{"id":202664,"date":"2025-03-18T17:18:35","date_gmt":"2025-03-18T17:18:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202664"},"modified":"2025-03-18T17:18:40","modified_gmt":"2025-03-18T17:18:40","slug":"draw-a-valid-alternative-lewis-structure-for-this-ions-with-one-br-o-and-one-bro-bond","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/18\/draw-a-valid-alternative-lewis-structure-for-this-ions-with-one-br-o-and-one-bro-bond\/","title":{"rendered":"Draw a valid alternative lewis structure for this ions with one Br-O and one Br=O bond"},"content":{"rendered":"\n

BrO2- ion<\/p>\n\n\n\n

a) Draw a valid alternative lewis structure for this ions with one Br-O and one Br=O bond<\/p>\n\n\n\n

b) Determine the formal charge on each atom in the new structure<\/p>\n\n\n\n

c) Based on the formal charges you determine whih is the best and what criateria are used with formal charge to determine the best structure ?<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

(a) Alternative Lewis Structure for BrO\u2082\u207b Ion<\/strong><\/h3>\n\n\n\n

A valid alternative Lewis structure of the BrO\u2082\u207b<\/strong> ion consists of:<\/p>\n\n\n\n

    \n
  • One Br\u2013O (single bond)<\/strong><\/li>\n\n\n\n
  • One Br=O (double bond)<\/strong><\/li>\n\n\n\n
  • The total valence electrons: 20<\/strong>\n
      \n
    • Br = 7<\/li>\n\n\n\n
    • O (each) = 6 \u00d7 2 = 12<\/li>\n\n\n\n
    • Extra electron due to negative charge = 1<\/li>\n\n\n\n
    • Total = 7 + 12 + 1 = 20 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

      Structure Details:<\/strong><\/p>\n\n\n\n

        \n
      • Bromine (Br) is the central atom.<\/strong><\/li>\n\n\n\n
      • One oxygen forms a single bond (Br\u2013O)<\/strong> and has three lone pairs<\/strong>.<\/li>\n\n\n\n
      • The other oxygen forms a double bond (Br=O)<\/strong> and has two lone pairs<\/strong>.<\/li>\n\n\n\n
      • Bromine has one lone pair<\/strong> to maintain 20 total electrons.<\/li>\n<\/ul>\n\n\n\n
        \n\n\n\n

        (b) Formal Charge Calculation<\/strong><\/h3>\n\n\n\n

        The formal charge formula: FC=(Valence electrons)\u2212(Non-bonding electrons)\u2212(Bonding electrons)2FC = \\text{(Valence electrons)} – \\text{(Non-bonding electrons)} – \\frac{\\text{(Bonding electrons)}}{2}<\/p>\n\n\n\n

          \n
        1. Bromine (Br):<\/strong>\n
            \n
          • Valence electrons = 7<\/li>\n\n\n\n
          • Non-bonding electrons = 2<\/li>\n\n\n\n
          • Bonding electrons = 8 (1 single + 1 double bond)<\/li>\n\n\n\n
          • Formal charge<\/strong> = 7 – 2 – (8\/2) = +1<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Oxygen (Br=O, double bond):<\/strong>\n
              \n
            • Valence electrons = 6<\/li>\n\n\n\n
            • Non-bonding electrons = 4<\/li>\n\n\n\n
            • Bonding electrons = 4<\/li>\n\n\n\n
            • Formal charge<\/strong> = 6 – 4 – (4\/2) = 0<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Oxygen (Br\u2013O, single bond):<\/strong>\n
                \n
              • Valence electrons = 6<\/li>\n\n\n\n
              • Non-bonding electrons = 6<\/li>\n\n\n\n
              • Bonding electrons = 2<\/li>\n\n\n\n
              • Formal charge<\/strong> = 6 – 6 – (2\/2) = -1<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n
                \n\n\n\n

                (c) Best Lewis Structure & Explanation<\/strong><\/h3>\n\n\n\n

                Criteria for the best Lewis structure:<\/strong><\/p>\n\n\n\n

                  \n
                1. Formal charge minimization<\/strong> \u2013 The sum of formal charges should equal the ion\u2019s charge (-1), and individual charges should be close to zero.<\/li>\n\n\n\n
                2. Electronegativity consideration<\/strong> \u2013 Negative formal charges should reside on more electronegative atoms (oxygen is more electronegative than bromine).<\/li>\n\n\n\n
                3. Octet rule compliance<\/strong> \u2013 All atoms should follow the octet rule when possible.<\/li>\n<\/ol>\n\n\n\n

                  Which is the Best Structure?<\/strong><\/h4>\n\n\n\n

                  The best Lewis structure<\/strong> for BrO\u2082\u207b is one where both oxygen atoms are singly bonded to bromine<\/strong>, and bromine carries a -1 formal charge<\/strong> instead of oxygen. This structure minimizes formal charge on Br and places the negative charge on the less electronegative element (Br)<\/strong> rather than oxygen.<\/p>\n\n\n\n

                  While the alternative structure with one Br=O and one Br\u2013O<\/strong> is valid, it is less stable because it places a +1 charge on Br<\/strong> and -1 on oxygen<\/strong>, which is less favorable.<\/p>\n\n\n\n

                  Thus, the best structure has:<\/p>\n\n\n\n

                    \n
                  • Two Br\u2013O single bonds<\/strong><\/li>\n\n\n\n
                  • A lone pair on Br<\/strong><\/li>\n\n\n\n
                  • -1 formal charge on Br (instead of oxygen)<\/strong><\/li>\n<\/ul>\n\n\n\n

                    This structure aligns better with formal charge minimization and electronegativity principles.<\/p>\n\n\n\n

                    \n\n\n\n

                    Generated Image<\/strong><\/h3>\n\n\n\n

                    Now, I’ll generate an image of the alternative Lewis structure.<\/p>\n\n\n\n

                    Here is the generated Lewis structure for the BrO\u2082\u207b ion. Let me know if you need any modifications or additional explanations!<\/p>\n\n\n\n

                    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                    BrO2- ion a) Draw a valid alternative lewis structure for this ions with one Br-O and one Br=O bond b) Determine the formal charge on each atom in the new structure c) Based on the formal charges you determine whih is the best and what criateria are used with formal charge to determine the best […]<\/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-202664","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202664","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=202664"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202664\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202664"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202664"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}