{"id":217602,"date":"2025-05-22T07:47:08","date_gmt":"2025-05-22T07:47:08","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217602"},"modified":"2025-05-22T07:47:12","modified_gmt":"2025-05-22T07:47:12","slug":"draw-a-molecular-orbital-energy-level-diagram-for-the-gaseous-heteronuclear-diatomic-molecule-boron-nitride-bn","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/draw-a-molecular-orbital-energy-level-diagram-for-the-gaseous-heteronuclear-diatomic-molecule-boron-nitride-bn\/","title":{"rendered":"Draw a molecular orbital energy-level diagram for the gaseous heteronuclear diatomic molecule boron nitride, BN."},"content":{"rendered":"\n

Draw a molecular orbital energy-level diagram for the gaseous heteronuclear diatomic molecule boron nitride, BN. How does it differ from that for C2?<\/p>\n\n\n\n

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

\u2705 Molecular Orbital (MO) Energy-Level Diagram for BN (Boron Nitride)<\/h3>\n\n\n\n

Molecular Formula: BN<\/h4>\n\n\n\n
    \n
  • Boron (B): Atomic number 5<\/li>\n\n\n\n
  • Nitrogen (N): Atomic number 7<\/li>\n\n\n\n
  • Total electrons in BN: 5 (B) + 7 (N) = 12 electrons<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    \ud83e\uddec MO Energy-Level Diagram for BN:<\/h3>\n\n\n\n

    Because BN is a heteronuclear diatomic molecule, the energy levels of atomic orbitals from B and N differ (N is more electronegative, so its atomic orbitals are lower in energy). The MO diagram is based on the same general pattern as for B\u2082, C\u2082, and N\u2082, with some asymmetry due to different atomic orbital energies.<\/p>\n\n\n\n

    BN MO Diagram:<\/h4>\n\n\n\n

    (Molecular orbitals filled with 12 electrons)<\/p>\n\n\n\n

      \n
    1. \u03c3(1s)<\/strong> \u2013 2 electrons<\/li>\n\n\n\n
    2. \u03c3(1s)<\/em>* \u2013 2 electrons<\/li>\n\n\n\n
    3. \u03c3(2s)<\/strong> \u2013 2 electrons<\/li>\n\n\n\n
    4. \u03c3(2s)<\/em>* \u2013 2 electrons<\/li>\n\n\n\n
    5. \u03c0(2p_x) = \u03c0(2p_y)<\/strong> \u2013 4 electrons<\/li>\n\n\n\n
    6. No electrons in \u03c0(2p) or \u03c3(2p_z)<\/em>*<\/li>\n<\/ol>\n\n\n\n

      Therefore, the electron configuration is:
      (\u03c31s)\u00b2 (\u03c31s)\u00b2 (\u03c32s)\u00b2 (\u03c3<\/em>2s)\u00b2 (\u03c02p_x = \u03c02p_y)\u2074<\/p>\n\n\n\n

      \n\n\n\n

      \ud83d\udd01 How BN’s MO Diagram Differs from C\u2082<\/h3>\n\n\n\n

      C\u2082 (Carbon Dimer):<\/strong><\/p>\n\n\n\n

        \n
      • Each carbon atom contributes 6 electrons, so C\u2082 has 12 electrons\u2014same as BN.<\/li>\n\n\n\n
      • However, because C\u2082 is homonuclear, the energy levels of the atomic orbitals are the same.<\/li>\n\n\n\n
      • In C\u2082, due to the mixing of s and p orbitals (sp mixing), the order of 2p MOs is:
        \u03c0(2p_x) = \u03c0(2p_y) < \u03c3(2p_z)<\/li>\n<\/ul>\n\n\n\n

        BN Difference:<\/strong><\/p>\n\n\n\n

          \n
        • BN is heteronuclear<\/em> \u2192 no perfect orbital symmetry.<\/li>\n\n\n\n
        • Nitrogen\u2019s atomic orbitals are lower in energy due to higher electronegativity.<\/li>\n\n\n\n
        • Energy asymmetry leads to polar covalent bonding<\/strong>.<\/li>\n\n\n\n
        • The MO diagram is asymmetric<\/strong>, with more contribution from nitrogen to lower-energy orbitals.<\/li>\n\n\n\n
        • Bonding molecular orbitals are closer to nitrogen, antibonding closer to boron.<\/li>\n\n\n\n
        • This results in bond polarity<\/strong>, unlike the nonpolar C\u2082 molecule.<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          \ud83e\uddea Summary:<\/h3>\n\n\n\n
            \n
          • BN and C\u2082 have the same number of electrons (12), so they fill orbitals similarly.<\/li>\n\n\n\n
          • However, BN\u2019s MOs are asymmetric<\/strong> due to differing atom electronegativities.<\/li>\n\n\n\n
          • This causes unequal sharing of electrons<\/strong> and partial ionic character in BN, unlike the symmetrical and purely covalent C\u2082.<\/li>\n<\/ul>\n\n\n\n

            \ud83e\uddee Bond Order for BN<\/strong> = (8 bonding – 4 antibonding) \u00f7 2 = 2<\/strong><\/p>\n\n\n\n

            \u2705 Final Answer:
            BN\u2019s MO diagram resembles that of C\u2082 in orbital filling but differs due to the asymmetric energy levels<\/strong> caused by different electronegativities.<\/p>\n\n\n\n

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

            Draw a molecular orbital energy-level diagram for the gaseous heteronuclear diatomic molecule boron nitride, BN. How does it differ from that for C2? The Correct Answer and Explanation is: \u2705 Molecular Orbital (MO) Energy-Level Diagram for BN (Boron Nitride) Molecular Formula: BN \ud83e\uddec MO Energy-Level Diagram for BN: Because BN is a heteronuclear diatomic molecule, […]<\/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-217602","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217602","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=217602"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217602\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217602"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217602"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217602"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}