The element $\mathbf M$ of the first transition series dissolves in dilute sulfuric acid and crystallizes to obtain a hydrated sulfate crystal $\mathbf A$, in which the mass fraction of $\mathbf M$ is 20.97%. Element $\mathbf M$ has only one non-radioactive stable isotope.

The THF solvate of the simple chloride of $\mathbf M$ reacts with 1-lithiated bicyclo[2.2.1]heptane in pentane to produce $\mathbf M$ element and a mononuclear, electrically neutral complex $\mathbf D$ with a carbon content of 76.50%.

Reduction of $\mathbf D$ with lithium element in tetrahydrofuran yields a salt $\mathbf E$ with a cation-anion ratio of 1:1, while oxidation of $\mathbf D$ with silver tetrafluoroborate yields a salt $\mathbf F$ with a cation-anion ratio of 1:1.

Regarding the above content, there are the following conclusions:

Conclusion 1: Isotope $\mathbf M_2$ whose mass number is 1 greater than that of the stable isotope of $\mathbf M$ emits β rays when decaying.

Conclusion 2: The molecular weight of $\mathbf D$ is 583.77.

Conclusion 3: $\mathbf D$, $\mathbf E$, and $\mathbf F$ all contain $\mathbf M$ with an octahedral coordination configuration.

Conclusion 4: The theoretically calculated spin-only magnetic moment of $\mathbf E$ is 2.83 B.M..

Conclusion 5: $\mathbf F$ is a diamagnetic substance.

Select the option corresponding to all the correct conclusions.
第一过渡系的金属元素$\mathbf M$的单质溶于稀硫酸并结晶能得到水合硫酸盐晶体$\mathbf A$，其中$\mathbf M$的质量分数为20.97%。元素$\mathbf M$只有一种无放射性的稳定同位素。

$\mathbf M$的简单氯化物的THF溶剂合物在戊烷中与1-锂代双环[2.2.1]庚烷发生反应，生成$\mathbf M$单质和含碳量为76.50%的单核电中性配合物$\mathbf D$。

在四氢呋喃中用锂单质还原$\mathbf D$即可得到阴阳离子比1:1的盐$\mathbf E$，而用四氟合硼酸银氧化$\mathbf D$则生成阴阳离子比1:1的盐$\mathbf F$。

关于上述内容，有以下一些结论：

结论1：质量数比$\mathbf M$的稳定同位素大1的同位素$\mathbf M_2$衰变时放出β射线。

结论2：$\mathbf D$的分子量为583.77。

结论3：$\mathbf D$、$\mathbf E$、$\mathbf F$中均含有八面体配位构型的$\mathbf M$.

结论4：理论计算$\mathbf E$的唯自旋磁矩为2.83 B.M.。

结论5：$\mathbf F$是抗磁性物质。

选择对应于所有正确结论的选项。
All other options are incorrect
Option 1, Option 2, Option 3
Option 1, Option 2, Option 4
Option 1, Option 2, Option 5
Option 1, Option 3, Option 4
Option 1, Option 3, Option 5
Option 1, Option 4, Option 5
Option 2, Option 3, Option 4
Option 2, Option 3, Option 5
Option 2, Option 4, Option 5
Option 3, Option 4, Option 5
其他选项均不正确
选项1、选项2、选项3
选项1、选项2、选项4
选项1、选项2、选项5
选项1、选项3、选项4
选项1、选项3、选项5
选项1、选项4、选项5
选项2、选项3、选项4
选项2、选项3、选项5
选项2、选项4、选项5
选项3、选项4、选项5
Only $Sc$, $Mn$, and $Co$ are elements in the first transition series that have only one stable isotope. Examine the mass fraction of the metal in their hydrated sulfates as follows:

- If $\mathbf M$ is $Sc$: $Sc$ usually forms +3 valent ions in acid, so the salt is $Sc_2(SO_4)_3·nH_2O$. The relative molecular mass $M_r(\mathbf A) = 2 \times 44.96 \div 0.2097 = 428.8$, the number of moles of water $n = (428.8-(2 \times 44.96 + 3 \times (32.07 + 4 \times 16.00))) \div 18.02 = 2.81$, which is not an integer, so $Sc$ is excluded.
- If $\mathbf M$ is $Mn$: $Mn$ usually forms +2 valent ions in dilute sulfuric acid, so the salt is $MnSO_4·nH_2O$. $M_r(\mathbf A) = 54.94 \div 0.2097 = 262.0$, the number of moles of water $n = (262.0-(54.94 + 32.07 + 4 \times 16.00)) \div 18.02 = 6.16$, which is not an integer, so $Mn$ is excluded.
- If $\mathbf M$ is $Co$: $Co$ forms +2 valent ions in dilute sulfuric acid, so the salt is $CoSO_4·nH_2O$. $M_r(\mathbf A) = 58.93 \div 0.2097 = 281.0$, the number of moles of water $n = (281.0-(58.93 + 32.07 + 4 \times 16.00)) \div 18.02 = 7.00$, which is an integer, so it may be $Co$.

