{"id":224133,"date":"2025-06-02T17:58:20","date_gmt":"2025-06-02T17:58:20","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=224133"},"modified":"2025-06-02T17:58:22","modified_gmt":"2025-06-02T17:58:22","slug":"when-you-draw-the-lewis-structure-for-ch3och2ch3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/02\/when-you-draw-the-lewis-structure-for-ch3och2ch3\/","title":{"rendered":"When you draw the Lewis Structure for CH3OCH2CH3"},"content":{"rendered":"\n<p>When you draw the Lewis Structure for CH3OCH2CH3, what is the total number of lone pair electrons? A 0 B 3 C 4 D 2 E 1<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct answer is: <strong>C) 4<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The molecule <strong>CH\u2083OCH\u2082CH\u2083<\/strong> is called <strong>ethyl methyl ether<\/strong>, a common ether. To determine the number of lone pair electrons in the Lewis structure, we follow these steps:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Count total valence electrons<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong>: 4 valence electrons \u00d7 3 = <strong>12<\/strong><\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong>: 1 valence electron \u00d7 8 = <strong>8<\/strong><\/li>\n\n\n\n<li><strong>Oxygen (O)<\/strong>: 6 valence electrons \u00d7 1 = <strong>6<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Total valence electrons<\/strong> = 12 + 8 + 6 = <strong>26 electrons<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Build the molecular skeleton<\/strong><\/h3>\n\n\n\n<p>The structure of <strong>CH\u2083OCH\u2082CH\u2083<\/strong> is:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>CH3\u2014O\u2014CH2\u2014CH3\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A methyl group (CH\u2083) is bonded to an oxygen atom.<\/li>\n\n\n\n<li>The oxygen is also bonded to an ethyl group (CH\u2082CH\u2083).<\/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>Step 3: Assign bonds and lone pairs<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Bonds<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Each <strong>single bond<\/strong> accounts for <strong>2 electrons<\/strong>.<\/li>\n\n\n\n<li>There are:\n<ul class=\"wp-block-list\">\n<li>8 single bonds involving H\u2013C (3 in CH\u2083, 2 in CH\u2082, 3 in CH\u2083): 8 bonds = <strong>16 electrons<\/strong><\/li>\n\n\n\n<li>2 C\u2013C single bonds = <strong>2 bonds = 4 electrons<\/strong><\/li>\n\n\n\n<li>1 C\u2013O single bond = <strong>1 bond = 2 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Total bonding electrons = 16 + 4 + 2 = <strong>22 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Remaining electrons<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Total valence electrons: 26<\/li>\n\n\n\n<li>Used in bonds: 22<\/li>\n\n\n\n<li><strong>Remaining: 4 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>These remaining <strong>4 electrons<\/strong> are placed as <strong>lone pairs<\/strong> on the <strong>oxygen atom<\/strong>, since all other atoms (C, H) are satisfied with their bonds.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Oxygen needs 8 electrons to complete its octet.<\/li>\n\n\n\n<li>It has 2 bonds (to C atoms), totaling 4 shared electrons.<\/li>\n\n\n\n<li>It needs 4 more electrons \u2192 2 lone pairs \u2192 <strong>4 lone pair electrons<\/strong>.<\/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>Final Result:<\/strong><\/h3>\n\n\n\n<p>Only the <strong>oxygen atom<\/strong> contributes <strong>1 pair + 1 pair = 2 lone pairs = 4 lone pair electrons<\/strong>.<\/p>\n\n\n\n<p><strong>Answer: C) 4<\/strong><\/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-banner5-36.jpeg\" alt=\"\" class=\"wp-image-224134\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>When you draw the Lewis Structure for CH3OCH2CH3, what is the total number of lone pair electrons? A 0 B 3 C 4 D 2 E 1 The Correct Answer and Explanation is: The correct answer is: C) 4 Explanation: The molecule CH\u2083OCH\u2082CH\u2083 is called ethyl methyl ether, a common ether. To determine the number [&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-224133","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224133","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=224133"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/224133\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=224133"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=224133"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=224133"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}