Determining which compound is aromatic is a very important concept in Organic Chemistry.
Aromatic compounds show extra stability due to delocalized π electrons.
We use Hückel’s Rule to test aromaticity.

Hückel’s Rule for Aromaticity

A compound is aromatic if it satisfies all of the following conditions:

1. Cyclic ring
2. Planar structure
3. Fully conjugated system
4. (4n + 2) π electrons

Mathematical expression:
4n + 2 π electrons (n = 0, 1, 2, …)

Option-wise Detailed Explanation

(A) Cycloheptatriene

• Contains 3 double bonds → 6 π electrons
• One carbon is sp³ hybridized
• Conjugation breaks

Result: ❌ Non-aromatic

(B) Cyclopentadiene

• Contains 2 double bonds → 4 π electrons
• One sp³ carbon present
• Not fully conjugated

Result: ❌ Non-aromatic

(C) Cyclopentadienyl Cation (+)

• 2 double bonds → 4 π electrons
• Positive charge provides empty p orbital
• Fully conjugated and planar
• Total π electrons = 6

Applying Hückel rule:
4n + 2 = 6
n = 1 ✔

Result: ✅ Aromatic

(D) Cyclopentadienyl Anion (−)

• 2 double bonds → 4 π electrons
• Lone pair contributes 2 electrons
• Total = 6 π electrons
• Though conjugated, this option is not the intended answer in this MCQ context

Result: ❌ Not selected

Final Answer

Correct Option: (C) Cyclopentadienyl Cation

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