{"id":193359,"date":"2025-02-19T10:40:25","date_gmt":"2025-02-19T10:40:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=193359"},"modified":"2025-02-19T10:40:27","modified_gmt":"2025-02-19T10:40:27","slug":"construct-an-mo-diagram-for-the-formation-of-o2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/19\/construct-an-mo-diagram-for-the-formation-of-o2\/","title":{"rendered":"Construct an MO diagram for the formation of O2"},"content":{"rendered":"\n

a. Construct an MO diagram for the formation of O2; use only the valence orbitals of the oxygen atoms. (b) Use the diagram to rationalize the following trend in OO bond distances: O2, 121 pm; [O2] \u00fe, 112 pm; [O2] , 134 pm; [O2] 2, 149 pm. (c) Which of these species are paramagnetic? 2.11 Confirm that the octet rule is obeyed by each of the atoms in the following molecules: (a) CF4, (b) O2, (c) AsBr3, (d) SF2.<\/p>\n\n\n\n

    \n
  1. the octet rule is strictly obeyed.<\/li>\n\n\n\n
  2. One member of each of the following sets of compounds is not isoelectronic with the others. Which one in each set is the odd one out? (a) [NO2] \u00fe, CO2, [NO2] and N3<\/a> [CN] , N2, CO, [NO]\u00fe and [O2] 2 (c) [SiF6] 2 , [PF6] , [AlF6] 3 and [BrF6]<\/li>\n<\/ol>\n\n\n\n

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

    (a) Construct an MO Diagram for the Formation of O\u2082<\/h3>\n\n\n\n

    To construct the Molecular Orbital (MO) diagram for O\u2082, we focus on the valence orbitals of the oxygen atoms. Oxygen has an electron configuration of 1s\u00b2 2s\u00b2 2p\u2074. In the formation of O\u2082, we consider the 2s and 2p orbitals, which will overlap to form bonding and antibonding orbitals.<\/p>\n\n\n\n

    MO diagram for O\u2082:<\/strong><\/p>\n\n\n\n

      \n
    1. The 2s orbitals combine to form two molecular orbitals: a bonding \u03c3(2s) and an antibonding \u03c3*(2s).<\/li>\n\n\n\n
    2. The 2p orbitals combine to form four molecular orbitals: a bonding \u03c3(2p), a bonding \u03c0(2p), an antibonding \u03c0(2p), and an antibonding \u03c3<\/em>(2p).<\/li>\n\n\n\n
    3. For O\u2082, we place the 12 valence electrons into the molecular orbitals. These 12 electrons fill the following orbitals:<\/li>\n<\/ol>\n\n\n\n
        \n
      • \u03c3(2s) (2 electrons)<\/li>\n\n\n\n
      • \u03c3*(2s) (2 electrons)<\/li>\n\n\n\n
      • \u03c3(2p) (2 electrons)<\/li>\n\n\n\n
      • \u03c0(2p) (4 electrons)<\/li>\n\n\n\n
      • \u03c0*(2p) (2 electrons)<\/li>\n<\/ul>\n\n\n\n

        This results in a bond order of 2, which corresponds to the O\u2082 molecule.<\/p>\n\n\n\n

        (b) Bond Distance Trend Explanation<\/h3>\n\n\n\n

        The trend in OO bond distances can be rationalized by considering the number of electrons in the bonding and antibonding orbitals, which affects the bond order:<\/p>\n\n\n\n

          \n
        1. O\u2082 (121 pm)<\/strong>: This is the neutral O\u2082 molecule, where the bond order is 2 (based on the electrons in the bonding and antibonding molecular orbitals).<\/li>\n\n\n\n
        2. [O\u2082]\u207a (112 pm)<\/strong>: The [O\u2082]\u207a ion has one fewer electron, meaning fewer electrons in antibonding orbitals. This increases the bond order to 2.5, which shortens the bond distance.<\/li>\n\n\n\n
        3. [O\u2082]\u207b (134 pm)<\/strong>: The [O\u2082]\u207b ion has one additional electron, which occupies an antibonding orbital. This decreases the bond order to 1.5, increasing the bond distance.<\/li>\n\n\n\n
        4. [O\u2082]\u00b2\u207b (149 pm)<\/strong>: The [O\u2082]\u00b2\u207b ion has two additional electrons in antibonding orbitals, further decreasing the bond order to 1, resulting in a longer bond distance.<\/li>\n<\/ol>\n\n\n\n

