{"id":19700,"date":"2025-06-13T20:59:34","date_gmt":"2025-06-13T20:59:34","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=19700"},"modified":"2025-06-13T20:59:36","modified_gmt":"2025-06-13T20:59:36","slug":"draw-the-lewis-structure-for-the-ethylene-c2h4-molecule","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-lewis-structure-for-the-ethylene-c2h4-molecule\/","title":{"rendered":"Draw the Lewis structure for the ethylene (C2H4) molecule"},"content":{"rendered":"\n<p>Draw the Lewis structure for the ethylene (C2H4) molecule. Be sure to include all resonance structures that satisfy the octet rule.<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Lewis Structure for Ethylene (C\u2082H\u2084)<\/h3>\n\n\n\n<p><strong>Correct Lewis Structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   H      H\n    \\    \/\n     C==C\n    \/    \\\n   H      H\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon (C) atom forms <strong>three sigma bonds<\/strong> (two with hydrogen and one with another carbon) and <strong>one pi bond<\/strong> (in the carbon-carbon double bond).<\/li>\n\n\n\n<li>Each hydrogen (H) forms <strong>one single bond<\/strong> with a carbon atom.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation (300 Words):<\/h3>\n\n\n\n<p>Ethylene, or <strong>C\u2082H\u2084<\/strong>, is the simplest alkene, featuring a <strong>double bond<\/strong> between two carbon atoms. The total number of valence electrons must be determined first to draw its Lewis structure correctly.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> has 4 valence electrons, and there are 2 carbon atoms:<br>4 \u00d7 2 = 8 electrons<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong> has 1 valence electron, and there are 4 hydrogen atoms:<br>1 \u00d7 4 = 4 electrons<\/li>\n\n\n\n<li><strong>Total valence electrons = 8 + 4 = 12 electrons<\/strong><\/li>\n<\/ul>\n\n\n\n<p>We use these 12 electrons to form bonds and satisfy the <strong>octet rule<\/strong> for carbon and the <strong>duet rule<\/strong> for hydrogen.<\/p>\n\n\n\n<p>Each carbon atom bonds with two hydrogen atoms and shares a double bond with the other carbon. This arrangement uses:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>4 single bonds between carbon and hydrogen = 4 \u00d7 2 = 8 electrons<\/li>\n\n\n\n<li>1 double bond between carbon atoms = 4 electrons<\/li>\n\n\n\n<li>Total used: 8 + 4 = 12 electrons \u2014 all valence electrons accounted for.<\/li>\n<\/ul>\n\n\n\n<p>In this structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each <strong>hydrogen<\/strong> has 2 electrons (1 bond) \u2014 satisfying the duet rule.<\/li>\n\n\n\n<li>Each <strong>carbon<\/strong> has 4 bonds (8 electrons) \u2014 satisfying the octet rule.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Resonance Structures:<\/h3>\n\n\n\n<p>Ethylene does <strong>not<\/strong> have any valid resonance structures. The double bond is localized between the two carbon atoms. Unlike molecules like ozone (O\u2083) or benzene (C\u2086H\u2086), where electrons can be delocalized over multiple atoms, ethylene\u2019s \u03c0 electrons are <strong>fixed<\/strong> between the two carbon atoms, forming a <strong>localized<\/strong> double bond. Therefore, <strong>only one valid Lewis structure<\/strong> satisfies the octet rule.<\/p>\n\n\n\n<p>Thus, the single Lewis structure above accurately represents ethylene.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for the ethylene (C2H4) molecule. Be sure to include all resonance structures that satisfy the octet rule. The correct answer and explanation is: Lewis Structure for Ethylene (C\u2082H\u2084) Correct Lewis Structure: Explanation (300 Words): Ethylene, or C\u2082H\u2084, is the simplest alkene, featuring a double bond between two carbon atoms. The total [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-19700","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19700","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=19700"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19700\/revisions"}],"predecessor-version":[{"id":19701,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19700\/revisions\/19701"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=19700"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=19700"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=19700"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}