Adding a slightly excessive amount of concentrated oxalic acid solution to a solution of nitrate $\rm \bf M\rm(NO_3)_3$ produces no gas. The precipitated precipitate is washed with an ethanol-water mixed solvent to obtain white crystals $\mathbf{A}$. The decomposition products of $\mathbf{A}$ are not completely the same under different atmospheres.

Thermogravimetric analysis of $\mathbf{A}$ in an argon atmosphere shows a weight loss of $20.9 \%$ when heated from $\rm 90\degree C$ to $\rm 300\degree C$, and a further weight loss of $23.5 \%$ when heated to $\rm 500\degree C$. The valence of $\bf M$ does not change in this step; further heating to $\rm 800\degree C$ results in a further weight loss of $4.1 \%$, and the residual solid is a black powder, the composition of which is a mixture of the oxide $\bf B$ $(\bf M\rm O_2)$ and another substance.

Heating $\mathbf{A}$ in air to $\rm 450\degree C$ results in a weight loss of $49.7 \%$, and the only remaining solid is oxide $\bf B$.

The following statements are correct:
在硝酸盐$\rm \bf M\rm(NO_3)_3$ 的溶液中加入稍过量的草酸浓溶液, 无气体产生。将析出的沉淀用乙醇-水混合溶剂洗涤, 得到白色结晶 $\mathbf{A}$。在不同的气氛下, $\mathbf{A}$ 的分解产物不完全相同。 

$\mathbf{A}$ 在氩气气氛中的热重分析显示, 从 $\rm 90\degree C$ 加热至 $\rm 300\degree C$, 失重 $20.9 \%$,  继续加热至 $\rm 500\degree C$, 再失重 $23.5 \%$，该步中$\bf M$化合价无变化; 继续加热至 $\rm 800\degree C$, 再失重 $4.1 \%$, 残留固体为黑色粉末, 其成分为氧化物 $\bf B$ $(\bf M\rm O_2)$ 与另一种物质的混合物。

$\mathbf{A}$在空气中加热至 $\rm 450\degree C$失重 $49.7 \%$, 唯一残留固体为氧化物$\bf B$。

下列说法正确的是：
All other options are incorrect
$\bf M$ ground state has a single electron
The relative molecular mass of $\mathbf{A}$ is less than 660.
The black powder does not contain any elemental substance.
$\mathbf{A}$ The second step of weight loss in an argon atmosphere only produces one kind of gas.
$\mathbf{A}$ The solid remaining after the second-step weight loss in an argon atmosphere is a pure substance.
$\mathbf{A}$ In the solid remaining after the second step of weight loss in an argon atmosphere, the chemical formula of the M-containing substance contains 8 atoms.
其他选项均不正确
$\bf M$基态有一个单电子
$\mathbf{A}$的相对分子质量小于660
黑色粉末中不含有单质
$\mathbf{A}$ 在氩气气氛的第二步失重只产生一种气体
$\mathbf{A}$ 在氩气气氛的第二步失重剩余的固体为纯净物
$\mathbf{A}$ 在氩气气氛的第二步失重剩余的固体中，含$ \rm M$的物质化学式含有8个原子
The addition of a slightly excessive amount of concentrated oxalic acid solution to a solution of nitrate $\rm \bf M\rm(NO_3)_3$ produces no gas, so no redox reaction occurs;

<Checkpoint>[Precipitate formation without redox reaction](1)</Checkpoint>

The precipitate formed should be hydrated oxalate, and the valence of $\bf M$ should still be +3, so $\bf A$ can be written as $\bf M\rm_2(C_2O_4)_3\cdot nH_2O$.

<Checkpoint>[The precipitate formed should be hydrated oxalate, and $\bf A$ can be written as $\bf M\rm_2(C_2O_4)_3\cdot nH_2O$](1)</Checkpoint>

$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$ decomposes in air, producing only oxide $\bf M\rm O_2$ and losing $\rm 49.7\%$ of its weight, so the following equation can be written:

$\rm \frac{x+32}{x+264.1+18n}=1-0.497$

$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$ loses weight in argon, and the first step is likely to be the loss of water of crystallization; therefore, the following equation is written:

<Checkpoint>[$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$ loses weight in argon, and the first step is likely to be the loss of water of crystallization](1)</Checkpoint>

$\rm \frac{18n}{x+264.1+18n}=0.209$

The simultaneous equations can be solved to get $\rm n=8.07\approx 8$. The solution is not an integer because of errors in the thermogravimetric analysis.

<Checkpoint>[Simultaneous equations can be solved to get the number of water of crystallization $\rm n=8.07\approx 8$](1)</Checkpoint>

Substituting $\rm n=8$, the solution is $\rm x=143.6$. Near the atomic weight of 144, considering that $\bf M$ has a stable +4 valence oxide, it should be $\rm Ce$. The electronic configuration of $\rm Ce$ is $\rm 4f^15d^16s^2$, with two unpaired electrons, so option B is incorrect.

<Checkpoint>[Considering that $\bf M$ has a stable +4 valence oxide, it should be $\rm Ce$](1)</Checkpoint>

Therefore, the chemical formula of $\bf A$ is $\rm Ce_2(C_2O_4)_3\cdot 8H_2O$. The relative molecular mass is 688.4, so option C is incorrect.

