{"id":184420,"date":"2025-01-20T20:34:38","date_gmt":"2025-01-20T20:34:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=184420"},"modified":"2025-01-20T20:34:40","modified_gmt":"2025-01-20T20:34:40","slug":"predict-the-molecular-geometry-of-each-interior-atom-in-acetic-acid-ch3cooh","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/20\/predict-the-molecular-geometry-of-each-interior-atom-in-acetic-acid-ch3cooh\/","title":{"rendered":"Predict the molecular geometry of each interior atom in acetic acid CH3COOH"},"content":{"rendered":"\n<p>Predict the molecular geometry of each interior atom in acetic acid CH3COOH. Drag the appropriate labels to their respective targets. linear trigonal planar bent tetrahedral Atom Carbon (left) Number of Electron Groups 4 3 Number of Lone Pairs 0 0 2 Molecular Geometry  Carbon (right) Oxygen<\/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 determine the molecular geometry of each interior atom in acetic acid (CH\u2083COOH), we examine the structure and consider the number of electron groups and lone pairs around each atom.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>Carbon (left) \u2013 CH\u2083 group<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Number of Electron Groups<\/strong>: 4 (three single bonds to hydrogen and one single bond to the adjacent carbon).<\/li>\n\n\n\n<li><strong>Number of Lone Pairs<\/strong>: 0.<\/li>\n\n\n\n<li><strong>Molecular Geometry<\/strong>: <strong>Tetrahedral<\/strong>. This carbon atom is sp\u00b3 hybridized, with its bonds arranged in a tetrahedral geometry to minimize electron repulsion.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Carbon (right) \u2013 COOH group<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Number of Electron Groups<\/strong>: 3 (one double bond to oxygen, one single bond to the other oxygen, and one single bond to the adjacent carbon).<\/li>\n\n\n\n<li><strong>Number of Lone Pairs<\/strong>: 0.<\/li>\n\n\n\n<li><strong>Molecular Geometry<\/strong>: <strong>Trigonal planar<\/strong>. This carbon atom is sp\u00b2 hybridized. The three groups are arranged in a planar triangular shape, ensuring minimal electron pair repulsion.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Oxygen (in C=O double bond)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Number of Electron Groups<\/strong>: 3 (one double bond to carbon and two lone pairs).<\/li>\n\n\n\n<li><strong>Number of Lone Pairs<\/strong>: 2.<\/li>\n\n\n\n<li><strong>Molecular Geometry<\/strong>: <strong>Trigonal planar<\/strong>. The oxygen atom is sp\u00b2 hybridized. Its lone pairs and the bonding electrons create a planar arrangement.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">4. <strong>Oxygen (in OH group)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Number of Electron Groups<\/strong>: 4 (one single bond to carbon and two lone pairs).<\/li>\n\n\n\n<li><strong>Number of Lone Pairs<\/strong>: 2.<\/li>\n\n\n\n<li><strong>Molecular Geometry<\/strong>: <strong>Bent<\/strong>. The oxygen atom is sp\u00b3 hybridized. The two lone pairs force the molecular geometry into a bent shape.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (left): <strong>Tetrahedral<\/strong>.<\/li>\n\n\n\n<li>Carbon (right): <strong>Trigonal planar<\/strong>.<\/li>\n\n\n\n<li>Oxygen (C=O): <strong>Trigonal planar<\/strong>.<\/li>\n\n\n\n<li>Oxygen (OH): <strong>Bent<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The molecular geometry is determined by the VSEPR (Valence Shell Electron Pair Repulsion) theory, which predicts that electron groups around a central atom will arrange themselves to minimize repulsion. Lone pairs take up more space than bonding pairs, which affects the observed geometry.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Predict the molecular geometry of each interior atom in acetic acid CH3COOH. Drag the appropriate labels to their respective targets. linear trigonal planar bent tetrahedral Atom Carbon (left) Number of Electron Groups 4 3 Number of Lone Pairs 0 0 2 Molecular Geometry Carbon (right) Oxygen The Correct Answer and Explanation is : To determine [&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-184420","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184420","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=184420"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184420\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184420"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184420"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184420"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}