The carbonyl compound $\mathbf{A}$ of a widely distributed common metallic element $\mathbf M$ can react with $\mathrm{N}_{2}\mathrm{O}_{4}$ to obtain a compound $\mathbf B$ of the uncommon +1 valence $\mathbf M$ and release a colorless gas. $\mathbf{A}$ can also react directly with bromine to obtain another +1 valence compound $\mathbf C$ of $\mathbf M$. Electrolysis of an aqueous solution of the common valence acetate of $\mathbf M$ can yield a multinuclear complex whose cation $\mathbf{D}$ contains a 3-coordinate oxygen atom and the mass fraction of $\mathbf M$ is $30.80\%$.

Which of the following options is correct?
分布广泛的常见金属元素$\mathbf M$的羰基化合物 $\mathbf{A}$ 可与 $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ 反应，得到不常见的 $+ 1$ 价$\mathbf M$的化合物$\mathbf B$并放出无色气体。 $\mathbf{ A }$ 也可以直接与溴反应得另一种 $+ 1$ 价$\mathbf M$的化合物 $\mathbf{ C }$ 。电解$\mathbf M$的常见价态乙酸盐水溶液可以得到一种多核配合物，其阳离子  $\mathbf{ D }$  中包含一个3配位氧原子且$\mathbf M$的质量分数为 $3 0 . 8 0 \%$ 

以下选项正确的是
$\mathbf M$ is located in Group IB
The molecular formula of $\mathbf A$ can be expressed as $\mathbf M \rm {(CO)_6}$
$\mathbf A$ contains bridging carbonyl ligands.
$\mathbf{ A }$ can react with $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ to produce $\mathbf B$ while releasing $\mathrm{CO}$
In the periodic table, the metal $\mathbf {M’}$ located directly below and adjacent to $\mathbf M$ has stable non-radioactive isotopes.
The reaction of the highest-valent oxyanion of $\mathbf M$ with sulfite yields an unstable oxyanion containing $\mathbf M$. The sodium salt of this oxyanion can disproportionate in aqueous solution. Balance the disproportionation equation and simplify the coefficients to the simplest integer ratio, where the sum of the coefficients on the left side of the equation is greater than the sum of the coefficients on the right side.
The valence of cation $\mathbf D$ is $+1$.
All of the above options are incorrect.
$\mathbf M$位于第IB族
$\mathbf A$的分子式可以表达为$\mathbf M \rm {(CO)_6}$
$\mathbf A$中存在桥接的羰基配体
$\mathbf{ A }$ 可与 $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ 反应生成$\mathbf B$的过程中放出$\mathrm{CO}$
在元素周期表中位于$\mathbf M$正下方且相邻的金属$\mathbf {M’}$有稳定的非放射性同位素
用$\mathbf M$的最高价含氧酸根和亚硫酸根反应，可得一种含$\mathbf M$的不稳定含氧酸根，这种含氧酸根的钠盐可在水溶液中歧化，将歧化方程式配平并将系数化为最简整数比，方程式左边系数和大于右边系数和。
阳离子$\mathbf D$的价态为$+1$价
以上选项均不正确
The question mentions that $\mathbf D$ is a polynuclear coordination complex cation, containing a 3-coordinate oxygen atom $\mathrm{（μ_3-O）}$, and the mass fraction of $\mathbf M$ in $\mathbf D$ is 30.80%. Polynuclear metal acetate complexes with $\mathrm {μ₃-O}$ bridging oxygen atoms are a classic structure, and their general formula is usually $\mathrm{ [M_3(μ_3-O)(μ_2-CH_3COO)_6]^{n+}}$. The molar mass of cation $\mathbf D$ $ = 3 × Ar(\mathbf M) + 370.26$, thus we get the equation:

$$ 3 ×Ar(\mathbf M)=0.308×( 3 × Ar(\mathbf M) + 370.26)$$

Solving for the mass of $\mathbf M$ gives 54.93, and the metal element corresponding to this mass is $\mathrm{Mn}$, so A is incorrect.

<Checkpoint>[$\mathbf M$ is $\mathrm{Mn}$](1)</Checkpoint>

The molecular formula of the carbonyl compound $\mathbf{ A }$ of the metal element $\mathbf M$ should be $\mathrm{Mn_2(CO)_{10}}$, which is axially connected by two $\mathrm{(CO)_5Mn}$ subunits through a $\mathrm {Mn-Mn}$ metal bond, without bridging carbonyl ligands, so B and C are incorrect.

