{"id":189366,"date":"2025-02-09T06:34:04","date_gmt":"2025-02-09T06:34:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189366"},"modified":"2025-02-09T06:34:07","modified_gmt":"2025-02-09T06:34:07","slug":"draw-the-most-appropriate-lewis-structures-for-bef2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/09\/draw-the-most-appropriate-lewis-structures-for-bef2\/","title":{"rendered":"Draw The Most Appropriate Lewis Structure(S) For BeF2"},"content":{"rendered":"\n
    \n
  1. Draw The Most Appropriate Lewis Structure(S) For BeF2. What Is The Electron Domain Geometry And The Molecular Geometry? Does The Molecule Have A Dipole? What Is The Hybridization Of The Central Atom? How Many O And It Bonds Are There?<\/li>\n<\/ol>\n\n\n\n

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

    Lewis Structure for BeF2<\/h3>\n\n\n\n

    To draw the Lewis structure for beryllium fluoride (BeF\u2082), follow these steps:<\/p>\n\n\n\n

      \n
    1. Determine the total number of valence electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
        \n
      • Beryllium (Be) has 2 valence electrons (group 2 element).<\/li>\n\n\n\n
      • Fluorine (F) has 7 valence electrons, and there are two fluorine atoms.<\/li>\n\n\n\n
      • Total electrons = 2 (from Be) + 7 \u00d7 2 (from F) = 16 electrons.<\/li>\n<\/ul>\n\n\n\n
          \n
        1. Place the central atom (Be) and surrounding atoms (F):<\/strong><\/li>\n<\/ol>\n\n\n\n
            \n
          • Place Be in the center because it is less electronegative than F.<\/li>\n\n\n\n
          • Connect each F to Be with a single bond, which uses 4 electrons (2 electrons per bond).<\/li>\n\n\n\n
          • The remaining 12 electrons will be placed as lone pairs on the fluorine atoms.<\/li>\n<\/ul>\n\n\n\n
              \n
            1. Distribute the electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
                \n
              • After bonding, 12 electrons remain. Assign these electrons as lone pairs to the fluorine atoms.<\/li>\n\n\n\n
              • Each fluorine will get three lone pairs, leaving each fluorine atom with a complete octet.<\/li>\n<\/ul>\n\n\n\n

                Thus, the Lewis structure for BeF\u2082 is:<\/p>\n\n\n\n

                   F-Be-F<\/code><\/pre>\n\n\n\n
                With each F atom having 3 lone pairs.<\/p>\n\n\n\n
                Electron Domain Geometry and Molecular Geometry<\/h3>\n\n\n\n
                  \n
                • Electron Domain Geometry:<\/strong> Be has two bonds and no lone pairs around it, which gives an electron domain geometry<\/strong> of linear<\/strong>.<\/li>\n\n\n\n
                • Molecular Geometry:<\/strong> Since there are no lone pairs on the central atom (Be), the molecular geometry<\/strong> is also linear<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                  Dipole Moment<\/h3>\n\n\n\n
                    \n
                  • Dipole Moment:<\/strong> The molecule does not have a dipole moment because BeF\u2082 is symmetric, with two fluorine atoms on opposite sides of the beryllium atom. The dipoles created by each Be-F bond cancel each other out, leading to no overall dipole moment.<\/li>\n<\/ul>\n\n\n\n
                    Hybridization of the Central Atom<\/h3>\n\n\n\n
                      \n
                    • Hybridization:<\/strong> The central atom Be has two bonding regions (two Be-F bonds). Therefore, the hybridization is sp<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                      O and I Bonds<\/h3>\n\n\n\n
                        \n
                      • In BeF\u2082, the molecule consists of 2 single bonds<\/strong> between the beryllium and fluorine atoms (I-type bonds). There are no O bonds in this structure.<\/li>\n<\/ul>\n\n\n\n
                        Summary<\/h3>\n\n\n\n
                          \n
                        • Lewis Structure:<\/strong> Be-F-Be with lone pairs on fluorine.<\/li>\n\n\n\n
                        • Electron Domain Geometry:<\/strong> Linear.<\/li>\n\n\n\n
                        • Molecular Geometry:<\/strong> Linear.<\/li>\n\n\n\n
                        • Dipole Moment:<\/strong> No dipole moment.<\/li>\n\n\n\n
                        • Hybridization:<\/strong> sp.<\/li>\n\n\n\n
                        • Bonds:<\/strong> 2 I-type (Be-F) bonds, no O bonds.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"
                          The Correct Answer and Explanation is : Lewis Structure for BeF2 To draw the Lewis structure for beryllium fluoride (BeF\u2082), follow these steps: Thus, the Lewis structure for BeF\u2082 is: With each F atom having 3 lone pairs. Electron Domain Geometry and Molecular Geometry Dipole Moment Hybridization of the Central Atom O and I Bonds […]<\/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-189366","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189366","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=189366"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189366\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189366"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189366"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189366"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}