{"id":213942,"date":"2025-05-13T06:04:27","date_gmt":"2025-05-13T06:04:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=213942"},"modified":"2025-05-13T06:04:29","modified_gmt":"2025-05-13T06:04:29","slug":"draw-the-lewis-structure-for-brcl3-and-answer-the-following-question","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/draw-the-lewis-structure-for-brcl3-and-answer-the-following-question\/","title":{"rendered":"Draw the Lewis structure for BrCl3 and answer the following question"},"content":{"rendered":"\n<p>Draw the Lewis structure for BrCl3 and answer the following question. What are the approximate bond angles about the central atom?<\/p>\n\n\n\n<p>Select all that apply.<\/p>\n\n\n\n<p>no v 60\u00b0<br>yes 90\u00b0<br>no 1090<br>yes 120\u00b0<br>no v 180<\/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>To answer this question, let\u2019s first <strong>draw the Lewis structure for BrCl\u2083 (bromine trichloride)<\/strong> and analyze its geometry to determine the approximate bond angles around the central atom.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Count Valence Electrons<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Br (Bromine)<\/strong> has 7 valence electrons.<\/li>\n\n\n\n<li><strong>Cl (Chlorine)<\/strong> has 7 valence electrons.<\/li>\n\n\n\n<li>There are <strong>3 Cl atoms<\/strong>, so:<br>Total = 7 (from Br) + 3 \u00d7 7 (from 3 Cl) = <strong>28 valence electrons<\/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\"><strong>Step 2: Choose Central Atom and Arrange Electrons<\/strong><\/h3>\n\n\n\n<p>Bromine is less electronegative than chlorine, so <strong>Br<\/strong> will be the central atom.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Connect each Cl atom to Br with a single bond. This uses up <strong>6 electrons<\/strong> (3 bonds \u00d7 2 electrons each).<\/li>\n\n\n\n<li>Distribute the remaining <strong>22 electrons<\/strong> to satisfy the octets of the Cl atoms first (6 electrons on each of the 3 Cl atoms = 18 electrons).<\/li>\n\n\n\n<li>Leftover: 22 &#8211; 18 = <strong>4 electrons<\/strong>, which go on the central atom Br as <strong>2 lone pairs<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>So, the central atom <strong>Br<\/strong> is surrounded by:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>3 bonding pairs (Br\u2013Cl bonds)<\/strong><\/li>\n\n\n\n<li><strong>2 lone pairs<\/strong><\/li>\n<\/ul>\n\n\n\n<p>This gives a <strong>total of 5 electron domains<\/strong> (regions of electron density).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Determine Electron Geometry<\/strong><\/h3>\n\n\n\n<p>With 5 electron domains, the electron geometry is <strong>trigonal bipyramidal<\/strong>.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In trigonal bipyramidal geometry:\n<ul class=\"wp-block-list\">\n<li>Axial bonds are <strong>90\u00b0<\/strong> from equatorial positions.<\/li>\n\n\n\n<li>Equatorial bonds are <strong>120\u00b0<\/strong> apart from each other.<\/li>\n<\/ul>\n<\/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\"><strong>Step 4: Determine Molecular Geometry<\/strong><\/h3>\n\n\n\n<p>Since 2 of the 5 electron domains are <strong>lone pairs<\/strong>, the molecular geometry is <strong>T-shaped<\/strong> (not trigonal planar or linear).<\/p>\n\n\n\n<p>In a T-shaped molecule:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The lone pairs occupy <strong>equatorial positions<\/strong> to minimize repulsion.<\/li>\n\n\n\n<li>The bonding pairs occupy <strong>2 axial<\/strong> and <strong>1 equatorial<\/strong> position.<\/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\"><strong>Step 5: Approximate Bond Angles<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>bond angle between axial and equatorial positions<\/strong> is about <strong>90\u00b0<\/strong>.<\/li>\n\n\n\n<li>The <strong>bond angle between equatorial atoms<\/strong> (if bonded) would be <strong>120\u00b0<\/strong>, but here the lone pairs occupy equatorial positions.<\/li>\n<\/ul>\n\n\n\n<p>So, the actual bond angles in BrCl\u2083 are approximately:<\/p>\n\n\n\n<p>\u2705 <strong>90\u00b0<\/strong><br>\u2705 <strong>120\u00b0<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Yes<\/strong>: 90\u00b0<\/li>\n\n\n\n<li><strong>Yes<\/strong>: 120\u00b0<\/li>\n\n\n\n<li><strong>No<\/strong>: 60\u00b0<\/li>\n\n\n\n<li><strong>No<\/strong>: 109\u00b0<\/li>\n\n\n\n<li><strong>No<\/strong>: 180\u00b0<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>These values come from the <strong>electron domain geometry (trigonal bipyramidal)<\/strong> and the resulting <strong>molecular shape (T-shaped)<\/strong> of BrCl\u2083.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for BrCl3 and answer the following question. What are the approximate bond angles about the central atom? Select all that apply. no v 60\u00b0yes 90\u00b0no 1090yes 120\u00b0no v 180 The Correct Answer and Explanation is: To answer this question, let\u2019s first draw the Lewis structure for BrCl\u2083 (bromine trichloride) and analyze [&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-213942","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213942","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=213942"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/213942\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=213942"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=213942"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=213942"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}