<Checkpoint>[$\mathbf{ A }$ is $\mathrm{Mn_2(CO)_{10}}$](1)</Checkpoint>

$\mathbf{ A }$ can react with $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ to obtain a compound $\mathbf B$ of the uncommon $+ 1$ valence $\mathbf M$ and release a colorless gas. The metal valence in $\mathrm{Mn_2(CO)_{10}}$ is 0, and it is oxidized in the reaction. The equation is
$\mathrm{ 2 N_2O_4+Mn_2(CO)_{10}=2 O_2NOMn(CO)_5+2NO}$

$\mathbf{ B }$ is $\mathrm{O_2NOMn(CO)_5}$. In the process of $\mathbf{ A }$ reacting with $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ to generate $\mathbf B$, $\mathrm{NO}$ is released, so D is incorrect.

<Checkpoint>[$\mathbf{ B }$ is $\mathrm{O_2NOMn(CO)_5}$](1)</Checkpoint>

$\mathbf{ A }$ can also directly react with bromine to obtain another $+ 1$ valence compound $\mathbf{ C }$ of $\mathbf M$. The equation is
$\mathrm{ Br_2+Mn_2(CO)_{10}=2 BrMn(CO)_5}$

$\mathbf{ C }$ is $\mathrm{BrMn(CO)_5}$

The metal $\mathbf {M’}$ located directly below $\mathbf M$ and adjacent to it in the periodic table is $\mathrm {Tc}$, which is the first metal prepared by artificial methods. It has no stable non-radioactive isotopes, only radioactive isotopes, so E is incorrect.

<Checkpoint>[$\mathbf {M’}$ is $\mathrm {Tc}$, has no stable non-radioactive isotopes](1)</Checkpoint>

Reacting the highest-valent oxyacid radical of $\mathbf M$ with sulfite can obtain an unstable oxyacid radical containing $\mathbf M$. In the highest-valent oxyacid radical, $\mathrm {Mn}$ has a $+7$ valence, and after reacting with sulfite, it becomes $+5$ valence. The sodium salt of this oxyacid radical can disproportionate in aqueous solution. The equation is
$$\mathrm{2Na_3MnO_4+2H_2O=Na_2MnO_4+MnO_2+4NaOH}$$, the sum of coefficients on the left side is less than the sum of coefficients on the right side, so F is incorrect.

<Checkpoint>[Disproportionation equation is $$\mathrm{2Na_3MnO_4+2H_2O=Na_2MnO_4+MnO_2+4NaOH}$$](1)</Checkpoint>

The structure of cation $\mathbf D$ is shown in the figure

<MultiModal>![Structure of cation $\mathbf D$, the structural formula is $\mathrm{ [Mn_3(μ_3-O)(μ_2-CH_3COO)_6]^{+}}$, $\mathrm{O-O}$ represents the acetate bidentate ligand, which contains 3 $\mathrm{Mn}$, 1 $\mathrm{μ_3-O}$ and 6 acetate ligands. Each $\mathrm{Mn}$ is connected to 4 $\mathrm O$ of acetate and $\mathrm{μ_3-O}$, and the valence is $+1$](/media/uploads/a04e52af-33f0-4071-b3bf-99f15fc6622d.png)</MultiModal>

The structural formula of cation $\mathbf D$ is $\mathrm{ [Mn_3(μ_3-O)(μ_2-CH_3COO)_6]^{+}}$, $\mathrm{O-O}$ represents the acetate bidentate ligand, which contains 3 $\mathrm{Mn}$, 1 $\mathrm{μ_3-O}$ and 6 acetate ligands. Each $\mathrm{Mn}$ is connected to 4 $\mathrm O$ of acetate and $\mathrm{μ_3-O}$, and the valence is $+1$

<Checkpoint>[The structural formula of cation $\mathbf D$ is $\mathrm{ [Mn_3(μ_3-O)(μ_2-CH_3COO)_6]^{+}}$](1)</Checkpoint>
题目中提到$\mathbf D$是一个多核配合物阳离子,包含一个3配位氧原子$\mathrm{（μ_3-O）}$，并且$\mathbf D$中$\mathbf M$的质量分数为 30.80%。具有$\mathrm {μ₃-O}$桥联氧原子的多核金属乙酸盐配合物是一个经典结构，其通式通常为$\mathrm{ [M_3(μ_3-O)(μ_2-CH_3COO)_6]^{n+}}$,阳离子$\mathbf D$的摩尔质量$ = 3 ×  Ar(\mathbf M) + 370.26$，因此得到方程：

