The following operations were performed on the powdered metal $\mathbf{M}$:

1. Step 1: After heating, the powder slowly dissolves in concentrated sulfuric acid, producing a gas $\mathbf{A}$ with a pungent odor.  
2. Step 2: A small amount of the solution is mixed with pyridine, and hydrogen peroxide solution is slowly added under low-temperature conditions. The solution turns blue, forming a mononuclear pentagonal pyramidal complex $\mathbf{B}$. If pyridine is replaced with bipyridine, the product undergoes only minor changes.  
3. Step 3: KOH solution is added to the sulfuric acid solution of $\mathbf{M}$ until nearly neutral, resulting in a gray-green precipitate $\mathbf{C}$.  
4. Step 4: Continued addition of KOH solution dissolves the precipitate, forming a colored ion $\mathbf{D}$ in the solution.  
5. Step 5: Hydrogen peroxide is added to the solution, causing a color change. Cooling and crystallization yield a solid $\mathbf{E}$.  
6. Step 6: An appropriate amount of solid $\mathbf{E}$ is dissolved in a relatively concentrated sulfuric acid solution, producing a red solution containing two ions $\mathbf{F}$ and $\mathbf{G}$, both of which appear red. $\mathbf{F}$ and $\mathbf{G}$ are trinuclear and tetranuclear oxoacid ions, respectively, each carrying a double negative charge.  

Please select the correct option.
对金属单质$\mathbf{M}$的粉末进行以下操作：

1. 第一步：加热后，粉末缓慢溶解于浓硫酸，并生成具有刺激性气味的气体$\mathbf{A}$。
2. 第二步：取少量溶液与吡啶混合，在低温条件下缓慢加入过氧化氢溶液，溶液变为蓝色，生成单核五角锥结构配合物$\mathbf{B}$。若将吡啶换为联吡啶，产物仅发生了较小的变化。
3. 第三步：向$\mathbf{M}$溶于硫酸的溶液加入KOH溶液至近中性，出现灰绿色沉淀$\mathbf{C}$。
4. 第四步：继续加入KOH溶液，沉淀溶解，溶液中生成一种显色离子$\mathbf{D}$。
5. 第五步：向溶液中加入过氧化氢，溶液颜色发生变化。冷却结晶得到固体$\mathbf{E}$。
6. 第六步：将适量固体$\mathbf{E}$溶于较浓的硫酸溶液中，得到红色溶液，其中有两种离子$\mathbf{F}$，$\mathbf{G}$均显红色；$\mathbf{F}$和$\mathbf{G}$分别为三核和四核酸根离子，均带两个负电荷。

请选择正确的选项。
In the process of dissolving $\mathbf{M}$ in sulfuric acid, the molar ratio of the generated gas $\mathbf{A}$ to $\mathbf{M}$ is 1:2
When pyridine in the second step is replaced with bipyridine, the structure of $\mathbf{B}$ transforms into a mononuclear pentagonal bipyramidal complex, featuring one independent oxygen atom coordination, two peroxide ligand coordinations, and two N coordinations. The coordinating atoms in the equatorial plane consist of 2 N atoms and 3 O atoms from peroxide ligands.
The color of $\mathbf{D}$ to $\mathbf{E}$ changes from green to orange-yellow
The sum of the oxygen atoms in $\mathbf{F}$ and $\mathbf{G}$ is 25.
The ionic structure symmetry of $\mathbf{D}$ belongs to the same point group as $Ni(CO)_4$.
The electron configuration of the $\mathbf{M}$ atom's extranuclear electrons conforms to the general formula $nd^5(n+1)s^1$.
$\mathbf{M}$溶于硫酸的过程中，生成的$\mathbf{A}$气体与$\mathbf{M}$的摩尔比为1：2
若将第二步中的吡啶换为联吡啶，则$\mathbf{B}$的结构变为单核五角双锥配合物，其中存在一个独立的氧原子配位、两个过氧根配位和两个N配位，锥平面上的配位原子为2个N和3个过氧根的O
$\mathbf{D}$到$\mathbf{E}$的颜色由绿色变为橙黄色
$\mathbf{F}$与$\mathbf{G}$所含的氧原子加和为25
$\mathbf{D}$的离子结构对称性与$Ni(CO)_4$为同一点群
$\mathbf{M}$原子的原子核外电子排布式符合通式$nd^5(n+1)s^1$
Based on the comprehensive analysis of the question's nature, it is determined that $\mathbf{M}$ is Cr.  

