Q.56 Assuming the molecule shown below is aromatic, the value of “n” according to
“Hückel’s rule” is

The given polycyclic conjugated system contains a total of 10 π‑electrons, so for Hückel’s rule $4n+2=\pi \text{‑electrons}$ the value of $n$ is 2.

Understanding the molecule and Hückel’s rule

Hückel’s rule states that a planar, cyclic, fully conjugated molecule is aromatic when it has $4n+2$ π‑electrons, where $n$ is a non‑negative integer (0, 1, 2, 3…).​

In the problem, the drawn structure is a fused‑ring system that forms a larger conjugated loop around the perimeter. Each double bond contributes 2 π‑electrons, so counting the double bonds along this outer loop gives 5 double bonds, i.e. $5\times 2=10$ π‑electrons in the conjugated circuit.​

Calculating the value of $n$

Use Hückel’s rule:

$$4n+2=\text{number of }\pi \text{-electrons}$$

Here, the conjugated loop has 10 π‑electrons, so:

$$4n+2=10$$$$4n=8$$$$n=2$$

Thus, assuming the system is planar and fully conjugated (as the question states), it is aromatic with $n=2$.​

Typical options and why only $n=2$ is correct

Such CSIR‑NET/competitive questions usually give options like
(a) 1, (b) 2, (c) 3, (d) 4.

• Option $n=1$: Would correspond to 6 π‑electrons $(4\times 1+2)$; the given structure clearly has more than 3 double bonds, so this is too low.​

• Option $n=2$: Gives 10 π‑electrons $(4\times 2+2=10)$, which matches the counted π‑electrons in the conjugated loop, so this is the correct value.​

• Option $n=3$: Would require 14 π‑electrons, i.e. 7 double bonds in the loop; the drawn molecule does not have that many double bonds in its perimeter, so this is incorrect.​

• Option $n=4$: Would correspond to 18 π‑electrons, far more than present in the depicted molecule, so it is also incorrect.​

Therefore, the correct answer is $n=2$.

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SEO‑optimized introduction

Determining the value of $n$ according to Hückel’s rule is essential for identifying aromatic molecules in CSIR NET and other competitive chemistry exams. This polycyclic conjugated system problem illustrates how to count π‑electrons, apply the $4n+2$ rule, and select the correct option for $n$ by analyzing a fused aromatic ring structure with 10 π‑electrons.​