$$ 3 ×Ar(\mathbf M)=0.308×( 3 ×  Ar(\mathbf M) + 370.26)$$

解得$\mathbf M$的质量为54.93，符合这个质量的金属元素是$\mathrm{Mn}$，A错误

<Checkpoint>[$\mathbf M$是$\mathrm{Mn}$](1)</Checkpoint>

金属元素$\mathbf M$的羰基化合物 $\mathbf{ A }$的分子式应该为$\mathrm{Mn_2(CO)_{10}}$，由两个$\mathrm{(CO)_5Mn}$亚基通过$\mathrm {Mn-Mn}$金属键轴向连接，没有桥接的羰基配体，B、C错误

<Checkpoint>[$\mathbf{ A }$为$\mathrm{Mn_2(CO)_{10}}$](1)</Checkpoint>

$\mathbf{ A }$ 可与 $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ 反应，得到不常见的 $+ 1$ 价$\mathbf M$的化合物$\mathbf B$并放出无色气体。 $\mathrm{Mn_2(CO)_{10}}$中金属价态为0，反应中被氧化，方程式为
$\mathrm{ 2 N_2O_4+Mn_2(CO)_{10}=2 O_2NOMn(CO)_5+2NO}$

$\mathbf{ B }$为$\mathrm{O_2NOMn(CO)_5}$,$\mathbf{ A }$ 可与 $\mathrm { N } _ { 2 } \mathrm { O } _ { 4 }$ 反应生成$\mathbf B$的过程中放出$\mathrm{NO}$，D错误

<Checkpoint>[$\mathbf{ B }$为$\mathrm{O_2NOMn(CO)_5}$](1)</Checkpoint>

$\mathbf{ A }$ 也可以直接与溴反应得另一种 $+ 1$ 价$\mathbf M$的化合物 $\mathbf{ C }$，方程式为
$\mathrm{ Br_2+Mn_2(CO)_{10}=2 BrMn(CO)_5}$

$\mathbf{ C }$为$\mathrm{BrMn(CO)_5}$

在元素周期表中位于$\mathbf M$正下方且相邻的金属$\mathbf {M’}$为$\mathrm {Tc}$，是第一个用人工方法制得的金属，没有稳定的非放射性同位素，只有放射性同位素，E错误

<Checkpoint>[$\mathbf {M’}$为$\mathrm {Tc}$，没有稳定的非放射性同位素](1)</Checkpoint>

用$\mathbf M$的最高价含氧酸根和亚硫酸根反应，可得一种含$\mathbf M$的不稳定含氧酸根，最高价含氧酸根中$\mathrm {Mn}$为$+7$价，和亚硫酸根反应后变为$+5$价。这种含氧酸根的钠盐可在水溶液中歧化，方程式为
$$\mathrm{2Na_3MnO_4+2H_2O=Na_2MnO_4+MnO_2+4NaOH}$$，左边系数和小于右边系数和，F错误

<Checkpoint>[歧化方程式为 $$\mathrm{2Na_3MnO_4+2H_2O=Na_2MnO_4+MnO_2+4NaOH}$$](1)</Checkpoint>

阳离子$\mathbf D$的结构如图所示

<MultiModal>![阳离子$\mathbf D$的结构，结构式为$\mathrm{ [Mn_3(μ_3-O)(μ_2-CH_3COO)_6]^{+}}$，$\mathrm{O-O}$代表乙酸根双齿配体，其中存在3个$\mathrm{Mn}$,1个$\mathrm{μ_3-O}$和6个乙酸根配体，每个$\mathrm{Mn}$和4个乙酸根的$\mathrm O$以及$\mathrm{μ_3-O}$相连，价态为$+1$价](/media/uploads/a04e52af-33f0-4071-b3bf-99f15fc6622d.png)</MultiModal>

阳离子$\mathbf D$的结构式为$\mathrm{ [Mn_3(μ_3-O)(μ_2-CH_3COO)_6]^{+}}$，$\mathrm{O-O}$代表乙酸根双齿配体，其中存在3个$\mathrm{Mn}$,1个$\mathrm{μ_3-O}$和6个乙酸根配体，每个$\mathrm{Mn}$和4个乙酸根的$\mathrm O$以及$\mathrm{μ_3-O}$相连，价态为$+1$价

<Checkpoint>[阳离子$\mathbf D$的结构式为$\mathrm{ [Mn_3(μ_3-O)(μ_2-CH_3COO)_6]^{+}}$](1)</Checkpoint>