Shape and bond angle in `ClO_(4)^(-)` ion is

Options

A. planar trigonal , `109^@ 28^(1)`
B. tetrahedral , `109^(@)28^(1)`
C. pyramidal `105^@`
D. v-shape , `118^@`

Answer

To determine the shape and bond angle of the `ClO4^(-)` ion, we can follow these steps:

### Step 1: Determine the Valence Electrons
Chlorine (Cl) is in Group 17 (VIIA) of the periodic table and has 7 valence electrons. Each oxygen (O) atom has 6 valence electrons. In `ClO4^(-)`, we have one chlorine atom and four oxygen atoms, plus one additional electron due to the negative charge.

- Chlorine: 7 electrons
- Oxygen: 4 × 6 = 24 electrons
- Additional electron due to the negative charge: +1 electron

Total valence electrons = 7 + 24 + 1 = 32 electrons.

### Step 2: Draw the Lewis Structure
In the Lewis structure of `ClO4^(-)`, chlorine is the central atom. It forms:
- One single bond with one oxygen atom (which carries a negative charge).
- Three double bonds with the other three oxygen atoms.

This arrangement uses all 32 valence electrons.

### Step 3: Determine the Molecular Geometry
Using VSEPR (Valence Shell Electron Pair Repulsion) theory, we consider the number of bonding pairs and lone pairs around the central chlorine atom.

- There are 4 bonding pairs (one single bond and three double bonds).
- There are no lone pairs on the chlorine atom.

The molecular geometry with 4 bonding pairs and no lone pairs is tetrahedral.

### Step 4: Calculate the Bond Angle
In a tetrahedral geometry, the ideal bond angle is approximately 109.5 degrees. However, due to the presence of different types of bonds (single and double), the actual bond angle might be slightly adjusted. For `ClO4^(-)`, the bond angle is approximately 109 degrees and 28 minutes.

### Final Answer
The shape of the `ClO4^(-)` ion is tetrahedral, and the bond angle is approximately 109 degrees and 28 minutes.

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