{"id":194561,"date":"2025-02-24T09:12:17","date_gmt":"2025-02-24T09:12:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=194561"},"modified":"2025-02-24T09:12:19","modified_gmt":"2025-02-24T09:12:19","slug":"the-electron-domain-and-molecular-geometry-of-bro2-is","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/24\/the-electron-domain-and-molecular-geometry-of-bro2-is\/","title":{"rendered":"The electron domain and molecular geometry of BrO2-is"},"content":{"rendered":"\n

The electron domain and molecular geometry of BrO2-is _<\/em><\/strong>. A)tetrahedral, bent B)trigonal pyramidal, linear C)tetrahedral, trigonal planar<\/p>\n\n\n\n

D)trigonal pyramidal, seesaw E)trigonal planar, trigonal planar<\/p>\n\n\n\n

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

The correct answer is:<\/p>\n\n\n\n

A) Tetrahedral, Bent<\/strong><\/p>\n\n\n\n

Explanation:<\/strong><\/h3>\n\n\n\n

To determine the electron domain and molecular geometry of BrO\u2082\u207b (bromite ion)<\/strong>, we follow these steps:<\/p>\n\n\n\n

    \n
  1. Determine the Total Valence Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Bromine (Br):<\/strong> 7 valence electrons<\/li>\n\n\n\n
    • Oxygen (O):<\/strong> Each oxygen has 6 valence electrons, so two oxygens contribute 12 electrons<\/strong><\/li>\n\n\n\n
    • Negative charge (-1):<\/strong> Adds 1 extra electron<\/strong><\/li>\n\n\n\n
    • Total valence electrons:<\/strong> (7 + 12 + 1 = 20) electrons<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Draw the Lewis Structure:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • Place Br<\/strong> as the central atom.<\/li>\n\n\n\n
        • Distribute the oxygens around Br.<\/li>\n\n\n\n
        • Connect each oxygen atom<\/strong> to Br with a single bond.<\/li>\n\n\n\n
        • Complete the octets<\/strong> of oxygen first.<\/li>\n\n\n\n
        • Assign the remaining electrons to Br as lone pairs.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Determine the Electron Domain Geometry:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • Br has two bonded atoms (O) and two lone pairs.<\/strong><\/li>\n\n\n\n
            • This gives four electron domains<\/strong> around Br.<\/li>\n\n\n\n
            • According to VSEPR (Valence Shell Electron Pair Repulsion) theory<\/strong>, a molecule with four electron domains<\/strong> adopts a tetrahedral electron domain geometry<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Determine the Molecular Geometry:<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • The presence of two lone pairs<\/strong> affects the molecular shape.<\/li>\n\n\n\n
                • Lone pairs repel more strongly than bonding pairs, pushing the oxygen atoms down<\/strong> and creating a bent shape<\/strong> (similar to water, H\u2082O).<\/li>\n\n\n\n
                • This results in a bent molecular geometry<\/strong> instead of a tetrahedral shape.<\/li>\n<\/ul>\n\n\n\n

                  Conclusion:<\/strong><\/h3>\n\n\n\n
                    \n
                  • Electron Domain Geometry:<\/strong> Tetrahedral<\/strong><\/li>\n\n\n\n
                  • Molecular Geometry:<\/strong> Bent<\/strong><\/li>\n\n\n\n
                  • This corresponds to Option A<\/strong>.<\/li>\n<\/ul>\n\n\n\n

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

                    \"\"<\/figure>\n\n\n\n

                    Here is the 3D molecular structure representation of the bromite ion (BrO\u2082\u207b). It visually demonstrates the tetrahedral electron domain geometry and the bent molecular shape due to the presence of two lone pairs on bromine. Let me know if you need any modifications!<\/p>\n","protected":false},"excerpt":{"rendered":"

                    The electron domain and molecular geometry of BrO2-is _. A)tetrahedral, bent B)trigonal pyramidal, linear C)tetrahedral, trigonal planar D)trigonal pyramidal, seesaw E)trigonal planar, trigonal planar The Correct Answer and Explanation is : The correct answer is: A) Tetrahedral, Bent Explanation: To determine the electron domain and molecular geometry of BrO\u2082\u207b (bromite ion), we follow these steps: […]<\/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-194561","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194561","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=194561"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/194561\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=194561"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=194561"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=194561"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}