{"id":190747,"date":"2025-02-13T07:57:26","date_gmt":"2025-02-13T07:57:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190747"},"modified":"2025-02-13T07:57:28","modified_gmt":"2025-02-13T07:57:28","slug":"draw-a-molecular-orbital-diagram-for-f22","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/13\/draw-a-molecular-orbital-diagram-for-f22\/","title":{"rendered":"Draw a molecular orbital diagram for F2^2-"},"content":{"rendered":"\n

Draw a molecular orbital diagram for F2^2- . Calculate the bond order and magnetic behavior.<\/p>\n\n\n\n

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

Molecular Orbital (MO) Diagram for (F_2^{2-})<\/h3>\n\n\n\n
    \n
  1. Atomic Orbitals Combination<\/strong>:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Fluorine has an atomic number of 9, so its electron configuration is (1s^2 2s^2 2p^5).<\/li>\n\n\n\n
    • In the molecular orbital diagram for diatomic fluorine, atomic orbitals combine to form molecular orbitals in the following order: [
      \\sigma(2s) < \\sigma^(2s) < \\pi(2p) < \\sigma(2p) < \\pi^<\/em>(2p) < \\sigma^*(2p)
      ]<\/li>\n\n\n\n
    • This order follows from the energy interactions where, for elements with atomic numbers greater than 7, the (\\sigma(2p)) orbital is lower in energy than the (\\pi(2p)) orbitals.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Electron Filling in (F_2^{2-})<\/strong>:<\/li>\n<\/ol>\n\n\n\n
          \n
        • The neutral (F_2) molecule has 18 valence electrons ((2 \\times 9) from fluorine atoms, minus 2 from the filled (1s) core orbitals).<\/li>\n\n\n\n
        • The (2-) charge adds two more electrons, making a total of 20 valence electrons<\/strong>.<\/li>\n\n\n\n
        • These electrons are distributed among the molecular orbitals as follows:\n
            \n
          • ( \\sigma(2s) )<\/strong>: 2 electrons<\/li>\n\n\n\n
          • ( \\sigma^*(2s) )<\/strong>: 2 electrons<\/li>\n\n\n\n
          • ( \\pi(2p) )<\/strong>: 4 electrons<\/li>\n\n\n\n
          • ( \\sigma(2p) )<\/strong>: 2 electrons<\/li>\n\n\n\n
          • ( \\pi^*(2p) )<\/strong>: 4 electrons<\/li>\n\n\n\n
          • ( \\sigma^*(2p) )<\/strong>: 2 electrons<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
            \n\n\n\n

            Bond Order Calculation<\/h3>\n\n\n\n

            Bond order is given by:<\/p>\n\n\n\n

            [
            \\text{Bond Order} = \\frac{\\text{Bonding electrons} – \\text{Antibonding electrons}}{2}
            ]<\/p>\n\n\n\n

              \n
            • Bonding electrons<\/strong>: (2+4+2 = 8)<\/li>\n\n\n\n
            • Antibonding electrons<\/strong>: (2+4+2 = 8)<\/li>\n<\/ul>\n\n\n\n

              [
              \\text{Bond Order} = \\frac{8 – 8}{2} = 0
              ]<\/p>\n\n\n\n

              Since the bond order is zero<\/strong>, the (F_2^{2-}) ion does not exist<\/strong> as a stable molecule.<\/p>\n\n\n\n

              \n\n\n\n

              Magnetic Behavior<\/h3>\n\n\n\n

              A molecule is paramagnetic<\/strong> if it has unpaired electrons and diamagnetic<\/strong> if all electrons are paired.<\/p>\n\n\n\n

                \n
              • In (F_2^{2-}), all electrons are paired.<\/li>\n\n\n\n
              • Thus, (F_2^{2-}) is diamagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n

                Conclusion<\/h3>\n\n\n\n
                  \n
                • MO Diagram<\/strong>: Shows complete filling of bonding and antibonding orbitals.<\/li>\n\n\n\n
                • Bond Order<\/strong>: 0<\/strong> (indicating instability).<\/li>\n\n\n\n
                • Magnetic Behavior<\/strong>: Diamagnetic<\/strong> (all electrons are paired).<\/li>\n\n\n\n
                • (F_2^{2-}) is not a stable molecule<\/strong> as its bond order is zero, meaning no net bonding interaction exists between fluorine atoms.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                  Draw a molecular orbital diagram for F2^2- . Calculate the bond order and magnetic behavior. The Correct Answer and Explanation is : Molecular Orbital (MO) Diagram for (F_2^{2-}) Bond Order Calculation Bond order is given by: [\\text{Bond Order} = \\frac{\\text{Bonding electrons} – \\text{Antibonding electrons}}{2}] [\\text{Bond Order} = \\frac{8 – 8}{2} = 0] Since the bond […]<\/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-190747","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190747","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=190747"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190747\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190747"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190747"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190747"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}