Option F provides a general formula containing only valence electrons, while the description refers to the extranuclear electrons of the atom. A complete electron configuration should be written, so Option F is incorrect.  

<Checkpoint>[The general formula given in Option F contains only valence electrons and is incorrect](1)</Checkpoint>  

The reaction equation for $\mathbf{M}$ dissolving in concentrated sulfuric acid:  

$Cr + 6H_2SO_4 (conc.) \rightarrow Cr (SO_4)_3 + 3SO_2 + 6H_2O$  

<Checkpoint>[$Cr + 6H_2SO_4 (conc.) \rightarrow Cr (SO_4)_3 + 3SO_2 + 6H_2O$](1)</Checkpoint>  

**Second Step**  

Combining knowledge of elemental chemistry, the generated blue complex should be $CrO(O_2)_2(py)$.  

<Checkpoint>[The blue complex is $CrO(O_2)_2(py)$](1)</Checkpoint>  

The structure of $\mathbf{B}$ consists of one independent oxygen atom coordinated at the apex of the cone, two bidentate peroxide ligands, and one pyridine nitrogen atom coordinated at the base of the cone.  

When replaced with a bipyridine ligand, the coordination changes from one N to two N atoms. Compared to the structure of $\mathbf{B}$, the additional N forms the other apex of the pentagonal bipyramid.  

Since the question states that the product undergoes only minor changes, the addition of one N directly coordinating on the opposite side to form the apex of the structure is the least altered scenario. Moreover, peroxide coordination cannot form a 90-degree angle, making it impossible for one oxygen to lie in the pentagonal plane while the other occupies the apex.  

Therefore, Option B is incorrect.  

<Checkpoint>[The structure of $\mathbf{B}$ consists of one independent oxygen atom coordinated at the apex of the cone, two bidentate peroxide ligands, and one pyridine nitrogen atom coordinated at the base of the cone.](1)</Checkpoint>  

<Checkpoint>[When replaced with a bipyridine ligand, the coordination changes from one N to two N atoms. Compared to the structure of $\mathbf{B}$, the additional N forms the other apex of the pentagonal bipyramid.](1)</Checkpoint>  

<Checkpoint>[Peroxide coordination cannot form a 90-degree angle, making it impossible for one oxygen to lie in the pentagonal plane while the other occupies the apex.](1)</Checkpoint>  

**Third Step**: The gray-green precipitate $\mathbf{C}$ is $Cr(OH)_3$.  

<Checkpoint>[The gray-green precipitate $\mathbf{C}$ is $Cr(OH)_3$](1)</Checkpoint>  

**Fourth Step**: Upon further addition of KOH solution, the precipitate dissolves, and the colored ion $\mathbf{D}$ generated in the solution is: $Cr(OH)_4^-$, appearing green and belonging to the Td point group. The option $Ni(CO)_4$ also belongs to the Td point group, so Option E is correct.  

<Checkpoint>[$Cr(OH)_4^-$, appearing green, belongs to the Td point group](1)</Checkpoint>  

<Checkpoint>[$Ni(CO)_4$ belongs to the Td point group, so Option E is correct](1)</Checkpoint>  

**Fifth Step**: Upon adding hydrogen peroxide to the solution, the color changes. Cooling and crystallization yield solid $\mathbf{E}$. Under alkaline conditions, the product should be: $K_2CrO_4$, appearing yellow.  

<Checkpoint>[$\mathbf{E}$ is $K_2CrO_4$, appearing yellow](1)</Checkpoint>  

**Sixth Step**: $\mathbf{F}$ and $\mathbf{G}$ are tri- and tetra-nuclear oxo-chromate ions, respectively, each carrying a double negative charge. The process involves no change in oxidation state: Cr remains +6, and O remains -2. Based on valence states, we obtain:  

$\mathbf{F}$: $Cr_3O_{10}^{2-}$, $\mathbf{G}$: $Cr_4O_{13}^{2-}$, with the total number of oxygen atoms summing to 23.  

