**M** is an element. The high-valent oxide **A** of **M** is orange-yellow and can be obtained by the thermal decomposition of salt **B** containing **M** at 43.55% (retaining four significant figures). **A** serves as an important catalyst in industrial production. When **A** is fused with oxalic acid, it forms a deep blue compound **C** while releasing gas. **C** adopts a rutile structure at room temperature. The reduction of **C** with hydrogen yields oxide **D**, which contains **M** at 63.16% (retaining four significant figures). Further reduction of **C** produces the elemental form of **M**. The direct reaction of **M** with chlorine gas yields **E**, which is unstable in air. Under a CO atmosphere, **E** is reduced by Na in pyridine to form the highly symmetric anionic salt **F**. Upon acidification of **F**, a volatile blue-green crystal **G** is obtained. A similar low-temperature compound **H**, containing **M** at 23.25% (retaining four significant figures), belongs to the $O_h$ point group.  

From the following options, select the correct statement.
**M**是一个元素。**M**的高价氧化物**A**为橘黄色，可以由含**M**43.55%（保留四位有效数字）的盐**B**热分解得到，**A**在工业生产中是重要的催化剂。**A**与草酸共熔，生成深蓝色的**C**，并放出气体，**C**在室温下为金红石结构。用氢气还原**C**，可以得到含**M** 63.16%（保留四位有效数字）的氧化物**D**。**C**进一步还原可得到**M**的单质,**M**直接与氯气作用，得到在空气中不稳定的**E**,**E**在CO气氛下，在吡啶中被Na还原，得到阴离子对称性很高的盐**F**，**F**酸化后得到可挥发的蓝绿色晶体**G**，与**G**类似的低温化合物**H**含**M** 23.25%（保留四位有效数字），为$O_h$点群。
从下列选项中选出正确的说法。
**B** has three more elements than **A**
In substance **D**, the ratio of the number of atoms of the element with the highest atomic count to that of the element with the lowest atomic count is greater than 2.
**F** molecular weight less than 200
**G** contains one more type of element than **H**.
The sum of the valence states of the metal elements in **C** and **E** is greater than 7.
In this problem, there are a total of 5 valence states for the metal elements among all substances.
**B**比**A**多三种元素
**D**这个物质中原子数最多的元素与原子数最少元素的原子数比大于2
**F**分子量小于200
**G**包含的元素种类比**H**多一种
**C**和**E**中所含金属元素价态之和大于7
本题目中所有物质中，金属元素共有5种价态
This question provides hints in multiple places, and can be approached from the color of the substances or any mass fraction.

Based on the color of high-valent oxides, chromium trioxide and vanadium pentoxide can be initially screened. Among them, vanadium pentoxide is a more suitable industrial catalyst (sulfuric acid production) as required by the question. According to the mass fraction, if the metal is vanadium, the molecular weight of the salt is 116.97, corresponding to ammonium vanadate, while chromium does not have a corresponding one.

It can be inferred that A and B, A is $V_2O_5$;

<Checkpoint>[**A** is $V_2O_5$](1)</Checkpoint>

B is $NH_4VO_3$; Thermal decomposition of ammonium vanadate is a common method for producing vanadium pentoxide, which verifies the correctness of the calculated result.

<Checkpoint>[**B** is $NH_4VO_3$](1)</Checkpoint>

Based on the color, rutile structure, and common chemical knowledge, it can be inferred that C is $VO_2$;

<Checkpoint>[**C** is $VO_2$](1)</Checkpoint>

Known as an oxide, based on the mass fraction and reduction conditions, it can be inferred that the ratio of vanadium to oxygen is 1:1.857, so D is $V_7O_{13}$;

<Checkpoint>[**D** is $V_7O_{13}$](1)</Checkpoint>

Based on common knowledge of elemental reactions, E is $VCl_3$;

<Checkpoint>[**E** is $VCl_3$](1)</Checkpoint>

Reduced by carbon monoxide atmosphere and then by sodium, F is $NaV(CO)_6$;

<Checkpoint>[**F** is $NaV(CO)_6$](1)</Checkpoint>

After acidification, based on common chemical knowledge and volatility, it can be known that G is $V(CO)_6$;

<Checkpoint>[**G** is $V(CO)_6$](1)</Checkpoint>

Based on similarity, point group, and mass fraction, it can be known that H is $V(N_2)_6$; Here, the mass fractions of $V(CO)_6$ and $V(N_2)_6$ both match, but vanadium hexacarbonyl has already been used for the previous substance.

<Checkpoint>[ H is $V(N_2)_6$  ](1)</Checkpoint>

In summary, analyzing the options, it is easy to know that only D is correct. This question focuses on the correct deduction of all substances.
此题多处进行提示，可以从物质颜色或任一质量分数入手。

根据高价氧化物颜色，可初步筛选出三氧化铬和五氧化二钒，其中五氧化二钒是更符合题目要求的工业催化剂（硫酸生产）。根据质量分数，如果金属是钒，则盐的分子量为116.97，对应钒酸铵，而铬没有对应的。

可以推断A和B，A是$V_2O_5$；

<Checkpoint>[**A**是$V_2O_5$](1)</Checkpoint>

B是$NH_4VO_3$；钒酸铵的热分解是生产五氧化二钒的常见方式，验证了由计算得到结果的正确性。

<Checkpoint>[**B**是$NH_4VO_3$](1)</Checkpoint>

根据颜色和金红石结构以及化学常识，可以推断C是$VO_2$;

<Checkpoint>[**C**是$VO_2$](1)</Checkpoint>  

已知是氧化物，根据质量分数和还原条件，可以推知钒和氧的比例是1:1.857，所以D是$V_7O_{13}$;

<Checkpoint>[**D**是$V_7O_{13}$](1)</Checkpoint> 

元素反应常识，E是$VCl_3$;

<Checkpoint>[**E**是$VCl_3$](1)</Checkpoint>

一氧化碳还原气氛，被钠还原，得到F是$NaV(CO)_6$;

<Checkpoint>[**F**是$NaV(CO)_6$](1)</Checkpoint>

酸化后，根据化学常识以及挥发性，可知G是$V(CO)_6$;  

<Checkpoint>[**G**是$V(CO)_6$](1)</Checkpoint>

根据相似性，点群和质量分数，可知H是$V(N_2)_6$ ;此处质量分数$V(CO)_6$和$V(N_2)_6$都符合，但六羰基钒是前一个物质已经使用过了。

<Checkpoint>[ H是$V(N_2)_6$  ](1)</Checkpoint>

综上所述，分析选项，易知只有D是正确的。此题注重对所有物质的正确推断。