A certain mixture $\mathbf{A}$ is purple-red and contains two common transition metal elements $\mathbf{M, N}$, with the relative atomic mass of $\mathbf{M}$ being smaller. The concentrations of these two metals in $\mathbf{A}$ are close. Concentrated hydrochloric acid is added to $\mathbf{A}$ and gently heated to obtain a blue solution $\mathbf{B}$ and a gas $\mathbf{C}$. If sodium hydroxide solution is added to $\mathbf{A}$ with stirring, a brownish-red precipitate $\mathbf{D}$ and a green solution $\mathbf{E}$ are obtained. When excess sulfur dioxide is bubbled into $\mathbf{A}$, the solution eventually turns into a pink solution $\mathbf{F}$. Ammonia water is quickly poured into $\mathbf{F}$ to obtain a white precipitate $\mathbf{G}$ and a brownish-yellow solution $\mathbf{H}$. $\mathbf{G}$ slowly turns into a brown precipitate in the air. When $\mathbf{D}$ and $\mathbf{G}$ are mixed and dilute sulfuric acid is added, solution $\mathbf{A}$ is obtained again.

Note: The solution in this problem may contain multiple major ions (ignoring hydrolysis reactions).

Which of the following statements is correct?
某混合液 $\mathbf{A}$ 为紫红色，含有两种常见副族金属元素$\mathbf{M,N}$且相对原子质量$\mathbf{M}$更小，这两种金属在$\mathbf{A}$中的浓度接近。向 $\mathbf{A}$ 中加入浓盐酸并微热得蓝色溶液 $\mathbf{B}$ 和气体 $\mathbf{C}$ 。若于 $\mathbf{A}$ 中加入氢氧化钠溶液搅拌得棕褐色沉淀 $\mathbf{D}$ 和绿色溶液 $\mathbf{E}$ 。向 $\mathbf{A}$ 中通入过量二氧化硫则溶液最后变为粉红色溶液 $\mathbf{F}$ 。向 $\mathbf{F}$ 中迅速倒入氨水得白色沉淀 $\mathbf{G}$ 和棕黄色溶液 $\mathbf{H}$ 。 $\mathbf{G}$ 在空气中缓慢转变为棕色沉淀。将 $\mathbf{D}$ 与 $\mathbf{G}$ 混合后加入稀硫酸又得溶液 $\mathbf{A}$ 。

注意：本题目溶液中可能含有多种主要离子（忽略水解反应）。

下列说法正确的是：
The relative atomic mass of $\mathbf{N}$ is less than 58.
The formula weight of a certain major ion contained in **B** is greater than 200.
**C** is a paramagnetic gas.
**D** is completely insoluble in dilute hydrochloric acid.
Solution **E** contains $\mathbf{M}$ and $\mathbf{N}$
**F** solution contains only $\mathbf{M}$ or only $\mathbf{N}$.
The central atom of the major ions in **H** has three unpaired electrons.
None of the above options are correct.
$\mathbf{N}$的相对原子质量小于58
**B**中含有的某种主要离子的式量大于200
**C**为顺磁性气体
**D**完全不溶于稀盐酸
**E**溶液中含$\mathbf{M}$和$\mathbf{N}$
**F**溶液中只含$\mathbf{M}$或只含$\mathbf{N}$
**H**中的主要离子不存在Jahn-Teller畸变
以上选项均不正确
**A** is a purple-red solution, most likely containing permanganate, indicating that one of the metals is **Mn**;

<Checkpoint>[Purple-red solution most likely contains permanganate, one of the metals is **Mn**](1)</Checkpoint>

Adding concentrated hydrochloric acid to **A** and heating it will cause the permanganate to oxidize chloride ions to produce $\mathrm{Mn^{2+}}$ and chlorine gas **C**, which is light pink; however, the solution is blue, meaning that there are chloride complexes or hydrated ions of other metal ions that are colored, possibly $\mathrm{Cu^{2+}}$, $\mathrm{Co^{2+}}$ or $\mathrm{Cr^{3+}}$. Chlorine gas is a diamagnetic molecule, so option C is incorrect.

