Elements **A, B, and C** are in the same period with increasing atomic numbers and are adjacent in the periodic table. They can all form ions with the general formula $\mathrm{XO_m ^{n-}}$. The value of m is the same for these three ions, and the value of n is also the same. None of these three ions can exist in large quantities in acidic aqueous solutions. It is known that the mass fraction of these three elements in $\mathrm{XO_m ^{n-}}$ is strictly less than $50\%$.

Regarding these three elements, which of the following statements is correct?
同周期元素**A,B,C**原子序数依次增大，且在元素周期表中位置相邻。它们均能形成通式为$\mathrm{XO_m ^{n-}}$的离子，这三种离子的m取值相同，n取值也相同，且这三种离子均不能在酸性的水溶液中大量存在。已知这三种元素在$\mathrm{XO_m ^{n-}}$的质量分数均严格小于$50\%$。

关于这三种元素，下列说法正确的是：
Element **A** has paired valence electrons.
The $\mathrm{AO_m^{n-}}$ ion of element **A** is colorless.
Acidification of the $\mathrm{BO_m^{n-}}$ ion of element **B** releases bubbles.
Acidification of $\mathrm{BO_m^{n-}}$ ions of element **B** can produce a white precipitate.
The standard electrode potential of the half-reaction for the oxidation of element **C** in the $\mathrm{CO_m^{n-}}$ ion under acidic conditions is greater than $2.0 V$.
Ions of element **M** with the formula $\mathrm{MO_m^{n-}}$ can be used to test for phosphate ions, then element **M** belongs to the same group as element **C**.
元素**A**具有成对的价电子。
元素**A**的$\mathrm{AO_m^{n-}}$离子不显色。
元素**B**的$\mathrm{BO_m^{n-}}$离子酸化后会放出气泡。
元素**B**的$\mathrm{BO_m^{n-}}$离子酸化后可生成白色沉淀。
元素**C**的$\mathrm{CO_m^{n-}}$离子酸性条件下氧化半反应的标准电极电势大于$2.0 V$。
元素**M**形成的离子$\mathrm{MO_m^{n-}}$可用于检验磷酸根离子，则元素**M**与元素**C**同族。
First forming $\mathrm{XO_m^{n-}}$ implies that these three adjacent elements in the same period have the same valence, which basically locks it down to a transition metal, because it is difficult for three consecutive main group nonmetals to have the same valence.

<Checkpoint>[It is difficult for three consecutive main group nonmetals to have the same valence](1)</Checkpoint>

To verify this hypothesis, consider the values of $m$ and $n$:
When $m$ is 1, the relative atomic mass of **X** must be less than 16, and it is impossible to have three consecutive $\mathrm{XO^{n-}}$.

When m is 2, the relative atomic mass of **X** must be less than 32. At this time, because the atomic mass of sulfur is 32.06, it is impossible. The most likely candidates are **Al, Si, P**, but since n can only be 1 (metaaluminate) at this time, and there is no such species as $\mathrm{SiO_2^-}$, m cannot be 2.

When m is 3, the relative atomic mass of **X** must be less than 48, and the element must be before **Ti**. Consider **P, S, Cl**. When n is 1, sulfur does not satisfy; when n is 2, chlorine does not satisfy; when n is 3, sulfur does not satisfy; when it is greater than 3, phosphorus does not satisfy. Therefore, m does not take 3 either.

<Checkpoint>[m cannot take values 1, 2, 3](2)</Checkpoint>

When m is 4, the relative atomic mass of **X** must be less than 64, and the most likely candidates are transition metals before **Cu**. Consider n being 1, but $\mathrm{MnO_4^-}$ is stable to acid, which does not meet the requirements, so n takes 2.

