{"id":215741,"date":"2025-05-17T17:14:21","date_gmt":"2025-05-17T17:14:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=215741"},"modified":"2025-05-19T11:21:02","modified_gmt":"2025-05-19T11:21:02","slug":"given-the-molecule-ch2chcn-please-draw-a-lewis-structure-for-this-compound-with-correct-geometry-and-label-all-sigma-and-pi-bonds-with-the-orbitals-it-is-made-from","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/17\/given-the-molecule-ch2chcn-please-draw-a-lewis-structure-for-this-compound-with-correct-geometry-and-label-all-sigma-and-pi-bonds-with-the-orbitals-it-is-made-from\/","title":{"rendered":"Given the molecule CH2CHCN Please draw a Lewis structure for this compound with CORRECT GEOMETRY and label all sigma and pi bonds with the orbitals it is made from (what orbitals are overlapping to form the bond)."},"content":{"rendered":"\n<p>Given the molecule CH2CHCN Please draw a Lewis structure for this compound with CORRECT GEOMETRY and label all sigma and pi bonds with the orbitals it is made from (what orbitals are overlapping to form the bond). You may draw the hybridized sigma orbitals as sticks and the unhybridized p-orbitals as lobes for clarity. For this question, you DO need to draw out the orbitals.<\/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<p>To address this question about <strong>CH\u2082CHCN (acrylonitrile)<\/strong>, we\u2019ll first draw its <strong>Lewis structure<\/strong> with correct <strong>molecular geometry<\/strong>, then show <strong>orbital overlaps<\/strong> for <strong>sigma (\u03c3) and pi (\u03c0) bonds<\/strong>, and explain all the hybridization involved.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Lewis Structure of CH\u2082CHCN<\/strong><\/h3>\n\n\n\n<p>Molecular formula: CH\u2082=CH\u2013C\u2261N<\/p>\n\n\n\n<p>This molecule is called <strong>acrylonitrile<\/strong> and consists of:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A terminal CH\u2082 group,<\/li>\n\n\n\n<li>A double bond to a CH (vinyl),<\/li>\n\n\n\n<li>A single bond to a carbon triple-bonded to nitrogen.<\/li>\n<\/ul>\n\n\n\n<p><strong>Lewis structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>     H      H\n      \\    \/\n       C==C\u2014C\u2261N\n      \/        \\\n     H          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\"><strong>Step 2: Geometry and Hybridization<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C1 (leftmost CH\u2082):<\/strong> sp\u00b2 hybridized \u2192 trigonal planar<\/li>\n\n\n\n<li><strong>C2 (central CH):<\/strong> sp\u00b2 hybridized \u2192 trigonal planar<\/li>\n\n\n\n<li><strong>C3 (triple bond carbon):<\/strong> sp hybridized \u2192 linear<\/li>\n\n\n\n<li><strong>N (nitrile nitrogen):<\/strong> sp hybridized \u2192 linear<\/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\"><strong>Step 3: Orbital Overlaps (\u03c3 and \u03c0 bonds)<\/strong><\/h3>\n\n\n\n<p>Let\u2019s break it down by bond:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>1. CH\u2082=C (C1=C2):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 bond<\/strong> between C1 and C2: <strong>sp\u00b2 (C1) \u2013 sp\u00b2 (C2)<\/strong><\/li>\n\n\n\n<li><strong>\u03c0 bond 1<\/strong>: formed by overlap of <strong>unhybridized p-orbitals<\/strong> from both sp\u00b2 C atoms<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>2. C2\u2013C3 single bond:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 bond<\/strong> between C2 and C3: <strong>sp\u00b2 (C2) \u2013 sp (C3)<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>3. C3\u2261N (triple bond):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 bond<\/strong>: <strong>sp (C3) \u2013 sp (N)<\/strong><\/li>\n\n\n\n<li><strong>2 \u03c0 bonds<\/strong>: two sets of <strong>unhybridized p-orbital overlaps<\/strong>, one from each atom<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>4. C\u2013H bonds (for C1 and C2):<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 bonds<\/strong>: <strong>sp\u00b2 (C) \u2013 1s (H)<\/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\"><strong>Step 4: Orbital Diagram Drawing<\/strong><\/h3>\n\n\n\n<p>In the <strong>drawing<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hybridized \u03c3 bonds<\/strong> are shown as <strong>sticks<\/strong> (e.g., sp\u00b2\u2013sp\u00b2, sp\u00b2\u20131s).<\/li>\n\n\n\n<li><strong>Unhybridized p-orbitals<\/strong> are shown as <strong>dumbbell lobes<\/strong> forming <strong>\u03c0 bonds<\/strong> above and below the molecular plane.<\/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\"><strong>Explanation (300+ words)<\/strong><\/h3>\n\n\n\n<p>The molecule CH\u2082CHCN, or acrylonitrile, contains a combination of single, double, and triple bonds that require an understanding of molecular geometry and hybridization. To determine orbital interactions, we first examine the bonding and geometry around each atom.<\/p>\n\n\n\n<p>The first carbon (C1), part of the CH\u2082 group, is connected to two hydrogens and one carbon via a double bond. This means it has three regions of electron density, implying <strong>sp\u00b2 hybridization<\/strong> and <strong>trigonal planar geometry<\/strong>. The central carbon (C2) also forms three sigma bonds\u2014one with C1, one with C3 (a single bond), and one with a hydrogen. This atom is also <strong>sp\u00b2 hybridized<\/strong>, maintaining a planar structure and allowing p-orbital overlap for the \u03c0 bond between C1 and C2.<\/p>\n\n\n\n<p>The third carbon (C3) is part of a triple bond with a nitrogen atom. This atom has two areas of electron density (one sigma bond to C2 and one to nitrogen), indicating <strong>sp hybridization<\/strong> and <strong>linear geometry<\/strong>. The nitrogen also uses sp orbitals to form the sigma bond and retains a lone pair in another sp orbital, with two unhybridized p-orbitals for \u03c0 bonding.<\/p>\n\n\n\n<p>Pi bonds are formed from <strong>side-by-side overlap<\/strong> of <strong>unhybridized p orbitals<\/strong>, and each pi bond exists above and below the internuclear axis. The triple bond between C3 and N has one \u03c3 bond (sp\u2013sp overlap) and two \u03c0 bonds (p\u2013p overlap).<\/p>\n\n\n\n<p>Each C\u2013H bond is formed via <strong>sp\u00b2\u20131s<\/strong> overlap, contributing to sigma bonds only.<\/p>\n\n\n\n<p>Understanding these orbital interactions explains not only molecular shape but also the electron density distribution and reactivity patterns of acrylonitrile in organic chemistry.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Given the molecule CH2CHCN Please draw a Lewis structure for this compound with CORRECT GEOMETRY and label all sigma and pi bonds with the orbitals it is made from (what orbitals are overlapping to form the bond). You may draw the hybridized sigma orbitals as sticks and the unhybridized p-orbitals as lobes for clarity. For [&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-215741","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215741","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=215741"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/215741\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=215741"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=215741"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=215741"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}