Recently, the synthesis and characterization of the anti-aromatic carbon allotrope – cyclo[16]carbon ($\rm C_{16}$) molecule has been achieved using tip-induced surface chemistry methods. The atomic bonding sequence of the cyclic $\rm C_{16}$ element is similar to a regular hexadecagon.

It has been proposed that the ground state electronic state of $\rm C_{16}$ corresponding to the orbital pattern may be a state with two sets of aromatic conjugated systems, with all carbon-carbon bonds being completely equal in length (structure 1); however, studies using atomic force microscopy and scanning tunneling microscopy suggest that $\rm C_{16}$ exhibits significant bond length alternation (structure 2).

The following statements are made:

Statement 1. Suppose the delocalized $\Pi$ bond types of structure 1 are $\Pi_{16}^{\rm ab}$ and $\Pi_{16}^{\rm cd}$ ($\rm ab,cd$ represent two-digit numbers, $\rm a,c$ are tens digits, and $\rm b,d$ are ones digits), then $0.34<\rm \frac ab+\frac cd<0.38$.

Statement 2. The sum of the number of rotational axes, mirror planes, and centers of symmetry of structure 1 is 34.

Statement 3. The sum of the number of rotational axes, mirror planes, and centers of symmetry of structure 2 is 18.

Butadiene molecules have a rectangular structure in the ground state, and their HOMO and LUMO orbitals are shown in the figure below:

<MultiModal>![The image is a scientific diagram mainly showing molecular orbital structures. The left side of the image is a simple two-dimensional molecular structure consisting of a side view of a rectangular skeleton with four vertices, with a parallel double bond on each of the upper and lower sides. The right side of the image is divided into two parts, labeled "HOMO" and "LUMO" respectively. In the "HOMO" part, the top is a three-dimensional molecular orbital model, which is a side view of a rectangular skeleton, and four dumbbell-shaped orbitals extend from the four vertices of the skeleton. Among these four orbitals, the two orbitals in the left rear and right rear are black on top and white on the bottom, and the two orbitals in the left front and right front are white on top and black on the bottom. Below the "HOMO" text is a two-dimensional simplified representation consisting of four adjacent circles forming a square, where the upper left and upper right circles are shaded, and the lower left and lower right are hollow. In the "LUMO" part, the top is also a three-dimensional molecular orbital model, the skeleton and orbital arrangement are similar to the "HOMO" part, but the shading pattern is different: the two orbitals in the left rear and left front are black on top and white on the bottom, and the right rear and right front are white on top and black on the bottom. Below the "LUMO" text is its corresponding two-dimensional simplified representation, also consisting of four adjacent circles forming a square, where the upper left and lower left circles are shaded, and the upper right and lower right are hollow.](/media/uploads/a116fd99-793e-4a9b-8960-039890d122c8.png)</MultiModal>

Statement 4. The schematic diagram of the HOMO orbital of $\rm C_{16}$ of structure 2 is as follows:
<MultiModal>![16 circles are connected to each other, starting from the top left clockwise, followed by 4 adjacent black circles, 4 adjacent white circles, 4 adjacent black circles, and 4 adjacent white circles. All circles are approximately the same size, shape, and arrangement distance.](/media/uploads/7636740a-d0f5-493e-a9b0-18041249de09.png)</MultiModal>

Statement 5. If the $\rm C_{16}$ of structure 2 is reduced to $\rm C_{16}^-$, the bond length of the original long carbon-carbon bond shortens, and the bond length of the original short carbon-carbon bond increases.

What is the sum of the numbers of the correct statements?
近日，人们利用针尖诱导的表面化学方法实现了反芳香性碳同素异形体——环十六碳（cyclo[16]carbon, $\rm C_{16}$）分子的合成和表征。环状的 $\rm C_{16}$ 单质原子键连顺序类似正十六边形。

人们曾提出， $\rm C_{16}$ 的轨道模式对应的基态电子态可能是具有两套芳香共轭体系的状态，所有碳碳键完全等长（结构1)；但是，原子力显微镜、扫描隧道显微镜的研究认为， $\rm C_{16}$ 具有显著的键长交替现象（结构2）。

有如下说法：

说法1.设结构1具有的离域 $\Pi$ 键类型为 $\Pi_{16}^{\rm ab}$ 和 $\Pi_{16}^{\rm cd}$（$\rm ab,cd$ 表示两位数，$\rm a,c$ 为十位，$\rm b,d$ 为个位），则 $0.34<\rm \frac ab+\frac cd<0.38$。

