Gas **A**, which is very important to industry and agriculture, has rich chemical properties. Before World War I, German chemist Haber optimized the conditions for the direct reaction of elements to produce **A**, making it possible for large-scale industrial production. Electrolysis of $HF$ salts of **A** and $HF$ melts can yield a colorless and odorless stable gas **B**. In addition to this method, first used by the German OttoRuff group, controlled fluorination of **A** over a copper catalyst can also produce **B**; however, if the temperature is not properly controlled, the generated **B** will fluorinate the copper catalyst and transform itself into **C**. At room temperature and above, **C** easily dissociates and can be regarded as a dimer of a certain free radical. In 1966, two research groups independently discovered gas **D**. One method of preparation is the discharge oxidation of **B** with oxygen at -196 °C; the other is the reaction of **E** with the single-electron oxidant $IrF_6$, where the byproduct **F** contains diatomic cations and octahedral coordinated anions. The cation of **F** can be reduced to gas **G**, which is easily oxidized in air to **H** and reaches equilibrium with the dimer of **H**, exhibiting a reddish-brown color. Direct fluorination of **G** or reaction of **G** with $AgF_2$ can yield the substrate **E** used to generate **D**. Direct fluorination of **B** in the presence of $SbF_3$ can yield salt **I**, whose cation is isoelectronic with **D**.

Deduce the substances represented by the above symbols and judge whether the following statements are correct or not.

1. The molecular weight of **C** is greater than 100

2. The molecular weight of **A** is greater than 20

3. **D** and **E** contain the same elements

4. **F** contains three elements

5. **G** is paramagnetic

6. The dipole moment of **A** is less than **B**

7. The cation of **I** has $T_d$ symmetry

M is the sum of the squares of the serial numbers corresponding to all correct statements, and N is the square of the sum of the serial numbers corresponding to all incorrect statements. Calculate the value of M/N and select the option with an error within 0.001 in the options.
对工农业非常重要的气体**A**有丰富的化学性质。第一次世界大战前，德国化学家Haber优化了直接由单质反应生成**A**的条件，使其可工业化大规模生产。电解**A**的$HF$盐和$HF$的熔融物体系可以得到无色无臭的稳定气体**B**。除了这个首先由德国OttoRuff小组使用的办法，在铜催化剂上控制氟化**A**也能得到**B**；但若温度控制不当，生成的**B**将会氟化铜催化剂，自身转化为**C**。在室温及以上**C**容易解离，可将其看作某自由基的二聚体。1966年两个研究小组分别发现了气体**D**。制备方法之一是在-196 °C 下氧气放电氧化**B**；另一是用单电子氧化剂 $IrF_6$ 与**E**反应，副产物**F**含有双原子阳离子和八面体配阴离子。**F**的阳离子能被还原为气体**G**，**G** 在空气中很容易被氧化成**H**，并与 **H**的二聚体达成平衡，显出红棕色。直接氟化**G**或使用 $AgF_2$ 与**G**作用均可得到生成**D**所用的底物**E**。在 $SbF_3$ 存在下直接氟化**B**可以得到盐**I**，其阳离子与**D**互为等电子体。

推断以上符号代表的物质，判断以下说法正确与否。

1.**C**的分子量大于100

2.**A**的分子量大于20

3.**D**和**E**所含元素相同

4.**F**中有三种元素

5.**G**具有顺磁性

6.**A**偶极矩小于**B**

7.**I**的阳离子具有$T_d$对称性

M为所有正确说法对应序号的平方和，N为所有错误说法对应序号的和的平方，计算M/N的值并在选项中选出误差在0.001之内的选项。
All other options are incorrect
0.491
0.420
0.583
0. 085
3.333
0.785
其他选项均不正确
0.491
0.420
0.583
0.085
3.333
0.785
The method of producing **A** from elements was optimized by German chemist Haber before World War I, and **A** is used in industry and agriculture, so **A** is $NH_3$. The molecular weight of ammonia is 17, so statement 2 is incorrect.

<Checkpoint>[**A** is $NH_3$](1)</Checkpoint>

**B** is obtained by electrolysis of the $HF$ salt of ammonia, i.e., ammonium fluoride, or by fluorination of ammonia, indicating that it is a fluorine-substituted ammonia. **B** will convert itself into **C** with copper fluoride catalyst, and since **C** is easily dissociated and can be regarded as a dimer of free radicals, **C** is $N_2F_4$. The dissociated $NF_2$ free radical is stabilized by the strong electron-withdrawing effect of fluorine and the conjugation of lone pair electrons. **C** is the product of **B** fluorinating the catalyst, so **B** is $NF_3$.

<Checkpoint>[**B** is $NF_3$](1)</Checkpoint>

<Checkpoint>[**C** is $N_2F_4$](1)</Checkpoint>

The molecular weight of **C** is 104, which is greater than 100, so statement 1 is correct.

The lone pair electrons of $NH_3$ are in the same direction as the bond moment, while $NF_3$ is the opposite, so the dipole moment of **A** is greater than that of **B**, so statement 6 is incorrect.

<Checkpoint>[Statement 6 is incorrect](1)</Checkpoint>

**D** is obtained by oxidizing **B** with oxygen, which can only be $ONF_3$.

<Checkpoint>[**D** is $ONF_3$](1)</Checkpoint>

The cation of **F** can be reduced to the gas **G**, and **G** is easily oxidized in the air to **H**, and is in equilibrium with the dimer, showing reddish-brown color, so **H** is $NO_2$, and **G** is $NO$.

