{"id":197868,"date":"2025-03-07T18:48:43","date_gmt":"2025-03-07T18:48:43","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=197868"},"modified":"2025-03-07T18:48:45","modified_gmt":"2025-03-07T18:48:45","slug":"draw-the-best-lewis-structure-for-bro2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/07\/draw-the-best-lewis-structure-for-bro2\/","title":{"rendered":"Draw the best Lewis structure for BrO2"},"content":{"rendered":"\n<p>Draw the best Lewis structure for BrO2\u2014.<\/p>\n\n\n\n<p>a). What is the Br-O bond order?<\/p>\n\n\n\n<p>b). What is the O-Br-O bond angle?<\/p>\n\n\n\n<p>c). How many lone pairs are on the central atom?<\/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>To draw the best Lewis structure for the <strong>BrO\u2082\u207b<\/strong> ion, we need to follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Count the total valence electrons<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bromine (Br) is in Group 17, so it has <strong>7 valence electrons<\/strong>.<\/li>\n\n\n\n<li>Oxygen (O) is in Group 16, so each oxygen atom has <strong>6 valence electrons<\/strong>. Since there are two oxygens, that gives a total of <strong>12 valence electrons<\/strong>.<\/li>\n\n\n\n<li>The negative charge (\u207b) means there is <strong>1 extra electron<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>So, the total number of valence electrons is:<br>[<br>7 (\\text{Br}) + 12 (\\text{O}_2) + 1 (\\text{charge}) = 20 \\, \\text{electrons}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Construct the skeleton of the molecule<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bromine is the central atom because it is less electronegative than oxygen. Each oxygen will be bonded to bromine with a single bond.<\/li>\n\n\n\n<li>Connect Br to each O atom with single bonds, which will use <strong>4 electrons<\/strong> (2 electrons per bond).<\/li>\n<\/ul>\n\n\n\n<p>Now, we have <strong>16 electrons left<\/strong> to place on the atoms.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Distribute the remaining electrons<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Oxygen atoms need to complete their octet. Each oxygen will have <strong>3 lone pairs<\/strong> of electrons (6 electrons) around it, using up <strong>12 electrons<\/strong>.<\/li>\n\n\n\n<li>This leaves <strong>4 electrons<\/strong>, which will go to the bromine atom as lone pairs.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Check the octet rule and formal charges<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each oxygen atom has <strong>6 electrons<\/strong> in lone pairs and <strong>2 electrons<\/strong> in the bond, satisfying the octet rule.<\/li>\n\n\n\n<li>The bromine atom will have <strong>2 lone pairs<\/strong> and <strong>2 bonds<\/strong> to oxygen, giving it <strong>8 electrons<\/strong> in total, which satisfies the octet rule.<\/li>\n\n\n\n<li>The negative charge is accommodated by the oxygen atoms since oxygen is more electronegative than bromine.<\/li>\n<\/ul>\n\n\n\n<p>The resulting Lewis structure looks like this:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>     :O:  \n      |   \n  :Br:O:\u207b\n      |  \n     :O:<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer to the Questions:<\/h3>\n\n\n\n<p>a) <strong>Bond order of Br-O:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The Br-O bond order can be determined by considering the total number of bonds between bromine and the oxygen atoms. Since there are 2 single bonds and no resonance structures to suggest double or triple bonds, the bond order is <strong>1<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>b) <strong>O-Br-O bond angle:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The BrO\u2082\u207b ion adopts a <strong>bent<\/strong> shape, similar to the structure of H\u2082O. This is due to the lone pairs on the central bromine atom. The angle between the two oxygen atoms around the bromine atom is around <strong>109.5\u00b0<\/strong>, which is typical of a <strong>bent<\/strong> molecular geometry (in the case of a sp\u00b3 hybridized central atom).<\/li>\n<\/ul>\n\n\n\n<p>c) <strong>Lone pairs on the central atom:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bromine has <strong>2 lone pairs<\/strong> of electrons on it in the best Lewis structure for BrO\u2082\u207b.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The structure follows basic principles like the octet rule for each atom and the minimization of formal charges. The lone pairs on the oxygen atoms contribute to the negative charge, while the central bromine atom follows the octet rule. The bond angles are influenced by the repulsion between the lone pairs on the bromine atom. The geometry is bent due to these lone pairs, resulting in a bond angle of about <strong>109.5\u00b0<\/strong>, typical for a trigonal pyramidal or bent structure with lone pairs.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the best Lewis structure for BrO2\u2014. a). What is the Br-O bond order? b). What is the O-Br-O bond angle? c). How many lone pairs are on the central atom? The correct answer and explanation is : To draw the best Lewis structure for the BrO\u2082\u207b ion, we need to follow these steps: Step [&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-197868","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197868","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=197868"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197868\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197868"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197868"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197868"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}