{"id":159464,"date":"2024-10-31T18:50:07","date_gmt":"2024-10-31T18:50:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=159464"},"modified":"2024-10-31T18:50:15","modified_gmt":"2024-10-31T18:50:15","slug":"draw-a-lewis-structure-for-the-bicarbonate-ion-hco3%e2%88%92sodium-bicarbonate-is-the-main-ingredient-of-baking-soda","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/10\/31\/draw-a-lewis-structure-for-the-bicarbonate-ion-hco3%e2%88%92sodium-bicarbonate-is-the-main-ingredient-of-baking-soda\/","title":{"rendered":"Draw a Lewis structure for the bicarbonate ion, HCO3\u2212(sodium bicarbonate is the main ingredient of baking soda)"},"content":{"rendered":"\n<p>Draw a Lewis structure for the bicarbonate ion, HCO3\u2212(sodium bicarbonate is the main ingredient of baking soda). Show all resonance forms and nonzero formal charges. What is the bond order of each C-O bond?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To draw the Lewis structure for the bicarbonate ion (HCO\u2083\u207b), we follow these steps:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count Valence Electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon (C) has 4 valence electrons.<\/li>\n\n\n\n<li>Oxygen (O) has 6 valence electrons, and there are three O atoms.<\/li>\n\n\n\n<li>Hydrogen (H) has 1 valence electron.<\/li>\n\n\n\n<li>The bicarbonate ion has a -1 charge, which adds 1 extra electron.<\/li>\n\n\n\n<li>Total = (4 + (3 \\times 6) + 1 + 1 = 22) valence electrons.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Draw the Skeletal Structure<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon is the central atom, bonded to one hydrogen atom and three oxygen atoms.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Distribute Electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initially, connect the C atom to one O with a double bond and the other two O atoms with single bonds.<\/li>\n\n\n\n<li>Place the hydrogen atom bonded to one of the singly bonded oxygen atoms.<\/li>\n\n\n\n<li>Distribute the remaining electrons to complete the octets of the oxygen atoms.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Formal Charges<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For the central carbon atom: (4 &#8211; (0 + 0 + 0) = 4) (4 is for its valence electrons, no lone pairs, and it shares 4 electrons).<\/li>\n\n\n\n<li>For the doubly bonded oxygen: (6 &#8211; (4 + 0) = 2) (6 for valence, 4 for bonding pairs).<\/li>\n\n\n\n<li>For each singly bonded oxygen: (6 &#8211; (2 + 1) = 3).<\/li>\n\n\n\n<li>For hydrogen: (1 &#8211; (0 + 0) = 1).<\/li>\n\n\n\n<li>The sum of formal charges: (-1) from the overall ion and (0) from the hydrogen.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Resonance Forms<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The bicarbonate ion exhibits resonance, where the double bond can be between carbon and any of the three oxygen atoms. Therefore, three resonance structures can be drawn.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Bond Order<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The bond order for each C-O bond in the bicarbonate ion can be calculated as follows:<\/li>\n\n\n\n<li>For the double-bonded C=O bond, the bond order is (2).<\/li>\n\n\n\n<li>For the single-bonded C-O bonds, the bond order is (1).<\/li>\n<\/ul>\n\n\n\n<p>Given the resonance structures, each C-O bond effectively has a bond order of (1.33) (the average of one double bond and two single bonds).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary<\/h3>\n\n\n\n<p>The bicarbonate ion (HCO\u2083\u207b) exhibits resonance with three major forms and formal charges indicating that one oxygen has a -1 charge, while the carbon maintains a neutral charge. The bond order reflects that the C-O bonds are not purely single or double but a mix due to resonance, leading to an average bond order of 1.33.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Lewis Structure Diagram<\/h3>\n\n\n\n<p>Here&#8217;s the Lewis structure representation for clarity:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>   O\n   ||\n H-C-O\u207b\n   |\n   O\u207b<\/code><\/pre>\n\n\n\n<p>This structure can shift between different resonance forms where the double bond is shared among the three oxygen atoms.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Draw a Lewis structure for the bicarbonate ion, HCO3\u2212(sodium bicarbonate is the main ingredient of baking soda). Show all resonance forms and nonzero formal charges. What is the bond order of each C-O bond? The Correct Answer and Explanation is: To draw the Lewis structure for the bicarbonate ion (HCO\u2083\u207b), we follow these steps: Bond [&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-159464","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/159464","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=159464"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/159464\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=159464"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=159464"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=159464"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}