$\mathrm{Fe}(\mathrm{CO})_{4}$ is an unstable molecule, and crystals cannot be isolated for structure determination. Its molecular structure can only be studied through infrared spectroscopy, theoretical calculations, and other methods. Calculations show that the ground state $\mathrm{Fe}(\mathrm{CO})_{4}$ molecule has 4 unpaired electrons at normal pressure; under high pressure, it transforms into a diamagnetic molecule.
The following statements are made:

1. $\mathrm{Fe}(\mathrm{CO})_{4}$ satisfies the EAN rule
2. Under high pressure, the 4s orbital in the valence electron configuration of $\mathrm{Fe}(\mathrm{CO})_{4}$ is not filled with electrons
3. Under normal pressure, the coordination polyhedron of $\mathrm{Fe}(\mathrm{CO})_{4}$ is a regular tetrahedron
4. Applying pressure does not change the coordination polyhedron type of $\mathrm{Fe}(\mathrm{CO})_{4}$

The following option contains all correct statements and the most correct statements:
$\mathrm{Fe}(\mathrm{CO})_{4}$ 是一种不稳定的分子, 无法分离出晶体进行结构测定, 只能通过红外光谱、理论计算等方式研究其分子结构。计算表明, 常压下基态的 $\mathrm{Fe}(\mathrm{CO})_{4}$ 分子具有 4 个单电子; 高压下,它转化为抗磁性分子。
有以下说法：

1. $\mathrm{Fe}(\mathrm{CO})_{4}$ 满足 EAN 规则
2. 高压下$\mathrm{Fe}(\mathrm{CO})_{4}$的价电子组态中4s轨道上未填充电子
3. 常压下$\mathrm{Fe}(\mathrm{CO})_{4}$的配位多面体为正四面体
4. 加压不会改变$\mathrm{Fe}(\mathrm{CO})_{4}$的配位多面体类型

以下选项中说法全部正确且正确说法最多的为：
All other options are incorrect
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
其他选项均不正确
1
2
3
4
1,2
1,3
1,4
2,3
2,4
3,4
1,2,3
1,2,4
1,3,4
2,3,4
1,2,3,4
The number of electrons around iron is: $8+4 \times 2=16$, which does not satisfy the EAN rule, statement 1 is incorrect.

<Checkpoint>[The number of electrons around iron is 16, which does not conform to the EAN rule](1)</Checkpoint>

Under high pressure, $\mathrm{Fe}(\mathrm{CO})_{4}$ transforms into a diamagnetic molecule, indicating that all electrons are paired. Therefore, the electronic configuration is $3d^84s^0$, and the $4s$ orbital is not filled with electrons, statement 2 is correct.

<Checkpoint>[Under high pressure, the electronic configuration of $\mathrm{Fe}(\mathrm{CO})_{4}$ is $3d^84s^0$](1)</Checkpoint>

Under normal pressure, $\mathrm{Fe}(\mathrm{CO})_{4}$ has 4 unpaired electrons, corresponding to the high-spin ${d}^{6}$ configuration. The corresponding crystal field (or ligand field) splitting energy should be small, so it should be a tetrahedral type; the ${d}^{6}$ electronic configuration is asymmetrically filled in the electron orbitals of the regular tetrahedral field, and Jahn-Teller distortion will occur to transform into a distorted tetrahedron, statement 3 is incorrect.

<Checkpoint>[Under normal pressure, $\mathrm{Fe}(\mathrm{CO})_{4}$ corresponds to the high-spin ${d}^{6}$ configuration, which is asymmetrically filled in the electron orbitals of the regular tetrahedral field, and Jahn-Teller distortion will occur to transform into a distorted tetrahedron](1)</Checkpoint>

Under high pressure, it is diamagnetic, corresponding to the low-spin ${d}^{8}$ configuration. The corresponding crystal field (or ligand field) splitting energy is large, so it should be a square planar type, statement 4 is incorrect.

<Checkpoint>[Under high pressure, the electronic configuration of $\mathrm{Fe}(\mathrm{CO})_{4}$ is $3d^84s^0$, the corresponding crystal field (or ligand field) splitting energy is large, so it should be a square planar type](1)</Checkpoint>
铁的周围电子数为: $8+4 \times 2=16$，不满足EAN规则，说法1错误。

<Checkpoint>[铁的周围电子数为16，不符合EAN规则](1)</Checkpoint>

高压下$\mathrm{Fe}(\mathrm{CO})_{4}$转变为抗磁性分子，说明电子全部配对，因此电子构型为$3d^84s^0$，$4s$轨道未填充电子，说法2正确

<Checkpoint>[高压下$\mathrm{Fe}(\mathrm{CO})_{4}$电子构型为$3d^84s^0$](1)</Checkpoint>

常压下 $\mathrm{Fe}(\mathrm{CO})_{4}$ 具有 4 个单电子, 对应高自旋 ${d}^{6}$ 组态, 对应的晶体场(或配位场)分裂能应较小, 故应为四面体类型; ${d}^{6}$ 电子组态在正四面体场的电子轨道中不对称填充, 会发生 Jahn-Teller 畸变转化为变形四面体，说法3错误。

<Checkpoint>[常压下 $\mathrm{Fe}(\mathrm{CO})_{4}$对应高自旋 ${d}^{6}$ 组态，在正四面体场的电子轨道中不对称填充, 会发生 Jahn-Teller 畸变转化为变形四面体](1)</Checkpoint>

高压下为抗磁性,对应低自旋 ${d}^{8}$ 组态,对应的晶体场(或配位场)分裂能较大,故应为平面四方类型，说法4错误。

<Checkpoint>[高压下$\mathrm{Fe}(\mathrm{CO})_{4}$电子构型为$3d^84s^0$，对应的晶体场(或配位场)分裂能较大,故应为平面四方类型](1)</Checkpoint>