{"id":163845,"date":"2024-11-09T07:14:26","date_gmt":"2024-11-09T07:14:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=163845"},"modified":"2024-11-09T07:14:28","modified_gmt":"2024-11-09T07:14:28","slug":"draw-the-lewis-structure-for-the-ccl4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/11\/09\/draw-the-lewis-structure-for-the-ccl4\/","title":{"rendered":"Draw the Lewis structure for the CCl4"},"content":{"rendered":"\n<p>Draw the Lewis structure for the CCl4<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To draw the Lewis structure of carbon tetrachloride (CCl\u2084), follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Count the total number of valence electrons:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> is in Group 14 of the periodic table, so it has 4 valence electrons.<\/li>\n\n\n\n<li><strong>Chlorine (Cl)<\/strong> is in Group 17, so each chlorine atom has 7 valence electrons. Since there are four chlorine atoms, this contributes ( 4 \\times 7 = 28 ) valence electrons.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the total number of valence electrons is:<br>[<br>4 \\, (\\text{from C}) + 28 \\, (\\text{from 4 Cl}) = 32 \\, \\text{valence electrons}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">2. Determine the central atom:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) is the least electronegative atom (except for hydrogen), so it will be the central atom. Chlorine atoms will surround the central carbon atom.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">3. Create bonds between the atoms:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon will form single bonds with each of the four chlorine atoms. Each single bond consists of 2 electrons. With 4 bonds, the total number of electrons used for bonding is:<br>[<br>4 \\times 2 = 8 \\, \\text{electrons}<br>]<br>This leaves us with ( 32 &#8211; 8 = 24 ) electrons to distribute as lone pairs.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">4. Distribute the remaining electrons:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place 3 lone pairs of electrons on each chlorine atom to complete their octet. Since each chlorine atom already shares 2 electrons with carbon (via the single bond), each chlorine atom needs 6 more electrons (3 lone pairs) to complete its octet.<\/li>\n<\/ul>\n\n\n\n<p>Now, each chlorine atom has 8 electrons (2 from the bond and 6 as lone pairs).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">5. Check the structure:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon has 4 single bonds, and each chlorine atom has 3 lone pairs. Carbon has no lone pairs, and all atoms have complete octets.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the final Lewis structure of CCl\u2084 is:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>    Cl\n    |\nCl\u2014C\u2014Cl\n    |\n    Cl<\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> is the central atom and forms four single bonds with the four <strong>Chlorine (Cl)<\/strong> atoms. Each bond represents 2 electrons.<\/li>\n\n\n\n<li><strong>Chlorine (Cl)<\/strong> atoms each have three lone pairs (6 electrons) around them, ensuring that each chlorine atom has a full octet.<\/li>\n\n\n\n<li>The structure is tetrahedral, as the four bonds to chlorine atoms form a 3D arrangement with 109.5\u00b0 bond angles. This molecular geometry minimizes electron pair repulsion, in accordance with the <strong>VSEPR (Valence Shell Electron Pair Repulsion) theory<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>In conclusion, CCl\u2084 is a nonpolar molecule due to its symmetric tetrahedral shape, where the dipoles from the C\u2013Cl bonds cancel each other out.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for the CCl4 The Correct Answer and Explanation is: To draw the Lewis structure of carbon tetrachloride (CCl\u2084), follow these steps: 1. Count the total number of valence electrons: Thus, the total number of valence electrons is:[4 \\, (\\text{from C}) + 28 \\, (\\text{from 4 Cl}) = 32 \\, \\text{valence electrons}] [&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-163845","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/163845","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=163845"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/163845\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=163845"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=163845"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=163845"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}