{"id":256489,"date":"2025-07-16T09:45:17","date_gmt":"2025-07-16T09:45:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256489"},"modified":"2025-07-16T09:45:20","modified_gmt":"2025-07-16T09:45:20","slug":"for-the-ibr2-lewis-structure-what-is-the-number-of-pairs-of-nonbonding-electrons-on-the-central-iodine-atom","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/for-the-ibr2-lewis-structure-what-is-the-number-of-pairs-of-nonbonding-electrons-on-the-central-iodine-atom\/","title":{"rendered":"For the IBr2- Lewis structure, what is the number of pairs of nonbonding electrons on the central iodine atom"},"content":{"rendered":"\n<p>For the IBr2- Lewis structure, what is the number of pairs of nonbonding electrons on 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-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p><strong>Correct Answer:<\/strong> 3 pairs of nonbonding electrons<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>To determine the number of nonbonding electron pairs (lone pairs) on the central iodine atom in the IBr\u2082\u207b ion, we start by drawing its Lewis structure using the following steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count total valence electrons:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Iodine (I) has 7 valence electrons.<\/li>\n\n\n\n<li>Each bromine (Br) atom also has 7 valence electrons.<\/li>\n\n\n\n<li>The negative charge (\u207b1) adds 1 more electron.<\/li>\n\n\n\n<li>Total = 7 (I) + 7 (Br) \u00d7 2 + 1 (charge) = 7 + 14 + 1 = <strong>22 valence electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Choose the central atom:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Iodine is less electronegative than bromine and can expand its octet, so it is the central atom.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Form single bonds:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Connect iodine to two bromine atoms with single bonds.<\/li>\n\n\n\n<li>Each bond uses 2 electrons, so 2 bonds use 4 electrons.<\/li>\n\n\n\n<li>Remaining electrons = 22 &#8211; 4 = <strong>18 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Distribute remaining electrons:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Complete the octet on the bromine atoms first. Each Br needs 6 more electrons to complete its octet.<\/li>\n\n\n\n<li>Two Br atoms \u00d7 6 electrons = 12 electrons.<\/li>\n\n\n\n<li>Remaining electrons = 18 &#8211; 12 = <strong>6 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Place remaining electrons on iodine:<\/strong>\n<ul class=\"wp-block-list\">\n<li>The remaining 6 electrons are placed on iodine as lone pairs.<\/li>\n\n\n\n<li>6 electrons = 3 pairs<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>So, the central iodine atom has <strong>3 lone pairs<\/strong> (nonbonding pairs).<\/p>\n\n\n\n<p>These lone pairs influence the shape of the molecule. With 2 bonding pairs and 3 lone pairs, the electron geometry is <strong>trigonal bipyramidal<\/strong>, but the molecular shape is <strong>linear<\/strong>, since the 3 lone pairs occupy the equatorial positions and cancel out their repulsions, leaving the two bromine atoms 180 degrees apart.<\/p>\n\n\n\n<p>Therefore, iodine in IBr\u2082\u207b has <strong>3 pairs of nonbonding electrons<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner6-633.jpeg\" alt=\"\" class=\"wp-image-256490\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>For the IBr2- Lewis structure, what is the number of pairs of nonbonding electrons on the central iodine atom The Correct Answer and Explanation is: Correct Answer: 3 pairs of nonbonding electrons Explanation: To determine the number of nonbonding electron pairs (lone pairs) on the central iodine atom in the IBr\u2082\u207b ion, we start by [&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 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