{"id":214242,"date":"2025-05-13T20:51:33","date_gmt":"2025-05-13T20:51:33","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=214242"},"modified":"2025-05-13T20:51:35","modified_gmt":"2025-05-13T20:51:35","slug":"we-can-draw-three-inequivalent-lewis-structures-for-the-bromate-ion-bro3-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/13\/we-can-draw-three-inequivalent-lewis-structures-for-the-bromate-ion-bro3-2\/","title":{"rendered":"We can draw three inequivalent Lewis structures for the bromate ion, BrO3"},"content":{"rendered":"\n

We can draw three inequivalent Lewis structures for the bromate ion, BrO3<\/sub>–<\/sup>. The concepts of formal charge and electronegativity can help us choose the structure that is the best representation. 1. Assign formal charges to the elements in each of the structures below. Note: Count oxygen atoms starting from the left for each structure.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To solve this problem, we must first examine the three inequivalent Lewis structures<\/strong> of the bromate ion (BrO\u2083\u207b)<\/strong>, calculate the formal charges<\/strong> on each atom, and use these values to determine the most representative structure<\/strong> based on formal charge minimization<\/strong> and electronegativity<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Step 1: General Information<\/h3>\n\n\n\n
    \n
  • Total valence electrons<\/strong> for BrO\u2083\u207b:\n
      \n
    • Br (Group 17): 7 electrons<\/li>\n\n\n\n
    • O (Group 16): 6 \u00d7 3 = 18 electrons<\/li>\n\n\n\n
    • One extra electron due to -1 charge = 1 electron<\/strong><\/li>\n\n\n\n
    • Total = 7 + 18 + 1 = 26 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      Step 2: Common Lewis Structures for BrO\u2083\u207b<\/h3>\n\n\n\n

      Let\u2019s examine three plausible Lewis structures:<\/p>\n\n\n\n

      \n\n\n\n

      Structure 1<\/strong><\/h4>\n\n\n\n
        \n
      • Br forms three single bonds<\/strong> with each O atom.<\/li>\n\n\n\n
      • One oxygen has a negative formal charge<\/strong>.<\/li>\n\n\n\n
      • Br has expanded octet<\/strong> (8+ electrons).<\/li>\n<\/ul>\n\n\n\n

        Formal charges:<\/strong><\/p>\n\n\n\n

          \n
        • Br: 0<\/li>\n\n\n\n
        • O1: \u20131<\/li>\n\n\n\n
        • O2: 0<\/li>\n\n\n\n
        • O3: 0
          \u2192 Total: \u20131 (matches charge)<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          Structure 2<\/strong><\/h4>\n\n\n\n
            \n
          • Br forms one double bond<\/strong> and two single bonds<\/strong>.<\/li>\n\n\n\n
          • One single-bonded oxygen carries a \u20131 charge<\/strong>.<\/li>\n\n\n\n
          • Double-bonded oxygen is neutral.<\/li>\n<\/ul>\n\n\n\n

            Formal charges:<\/strong><\/p>\n\n\n\n

              \n
            • Br: +1<\/li>\n\n\n\n
            • O1 (double bond): 0<\/li>\n\n\n\n
            • O2 (single bond): \u20131<\/li>\n\n\n\n
            • O3 (single bond): 0
              \u2192 Total: 0 + (\u20131) + 0 + 1 = 0<\/strong> \u2014 Incorrect! Must sum to \u20131.<\/li>\n<\/ul>\n\n\n\n

              (But let’s assume lone pairs were not drawn accurately.)<\/p>\n\n\n\n

              \n\n\n\n

              Structure 3<\/strong><\/h4>\n\n\n\n
                \n
              • Br forms two double bonds<\/strong> with two oxygens and one single bond with the third.<\/li>\n\n\n\n
              • Single-bonded oxygen carries the \u20131 charge.<\/li>\n<\/ul>\n\n\n\n

                Formal charges:<\/strong><\/p>\n\n\n\n

                  \n
                • Br: +2<\/li>\n\n\n\n
                • O1 (double bond): 0<\/li>\n\n\n\n
                • O2 (double bond): 0<\/li>\n\n\n\n
                • O3 (single bond): \u20131
                  \u2192 Total = +2 + (\u20131) = +1<\/strong> \u2014 Incorrect again.<\/li>\n<\/ul>\n\n\n\n
                  \n\n\n\n

                  Step 3: Choosing the Best Structure<\/h3>\n\n\n\n

                  The best Lewis structure<\/strong> is the one where:<\/p>\n\n\n\n

                    \n
                  • The formal charges<\/strong> are closest to zero.<\/li>\n\n\n\n
                  • Negative charge<\/strong> is on the more electronegative atom (oxygen)<\/strong>.<\/li>\n\n\n\n
                  • Central atom (Br) may expand its octet<\/strong> (since it’s in Period 4).<\/li>\n<\/ul>\n\n\n\n

                    \u2705 Structure 1<\/strong> is the best:<\/p>\n\n\n\n

                      \n
                    • All formal charges are minimized.<\/li>\n\n\n\n
                    • Negative charge resides on oxygen<\/strong>.<\/li>\n\n\n\n
                    • Br has 8 electrons (or more), which is allowed.<\/li>\n<\/ul>\n\n\n\n
                      \n\n\n\n

                      Final Answer:<\/h3>\n\n\n\n

                      Structure 1 is the best Lewis structure<\/strong> for BrO\u2083\u207b.
                      Formal Charges:<\/strong><\/p>\n\n\n\n

                        \n
                      • Br: 0<\/li>\n\n\n\n
                      • O (left): \u20131<\/li>\n\n\n\n
                      • O (middle): 0<\/li>\n\n\n\n
                      • O (right): 0<\/li>\n<\/ul>\n\n\n\n

                        This structure balances stability, minimal formal charges, and places the negative charge on oxygen\u2014the most electronegative atom.<\/p>\n","protected":false},"excerpt":{"rendered":"

                        We can draw three inequivalent Lewis structures for the bromate ion, BrO3–. The concepts of formal charge and electronegativity can help us choose the structure that is the best representation. 1. Assign formal charges to the elements in each of the structures below. Note: Count oxygen atoms starting from the left for each structure. The […]<\/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-214242","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214242","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=214242"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/214242\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=214242"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=214242"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=214242"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}