To draw the Lewis structure for ICl4, start by counting the total number of valence electrons. Iodine (I) has 7 valence electrons, and each chlorine (Cl) atom has 7 valence electrons. Since there are four chlorine atoms, we add:
7 (from I) + 4 * 7 (from Cl) = 35 valence electrons.
Next, place the iodine atom in the center and surround it with the four chlorine atoms. Each Cl will form a single bond with I, using 8 electrons in total for the bonds. This leaves us with 27 valence electrons. Now, each chlorine needs 6 more electrons to complete its octet, which uses up 24 electrons (6 * 4), leaving us with 3 electrons. You can place these 3 leftover electrons as a lone pair on the iodine.
The final Lewis structure shows iodine in the center with four single bonds to chlorine and a lone pair of electrons.
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