{"id":201354,"date":"2025-03-15T08:57:15","date_gmt":"2025-03-15T08:57:15","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201354"},"modified":"2025-03-15T08:57:17","modified_gmt":"2025-03-15T08:57:17","slug":"draw-a-molecular-orbital-diagram-for-the-neutral-c2-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/draw-a-molecular-orbital-diagram-for-the-neutral-c2-molecule\/","title":{"rendered":"Draw a molecular orbital diagram for the neutral C2 molecule"},"content":{"rendered":"\n<p>Draw a molecular orbital diagram for the neutral C2 molecule. Label all atomic orbitals and molecular orbitals. Plausible Lewis structures are :C=C: and \u2022C\u2261C\u2022. Use your molecular orbital diagram to explain which Lewis structure is a better representation.<\/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<p>The <strong>molecular orbital diagram<\/strong> for the neutral C\u2082 molecule (diatomic carbon) is constructed using the atomic orbitals of carbon and the interactions that lead to the formation of molecular orbitals (MOs). The key is understanding how atomic orbitals from each carbon atom combine to form bonding and antibonding molecular orbitals. Carbon has 2s and 2p atomic orbitals, and the resulting molecular orbitals are derived from combinations of these orbitals.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Atomic Orbitals:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon atom has 2s and 2p orbitals available for bonding.<\/li>\n\n\n\n<li>The 2s orbitals combine to form a bonding \u03c3 (2s) and an antibonding \u03c3* (2s).<\/li>\n\n\n\n<li>The 2p orbitals form \u03c0 and \u03c3 molecular orbitals depending on their orientation.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Molecular Orbitals:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The \u03c3 and \u03c0 orbitals can be classified as bonding and antibonding based on whether the atomic orbitals interact constructively or destructively.<\/li>\n\n\n\n<li>The bonding molecular orbitals (MO) result from constructive interference of the atomic orbitals.<\/li>\n\n\n\n<li>The antibonding MOs result from destructive interference.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Molecular Orbitals of C\u2082:<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 (2s)<\/strong>: Bonding orbital formed by the overlap of 2s orbitals from each carbon atom.<\/li>\n\n\n\n<li><strong>\u03c3* (2s)<\/strong>: Antibonding orbital formed by the out-of-phase overlap of the 2s orbitals.<\/li>\n\n\n\n<li><strong>\u03c0 (2p)<\/strong>: Two degenerate bonding orbitals formed by the sideways overlap of 2p orbitals from each carbon atom.<\/li>\n\n\n\n<li><strong>\u03c0* (2p)<\/strong>: Antibonding orbitals formed by the out-of-phase overlap of the 2p orbitals.<\/li>\n\n\n\n<li><strong>\u03c3 (2p)<\/strong>: A bonding orbital formed by the head-on overlap of 2p orbitals.<\/li>\n\n\n\n<li><strong>\u03c3* (2p)<\/strong>: An antibonding orbital formed by the out-of-phase overlap of 2p orbitals.<\/li>\n<\/ul>\n\n\n\n<p>The electron configuration for C\u2082 is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10 electrons (5 pairs) to distribute in these MOs, filling the lowest energy orbitals first.<\/li>\n<\/ul>\n\n\n\n<p>The <strong>Lewis structure<\/strong> of C\u2082 with a <strong>C=C<\/strong> bond is favored. The bonding electrons in the molecular orbital diagram result in a bond order of 2 (indicating a double bond between the two carbon atoms), which matches the Lewis structure of <strong>C=C: (double bond)<\/strong>. In contrast, the <strong>C\u2261C<\/strong> (triple bond) structure would require an odd number of electrons and would not correspond to the ground-state configuration of C\u2082.<\/p>\n\n\n\n<p>Thus, the <strong>C=C<\/strong> structure is a better representation, supported by the molecular orbital diagram&#8217;s bonding order and electron configuration. The <strong>C\u2261C<\/strong> structure would imply a bond order of 3, but the distribution of electrons in the MOs does not support this. Therefore, the neutral carbon molecule (C\u2082) prefers the <strong>C=C<\/strong> structure over the <strong>C\u2261C<\/strong> structure.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw a molecular orbital diagram for the neutral C2 molecule. Label all atomic orbitals and molecular orbitals. Plausible Lewis structures are :C=C: and \u2022C\u2261C\u2022. Use your molecular orbital diagram to explain which Lewis structure is a better representation. The correct answer and explanation is : The molecular orbital diagram for the neutral C\u2082 molecule (diatomic [&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-201354","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201354","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=201354"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201354\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}