{"id":198746,"date":"2025-03-10T10:08:12","date_gmt":"2025-03-10T10:08:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=198746"},"modified":"2025-03-10T10:08:15","modified_gmt":"2025-03-10T10:08:15","slug":"draw-a-molecular-orbital-diagram-and-use-it-to-determine-which-of-the-following-is-paramagnetic","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/10\/draw-a-molecular-orbital-diagram-and-use-it-to-determine-which-of-the-following-is-paramagnetic\/","title":{"rendered":"Draw a molecular orbital diagram and use it to determine which of the following is paramagnetic"},"content":{"rendered":"\n

Draw a molecular orbital diagram and use it to determine which of the following is paramagnetic. F22+ Ne22+ O22- O22+ None of the above is paramagnetic. Draw a molecular orbital diagram and use it to determine which of the following is paramagnetic.
F22+
Ne22+
O22-
O22+
None of the above is paramagnetic.<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

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

Among the given species\u2014F\u2082\u00b2\u207a, Ne\u2082\u00b2\u207a, O\u2082\u00b2\u207b, and O\u2082\u00b2\u207a\u2014the paramagnetic species is F\u2082\u00b2\u207a. This conclusion is derived through molecular orbital (MO) theory, which helps predict the magnetic properties of molecules based on their electron configurations.<\/p>\n\n\n\n

Molecular Orbital Theory Overview<\/strong><\/p>\n\n\n\n

MO theory describes the behavior of electrons in molecules by combining atomic orbitals to form molecular orbitals, which are classified as bonding, antibonding, or non-bonding. Electrons in bonding orbitals stabilize the molecule, while those in antibonding orbitals destabilize it. The bond order, calculated as half the difference between the number of bonding and antibonding electrons, indicates the strength and stability of a bond:<\/p>\n\n\n\n

[ \\text{Bond Order} = \\frac{(\\text{Number of Bonding Electrons}) – (\\text{Number of Antibonding Electrons})}{2} ]<\/p>\n\n\n\n

A bond order greater than zero suggests a stable bond. Additionally, the presence of unpaired electrons in molecular orbitals renders a molecule paramagnetic, meaning it is attracted to magnetic fields. Conversely, molecules with all electrons paired are diamagnetic and are slightly repelled by magnetic fields.<\/p>\n\n\n\n

