $\rm Mr(\bf A\rm) = 115.3$, no reasonable metal.

x=3,y=2, $\rm Mr(\bf A\rm) = 190.0$, which is $\rm Os$.

The others have no reasonable solutions.

Therefore, element $\bf A$ is $\rm Os$. Then, $\bf B$ is $\rm OsF_6$, $\bf C$ is $\rm OsO_2$, $\bf D$ is $\rm OsO_4$, $\bf F$ is $\rm OsO_2F_3$.

<Checkpoint>[Based on the mass fraction deduction, element $\bf A$ is $\rm Os$](2)</Checkpoint>

<Checkpoint>[$\bf B$ is $\rm OsF_6$](1)</Checkpoint>

<Checkpoint>[$\bf D$ is $\rm OsO_4$](1)</Checkpoint>

<Checkpoint>[$\bf C$ is $\rm OsO_2$](1)</Checkpoint>

<Checkpoint>[$\bf F$ is $\rm OsO_2F_3$](1)</Checkpoint>

Based on the properties, options B and C are both incorrect.

$\rm OsO_4$ reacts with equimolar $\rm KNO_2$ in potassium hydroxide to produce a 2:1 type potassium salt. It is easy to think that this reaction is a redox reaction. Since it is an equimolar reaction, nitrite is oxidized to nitrate, so $\rm Os(VIII)$ is reduced to $\rm Os(VI)$; the common divalent anion of $\rm Os(VI)$ under alkaline conditions is $\rm OsO_2(OH)_4^{2-}$. Therefore, $\bf E$ is $\rm K_2[OsO_2(OH)_4]$. Therefore, balance the chemical equation of the reaction:

<Checkpoint>[Nitrite is oxidized to nitrate, so $\rm Os(VIII)$ is reduced to $\rm Os(VI)$](1)</Checkpoint>

<Checkpoint>[$\bf E$ is $\rm K_2[OsO_2(OH)_4]$](1)</Checkpoint>

$\rm OsO_4 + KNO_2 + H_2O + 2KOH = K_2[OsO_2(OH)_4] + KNO_3$

<Checkpoint>[$\rm OsO_4 + KNO_2 + H_2O + 2KOH = K_2[OsO_2(OH)_4] + KNO_3$](1)</Checkpoint>

The sum of the coefficients is 7, so option D is incorrect.

$\rm K_2[OsO_2(OH)_4]$ releases $\rm NO$ in nitric acid, producing $\rm OsO_4$. Balance the redox equation:

$\rm 3K_2[OsO_2(OH)_4] + 8HNO_3 = 3OsO_4 + 2NO + 6KNO_3 + 10H_2O$

<Checkpoint>[$\rm 3K_2[OsO_2(OH)_4] + 8HNO_3 = 3OsO_4 + 2NO + 6KNO_3 + 10H_2O$ ](1)</Checkpoint>

The sum of the coefficients has one prime number, so option E is incorrect.

According to the Jahn-Teller effect, the electronic configuration of $\rm Os(VI)$ is $\rm t_{2g}^2e_g^0$, which will cause distortion and is not a regular octahedron, so option F is incorrect.

<Checkpoint>[According to the Jahn-Teller effect, the electronic configuration of $\rm Os(VI)$ is $\rm t_{2g}^2e_g^0$, which will cause distortion and is not a regular octahedron](1)</Checkpoint>

According to the bond length data of $\rm OsO_2F_3$, in this polymer structure, $\rm Os$ exists in two environments: one is connected to five $\rm F$ and one $\rm O$, and the other is connected to three $\rm F$ and three $\rm O$. Therefore, the valences of the two $\rm Os$ are $\rm Os(VI), Os(VIII)$ respectively. Obviously, $\rm Os(VII)$ does not have unpaired electrons. The terminal $\rm Os=O$ bond of $\rm Os(VI)$ is much shorter than the $\rm Os-F$ bond, so it can be regarded as a flattened octahedral field coordination. In the case of two d electrons, the lowest energy $\rm d_{xy}$ is filled, and there are no unpaired electrons. Therefore, the magnetic moment of this compound is almost 0, so option G is correct.

<Checkpoint>[In $\rm OsO_2F_3$, the valences of the two $\rm Os$ are $\rm Os(VI),Os(VIII)$](1)</Checkpoint>

<Checkpoint>[The terminal $\rm Os=O$ bond of $\rm Os(VI)$ is much shorter than the $\rm Os-F$ bond, so it can be regarded as a flattened octahedral field coordination](1)</Checkpoint>

<Checkpoint>[$\rm Os(VI)$ has two d electrons, filling the lowest energy $\rm d_{xy}$, with no unpaired electrons](1)</Checkpoint>

<Checkpoint>[The magnetic moment of the compound $\rm OsO_2F_3$ is almost 0](1)</Checkpoint>
根据题目信息可知很明显$\bf B$为六氟化物，$\bf C$为二氧化物，$\bf D$为最高价氧化物。

