{"id":183887,"date":"2025-01-17T08:02:17","date_gmt":"2025-01-17T08:02:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183887"},"modified":"2025-01-17T08:02:19","modified_gmt":"2025-01-17T08:02:19","slug":"26-lewis-structure-show-all-resonance-structures-if-applicable-any-polar-bonds-in-the-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/26-lewis-structure-show-all-resonance-structures-if-applicable-any-polar-bonds-in-the-molecule\/","title":{"rendered":"26 Lewis Structure (show all resonance structures if applicable) Any polar bonds in the molecule"},"content":{"rendered":"\n

CHBr3 Total # of Valence Electrons: 26 Lewis Structure (show all resonance structures if applicable) Any polar bonds in the molecule? Yes No 8. OF2 Total # of Valence Electrons: 20 Lewis Structure (show all resonance structures if applicable) Any polar bonds in the molecule? Yes No 3-D Model Sketch Molecular Shape: Tetrahedral Molecular Polarity: 3-D Model Sketch Molecular Shape: Bent Molecular Polarity: Polar Polar Bond Angles Non-Polar 109.5\u00c3\u201a\u00c2\u00b0 Bond Angles 109.5\u00c3\u201a\u00c2\u00b0 Non-Polar<\/p>\n\n\n\n

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

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

Answers:<\/h3>\n\n\n\n

7. CHBr\u2083 (Bromoform)<\/strong><\/h4>\n\n\n\n
    \n
  • Total # of Valence Electrons<\/strong>: 26<\/li>\n\n\n\n
  • Carbon (C): 4 electrons<\/li>\n\n\n\n
  • Hydrogen (H): 1 electron<\/li>\n\n\n\n
  • Bromine (Br): 7 electrons \u00d7 3 = 21 electrons<\/li>\n\n\n\n
  • Total<\/strong>: 4 + 1 + 21 = 26 valence electrons<\/strong><\/li>\n\n\n\n
  • Lewis Structure<\/strong>:<\/li>\n\n\n\n
  • Central carbon atom single-bonded to one hydrogen atom and three bromine atoms.<\/li>\n\n\n\n
  • Bromine atoms each have 6 lone electrons (3 lone pairs).<\/li>\n\n\n\n
  • No resonance structures exist because the bonds are fixed.<\/li>\n\n\n\n
  • Any Polar Bonds in the Molecule?<\/strong>: Yes<\/strong><\/li>\n\n\n\n
  • C\u2013Br bonds are polar because bromine is more electronegative than carbon.<\/li>\n\n\n\n
  • Molecular Shape<\/strong>: Tetrahedral.<\/li>\n\n\n\n
  • Molecular Polarity<\/strong>: Polar<\/strong> due to the asymmetrical distribution of electron density (bromine atoms create a dipole moment).<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    8. OF\u2082 (Oxygen Difluoride)<\/strong><\/h4>\n\n\n\n
      \n
    • Total # of Valence Electrons<\/strong>: 20<\/li>\n\n\n\n
    • Oxygen (O): 6 electrons<\/li>\n\n\n\n
    • Fluorine (F): 7 electrons \u00d7 2 = 14 electrons<\/li>\n\n\n\n
    • Total<\/strong>: 6 + 14 = 20 valence electrons<\/strong><\/li>\n\n\n\n
    • Lewis Structure<\/strong>:<\/li>\n\n\n\n
    • Central oxygen atom single-bonded to two fluorine atoms.<\/li>\n\n\n\n
    • Oxygen has 4 lone electrons (2 lone pairs).<\/li>\n\n\n\n
    • Fluorine atoms each have 6 lone electrons (3 lone pairs).<\/li>\n\n\n\n
    • No resonance structures exist because the bonds are fixed.<\/li>\n\n\n\n
    • Any Polar Bonds in the Molecule?<\/strong>: Yes<\/strong><\/li>\n\n\n\n
    • O\u2013F bonds are polar because fluorine is more electronegative than oxygen.<\/li>\n\n\n\n
    • Molecular Shape<\/strong>: Bent.<\/li>\n\n\n\n
    • Molecular Polarity<\/strong>: Polar<\/strong> due to the asymmetrical shape and dipole moments.<\/li>\n\n\n\n
    • Bond Angles<\/strong>: Approximately 109.5\u00b0<\/strong>, as determined by the tetrahedral electron geometry.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

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

      The Lewis structures and polarity are determined by analyzing the total number of valence electrons and the bonding\/geometry of the molecules:<\/p>\n\n\n\n

        \n
      1. CHBr\u2083<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • Carbon forms four single bonds with hydrogen and bromine atoms, creating a tetrahedral shape.<\/li>\n\n\n\n
        • Bromine\u2019s higher electronegativity makes the molecule polar because the dipoles don\u2019t cancel out.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. OF\u2082<\/strong>:<\/li>\n<\/ol>\n\n\n\n
              \n
            • Oxygen, with two lone pairs and two bonds to fluorine, forms a bent molecular shape.<\/li>\n\n\n\n
            • Fluorine\u2019s high electronegativity creates strong dipole moments, making the molecule polar.<\/li>\n<\/ul>\n\n\n\n

              Both molecules exhibit polar bonds, but their overall polarity arises from their 3D geometry.<\/p>\n","protected":false},"excerpt":{"rendered":"

              CHBr3 Total # of Valence Electrons: 26 Lewis Structure (show all resonance structures if applicable) Any polar bonds in the molecule? Yes No 8. OF2 Total # of Valence Electrons: 20 Lewis Structure (show all resonance structures if applicable) Any polar bonds in the molecule? Yes No 3-D Model Sketch Molecular Shape: Tetrahedral Molecular Polarity: […]<\/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-183887","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183887","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=183887"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183887\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183887"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183887"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183887"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}