{"id":249592,"date":"2025-07-09T17:37:28","date_gmt":"2025-07-09T17:37:28","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=249592"},"modified":"2025-07-09T17:37:30","modified_gmt":"2025-07-09T17:37:30","slug":"describe-the-bonding-in-the-c2-2-ion-in-terms-of-molecular-orbital-theory-and-compare-the-bond-order-to-that-of-c2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/09\/describe-the-bonding-in-the-c2-2-ion-in-terms-of-molecular-orbital-theory-and-compare-the-bond-order-to-that-of-c2\/","title":{"rendered":"Describe the bonding in the C2 2- ion in terms of molecular orbital theory and compare the bond order to that of C2."},"content":{"rendered":"\n

Describe the bonding in the C2 2- ion in terms of molecular orbital theory and compare the bond order to that of C2.<\/p>\n\n\n\n

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

The bonding in the C\u2082\u00b2\u207b ion<\/strong> (C\u2082 with two extra electrons) can be explained using molecular orbital theory, which describes bonding in terms of the combination of atomic orbitals from the individual atoms to form molecular orbitals. The two carbon atoms in C\u2082\u00b2\u207b are held together by molecular orbitals that result from the combination of the atomic orbitals of each carbon.<\/p>\n\n\n\n

Electron Configuration in Molecular Orbitals<\/h3>\n\n\n\n

In molecular orbital theory, the atomic orbitals combine to form bonding and antibonding molecular orbitals. The molecular orbitals for diatomic molecules like C\u2082 are:<\/p>\n\n\n\n

    \n
  • \u03c3(1s)<\/strong> and \u03c3(1s)<\/em>* (bonding and antibonding orbitals from the 1s orbitals)<\/li>\n\n\n\n
  • \u03c3(2s)<\/strong> and \u03c3(2s)<\/em>* (bonding and antibonding orbitals from the 2s orbitals)<\/li>\n\n\n\n
  • \u03c0(2px)<\/strong> and \u03c0(2py)<\/strong> (bonding orbitals from the 2p orbitals)<\/li>\n\n\n\n
  • \u03c0(2px)<\/em>* and \u03c0(2py)<\/em>* (antibonding orbitals from the 2p orbitals)<\/li>\n\n\n\n
  • \u03c3(2pz)<\/strong> (bonding orbital from the 2p orbitals)<\/li>\n\n\n\n
  • \u03c3(2pz)<\/em>* (antibonding orbital from the 2p orbitals)<\/li>\n<\/ul>\n\n\n\n

    For C\u2082<\/strong> (the neutral molecule), each carbon atom has 4 valence electrons, so in C\u2082, there are a total of 8 electrons to distribute in the molecular orbitals.<\/p>\n\n\n\n

    The electron configuration for C\u2082<\/strong> is:<\/p>\n\n\n\n

      \n
    • \u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2px)\u00b2, \u03c0(2py)\u00b2 (6 bonding electrons)<\/li>\n\n\n\n
    • \u03c0*(2px)\u00b9, \u03c0*(2py)\u00b9 (2 antibonding electrons)<\/li>\n<\/ul>\n\n\n\n

      The bond order for C\u2082<\/strong> is calculated as:Bond Order=(Number of bonding electrons) – (Number of antibonding electrons)2\\text{Bond Order} = \\frac{\\text{(Number of bonding electrons) – (Number of antibonding electrons)}}{2}Bond Order=2(Number of bonding electrons) – (Number of antibonding electrons)\u200bBond Order for C\u2082=8\u221242=2\\text{Bond Order for C\u2082} = \\frac{8 – 4}{2} = 2Bond Order for C\u2082=28\u22124\u200b=2<\/p>\n\n\n\n

      Bonding in C\u2082\u00b2\u207b<\/h3>\n\n\n\n

      For C\u2082\u00b2\u207b<\/strong>, there are two extra electrons, so the electron configuration becomes:<\/p>\n\n\n\n

        \n
      • \u03c3(1s)\u00b2, \u03c3*(1s)\u00b2, \u03c3(2s)\u00b2, \u03c3*(2s)\u00b2, \u03c0(2px)\u00b2, \u03c0(2py)\u00b2 (6 bonding electrons)<\/li>\n\n\n\n
      • \u03c0*(2px)\u00b2, \u03c0*(2py)\u00b2 (4 antibonding electrons)<\/li>\n<\/ul>\n\n\n\n

        Now, the bond order for C\u2082\u00b2\u207b<\/strong> is:Bond Order for C\u2082\u00b2\u207b=8\u221262=1\\text{Bond Order for C\u2082\u00b2\u207b} = \\frac{8 – 6}{2} = 1Bond Order for C\u2082\u00b2\u207b=28\u22126\u200b=1<\/p>\n\n\n\n

        Comparison of Bond Orders<\/h3>\n\n\n\n
          \n
        • C\u2082<\/strong> has a bond order of 2<\/strong>, indicating a relatively strong bond between the two carbon atoms.<\/li>\n\n\n\n
        • C\u2082\u00b2\u207b<\/strong> has a bond order of 1<\/strong>, which indicates a weaker bond due to the additional electrons that fill antibonding orbitals, reducing the overall bonding strength.<\/li>\n<\/ul>\n\n\n\n

          Thus, the bond in C\u2082\u00b2\u207b is weaker compared to the bond in C\u2082, as the extra electrons in antibonding orbitals result in a lower bond order and, consequently, a less stable molecule.<\/p>\n\n\n\n

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

          Describe the bonding in the C2 2- ion in terms of molecular orbital theory and compare the bond order to that of C2. The Correct Answer and Explanation is: The bonding in the C\u2082\u00b2\u207b ion (C\u2082 with two extra electrons) can be explained using molecular orbital theory, which describes bonding in terms of the combination […]<\/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-249592","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249592","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=249592"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249592\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=249592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=249592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=249592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}