<Checkpoint>[$\bf B$为六氟化物，$\bf C$为二氧化物，$\bf D$为最高价氧化物](1)</Checkpoint>

$\bf B,D$等摩尔反应得到$\bf F$，可知$\bf F$含有$\rm F,O$元素。设$\bf F$的化学式为$\rm \bf A\rm F_xO_y$，对x,y的取值进行尝试：

<Checkpoint>[$\bf F$含有$\rm F,O$元素](1)</Checkpoint>

x=y=1, $\rm Mr(\bf A\rm) = 74.7$，无合理金属。

x=1,y=2, $\rm Mr(\bf A\rm) = 108.9$，无合理金属。

x=y=2, $\rm Mr(\bf A\rm) = 149.4$，无合理金属。

x=2,y=1, $\rm Mr(\bf A\rm) = 115.3$，无合理金属。

x=3,y=2, $\rm Mr(\bf A\rm) = 190.0$，为$\rm Os$。

其余均无合理解。

因此元素$\bf A$为$\rm Os$。则，$\bf B$为$\rm OsF_6$，$\bf C$为$\rm OsO_2$，$\bf D$为$\rm OsO_4$，$\bf F$为$\rm OsO_2F_3$。

<Checkpoint>[根据质量分数推断，元素$\bf A$为$\rm Os$](2)</Checkpoint>

<Checkpoint>[$\bf B$为$\rm OsF_6$](1)</Checkpoint>

<Checkpoint>[$\bf D$为$\rm OsO_4$](1)</Checkpoint>

<Checkpoint>[$\bf C$为$\rm OsO_2$](1)</Checkpoint>

<Checkpoint>[$\bf F$为$\rm OsO_2F_3$](1)</Checkpoint>

根据性质可知选项B,C均错误。

$\rm OsO_4$与等摩尔$\rm KNO_2$在氢氧化钾中反应产生2：1型钾盐，比较容易想到该反应为氧化还原反应，由于等摩尔反应，亚硝酸根被氧化为硝酸根，则$\rm Os(VII)$降低为$\rm Os(VI)$；$\rm Os(VI)$碱性条件下常见的二价阴离子为$\rm OsO_2(OH)_4^{2-}$。故$\bf E$为$\rm K_2[OsO_2(OH)_4]$。因此配平反应化学方程式：

<Checkpoint>[亚硝酸根被氧化为硝酸根，则$\rm Os(VII)$降低为$\rm Os(VI)$](1)</Checkpoint>

<Checkpoint>[$\bf E$为$\rm K_2[OsO_2(OH)_4]$](1)</Checkpoint>

$\rm OsO_4 + KNO_2 + H_2O + 2KOH = K_2[OsO_2(OH)_4] + KNO_3$

<Checkpoint>[$\rm OsO_4 + KNO_2 + H_2O + 2KOH = K_2[OsO_2(OH)_4] + KNO_3$](1)</Checkpoint>

系数和为7，选项D错误。

$\rm K_2[OsO_2(OH)_4]$在硝酸中放出$\rm NO$，产生$\rm OsO_4$，配平氧化还原方程式：

$\rm 3K_2[OsO_2(OH)_4] + 8HNO_3 = 3OsO_4 + 2NO + 6KNO_3 + 10H_2O$ 

<Checkpoint>[$\rm 3K_2[OsO_2(OH)_4] + 8HNO_3 = 3OsO_4 + 2NO + 6KNO_3 + 10H_2O$ ](1)</Checkpoint>

系数和有一个为质数，选项E错误。

根据Jahn-Taller效应，$\rm Os(VI)$电子组态为$\rm t_{2g}^2e_g^0$，会发生畸变，不为正八面体，选项F错误。

<Checkpoint>[根据Jahn-Taller效应，$\rm Os(VI)$电子组态为$\rm t_{2g}^2e_g^0$，会发生畸变，不为正八面体](1)</Checkpoint>

根据$\rm OsO_2F_3$的键长数据可知，该高分子聚合结构中$\rm Os$存在两种环境，一种与五个$\rm F$和一个$\rm O$相连，另一种与三个$\rm F$和三个$\rm O$相连，因此两种$\rm Os$的价态分别为$\rm Os(VI),Os(VIII)$。很明显$\rm Os(VII)$ 不存在单电子，$\rm Os(VI)$的端基$\rm Os=O$键远短于$\rm Os-F$键，因此可看作压扁的八面体场配位，在两个d电子情况下填充能量最低的$\rm d_{xy}$，无单电子。因此该化合物磁矩几乎为0，选项G正确。

<Checkpoint>[$\rm OsO_2F_3$中，两种$\rm Os$的价态分别为$\rm Os(VI),Os(VIII)$](1)</Checkpoint>

<Checkpoint>[$\rm Os(VI)$的端基$\rm Os=O$键远短于$\rm Os-F$键，因此可看作压扁的八面体场配位](1)</Checkpoint>

<Checkpoint>[$\rm Os(VI)$存在两个d电子，填充能量最低的$\rm d_{xy}$，无单电子](1)</Checkpoint>

<Checkpoint>[化合物$\rm OsO_2F_3$的磁矩几乎为0](1)</Checkpoint>