{"id":230523,"date":"2025-06-09T16:51:14","date_gmt":"2025-06-09T16:51:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=230523"},"modified":"2025-06-09T16:51:17","modified_gmt":"2025-06-09T16:51:17","slug":"draw-a-lewis-structure-for-the-urea-molecule-conh22","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/09\/draw-a-lewis-structure-for-the-urea-molecule-conh22\/","title":{"rendered":"Draw a Lewis structure for the urea molecule, CO(NH2)2"},"content":{"rendered":"\n<div class=\"wp-block-group is-vertical is-layout-flex wp-container-core-group-is-layout-fe9cc265 wp-block-group-is-layout-flex\">\n<p><br>Draw a Lewis structure for the urea molecule, CO(NH2)2. Draw the Lewis dot structure for urea. Include all lone pairs of electrons. Part B Predict its geometric shape around the C atom with the VSEPR theory. Predict its geometric shape around the N atom with the VSEPR theory. tetrahedral trigonal pyramidal bent trigonal planar Part C Predict its geometric shape around the N atom with the VSEPR theory. Predict its geometric shape around the C atom with the VSEPR theory. tetrahedral trigonal pyramidal bent trigonal planar<\/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<\/div>\n\n\n\n<p><strong>Correct Answers:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Part A<\/strong>: Lewis structure of urea, CO(NH\u2082)\u2082, includes all atoms and lone pairs.<\/li>\n\n\n\n<li><strong>Part B<\/strong>:\n<ul class=\"wp-block-list\">\n<li><strong>Geometric shape around the C atom<\/strong>: <strong>Trigonal planar<\/strong><\/li>\n\n\n\n<li><strong>Geometric shape around the N atom<\/strong>: <strong>Trigonal pyramidal<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Part C<\/strong>:\n<ul class=\"wp-block-list\">\n<li><strong>Geometric shape around the N atom<\/strong>: <strong>Trigonal pyramidal<\/strong><\/li>\n\n\n\n<li><strong>Geometric shape around the C atom<\/strong>: <strong>Trigonal planar<\/strong><\/li>\n<\/ul>\n<\/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 <\/strong><\/h3>\n\n\n\n<p><strong>Urea<\/strong>, with the formula <strong>CO(NH\u2082)\u2082<\/strong>, consists of a central <strong>carbon atom<\/strong> double bonded to an <strong>oxygen atom<\/strong> and singly bonded to two <strong>amino groups (\u2013NH\u2082)<\/strong>. To analyze its geometry, we first draw its <strong>Lewis structure<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Lewis Structure of Urea:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong>: 4 valence electrons<\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong>: 6 valence electrons<\/li>\n\n\n\n<li><strong>Nitrogen (N)<\/strong>: 5 valence electrons \u00d7 2 = 10<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong>: 1 valence electron \u00d7 4 = 4<\/li>\n<\/ol>\n\n\n\n<p><strong>Total = 4 + 6 + 10 + 4 = 24 valence electrons<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon forms four bonds: a double bond with oxygen and single bonds with two nitrogen atoms.<\/li>\n\n\n\n<li>Each nitrogen forms three bonds: one with carbon and two with hydrogen.<\/li>\n\n\n\n<li>Oxygen has two lone pairs; each nitrogen has one lone pair.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Molecular Geometry (VSEPR Theory)<\/strong><\/h3>\n\n\n\n<p><strong>VSEPR (Valence Shell Electron Pair Repulsion) Theory<\/strong> states that electron pairs around a central atom will arrange themselves to minimize repulsion.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Carbon Atom:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The carbon is bonded to three atoms: one oxygen and two nitrogens.<\/li>\n\n\n\n<li>It has <strong>no lone pairs<\/strong>.<\/li>\n\n\n\n<li>With three electron regions, the shape is <strong>trigonal planar<\/strong> (bond angles ~120\u00b0).<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Nitrogen Atom:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each nitrogen is bonded to <strong>three atoms<\/strong> (one C and two H).<\/li>\n\n\n\n<li>Each nitrogen has <strong>one lone pair<\/strong>, giving it four electron regions.<\/li>\n\n\n\n<li>The molecular geometry is <strong>trigonal pyramidal<\/strong>, similar to ammonia (NH\u2083).<\/li>\n<\/ul>\n\n\n\n<p>In conclusion, VSEPR theory predicts a <strong>trigonal planar<\/strong> geometry around the carbon atom due to three bonding groups and no lone pairs, and a <strong>trigonal pyramidal<\/strong> geometry around each nitrogen due to three bonds and one lone pair. These shapes help determine urea\u2019s structural and chemical behavior, such as hydrogen bonding and polarity.<\/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-banner4-762.jpeg\" alt=\"\" class=\"wp-image-230524\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw a Lewis structure for the urea molecule, CO(NH2)2. Draw the Lewis dot structure for urea. Include all lone pairs of electrons. Part B Predict its geometric shape around the C atom with the VSEPR theory. Predict its geometric shape around the N atom with the VSEPR theory. tetrahedral trigonal pyramidal bent trigonal planar Part [&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-230523","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230523","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=230523"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/230523\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=230523"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=230523"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=230523"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}