{"id":232991,"date":"2025-06-12T12:33:31","date_gmt":"2025-06-12T12:33:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=232991"},"modified":"2025-06-12T12:33:33","modified_gmt":"2025-06-12T12:33:33","slug":"draw-the-molecular-orbital-diagram-of-b2-and-answer-the-following-questions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/12\/draw-the-molecular-orbital-diagram-of-b2-and-answer-the-following-questions\/","title":{"rendered":"Draw the Molecular Orbital Diagram of B2+ and answer the following questions"},"content":{"rendered":"\n<p> (a) How many valence electron are in B2+? (b) How many valence electrons of B2+ are in bonding molecular orbitals? (c) How many valence electrons of B2+ are in anti-bonding molecular orbitals? (d) What is the bond order of B2+? (e) If B2+ is a stable chemical species, what is it&#8217;s magnetism? (f) What type of molecular orbital contains the valence electron with the highest energy in B2+?<\/p>\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<h3 class=\"wp-block-heading\">\u269b Molecular Orbital Diagram for B\u2082\u207a<\/h3>\n\n\n\n<p>For elements with atomic number \u2264 7 (like boron), the MO diagram is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">scssCopyEdit<code>          \u03c3*(2p)\n         \u03c0*(2p) \u03c0*(2p)\n         \u03c0(2p)  \u03c0(2p)\n          \u03c3(2p)\n          \u03c3*(2s)\n          \u03c3(2s)\n<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">(a) How many valence electrons are in B\u2082\u207a?<\/h3>\n\n\n\n<p>B atom: 5 electrons \u2192 2 B atoms = 10 electrons<br>B\u2082\u207a loses 1 electron \u2192 <strong>9 valence electrons<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(b) How many valence electrons are in bonding molecular orbitals?<\/h3>\n\n\n\n<p>Bonding MOs: \u03c3(2s), \u03c0(2p), \u03c3(2p)<br>Filled as:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c3(2s): 2 e\u207b<\/li>\n\n\n\n<li>\u03c0(2p): 4 e\u207b (2 in each \u03c0 orbital)<\/li>\n\n\n\n<li>\u03c3(2p): 1 e\u207b<br>\u2192 <strong>7 electrons in bonding orbitals<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">(c) How many valence electrons are in anti-bonding molecular orbitals?<\/h3>\n\n\n\n<p>Anti-bonding MOs: \u03c3*(2s), \u03c0*(2p)<br>Filled as:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c3*(2s): 2 e\u207b<\/li>\n\n\n\n<li>\u03c0*(2p): 0 e\u207b<br>\u2192 <strong>2 electrons in anti-bonding orbitals<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">(d) What is the bond order of B\u2082\u207a?<\/h3>\n\n\n\n<p>Bond order = (Bonding e\u207b &#8211; Antibonding e\u207b) \u00f7 2<br>= (7 &#8211; 2) \u00f7 2 = <strong>2.5<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(e) What is its magnetism?<\/h3>\n\n\n\n<p>Unpaired electrons = 1 (in \u03c3(2p)) \u2192 <strong>Paramagnetic<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">(f) What type of MO contains the highest-energy valence electron?<\/h3>\n\n\n\n<p>Highest energy MO occupied = <strong>\u03c3(2p)<\/strong><\/p>\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<\/h3>\n\n\n\n<p>The molecular orbital theory provides a more nuanced understanding of bonding in diatomic molecules by considering the combination of atomic orbitals to form molecular orbitals (MOs). In the case of B\u2082\u207a (a boron-boron cation), we begin by counting the total number of valence electrons. Each boron atom contributes 5 electrons, so neutral B\u2082 has 10. Since B\u2082\u207a has lost one electron, it contains 9 valence electrons.<\/p>\n\n\n\n<p>The MO diagram for B\u2082\u207a follows the pattern used for lighter diatomic molecules (atomic numbers \u2264 7), where the energy ordering of the MOs is: \u03c3(2s), \u03c3*(2s), \u03c0(2p), \u03c3(2p), \u03c0*(2p), \u03c3*(2p). Electrons fill from the lowest energy levels upward according to the Aufbau principle, and obey Hund\u2019s rule and the Pauli exclusion principle.<\/p>\n\n\n\n<p>In B\u2082\u207a, the electrons fill up to the \u03c3(2p) orbital. We find two electrons in \u03c3(2s), two in \u03c3*(2s), four in the degenerate \u03c0(2p) orbitals, and one in the \u03c3(2p). This results in <strong>7 electrons in bonding MOs<\/strong> and <strong>2 in anti-bonding MOs<\/strong>. The bond order, calculated as (bonding \u2013 anti-bonding)\/2, gives <strong>2.5<\/strong>, suggesting a fairly strong bond despite the loss of one electron.<\/p>\n\n\n\n<p>Importantly, the presence of an unpaired electron in the \u03c3(2p) orbital implies <strong>paramagnetism<\/strong>, meaning B\u2082\u207a is attracted to a magnetic field. This characteristic differentiates it from diamagnetic species, which have all electrons paired. Finally, the highest energy molecular orbital occupied in B\u2082\u207a is the <strong>\u03c3(2p)<\/strong> orbital, which houses the unpaired electron and defines much of the ion\u2019s chemical behavior.<\/p>\n\n\n\n<p>This MO analysis helps predict not only bond strength but also magnetic properties, reactivity, and relative stability of molecules and ions like B\u2082\u207a.<\/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-banner5-392.jpeg\" alt=\"\" class=\"wp-image-232992\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>(a) How many valence electron are in B2+? (b) How many valence electrons of B2+ are in bonding molecular orbitals? (c) How many valence electrons of B2+ are in anti-bonding molecular orbitals? (d) What is the bond order of B2+? (e) If B2+ is a stable chemical species, what is it&#8217;s magnetism? (f) What type [&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-232991","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232991","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=232991"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/232991\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=232991"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=232991"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=232991"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}