{"id":201321,"date":"2025-03-15T08:31:25","date_gmt":"2025-03-15T08:31:25","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201321"},"modified":"2025-03-15T08:31:28","modified_gmt":"2025-03-15T08:31:28","slug":"what-is-the-molecular-geometry-at-the-central-atom-in-ch%e2%82%82br%e2%82%82","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/what-is-the-molecular-geometry-at-the-central-atom-in-ch%e2%82%82br%e2%82%82\/","title":{"rendered":"What is the molecular geometry at the central atom in CH\u2082Br\u2082"},"content":{"rendered":"\n<p>What is the molecular geometry at the central atom in CH\u2082Br\u2082?<br>A. trigonal planar<br>B. trigonal pyramidal<br>C. square planar<br>D. tetrahedral<br>E. None of these<\/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>The molecular geometry at the central atom in <strong>CH\u2082Br\u2082<\/strong> is <strong>tetrahedral<\/strong>.<\/p>\n\n\n\n<p>To understand why, let&#8217;s break it down step by step:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Draw the structure of CH\u2082Br\u2082<\/h3>\n\n\n\n<p>In CH\u2082Br\u2082 (dibromomethane), the central atom is carbon (C), which is bonded to two hydrogen (H) atoms and two bromine (Br) atoms. Therefore, the structure can be written as:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C is the central atom.<\/li>\n\n\n\n<li>It has four single bonds: two to hydrogen atoms and two to bromine atoms.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Count the number of electron pairs around the central atom<\/h3>\n\n\n\n<p>Carbon (C) in CH\u2082Br\u2082 is in Group 14 of the periodic table, so it has 4 valence electrons. In this molecule, each single bond involves one of these valence electrons. Since C is bonded to 4 other atoms (2 hydrogens and 2 bromines), it has 4 bonding pairs of electrons.<\/p>\n\n\n\n<p>There are no lone pairs on the carbon atom. Therefore, the central carbon has 4 bonding pairs of electrons, and no lone pairs.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine the electron pair geometry<\/h3>\n\n\n\n<p>The electron pair geometry depends on the number of bonding pairs around the central atom. According to the VSEPR (Valence Shell Electron Pair Repulsion) theory:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>4 bonding pairs of electrons around a central atom will arrange themselves in a <strong>tetrahedral<\/strong> geometry to minimize electron-electron repulsion.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Conclusion \u2013 Molecular Geometry<\/h3>\n\n\n\n<p>Since there are no lone pairs of electrons on the carbon atom, the <strong>molecular geometry<\/strong> is the same as the electron pair geometry, which is <strong>tetrahedral<\/strong>.<\/p>\n\n\n\n<p>Thus, the correct answer is:<\/p>\n\n\n\n<p><strong>D. tetrahedral<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Why not the other options?<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>A. Trigonal planar<\/strong>: This geometry occurs with 3 bonding pairs of electrons and no lone pairs, which is not the case for CH\u2082Br\u2082.<\/li>\n\n\n\n<li><strong>B. Trigonal pyramidal<\/strong>: This geometry occurs when there are 3 bonding pairs and 1 lone pair, which again, is not the case here.<\/li>\n\n\n\n<li><strong>C. Square planar<\/strong>: This geometry occurs in molecules with 4 bonding pairs and 2 lone pairs (e.g., XeF\u2084), not in CH\u2082Br\u2082.<\/li>\n\n\n\n<li><strong>E. None of these<\/strong>: As we&#8217;ve seen, the correct answer is tetrahedral, so this option is incorrect.<\/li>\n<\/ul>\n\n\n\n<p>Therefore, <strong>tetrahedral<\/strong> is the correct molecular geometry for CH\u2082Br\u2082.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the molecular geometry at the central atom in CH\u2082Br\u2082?A. trigonal planarB. trigonal pyramidalC. square planarD. tetrahedralE. None of these The correct answer and explanation is : The molecular geometry at the central atom in CH\u2082Br\u2082 is tetrahedral. To understand why, let&#8217;s break it down step by step: Step 1: Draw the structure of [&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-201321","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201321","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=201321"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201321\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201321"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201321"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201321"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}