<Checkpoint>[**G** is $NO$](1)</Checkpoint>

<Checkpoint>[**H** is $NO_2$](1)</Checkpoint>

NO has a single electron and is a paramagnetic gas, so statement 5 is correct.

Then the cation of **F** is the nitrosyl cation, which conforms to the description of a diatomic cation. Since $IrF_6$ is a single-electron oxidant, and its oxidation of **E** can generate $ONF_3$, and at the same time generate **F** as a byproduct, the anion of **F** is $IrF_6^-$, which conforms to the description of the octahedral coordinated anion in the question, so **F** is $[NO][IrF_6]$.

<Checkpoint>[**F** is $[NO][IrF_6]$](2)</Checkpoint>

**F** contains 4 kinds of elements, so statement 4 is incorrect.

Fluorinating $NO$ can obtain **E**, indicating that **E** is a fluoride or oxyfluoride of $N$. And **E** can be oxidized to $ONF_3$ by $IrF_6$, indicating that it is the oxyfluoride $ONF$.

<Checkpoint>[**E** is $ONF$](1)</Checkpoint>

**D** and **E** contain the same elements, so statement 3 is correct.

Direct fluorination of $NF_3$ in the direct presence of $SbF_3$ yields the salt **I**. From common sense, the anion is $SbF_6^-$, and the cation contains only nitrogen and fluorine, so it can only be $NF_4^+$. Therefore, **I** is $[NF_4][SbF_6]$. This conforms to the description that its cation and $ONF_3$ are isoelectronic.

<Checkpoint>[I is $[NF_4][SbF_6]$](1)</Checkpoint>

Therefore, the cation of **I** has $T_d$ symmetry, so statement 7 is correct.

<Checkpoint>[Statement 7 is correct](1)</Checkpoint>

In summary, the correct statements are 1, 3, 5, 7, and the incorrect statements are 2, 4, 6.

$M=1+9+25+49=84$

$N=(2+4+6)^2=144$

$M/N=0.583$

Therefore, choose D
由一战前德国化学家Haber优化了由单质生产**A**的方法，且**A**用于工农业，可知**A**是$NH_3$。氨气分子量17，说法2错误。

<Checkpoint>[**A**是$NH_3$](1)</Checkpoint>

**B**由氨气的$HF$盐即氟化铵电解得到，或者由氨气的氟化得到，可知是氟取代的氨气。**B**会氟化铜催化剂自身转化为**C**，由**C**容易解离且可视为自由基的二聚体可知**C**为$N_2F_4$，解离出的$NF_2$自由基由于氟的强吸电子作用和孤对电子的共轭而稳定。**C**是**B**氟化催化剂的产物，故**B**是$NF_3$。

<Checkpoint>[**B**是$NF_3$](1)</Checkpoint>


<Checkpoint>[**C**是$N_2F_4$](1)</Checkpoint>

**C**分子量104，大于100，说法1正确。

$NH_3$的孤对电子与键矩方向相同，而$NF_3$相反，所以**A**的偶极矩大于**B**，说法6错误。

<Checkpoint>[说法6错误](1)</Checkpoint>

**D**由**B**被氧气氧化得到，只能是$ONF_3$。

<Checkpoint>[**D**是$ONF_3$](1)</Checkpoint>

**F**的阳离子可以还原为气体**G**，**G**在空气中易氧化为**H**，并与二聚体平衡，显红棕色，可知**H**是$NO_2$，**G**是$NO$。

<Checkpoint>[**G**是$NO$](1)</Checkpoint>

<Checkpoint>[**H**是$NO_2$](1)</Checkpoint>

NO有单电子，是顺磁性气体，说法5正确。

则**F**的阳离子是亚硝酰正离子，符合双原子阳离子的描述。由$IrF_6$是单电子氧化剂，其氧化**E**可生成$ONF_3$，同时生成**F**副产物可知，**F**的阴离子为$IrF_6^-$，符合题目八面体配阴离子的描述，故**F**为$[NO][IrF_6]$。

<Checkpoint>[**F**是$[NO][IrF_6]$](2)</Checkpoint>

**F**中有4种元素，说法4错误。

氟化$NO$可以得到**E**，说明**E**是$N$的氟化物或氟氧化物。而**E**可以被$IrF_6$氧化为$ONF_3$，说明是氟氧化物$ONF$。

<Checkpoint>[**E**是$ONF$](1)</Checkpoint>

**D**和**E**所含元素相同，说法3正确。

在$SbF_3$直接存在下，直接氟化$NF_3$得到盐**I**，由常识可知阴离子为$SbF_6^-$，阳离子只含氮氟，只能为$NF_4^+$。故**I**为$[NF_4][SbF_6]$。符合题目其阳离子与$ONF_3$为等电子体的描述。

<Checkpoint>[I为$[NF_4][SbF_6]$](1)</Checkpoint>

故**I**的阳离子为$T_d$对称性，说法7正确。

<Checkpoint>[说法7正确](1)</Checkpoint>

综上所述，正确的说法有1,3,5,7，错误的说法有2,4,6。

$M=1+9+25+49=84$

$N=(2+4+6)^2=144$

$M/N=0.583$

故选择D