{"id":250938,"date":"2025-07-10T13:28:23","date_gmt":"2025-07-10T13:28:23","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=250938"},"modified":"2025-07-10T13:28:25","modified_gmt":"2025-07-10T13:28:25","slug":"draw-the-lewis-structure-of-bromoform-chbr3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/10\/draw-the-lewis-structure-of-bromoform-chbr3\/","title":{"rendered":"Draw the Lewis structure of bromoform (CHBr3)."},"content":{"rendered":"\n<p>Draw the Lewis structure of bromoform (CHBr3). 2. What is the hybridization type of the carbon atom? 3. What orbitals are involved in the formation of the C-H bond? 4. What orbitals are involved in the formation of the C-Br bond? 5. What is the electron geometry around the carbon atom? 2. Corrected_text: 1. Draw the Lewis diagram of hydroxylamine (HNOH). 2. What is the hybridization type of the N and O atoms? 3. What is the electron geometry around the N and O atoms? 4. What orbitals are involved in the formation of the N-O bond? Title_with_topic: Lewis Structures and Hybridization in Bromoform and Hydroxylamine<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Lewis Structure of Bromoform (CHBr3):<\/h3>\n\n\n\n<p><strong>Step-by-step:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) is the central atom, with 4 valence electrons.<\/li>\n\n\n\n<li>Hydrogen (H) and Bromine (Br) are bonded to Carbon.<\/li>\n\n\n\n<li>Bromine (Br) atoms each have 7 valence electrons.<\/li>\n\n\n\n<li>Hydrogen (H) has 1 valence electron.<\/li>\n<\/ul>\n\n\n\n<p>The structure of <strong>CHBr3<\/strong> can be drawn as:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\"><code>    H<br>        |<br>Br - C - Br-O<br>        |<br>        Br<br><\/code><\/pre>\n\n\n\n<p>In this structure, the carbon atom is bonded to three bromine atoms and one hydrogen atom. Each bond to bromine is a single bond, and the bromines each have three lone pairs of electrons. The hydrogen atom is bonded to carbon with a single bond. Carbon completes its octet by sharing electrons with the three bromines and the hydrogen atom.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. Hybridization of the Carbon Atom in Bromoform (CHBr3):<\/h3>\n\n\n\n<p>The carbon atom in CHBr3 forms four bonds (three single C-Br bonds and one C-H bond). Since there are four bonding regions around the carbon atom, the carbon undergoes <strong>sp\u00b3 hybridization<\/strong>. The four sp\u00b3 hybrid orbitals form sigma bonds with the three bromines and the hydrogen atom.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">3. Orbitals Involved in the Formation of the C-H Bond:<\/h3>\n\n\n\n<p>The <strong>C-H<\/strong> bond is a sigma bond formed between an sp\u00b3 hybrid orbital of the carbon atom and the 1s orbital of the hydrogen atom. The overlap of these orbitals creates the sigma bond between carbon and hydrogen.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">4. Orbitals Involved in the Formation of the C-Br Bond:<\/h3>\n\n\n\n<p>The <strong>C-Br<\/strong> bond is also a sigma bond. The bonding involves the overlap between an sp\u00b3 hybrid orbital of the carbon atom and a p orbital of the bromine atom. Bromine uses one of its unhybridized p orbitals to overlap with the sp\u00b3 hybrid orbital of carbon.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">5. Electron Geometry Around the Carbon Atom in Bromoform (CHBr3):<\/h3>\n\n\n\n<p>The electron geometry around the carbon atom is <strong>tetrahedral<\/strong>. This is because there are four regions of electron density (three C-Br bonds and one C-H bond), resulting in a tetrahedral shape.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">1. Lewis Structure of Hydroxylamine (HNOH):<\/h3>\n\n\n\n<p><strong>Step-by-step:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen (N) is the central atom, with 5 valence electrons.<\/li>\n\n\n\n<li>Oxygen (O) has 6 valence electrons, and hydrogen (H) has 1 valence electron.<\/li>\n\n\n\n<li>Nitrogen bonds to both a hydrogen atom and an oxygen atom.<\/li>\n<\/ul>\n\n\n\n<p>The structure of <strong>HNOH<\/strong> is:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>     H\n     |\nH - N - O\n     |\n     H\n<\/code><\/pre>\n\n\n\n<p>In this structure, nitrogen is bonded to one hydrogen atom, one oxygen atom, and another hydrogen atom. The oxygen atom is also bonded to a hydrogen atom. Nitrogen completes its octet by forming two single bonds, while oxygen has two lone pairs of electrons.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. Hybridization of Nitrogen (N) and Oxygen (O) Atoms in Hydroxylamine (HNOH):<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>nitrogen<\/strong> atom has three regions of electron density (two N-H bonds and one N-O bond), so it undergoes <strong>sp\u00b2 hybridization<\/strong>. One of its sp\u00b2 orbitals forms a sigma bond with hydrogen, and another forms a sigma bond with oxygen.<\/li>\n\n\n\n<li>The <strong>oxygen<\/strong> atom in HNOH has two regions of electron density (one O-H bond and one lone pair), so it undergoes <strong>sp\u00b3 hybridization<\/strong>. The sp\u00b3 orbital forms a sigma bond with the nitrogen atom, and the other holds a lone pair.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3. Electron Geometry Around the Nitrogen (N) and Oxygen (O) Atoms:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The electron geometry around the <strong>nitrogen<\/strong> atom is <strong>trigonal planar<\/strong>, as there are three regions of electron density (two N-H bonds and one N-O bond).<\/li>\n\n\n\n<li>The electron geometry around the <strong>oxygen<\/strong> atom is <strong>tetrahedral<\/strong>, as it has two regions of electron density (one O-H bond and one lone pair).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">4. Orbitals Involved in the Formation of the N-O Bond:<\/h3>\n\n\n\n<p>The <strong>N-O<\/strong> bond in hydroxylamine is a sigma bond formed by the overlap of an sp\u00b2 hybrid orbital from nitrogen and an sp\u00b3 hybrid orbital from oxygen.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-22.jpeg\" alt=\"\" class=\"wp-image-250940\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure of bromoform (CHBr3). 2. What is the hybridization type of the carbon atom? 3. What orbitals are involved in the formation of the C-H bond? 4. What orbitals are involved in the formation of the C-Br bond? 5. What is the electron geometry around the carbon atom? 2. Corrected_text: 1. Draw [&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-250938","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250938","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=250938"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/250938\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=250938"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=250938"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=250938"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}