{"id":201353,"date":"2025-06-16T10:01:04","date_gmt":"2025-06-16T10:01:04","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201353"},"modified":"2025-06-16T10:01:07","modified_gmt":"2025-06-16T10:01:07","slug":"draw-a-molecular-orbital-diagram-for-the-neutral-c2-molecule-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/draw-a-molecular-orbital-diagram-for-the-neutral-c2-molecule-2\/","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<h3 class=\"wp-block-heading\"><strong>Molecular Orbital Diagram for the C\u2082 Molecule<\/strong><\/h3>\n\n\n\n<p><strong>Step 1: Atomic Orbitals of Carbon<\/strong><br>Each carbon atom has 6 electrons:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1s\u00b2<\/li>\n\n\n\n<li>2s\u00b2<\/li>\n\n\n\n<li>2p\u00b2<\/li>\n<\/ul>\n\n\n\n<p>So for two carbon atoms, we have a total of <strong>12 electrons<\/strong> to place in molecular orbitals.<\/p>\n\n\n\n<p><strong>Step 2: Molecular Orbital Energy Levels for C\u2082<\/strong><br>For elements up to oxygen in the periodic table, the 2p orbital ordering is as follows due to s-p mixing:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">CopyEdit<code>\u03c31s &lt; \u03c31s* &lt; \u03c32s &lt; \u03c32s* &lt; \u03c02p &lt; \u03c32p &lt; \u03c02p* &lt; \u03c32p*\n<\/code><\/pre>\n\n\n\n<p><strong>Step 3: Electron Configuration in Molecular Orbitals<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>\u03c31s (2 electrons)<\/li>\n\n\n\n<li>\u03c31s* (2 electrons)<\/li>\n\n\n\n<li>\u03c32s (2 electrons)<\/li>\n\n\n\n<li>\u03c32s* (2 electrons)<\/li>\n\n\n\n<li>\u03c02p (4 electrons)<\/li>\n\n\n\n<li>\u03c32p (0 electrons)<\/li>\n<\/ol>\n\n\n\n<p>So the molecular orbital filling is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">markdownCopyEdit<code>\u03c31s\u00b2 \u03c31s*\u00b2 \u03c32s\u00b2 \u03c32s*\u00b2 \u03c02p\u2074\n<\/code><\/pre>\n\n\n\n<p><strong>Bond Order Calculation:<\/strong><br>Bond order = \u00bd [(number of bonding electrons) \u2013 (number of antibonding electrons)]<br>= \u00bd [(2 + 2 + 4) \u2013 (2 + 2)]<br>= \u00bd [8 \u2013 4]<br>= <strong>2<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Which Lewis Structure Is More Accurate?<\/strong><\/h3>\n\n\n\n<p>There are two plausible Lewis structures:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>:C=C:<\/strong> (double bond between carbons)<\/li>\n\n\n\n<li><strong>\u2022C\u2261C\u2022<\/strong> (triple bond with one unpaired electron on each carbon)<\/li>\n<\/ol>\n\n\n\n<p>From the MO diagram, we find:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>bond order is 2<\/strong>, not 3.<\/li>\n\n\n\n<li>All electrons are <strong>paired<\/strong> in the diagram\u2014no unpaired electrons.<\/li>\n\n\n\n<li>Therefore, <strong>C\u2082 is a diamagnetic molecule<\/strong>, not paramagnetic.<\/li>\n<\/ul>\n\n\n\n<p>This rules out the structure with unpaired electrons (\u2022C\u2261C\u2022). The more accurate representation is:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>:C=C:<\/strong><\/h3>\n\n\n\n<p>This agrees with the bond order of 2 from the MO diagram and shows no unpaired electrons. While it might seem odd that two \u03c0 bonds (from \u03c02p orbitals) form before a \u03c32p bond, this is consistent with molecular orbital theory for lighter diatomic molecules like C\u2082.<\/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 : Molecular Orbital Diagram for the C\u2082 Molecule Step 1: Atomic [&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-201353","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201353","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=201353"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201353\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201353"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201353"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201353"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}