{"id":206961,"date":"2025-04-17T14:46:27","date_gmt":"2025-04-17T14:46:27","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=206961"},"modified":"2025-04-17T14:46:33","modified_gmt":"2025-04-17T14:46:33","slug":"what-is-the-hybridization-of-the-carbon-in-hcn","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/17\/what-is-the-hybridization-of-the-carbon-in-hcn\/","title":{"rendered":"What is the hybridization of the carbon in HCN"},"content":{"rendered":"\n<p>What is the hybridization of the carbon in HCN? A. sp B. sp C. sp3 D. sp3d<br>A. sp<br>B. sp2<br>C. sp\u00b3<br>D. sp\u00b3d<br>E. sp\u00b3d2<\/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><strong>Correct Answer: A. sp<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (\u2248300 words):<\/strong><\/h3>\n\n\n\n<p>To determine the <strong>hybridization of the carbon atom in hydrogen cyanide (HCN)<\/strong>, we need to analyze the molecule\u2019s structure and bonding.<\/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: Lewis Structure of HCN<\/strong><\/h3>\n\n\n\n<p>HCN consists of three atoms: <strong>hydrogen (H)<\/strong>, <strong>carbon (C)<\/strong>, and <strong>nitrogen (N)<\/strong>. The correct Lewis structure is:<\/p>\n\n\n\n<p><strong>H\u2013C\u2261N<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hydrogen forms <strong>one single bond<\/strong> (H\u2013C).<\/li>\n\n\n\n<li>Carbon forms a <strong>single bond with H<\/strong> and a <strong>triple bond with N<\/strong>.<\/li>\n\n\n\n<li>Nitrogen has a lone pair of electrons.<\/li>\n<\/ul>\n\n\n\n<p>So, the carbon is bonded to <strong>two atoms<\/strong>: one hydrogen and one nitrogen. Even though the triple bond with nitrogen includes multiple electron pairs, <strong>it still counts as one region of electron density<\/strong> in terms of hybridization.<\/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 2: Electron Domain Geometry<\/strong><\/h3>\n\n\n\n<p>Hybridization depends on the <strong>number of regions of electron density<\/strong> (bonding domains and lone pairs) around the atom. In this case:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The carbon has <strong>two regions of electron density<\/strong> (H and N).<\/li>\n<\/ul>\n\n\n\n<p>Thus, carbon in HCN has a <strong>linear geometry<\/strong> and requires <strong>two hybrid orbitals<\/strong>.<\/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 3: Type of Hybridization<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>2 regions of electron density \u2192 sp hybridization<\/strong><\/li>\n\n\n\n<li><strong>3 regions \u2192 sp\u00b2<\/strong><\/li>\n\n\n\n<li><strong>4 regions \u2192 sp\u00b3<\/strong><\/li>\n\n\n\n<li><strong>5 regions \u2192 sp\u00b3d<\/strong><\/li>\n\n\n\n<li><strong>6 regions \u2192 sp\u00b3d\u00b2<\/strong><\/li>\n<\/ul>\n\n\n\n<p>Therefore, with two regions of electron density, <strong>carbon in HCN is sp hybridized<\/strong>.<\/p>\n\n\n\n<p>This hybridization allows the carbon to form:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One <strong>\u03c3 bond with hydrogen<\/strong> using an sp orbital.<\/li>\n\n\n\n<li>One <strong>\u03c3 bond with nitrogen<\/strong> using the other sp orbital.<\/li>\n\n\n\n<li>Two <strong>\u03c0 bonds<\/strong> (from the triple bond) are formed from the unhybridized <strong>p orbitals<\/strong> on carbon and nitrogen.<\/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>Conclusion<\/strong><\/h3>\n\n\n\n<p>The correct hybridization of the carbon atom in HCN is <strong>sp<\/strong>, which corresponds to option:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p><strong>A. sp<\/strong> \u2705<\/p>\n<\/blockquote>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/04\/image-117.png\" alt=\"\" class=\"wp-image-206962\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>What is the hybridization of the carbon in HCN? A. sp B. sp C. sp3 D. sp3dA. spB. sp2C. sp\u00b3D. sp\u00b3dE. sp\u00b3d2 The correct answer and explanation is : Correct Answer: A. sp Explanation (\u2248300 words): To determine the hybridization of the carbon atom in hydrogen cyanide (HCN), we need to analyze the molecule\u2019s structure [&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-206961","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206961","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=206961"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/206961\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=206961"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=206961"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=206961"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}