{"id":188411,"date":"2025-02-07T02:59:32","date_gmt":"2025-02-07T02:59:32","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=188411"},"modified":"2025-02-07T02:59:34","modified_gmt":"2025-02-07T02:59:34","slug":"electron-group-arrangement","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/07\/electron-group-arrangement\/","title":{"rendered":"Electron group arrangement"},"content":{"rendered":"\n<p>2-D Lewis structure 3-D Lewis structure IF4+ Atom # of valence e per atom Total # of alence e v Te FITO ion Total (0.25 pt for electron accounting) F: 5 (0.75 pt) (0.25 pt) Electron group arrangement (0.25 pt): Molecular shape &amp; class (0.25 pt ea): Shape: Class (AXmE.): Ideal Bond Angle(s) (list all angles) (0.25 pt): Polar or Non-Polar (0.25 pt): (if polar, show dipole moment in the 3D structure)\u00a0<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-111.png\" alt=\"\" class=\"wp-image-188412\"\/><\/figure>\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\">Lewis Structure of IF\u2084\u207a (Iodine Tetrafluoride Ion)<\/h3>\n\n\n\n<p><strong>Step 1: Determining Valence Electrons<\/strong><\/p>\n\n\n\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>Fluorine (F) is in group 17 as well, so each fluorine atom has 7 valence electrons.<\/li>\n\n\n\n<li>Since we have 4 fluorine atoms, the total valence electrons from fluorine atoms is 4 \u00d7 7 = 28.<\/li>\n\n\n\n<li>The positive charge of the ion (IF\u2084\u207a) means it has lost 1 electron, so we subtract 1 from the total count.<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Valence Electrons:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>7 (I) + 28 (4 F) \u2212 1 (charge) = 34 valence electrons.<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 2: Drawing the 2D Lewis Structure<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The iodine atom will be the central atom because it is less electronegative than fluorine.<\/li>\n\n\n\n<li>Each fluorine atom will form a single bond with iodine, using 2 electrons per bond.<\/li>\n\n\n\n<li>After placing bonds, distribute the remaining electrons to fulfill the octet rule for fluorine atoms.<\/li>\n\n\n\n<li>There will be no lone pairs on iodine in this structure because iodine has expanded its valence shell to accommodate more than 8 electrons due to its position in period 5 of the periodic table.<\/li>\n<\/ul>\n\n\n\n<p>The structure looks like this:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>    F\n    |\nF - I - F\n    |\n    F<\/code><\/pre>\n\n\n\n<p><strong>Step 3: Electron Group Arrangement<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Iodine has 4 bonding pairs of electrons (one for each F atom) and no lone pairs. So, the electron group arrangement around iodine is <strong>tetrahedral<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 4: Molecular Shape &amp; Class<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular shape of IF\u2084\u207a is <strong>tetrahedral<\/strong> because there are 4 bonding pairs and no lone pairs on the central iodine atom.<\/li>\n\n\n\n<li>The class is <strong>AX\u2084<\/strong>, where A is the central atom (iodine), and X represents the fluorine atoms.<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 5: Ideal Bond Angles<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The ideal bond angles for a tetrahedral geometry are approximately <strong>109.5\u00b0<\/strong> between all the bonds.<\/li>\n<\/ul>\n\n\n\n<p><strong>Step 6: Polarity<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule <strong>IF\u2084\u207a<\/strong> is <strong>non-polar<\/strong>. This is because the 4 fluorine atoms are symmetrically arranged around the central iodine atom in a tetrahedral shape, and the dipoles cancel out.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron Accounting:<\/strong> Each bond involves two electrons. The iodine has 4 bonds (8 electrons around it) plus any lone pairs if there were any. The fluorine atoms each have 3 lone pairs of electrons and one bonding electron shared with iodine. The net charge affects the electron count, leading to a subtraction of one electron for the positive charge.<\/li>\n\n\n\n<li><strong>Geometry:<\/strong> The electron groups around iodine cause it to adopt a tetrahedral arrangement for minimal repulsion between the electron pairs.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>2-D Lewis structure 3-D Lewis structure IF4+ Atom # of valence e per atom Total # of alence e v Te FITO ion Total (0.25 pt for electron accounting) F: 5 (0.75 pt) (0.25 pt) Electron group arrangement (0.25 pt): Molecular shape &amp; class (0.25 pt ea): Shape: Class (AXmE.): Ideal Bond Angle(s) (list all [&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-188411","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188411","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=188411"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188411\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188411"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188411"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188411"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}