{"id":145276,"date":"2024-09-26T17:10:41","date_gmt":"2024-09-26T17:10:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=145276"},"modified":"2024-09-26T17:10:44","modified_gmt":"2024-09-26T17:10:44","slug":"draw-the-lewis-structure-of-bh3-and-then-determine-its-electrondomain-and-molecular-geometries","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/09\/26\/draw-the-lewis-structure-of-bh3-and-then-determine-its-electrondomain-and-molecular-geometries\/","title":{"rendered":"Draw the lewis structure of BH3 and then determine its electrondomain and molecular geometries"},"content":{"rendered":"\n<p>Draw the lewis structure of BH3 and then determine its electron<br>domain and molecular geometries.<br>a) trigonal \/ planar<br>b) trigonal \/ tetrahedral<br>c) trigonal planar\/ tetrahedral<br>d) trigonal \/ bent (120<\/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 Lewis structure, electron domain geometry, and molecular geometry of borane (BH\u2083), let&#8217;s go through each step carefully.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Draw the Lewis Structure of BH\u2083<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Boron (B)<\/strong> is in Group 13 of the periodic table and has 3 valence electrons.<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong> is in Group 1 and has 1 valence electron. Since there are three hydrogens, the total number of valence electrons for BH\u2083 is: [<br>3 (\\text{from B}) + 3 \\times 1 (\\text{from each H}) = 6 \\text{ total valence electrons}<br>]<\/li>\n\n\n\n<li>In the Lewis structure, the boron atom will be placed at the center with three hydrogen atoms bonded to it, using all 6 electrons to form three B-H single bonds. Each hydrogen atom achieves its stable duet, while boron has only six electrons (an incomplete octet), which is typical for boron.<\/li>\n<\/ul>\n\n\n\n<p>The Lewis structure looks like this:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>  H\n  |\nH-B-H<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine Electron Domain Geometry<\/h3>\n\n\n\n<p>The electron domain geometry is based on the regions of electron density (bonding or lone pairs) around the central atom. In BH\u2083, boron has 3 bonding pairs of electrons and no lone pairs.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron domain geometry<\/strong> is determined by the number of electron pairs around the central atom. Since there are 3 bonding pairs, the electron domain geometry is <strong>trigonal planar<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Determine Molecular Geometry<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molecular geometry<\/strong> is determined by the positions of the atoms, not the electron pairs. Since BH\u2083 has no lone pairs and the 3 hydrogen atoms are arranged symmetrically around the boron atom, the molecular geometry is also <strong>trigonal planar<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Answer<\/h3>\n\n\n\n<p>The correct answer is <strong>(a) trigonal \/ planar<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Boron in BH\u2083 has three bonding pairs and no lone pairs, resulting in 3 regions of electron density. According to VSEPR (Valence Shell Electron Pair Repulsion) theory, the arrangement that minimizes repulsion between these regions is a trigonal planar shape.<\/li>\n\n\n\n<li>In a <strong>trigonal planar<\/strong> geometry, the bond angles are approximately <strong>120\u00b0<\/strong>, as the atoms and electron pairs spread out evenly in a flat, triangular shape around the central atom.<\/li>\n\n\n\n<li>Boron does not obey the octet rule here, as it only has 6 valence electrons, which is common for elements in Group 13.<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Draw the lewis structure of BH3 and then determine its electrondomain and molecular geometries.a) trigonal \/ planarb) trigonal \/ tetrahedralc) trigonal planar\/ tetrahedrald) trigonal \/ bent (120 The Correct Answer and Explanation is : To determine the Lewis structure, electron domain geometry, and molecular geometry of borane (BH\u2083), let&#8217;s go through each step carefully. Step [&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-145276","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/145276","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=145276"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/145276\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=145276"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=145276"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=145276"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}