<Checkpoint>[$\mathrm{MnO_4^-}$ is stable to acid, which does not meet the requirements, so n takes 2](1)</Checkpoint>

It is easy to associate with chromate ion $\mathrm{CrO_4^{2-}}$, which is unstable under acidic conditions and transforms into dichromate ion; the adjacent element can also form manganate ion $\mathrm{MnO_4^{2-}}$, which is unstable under acidic conditions and undergoes self-disproportionation; the element adjacent to it is iron, and $\mathrm{FeO_4^{2-}}$ is ferrate ion, which is unstable under acidic conditions and can oxidize water. Therefore, the elements **Cr, Mn, Fe** meet the requirements of the question. **A** is **Cr**, **B** is **Mn**, **C** is **Fe**.

<Checkpoint>[$\mathrm{CrO_4^{2-}}$, $\mathrm{MnO_4^{2-}}$, $\mathrm{FeO_4^{2-}}$ are all unstable to acid](1)</Checkpoint>

<Checkpoint>[**A** is **Cr**, **B** is **Mn**, **C** is **Fe**.](2)</Checkpoint>

Therefore, we begin to judge the options:

A. **Cr** has a half-filled orbital and does not have paired valence electrons, A is incorrect.

<Checkpoint>[**Cr** has a half-filled orbital and does not have paired valence electrons, A is incorrect.](1)</Checkpoint>

B. $\mathrm{CrO_4^{2-}}$ appears yellow, B is incorrect.

<Checkpoint>[$\mathrm{CrO_4^{2-}}$ appears yellow, B is incorrect.](1)</Checkpoint>

C. The disproportionation reaction of $\mathrm{MnO_4^{2-}}$ under acidic conditions is

$\mathrm{3MnO_4^{2-}+4H^+=2MnO_4^-+MnO_2+2H_2O}$

<Checkpoint>[$\mathrm{3MnO_4^{2-}+4H^+=2MnO_4^-+MnO_2+2H_2O}$](1)</Checkpoint>

Disproportionation does not produce gas, C is incorrect.

<Checkpoint>[Disproportionation does not produce gas, C is incorrect.](0.5)</Checkpoint>

D. The disproportionation of $\mathrm{MnO_4^{2-}}$ produces a brownish-black manganese dioxide precipitate, not white, D is incorrect.

<Checkpoint>[The disproportionation of $\mathrm{MnO_4^{2-}}$ produces a brownish-black manganese dioxide precipitate, not white, D is incorrect.](0.5)</Checkpoint>

E. $\mathrm{FeO_4^{2-}}$ ion has strong oxidizing properties, and the standard electrode potential of $\mathrm{FeO_4^{2-}/Fe^{3+}}$ under acidic conditions is $2.2 V$, E is correct.

<Checkpoint>[The standard electrode potential of $\mathrm{FeO_4^{2-}/Fe^{3+}}$ under acidic conditions is $2.2 V$, E is correct](1)</Checkpoint>

F. The $\mathrm{MO_4^{2-}}$ ion used to test for phosphate is molybdate ion $\mathrm{MoO_4^{2-}}$ ion, which belongs to the same group as **Cr** and not **Fe**, F is incorrect.

<Checkpoint>[The $\mathrm{MO_4^{2-}}$ ion used to test for phosphate is molybdate ion $\mathrm{MoO_4^{2-}}$ ion, which belongs to the same group as **Cr** and not **Fe**, F is incorrect](1)</Checkpoint>

In summary, option E is correct.
首先形成$\mathrm{XO_m^{n-}}$意味着这三种同周期相邻元素均具有同一种化合价，基本可以锁定是副族金属，因为连续三个主族非金属难以存在同价可能性。

<Checkpoint>[连续三个主族非金属难以存在同价可能性](1)</Checkpoint>

为验证这一假设，考虑$m,n$的取值：
$m$取1时，**X**的相对原子质量一定小于16，不可能存在连续三个$\mathrm{XO^{n-}}$。

m取2时，**X**的相对原子质量一定小于32，此时因为硫原子为32.06，故不可能，最有可能的为**Al,Si,P**，但因为此时n只能取1（偏铝酸根），而不存在$\mathrm{SiO_2^-}$这种物种，故m不能取2。

m取3时，**X**的相对原子质量一定小于48，此时元素一定在**Ti**之前。考虑**P,S,Cl**，当n为1时，硫元素不满足；n为2时，氯元素不满足，n为3时，硫元素不满足，大于3时磷元素不满足，因此m也不取3.