<Checkpoint>[$\bf A$ chemical formula is $\rm Ce_2(C_2O_4)_3\cdot 8H_2O$](1)</Checkpoint>

The amount of weight loss in the second step of $\rm Ce_2(C_2O_4)_3\cdot 8H_2O$ in argon is $\rm 688.4\times0.235=162$, which should be $\rm 3CO_2 + CO$ (the difference in molecular weight is 2, which is the difference between the theoretical and actual weight loss). Since $\rm CO$ is produced, a redox reaction should have occurred. Since the valence of $\rm Ce$ has not changed, only elemental carbon can be produced.

<Checkpoint>[The amount of weight loss in the second step of $\rm Ce_2(C_2O_4)_3\cdot 8H_2O$ in argon is 162, which should be $\rm 3CO_2 + CO$](1)</Checkpoint>

<Checkpoint>[The valence of $\rm Ce$ has not changed, so only elemental carbon can be produced](1)</Checkpoint>

Therefore, the equation can be balanced:

$\rm Ce_2(C_2O_4)_3 = 3CO_2 + CO + Ce_2O_2(CO_3) + 2C$

<Checkpoint>[ $\rm Ce_2(C_2O_4)_3 = 3CO_2 + CO + Ce_2O_2(CO_3) + 2C$](1)</Checkpoint>

Therefore, the solid produced in the second step is $\rm Ce_2O_2(CO_3) + 2C$, and the black solid in the third step is $\rm CeO_2 + C$.

<Checkpoint>[The solid produced in the second step is $\rm Ce_2O_2(CO_3) + 2C$, and the black solid in the third step is $\rm CeO_2 + C$](1)</Checkpoint>

Therefore, options D, E, and F are all incorrect, and option G is correct.
硝酸盐$\rm \bf M\rm(NO_3)_3$ 的溶液中加入稍过量的草酸浓溶液, 无气体产生，因此未发生氧化还原反应；

<Checkpoint>[产生沉淀未发生氧化还原反应](1)</Checkpoint>

产生的沉淀应当为水合草酸盐，$\bf M$应当仍旧为+3价态，因此$\bf A$可写为$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$。

<Checkpoint>[产生的沉淀应当为水合草酸盐，$\bf A$可写为$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$](1)</Checkpoint>

$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$在空气中分解，只生成氧化物$\bf M\rm O_2$且失重$\rm 49.7\%$，因此可列出方程：

$\rm \frac{x+32}{x+264.1+18n}=1-0.497$

$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$在氩气中失重，第一步大概率为失去结晶水；因此列出方程：

<Checkpoint>[$\bf M\rm_2(C_2O_4)_3\cdot nH_2O$在氩气中失重，第一步大概率为失去结晶水](1)</Checkpoint>

$\rm \frac{18n}{x+264.1+18n}=0.209$

联立可解的$\rm n=8.07\approx 8$，解出非整数是因为热重分析有误差。

<Checkpoint>[联立可解结晶水个数$\rm n=8.07\approx 8$](1)</Checkpoint>

代入$\rm n=8$，解的$\rm x=143.6$，在原子量144附近，考虑到$\bf M$存在稳定的+4价氧化物，应当为$\rm Ce$。$\rm Ce$电子组态为$\rm 4f^15d^16s^2$，有两个单电子，选项B错误。

<Checkpoint>[考虑到$\bf M$存在稳定的+4价氧化物，应当为$\rm Ce$](1)</Checkpoint>

因此$\bf A$化学式为$\rm Ce_2(C_2O_4)_3\cdot 8H_2O$。相对分子量为688.4，选项C错误。

<Checkpoint>[$\bf A$化学式为$\rm Ce_2(C_2O_4)_3\cdot 8H_2O$](1)</Checkpoint>

$\rm Ce_2(C_2O_4)_3\cdot 8H_2O$在氩气中第二步失重的量为$\rm 688.4\times0.235=162$，应当为$\rm 3CO_2 + CO$（分子量差距2，为理论失重与实际失重的差异）。由于产生了$\rm CO$，应当发生了氧化还原反应，由于$\rm Ce$化合价未变化，因此只能生成碳单质。

<Checkpoint>[$\rm Ce_2(C_2O_4)_3\cdot 8H_2O$在氩气中第二步失重的量为162，应当为$\rm 3CO_2 + CO$](1)</Checkpoint>

<Checkpoint>[$\rm Ce$化合价未变化，因此只能生成碳单质](1)</Checkpoint>

因此可以配平方程式：

$\rm Ce_2(C_2O_4)_3 = 3CO_2 + CO + Ce_2O_2(CO_3) + 2C$

<Checkpoint>[ $\rm Ce_2(C_2O_4)_3 = 3CO_2 + CO + Ce_2O_2(CO_3) + 2C$](1)</Checkpoint>

因此第二步产生的固体为$\rm Ce_2O_2(CO_3) + 2C$，第三步黑色固体为$\rm CeO_2 + C$。

<Checkpoint>[第二步产生的固体为$\rm Ce_2O_2(CO_3) + 2C$，第三步黑色固体为$\rm CeO_2 + C$](1)</Checkpoint>

因此选项D,E,F均错误，选项G正确。