{"id":189062,"date":"2025-02-08T17:24:00","date_gmt":"2025-02-08T17:24:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=189062"},"modified":"2025-02-08T17:24:03","modified_gmt":"2025-02-08T17:24:03","slug":"draw-the-lewis-structure-of-chcch-and-then-choose-the-appropriate-set-of-hybridization-states-for-the-three-central-atoms","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/08\/draw-the-lewis-structure-of-chcch-and-then-choose-the-appropriate-set-of-hybridization-states-for-the-three-central-atoms\/","title":{"rendered":"Draw the Lewis structure of CH,CCH and then choose the appropriate set of hybridization states for the three central atoms"},"content":{"rendered":"\n<p>Draw the Lewis structure of CH,CCH and then choose the appropriate set of hybridization states for the three central atoms. Your answer choice is independent of the orientation of your drawn structure.<\/p>\n\n\n\n<p>A) sp\/sp\/sp<br>B) sp\u00b3\/sp\/sp<br>C) sp\u00b3\/sp\u00b2\/sp\u00b2<br>D) sp\u00b2\/sp\/sp\u00b3<br>E) sp\/sp\/sp\u00b3<\/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>To address this question, let&#8217;s break down the molecule and structure.<\/p>\n\n\n\n<p>The molecule given is <strong>CH\u2082CCH<\/strong>, which is a propene-like structure, also known as <strong>propylene<\/strong>. The structure of <strong>CH\u2082CCH<\/strong> consists of three carbon atoms and a set of hydrogens. The central atoms are the two carbons (C2 and C3) in the middle, and C1 is on the left.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Draw the Lewis Structure<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Carbon 1 (C1)<\/strong>: This carbon is bonded to two hydrogen atoms (H) and is double-bonded to Carbon 2 (C2).<\/li>\n\n\n\n<li><strong>Carbon 2 (C2)<\/strong>: This carbon is double-bonded to Carbon 1 (C1) and single-bonded to Carbon 3 (C3), which is in turn bonded to another hydrogen atom.<\/li>\n\n\n\n<li><strong>Carbon 3 (C3)<\/strong>: This carbon is single-bonded to Carbon 2 (C2) and bonded to one hydrogen atom.<\/li>\n<\/ol>\n\n\n\n<p>The structure can be represented as:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   H   H\n   |   |\nH-C=C-C-H\n   |  \n   H<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the Hybridization States of the Central Atoms<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon 1 (C1)<\/strong>: This carbon is involved in a <strong>double bond<\/strong> with C2 and has two single bonds with hydrogens. This suggests that it has <strong>sp<\/strong> hybridization, as it forms two sigma bonds (one to C2, one to H) and one pi bond in the double bond with C2.<\/li>\n\n\n\n<li><strong>Carbon 2 (C2)<\/strong>: This carbon is involved in a <strong>double bond<\/strong> with C1 and a <strong>single bond<\/strong> with C3. This carbon forms three sigma bonds and one pi bond, suggesting that it has <strong>sp\u00b2<\/strong> hybridization.<\/li>\n\n\n\n<li><strong>Carbon 3 (C3)<\/strong>: This carbon forms a <strong>single bond<\/strong> with C2 and a <strong>single bond<\/strong> with hydrogen. It forms two sigma bonds and no pi bonds, suggesting <strong>sp\u00b3<\/strong> hybridization.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Answer Choice<\/h3>\n\n\n\n<p>Based on the analysis above, the appropriate set of hybridizations for the central atoms is:<\/p>\n\n\n\n<p><strong>D) sp\u00b2\/sp\/sp\u00b3<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon 1 (sp)<\/strong>: A double bond with C2 (one pi and one sigma bond) and two single bonds (one to H, one to C2).<\/li>\n\n\n\n<li><strong>Carbon 2 (sp\u00b2)<\/strong>: A double bond with C1 (one pi and one sigma bond) and a single bond with C3.<\/li>\n\n\n\n<li><strong>Carbon 3 (sp\u00b3)<\/strong>: A single bond to C2 and a single bond to hydrogen, forming two sigma bonds.<\/li>\n<\/ul>\n\n\n\n<p>This configuration matches the structure of CH\u2082CCH, and thus the correct choice is <strong>D) sp\u00b2\/sp\/sp\u00b3<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure of CH,CCH and then choose the appropriate set of hybridization states for the three central atoms. Your answer choice is independent of the orientation of your drawn structure. A) sp\/sp\/spB) sp\u00b3\/sp\/spC) sp\u00b3\/sp\u00b2\/sp\u00b2D) sp\u00b2\/sp\/sp\u00b3E) sp\/sp\/sp\u00b3 The Correct Answer and Explanation is : To address this question, let&#8217;s break down the molecule and [&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-189062","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189062","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=189062"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/189062\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=189062"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=189062"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=189062"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}