{"id":185621,"date":"2025-01-22T18:54:11","date_gmt":"2025-01-22T18:54:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=185621"},"modified":"2025-01-22T18:54:14","modified_gmt":"2025-01-22T18:54:14","slug":"draw-the-lewis-structure-for-brcl-4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/22\/draw-the-lewis-structure-for-brcl-4\/","title":{"rendered":"Draw the Lewis structure for BrCl 4"},"content":{"rendered":"\n<p>Draw the Lewis structure for BrCl 4 -. The molecular geometry (molecular shape) and the hybridization of the central atom<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Square Planar, sp3d2<\/li>\n\n\n\n<li>Octahedral, sp3d2<\/li>\n\n\n\n<li>Trigonal bipyramidal, sp3d<\/li>\n\n\n\n<li>Tetrahedral, sp3<\/li>\n<\/ol>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Answer:<\/h3>\n\n\n\n<p><strong>1. Square Planar, sp\u00b3d\u00b2<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>To determine the Lewis structure, molecular geometry, and hybridization of <strong>BrCl\u2084\u207b<\/strong>, let\u2019s proceed step by step:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Lewis Structure<\/strong>:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Central Atom<\/strong>: Bromine (Br) is the central atom because it is less electronegative than chlorine.<\/li>\n\n\n\n<li><strong>Valence Electrons<\/strong>: Bromine has 7 valence electrons, and each of the four chlorines also contributes 7 electrons. Additionally, there is one extra electron due to the negative charge.<\/li>\n\n\n\n<li><strong>Total Valence Electrons<\/strong>: 7+(4\u00d77)+1=367 + (4 \\times 7) + 1 = 36.<\/li>\n\n\n\n<li><strong>Bond Formation<\/strong>: Bromine forms single bonds with the four chlorines, which use up 4\u00d72=84 \\times 2 = 8 electrons for bonding. This leaves 36\u22128=2836 &#8211; 8 = 28 electrons.<\/li>\n\n\n\n<li><strong>Lone Pairs<\/strong>: The remaining electrons are distributed as lone pairs. Each chlorine gets 3 lone pairs (4\u00d76=244 \\times 6 = 24), leaving 4 electrons, which are placed as 2 lone pairs on the bromine atom.<\/li>\n<\/ul>\n\n\n\n<p>The Lewis structure shows Br in the center with four single bonds to Cl and two lone pairs on Br.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Molecular Geometry<\/strong>:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The electron domain geometry for 6 regions of electron density (4 bonding pairs + 2 lone pairs) is <strong>octahedral<\/strong>.<\/li>\n\n\n\n<li>However, the two lone pairs occupy axial positions to minimize electron repulsion, leaving the four chlorine atoms in a <strong>square planar<\/strong> arrangement.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Hybridization<\/strong>:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bromine forms 6 orbitals to accommodate the 4 bonds and 2 lone pairs.<\/li>\n\n\n\n<li>This requires the mixing of one ss, three pp, and two dd orbitals, leading to <strong>sp\u00b3d\u00b2 hybridization<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<p>The molecular shape of <strong>BrCl\u2084\u207b<\/strong> is <strong>square planar<\/strong>, and the hybridization of the central bromine atom is <strong>sp\u00b3d\u00b2<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for BrCl 4 -. The molecular geometry (molecular shape) and the hybridization of the central atom The correct answer and explanation is: Correct Answer: 1. Square Planar, sp\u00b3d\u00b2 Explanation: To determine the Lewis structure, molecular geometry, and hybridization of BrCl\u2084\u207b, let\u2019s proceed step by step: 1. Lewis Structure: The Lewis structure [&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-185621","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185621","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=185621"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/185621\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=185621"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=185621"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=185621"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}