{"id":227915,"date":"2025-06-06T15:19:47","date_gmt":"2025-06-06T15:19:47","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=227915"},"modified":"2025-06-06T15:19:49","modified_gmt":"2025-06-06T15:19:49","slug":"draw-the-lewis-structure-for-cocl","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/06\/draw-the-lewis-structure-for-cocl\/","title":{"rendered":"Draw the Lewis structure for COCl"},"content":{"rendered":"\n<p>Draw the Lewis structure for COCl?, including lone pairs. Select Draw Rings More Erase C O Cl :O: :: :Cl\u2014C\u2014Cl: What is the molecular shape of COCl?? ? bent ? linear ? trigonal pyramidal ? tetrahedral ? trigonal planar What is the Cl-C-Cl bond angle? ? 180\u00b0 ? 109.5\u00b0 ? 120\u00b0 ? &lt;109.5\u00b0 The C-Cl bond in COCl? is The Lewis structures of four compounds are given. ::O=S=O:: ::O=C=O:: H H :Cl: :Cl: C C :Cl: :Cl: H H :Cl: Which of these molecules are polar? ? CH?Cl? ? PCl? ? SO? ? CO? The molecule COCl? is<\/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\">Correct Answers:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Lewis Structure for COCl\u2082 (Phosgene):<\/strong><br>The Lewis structure is: mathematicaCopyEdit<code> :O: || Cl\u2013C\u2013Cl .. ..<\/code>\n<ul class=\"wp-block-list\">\n<li>Carbon (C) is the central atom, forming <strong>two single bonds<\/strong> with chlorine (Cl) atoms and <strong>a double bond<\/strong> with the oxygen (O) atom.<\/li>\n\n\n\n<li>The oxygen atom has <strong>two lone pairs<\/strong>.<\/li>\n\n\n\n<li>Each chlorine atom has <strong>three lone pairs<\/strong>.<\/li>\n\n\n\n<li>Carbon has <strong>no lone pairs<\/strong> (8 electrons total, fulfilling the octet rule).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Molecular Shape:<\/strong><br>\u2192 <strong>Trigonal planar<\/strong><\/li>\n\n\n\n<li><strong>Cl\u2013C\u2013Cl Bond Angle:<\/strong><br>\u2192 <strong>120\u00b0<\/strong><\/li>\n\n\n\n<li><strong>The C\u2013Cl bond in COCl\u2082 is:<\/strong><br>\u2192 <strong>Polar<\/strong><\/li>\n\n\n\n<li><strong>Which of the following molecules are polar?<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>CH\u2082Cl\u2082<\/strong> \u2705 (Polar \u2014 asymmetrical dipole moments)<\/li>\n\n\n\n<li><strong>PCl\u2083<\/strong> \u2705 (Polar \u2014 lone pair on P causes a trigonal pyramidal shape)<\/li>\n\n\n\n<li><strong>SO\u2082<\/strong> \u2705 (Polar \u2014 bent shape)<\/li>\n\n\n\n<li><strong>CO\u2082<\/strong> \u274c (Non-polar \u2014 linear with equal and opposite dipoles)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>The molecule COCl\u2082 is:<\/strong><br>\u2192 <strong>Polar<\/strong><\/li>\n<\/ol>\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><strong>COCl\u2082<\/strong>, or <strong>phosgene<\/strong>, consists of a central carbon atom bonded to two chlorine atoms and one oxygen atom via a double bond. The <strong>Lewis structure<\/strong> reflects this with a <strong>double bond to oxygen<\/strong> (O=C) and <strong>single bonds to each Cl<\/strong>. Oxygen has <strong>two lone pairs<\/strong>, and each chlorine atom has <strong>three lone pairs<\/strong>.<\/p>\n\n\n\n<p>The carbon atom in COCl\u2082 has <strong>three electron domains<\/strong> (two single bonds and one double bond), which adopt a <strong>trigonal planar geometry<\/strong> to minimize electron-pair repulsion, according to <strong>VSEPR theory<\/strong> (Valence Shell Electron Pair Repulsion). This results in <strong>bond angles of approximately 120\u00b0<\/strong>, which is typical of trigonal planar molecules.<\/p>\n\n\n\n<p>Each <strong>C\u2013Cl bond<\/strong> is <strong>polar<\/strong> due to the difference in electronegativity between carbon (2.55) and chlorine (3.16). Similarly, the <strong>C=O bond<\/strong> is highly polar, with oxygen (3.44) being more electronegative than carbon. While the molecule has a symmetrical shape, the differing strengths and directions of the bond dipoles (especially the stronger C=O dipole) cause <strong>COCl\u2082 to have a net dipole moment<\/strong>, making the molecule <strong>polar overall<\/strong>.<\/p>\n\n\n\n<p>Regarding the comparison molecules:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>CO\u2082<\/strong> is linear and symmetrical, making it <strong>non-polar<\/strong> despite polar bonds.<\/li>\n\n\n\n<li><strong>SO\u2082<\/strong> is <strong>bent<\/strong> due to lone pairs on sulfur, making it <strong>polar<\/strong>.<\/li>\n\n\n\n<li><strong>PCl\u2083<\/strong> has a <strong>trigonal pyramidal<\/strong> shape due to a lone pair on phosphorus, creating an <strong>asymmetrical<\/strong> charge distribution \u2014 hence <strong>polar<\/strong>.<\/li>\n\n\n\n<li><strong>CH\u2082Cl\u2082<\/strong> (dichloromethane) is <strong>tetrahedral<\/strong>, and the unequal distribution of Cl and H atoms makes it <strong>polar<\/strong> as well.<\/li>\n<\/ul>\n\n\n\n<p>In summary, COCl\u2082 is a <strong>polar, trigonal planar molecule<\/strong> with <strong>120\u00b0 Cl\u2013C\u2013Cl bond angles<\/strong> and <strong>polar C\u2013Cl bonds<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-gallery has-nested-images columns-default is-cropped wp-block-gallery-1 is-layout-flex wp-block-gallery-is-layout-flex\">\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" data-id=\"227916\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-544.jpeg\" alt=\"\" class=\"wp-image-227916\"\/><\/figure>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" data-id=\"227917\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-122.jpeg\" alt=\"\" class=\"wp-image-227917\"\/><\/figure>\n<\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for COCl?, including lone pairs. Select Draw Rings More Erase C O Cl :O: :: :Cl\u2014C\u2014Cl: What is the molecular shape of COCl?? ? bent ? linear ? trigonal pyramidal ? tetrahedral ? trigonal planar What is the Cl-C-Cl bond angle? ? 180\u00b0 ? 109.5\u00b0 ? 120\u00b0 ? &lt;109.5\u00b0 The C-Cl [&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-227915","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227915","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=227915"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227915\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227915"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227915"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227915"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}