{"id":217880,"date":"2025-05-22T15:46:54","date_gmt":"2025-05-22T15:46:54","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=217880"},"modified":"2025-05-22T15:46:56","modified_gmt":"2025-05-22T15:46:56","slug":"which-has-the-highest-bond-order","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/22\/which-has-the-highest-bond-order\/","title":{"rendered":"Which has the highest bond order"},"content":{"rendered":"\n<p class=\"wp-block-paragraph\">Draw molecular orbital diagrams for O2-, O22-, and O2. Which has the highest bond order? Which would be paramagnetic, and which would be diamagnetic? Can you draw good dot structures that correspond to each of these ions or molecules?<\/p>\n\n\n\n<p class=\"wp-block-paragraph\"><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p class=\"wp-block-paragraph\">To compare <strong>O\u2082<\/strong>, <strong>O\u2082\u207b<\/strong>, and <strong>O\u2082\u00b2\u207b<\/strong>, we will:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Draw molecular orbital (MO) diagrams<\/strong><\/li>\n\n\n\n<li><strong>Determine bond orders<\/strong><\/li>\n\n\n\n<li><strong>Classify each as paramagnetic or diamagnetic<\/strong><\/li>\n\n\n\n<li><strong>Provide Lewis (dot) structures<\/strong><\/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\">1. <strong>Molecular Orbital Diagrams and Bond Order<\/strong><\/h3>\n\n\n\n<p class=\"wp-block-paragraph\">For oxygen and other elements with atomic numbers \u2265 8, the <strong>molecular orbital energy ordering<\/strong> is:<br>\u03c3(1s), \u03c3*(1s), \u03c3(2s), \u03c3*(2s), <strong>\u03c3(2p_z)<\/strong>, <strong>\u03c0(2p_x) = \u03c0(2p_y)<\/strong>, <em>\u03c0(2p_x) = \u03c0<\/em>(2p_y)**, <em>\u03c3(2p_z)<\/em>*<\/p>\n\n\n\n<p class=\"wp-block-paragraph\">Oxygen has 8 electrons, so O\u2082 has 16 total valence electrons (8 each).<br>We fill MOs in the order above.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">\u27a4 <strong>O\u2082 (16 e\u207b total):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bonding orbitals: \u03c3(2s), \u03c3(2p_z), \u03c0(2p_x), \u03c0(2p_y) = <strong>8 electrons<\/strong><\/li>\n\n\n\n<li>Antibonding orbitals: \u03c3*(2s), \u03c0*(2p_x), \u03c0*(2p_y) = <strong>4 electrons<\/strong><\/li>\n\n\n\n<li><strong>Bond order = (bonding &#8211; antibonding) \/ 2 = (10 &#8211; 6) \/ 2 = 2<\/strong><\/li>\n\n\n\n<li><strong>Paramagnetic<\/strong> (2 unpaired electrons in \u03c0* orbitals)<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">\u27a4 <strong>O\u2082\u207b (17 e\u207b):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Add 1 e\u207b to \u03c0*(2p_x) or \u03c0*(2p_y):<\/li>\n\n\n\n<li><strong>Bond order = (10 &#8211; 7) \/ 2 = 1.5<\/strong><\/li>\n\n\n\n<li><strong>Paramagnetic<\/strong> (1 unpaired electron)<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">\u27a4 <strong>O\u2082\u00b2\u207b (18 e\u207b):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Add another e\u207b to complete both \u03c0* orbitals<\/li>\n\n\n\n<li><strong>Bond order = (10 &#8211; 8) \/ 2 = 1<\/strong><\/li>\n\n\n\n<li><strong>Diamagnetic<\/strong> (all electrons paired)<\/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\">2. <strong>Lewis Dot Structures<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>O\u2082<\/strong> \u2192 O=O with one unpaired electron on each O (diradical); hard to represent well with dots<\/li>\n\n\n\n<li><strong>O\u2082\u207b<\/strong> \u2192 O\u2013O\u207b with a single bond and a lone electron (radical ion)<\/li>\n\n\n\n<li><strong>O\u2082\u00b2\u207b<\/strong> \u2192 O\u2013O\u00b2\u207b with a single bond, all electrons paired<\/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\">3. <strong>Summary<\/strong><\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Molecule\/Ion<\/th><th>Bond Order<\/th><th>Magnetic Property<\/th><th>Dot Structure Summary<\/th><\/tr><\/thead><tbody><tr><td>O\u2082<\/td><td>2<\/td><td>Paramagnetic<\/td><td>O=O with unpaired electrons<\/td><\/tr><tr><td>O\u2082\u207b<\/td><td>1.5<\/td><td>Paramagnetic<\/td><td>O\u2013O\u207b with one unpaired e\u207b<\/td><\/tr><tr><td>O\u2082\u00b2\u207b<\/td><td>1<\/td><td>Diamagnetic<\/td><td>O\u2013O\u00b2\u207b, all electrons paired<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Highest bond order:<\/strong> <strong>O\u2082 (2)<\/strong><\/li>\n\n\n\n<li><strong>Paramagnetic:<\/strong> <strong>O\u2082 and O\u2082\u207b<\/strong><\/li>\n\n\n\n<li><strong>Diamagnetic:<\/strong> <strong>O\u2082\u00b2\u207b<\/strong><\/li>\n<\/ul>\n\n\n\n<p class=\"wp-block-paragraph\">MO theory explains both bond strength and magnetism, which dot structures alone can&#8217;t fully convey\u2014especially for O\u2082.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-95.png\" alt=\"\" class=\"wp-image-217881\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw molecular orbital diagrams for O2-, O22-, and O2. Which has the highest bond order? Which would be paramagnetic, and which would be diamagnetic? Can you draw good dot structures that correspond to each of these ions or molecules? The Correct Answer and Explanation is: To compare O\u2082, O\u2082\u207b, and O\u2082\u00b2\u207b, we will: 1. Molecular [&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-217880","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217880","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=217880"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/217880\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=217880"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=217880"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=217880"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}