{"id":189304,"date":"2025-02-09T04:23:25","date_gmt":"2025-02-09T04:23:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189304"},"modified":"2025-02-09T04:23:28","modified_gmt":"2025-02-09T04:23:28","slug":"construct-the-molecular-orbital-diagram-for-h2-and-then-identify-the-bond-order-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/09\/construct-the-molecular-orbital-diagram-for-h2-and-then-identify-the-bond-order-2\/","title":{"rendered":"Construct the molecular orbital diagram for H2 and then identify the bond order"},"content":{"rendered":"\n

Construct the molecular orbital diagram for H2 and then identify the bond order. 1S Bond order<\/p>\n\n\n\n

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

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

The molecular orbital (MO) diagram for the hydrogen molecule (H\u2082) can be constructed by considering the atomic orbitals of the two hydrogen atoms and how they combine to form molecular orbitals. In this case, hydrogen atoms each have a 1s atomic orbital.<\/p>\n\n\n\n

Steps to Construct the MO Diagram for H\u2082:<\/h3>\n\n\n\n
    \n
  1. Atomic Orbitals (AO):<\/strong>
    Each hydrogen atom has a 1s atomic orbital. These atomic orbitals will combine to form molecular orbitals.<\/li>\n\n\n\n
  2. Combination of Orbitals:<\/strong>
    When two atomic orbitals overlap, they form two molecular orbitals:<\/li>\n<\/ol>\n\n\n\n
      \n
    • Bonding Molecular Orbital (\u03c3\u2081s)<\/strong>: This is formed when the atomic orbitals combine constructively, resulting in electron density between the two nuclei.<\/li>\n\n\n\n
    • Antibonding Molecular Orbital (\u03c3*\u2081s)<\/strong>: This is formed when the atomic orbitals combine destructively, creating a region of reduced electron density between the nuclei, leading to an antibonding interaction.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Energy Level Arrangement:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • The bonding molecular orbital (\u03c3\u2081s) is lower in energy than the atomic orbitals.<\/li>\n\n\n\n
        • The antibonding molecular orbital (\u03c3*\u2081s) is higher in energy than the atomic orbitals.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Electron Filling:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • Each hydrogen atom has one electron, so in H\u2082, there are two electrons to place in the molecular orbitals.<\/li>\n\n\n\n
            • The two electrons will occupy the lower energy bonding molecular orbital (\u03c3\u2081s).<\/li>\n<\/ul>\n\n\n\n

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

              Bond order is calculated using the formula:<\/p>\n\n\n\n

              [
              \\text{Bond Order} = \\frac{1}{2} \\left( \\text{Number of electrons in bonding MOs} – \\text{Number of electrons in antibonding MOs} \\right)
              ]<\/p>\n\n\n\n

              For H\u2082:<\/p>\n\n\n\n

                \n
              • There are 2 electrons in the bonding molecular orbital (\u03c3\u2081s).<\/li>\n\n\n\n
              • There are 0 electrons in the antibonding molecular orbital (\u03c3*\u2081s).<\/li>\n<\/ul>\n\n\n\n

                Thus, the bond order is:<\/p>\n\n\n\n

                [
                \\text{Bond Order} = \\frac{1}{2} (2 – 0) = 1
                ]<\/p>\n\n\n\n

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

                The bond order of 1 indicates that the H\u2082 molecule has a single bond between the two hydrogen atoms. This is consistent with the fact that H\u2082 forms a stable molecule with a single covalent bond between its atoms. The presence of two electrons in the bonding molecular orbital contributes to the attraction between the nuclei, leading to a stable bond. The antibonding orbital is empty in the ground state of the molecule, so it does not affect the bond strength. Therefore, the bond order of 1 is indicative of a single, stable covalent bond in the H\u2082 molecule.<\/p>\n","protected":false},"excerpt":{"rendered":"

                Construct the molecular orbital diagram for H2 and then identify the bond order. 1S Bond order The Correct Answer and Explanation is : The molecular orbital (MO) diagram for the hydrogen molecule (H\u2082) can be constructed by considering the atomic orbitals of the two hydrogen atoms and how they combine to form molecular orbitals. In […]<\/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-189304","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189304","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=189304"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189304\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189304"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189304"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189304"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}