{"id":212098,"date":"2025-05-05T15:27:27","date_gmt":"2025-05-05T15:27:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=212098"},"modified":"2025-05-05T15:27:29","modified_gmt":"2025-05-05T15:27:29","slug":"for-ch2chch3-draw-an-appropriate-lewis-structure","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/05\/for-ch2chch3-draw-an-appropriate-lewis-structure\/","title":{"rendered":"For CH2CHCH3 draw an appropriate Lewis structure"},"content":{"rendered":"\n<p>For CH2CHCH3 draw an appropriate Lewis structure.<\/p>\n\n\n\n<p>Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To draw the <strong>Lewis structure<\/strong> for <strong>CH\u2082CHCH\u2083 (propene)<\/strong>, follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Correct Lewis Structure of CH\u2082CHCH\u2083 (Propene):<\/strong><\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>     H   H\n      \\ \/\n       C = C \u2014 C \u2014 H\n      \/     \\\n     H       H\n             |\n             H\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The leftmost carbon (CH\u2082) is <strong>sp\u00b2-hybridized<\/strong>, double bonded to the middle carbon.<\/li>\n\n\n\n<li>The middle carbon (CH) is also <strong>sp\u00b2-hybridized<\/strong>, forming a double bond with the first carbon and a single bond with the third carbon.<\/li>\n\n\n\n<li>The third carbon (CH\u2083) is <strong>sp\u00b3-hybridized<\/strong>, bonded to three hydrogen atoms.<\/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\">\ud83d\udd0d <strong>Step-by-Step Explanation (300 words)<\/strong><\/h3>\n\n\n\n<p><strong>1. Determine the total number of valence electrons:<\/strong><br>Each atom contributes:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 4 valence electrons \u00d7 3 = <strong>12<\/strong><\/li>\n\n\n\n<li>Hydrogen (H): 1 valence electron \u00d7 6 = <strong>6<\/strong><\/li>\n\n\n\n<li><strong>Total = 18 valence electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>2. Identify the molecular skeleton:<\/strong><br>The name CH\u2082CHCH\u2083 implies:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One <strong>double bond<\/strong> between the first and second carbon (CH\u2082=CH)<\/li>\n\n\n\n<li>One <strong>single bond<\/strong> between the second and third carbon (CH\u2013CH\u2083)<\/li>\n<\/ul>\n\n\n\n<p>The carbon chain looks like: <strong>CH\u2082=CH\u2013CH\u2083<\/strong><\/p>\n\n\n\n<p><strong>3. Draw bonds between atoms:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place a <strong>double bond<\/strong> between the first and second carbon atoms.<\/li>\n\n\n\n<li>Place a <strong>single bond<\/strong> between the second and third carbon atoms.<\/li>\n\n\n\n<li>Attach enough hydrogen atoms to each carbon so that each carbon has 4 total bonds.<\/li>\n<\/ul>\n\n\n\n<p><strong>4. Place the remaining electrons:<\/strong><br>Hydrogen atoms can only form <strong>one bond<\/strong>, so no lone pairs are added to H.<br>Carbon atoms need <strong>four bonds<\/strong> to satisfy the octet rule.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>first carbon (CH\u2082)<\/strong> has two single bonds to H and a double bond to the second carbon.<\/li>\n\n\n\n<li>The <strong>second carbon (CH)<\/strong> has a double bond to the first carbon, a single bond to the third carbon, and a single bond to one H.<\/li>\n\n\n\n<li>The <strong>third carbon (CH\u2083)<\/strong> has three single bonds to H and one to the second carbon.<\/li>\n<\/ul>\n\n\n\n<p><strong>5. Lone pairs:<\/strong><br>There are <strong>no lone pairs<\/strong> on carbon or hydrogen in this molecule, as all valence electrons are used in bonding.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>For CH2CHCH3 draw an appropriate Lewis structure. Draw the molecule by placing atoms on the grid and connecting them with bonds. Include all lone pairs of electrons. The Correct Answer and Explanation is: To draw the Lewis structure for CH\u2082CHCH\u2083 (propene), follow these steps: \u2705 Correct Lewis Structure of CH\u2082CHCH\u2083 (Propene): \ud83d\udd0d Step-by-Step Explanation (300 [&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-212098","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212098","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=212098"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/212098\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=212098"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=212098"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=212098"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}