Therefore, the metal $\mathbf M$ is cobalt ($Co$), and its only stable isotope is $^{59}Co$.

<Checkpoint>[The metal element $\mathbf M$ is cobalt](3)</Checkpoint>

The only simple chloride of $Co$ commonly used for synthesis is $CoCl_2$. The reaction of $CoCl_2$ with 1-lithiated bicyclo[2.2.1]heptane (denoted as Li-R) to produce $Co$ element indicates that $Co$ may have undergone a disproportionation reaction, simultaneously producing $Co$ element and $CoR_x$. $\mathbf D$ may be $CoR_x$, with the molecular formula $CoC_{7x}H_{11x}$. List the mass fraction equation:

$$
\frac{7x\times12.01}{58.93+7x\times12.01+11x\times1.008}=76.50\%
$$

Solving for: $x = 3.999 \approx 4$, so $\mathbf D$ is $CoR_4$.

<Checkpoint>[$\mathbf D$ is $CoR_4$](3)</Checkpoint>

Lithium element is a strong reducing agent, and the product $\mathbf E$ formed by reducing $\mathbf D$ should be a lithium salt with $\mathbf D$ carrying several negative charges as the anion. According to the 1:1 ratio of cations to anions, it can be inferred that $\mathbf E$ is $[Li(THF)_x][CoR_4]$

<Checkpoint>[$\mathbf E$ is $[Li(THF)_x][CoR_4]$](1)</Checkpoint>

+1 valent silver ion is a moderately strong oxidizing agent, and the product $\mathbf F$ formed by AgBF_4 oxidizing $\mathbf D$ should be a tetrafluoroborate salt with $\mathbf D$ carrying several positive charges as the cation. According to the 1:1 ratio of cations to anions, it can be inferred that $\mathbf F$ is $[CoR_4][BF_4]$

<Checkpoint>[$\mathbf F$ is $[CoR_4][BF_4]$](1)</Checkpoint>

The isotope $\mathbf M_2$ whose mass number is 1 greater than the stable isotope of $\mathbf M$ is $^{60}Co$. $^{60}Co$ undergoes β decay and emits β rays. Conclusion 1 is correct.

<Checkpoint>[The isotope $\mathbf M_2$ whose mass number is 1 greater than the stable isotope of $\mathbf M$ is $^{60}Co$. $^{60}Co$ undergoes β decay and emits β rays.](1)</Checkpoint>

The molecular weight of $\mathbf D$: $M_r(\mathbf D)=58.93+4\times(7\times12.01+11\times1.008)=439.56 \neq 583.77$. Conclusion 2 is incorrect.

<Checkpoint>[The molecular weight of $\mathbf D$: $M_r(\mathbf D)=58.93+4\times(7\times12.01+11\times1.008)=439.56 \neq 583.77$](1)</Checkpoint>

$Co$ in $\mathbf D$, $\mathbf E$, and $\mathbf F$ all have a tetrahedral coordination configuration. Conclusion 3 is incorrect.

<Checkpoint>[$Co$ in $\mathbf D$, $\mathbf E$, and $\mathbf F$ all have a tetrahedral coordination configuration.](1)</Checkpoint>

$Co$ in $\mathbf E$ is +3 valent, and its electronic configuration in a tetrahedral field is $(e)^4(t_2)^2$, with 2 unpaired electrons. The theoretically calculated spin-only magnetic moment of $\mathbf E$ is $\sqrt{2\times(2+2)}=2.83 \mathrm{~B.M.}$ Conclusion 4 is correct.

<Checkpoint>[$Co$ in $\mathbf E$ is +3 valent, and its electronic configuration in a tetrahedral field is $(e)^4(t_2)^2$, with 2 unpaired electrons. The theoretically calculated spin-only magnetic moment of $\mathbf E$ is $\sqrt{2\times(2+2)}=2.83 \mathrm{~B.M.}$](1)</Checkpoint>

$Co$ in $\mathbf F$ is +5 valent, and its electronic configuration in a tetrahedral field is $(e)^4(t_2)^0$, with no unpaired electrons, so it is diamagnetic. Conclusion 5 is correct.