<Checkpoint>[m不能取值1，2，3](2)</Checkpoint>

m取4时，**X**的相对原子质量一定小于64，此时最有可能的为**Cu**之前的副族金属。考虑n取1，但$\mathrm{MnO_4^-}$对酸稳定，不满足题意，从而n取2。

<Checkpoint>[$\mathrm{MnO_4^-}$对酸稳定，不满足题意，从而n取2](1)</Checkpoint>

容易联想到铬酸根离子$\mathrm{CrO_4^{2-}}$，在酸性条件下不稳定，转化为重铬酸根离子；与之相邻的元素刚好也可以形成锰酸根离子$\mathrm{MnO_4^{2-}}$，酸性条件下不稳定发生自我歧化；再与之相邻的元素为铁，$\mathrm{FeO_4^{2-}}$为高铁酸根离子，酸性条件下不稳定可氧化水，从而元素**Cr,Mn,Fe**满足题目要求。**A**为**Cr**,**B**为**Mn**,**C**为**Fe**。

<Checkpoint>[$\mathrm{CrO_4^{2-}}$,$\mathrm{MnO_4^{2-}}$,$\mathrm{FeO_4^{2-}}$均对酸不稳定](1)</Checkpoint>

<Checkpoint>[**A**为**Cr**,**B**为**Mn**,**C**为**Fe**。](2)</Checkpoint>

从而我们开始判断选项：

A. **Cr**为半充满轨道，不具有成对的价电子，A错误。

<Checkpoint>[**Cr**为半充满轨道，不具有成对的价电子，A错误。](1)</Checkpoint>

B. $\mathrm{CrO_4^{2-}}$显黄色，B错误。

<Checkpoint>[$\mathrm{CrO_4^{2-}}$显黄色，B错误。](1)</Checkpoint>

C. $\mathrm{MnO_4^{2-}}$酸性条件下的歧化反应为

$\mathrm{3MnO_4^{2-}+4H^+=2MnO_4^-+MnO_2+2H_2O}$

<Checkpoint>[$\mathrm{3MnO_4^{2-}+4H^+=2MnO_4^-+MnO_2+2H_2O}$](1)</Checkpoint>

歧化不生成气体，C错误。

<Checkpoint>[歧化不生成气体，C错误。](0.5)</Checkpoint>

D. $\mathrm{MnO_4^{2-}}$歧化生成棕黑色的二氧化锰沉淀而非白色，D错误。

<Checkpoint>[$\mathrm{MnO_4^{2-}}$歧化生成棕黑色的二氧化锰沉淀而非白色，D错误。](0.5)</Checkpoint>

E. $\mathrm{FeO_4^{2-}}$离子具有强氧化性，酸性条件下$\mathrm{FeO_4^{2-}/Fe^{3+}}$的标准电极电势为$2.2 V$，E正确。

<Checkpoint>[酸性条件下$\mathrm{FeO_4^{2-}/Fe^{3+}}$的标准电极电势为$2.2 V$，E正确](1)</Checkpoint>

F. 用于检验磷酸根的$\mathrm{MO_4^{2-}}$离子为钼酸根离子$\mathrm{MoO_4^{2-}}$离子，与**Cr**同族而非**Fe**，F错误。

<Checkpoint>[用于检验磷酸根的$\mathrm{MO_4^{2-}}$离子为钼酸根离子$\mathrm{MoO_4^{2-}}$离子，与**Cr**同族而非**Fe**，F错误](1)</Checkpoint>

综上，E选项正确。