All the four sigma bonds in perchlorate ion are

Options

A. `sp^(3)-sp^(3)` bond
B. `sp^(3) -p` bond
C. `sp^(3)-sp^(2)` bond
D. `sp^(2)-p` bond

Explanation

To determine the nature of the four sigma bonds in the perchlorate ion (ClO₄⁻), we will follow these steps:

### Step 1: Identify the Structure of the Perchlorate Ion
The perchlorate ion (ClO₄⁻) consists of one chlorine atom (Cl) centrally located and four oxygen atoms (O) surrounding it. The chlorine atom is bonded to the four oxygen atoms.

**Hint:** Remember that the central atom in a polyatomic ion is usually the atom with the lowest electronegativity.

### Step 2: Determine the Valence Electrons
Chlorine has 7 valence electrons, and each oxygen has 6 valence electrons. Since there are four oxygen atoms, the total number of valence electrons contributed by oxygen is 4 × 6 = 24. The ion has an overall negative charge (-1), which adds one more electron, making the total number of valence electrons:

\[ 7 \text{ (from Cl)} + 24 \text{ (from 4 O)} + 1 \text{ (for -1 charge)} = 32 \text{ valence electrons} \]

**Hint:** Count the total number of valence electrons carefully, considering both the atoms and the charge of the ion.

### Step 3: Draw the Lewis Structure
In the Lewis structure of ClO₄⁻, chlorine forms one double bond with one oxygen atom and single bonds with the other three oxygen atoms. Each single bond contributes one sigma bond, while the double bond consists of one sigma bond and one pi bond.

**Hint:** Use the octet rule to ensure that each atom (especially the oxygen atoms) achieves a stable electron configuration.

### Step 4: Identify the Sigma Bonds
In the perchlorate ion:
- The double bond between Cl and one O contributes 1 sigma bond.
- The three single bonds between Cl and the other three O atoms each contribute 1 sigma bond.

Thus, the total number of sigma bonds in ClO₄⁻ is:

\[ 1 \text{ (from the double bond)} + 3 \text{ (from the three single bonds)} = 4 \text{ sigma bonds} \]

**Hint:** Remember that a sigma bond is formed by the head-on overlap of orbitals, while pi bonds are formed by the side-on overlap.

### Step 5: Determine the Hybridization
The central chlorine atom in ClO₄⁻ is sp³ hybridized because it forms four sigma bonds. The geometry of the ion is tetrahedral due to this hybridization.

**Hint:** Hybridization can often be determined by counting the number of sigma bonds and lone pairs around the central atom.

### Conclusion
All four sigma bonds in the perchlorate ion (ClO₄⁻) are formed from sp³ hybridized orbitals of chlorine and p orbitals of oxygen.

**Final Answer:** The four sigma bonds in the perchlorate ion are sp³ hybridized.

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