{"id":43237,"date":"2025-06-29T16:13:18","date_gmt":"2025-06-29T16:13:18","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43237"},"modified":"2025-06-29T16:13:19","modified_gmt":"2025-06-29T16:13:19","slug":"in-the-lewis-structure-for-icl2-how-many-lone-pairs-of-electrons-are-around-the-central-iodine-atom","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/in-the-lewis-structure-for-icl2-how-many-lone-pairs-of-electrons-are-around-the-central-iodine-atom\/","title":{"rendered":"In the Lewis structure for ICl2, how many lone pairs of electrons are around the central iodine atom"},"content":{"rendered":"\n<p>In the Lewis structure for ICl2, how many lone pairs of electrons are around the central iodine atom?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>In the Lewis structure for ICl2, there are three lone pairs of electrons around the central iodine atom. Let&#8217;s break this down step by step.<\/p>\n\n\n\n<p>Iodine (I) is in Group 17 of the periodic table, which means it has 7 valence electrons. Each chlorine (Cl) atom also has 7 valence electrons. ICl2 consists of one iodine atom and two chlorine atoms, which means the iodine atom is bonded to two chlorine atoms.<\/p>\n\n\n\n<p>To determine the number of lone pairs on iodine, first count the total number of valence electrons:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Iodine: 7 valence electrons<\/li>\n\n\n\n<li>Chlorine: 7 valence electrons \u00d7 2 = 14 valence electrons<\/li>\n<\/ul>\n\n\n\n<p>So, in total, there are 7 + 14 = 21 valence electrons to distribute in the molecule.<\/p>\n\n\n\n<p>Next, we place the iodine in the center and the chlorine atoms around it. Since iodine forms two single bonds with the chlorine atoms, this consumes 2 electrons for each bond, totaling 4 electrons (2 electrons per bond \u00d7 2 bonds).<\/p>\n\n\n\n<p>After bonding, 21 total valence electrons \u2212 4 electrons used in bonding = 17 electrons remain to be placed as lone pairs. These electrons are placed around the iodine and chlorine atoms.<\/p>\n\n\n\n<p>Each chlorine atom will complete its octet by having three lone pairs, using up 6 electrons (3 lone pairs \u00d7 2 chlorine atoms). This leaves 11 electrons to be placed on iodine.<\/p>\n\n\n\n<p>Now, iodine can use its remaining electrons to form lone pairs. Since iodine is in the third period or below, it can accommodate more than 8 electrons, and in this case, it will have three lone pairs. Thus, there are three lone pairs on iodine in ICl2.<\/p>\n\n\n\n<p>The remaining 2 electrons (from the 11) form a single bond between iodine and chlorine, confirming the overall structure and bonding.<\/p>\n\n\n\n<p>Therefore, iodine has three lone pairs of electrons in ICl2.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-448.jpeg\" alt=\"\" class=\"wp-image-43238\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-448.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-448-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-448-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-448-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>In the Lewis structure for ICl2, how many lone pairs of electrons are around the central iodine atom? The Correct Answer and Explanation is: In the Lewis structure for ICl2, there are three lone pairs of electrons around the central iodine atom. Let&#8217;s break this down step by step. Iodine (I) is in Group 17 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-43237","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43237","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=43237"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43237\/revisions"}],"predecessor-version":[{"id":43239,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43237\/revisions\/43239"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43237"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43237"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43237"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}