{"id":199267,"date":"2025-03-11T03:30:44","date_gmt":"2025-03-11T03:30:44","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199267"},"modified":"2025-03-11T03:31:04","modified_gmt":"2025-03-11T03:31:04","slug":"determine-the-electron-geometry-eg-and-molecular-geometry-mg-of-icl3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/11\/determine-the-electron-geometry-eg-and-molecular-geometry-mg-of-icl3\/","title":{"rendered":"Determine the electron geometry (eg) and molecular geometry (mg) of ICl3"},"content":{"rendered":"\n<p>Determine the electron geometry (eg) and molecular geometry (mg) of ICl3.<\/p>\n\n\n\n<p>a<\/p>\n\n\n\n<p>eg = trigonal bipyramidal, mg = T-shape<\/p>\n\n\n\n<p>b<\/p>\n\n\n\n<p>eg = trigonal bipyramidal, mg = bent<\/p>\n\n\n\n<p>c<\/p>\n\n\n\n<p>eg = tetrahedral, mg = trigonal pyramidal<\/p>\n\n\n\n<p>d<\/p>\n\n\n\n<p>eg = trigonal planar, mg = trigonal planar<\/p>\n\n\n\n<p>e<\/p>\n\n\n\n<p>eg = trigonal planar, mg = tetrahedral<\/p>\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<p>The correct answer is:<\/p>\n\n\n\n<p><strong>a) eg = trigonal bipyramidal, mg = T-shape<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>Iodine trichloride (<strong>ICl\u2083<\/strong>) consists of an iodine (<strong>I<\/strong>) atom bonded to three chlorine (<strong>Cl<\/strong>) atoms. To determine its electron geometry (EG) and molecular geometry (MG), we follow these steps:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. Valence Electrons Calculation:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Iodine (I)<\/strong> has <strong>7<\/strong> valence electrons.<\/li>\n\n\n\n<li><strong>Each Chlorine (Cl)<\/strong> has <strong>7<\/strong> valence electrons.<\/li>\n\n\n\n<li>Total electrons from ICl\u2083:<\/li>\n\n\n\n<li>Iodine: <strong>7<\/strong><\/li>\n\n\n\n<li>Chlorine (3 \u00d7 7) = <strong>21<\/strong><\/li>\n\n\n\n<li>Total: <strong>28 valence electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>2. Lewis Structure and Electron Regions:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Iodine is the central atom.<\/li>\n\n\n\n<li>It forms three <strong>single bonds<\/strong> with three chlorine atoms.<\/li>\n\n\n\n<li>Iodine has <strong>two lone pairs<\/strong> of electrons.<\/li>\n\n\n\n<li>This gives a total of <strong>five<\/strong> electron regions around iodine (3 bonding pairs + 2 lone pairs).<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>3. Electron Geometry (EG):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A molecule with <strong>five<\/strong> electron regions follows the <strong>trigonal bipyramidal<\/strong> electron geometry.<\/li>\n\n\n\n<li>This is because the <strong>VSEPR (Valence Shell Electron Pair Repulsion) Theory<\/strong> states that electron domains arrange themselves to minimize repulsion.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>4. Molecular Geometry (MG):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The presence of <strong>two lone pairs<\/strong> affects the molecular shape.<\/li>\n\n\n\n<li>The ideal <strong>trigonal bipyramidal<\/strong> structure has five positions:<\/li>\n\n\n\n<li><strong>Two axial positions (above and below the central atom)<\/strong><\/li>\n\n\n\n<li><strong>Three equatorial positions (around the middle plane)<\/strong><\/li>\n\n\n\n<li>Lone pairs preferentially occupy <strong>equatorial positions<\/strong> to minimize repulsion.<\/li>\n\n\n\n<li>This leaves the three chlorine atoms in a <strong>T-shape<\/strong> molecular geometry.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>5. Final Answer:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Geometry (EG) = Trigonal Bipyramidal<\/strong><\/li>\n\n\n\n<li><strong>Molecular Geometry (MG) = T-shape<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Thus, the correct answer is <strong>(a) eg = trigonal bipyramidal, mg = T-shape.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Determine the electron geometry (eg) and molecular geometry (mg) of ICl3. a eg = trigonal bipyramidal, mg = T-shape b eg = trigonal bipyramidal, mg = bent c eg = tetrahedral, mg = trigonal pyramidal d eg = trigonal planar, mg = trigonal planar e eg = trigonal planar, mg = tetrahedral The correct answer [&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-199267","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199267","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=199267"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199267\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199267"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199267"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199267"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}