{"id":198999,"date":"2025-03-10T16:05:05","date_gmt":"2025-03-10T16:05:05","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=198999"},"modified":"2025-03-10T16:05:08","modified_gmt":"2025-03-10T16:05:08","slug":"draw-the-complete-orbital-diagram-for-formamide-hconh2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/10\/draw-the-complete-orbital-diagram-for-formamide-hconh2\/","title":{"rendered":"Draw the complete orbital diagram for formamide HCONH2"},"content":{"rendered":"\n<p>Draw the complete orbital diagram for formamide HCONH2. Clearly state the hybridization of each atom, clearly state the orbitals being drawn, and show the hybridization of each heteroatom.<\/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>Orbital Diagram and Hybridization of Formamide (HCONH\u2082)<\/strong><\/h3>\n\n\n\n<p>Formamide (HCONH\u2082) has the molecular formula <strong>HCONH\u2082<\/strong> and consists of the following atoms: carbon (C), oxygen (O), nitrogen (N), and hydrogen (H). To construct the complete orbital diagram, we must analyze the hybridization and bonding within the molecule.<\/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: Hybridization of Each Atom<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon in formamide forms <strong>three sigma (\u03c3) bonds<\/strong>: one with oxygen (O) and two with nitrogen (N) and hydrogen (H).<\/li>\n\n\n\n<li>It also forms <strong>one \u03c0-bond<\/strong> with oxygen due to the C=O bond.<\/li>\n\n\n\n<li>Carbon has <strong>sp\u00b2 hybridization<\/strong> (one s orbital + two p orbitals \u2192 three sp\u00b2 hybrid orbitals).<\/li>\n\n\n\n<li>The unhybridized <strong>p orbital<\/strong> is used to form the \u03c0 bond in the <strong>C=O<\/strong> double bond.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Oxygen (O)<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Oxygen forms <strong>one sigma bond<\/strong> (\u03c3) with carbon and a <strong>\u03c0-bond<\/strong> using a p orbital.<\/li>\n\n\n\n<li>It also has <strong>two lone pairs<\/strong>.<\/li>\n\n\n\n<li>Oxygen is <strong>sp\u00b2 hybridized<\/strong>: it uses three sp\u00b2 orbitals for the lone pairs and the sigma bond, while one unhybridized p orbital participates in \u03c0-bonding.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Nitrogen (N)<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen forms <strong>two sigma bonds<\/strong> (one with carbon, one with hydrogen).<\/li>\n\n\n\n<li>It also has a <strong>lone pair<\/strong> of electrons.<\/li>\n\n\n\n<li>Due to resonance, nitrogen is <strong>sp\u00b2 hybridized<\/strong> (rather than sp\u00b3). The lone pair occupies one of the sp\u00b2 orbitals.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Hydrogen (H)<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each hydrogen forms a single sigma bond using <strong>1s orbitals<\/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 2: Orbital Representation<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Sigma (\u03c3) bonds<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C=O sigma bond: sp\u00b2 (C) \u2013 sp\u00b2 (O)<\/li>\n\n\n\n<li>C-N sigma bond: sp\u00b2 (C) \u2013 sp\u00b2 (N)<\/li>\n\n\n\n<li>N-H sigma bonds: sp\u00b2 (N) \u2013 1s (H)<\/li>\n\n\n\n<li>C-H sigma bond: sp\u00b2 (C) \u2013 1s (H)<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Pi (\u03c0) bonds<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C=O \u03c0 bond: <strong>p (C) \u2013 p (O)<\/strong> (unhybridized orbitals).<\/li>\n\n\n\n<li>Nitrogen&#8217;s lone pair occupies an sp\u00b2 orbital, allowing delocalization.<\/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 Diagram Representation<\/strong><\/h3>\n\n\n\n<p>The <strong>orbital diagram<\/strong> consists of:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sigma (\u03c3) framework<\/strong>: Shows the overlap of sp\u00b2 hybrid orbitals forming single bonds.<\/li>\n\n\n\n<li><strong>Pi (\u03c0) framework<\/strong>: Shows the overlap of unhybridized p orbitals in the C=O bond.<\/li>\n\n\n\n<li><strong>Lone Pairs<\/strong>: Represented in sp\u00b2 hybrid orbitals on nitrogen and oxygen.<\/li>\n<\/ul>\n\n\n\n<p>The resonance between <strong>C=O and C-N<\/strong> results in partial double bond character, stabilizing the molecule.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the complete orbital diagram for formamide HCONH2. Clearly state the hybridization of each atom, clearly state the orbitals being drawn, and show the hybridization of each heteroatom. The correct answer and explanation is : Orbital Diagram and Hybridization of Formamide (HCONH\u2082) Formamide (HCONH\u2082) has the molecular formula HCONH\u2082 and consists of the following atoms: [&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-198999","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198999","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=198999"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/198999\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=198999"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=198999"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=198999"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}