<Checkpoint>[Permanganate reduces chloride ions to produce $\mathrm{Mn^{2+}}$ and gas **C** $\mathrm{Cl_2}$](1)</Checkpoint>

<Checkpoint>[The presence of chloride complexes or hydrated ions of other metal ions gives a blue color, possibly $\mathrm{Cu^{2+}}$, $\mathrm{Co^{2+}}$ or $\mathrm{Cr^{3+}}$](1)</Checkpoint>

Adding sodium hydroxide solution to **A** will convert the permanganate into manganate and manganese dioxide, with manganate being a green solution **E** and manganese dioxide being a brown precipitate **D**.

<Checkpoint>[Permanganate in alkaline conditions will be converted to green solution **E** $\mathrm{MnO_4^{2-}}$ and brown precipitate **D** $\mathrm{MnO_2}$](1)</Checkpoint>

Passing sulfur dioxide into **A** to generate **F** will reduce the permanganate to $\mathrm{Mn^{2+}}$. Since solution **F** is pink, it indicates that the reduced state of the other metal ion is also pink or very lightly colored; $\mathrm{Co^{2+}}$ is exactly pink, so it is guessed that the other metal is **Co**.

<Checkpoint>[The reduced state of the other metal ion is also pink or very lightly colored](1)</Checkpoint>

<Checkpoint>[Guessing that the other metal is **Co**](1)</Checkpoint>

Adding sodium hydroxide solution to **A** converts the other metal ion into a brown precipitate **D**. If the metal is $\mathrm{Cu^{2+}}$, its hydroxide should be blue; if the metal is $\mathrm{Cr^{3+}}$, its hydroxide is gray-green; if the metal is **Co**, the generated cobalt(III) hydroxide is exactly brown, confirming the above guess; therefore, it is determined that **M** is **Mn** and **N** is **Co**. The molecular weight of **Co** is greater than 58, so option A is incorrect.

<Checkpoint>[$\mathrm{Co(OH)_3}$ is brown](1)</Checkpoint>

<Checkpoint>[**M** is **Mn**, **N** is **Co**](1)</Checkpoint>

At this time, the colored chloride complex in **B** is $\mathrm{CoCl_4^{2-}}$, with a formula weight of 200.9, which is greater than 200, so option B is correct.

<Checkpoint>[The colored chloride complex in **B** is $\mathrm{CoCl_4^{2-}}$](1)</Checkpoint>

The brown precipitate **D** contains manganese dioxide and cobalt(III) hydroxide. Cobalt(III) hydroxide is soluble in dilute hydrochloric acid, so option D is incorrect.

<Checkpoint>[The brown precipitate **D** contains $\mathrm{MnO_2}$ and $\mathrm{Co(OH)_3}$](1)</Checkpoint>

Solution **E** only contains manganate, and $\mathrm{Co^{2+}}$ is converted into cobalt(III) hydroxide, so option E is incorrect.

<Checkpoint>[Solution **E** only contains $\mathrm{MnO_4^{2-}}$](1)</Checkpoint>

**F** contains $\mathrm{Mn^{2+}}$ and $\mathrm{Co^{2+}}$, both of which are pink, so option F is incorrect.

<Checkpoint>[**F** contains $\mathrm{Mn^{2+}}$ and $\mathrm{Co^{2+}}$](1)</Checkpoint>

Quickly pouring ammonia into **F** yields a white precipitate **G** and a brownish-yellow solution **H**. The white precipitate **G** is $\mathrm{Mn(OH)_2}$, so **H** is the ammine complex of $\mathrm{Co^{2+}}$, $\mathrm{Co(NH_3)_6^{2+}}$, which has a $\mathrm{d^7}$ configuration. Whether it is low-spin or high-spin, there must be Jahn-Teller distortion, so option G is incorrect.

<Checkpoint>[White precipitate **G** is $\mathrm{Mn(OH)_2}$](1)</Checkpoint>

<Checkpoint>[**H** is $\mathrm{Co(NH_3)_6^{2+}}$, which has a $\mathrm{d^7}$ configuration, there must be Jahn-Teller distortion](1)</Checkpoint>

In summary, option B is correct.
**A**为紫红色溶液，很大概率含有高锰酸根，从而金属中有一个为**Mn**；

<Checkpoint>[紫红色溶液大概率含有高锰酸根，金属中有一个为**Mn**](1)</Checkpoint>

**A**加入浓盐酸并加热，高锰酸根会氧化氯离子生成$\mathrm{Mn^{2+}}$和气体**C**氯气，为浅粉色；但该溶液为蓝色，意味着存在其他金属离子的氯配合物或水合离子显色，可能为$\mathrm{Cu^{2+}}$，$\mathrm{Co^{2+}}$或者$\mathrm{Cr^{3+}}$。氯气为抗磁性分子，选项C错误。