Analysis of Each Species<\/strong><\/p>\n\n\n\n

    \n
  1. F\u2082\u00b2\u207a (Dication of Fluorine)<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Total Electrons:<\/strong> 14 (each fluorine atom has 9 electrons; removing two electrons for the 2+ charge results in 14 electrons)\ue206<\/li>\n\n\n\n
    • MO Configuration:<\/strong> \u03c3(2s)\u00b2, \u03c3(2s)\u00b2, \u03c3(2p)\u00b2, \u03c0(2p)\u2074, \u03c0<\/em>(2p)\u00b2<\/li>\n\n\n\n
    • Bond Order Calculation:<\/strong>\n
        \n
      • Bonding Electrons: 2 (\u03c32s) + 2 (\u03c32p) + 4 (\u03c02p) = 8<\/li>\n\n\n\n
      • Antibonding Electrons: 2 (\u03c32s) + 2 (\u03c0<\/em>2p) = 4<\/li>\n\n\n\n
      • Bond Order = (8 – 4) \/ 2 = 2<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Magnetic Property:<\/strong> The \u03c0*(2p) molecular orbitals each contain one unpaired electron, leading to two unpaired electrons in total. Therefore, F\u2082\u00b2\u207a is paramagnetic.<\/li>\n<\/ul>\n\n\n\n
          \n
        1. Ne\u2082\u00b2\u207a (Dication of Neon)<\/strong><\/li>\n<\/ol>\n\n\n\n
            \n
          • Total Electrons:<\/strong> 18 (each neon atom has 10 electrons; removing two electrons for the 2+ charge results in 18 electrons)\ue206<\/li>\n\n\n\n
          • MO Configuration:<\/strong> \u03c3(2s)\u00b2, \u03c3(2s)\u00b2, \u03c3(2p)\u00b2, \u03c0(2p)\u2074, \u03c0<\/em>(2p)\u2074, \u03c3*(2p)\u00b2<\/li>\n\n\n\n
          • Bond Order Calculation:<\/strong>\n
              \n
            • Bonding Electrons: 2 (\u03c32s) + 2 (\u03c32p) + 4 (\u03c02p) = 8<\/li>\n\n\n\n
            • Antibonding Electrons: 2 (\u03c32s) + 4 (\u03c0<\/em>2p) + 2 (\u03c3*2p) = 8<\/li>\n\n\n\n
            • Bond Order = (8 – 8) \/ 2 = 0<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Magnetic Property:<\/strong> All electrons are paired. However, a bond order of zero indicates that Ne\u2082\u00b2\u207a is not a stable molecule and does not exist under normal conditions.<\/li>\n<\/ul>\n\n\n\n
                \n
              1. O\u2082\u00b2\u207b (Peroxide Ion)<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • Total Electrons:<\/strong> 18 (each oxygen atom has 8 electrons; adding two electrons for the 2- charge results in 18 electrons)<\/li>\n\n\n\n
                • MO Configuration:<\/strong> \u03c3(2s)\u00b2, \u03c3(2s)\u00b2, \u03c3(2p)\u00b2, \u03c0(2p)\u2074, \u03c0<\/em>(2p)\u2074, \u03c3*(2p)\u00b2<\/li>\n\n\n\n
                • Bond Order Calculation:<\/strong>\n
                    \n
                  • Bonding Electrons: 2 (\u03c32s) + 2 (\u03c32p) + 4 (\u03c02p) = 8<\/li>\n\n\n\n
                  • Antibonding Electrons: 2 (\u03c32s) + 4 (\u03c0<\/em>2p) + 2 (\u03c3*2p) = 8<\/li>\n\n\n\n
                  • Bond Order = (8 – 8) \/ 2 = 0<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                  • Magnetic Property:<\/strong> All electrons are paired, indicating diamagnetism. However, a bond order of zero suggests that O\u2082\u00b2\u207b is not a stable molecule and does not exist under normal conditions.<\/li>\n<\/ul>\n\n\n\n
                      \n
                    1. O\u2082\u00b2\u207a (Dication of Oxygen)<\/strong><\/li>\n<\/ol>\n\n\n\n
                        \n
                      • Total Electrons:<\/strong> 14 (each oxygen atom has 8 electrons; removing two electrons for the 2+ charge results in 14 electrons)<\/li>\n\n\n\n
                      • MO Configuration:<\/strong> \u03c3(2s)\u00b2, \u03c3(2s)\u00b2, \u03c3(2p)\u00b2, \u03c0(2p)\u2074, \u03c0<\/em>(2p)\u00b2<\/li>\n\n\n\n
                      • Bond Order Calculation:<\/strong>\n
                          \n
                        • Bonding Electrons: 2 (\u03c32s) + 2 (\u03c32p) + 4 (\u03c02p) = 8<\/li>\n\n\n\n
                        • Antibonding Electrons: 2 (\u03c32s) + 2 (\u03c0<\/em>2p) = 4<\/li>\n\n\n\n
                        • Bond Order = (8 – 4) \/ 2 = 2<\/li>\n<\/ul>\n<\/li>\n\n\n\n
                        • Magnetic Property:<\/strong> All electrons are paired, indicating diamagnetism.<\/li>\n<\/ul>\n\n\n\n

                          Conclusion<\/strong><\/p>\n\n\n\n

                          Among the species analyzed, F\u2082\u00b2\u207a has unpaired electrons in its molecular orbitals, rendering it paramagnetic. The other species\u2014Ne\u2082\u00b2\u207a, O<\/p>\n","protected":false},"excerpt":{"rendered":"

                          Draw a molecular orbital diagram and use it to determine which of the following is paramagnetic. F22+ Ne22+ O22- O22+ None of the above is paramagnetic. Draw a molecular orbital diagram and use it to determine which of the following is paramagnetic. F22+Ne22+O22-O22+None of the above is paramagnetic. The correct answer and explanation is : […]<\/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-198746","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198746","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=198746"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198746\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=198746"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=198746"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=198746"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}