{"id":228716,"date":"2025-06-07T13:41:09","date_gmt":"2025-06-07T13:41:09","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=228716"},"modified":"2025-06-07T13:41:11","modified_gmt":"2025-06-07T13:41:11","slug":"chemical-bonding-drawing-the-mo-energy-diagram-for-a-period-2-homodiatom-draw-the-molecular-orbital-mo-electron-diagram-for-the-c2-molecule-be-sure-your-diagram-contains-all-of-the-electrons-in-the","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/07\/chemical-bonding-drawing-the-mo-energy-diagram-for-a-period-2-homodiatom-draw-the-molecular-orbital-mo-electron-diagram-for-the-c2-molecule-be-sure-your-diagram-contains-all-of-the-electrons-in-the\/","title":{"rendered":"CHEMICAL BONDING Drawing the MO energy diagram for a Period 2 homodiatom Draw the molecular orbital (MO) electron diagram for the C2 molecule: Be sure your diagram contains all of the electrons in the molecule, including any core electrons."},"content":{"rendered":"\n<p>CHEMICAL BONDING Drawing the MO energy diagram for a Period 2 homodiatom Draw the molecular orbital (MO) electron diagram for the C2 molecule: Be sure your diagram contains all of the electrons in the molecule, including any core electrons. Energy<\/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\">\u2705 <strong>MO Energy Diagram for C\u2082 (Carbon Dimer)<\/strong><\/h3>\n\n\n\n<pre class=\"wp-block-preformatted\">scssCopyEdit<code>                            \u03c3*(2p)\n                            \u2191   \u2191\n                        \u03c0*(2p) \u03c0*(2p)\n                        \u2191   \u2191\n                        \u03c0(2p)  \u03c0(2p)\n                        \u2191   \u2191\n                            \u03c3(2p)\n                            \u2191\n                          \u03c3*(2s)\n                          \u2191   \u2191\n                          \u03c3(2s)\n                          \u2191   \u2191\nCarbon Atom 1                        Carbon Atom 2\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Total electrons in C\u2082 = 12 (6 from each carbon)<\/li>\n\n\n\n<li>Electron configuration: 1s\u00b2 2s\u00b2 2p\u00b2 for each carbon<\/li>\n\n\n\n<li>Core electrons (1s) typically not shown in MO diagrams, but if needed:\n<ul class=\"wp-block-list\">\n<li>\u03c3(1s) \u2192 2 electrons<\/li>\n\n\n\n<li>\u03c3*(1s) \u2192 2 electrons<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>So including core electrons:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">scssCopyEdit<code>\u03c3(1s)      \u2191\u2193  \n\u03c3*(1s)     \u2191\u2193  \n\u03c3(2s)      \u2191\u2193  \n\u03c3*(2s)     \u2191\u2193  \n\u03c0(2p)      \u2191\u2193  \u2191\u2193  \n\u03c0*(2p)     (empty)  \n\u03c3(2p)      (empty)\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\">\ud83e\uddea Explanation<\/h3>\n\n\n\n<p>The molecular orbital (MO) theory explains bonding by combining atomic orbitals to form molecular orbitals. In the case of <strong>C\u2082<\/strong>, which consists of two carbon atoms (atomic number 6), each contributes 6 electrons for a total of 12.<\/p>\n\n\n\n<p>These electrons fill molecular orbitals in a specific order. For <strong>Period 2 homonuclear diatomic molecules<\/strong> like B\u2082, C\u2082, and N\u2082, the correct MO ordering (due to s-p mixing) is:<\/p>\n\n\n\n<p><em>\u03c3(1s) &lt; \u03c3(1s) &lt; \u03c3(2s) &lt; \u03c3<\/em>(2s) &lt; \u03c0(2p) &lt; \u03c3(2p) &lt; \u03c0*(2p) &lt; \u03c3*(2p)**<\/p>\n\n\n\n<p>This order changes in O\u2082 and F\u2082, where the \u03c3(2p) drops below \u03c0(2p) due to reduced s-p mixing.<\/p>\n\n\n\n<p>For C\u2082:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The first 4 electrons go into \u03c3(1s) and \u03c3*(1s) orbitals (core).<\/li>\n\n\n\n<li>The next 4 electrons fill \u03c3(2s) and \u03c3*(2s).<\/li>\n\n\n\n<li>The remaining 4 electrons occupy the <strong>\u03c0(2p) orbitals<\/strong> (2 electrons in each \u03c0 orbital, degenerate in energy).<\/li>\n<\/ul>\n\n\n\n<p>Notice:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The \u03c3(2p) and all antibonding \u03c0* and \u03c3* orbitals are empty.<\/li>\n\n\n\n<li>This gives <strong>bond order = (bonding electrons &#8211; antibonding electrons)\/2 = (8 &#8211; 4)\/2 = 2<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Thus, C\u2082 has a <strong>double bond<\/strong> and is <strong>diamagnetic<\/strong> (all electrons are paired). Interestingly, this contrasts with Lewis structures which suggest a single or triple bond. MO theory provides a more nuanced view, showing that the bonding involves primarily p-orbitals in \u03c0 interactions.<\/p>\n\n\n\n<p>This diagram and analysis explain the electronic structure and bonding nature of C\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\/06\/learnexams-banner8-123.jpeg\" alt=\"\" class=\"wp-image-228717\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>CHEMICAL BONDING Drawing the MO energy diagram for a Period 2 homodiatom Draw the molecular orbital (MO) electron diagram for the C2 molecule: Be sure your diagram contains all of the electrons in the molecule, including any core electrons. Energy The Correct Answer and Explanation is: \u2705 MO Energy Diagram for C\u2082 (Carbon Dimer) scssCopyEdit [&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-228716","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228716","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=228716"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/228716\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=228716"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=228716"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=228716"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}