{"id":223505,"date":"2025-06-01T19:21:20","date_gmt":"2025-06-01T19:21:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=223505"},"modified":"2025-06-01T19:21:22","modified_gmt":"2025-06-01T19:21:22","slug":"the-lewis-structure-of-ch3och2ch3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/the-lewis-structure-of-ch3och2ch3\/","title":{"rendered":"The Lewis structure of CH3OCH2CH3."},"content":{"rendered":"\n<p>The Lewis structure of CH3OCH2CH3. In the Lewis structure, valence electrons increase. A) 18 B) 20 C) 26 D) 32 E) 36 symbol I33 In the Lewis paired and sulfur atom, there C C; the \u00e2\u201a\u00ac bond CC and = unpaired electrons- F) $<br>The Lewis structure of CH3OCH2CH3. In the Lewis structure, valence electrons increase. A) 18 B) 20 C) 26 D) 32 E) 36 symbol I33 In the Lewis paired and sulfur atom, there C C; the \u00e2\u201a\u00ac bond CC and = unpaired electrons- F) $<\/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<p>To determine the <strong>correct Lewis structure<\/strong> and number of <strong>valence electrons<\/strong> in <strong>CH\u2083OCH\u2082CH\u2083 (ethyl methyl ether)<\/strong>, we follow a step-by-step approach. This molecule consists of <strong>carbon (C), hydrogen (H), and oxygen (O)<\/strong> atoms.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udd22 Step 1: Count Valence Electrons<\/h3>\n\n\n\n<p>We add the valence electrons from each atom:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> has 4 valence electrons \u00d7 3 atoms = <strong>12 e\u207b<\/strong><\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong> has 1 valence electron \u00d7 8 atoms = <strong>8 e\u207b<\/strong><\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong> has 6 valence electrons \u00d7 1 atom = <strong>6 e\u207b<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Total valence electrons = 12 + 8 + 6 = 26 electrons<\/strong><\/p>\n\n\n\n<p>\u2705 <strong>Correct answer: C) 26<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\uddea Step 2: Build the Structure<\/h3>\n\n\n\n<p>CH\u2083OCH\u2082CH\u2083 is an ether:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One <strong>methyl group (CH\u2083)<\/strong> on the left<\/li>\n\n\n\n<li>One <strong>ethyl group (CH\u2082CH\u2083)<\/strong> on the right<\/li>\n\n\n\n<li>An <strong>oxygen atom<\/strong> connects both sides<\/li>\n<\/ul>\n\n\n\n<p>Structure:<br><strong>CH\u2083\u2013O\u2013CH\u2082CH\u2083<\/strong><\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Connectivity:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>C (left) forms 4 bonds: 3 with H, 1 with O<\/li>\n\n\n\n<li>O forms 2 bonds: one with each C, and has 2 lone pairs<\/li>\n\n\n\n<li>Middle C (CH\u2082) forms 2 single bonds: one with O, one with the terminal C, and 2 Hs<\/li>\n\n\n\n<li>Terminal C (CH\u2083) forms 3 bonds with H and one with CH\u2082<\/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\">\ud83e\uddf2 Step 3: Distribute Electrons<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each bond represents <strong>2 shared electrons<\/strong><\/li>\n\n\n\n<li>We ensure all atoms (except H) follow the <strong>octet rule<\/strong><\/li>\n\n\n\n<li>Oxygen must have <strong>2 lone pairs<\/strong> (4 electrons) to complete its octet<\/li>\n\n\n\n<li>All 8 hydrogen atoms get 1 bond each (2 electrons total), satisfying their duet rule<\/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\">\u270d\ufe0f Final Notes<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The Lewis structure does not involve any double or triple bonds in this molecule.<\/li>\n\n\n\n<li>Oxygen contributes a major portion of the <strong>non-bonding (lone pair) electrons<\/strong><\/li>\n\n\n\n<li>All atoms are in a stable, neutral state in this structure<\/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\">\u2705 Answer: <strong>C) 26 valence electrons<\/strong><\/h3>\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-banner5-16.jpeg\" alt=\"\" class=\"wp-image-223506\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The Lewis structure of CH3OCH2CH3. In the Lewis structure, valence electrons increase. A) 18 B) 20 C) 26 D) 32 E) 36 symbol I33 In the Lewis paired and sulfur atom, there C C; the \u00e2\u201a\u00ac bond CC and = unpaired electrons- F) $The Lewis structure of CH3OCH2CH3. In the Lewis structure, valence electrons increase. [&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-223505","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223505","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=223505"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/223505\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=223505"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=223505"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=223505"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}