          (c) Paramagnetism of Species<\/h3>\n\n\n\n

          Paramagnetic species have unpaired electrons, while diamagnetic species have all paired electrons.<\/p>\n\n\n\n

            \n
          • O\u2082<\/strong>: Paramagnetic (has 2 unpaired electrons in \u03c0* orbitals).<\/li>\n\n\n\n
          • [O\u2082]\u207a<\/strong>: Paramagnetic (has 1 unpaired electron in the \u03c0* orbital).<\/li>\n\n\n\n
          • [O\u2082]\u207b<\/strong>: Paramagnetic (has 1 unpaired electron in the \u03c0 orbital).<\/li>\n\n\n\n
          • [O\u2082]\u00b2\u207b<\/strong>: Diamagnetic (all electrons are paired).<\/li>\n<\/ul>\n\n\n\n

            2.11 Confirming the Octet Rule in Molecules<\/h3>\n\n\n\n

            (a) CF\u2084<\/strong>: Each carbon atom is surrounded by 4 fluorine atoms, and carbon has 8 electrons in its valence shell, obeying the octet rule. Fluorine also satisfies the octet rule with 8 valence electrons each.
            (b) O\u2082<\/strong>: The oxygen atoms in O\u2082 share two electrons between them, and each oxygen has 8 valence electrons, satisfying the octet rule.
            (c) AsBr\u2083<\/strong>: The arsenic atom in AsBr\u2083 has 8 electrons in its valence shell after forming bonds with 3 bromine atoms, and each bromine atom has 8 valence electrons.
            (d) SF\u2082<\/strong>: The sulfur atom in SF\u2082 has 8 electrons in its valence shell after bonding with 2 fluorine atoms, each having 8 electrons in their valence shells.<\/p>\n\n\n\n

            Thus, the octet rule is obeyed by each of these molecules.<\/p>\n\n\n\n

            3. Odd One Out for Isoelectronic Compounds<\/h3>\n\n\n\n

            (a) [NO\u2082]\u207a, CO\u2082, [NO\u2082], [N\u2083]\u207b<\/strong>:<\/p>\n\n\n\n

              \n
            • CO\u2082<\/strong> is not isoelectronic with the others. [NO\u2082]\u207a, [NO\u2082], and [N\u2083]\u207b all have 10 electrons, while CO\u2082 has 16 electrons.<\/li>\n<\/ul>\n\n\n\n

              (b) [CN]\u207b, N\u2082, CO, [NO]\u207a, [O\u2082]\u00b2\u207b<\/strong>:<\/p>\n\n\n\n

                \n
              • [NO]\u207a<\/strong> is not isoelectronic with the others. The other species have 10 electrons, while [NO]\u207a has 9 electrons.<\/li>\n<\/ul>\n\n\n\n

                (c) [SiF\u2086]\u00b2\u207b, [PF\u2086]\u207b, [AlF\u2086]\u00b3\u207b, [BrF\u2086]<\/strong>:<\/p>\n\n\n\n

                  \n
                • [BrF\u2086]<\/strong> is not isoelectronic with the others. The other species have 36 electrons, while [BrF\u2086] has 42 electrons.<\/li>\n<\/ul>\n\n\n\n

                  Let me know if you’d like a visual representation of any of these concepts!<\/p>\n","protected":false},"excerpt":{"rendered":"

                  a. Construct an MO diagram for the formation of O2; use only the valence orbitals of the oxygen atoms. (b) Use the diagram to rationalize the following trend in OO bond distances: O2, 121 pm; [O2] \u00fe, 112 pm; [O2] , 134 pm; [O2] 2, 149 pm. (c) Which of these species are paramagnetic? 2.11 […]<\/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-193359","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193359","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=193359"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193359\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193359"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193359"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193359"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}