{"id":231751,"date":"2025-06-11T11:07:01","date_gmt":"2025-06-11T11:07:01","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231751"},"modified":"2025-06-11T11:07:03","modified_gmt":"2025-06-11T11:07:03","slug":"what-is-the-bond-order-for-o2-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/what-is-the-bond-order-for-o2-2\/","title":{"rendered":"What is the bond order for O2^-2"},"content":{"rendered":"\n

What is the bond order for O2^-2<\/p>\n\n\n\n

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

The bond order for O22\u2212\\text{O}_2^{2-}O22\u2212\u200b (the peroxide ion) is 1<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

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

To determine the bond order of a molecule or ion, we use molecular orbital (MO) theory<\/strong>. According to MO theory, atomic orbitals combine to form molecular orbitals that are spread over the entire molecule. Electrons are placed into these molecular orbitals in order of increasing energy.<\/p>\n\n\n\n

The bond order<\/strong> is calculated using the formula:Bond Order=12(Nb\u2212Na)\\text{Bond Order} = \\frac{1}{2} \\left( N_b – N_a \\right)Bond Order=21\u200b(Nb\u200b\u2212Na\u200b)<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • NbN_bNb\u200b = number of electrons in bonding<\/strong> molecular orbitals<\/li>\n\n\n\n
  • NaN_aNa\u200b = number of electrons in antibonding<\/strong> molecular orbitals<\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step-by-step:<\/h3>\n\n\n\n

    Oxygen has an atomic number of 8, so a neutral O2\\text{O}_2O2\u200b molecule has 16 electrons. The O22\u2212\\text{O}_2^{2-}O22\u2212\u200b ion has 2 extra electrons, giving a total of 18 electrons<\/strong>.<\/p>\n\n\n\n

    For diatomic molecules formed from elements with atomic number > 7 (like oxygen), the MO filling order is:\u03c31s,\u03c31s\u2217,\u03c32s,\u03c32s\u2217,\u03c32pz,\u03c02px=\u03c02py,\u03c02px\u2217=\u03c02py\u2217,\u03c32pz\u2217\\sigma_{1s}, \\sigma^*_{1s}, \\sigma_{2s}, \\sigma^*_{2s}, \\sigma_{2p_z}, \\pi_{2p_x} = \\pi_{2p_y}, \\pi^*_{2p_x} = \\pi^*_{2p_y}, \\sigma^*_{2p_z}\u03c31s\u200b,\u03c31s\u2217\u200b,\u03c32s\u200b,\u03c32s\u2217\u200b,\u03c32pz\u200b\u200b,\u03c02px\u200b\u200b=\u03c02py\u200b\u200b,\u03c02px\u200b\u2217\u200b=\u03c02py\u200b\u2217\u200b,\u03c32pz\u200b\u2217\u200b<\/p>\n\n\n\n

    Now distribute the 18 electrons into these orbitals:<\/p>\n\n\n\n

      \n
    • \u03c31s\\sigma_{1s}\u03c31s\u200b: 2 electrons (bonding)<\/li>\n\n\n\n
    • \u03c31s\u2217\\sigma^*_{1s}\u03c31s\u2217\u200b: 2 electrons (antibonding)<\/li>\n\n\n\n
    • \u03c32s\\sigma_{2s}\u03c32s\u200b: 2 electrons (bonding)<\/li>\n\n\n\n
    • \u03c32s\u2217\\sigma^*_{2s}\u03c32s\u2217\u200b: 2 electrons (antibonding)<\/li>\n\n\n\n
    • \u03c32pz\\sigma_{2p_z}\u03c32pz\u200b\u200b: 2 electrons (bonding)<\/li>\n\n\n\n
    • \u03c02px\\pi_{2p_x}\u03c02px\u200b\u200b, \u03c02py\\pi_{2p_y}\u03c02py\u200b\u200b: 4 electrons (2 each in bonding)<\/li>\n\n\n\n
    • \u03c02px\u2217\\pi^*_{2p_x}\u03c02px\u200b\u2217\u200b, \u03c02py\u2217\\pi^*_{2p_y}\u03c02py\u200b\u2217\u200b: 4 electrons (2 each in antibonding)<\/li>\n<\/ul>\n\n\n\n

      Tallying:<\/p>\n\n\n\n

        \n
      • Bonding electrons = 2 (\u03c31s) + 2 (\u03c32s) + 2 (\u03c32p_z) + 4 (\u03c02p_x and \u03c02p_y) = 10<\/li>\n\n\n\n
      • Antibonding electrons = 2 (\u03c31s) + 2 (\u03c3<\/em>2s) + 4 (\u03c02p_x and \u03c0<\/em>2p_y) = 8<\/li>\n<\/ul>\n\n\n\n

        Bond Order=12(10\u22128)=1\\text{Bond Order} = \\frac{1}{2} (10 – 8) = 1Bond Order=21\u200b(10\u22128)=1<\/p>\n\n\n\n

        Thus, the bond order of O22\u2212\\text{O}_2^{2-}O22\u2212\u200b is 1<\/strong>, indicating a single bond between the oxygen atoms.<\/p>\n\n\n\n

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

        What is the bond order for O2^-2 The Correct Answer and Explanation is: The bond order for O22\u2212\\text{O}_2^{2-}O22\u2212\u200b (the peroxide ion) is 1. Explanation: To determine the bond order of a molecule or ion, we use molecular orbital (MO) theory. According to MO theory, atomic orbitals combine to form molecular orbitals that are spread over […]<\/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-231751","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231751","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=231751"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231751\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231751"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231751"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231751"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}