{"id":189674,"date":"2025-02-10T07:06:52","date_gmt":"2025-02-10T07:06:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189674"},"modified":"2025-02-10T07:06:54","modified_gmt":"2025-02-10T07:06:54","slug":"draw-the-lewis-dot-structure-for-ici2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/10\/draw-the-lewis-dot-structure-for-ici2\/","title":{"rendered":"Draw the Lewis Dot Structure for ICI2"},"content":{"rendered":"\n<ol class=\"wp-block-list\">\n<li>Draw the Lewis Dot Structure for ICI2.<\/li>\n\n\n\n<li>What are the electron pair and molecular geometries for ICI\u2082?<\/li>\n\n\n\n<li>Draw the bond moments and overall dipole moment for ICI2<\/li>\n<\/ol>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Lewis Dot Structure for ICl\u2082\u207b<\/h3>\n\n\n\n<p>To draw the Lewis dot structure for ICl\u2082\u207b, follow these steps:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Step 1<\/strong>: Count the total number of valence electrons.<\/li>\n\n\n\n<li><strong>Iodine (I)<\/strong>: 7 valence electrons (Group 17)<\/li>\n\n\n\n<li><strong>Chlorine (Cl)<\/strong>: 7 valence electrons (Group 17)<\/li>\n\n\n\n<li>There are two chlorine atoms, so 2 \u00d7 7 = 14 valence electrons from chlorine.<\/li>\n\n\n\n<li>The negative charge adds one more electron, so the total number of valence electrons is:<br>(7 (\\text{from I}) + 14 (\\text{from Cl}) + 1 (\\text{charge}) = 22) valence electrons.<\/li>\n\n\n\n<li><strong>Step 2<\/strong>: Connect the atoms.<\/li>\n\n\n\n<li>Place iodine (I) in the center as it is less electronegative than chlorine (Cl).<\/li>\n\n\n\n<li>Attach the two chlorine atoms to iodine with single bonds, each contributing 2 electrons.<\/li>\n\n\n\n<li><strong>Step 3<\/strong>: Distribute the remaining electrons.<\/li>\n\n\n\n<li>After placing two single bonds (2 \u00d7 2 electrons = 4 electrons), we have 18 electrons left.<\/li>\n\n\n\n<li>Place lone pairs of electrons on each chlorine atom, giving each chlorine 6 more electrons (total of 8 electrons around chlorine).<\/li>\n\n\n\n<li>After placing 12 electrons on chlorine, we are left with 6 electrons.<\/li>\n\n\n\n<li>Place the remaining 6 electrons on iodine as three lone pairs.<\/li>\n<\/ul>\n\n\n\n<p>The final structure will look like this:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>    Cl - I - Cl\n         |\n      (3 lone pairs on I)<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">2. Electron Pair and Molecular Geometries for ICl\u2082\u207b<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Pair Geometry<\/strong>: ICl\u2082\u207b has 22 valence electrons. The iodine atom is the central atom with 2 bonding pairs (from chlorine atoms) and 3 lone pairs of electrons. Therefore, the electron pair geometry is <strong>trigonal bipyramidal<\/strong>.<\/li>\n\n\n\n<li><strong>Molecular Geometry<\/strong>: Since there are 3 lone pairs on iodine and 2 bonding pairs, the molecular geometry is <strong>linear<\/strong>. The lone pairs occupy the equatorial positions, which leaves the two chlorine atoms positioned 180\u00b0 apart, making the molecular geometry linear.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3. Bond Moments and Dipole Moment for ICl\u2082\u207b<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bond Moments<\/strong>: In ICl\u2082\u207b, each I-Cl bond is polar because chlorine is more electronegative than iodine, resulting in a partial negative charge on chlorine and a partial positive charge on iodine. Each I-Cl bond has a bond dipole moment pointing from iodine to chlorine.<\/li>\n\n\n\n<li><strong>Overall Dipole Moment<\/strong>: The molecular geometry of ICl\u2082\u207b is linear, which means the bond dipoles in the molecule will cancel each other out. Since the two I-Cl bonds are symmetrically arranged opposite each other, their dipole moments cancel each other, leading to <strong>no overall dipole moment<\/strong> for ICl\u2082\u207b.<\/li>\n<\/ul>\n\n\n\n<p>Thus, despite the individual polar bonds, the linear arrangement of ICl\u2082\u207b means the molecule does not have a net dipole moment.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The Correct Answer and Explanation is : 1. Lewis Dot Structure for ICl\u2082\u207b To draw the Lewis dot structure for ICl\u2082\u207b, follow these steps: The final structure will look like this: 2. Electron Pair and Molecular Geometries for ICl\u2082\u207b 3. Bond Moments and Dipole Moment for ICl\u2082\u207b Thus, despite the individual polar bonds, the linear [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-189674","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189674","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=189674"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189674\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189674"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189674"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189674"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}