{"id":201351,"date":"2025-03-15T08:54:51","date_gmt":"2025-03-15T08:54:51","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201351"},"modified":"2025-03-15T08:54:53","modified_gmt":"2025-03-15T08:54:53","slug":"draw-and-label-the-molecular-orbital-mo-diagrams-for-c2-and-c22","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/draw-and-label-the-molecular-orbital-mo-diagrams-for-c2-and-c22\/","title":{"rendered":"Draw and label the molecular orbital (MO) diagrams for C2 and C22\u2013"},"content":{"rendered":"\n<p>Draw and label the molecular orbital (MO) diagrams for C2 and C2<sup>2\u2013<\/sup>. Give the MO ground state electron configuration, the bond order, and the magnetism for each molecule. Which molecule has the stronger bond?<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Molecular Orbital (MO) Diagrams and Electron Configurations<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>C\u2082 Molecule (Neutral)<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular orbital diagram for the C\u2082 molecule can be drawn as follows:<ul><li>Carbon (C) has 6 electrons in its valence shell, so the C\u2082 molecule has a total of 12 electrons.<\/li><li>For homonuclear diatomic molecules like C\u2082, the molecular orbitals are filled in the following order:<\/li><li>\u03c3(1s), \u03c3<em>(1s), \u03c3(2s), \u03c3<\/em>(2s), \u03c0(2px), \u03c0(2py), \u03c0<em>(2px), \u03c0<\/em>(2py)<\/li><li>The filling of electrons in the molecular orbitals (MO) will be as follows:<\/li><\/ul><code>\u03c3(1s) 2 \u03c3*(1s) 2 \u03c3(2s) 2 \u03c3*(2s) 2 \u03c0(2px) 2 \u03c0(2py) 2 \u03c0*(2px) 0 \u03c0*(2py) 0<\/code> <strong>Ground State Electron Configuration for C\u2082<\/strong>:<\/li>\n\n\n\n<li>2 electrons in \u03c3(1s)<\/li>\n\n\n\n<li>2 electrons in \u03c3*(1s)<\/li>\n\n\n\n<li>2 electrons in \u03c3(2s)<\/li>\n\n\n\n<li>2 electrons in \u03c3*(2s)<\/li>\n\n\n\n<li>2 electrons in \u03c0(2px)<\/li>\n\n\n\n<li>2 electrons in \u03c0(2py) <strong>Bond Order Calculation<\/strong>:<br>Bond Order = (Number of bonding electrons \u2212 Number of antibonding electrons) \/ 2<br>Bond Order = (2 + 2 + 2 + 2 \u2212 2 \u2212 2) \/ 2 = 2 <strong>Magnetism<\/strong>:<br>Since there are no unpaired electrons, C\u2082 is <strong>diamagnetic<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li><strong>C\u2082\u00b2\u207b Molecule (C\u2082 with two extra electrons)<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For C\u2082\u00b2\u207b, two additional electrons are added to the \u03c0<em>(2px) and \u03c0<\/em>(2py) antibonding orbitals. The filling of the molecular orbitals is as follows: <code>\u03c3(1s) 2 \u03c3*(1s) 2 \u03c3(2s) 2 \u03c3*(2s) 2 \u03c0(2px) 2 \u03c0(2py) 2 \u03c0*(2px) 1 \u03c0*(2py) 1<\/code> <strong>Ground State Electron Configuration for C\u2082\u00b2\u207b<\/strong>:<\/li>\n\n\n\n<li>2 electrons in \u03c3(1s)<\/li>\n\n\n\n<li>2 electrons in \u03c3*(1s)<\/li>\n\n\n\n<li>2 electrons in \u03c3(2s)<\/li>\n\n\n\n<li>2 electrons in \u03c3*(2s)<\/li>\n\n\n\n<li>2 electrons in \u03c0(2px)<\/li>\n\n\n\n<li>2 electrons in \u03c0(2py)<\/li>\n\n\n\n<li>1 electron in \u03c0*(2px)<\/li>\n\n\n\n<li>1 electron in \u03c0*(2py) <strong>Bond Order Calculation<\/strong>:<br>Bond Order = (Number of bonding electrons \u2212 Number of antibonding electrons) \/ 2<br>Bond Order = (2 + 2 + 2 + 2 \u2212 2 \u2212 2 \u2212 1 \u2212 1) \/ 2 = 1 <strong>Magnetism<\/strong>:<br>Since C\u2082\u00b2\u207b has 2 unpaired electrons, 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\">Comparison of Bond Strengths<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C\u2082<\/strong> has a bond order of 2, and <strong>C\u2082\u00b2\u207b<\/strong> has a bond order of 1.<\/li>\n\n\n\n<li>A <strong>higher bond order<\/strong> indicates a stronger bond. Therefore, <strong>C\u2082<\/strong> has a stronger bond than <strong>C\u2082\u00b2\u207b<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C\u2082<\/strong> is diamagnetic, with a bond order of 2, and has a stronger bond than <strong>C\u2082\u00b2\u207b<\/strong>.<\/li>\n\n\n\n<li><strong>C\u2082\u00b2\u207b<\/strong> is paramagnetic, with a bond order of 1, and has a weaker bond than <strong>C\u2082<\/strong>.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Draw and label the molecular orbital (MO) diagrams for C2 and C22\u2013. Give the MO ground state electron configuration, the bond order, and the magnetism for each molecule. Which molecule has the stronger bond? The correct answer and explanation is : Molecular Orbital (MO) Diagrams and Electron Configurations Comparison of Bond Strengths Conclusion:<\/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-201351","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201351","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=201351"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201351\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201351"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201351"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201351"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}