Number of delocalised pi electrons present in perchlorate anion is 

Options

A. 8
B. 6
C. 4
D. 2

Answer

To determine the number of delocalized pi electrons in the perchlorate anion (ClO₄⁻), we can follow these steps:

### Step 1: Identify the structure of the perchlorate anion
The perchlorate anion has the formula ClO₄⁻. It consists of one chlorine atom bonded to four oxygen atoms. The structure can be represented with resonance forms, where chlorine is the central atom.

### Step 2: Draw the Lewis structure
In the Lewis structure of ClO₄⁻, chlorine is bonded to four oxygen atoms. Three of these oxygen atoms are connected via double bonds, and one oxygen atom carries a negative charge and is connected via a single bond. The negative charge is typically placed on one of the oxygen atoms.

### Step 3: Count the pi bonds
In the Lewis structure, each double bond consists of one sigma bond and one pi bond. In the case of ClO₄⁻, there are three Cl=O double bonds. Each double bond contributes one pair of pi electrons.

### Step 4: Consider the lone pair on the negatively charged oxygen
The negatively charged oxygen atom has three lone pairs of electrons. One of these lone pairs can participate in resonance, contributing to the delocalization of electrons.

### Step 5: Calculate the total number of delocalized pi electrons
- From the three Cl=O double bonds, we have 3 pi bonds contributing 3 pairs of pi electrons (3 x 2 = 6 electrons).
- The lone pair from the negatively charged oxygen contributes an additional 2 electrons.

So, the total number of delocalized pi electrons is:
\[ 6 \text{ (from pi bonds)} + 2 \text{ (from the lone pair)} = 8 \text{ delocalized pi electrons.} \]

### Final Answer
The number of delocalized pi electrons present in the perchlorate anion is **8**.

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