{"id":225566,"date":"2025-06-04T10:13:16","date_gmt":"2025-06-04T10:13:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225566"},"modified":"2025-06-04T10:13:18","modified_gmt":"2025-06-04T10:13:18","slug":"draw-the-lewis-structure-of-sbbr3-and-use-it-to-answer-the-following-questions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/draw-the-lewis-structure-of-sbbr3-and-use-it-to-answer-the-following-questions\/","title":{"rendered":"Draw the Lewis structure of SbBr3 and use it to answer the following questions"},"content":{"rendered":"\n

Draw the Lewis structure of SbBr3 and use it to answer the following questions: What is the electron domain geometry of this molecule? [Select ] What is the molecular geometry of this molecule? [Select ] Is this molecule polar? [Select ] What is the approximate value of the Br-Sb-Br bond angle in this molecule? [Select ] What is the bond order of the Sb-Br bonds in this molecule? [Select ]<\/p>\n\n\n\n

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

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

Here are the correct answers based on the Lewis structure of SbBr\u2083<\/strong> (antimony tribromide), followed by a detailed explanation:<\/p>\n\n\n\n

\n\n\n\n

Correct Answers<\/strong>:<\/h3>\n\n\n\n
    \n
  1. Electron domain geometry<\/strong>: Tetrahedral<\/strong><\/li>\n\n\n\n
  2. Molecular geometry<\/strong>: Trigonal pyramidal<\/strong><\/li>\n\n\n\n
  3. Is this molecule polar?<\/strong>: Yes<\/strong><\/li>\n\n\n\n
  4. Approximate Br-Sb-Br bond angle<\/strong>: <109.5\u00b0<\/strong><\/li>\n\n\n\n
  5. Bond order of Sb-Br bonds<\/strong>: 1<\/strong><\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

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

    To determine the geometry and other properties of SbBr\u2083<\/strong>, we start with its Lewis structure<\/strong>. Antimony (Sb), in Group 15 of the periodic table, has 5 valence electrons<\/strong>. Each bromine atom (Br), from Group 17, has 7 valence electrons<\/strong>. In SbBr\u2083<\/strong>, Sb forms three single covalent bonds<\/strong> with three Br atoms.<\/p>\n\n\n\n

    Step 1: Lewis Structure<\/strong><\/p>\n\n\n\n

      \n
    • Sb is the central atom.<\/li>\n\n\n\n
    • It forms three single bonds with Br atoms (each Br needs only one electron to complete its octet).<\/li>\n\n\n\n
    • This uses 3 of Sb’s 5 valence electrons.<\/li>\n\n\n\n
    • The remaining 2 electrons form a lone pair<\/strong> on Sb.<\/li>\n<\/ul>\n\n\n\n

      Step 2: Electron Domain Geometry<\/strong><\/p>\n\n\n\n

        \n
      • There are four electron domains (3 bonding pairs + 1 lone pair) around Sb.<\/li>\n\n\n\n
      • This corresponds to a tetrahedral<\/strong> electron domain geometry.<\/li>\n<\/ul>\n\n\n\n

        Step 3: Molecular Geometry<\/strong><\/p>\n\n\n\n

          \n
        • Because one of the electron domains is a lone pair, the molecular shape<\/strong> is not tetrahedral but trigonal pyramidal<\/strong> (like NH\u2083).<\/li>\n<\/ul>\n\n\n\n

          Step 4: Polarity<\/strong><\/p>\n\n\n\n

            \n
          • Due to the lone pair, the molecule is not symmetric, and the dipole moments do not cancel.<\/li>\n\n\n\n
          • Thus, SbBr\u2083 is polar<\/strong>.<\/li>\n<\/ul>\n\n\n\n

            Step 5: Bond Angle<\/strong><\/p>\n\n\n\n

              \n
            • Ideal tetrahedral bond angle is 109.5\u00b0<\/strong>.<\/li>\n\n\n\n
            • The lone pair exerts more repulsion, reducing the angle to slightly less than 109.5\u00b0<\/strong>, typically around 107\u00b0<\/strong>.<\/li>\n<\/ul>\n\n\n\n

              Step 6: Bond Order<\/strong><\/p>\n\n\n\n

                \n
              • Each Sb-Br bond is a single bond<\/strong>, so the bond order is 1<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n

                In summary, SbBr\u2083<\/strong> has a tetrahedral electron geometry, trigonal pyramidal molecular geometry, is polar, has Br-Sb-Br bond angles <109.5\u00b0, and each Sb-Br bond has a bond order of 1.<\/p>\n\n\n\n

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

                Draw the Lewis structure of SbBr3 and use it to answer the following questions: What is the electron domain geometry of this molecule? [Select ] What is the molecular geometry of this molecule? [Select ] Is this molecule polar? [Select ] What is the approximate value of the Br-Sb-Br bond angle in this molecule? [Select […]<\/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-225566","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225566","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=225566"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225566\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225566"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225566"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225566"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}