<Checkpoint>[$\mathbf{F}$: $Cr_3O_{10}^{2-}$, $\mathbf{G}$: $Cr_4O_{13}^{2-}$, with the total number of oxygen atoms summing to 23](1)</Checkpoint>  

In conclusion, the correct answer is **E**.
综合题目性质判断出$\mathbf{M}$为Cr

选项F中给出的通式仅包含价电子，而选项描述为原子核外电子，需要写出完整的电子排布式，因此选项F错误。

<Checkpoint>[选项F中给出的通式仅包含价电子，错误](1)</Checkpoint>

$\mathbf{M}$溶于浓硫酸的反应方程式：

$Cr + 6H_2SO_4 (浓) \rightarrow Cr (SO_4)_3 + 3SO_2 + 6H_2O$

<Checkpoint>[$Cr + 6H_2SO_4 (浓) \rightarrow Cr (SO_4)_3 + 3SO_2 + 6H_2O$](1)</Checkpoint>

第二步反应

结合元素知识分析生成的蓝色配合物应当为$CrO(O_2)_2(py)$

<Checkpoint>[蓝色配合物为$CrO(O_2)_2(py)$](1)</Checkpoint>

$\mathbf{B}$的结构为1个独立的氧原子配位并位于锥顶，2个过氧根双齿配体配位和一个吡啶N原子配位且均位于锥底。

改为联吡啶配位后，由1个N配位变为2个N配位，相比于$\mathbf{B}$的结构，另一个N形成五角双锥的另一个锥顶。

由于题目中说明产物仅发生了较小的变化，多出的一个N在另一侧直接配位形成结构的锥顶是改变最小的，且过氧根配位时无法形成90度角，因此无法达到其中一个氧在五角平面而另一个氧在锥顶的情况。

因此选项B错误。

<Checkpoint>[$\mathbf{B}$的结构为1个独立的氧原子配位并位于锥顶，2个过氧根双齿配体配位和一个吡啶N原子配位且均位于锥底。](1)</Checkpoint>

<Checkpoint>[改为联吡啶配位后，由1个N配位变为2个N配位，相比于$\mathbf{B}$的结构，另一个N形成五角双锥的另一个锥顶。](1)</Checkpoint>

<Checkpoint>[过氧根配位时无法形成90度角，因此无法达到其中一个氧在五角平面而另一个氧在锥顶的情况。](1)</Checkpoint>

第三步：灰绿色沉淀$\mathbf{C}$为$Cr(OH)_3$

<Checkpoint>[灰绿色沉淀$\mathbf{C}$为$Cr(OH)_3$](1)</Checkpoint>

第四步：继续加入KOH溶液，沉淀溶解，溶液中生成的显色离子$\mathbf{D}$为：$Cr(OH)_4^-$，呈绿色，为Td点群。选项中的$Ni(CO)_4$为Td点群，选项E正确。

<Checkpoint>[$Cr(OH)_4^-$，呈绿色，为Td点群](1)</Checkpoint>

<Checkpoint>[选项中的$Ni(CO)_4$为Td点群，选项E正确](1)</Checkpoint>

第五步：向溶液中加入过氧化氢，溶液颜色发生变化。冷却结晶得到固体$\mathbf{E}$。在碱性条件下生成的应当为：$K_2CrO_4$，呈黄色。

<Checkpoint>[$\mathbf{E}$为$K_2CrO_4$，呈黄色](1)</Checkpoint>

第六步：$\mathbf{F}$和$\mathbf{G}$分别为三核和四核酸根离子，均带两个负电荷。过程无氧化态变化，Cr为正六价，O始终为负二价，根据价态可得到：

$\mathbf{F}$：$Cr_3O_{10}^{2-}$, $\mathbf{G}$：$Cr_4O_{13}^{2-}$，氧原子数加和为23

<Checkpoint>[$\mathbf{F}$：$Cr_3O_{10}^{2-}$, $\mathbf{G}$：$Cr_4O_{13}^{2-}$，氧原子数加和为23](1)</Checkpoint>

综上，本题选E。