{"id":231801,"date":"2025-06-11T11:49:39","date_gmt":"2025-06-11T11:49:39","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=231801"},"modified":"2025-06-11T11:49:41","modified_gmt":"2025-06-11T11:49:41","slug":"a-nitrogen-atom-can-undergo-sp2-hybridization-when-it-becomes-part-of-a-carbon-nitrogen-double-bond-as-in-h2cnh-a-using-a-sketch","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/11\/a-nitrogen-atom-can-undergo-sp2-hybridization-when-it-becomes-part-of-a-carbon-nitrogen-double-bond-as-in-h2cnh-a-using-a-sketch\/","title":{"rendered":"A nitrogen atom can undergo sp2 hybridization when it becomes part of a carbon-nitrogen double bond, as in H2C=NH (a) Using a sketch"},"content":{"rendered":"\n<p>A nitrogen atom can undergo sp2 hybridization when it becomes part of a carbon-nitrogen double bond, as in H2C=NH (a) Using a sketch , show the electron configuration of sp2 hybridized nitrogen just before the overlapping occurs to make this double bond (b) Using sketches (and the analogy to the double bond in C2H4), describe the two bonds of the carbon-nitrogen double bond (c) Describe the geometry of H2C=NH (using a sketch that shows all expected bond angle)<\/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<h3 class=\"wp-block-heading\">Hybridization and Bonding in H\u2082C=NH (Iminomethane or Methanimine)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">(a) Electron Configuration of sp\u00b2 Hybridized Nitrogen Before Overlap<\/h4>\n\n\n\n<p>The ground-state electron configuration of nitrogen (atomic number 7) is:<\/p>\n\n\n\n<p><strong>1s\u00b2 2s\u00b2 2p\u00b3<\/strong><\/p>\n\n\n\n<p>In sp\u00b2 hybridization, one 2s orbital mixes with two 2p orbitals, leaving one unhybridized 2p orbital:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>sp\u00b2 hybrid orbitals:<\/strong> 3 orbitals (used for \u03c3 bonding and lone pair)<\/li>\n\n\n\n<li><strong>Unhybridized 2p orbital:<\/strong> 1 orbital (used for \u03c0 bonding)<\/li>\n<\/ul>\n\n\n\n<p><strong>Sketch (conceptual, text description):<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">yamlCopyEdit<code>Before hybridization:             After sp\u00b2 hybridization:\n\n2s  \u2191\u2193                            sp\u00b2  \u2191  \u2191  \u2191   (3 hybrid orbitals)\n2p  \u2191  \u2191  \u2191                       2p  \u2191         (1 unhybridized p orbital)\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One sp\u00b2 hybrid orbital contains the lone pair.<\/li>\n\n\n\n<li>Two sp\u00b2 orbitals form \u03c3 bonds (one with C, one with H).<\/li>\n\n\n\n<li>The unhybridized 2p orbital participates in \u03c0 bonding with carbon.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(b) Description of the Two Bonds in the C=N Double Bond<\/h4>\n\n\n\n<p>The carbon in H\u2082C=NH is also sp\u00b2 hybridized:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 bond:<\/strong> Formed by overlap of carbon&#8217;s sp\u00b2 hybrid orbital with nitrogen&#8217;s sp\u00b2 hybrid orbital.<\/li>\n\n\n\n<li><strong>\u03c0 bond:<\/strong> Formed by sideways overlap of unhybridized 2p orbitals from both carbon and nitrogen.<\/li>\n<\/ul>\n\n\n\n<p><strong>Analogy to C\u2082H\u2084 (Ethene):<\/strong><br>Just as in ethene, where each carbon forms a \u03c3 bond with one another using sp\u00b2 orbitals and a \u03c0 bond with p orbitals, here nitrogen and carbon behave similarly.<\/p>\n\n\n\n<p><strong>Sketch (conceptual):<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>       \u03c0 overlap (p orbital on N and C)\n        \u2191     \u2191\n\nC (sp\u00b2) \u2014 \u03c3 \u2014 N (sp\u00b2)\n       |\n    H  H\n<\/code><\/pre>\n\n\n\n<h4 class=\"wp-block-heading\">(c) Geometry of H\u2082C=NH<\/h4>\n\n\n\n<p>Because both carbon and nitrogen are sp\u00b2 hybridized, the geometry around both atoms is approximately <strong>trigonal planar<\/strong>.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bond angles: ~120\u00b0<\/li>\n\n\n\n<li>The lone pair on nitrogen slightly compresses the H\u2013N\u2013C angle to slightly less than 120\u00b0.<\/li>\n<\/ul>\n\n\n\n<p><strong>Sketch (conceptual):<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>       H\n        \\\n         C = N \u2014 H\n        \/\n      H\n\nBond angles ~120\u00b0\n<\/code><\/pre>\n\n\n\n<h4 class=\"wp-block-heading\">Summary Explanation (300 words)<\/h4>\n\n\n\n<p>In H\u2082C=NH, the carbon atom forms a double bond with nitrogen, requiring both atoms to adopt sp\u00b2 hybridization to maximize orbital overlap and minimize electron repulsion. Nitrogen\u2019s electron configuration of 1s\u00b2 2s\u00b2 2p\u00b3 is modified during hybridization: one 2s orbital and two 2p orbitals combine to form three sp\u00b2 hybrid orbitals, while one 2p orbital remains unhybridized. One sp\u00b2 hybrid orbital holds the nitrogen\u2019s lone pair, while the remaining two form sigma (\u03c3) bonds: one with carbon and one with hydrogen.<\/p>\n\n\n\n<p>Simultaneously, carbon, which also undergoes sp\u00b2 hybridization, uses one sp\u00b2 orbital to form a \u03c3 bond with nitrogen. The unhybridized p orbitals on both carbon and nitrogen lie parallel and overlap sideways, forming a pi (\u03c0) bond. This \u03c0 bond constitutes the second bond in the C=N double bond, analogous to the bonding in ethene (C\u2082H\u2084).<\/p>\n\n\n\n<p>The molecular geometry around both carbon and nitrogen is approximately trigonal planar due to sp\u00b2 hybridization, with ideal bond angles of 120\u00b0. However, the lone pair on nitrogen exerts greater repulsion, slightly decreasing the H\u2013N\u2013C bond angle. The planar structure allows maximum p orbital overlap, stabilizing the \u03c0 bond. This hybridization and bonding arrangement account for both the structure and reactivity of imines like H\u2082C=NH.<\/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-banner4-975.jpeg\" alt=\"\" class=\"wp-image-231802\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A nitrogen atom can undergo sp2 hybridization when it becomes part of a carbon-nitrogen double bond, as in H2C=NH (a) Using a sketch , show the electron configuration of sp2 hybridized nitrogen just before the overlapping occurs to make this double bond (b) Using sketches (and the analogy to the double bond in C2H4), describe [&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-231801","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231801","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=231801"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/231801\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=231801"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=231801"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=231801"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}