{"id":42518,"date":"2025-06-29T05:36:10","date_gmt":"2025-06-29T05:36:10","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42518"},"modified":"2025-06-29T05:36:12","modified_gmt":"2025-06-29T05:36:12","slug":"draw-the-lewis-structure-for-iodoethane","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-lewis-structure-for-iodoethane\/","title":{"rendered":"Draw the Lewis structure for iodoethane"},"content":{"rendered":"\n<p>Draw the Lewis structure for iodoethane<br>. Be certain you include any lone pairs.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To draw the Lewis structure of iodoethane (C\u2082H\u2085I), follow these steps:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Count the Total Number of Valence Electrons<\/h3>\n\n\n\n<p>Each element in the molecule contributes electrons based on its group in the periodic table:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carbon (C)<\/strong> is in Group 14, so each carbon atom has 4 valence electrons. Since there are two carbon atoms, the total from carbon is 2\u00d74=82 \\times 4 = 82\u00d74=8 electrons.<\/li>\n\n\n\n<li><strong>Hydrogen (H)<\/strong> is in Group 1, so each hydrogen atom has 1 valence electron. Since there are five hydrogen atoms, the total from hydrogen is 5\u00d71=55 \\times 1 = 55\u00d71=5 electrons.<\/li>\n\n\n\n<li><strong>Iodine (I)<\/strong> is in Group 17, so iodine has 7 valence electrons.<\/li>\n<\/ul>\n\n\n\n<p>So, the total number of valence electrons is:8\u2009(from&nbsp;C)+5\u2009(from&nbsp;H)+7\u2009(from&nbsp;I)=20\u2009electrons.8 \\, (\\text{from C}) + 5 \\, (\\text{from H}) + 7 \\, (\\text{from I}) = 20 \\, \\text{electrons}.8(from&nbsp;C)+5(from&nbsp;H)+7(from&nbsp;I)=20electrons.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Determine the Skeleton Structure<\/h3>\n\n\n\n<p>Iodoethane consists of an ethyl group (C\u2082H\u2085) attached to an iodine (I) atom. The skeleton structure is:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One carbon (C) atom bonded to three hydrogens (H) and the second carbon (C) atom.<\/li>\n\n\n\n<li>The second carbon is bonded to two hydrogens (H) and an iodine (I) atom.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Distribute Electrons<\/h3>\n\n\n\n<p>Now, we place single bonds between the atoms:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each C-H bond uses 2 electrons. With five C-H bonds, that consumes 5\u00d72=105 \\times 2 = 105\u00d72=10 electrons.<\/li>\n\n\n\n<li>The C-C bond between the two carbons uses 2 electrons.<\/li>\n<\/ul>\n\n\n\n<p>So far, we\u2019ve used 10+2=1210 + 2 = 1210+2=12 electrons. That leaves 20\u221212=820 &#8211; 12 = 820\u221212=8 electrons to be placed.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Assign Lone Pairs<\/h3>\n\n\n\n<p>The remaining 8 electrons should be placed as lone pairs:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon atom already has 8 electrons in its bonds, so no lone pairs are placed on the carbons.<\/li>\n\n\n\n<li>The iodine atom (I) has 7 valence electrons. After forming a single bond with carbon, it will have 6 remaining electrons, which are placed as 3 lone pairs around iodine.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 5: Check for Octet Rule<\/h3>\n\n\n\n<p>All atoms should satisfy the octet rule (except hydrogen, which needs only 2 electrons):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The two carbon atoms have 8 electrons each, fulfilling the octet rule.<\/li>\n\n\n\n<li>Each hydrogen atom has 2 electrons from its bond with carbon.<\/li>\n\n\n\n<li>The iodine atom has 8 electrons, including 3 lone pairs and the 1 electron from the bond with carbon.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Final Structure:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C\u2086H\u2081\u2082I<\/strong> with bonds between the two carbon atoms and iodine, with 3 lone pairs on the iodine atom.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the Lewis structure of iodoethane should look like this:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>   H   H\n   |   |\nH-C-C-I\n   |   |\n   H   H\n<\/code><\/pre>\n\n\n\n<p>The iodine atom has three lone pairs, and all atoms follow the octet rule, except for hydrogen which follows the duet rule.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-323.jpeg\" alt=\"\" class=\"wp-image-42519\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-323.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-323-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-323-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-323-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for iodoethane. Be certain you include any lone pairs. The Correct Answer and Explanation is: To draw the Lewis structure of iodoethane (C\u2082H\u2085I), follow these steps: Step 1: Count the Total Number of Valence Electrons Each element in the molecule contributes electrons based on its group in the periodic table: So, [&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-42518","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42518","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=42518"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42518\/revisions"}],"predecessor-version":[{"id":42520,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42518\/revisions\/42520"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42518"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42518"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42518"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}