<Checkpoint>[高锰酸根还原氯离子生成$\mathrm{Mn^{2+}}$和气体**C** $\mathrm{Cl_2}$](1)</Checkpoint>

<Checkpoint>[存在其他金属离子的氯配合物或水合离子显蓝色，可能为$\mathrm{Cu^{2+}}$，$\mathrm{Co^{2+}}$或者$\mathrm{Cr^{3+}}$](1)</Checkpoint>

**A**中加入氢氧化钠溶液，高锰酸根会转化为锰酸根和二氧化锰，锰酸根为绿色溶液**E**，二氧化锰为棕褐色沉淀**D**。

<Checkpoint>[高锰酸根碱性条件会转化为绿色溶液**E** $\mathrm{MnO_4^{2-}}$和棕褐色沉淀**D** $\mathrm{MnO_2}$](1)</Checkpoint>

**A**中通入二氧化硫生成**F**，高锰酸根会被还原为$\mathrm{Mn^{2+}}$，由于溶液**F**为粉红色，说明另一种金属离子被还原的状态也为粉红色或颜色很浅；$\mathrm{Co^{2+}}$刚好为粉红色，所以猜测另一种金属为**Co**。

<Checkpoint>[另一种金属离子被还原的状态也为粉红色或颜色很浅](1)</Checkpoint>

<Checkpoint>[猜测另一种金属为**Co**](1)</Checkpoint>

**A**中加入氢氧化钠溶液，另一种金属离子被转化为棕褐色沉淀**D**。若该金属为$\mathrm{Cu^{2+}}$，其氢氧化物应当为蓝色；若该金属为$\mathrm{Cr^{3+}}$，其氢氧化物为灰绿色；若金属为**Co**，生成的氢氧化钴(III)刚好为棕褐色，印证了上面的猜想；因此确定**M**为**Mn**，**N**为**Co**。**Co**的分子量大于58，选项A错误。

<Checkpoint>[$\mathrm{Co(OH)_3}$为棕褐色](1)</Checkpoint>

<Checkpoint>[**M**为**Mn**，**N**为**Co**](1)</Checkpoint>

此时**B**中的显色氯配合物为$\mathrm{CoCl_4^{2-}}$，式量为200.9，大于200，选项B正确。

<Checkpoint>[**B**中的显色氯配合物为$\mathrm{CoCl_4^{2-}}$](1)</Checkpoint>

棕褐色沉淀**D**包含二氧化锰和氢氧化钴(III)，氢氧化钴(III)溶于稀盐酸，选项D错误。

<Checkpoint>[棕褐色沉淀**D**包含$\mathrm{MnO_2}$和$\mathrm{Co(OH)_3}$](1)</Checkpoint>

**E**溶液中只含有锰酸根，$\mathrm{Co^{2+}}$被转化为氢氧化钴(III)，从而选项E错误。

<Checkpoint>[**E**溶液中只含有$\mathrm{MnO_4^{2-}}$](1)</Checkpoint>

**F**中含有$\mathrm{Mn^{2+}}$和$\mathrm{Co^{2+}}$，均为粉红色，选项F错误。

<Checkpoint>[**F**中含有$\mathrm{Mn^{2+}}$和$\mathrm{Co^{2+}}$](1)</Checkpoint>

向**F**中迅速倒入氨水得白色沉淀**G**和棕黄色溶液**H**，白色沉淀**G**为$\mathrm{Mn(OH)_2}$，从而**H**为$\mathrm{Co^{2+}}$的氨配合物$\mathrm{Co(NH_3)_6^{2+}}$，为$\mathrm{d^7}$构型，无论是低自旋还是高自旋，一定存在Jahn-Teller畸变，选项G错误。

<Checkpoint>[白色沉淀**G**为$\mathrm{Mn(OH)_2}$](1)</Checkpoint>

<Checkpoint>[**H**为$\mathrm{Co(NH_3)_6^{2+}}$，为$\mathrm{d^7}$构型，一定存在Jahn-Teller畸变](1)</Checkpoint>

综上，选项B正确。