{"id":230759,"date":"2025-06-10T05:29:36","date_gmt":"2025-06-10T05:29:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230759"},"modified":"2025-06-10T05:29:38","modified_gmt":"2025-06-10T05:29:38","slug":"show-the-molecular-orbital-diagram-for-bn-boron-nitride-calculate-its-bond-order-and-compare-its-stability-to-bc-boron-carbide-which-is-paramagnetic","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/show-the-molecular-orbital-diagram-for-bn-boron-nitride-calculate-its-bond-order-and-compare-its-stability-to-bc-boron-carbide-which-is-paramagnetic\/","title":{"rendered":"Show the molecular orbital diagram for BN (boron nitride), calculate its bond order and compare its stability to BC (boron carbide) – which is paramagnetic"},"content":{"rendered":"\n

Show the molecular orbital diagram for BN (boron nitride), calculate its bond order and compare its stability to BC (boron carbide) – which is paramagnetic? Which is diamagnetic<\/p>\n\n\n\n

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

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

To solve this, we analyze the molecular orbital (MO) diagrams for BN (boron nitride)<\/strong> and BC (boron carbide)<\/strong> and use electron configurations to determine bond order, magnetism, and stability.<\/p>\n\n\n\n

\n\n\n\n

1. MO Diagram Overview<\/strong><\/h3>\n\n\n\n
    \n
  • Left MO diagram (BN)<\/strong>: Resembles that of diatomic molecules like CO or NO, where nitrogen is more electronegative.<\/li>\n\n\n\n
  • Right MO diagram (BC)<\/strong>: Resembles earlier-period diatomic molecules like B\u2082 or C\u2082.<\/li>\n<\/ul>\n\n\n\n

    The key difference lies in the energy order of molecular orbitals:<\/p>\n\n\n\n

      \n
    • For BN (left): the ordering is affected by nitrogen\u2019s higher electronegativity, placing \u03c32p lower than \u03c02p.<\/li>\n\n\n\n
    • For BC (right): due to smaller electronegativity difference, \u03c02p < \u03c32p.<\/li>\n<\/ul>\n\n\n\n
      \n\n\n\n

      2. Electron Count<\/strong><\/h3>\n\n\n\n
        \n
      • Boron (B)<\/strong> = 5 electrons<\/li>\n\n\n\n
      • Nitrogen (N)<\/strong> = 7 electrons<\/li>\n\n\n\n
      • Carbon (C)<\/strong> = 6 electrons<\/li>\n<\/ul>\n\n\n\n

        So:<\/p>\n\n\n\n

          \n
        • BN<\/strong> = 5 + 7 = 12 electrons<\/li>\n\n\n\n
        • BC<\/strong> = 5 + 6 = 11 electrons<\/li>\n<\/ul>\n\n\n\n
          \n\n\n\n

          3. Bond Order Calculation<\/strong><\/h3>\n\n\n\n

          Bond order = (Number of bonding electrons \u2013 Number of antibonding electrons) \/ 2<\/strong><\/p>\n\n\n\n

          BN (12 electrons):<\/h4>\n\n\n\n

          Filling the MOs:
          \u03c31s\u00b2, \u03c31s\u00b2, \u03c32s\u00b2, \u03c32s<\/em>\u00b2, \u03c02p\u2074<\/p>\n\n\n\n

            \n
          • Bonding electrons: 2 (\u03c32s) + 4 (\u03c02p) = 6<\/li>\n\n\n\n
          • Antibonding electrons: 2 (\u03c32s*) = 2
            Bond order = (8 \u2013 2)\/2 = 3<\/strong><\/li>\n<\/ul>\n\n\n\n

            \u2192 BN has bond order = 3<\/strong>
            \u2192 All electrons paired \u2192 Diamagnetic<\/strong><\/p>\n\n\n\n

            BC (11 electrons):<\/h4>\n\n\n\n

            Same orbitals, but one fewer electron.
            \u03c31s\u00b2, \u03c31s\u00b2, \u03c32s\u00b2, \u03c32s<\/em>\u00b2, \u03c02p\u00b3<\/p>\n\n\n\n

              \n
            • Bonding electrons: 2 (\u03c32s) + 3 (\u03c02p) = 5<\/li>\n\n\n\n
            • Antibonding: 2 (\u03c32s*) = 2
              Bond order = (7 \u2013 2)\/2 = 2.5<\/strong><\/li>\n<\/ul>\n\n\n\n

              \u2192 BC has bond order = 2.5<\/strong>
              \u2192 Unpaired electron in \u03c02p \u2192 Paramagnetic<\/strong><\/p>\n\n\n\n

              \n\n\n\n

              4. Stability Comparison<\/strong><\/h3>\n\n\n\n
                \n
              • BN (bond order 3)<\/strong> is more stable<\/strong> than BC (bond order 2.5)<\/strong>.<\/li>\n\n\n\n
              • Diamagnetic (BN)<\/strong> indicates all electrons are paired \u2192 typically more stable.<\/li>\n\n\n\n
              • Paramagnetic (BC)<\/strong> has unpaired electrons \u2192 less stable.<\/li>\n<\/ul>\n\n\n\n
                \n\n\n\n

                Final Answer Summary<\/strong>:<\/h3>\n\n\n\n
                  \n
                • BN<\/strong>: Bond order = 3, Diamagnetic<\/strong>, More stable<\/strong><\/li>\n\n\n\n
                • BC<\/strong>: Bond order = 2.5, Paramagnetic<\/strong>, Less stable<\/strong><\/li>\n<\/ul>\n\n\n\n
                  \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                  Show the molecular orbital diagram for BN (boron nitride), calculate its bond order and compare its stability to BC (boron carbide) – which is paramagnetic? Which is diamagnetic The Correct Answer and Explanation is: To solve this, we analyze the molecular orbital (MO) diagrams for BN (boron nitride) and BC (boron carbide) and use electron […]<\/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-230759","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230759","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=230759"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230759\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230759"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230759"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230759"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}