{"id":22456,"date":"2025-06-16T17:33:24","date_gmt":"2025-06-16T17:33:24","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=22456"},"modified":"2025-06-16T17:33:29","modified_gmt":"2025-06-16T17:33:29","slug":"draw-the-lewis-structure-for-the-molecule-chcch3","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-lewis-structure-for-the-molecule-chcch3\/","title":{"rendered":"Draw the Lewis structure for the molecule CHCCH3"},"content":{"rendered":"\n<p>Draw the Lewis structure for the molecule CHCCH3. How many sigma and pi bonds does it contain?<br><\/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<p>To draw the Lewis structure for <strong>CHCCH\u2083<\/strong> (also written as <strong>CH\u2261C\u2013CH\u2083<\/strong>, known as <strong>propyne<\/strong>), 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\">\u2705 <strong>Lewis Structure of CHCCH\u2083 (Propyne)<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine total valence electrons<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Carbon (C): 4 valence electrons \u00d7 3 atoms = 12<\/li>\n\n\n\n<li>Hydrogen (H): 1 valence electron \u00d7 4 atoms = 4<\/li>\n\n\n\n<li><strong>Total = 12 + 4 = 16 valence electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Skeleton structure<\/strong>:<br>CH\u2261C\u2013CH\u2083\n<ul class=\"wp-block-list\">\n<li>The first C (leftmost) is bonded to one H.<\/li>\n\n\n\n<li>That C is triple bonded to the second C.<\/li>\n\n\n\n<li>The second C is single bonded to a third C (the CH\u2083 group).<\/li>\n\n\n\n<li>The third C is bonded to 3 H atoms.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Place bonds<\/strong>:\n<ul class=\"wp-block-list\">\n<li>C\u2013H (left) = 1 sigma bond<\/li>\n\n\n\n<li>C\u2261C = 1 sigma + 2 pi bonds<\/li>\n\n\n\n<li>C\u2013C (middle to methyl) = 1 sigma bond<\/li>\n\n\n\n<li>C\u2013H \u00d7 3 (on methyl) = 3 sigma bonds<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Assign electrons and check octets<\/strong>:<br>All carbon atoms follow the octet rule, and all hydrogen atoms have 2 electrons in their bond.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Sigma (\u03c3) and Pi (\u03c0) Bonds Count<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sigma (\u03c3) bonds<\/strong>:\n<ul class=\"wp-block-list\">\n<li>1 from C\u2013H (left)<\/li>\n\n\n\n<li>1 from C\u2261C<\/li>\n\n\n\n<li>1 from C\u2013C (single bond to CH\u2083)<\/li>\n\n\n\n<li>3 from C\u2013H (on methyl)<\/li>\n\n\n\n<li><strong>Total sigma bonds = 6<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Pi (\u03c0) bonds<\/strong>:\n<ul class=\"wp-block-list\">\n<li>2 from the triple bond between the two carbon atoms<\/li>\n\n\n\n<li><strong>Total pi bonds = 2<\/strong><\/li>\n<\/ul>\n<\/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 <strong>Final Answer<\/strong>:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Sigma (\u03c3) bonds<\/strong>: <strong>6<\/strong><\/li>\n\n\n\n<li><strong>Pi (\u03c0) bonds<\/strong>: <strong>2<\/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\">\ud83e\udde0 <strong>Explanation (300 words)<\/strong><\/h3>\n\n\n\n<p>To understand the bonding in CHCCH\u2083 (propyne), we begin by interpreting its structural formula. The molecule consists of three carbon atoms and four hydrogen atoms. The first two carbon atoms are connected by a triple bond, and the second carbon is connected to a methyl group (CH\u2083). The molecule looks like this:<br><strong>H\u2013C\u2261C\u2013CH\u2083<\/strong><\/p>\n\n\n\n<p>In Lewis structures, each bond is made of electrons shared between atoms. A <strong>sigma (\u03c3) bond<\/strong> is the first bond formed between any two atoms and involves head-on orbital overlap. <strong>Pi (\u03c0) bonds<\/strong> arise from the sideways overlap of p-orbitals and only exist when there are double or triple bonds.<\/p>\n\n\n\n<p>In a triple bond (as seen between the first and second carbon), there is one sigma bond and two pi bonds. The sigma bond gives the primary linkage, and the pi bonds contribute additional strength and rigidity to the bond. Between the second and third carbon atoms (C\u2013C single bond), there is just one sigma bond. The hydrogen atoms bonded to carbon contribute only sigma bonds.<\/p>\n\n\n\n<p>So, adding all:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>1 sigma (C\u2013H on first carbon)<\/li>\n\n\n\n<li>1 sigma (from the C\u2261C triple bond)<\/li>\n\n\n\n<li>2 pi (from the C\u2261C triple bond)<\/li>\n\n\n\n<li>1 sigma (C\u2013C single bond to methyl group)<\/li>\n\n\n\n<li>3 sigma (C\u2013H bonds on the methyl group)<\/li>\n<\/ul>\n\n\n\n<p>Hence, the molecule contains <strong>6 sigma bonds<\/strong> and <strong>2 pi bonds<\/strong>. These bond counts help in predicting the molecule\u2019s shape, reactivity, and properties like boiling point or bond strength.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for the molecule CHCCH3. How many sigma and pi bonds does it contain? The correct answer and explanation is: To draw the Lewis structure for CHCCH\u2083 (also written as CH\u2261C\u2013CH\u2083, known as propyne), follow these steps: \u2705 Lewis Structure of CHCCH\u2083 (Propyne) \u2705 Sigma (\u03c3) and Pi (\u03c0) Bonds Count \u2705 [&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-22456","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22456","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=22456"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22456\/revisions"}],"predecessor-version":[{"id":22457,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/22456\/revisions\/22457"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=22456"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=22456"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=22456"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}