{"id":230764,"date":"2025-06-10T05:43:59","date_gmt":"2025-06-10T05:43:59","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230764"},"modified":"2025-06-10T05:44:02","modified_gmt":"2025-06-10T05:44:02","slug":"use-the-template-below-to-construct-a-mo-diagram-for-the-bn","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/10\/use-the-template-below-to-construct-a-mo-diagram-for-the-bn\/","title":{"rendered":"Use the template below to construct a MO diagram for the BN"},"content":{"rendered":"\n<p>Use the template below to construct a MO diagram for the BN<br>molecule and use it to answer the following questions. In the completed diagram for BN<br>, the number of electrons in bonding orbitals is: in antibonding orbitals is: What is the bond order? If a fraction is needed, use a decimal number. BN<br>is Energy Atomic orbitals (Atom A) Molecular orbitals<\/p>\n\n\n\n<p>Atomic orbitals (Atom B)<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-330.png\" alt=\"\" class=\"wp-image-230765\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve the problem and construct the MO (molecular orbital) diagram for the BN\u207b ion, we first need to consider the total number of electrons in the BN\u207b molecule.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Determine Total Electrons<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Boron (B)<\/strong> has 5 electrons<\/li>\n\n\n\n<li><strong>Nitrogen (N)<\/strong> has 7 electrons<\/li>\n\n\n\n<li>The negative charge (<strong>\u207b<\/strong>) adds 1 extra electron<\/li>\n<\/ul>\n\n\n\n<p><strong>Total electrons = 5 (B) + 7 (N) + 1 (extra) = 13 electrons<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Use the Correct MO Energy Ordering<\/strong><\/h3>\n\n\n\n<p>Since BN is a <strong>heteronuclear diatomic molecule<\/strong> and relatively light (similar to molecules like B\u2082, C\u2082, N\u2082), it follows the MO ordering:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">scssCopyEdit<code>\u03c3(2s), \u03c3*(2s), \u03c0(2p), \u03c3(2p), \u03c0*(2p), \u03c3*(2p)\n<\/code><\/pre>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Fill the Molecular Orbitals with Electrons<\/strong><\/h3>\n\n\n\n<p>We now fill 13 electrons in order of increasing energy:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>\u03c3(2s)<\/strong> \u2192 2 electrons (bonding)<\/li>\n\n\n\n<li><em>\u03c3(2s)<\/em>* \u2192 2 electrons (antibonding)<\/li>\n\n\n\n<li><strong>\u03c0(2p)<\/strong> \u2192 4 electrons (2 in each \u03c0 orbital, bonding)<\/li>\n\n\n\n<li><strong>\u03c3(2p)<\/strong> \u2192 2 electrons (bonding)<\/li>\n\n\n\n<li><em>\u03c0(2p)<\/em>* \u2192 2 electrons (antibonding)<\/li>\n\n\n\n<li><em>\u03c3(2p)<\/em>* \u2192 1 electron (antibonding)<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Count Bonding and Antibonding Electrons<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bonding electrons:<\/strong><br>\u03c3(2s) = 2<br>\u03c0(2p) = 4<br>\u03c3(2p) = 2<br><strong>Total = 8<\/strong><\/li>\n\n\n\n<li><strong>Antibonding electrons:<\/strong><br>\u03c3*(2s) = 2<br>\u03c0*(2p) = 2<br>\u03c3*(2p) = 1<br><strong>Total = 5<\/strong><\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 5: Calculate Bond Order<\/strong><\/h3>\n\n\n\n<p>Bond&nbsp;Order=(Bonding&nbsp;electrons\u2212Antibonding&nbsp;electrons)2=8\u221252=32=1.5\\text{Bond Order} = \\frac{(\\text{Bonding electrons} &#8211; \\text{Antibonding electrons})}{2} = \\frac{8 &#8211; 5}{2} = \\frac{3}{2} = 1.5Bond&nbsp;Order=2(Bonding&nbsp;electrons\u2212Antibonding&nbsp;electrons)\u200b=28\u22125\u200b=23\u200b=1.5<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 6: Determine Magnetic Behavior<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Since BN\u207b has <strong>one unpaired electron<\/strong> (in the \u03c3*2p orbital), it is <strong>paramagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Final Answers<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bonding electrons:<\/strong> 8<\/li>\n\n\n\n<li><strong>Antibonding electrons:<\/strong> 5<\/li>\n\n\n\n<li><strong>Bond order:<\/strong> 1.5<\/li>\n\n\n\n<li><strong>BN\u207b is:<\/strong> Paramagnetic<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcd8 Explanation (Summary)<\/h3>\n\n\n\n<p>In molecular orbital (MO) theory, atomic orbitals combine to form bonding and antibonding molecular orbitals. For BN\u207b, the molecular orbitals are filled based on energy levels, following the Aufbau principle and Hund&#8217;s rule. Because it&#8217;s a light heteronuclear molecule, the MO energy order resembles that of B\u2082 and C\u2082. By distributing 13 electrons across these orbitals, we determine that 8 are bonding and 5 are antibonding. Using the bond order formula, we calculate a bond order of 1.5, suggesting moderate bond strength between B and N. The presence of an unpaired electron also makes BN\u207b paramagnetic.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-782.jpeg\" alt=\"\" class=\"wp-image-230766\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Use the template below to construct a MO diagram for the BNmolecule and use it to answer the following questions. In the completed diagram for BN, the number of electrons in bonding orbitals is: in antibonding orbitals is: What is the bond order? If a fraction is needed, use a decimal number. BNis Energy Atomic [&hellip;]<\/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-230764","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230764","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=230764"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230764\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230764"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230764"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}