To determine the Lewis structure for ICl₄ (iodine tetrachloride), we start by counting the total number of valence electrons. Iodine has 7 valence electrons and each chlorine atom has 7 valence electrons. Since there are four chlorine atoms, we have:
7 (I) + 4 × 7 (Cl) = 7 + 28 = 35 valence electrons.
Next, we place iodine in the center and surround it with the four chlorine atoms. Each I-Cl bond uses 2 electrons, accounting for 8 of the total:
35 – 8 = 27 valence electrons remaining.
We then distribute the remaining electrons to the chlorine atoms as lone pairs. Each chlorine atom needs three lone pairs to fulfill the octet rule. Hence, with 4 Cl atoms, we use:
4 × 6 = 24 electrons, leaving us with:
27 – 24 = 3 valence electrons.
These remaining three electrons can be placed on the iodine atom as one lone pair.
So, the Lewis structure of ICl₄ looks like this:
In terms of shape, ICl₄ adopts a square planar geometry. This is due to the presence of four bond pairs (I-Cl) and one lone pair around the central iodine atom, which arranges itself to minimize electron pair repulsion.
Regarding hybridization, the hybridization scheme for ICl₄ is sp³d². This indicates that one s orbital, three p orbitals, and two d orbitals from the iodine atom hybridize to form a total of five hybrid orbitals used for bond formation and lone pair accommodation.
In summary, the Lewis structure for ICl₄ includes one iodine atom bonded to four chlorine atoms with a lone pair on iodine, resulting in a square planar shape and sp³d² hybridization.