{"id":227544,"date":"2025-06-06T07:51:21","date_gmt":"2025-06-06T07:51:21","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=227544"},"modified":"2025-06-06T07:51:23","modified_gmt":"2025-06-06T07:51:23","slug":"below-is-the-lewis-structure-of-the-chloroform-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/below-is-the-lewis-structure-of-the-chloroform-molecule\/","title":{"rendered":"Below is the Lewis structure of the chloroform () molecule"},"content":{"rendered":"\n<p>Below is the Lewis structure of the chloroform (<br>) molecule. H :Cl:C:Cl: :Cl: Count the number of bonding pairs and the number of lone pairs around the bottom chlorine atom. bonding pairs: lone pairs:<\/p>\n\n\n\n<p><strong>The Correct Answer and Explanation is:<\/strong><sup data-fn=\"55570273-f77c-4b2f-9f30-b2ae7dda6be6\" class=\"fn\"><a id=\"55570273-f77c-4b2f-9f30-b2ae7dda6be6-link\" href=\"#55570273-f77c-4b2f-9f30-b2ae7dda6be6\">1<\/a><\/sup><\/p>\n\n\n\n\n\n<p><strong>Answer:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bonding pairs:<\/strong> 1<\/li>\n\n\n\n<li><strong>Lone pairs:<\/strong> 3<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>Chloroform, with the chemical formula <strong>CHCl\u2083<\/strong>, consists of one carbon atom bonded to one hydrogen atom and three chlorine atoms. Let&#8217;s analyze the bonding and electron arrangement specifically around the <strong>bottom chlorine atom<\/strong> in its <strong>Lewis structure<\/strong>.<\/p>\n\n\n\n<p>A <strong>Lewis structure<\/strong> shows how atoms are connected using <strong>bonding pairs<\/strong> (shared electrons between atoms forming covalent bonds) and <strong>lone pairs<\/strong> (non-bonding electrons localized on a single atom). Chlorine (Cl), which has 7 valence electrons (since it&#8217;s in Group 17 of the periodic table), tends to form <strong>one single covalent bond<\/strong> to complete its <strong>octet<\/strong> (8 electrons in total).<\/p>\n\n\n\n<p>In <strong>CHCl\u2083<\/strong>, each chlorine atom forms a <strong>single covalent bond with the central carbon atom<\/strong>, contributing <strong>1 bonding pair<\/strong> (2 electrons shared between Cl and C). After forming this bond, the chlorine atom still has 6 remaining valence electrons, which it keeps as <strong>3 lone pairs<\/strong> (6 electrons total).<\/p>\n\n\n\n<p>So, for the <strong>bottom chlorine atom<\/strong>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1 bonding pair<\/strong>: shared with carbon.<\/li>\n\n\n\n<li><strong>3 lone pairs<\/strong>: non-bonding electrons.<\/li>\n<\/ul>\n\n\n\n<p>To double-check:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chlorine starts with 7 valence electrons.<\/li>\n\n\n\n<li>It shares 1 electron with carbon (1 bonding pair = 2 electrons, 1 from Cl and 1 from C).<\/li>\n\n\n\n<li>That leaves it with 6 electrons as lone pairs.<\/li>\n\n\n\n<li>6 electrons = 3 lone pairs.<\/li>\n<\/ul>\n\n\n\n<p>This satisfies the <strong>octet rule<\/strong> (1 bond = 2 electrons + 6 electrons from lone pairs = 8 electrons total around Cl).<\/p>\n\n\n\n<p>This is the same for all three chlorine atoms in chloroform, but since the question asks only about the <strong>bottom chlorine<\/strong>, the answer remains:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Bonding pairs: 1<\/strong><\/li>\n\n\n\n<li><strong>Lone pairs: 3<\/strong><\/li>\n<\/ul>\n\n\n\n<p>This understanding is fundamental in predicting molecular geometry, reactivity, and polarity.<\/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-banner8-81.jpeg\" alt=\"\" class=\"wp-image-227545\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Below is the Lewis structure of the chloroform () molecule. H :Cl:C:Cl: :Cl: Count the number of bonding pairs and the number of lone pairs around the bottom chlorine atom. bonding pairs: lone pairs: The Correct Answer and Explanation is:1 Answer: Explanation Chloroform, with the chemical formula CHCl\u2083, consists of one carbon atom bonded to [&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-227544","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227544","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=227544"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227544\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227544"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227544"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227544"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}