What is the hybridization of the central atom In the periodate (IO4-) anion?
The periodate (IO4-) anion consists of one iodine atom (I) and four oxygen atoms (O).
To determine the hybridization of the central atom, we first need to calculate the total number of valence electrons present in the molecule.
Iodine is in group 7, so it has 7 valence electrons. Each oxygen atom is in group 6, so they each contribute 6 valence electrons.
Therefore, the total number of valence electrons is:
1 (I) + 4 (O) = 1 + 4(6) = 25
The central atom, iodine (I), bonds with all four oxygen atoms. Each bond consists of one sigma (σ) bond and two pi (π) bonds.
The hybridization of iodine in periodate can be determined using the formula:
Hybridization = number of sigma (σ) bonds + number of lone pairs
In the case of periodate, the iodine atom forms four sigma (σ) bonds and no lone pairs.
Therefore, the hybridization of the central atom (iodine) in the periodate (IO4-) anion is sp3.