{"id":202658,"date":"2025-03-17T18:58:53","date_gmt":"2025-03-17T18:58:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202658"},"modified":"2025-03-17T18:58:55","modified_gmt":"2025-03-17T18:58:55","slug":"show-a-lewis-structure-for-urea","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/17\/show-a-lewis-structure-for-urea\/","title":{"rendered":"Show a Lewis structure for urea"},"content":{"rendered":"\n<p>Show a Lewis structure for urea. CH4N2O. Both N\u2019s and the O are bonded to the C. The H\u2019s are bonded to the N\u2019s. None of the atoms has a formal charge.<\/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<p>The Lewis structure for urea (CH\u2084N\u2082O) can be drawn by considering the bonding and electron distribution for each atom. Here\u2019s a step-by-step explanation of how to draw the structure:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Count the total valence electrons.<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> is in group 14 and has 4 valence electrons.<\/li>\n\n\n\n<li><strong>Nitrogen (N)<\/strong> is in group 15 and has 5 valence electrons. Since there are two nitrogen atoms, the total is ( 5 \\times 2 = 10 ) valence electrons.<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong> is in group 1 and has 1 valence electron. Since there are four hydrogen atoms, the total is ( 1 \\times 4 = 4 ) valence electrons.<\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong> is in group 16 and has 6 valence electrons.<\/li>\n<\/ul>\n\n\n\n<p>The total number of valence electrons is:<br>[<br>4 \\text{ (from C)} + 10 \\text{ (from N)} + 4 \\text{ (from H)} + 6 \\text{ (from O)} = 24 \\text{ valence electrons.}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the skeletal structure.<\/h3>\n\n\n\n<p>From the molecular formula (CH\u2084N\u2082O), we know:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The carbon (C) is the central atom.<\/li>\n\n\n\n<li>Each nitrogen (N) is bonded to carbon (C).<\/li>\n\n\n\n<li>The hydrogens (H) are bonded to the nitrogen atoms.<\/li>\n\n\n\n<li>Oxygen (O) is also bonded to carbon.<\/li>\n<\/ul>\n\n\n\n<p>The skeletal structure is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C<\/strong> is the central atom, bonded to both <strong>N<\/strong> atoms and <strong>O<\/strong>.<\/li>\n\n\n\n<li>Each <strong>N<\/strong> atom is bonded to two <strong>H<\/strong> atoms.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Distribute the electrons to form bonds.<\/h3>\n\n\n\n<p>Each single bond consists of two electrons. We place single bonds between C and N, and between N and H. We also place a single bond between C and O.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C-N<\/strong> bond: 2 electrons.<\/li>\n\n\n\n<li><strong>N-H<\/strong> bonds: Each N has two H atoms, so 4 electrons per N are used for N-H bonds.<\/li>\n\n\n\n<li><strong>C=O<\/strong>: A double bond between C and O uses 4 electrons.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Fill the octet rule.<\/h3>\n\n\n\n<p>Now, distribute the remaining electrons to satisfy the octet rule for each atom:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C<\/strong>: The carbon has 4 bonds (two N-H bonds, one C-N bond, and one C=O bond), satisfying its octet.<\/li>\n\n\n\n<li><strong>N<\/strong>: Each nitrogen is involved in three bonds (C-N and N-H), which satisfies the octet.<\/li>\n\n\n\n<li><strong>O<\/strong>: Oxygen has two bonds (C=O) and two lone pairs, completing its octet.<\/li>\n\n\n\n<li><strong>H<\/strong>: Each hydrogen is bonded to nitrogen and satisfies its duet rule.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Verify the formal charges.<\/h3>\n\n\n\n<p>To ensure that the atoms have no formal charge:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The formal charge on <strong>C<\/strong> is 0 because it has 4 bonds and 0 lone pairs.<\/li>\n\n\n\n<li>The formal charge on <strong>N<\/strong> is 0 because each nitrogen has 3 bonds and 1 lone pair.<\/li>\n\n\n\n<li>The formal charge on <strong>O<\/strong> is 0 because it has 2 bonds and 2 lone pairs.<\/li>\n\n\n\n<li>The formal charge on each <strong>H<\/strong> is 0 because each hydrogen has 1 bond and no lone pairs.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Final Lewis Structure for Urea:<\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>   H2N - C - NH2\n        ||\n        O<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The structure has no formal charges on any of the atoms.<\/li>\n\n\n\n<li>Carbon (C) is bonded to two nitrogen atoms (N) and one oxygen atom (O).<\/li>\n\n\n\n<li>Nitrogen atoms (N) are each bonded to two hydrogen atoms (H).<\/li>\n\n\n\n<li>Oxygen (O) is double-bonded to the central carbon.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the Lewis structure for urea is complete, satisfying the octet rule for all atoms involved, with no formal charges.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Show a Lewis structure for urea. CH4N2O. Both N\u2019s and the O are bonded to the C. The H\u2019s are bonded to the N\u2019s. None of the atoms has a formal charge. The correct answer and explanation is : The Lewis structure for urea (CH\u2084N\u2082O) can be drawn by considering the bonding and electron distribution [&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-202658","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202658","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=202658"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202658\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202658"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202658"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202658"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}