The hybridization of chlorine in ${\mathrm{ClO}}^{-1},{\mathrm{ClO}}_2,{\mathrm{ClO}}_3^{-}$, and ${\mathrm{ClO}}_4^{-}$ions are;

To determine the hybridization of chlorine in the ions ClO^-, ClO₂^-, ClO₃^-, and ClO₄^-, we follow these steps:

Step 1: Determine the Total Valence Electrons for Chlorine

Chlorine (Cl) has 7 valence electrons. For each ion, the charge must be accounted for by adding or removing electrons accordingly.

Step 2: Analyze Each Ion

Let us calculate the number of bond pairs and lone pairs for each ion, and then determine the hybridization:

For ClO^-:

• Valence Electrons: Chlorine has 7 electrons, and the negative charge adds 1 more, making it 8 total electrons.
• Bonding: Chlorine forms a single bond with oxygen, using 2 electrons.
• Lone Pairs: 8 total electrons - 2 bonding electrons = 6 electrons left, which are arranged as 3 lone pairs.
• Total Pairs: 3 lone pairs + 1 bond pair = 4 pairs.
• Hybridization: Since there are 4 pairs, the hybridization is sp³.

For ClO₂^-:

• Valence Electrons: Chlorine has 7 electrons, and the negative charge adds 1 more, making it 8 total electrons.
• Bonding: Chlorine forms one double bond with one oxygen (4 electrons) and one single bond with another oxygen (2 electrons).
• Lone Pairs: 8 total electrons - 6 bonding electrons = 2 electrons left, forming 1 lone pair.
• Total Pairs: 1 lone pair + 2 bond pairs = 4 pairs.
• Hybridization: With 4 pairs, the hybridization of chlorine is sp³.

For ClO₃^-:

• Valence Electrons: Chlorine has 7 electrons, and the negative charge adds 1 more, making it 8 total electrons.
• Bonding: Chlorine forms three bonds with oxygen atoms (using 6 electrons).
• Lone Pairs: 8 total electrons - 6 bonding electrons = 2 electrons left, forming 1 lone pair.
• Total Pairs: 1 lone pair + 3 bond pairs = 4 pairs.
• Hybridization: Chlorine’s hybridization is sp³.

For ClO₄^-:

• Valence Electrons: Chlorine has 7 electrons, and the negative charge adds 1 more, making it 8 total electrons.
• Bonding: Chlorine forms four bonds with oxygen atoms, using all 8 electrons.
• Lone Pairs: No electrons are left for lone pairs.
• Total Pairs: 0 lone pairs + 4 bond pairs = 4 pairs.
• Hybridization: The hybridization of chlorine in ClO₄^- is sp³.

Summary of Hybridization

The hybridization of chlorine in each ion is as follows:

• ClO^-: sp³
• ClO₂^-: sp³
• ClO₃^-: sp³
• ClO₄^-: sp³

Note on ClO₄^- Hybridization

The ClO₄^- hybridization involves sp³ orbitals, allowing chlorine to form stable bonds with the four oxygen atoms. This is consistent with its tetrahedral geometry, where bond pairs are evenly spaced.

Final Answer: The hybridization of chlorine in ClO^-, ClO₂^-, ClO₃^-, and ClO₄^- is sp³ for all.