Let's solve the problem step by step: --- ## **Step 1: Identify the Mole** The molecule shown is cycloptatrienyl anion (often called the "tropylium anion"). It is a seven-membered ring with three double bonds and a pair of lone electrons (from the negative charge). --- ## **Step 2: Count the π Electrons** - Each double bond contributes **2 π electrons**. - There are **3 double bonds**: \(3 \times 2 = 6\) π electrons. - The lone pair on the carbon (from the negative charge) adds **2 more π electrons**. **Total π electrons:** \[ 6 + 2 = 8 \] --- ## **Step 3: Apply the Frost Circle (Polygon) Method** The Frost circle helps us determine the molecular orbitals for planar, cyclic, conjugated systems: - Draw a regular heptagon (7 sides) with one vertex pointing down. - Inscribe this polygon in a circle. - Each vertex represents a π molecular orbital. - The horizontal axis passing through the center is the reference for energy. #### **Energy Levels:** For 7-membered ring: - 1 orbital at the bottom (non-degenerate, lowest energy) - 2 pairs of degenerate orbitals (next higher energy) - 2 orbitals at the top (highest energy, degenerate) --- ## **Step 4: Fill the Molecular Orbitals** Fill the π electrons (8 total) from the lowest energy upwards, **two electrons per orbital**: 1. **Lowest MO:** 2 electrons (filled) 2. **Next degenerate pair:** 4 electrons (2 per orbital, filled) 3. **Next degenerate pair:** 2 electrons (each orbital gets 1 electron, both half-filled) So, the 8th electron goes into the next orbital, resulting in **two half-filled degenerate orbitals**. --- ## **Step 5: Aromaticity Analysis (Hückel's Rule)** - **Hückel's Rule:** A molecule is aromatic if it is planar, fully conjugated, and contains \(4n + 2\) π electrons (where \(n\) is a non-negative integer). - Here, π electrons = 8. - \(4n + 2 = 8 \implies n = 1.5\), which is **not** an integer. **Therefore:** - The molecule is **not aromatic**. - It is **antiaromatic** if it is planar and fully conjugated with \(4n\) π electrons (\(n=2\) here). - **8 π electrons fit \(4n\) (n=2): Antiaromatic.** --- ## **Step 6: Draw Frost Diagram and Label HOMO/LUMO** ``` o (highest energy, half-filled) o o (second highest, half-filled) o o (middle, filled) o o (lowest, filled) o (lowest, filled) ``` - **HOMO (Highest Occupied Molecular Orbital):** The two half-filled degenerate orbitals. - **LUMO (Lowest Unoccupied Molecular Orbital):** The next available level above the half-filled orbitals. --- ## **Step 7: Final Answer** ### **Summary Table** | Property | Value | |-----------------|---------------------------------------| | π electrons | 8 | | Frost Diagram | Fill from bottom, 2 electrons per MO | | HOMO | Two degenerate half-filled orbitals | | LUMO | Next higher unfilled orbital | | Aromatic? | **No** (not \(4n+2\) π electrons) | | Antiaromatic? | **Yes** (\(4n\) π electrons, planar) | | Nonaromatic? | No (because it's planar and conjugated)| --- ## **Conclusion** > **The molecule has 8 π electrons and is antiaromatic according to Hückel’s rule. The Frost circle shows two half-filled degenerate HOMOs. The molecule is not aromatic.** --- ### **Diagram (Frost Circle)** Below is a schematic Frost circle for a 7-membered ring, filled with 8 electrons: ``` o o o o o o o o ``` Fill electrons from the bottom up (2 per orbital), label the HOMO as the two half-filled orbitals, and the LUMO as the next empty orbital above. --- If you need an explicit energy diagram drawn or further clarification, let me know!