{"id":267075,"date":"2025-07-23T09:12:40","date_gmt":"2025-07-23T09:12:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=267075"},"modified":"2025-07-23T09:12:48","modified_gmt":"2025-07-23T09:12:48","slug":"draw-the-lewis-structure-of-bromine-pentafluoride-brf5","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/23\/draw-the-lewis-structure-of-bromine-pentafluoride-brf5\/","title":{"rendered":"Draw the Lewis structure of bromine pentafluoride, BrF5."},"content":{"rendered":"\n

Draw the Lewis structure of bromine pentafluoride, BrF5. Do not include brackets or formal charges. If the molecule exhibits resonance, draw only one (1) of the resonance forms. Hartree-Fock method can be used to help with the drawing. Edit drawing. What is the electron-pair geometry of bromine pentafluoride? What is the shape (molecular geometry) of bromine pentafluoride? What is the hybridization of the central atom? sp3d2<\/p>\n\n\n\n

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

Lewis Structure of BrF5:<\/h3>\n\n\n\n
    \n
  1. Determine the total number of valence electrons<\/strong>:\n
      \n
    • Bromine (Br) is in Group 17, so it has 7 valence electrons.<\/li>\n\n\n\n
    • Fluorine (F) is also in Group 17, and each fluorine atom has 7 valence electrons. Since there are 5 fluorine atoms, this contributes 5\u00d77=355 \\times 7 = 355\u00d77=35 valence electrons.<\/li>\n\n\n\n
    • Total valence electrons = 7(Br)+35(F)=427 (Br) + 35 (F) = 427(Br)+35(F)=42 valence electrons.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Arrange atoms<\/strong>:
      Bromine is the central atom, as it is less electronegative than fluorine.<\/li>\n\n\n\n
    • Connect the atoms with single bonds<\/strong>:
      Place a single bond between Br and each of the five F atoms. Each single bond represents 2 electrons. For 5 bonds, we use 5\u00d72=105 \\times 2 = 105\u00d72=10 electrons.<\/li>\n\n\n\n
    • Distribute the remaining electrons<\/strong>:
      After forming the bonds, we have 42\u221210=3242 – 10 = 3242\u221210=32 electrons left. These will be placed as lone pairs on the fluorine atoms first. Each fluorine atom needs 3 more pairs (6 electrons) to complete its octet. This uses up 5\u00d76=305 \\times 6 = 305\u00d76=30 electrons, leaving 2 electrons.<\/li>\n\n\n\n
    • Place the remaining electrons on the central atom (Br)<\/strong>:
      Bromine, being in Period 4, can accommodate more than 8 electrons. The 2 remaining electrons are placed as a lone pair on the bromine atom.<\/li>\n\n\n\n
    • Final Lewis Structure<\/strong>:
      Bromine is surrounded by five single bonds to fluorine atoms, with each fluorine having 3 lone pairs. Bromine also has one lone pair.<\/li>\n<\/ol>\n\n\n\n

      Electron-pair Geometry and Molecular Geometry:<\/h3>\n\n\n\n
        \n
      • Electron-pair Geometry<\/strong>: Bromine pentafluoride (BrF5) has 6 regions of electron density (5 bonding pairs and 1 lone pair). According to the VSEPR (Valence Shell Electron Pair Repulsion) theory, the electron-pair geometry around the central atom is octahedral<\/strong>.<\/li>\n\n\n\n
      • Molecular Geometry<\/strong>: The lone pair on bromine occupies one of the positions in the octahedral geometry, leaving the fluorine atoms in a square pyramidal<\/strong> arrangement. Therefore, the molecular geometry of BrF5 is square pyramidal<\/strong>.<\/li>\n<\/ul>\n\n\n\n

        Hybridization of the Central Atom:<\/h3>\n\n\n\n
          \n
        • The central bromine atom is involved in 5 bonds and 1 lone pair, which requires 6 orbitals. The hybridization of the central atom is sp3d2<\/strong>, which corresponds to an octahedral geometry.<\/li>\n<\/ul>\n\n\n\n

          Summary:<\/h3>\n\n\n\n
            \n
          • Electron-pair Geometry<\/strong>: Octahedral<\/li>\n\n\n\n
          • Molecular Geometry<\/strong>: Square pyramidal<\/li>\n\n\n\n
          • Hybridization<\/strong>: sp3d2<\/li>\n<\/ul>\n\n\n\n

            This structure reflects the bonding and geometry of BrF5 as derived from VSEPR theory and the hybridization concept.<\/p>\n\n\n\n

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

            Draw the Lewis structure of bromine pentafluoride, BrF5. Do not include brackets or formal charges. If the molecule exhibits resonance, draw only one (1) of the resonance forms. Hartree-Fock method can be used to help with the drawing. Edit drawing. What is the electron-pair geometry of bromine pentafluoride? What is the shape (molecular geometry) of […]<\/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-267075","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267075","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=267075"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/267075\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=267075"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=267075"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=267075"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}