<Checkpoint>[$Co$ in $\mathbf F$ is +5 valent, and its electronic configuration in a tetrahedral field is $(e)^4(t_2)^0$, with no unpaired electrons, so it is diamagnetic.](1)</Checkpoint>

In summary, choose option G.
第一过渡系元素中只有一种稳定同位素的元素只有$Sc$、$Mn$、$Co$，检验其水合硫酸盐中金属质量分数如下：

- 如果$\mathbf M$是$Sc$：$Sc$在酸中通常形成+3价离子，所以盐为$Sc_2(SO_4)_3·nH_2O$。相对分子质量$M_r(\mathbf A) = 2 \times 44.96 \div 0.2097 = 428.8$，水的摩尔数$n = (428.8-(2 \times 44.96 + 3 \times (32.07 + 4 \times 16.00))) \div 18.02 = 2.81$，不是整数，排除$Sc$。
- 如果$\mathbf M$是$Mn$：$Mn$在稀硫酸中通常形成+2价离子，所以盐为$MnSO_4·nH_2O$。$M_r(\mathbf A) = 54.94 \div 0.2097 = 262.0$，水的摩尔数$n = (262.0-(54.94 + 32.07 + 4 \times 16.00)) \div 18.02 = 6.16$，不是整数，排除$Mn$。
- 如果$\mathbf M$是$Co$：$Co$在稀硫酸中形成+2价离子，所以盐为$CoSO_4·nH_2O$。$M_r(\mathbf A) = 58.93 \div 0.2097 = 281.0$，水的摩尔数$n = (281.0-(58.93 + 32.07 + 4 \times 16.00)) \div 18.02 = 7.00$，是整数，可能是$Co$。

因此，金属$\mathbf M$为钴($Co$)，其唯一的稳定同位素是$^{59}Co$。

<Checkpoint>[金属元素$\mathbf M$为钴](3)</Checkpoint>

常用于合成的$Co$的简单氯化物仅有$CoCl_2$。$CoCl_2$与1-锂代双环[2.2.1]庚烷（记作Li-R）的反应中有$Co$单质生成表明$Co$可能发生了歧化反应，同时生成了$Co$单质和$CoR_x$。$\mathbf D$可能是$CoR_x$，分子式为$CoC_{7x}H_{11x}$，列出质量分数方程：

$$
\frac{7x\times12.01}{58.93+7x\times12.01+11x\times1.008}=76.50\%
$$

解得：$x = 3.999 \approx 4$，故$\mathbf D$是$CoR_4$。

<Checkpoint>[$\mathbf D$是$CoR_4$](3)</Checkpoint>

锂单质是强还原剂，还原$\mathbf D$生成的产物$\mathbf E$应该是以带若干负电荷的$\mathbf D$为阴离子的锂盐，根据阴阳离子比1:1可以推断$\mathbf E$是$[Li(THF)_x][CoR_4]$。

<Checkpoint>[$\mathbf E$是$[Li(THF)_x][CoR_4]$](1)</Checkpoint>

+1价银离子是中等强度的氧化剂，$AgBF_4$氧化$\mathbf D$生成的产物$\mathbf F$应该是以带若干正电荷的$\mathbf D$为阳离子的四氟合硼酸盐。根据阴阳离子比1:1可以推断$\mathbf F$是$[CoR_4][BF_4]$。

<Checkpoint>[$\mathbf F$是$[CoR_4][BF_4]$](1)</Checkpoint>

质量数比$\mathbf M$的稳定同位素大1的同位素$\mathbf M_2$是$^{60}Co$。$^{60}Co$发生β衰变放出β射线。结论1正确。

<Checkpoint>[$\mathbf M_2$是$^{60}Co$，$^{60}Co$发生β衰变放出β射线](1)</Checkpoint>

$\mathbf D$的分子量：$M_r(\mathbf D)=58.93+4\times(7\times12.01+11\times1.008)=439.56 \neq 583.77$。结论2错误。

<Checkpoint>[$\mathbf D$的分子量：$M_r(\mathbf D)=439.56 \neq 583.77$](1)</Checkpoint>

$\mathbf D$、$\mathbf E$、$\mathbf F$中$Co$均为四面体配位构型。结论3错误。

<Checkpoint>[$\mathbf D$、$\mathbf E$、$\mathbf F$中$Co$均为四面体配位构型](1)</Checkpoint>

$\mathbf E$中$Co$为+3价，在四面体场中电子组态为$(e)^4(t_2)^2$，有2个未成对电子，理论计算$\mathbf E$的唯自旋磁矩为$\sqrt{2\times(2+2)}=2.83 \mathrm{~B.M.}$。结论4正确。

<Checkpoint>[$\mathbf E$中$Co$为+3价四面体构型，$\mathbf E$的唯自旋磁矩为$2.83 \mathrm{~B.M.}$](1)</Checkpoint>

$\mathbf F$中$Co$为+5价，在四面体场中电子组态为$(e)^4(t_2)^0$，无未成对电子，故为抗磁性。结论5正确。

<Checkpoint>[$\mathbf F$中$Co$为+5价四面体构型，无未成对电子，为抗磁性](1)</Checkpoint>

综上所述，选择G选项。