说法2.结构1的旋转轴、镜面、对称中心个数之和为34。

说法3.结构2的旋转轴、镜面、对称中心个数之和为18。

环丁二烯分子在基态为长方形结构，其HOMO 轨道和LUMO轨道如下图所示：

<MultiModal>![该图像是一个科学示意图，主要展示了分子轨道结构。图像左侧是一个简单的二维分子结构，由一个包含四个顶点的侧视长方形骨架组成，其中上下两侧各有一条平行的双键。图像的右侧分成了两个部分，分别标记为“HOMO”和“LUMO”。在“HOMO”部分，上方是一个三维分子轨道模型，为侧视长方形骨架，骨架的四个顶点处分别伸出四个哑铃状的轨道。这四个轨道中，左后和右后的两个轨道是上黑下白，左前和右前的两个轨道是上白下黑的。在“HOMO”文字下方，是一个二维的简化表示，由四个相邻的圆圈组成一个正方形，其中左上和右上的圆圈是阴影填充的，左下和右下是空心的。在“LUMO”部分，上方同样是一个三维分子轨道模型，骨架和轨道排布与“HOMO”部分类似，但阴影填充的模式不同：左后和左前两个轨道是上黑下白，右后和右前是上白下黑。在“LUMO”文字下方，是其对应的二维简化表示，同样由四个相邻的圆圈组成一个正方形，其中左上和左下的圆圈是阴影填充的，右上和右下是空心的。](/media/uploads/a116fd99-793e-4a9b-8960-039890d122c8.png)</MultiModal>

说法4.结构2的 $\rm C_{16}$ 的HOMO轨道示意图如下：
<MultiModal>![16个圆相互连接，从顶部偏左顺时针出发，依次是4个相邻的黑色圆、4个相邻的白色圆、4个相邻的黑色圆、4个相邻的白色圆。所有圆的大小、形状和排列距离都大致相同。](/media/uploads/7636740a-d0f5-493e-a9b0-18041249de09.png)</MultiModal>

说法5.若将结构2的 $\rm C_{16}$ 还原为 $\rm C_{16}^-$，则原先长的碳碳键的键长缩短，原先短的碳碳键的键长增长。

正确说法序号之和为多少？

该图像是一个科学示意图，主要展示了分子轨道结构。图像左侧是一个简单的二维分子结构，由一个包含四个顶点的侧视长方形骨架组成，其中上下两侧各有一条平行的双键。图像的右侧分成了两个部分，分别标记为“HOMO”和“LUMO”。在“HOMO”部分，上方是一个三维分子轨道模型，为侧视长方形骨架，骨架的四个顶点处分别伸出四个哑铃状的轨道。这四个轨道中，左后和右后的两个轨道是上黑下白，左前和右前的两个轨道是上白下黑的。在“HOMO”文字下方，是一个二维的简化表示，由四个相邻的圆圈组成一个正方形，其中左上和右上的圆圈是阴影填充的，左下和右下是空心的。在“LUMO”部分，上方同样是一个三维分子轨道模型，骨架和轨道排布与“HOMO”部分类似，但阴影填充的模式不同：左后和左前两个轨道是上黑下白，右后和右前是上白下黑。在“LUMO”文字下方，是其对应的二维简化表示，同样由四个相邻的圆圈组成一个正方形，其中左上和左下的圆圈是阴影填充的，右上和右下是空心的。

16个圆相互连接，从顶部偏左顺时针出发，依次是4个相邻的黑色圆、4个相邻的白色圆、4个相邻的黑色圆、4个相邻的白色圆。所有圆的大小、形状和排列距离都大致相同。

16个圆相互连接，从顶部偏左顺时针出发，依次是2个相邻的黑色圆、2个相邻的白色圆、2个相邻的黑色圆、2个相邻的白色圆，以此类推。所有圆的大小、形状和排列距离都大致相同。
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All of the above options are incorrect.
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以上选项均不对
In structure 1, each carbon provides 2 p electrons. All carbon atoms are coplanar, assuming each carbon has a p orbital perpendicular to the molecular plane, the formed $\Pi$ bond is 16 electrons, anti-aromatic, and the p orbital of each carbon within the molecular plane can also form a $\Pi$ bond, which is also 16 electrons, anti-aromatic. To match the stated "two sets of aromatic conjugated systems", one conjugated system transfers 2 electrons to the other system, forming $\Pi_{16}^{14}$ and $\Pi_{16}^{18}$, and the calculation shows that statement 1 is correct.

<Checkpoint>[The types of delocalized $\Pi$ bonds possessed by structure 1 are $\Pi_{16}^{14}$ and $\Pi_{16}^{18}$](1)</Checkpoint>

Structure 1 is a planar regular hexadecagon, with 1 16-fold rotation axis, 16 2-fold rotation axes, 17 mirror planes, and 1 center of symmetry. The sum of the number of the above symmetry elements is 35, so statement 2 is incorrect.

<Checkpoint>[The sum of the number of rotation axes, mirror planes, and centers of symmetry in structure 1 is 35](1)</Checkpoint>

Structure 2 is a planar equiangular hexadecagon with alternating long and short sides, with 1 8-fold rotation axis, 8 2-fold rotation axes, 9 mirror planes, and 1 center of symmetry. The sum of the number of the above symmetry elements is 19, so statement 3 is incorrect.

