Write the Lewis structure for ICl4- and identify its shape and hybridization scheme
The Correct Answer and Explanation is:
To draw the Lewis structure for the ICl₄⁻ ion, follow these steps:
Step 1: Count Valence Electrons
- Iodine (I) is in group 17 (halogens) and has 7 valence electrons.
- Chlorine (Cl) is also in group 17 and each Cl atom has 7 valence electrons. Since there are four Cl atoms, they contribute 4 × 7 = 28 electrons.
- The charge on the ion is negative, meaning we add one more electron, for a total of 7 + 28 + 1 = 36 valence electrons.
Step 2: Skeleton Structure
Place the iodine atom in the center and connect it with single bonds to the four chlorine atoms. Each single bond represents 2 electrons.
- Iodine (I) has 4 single bonds to chlorine atoms, using 8 electrons (4 × 2 = 8).
Step 3: Distribute Remaining Electrons
After forming the bonds, we have 36 – 8 = 28 electrons remaining. These will be used to fill the octets of the chlorine atoms.
- Place 6 electrons (3 lone pairs) around each chlorine atom to complete their octet. This uses 24 electrons (4 × 6 = 24).
- We now have 4 electrons remaining. These will go on iodine.
Step 4: Place Remaining Electrons on Iodine
Place the remaining 4 electrons as two lone pairs on iodine.
Step 5: Formal Charges
- The formal charge on iodine: 7 valence electrons – 0 bonds – 4 lone electrons = +1.
- The formal charge on each chlorine: 7 valence electrons – 1 bond – 6 lone electrons = 0.
The Lewis structure of ICl₄⁻ is now complete, and the structure looks like this:
- Iodine in the center with four chlorine atoms around it, each chlorine having three lone pairs, and iodine having two lone pairs.
Step 6: Molecular Shape and Hybridization
- Electron Geometry: The electron geometry is octahedral because there are 6 regions of electron density (4 bonding pairs and 2 lone pairs).
- Molecular Geometry: The molecular geometry is square planar. The lone pairs on iodine are positioned opposite each other, so they do not affect the overall shape, and the four chlorine atoms form a square around the iodine atom.
- Hybridization: The hybridization of the iodine atom is sp³d², which corresponds to the octahedral electron geometry.
Summary:
The Lewis structure of ICl₄⁻ shows iodine at the center, with four single bonds to chlorine atoms and two lone pairs on iodine. The molecular geometry is square planar, and the hybridization of iodine is sp³d².
