What is the “AXE” description of the periodate (IO4-) anion? AXE 0 0
The Correct Answer and Explanation is:
The correct AXE description for the periodate (IO₄⁻) anion is AX₄E₀.
Explanation:
The “AXE” notation is a way to describe the geometry of a molecule based on the Valence Shell Electron Pair Repulsion (VSEPR) theory. Here is a step-by-step breakdown of how to determine the AXE description for the periodate anion (IO₄⁻):
- Understand the AXE Components:
- A represents the central atom in the molecule.
- X represents the number of atoms bonded to the central atom. The subscript next to X is this number.
- E represents the number of non-bonding electron pairs (lone pairs) on the central atom only. The subscript next to E is this number.
- Determine the Central Atom (A):
In the IO₄⁻ anion, Iodine (I) is the central atom because it is the least electronegative element and there is only one of it. - Count the Total Valence Electrons:
To find the number of lone pairs, we must first determine the Lewis structure by counting the total valence electrons.- Iodine (I) is in Group 17, so it has 7 valence electrons.
- Oxygen (O) is in Group 16, so each of the four atoms has 6 valence electrons (4 × 6 = 24).
- The anion has a negative one charge (-1), which means there is one extra electron.
- Total valence electrons = 7 (from I) + 24 (from 4 O’s) + 1 (from charge) = 32 electrons.
- Determine X (Number of Bonded Atoms):
The formula is IO₄⁻, which clearly shows that there are four oxygen atoms bonded to the central iodine atom.- Therefore, the value for X is 4.
- Determine E (Number of Lone Pairs on the Central Atom):
- Start by drawing a skeleton structure with the central iodine single-bonded to the four oxygen atoms. This uses 4 bonds × 2 electrons/bond = 8 electrons.
- Subtract these from the total: 32 – 8 = 24 electrons remaining.
- Distribute the remaining 24 electrons as lone pairs to the outer oxygen atoms to satisfy their octets. Each oxygen needs 6 more electrons (3 lone pairs). 4 oxygen atoms × 6 electrons = 24 electrons.
- After distributing these 24 electrons, there are no electrons left over to place on the central iodine atom.
- Therefore, the number of lone pairs on the central iodine atom is 0. The value for E is 0.
By combining these parts, we get the final AXE description: AX₄E₀. This notation corresponds to a tetrahedral electron geometry and a tetrahedral molecular geometry.