<Checkpoint>[The sum of the number of rotation axes, mirror planes, and centers of symmetry in structure 2 is 19](1)</Checkpoint>

Analyzing the frontier molecular orbital situation of cyclobutadiene, we can know that its HOMO and LUMO were originally non-bonding orbitals, and degeneracy occurred because the molecular structure changed from a regular polygon to a polygon with alternating long and short sides. In the HOMO, the orbital lobes on both sides of the short bond are the same, and the orbital lobes on both sides of the long bond are opposite; the LUMO is the opposite, with the orbital lobes on both sides of the short bond being opposite, and the orbital lobes on both sides of the long bond being the same. Analogous to cyclobutadiene, it is inferred that the HOMO of $\rm C_{16}$ should be two groups of two identical lobes appearing alternately, as shown in the figure below, so statement 4 is incorrect.

<MultiModal>![16 circles are connected to each other, starting from the top left clockwise, followed by 2 adjacent black circles, 2 adjacent white circles, 2 adjacent black circles, 2 adjacent white circles, and so on. All circles are approximately the same size, shape, and arrangement distance.](/media/uploads/a65e90a0-3e7a-42e4-866c-8655855fdc2c.png)</MultiModal>

<Checkpoint>[In the HOMO, the orbital lobes on both sides of the short bond are the same, and the orbital lobes on both sides of the long bond are opposite; the LUMO is the opposite, with the orbital lobes on both sides of the short bond being opposite, and the orbital lobes on both sides of the long bond being the same](2)</Checkpoint>

$\rm C_{16}$ is reduced to $\rm C_{16}^-$, that is, one electron is filled in the LUMO. In the LUMO, the orbital lobes on both sides of the short bond are opposite, and the orbital lobes on both sides of the long bond are the same, so the bond length of the original long carbon-carbon bond is shortened, and the bond length of the original short carbon-carbon bond is increased, so statement 5 is correct.

<Checkpoint>[The LUMO is bonding at the long bond and antibonding at the short bond](1)</Checkpoint>

In summary, the correct statements are 1 and 5, so the answer is option D.
在结构1中，每个碳提供2个p电子。所有碳原子共面，假设每个碳都有垂直于分子平面的p轨道，形成的 $\Pi$ 键为16电子，反芳香性，而每个碳在分子平面内的p轨道也能形成 $\Pi$ 键，也是16电子反芳香性的。为了契合题述的”两套芳香共轭体系”，一套共轭体系对另一套体系转移2个电子，形成 $\Pi_{16}^{14}$ 和 $\Pi_{16}^{18}$，计算可知说法1正确。

<Checkpoint>[结构1具有的离域 $\Pi$ 键类型为 $\Pi_{16}^{14}$ 和 $\Pi_{16}^{18}$](1)</Checkpoint>

结构1是平面的正十六边形，有1个16次旋转轴，16个2次旋转轴，17个镜面，1个对称中心，以上对称元素数目之和为35，说法2错误。

<Checkpoint>[结构1的旋转轴、镜面、对称中心个数之和为35](1)</Checkpoint>

结构2是平面的长短边交替的等角十六边形，有1个8次旋转轴，8个2次旋转轴，9个镜面，1个对称中心，以上对称元素数目之和为19，说法3错误。

<Checkpoint>[结构2的旋转轴、镜面、对称中心个数之和为19](1)</Checkpoint>

分析环丁二烯的前线分子轨道情况，可以知道其HOMO和LUMO原本为非键轨道，因分子结构从正多边形变为长短边交替的多边形而出现了去简并。HOMO中短键两侧轨道波瓣相同，长键两侧轨道波瓣相反；LUMO则是短键两侧轨道波瓣相反，长键两侧轨道波瓣相同。类比环丁二烯，推知 $\rm C_{16}$ 的HOMO应该是两种波瓣相同的两个一组交替出现，如下图，说法4错误。

<MultiModal>![16个圆相互连接，从顶部偏左顺时针出发，依次是2个相邻的黑色圆、2个相邻的白色圆、2个相邻的黑色圆、2个相邻的白色圆，以此类推。所有圆的大小、形状和排列距离都大致相同。](/media/uploads/a65e90a0-3e7a-42e4-866c-8655855fdc2c.png)</MultiModal>

<Checkpoint>[HOMO中短键两侧轨道波瓣相同，长键两侧轨道波瓣相反；LUMO则是短键两侧轨道波瓣相反，长键两侧轨道波瓣相同](2)</Checkpoint>

 $\rm C_{16}$ 被还原为 $\rm C_{16}^-$，即在LUMO填一个电子。LUMO中短键两侧轨道波瓣相反，长键两侧轨道波瓣相同，所以原先长的碳碳键的键长缩短，原先短的碳碳键的键长增长，说法5正确。

<Checkpoint>[LUMO在长键处是成键的，在短键处是反键的](1)</Checkpoint>

综上所述，正确的说法为1,5，答案选D选项。