{"id":43274,"date":"2025-06-29T17:09:18","date_gmt":"2025-06-29T17:09:18","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=43274"},"modified":"2025-06-29T17:09:19","modified_gmt":"2025-06-29T17:09:19","slug":"in-the-lewis-structure-for-icl2-how-many-lone-pairs-of-electrons-are-around-the-central-iodine-atom-2","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-2\/","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? Group of answer choices 4 1 0 2 3<\/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>To determine the number of lone pairs of electrons around the central iodine atom in the Lewis structure for ICl\u2082\u207b, follow these steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the total number of valence electrons:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Iodine (I) is in Group 17 of the periodic table, so it has 7 valence electrons.<\/li>\n\n\n\n<li>Each chlorine (Cl) atom is also in Group 17, so each chlorine contributes 7 valence electrons.<\/li>\n\n\n\n<li>The ion has a negative charge (ICl\u2082\u207b), which means it has gained one electron, adding 1 more valence electron.<\/li>\n\n\n\n<li>Therefore, the total number of valence electrons is: (7\u2009from\u00a0iodine)+2\u00d7(7\u2009from\u00a0chlorine)+1\u2009(from\u00a0the\u00a0negative\u00a0charge)=22\u2009electrons.(7 \\, \\text{from iodine}) + 2 \\times (7 \\, \\text{from chlorine}) + 1 \\, (\\text{from the negative charge}) = 22 \\, \\text{electrons}.(7from\u00a0iodine)+2\u00d7(7from\u00a0chlorine)+1(from\u00a0the\u00a0negative\u00a0charge)=22electrons.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Distribute the electrons in the structure:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Place iodine in the center since it is the least electronegative atom. Each chlorine atom will form a single bond with iodine.<\/li>\n\n\n\n<li>Two single bonds between iodine and the two chlorine atoms use 4 electrons (2 electrons per bond).<\/li>\n\n\n\n<li>The remaining 18 electrons (22 total electrons &#8211; 4 electrons used in bonds) will be placed as lone pairs.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Complete the octet for each atom:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Each chlorine atom needs 6 more electrons to complete its octet. So, place 3 lone pairs (6 electrons) around each chlorine atom.<\/li>\n\n\n\n<li>After placing these electrons, 12 electrons are used for the chlorines.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Place the remaining electrons around iodine:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Iodine now has 4 electrons involved in bonds, and it has 18 electrons remaining. Distribute these 18 electrons around the iodine atom.<\/li>\n\n\n\n<li>Iodine can hold more than 8 electrons in its valence shell because it is in period 5, and therefore can have an expanded octet. Place 3 lone pairs of electrons around iodine.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Thus, <strong>there are 3 lone pairs of electrons<\/strong> around the central iodine atom in ICl\u2082\u207b.<\/p>\n\n\n\n<p><strong>Correct answer: 3<\/strong><\/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-banner5-457.jpeg\" alt=\"\" class=\"wp-image-43275\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-457.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-457-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-457-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-457-768x768.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? Group of answer choices 4 1 0 2 3 The Correct Answer and Explanation is: To determine the number of lone pairs of electrons around the central iodine atom in the Lewis structure for ICl\u2082\u207b, follow these [&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-43274","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43274","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=43274"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43274\/revisions"}],"predecessor-version":[{"id":43276,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/43274\/revisions\/43276"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=43274"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=43274"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=43274"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}