Sodium cyclopentadienyl dicarbonylferrate(0) reacts with 2-chloroethyl methyl sulfide to generate compound $\mathbf{A}$ with the molecular formula $\mathrm{C_{10}H_{12}FeO_2S}$; $\mathbf{A}$ generates $\mathbf{B}$ upon heating. $\mathbf{B}$ exhibits two carbonyl stretching vibrational infrared absorption peaks. It is known that sodium cyclopentadienyl dicarbonylferrate(0) can be obtained by reacting an iron complex $\mathbf{X}$ that satisfies the 18-electron rule with sodium, and sodium cyclopentadienyl dicarbonylferrate(0) is the only product. The following statements are made:

1. The cyclopentadienyl dicarbonylferrate(0) anion carries 2 negative charges.
2. Iron in both $\mathbf{A}$ and $\mathbf{B}$ satisfies the 18-electron rule.
3. $\mathbf{B}$ has a bridging carbonyl structure.
4. $\mathbf{X}$ is a mononuclear complex.

What is the sum of the numbers of the correct statements?
环戊二烯基二羰基铁(0)酸钠与 2-氯乙基甲基硫醚反应，生成分子式为 $\mathrm{C_{10}H_{12}FeO_2S}$ 的化合物$\mathbf{A}$；$\mathbf{A}$在加热时生成$\mathbf{B}$。$\mathbf{B}$有2种羰基的伸缩振动红外吸收峰。已知环戊二烯基二羰基铁(0)酸钠可由满足18电子规则的铁配合物$\mathbf{X}$与钠反应而得，且环戊二烯基二羰基铁(0)酸钠是唯一产物。有如下说法：

1. 环戊二烯基二羰基铁(0)酸根带2个负电荷。
2. $\mathbf{A}$和$\mathbf{B}$中铁都满足18电子规则。
3. $\mathbf{B}$存在桥连羰基结构。
4. $\mathbf{X}$为单核配合物。

请问正确说法编号之和为多少？
2
5
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13
None of the above answers are correct.
2
5
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13
以上答案均不对
Sodium cyclopentadienyl dicarbonyl iron(0) satisfies the 18-electron rule. The charge carried by the negative ion is $18-8-5-2\times 2=1$. Statement 1 is incorrect.

<Checkpoint>[Cyclopentadienyl dicarbonyl iron(0) anion carries a 1- negative charge](0.5)</Checkpoint>

$\mathbf{A}$ is the product of the $\mathrm{S_N2}$ reaction of cyclopentadienyl dicarbonyl iron(0) anion with 2-chloroethyl methyl sulfide ($\mathrm{ClCH_2CH_2SCH_3}$), i.e., $\mathrm{(C_5H_5)(CO)_2FeCH_2CH_2SCH_3}$. If sulfur does not coordinate to iron, then there are $8+5+2\times2+1=18$ electrons around iron, satisfying the 18-electron rule.

<Checkpoint>[Iron in $\mathbf{A}$ satisfies the 18-electron rule](0.5)</Checkpoint>

$\mathbf{B}$ has two different carbon-oxygen bonds, and bridging carbonyls may exist. $\mathbf{B}$ is a dimer of $\mathbf{A}$. However, $\mathbf{A}$ has no driving force for dimerization, so this reaction pathway is excluded. In addition, $\beta$-elimination is uncommon when iron is the central metal, so it is excluded. Another possibility is that the alkyl ligand of $\mathbf{A}$ migrates to $\mathrm{CO}$ to form an acyl ligand, and the product is $\mathrm{(C_5H_5)(CO)Fe[C(O)CH_2CH_2SCH_3]}$. At this time, there are only 16 electrons around iron, and the sulfur atom of the acyl ligand can coordinate to iron, making iron satisfy the 18-electron rule, and the stability is improved. Statement 2 is correct, and statement 3 is incorrect.

<Checkpoint>[Alkyl ligand migrates to $\mathrm{CO}$](1)</Checkpoint>

<Checkpoint>[Sulfur atom of the acyl ligand coordinates to iron](1)</Checkpoint>

<Checkpoint>[Iron in $\mathbf{B}$ satisfies the 18-electron rule](0.5)</Checkpoint>

$\mathrm{Na[Fe(C_5H_5)(CO)_2]}$ is obtained by the reaction of sodium and $\mathbf{X}$, and the product is unique, so the simplest formula of $\mathbf{X}$ is $\mathrm{[Fe(C_5H_5)(CO)_2]}$. At this time, there are $8+5+2\times 2=17$ electrons around iron, which does not satisfy the 18-electron rule, so there is a metal-metal bond in $\mathbf{X}$, which should be $\mathrm{[Fe(C_5H_5)(CO)_2]_2}$. Statement 4 is incorrect.

<Checkpoint>[$\mathbf{X}$ is $\mathrm{[Fe(C_5H_5)(CO)_2]_2}$](0.5)</Checkpoint>

In summary, the correct statement is 2, choose option A.
环戊二烯基二羰基铁(0)酸钠中铁满足18电子规则，负离子带的电荷为$18-8-5-2\times 2=1$，说法1错误。

<Checkpoint>[环戊二烯基二羰基铁(0)酸根带1个负电荷](0.5)</Checkpoint>

$\mathbf{A}$是环戊二烯基二羰基铁(0)酸根对2-氯乙基甲基硫醚（$\mathrm{ClCH_2CH_2SCH_3}$）进行 $\mathrm{S_N2}$ 反应的产物，即$\mathrm{(C_5H_5)(CO)_2FeCH_2CH_2SCH_3}$。若硫不对铁配位，则铁周围有 $8+5+2\times2+1=18$ 电子，满足18电子规则。

<Checkpoint>[$\mathbf{A}$中铁满足18电子规则](0.5)</Checkpoint>

$\mathbf{B}$有两种不同的碳氧键，可能存在桥连羰基，$\mathbf{B}$为$\mathbf{A}$的二聚。但$\mathbf{A}$没有二聚的动力，排除该反应途径。另外$\beta$-消除在铁为中心金属时不常见，排除。还有一种可能是$\mathbf{A}$的烷基配体向$\mathrm{CO}$迁移，形成酰基配体，产物为$\mathrm{(C_5H_5)(CO)Fe[C(O)CH_2CH_2SCH_3]}$。此时铁周围仅16电子，酰基配体的硫原子可以向铁配位，使铁满足18电子规则，稳定性提升。说法2正确，说法3错误。

<Checkpoint>[烷基配体向$\mathrm{CO}$迁移](1)</Checkpoint>

<Checkpoint>[酰基配体的硫原子向铁配位](1)</Checkpoint>

<Checkpoint>[$\mathbf{B}$中铁满足18电子规则](0.5)</Checkpoint>

$\mathrm{Na[Fe(C_5H_5)(CO)_2]}$由钠和$\mathbf{X}$反应得到，且产物唯一，则$\mathbf{X}$的最简式为$\mathrm{[Fe(C_5H_5)(CO)_2]}$。此时铁周围有 $8+5+2\times 2=17$电子，不满足18电子规则，故$\mathbf{X}$存在金属-金属键，应为$\mathrm{[Fe(C_5H_5)(CO)_2]_2}$，说法4错误。

<Checkpoint>[$\mathbf{X}$为$\mathrm{[Fe(C_5H_5)(CO)_2]_2}$](0.5)</Checkpoint>

综上所述，正确的说法为2，选A选项。