{"id":202213,"date":"2025-03-17T09:26:35","date_gmt":"2025-03-17T09:26:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202213"},"modified":"2025-03-17T09:26:38","modified_gmt":"2025-03-17T09:26:38","slug":"draw-the-lewis-structure-for-ch3soch3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/17\/draw-the-lewis-structure-for-ch3soch3\/","title":{"rendered":"Draw the Lewis structure for CH3S(O)CH3"},"content":{"rendered":"\n<p>Draw the Lewis structure for CH3S(O)CH3, a neutral molecule.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Lewis Structure of CH\u2083S(O)CH\u2083 (Dimethyl Sulfoxide, DMSO)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Step 1: Count Valence Electrons<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 4 valence electrons \u00d7 2 = 8<\/li>\n\n\n\n<li>Hydrogen (H): 1 valence electron \u00d7 6 = 6<\/li>\n\n\n\n<li>Sulfur (S): 6 valence electrons \u00d7 1 = 6<\/li>\n\n\n\n<li>Oxygen (O): 6 valence electrons \u00d7 1 = 6<\/li>\n\n\n\n<li><strong>Total valence electrons<\/strong> = <strong>26<\/strong><\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Step 2: Arrange the Atoms<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon is less electronegative than sulfur, so the two carbon atoms are bonded to sulfur.<\/li>\n\n\n\n<li>Oxygen is more electronegative and is bonded to sulfur.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Step 3: Draw Single Bonds<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon is bonded to three hydrogen atoms and the sulfur atom.<\/li>\n\n\n\n<li>Sulfur is bonded to oxygen with a double bond (to satisfy the octet rule).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Step 4: Assign Lone Pairs<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Oxygen has two lone pairs.<\/li>\n\n\n\n<li>Sulfur has one lone pair (since it forms three bonds: one to each methyl group and one to oxygen).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Lewis Structure<\/strong><\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>        H   H\n        |   |\n    H\u2014C\u2014S\u2014O\n        |   ||\n        C   ..\n        |   \n        H  <\/code><\/pre>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (300 words)<\/strong><\/h3>\n\n\n\n<p>Dimethyl sulfoxide (DMSO, CH\u2083S(O)CH\u2083) is a neutral molecule with a sulfur-oxygen double bond and two methyl (CH\u2083) groups. The Lewis structure follows the octet rule, ensuring that all atoms have a stable electron configuration.<\/p>\n\n\n\n<p><strong>Sulfur\u2019s Bonding and Hybridization<\/strong><br>Sulfur in DMSO has an oxidation state of +2 and adopts a <strong>trigonal pyramidal<\/strong> geometry due to its lone pair. The S=O bond is <strong>polar<\/strong>, as oxygen is more electronegative than sulfur, resulting in a partial negative charge on oxygen and a partial positive charge on sulfur. The methyl groups provide non-polar characteristics to the molecule, making DMSO a useful solvent for both polar and non-polar compounds.<\/p>\n\n\n\n<p><strong>Resonance Considerations<\/strong><br>While sulfur can theoretically form an expanded octet, DMSO is best represented with a <strong>double bond between sulfur and oxygen<\/strong>, with no significant resonance forms involving single bonds. The S=O bond has some double bond character due to <strong>p-\u03c0 conjugation<\/strong>, where oxygen\u2019s lone pair interacts with sulfur\u2019s empty d-orbitals.<\/p>\n\n\n\n<p><strong>Molecular Polarity and Applications<\/strong><br>DMSO is <strong>polar<\/strong> due to the S=O dipole moment and is widely used as an industrial solvent because it dissolves both organic and inorganic compounds. It also has medical applications, such as in drug delivery systems.<\/p>\n\n\n\n<p>This Lewis structure accurately represents the bonding and electronic distribution in DMSO, ensuring all atoms satisfy the octet rule while reflecting the molecular geometry and polarity.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for CH3S(O)CH3, a neutral molecule. The correct answer and explanation is : Lewis Structure of CH\u2083S(O)CH\u2083 (Dimethyl Sulfoxide, DMSO) Final Lewis Structure Explanation (300 words) Dimethyl sulfoxide (DMSO, CH\u2083S(O)CH\u2083) is a neutral molecule with a sulfur-oxygen double bond and two methyl (CH\u2083) groups. The Lewis structure follows the octet rule, ensuring [&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-202213","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202213","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=202213"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202213\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202213"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202213"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202213"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}