{"id":224420,"date":"2025-06-03T05:53:31","date_gmt":"2025-06-03T05:53:31","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224420"},"modified":"2025-06-03T05:53:33","modified_gmt":"2025-06-03T05:53:33","slug":"draw-the-lewis-structure-of-bh3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/03\/draw-the-lewis-structure-of-bh3\/","title":{"rendered":"Draw the Lewis structure of BH3."},"content":{"rendered":"\n<p>Draw the Lewis structure of BH3.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/image-55.png\" alt=\"\" class=\"wp-image-224421\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Lewis Structure for BH\u2083:<\/h3>\n\n\n\n<p>To draw the correct Lewis structure for borane (BH\u2083):<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>Place the boron (B) atom in the center since hydrogen (H) can form only one bond and cannot be the central atom.<\/li>\n\n\n\n<li>Attach three hydrogen atoms to the boron atom using single bonds.<\/li>\n\n\n\n<li>Do <strong>not<\/strong> add lone pairs to hydrogen, as each hydrogen only needs 2 electrons (1 bond).<\/li>\n\n\n\n<li>Boron will have only 6 electrons (3 bonds), which is an exception to the octet rule.<\/li>\n<\/ol>\n\n\n\n<h4 class=\"wp-block-heading\">Lewis Structure:<\/h4>\n\n\n\n<pre class=\"wp-block-code\"><code>      H\n      |\nH \u2014 B \u2014 H\n<\/code><\/pre>\n\n\n\n<p>Each line represents a bonding pair of electrons.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>The Lewis structure of BH\u2083 (borane) illustrates the bonding arrangement between boron and hydrogen atoms. BH\u2083 is a molecule composed of one boron (B) atom and three hydrogen (H) atoms. To draw the Lewis structure, we begin by determining the number of valence electrons:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Boron (Group 13): 3 valence electrons<\/li>\n\n\n\n<li>Hydrogen (Group 1): 1 valence electron \u00d7 3 = 3<\/li>\n\n\n\n<li>Total = 3 (B) + 3 (H) = <strong>6 valence electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p>In the Lewis structure, the boron atom is placed at the center with three single bonds connecting it to three hydrogen atoms. Each single bond (B\u2013H) consists of 2 electrons. Thus, 3 bonds use all 6 available valence electrons.<\/p>\n\n\n\n<p>Hydrogen atoms follow the <strong>duet rule<\/strong>, meaning they are stable with 2 electrons (1 bond). Each H atom in BH\u2083 forms a single bond with B, which satisfies this requirement.<\/p>\n\n\n\n<p>Boron, however, is an <strong>exception to the octet rule<\/strong>. In BH\u2083, boron has only 6 electrons around it (3 single bonds), making it electron-deficient. It does not have a complete octet, but this is acceptable for boron compounds, especially in small molecules like BH\u2083.<\/p>\n\n\n\n<p>This electron deficiency makes BH\u2083 highly reactive; it often acts as a <strong>Lewis acid<\/strong>, meaning it can accept a pair of electrons to complete its octet. In chemistry, BH\u2083 is often seen in reactions where it forms adducts with Lewis bases.<\/p>\n\n\n\n<p>In summary, the Lewis structure for BH\u2083 is a simple molecule with three B\u2013H bonds and no lone pairs. It illustrates key concepts such as exceptions to the octet rule and the stability of hydrogen with only two electrons.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-187.jpeg\" alt=\"\" class=\"wp-image-224422\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure of BH3. The Correct Answer and Explanation is: Correct Lewis Structure for BH\u2083: To draw the correct Lewis structure for borane (BH\u2083): Lewis Structure: Each line represents a bonding pair of electrons. Explanation The Lewis structure of BH\u2083 (borane) illustrates the bonding arrangement between boron and hydrogen atoms. BH\u2083 is a [&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-224420","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224420","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=224420"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224420\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224420"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224420"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224420"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}