{"id":203383,"date":"2025-03-19T17:00:14","date_gmt":"2025-03-19T17:00:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=203383"},"modified":"2025-03-19T17:00:19","modified_gmt":"2025-03-19T17:00:19","slug":"what-is-the-lewis-structure-for-brf2-and-the-number-of-valence","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/19\/what-is-the-lewis-structure-for-brf2-and-the-number-of-valence\/","title":{"rendered":"what is the lewis structure for BrF2+ and the number of valence"},"content":{"rendered":"\n<p>what is the lewis structure for BrF2+ and the number of valence<\/p>\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\"><strong>Lewis Structure of BrF\u2082\u207a and Valence Electrons Calculation<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Valence Electrons Calculation:<\/strong><\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Bromine (Br) Contribution:<\/strong> Bromine belongs to Group 17 and has <strong>7 valence electrons<\/strong>.<\/li>\n\n\n\n<li><strong>Fluorine (F) Contribution:<\/strong> Each fluorine atom also belongs to Group 17 and has <strong>7 valence electrons<\/strong>. Since there are <strong>two fluorine atoms<\/strong>, the total contribution from fluorine is <strong>2 \u00d7 7 = 14 valence electrons<\/strong>.<\/li>\n\n\n\n<li><strong>Cation (+1) Effect:<\/strong> Since BrF\u2082\u207a has a <strong>+1 charge<\/strong>, it means one electron is lost from the total count.<\/li>\n<\/ol>\n\n\n\n<p><strong>Total valence electrons calculation:<\/strong> 7(Br)+(2\u00d77)(F)\u22121(for&nbsp;+1&nbsp;charge)=7+14\u22121=20&nbsp;valence&nbsp;electrons.7 (Br) + (2 \\times 7) (F) &#8211; 1 (\\text{for +1 charge}) = 7 + 14 &#8211; 1 = 20 \\text{ valence electrons}.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Lewis Structure Construction:<\/strong><\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Central Atom Selection:<\/strong> Bromine is less electronegative than fluorine, so it is the <strong>central atom<\/strong>.<\/li>\n\n\n\n<li><strong>Bonding:<\/strong> Two fluorine atoms are bonded to bromine via <strong>single bonds<\/strong>. This uses up <strong>2 \u00d7 2 = 4 electrons<\/strong>.<\/li>\n\n\n\n<li><strong>Remaining Electrons:<\/strong> After bonding, <strong>16 electrons<\/strong> remain.<\/li>\n\n\n\n<li><strong>Lone Pairs Assignment:<\/strong>\n<ul class=\"wp-block-list\">\n<li>Fluorine atoms need <strong>3 lone pairs<\/strong> each (total <strong>2 \u00d7 6 = 12 electrons<\/strong>).<\/li>\n\n\n\n<li>The remaining <strong>4 electrons<\/strong> are placed as <strong>two lone pairs<\/strong> on bromine.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Molecular Geometry &amp; Formal Charge:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Formal Charge Calculation:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Br:<\/strong> 7\u2212(4+2)=+17 &#8211; (4 + 2) = +1<\/li>\n\n\n\n<li><strong>F:<\/strong> 7\u2212(6+1)=07 &#8211; (6 + 1) = 0<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Structure Shape:<\/strong> <strong>Bent (T-Shaped)<\/strong> due to lone pairs and the AX\u2082E\u2083 arrangement (based on VSEPR theory).<\/li>\n<\/ul>\n\n\n\n<p>Now, generating the image for the <strong>Lewis structure of BrF\u2082\u207a<\/strong>:<\/p>\n\n\n\n<p>Here is the Lewis structure of BrF\u2082\u207a. The diagram shows bromine in the center with two fluorine atoms bonded via single bonds, two lone pairs on bromine, and a +1 charge indicated. The molecular geometry follows a T-shape due to the electron repulsions. Let me know if you need further explanations!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-1066.png\" alt=\"\" class=\"wp-image-203384\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>what is the lewis structure for BrF2+ and the number of valence The correct answer and explanation is: Lewis Structure of BrF\u2082\u207a and Valence Electrons Calculation Valence Electrons Calculation: Total valence electrons calculation: 7(Br)+(2\u00d77)(F)\u22121(for&nbsp;+1&nbsp;charge)=7+14\u22121=20&nbsp;valence&nbsp;electrons.7 (Br) + (2 \\times 7) (F) &#8211; 1 (\\text{for +1 charge}) = 7 + 14 &#8211; 1 = 20 \\text{ valence [&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-203383","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203383","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=203383"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203383\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203383"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203383"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203383"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}