{"id":182171,"date":"2025-01-13T13:10:41","date_gmt":"2025-01-13T13:10:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=182171"},"modified":"2025-01-13T13:10:43","modified_gmt":"2025-01-13T13:10:43","slug":"which-is-a-correct-lewis-structure-for-carbonic-acid","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/13\/which-is-a-correct-lewis-structure-for-carbonic-acid\/","title":{"rendered":"Which is a correct Lewis structure for carbonic acid"},"content":{"rendered":"\n<p>Which is a correct Lewis structure for carbonic acid, H2CO3?<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-149.png\" alt=\"\" class=\"wp-image-182172\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>The correct Lewis structure for carbonic acid (H\u2082CO\u2083) features a central carbon atom double-bonded to one oxygen atom and single-bonded to two hydroxyl groups (\u2013OH). Each oxygen atom in the hydroxyl groups carries two lone pairs of electrons, while the oxygen double-bonded to carbon has two lone pairs. This arrangement satisfies the octet rule for all atoms involved.<\/p>\n\n\n\n<p><strong>Steps to Draw the Lewis Structure:<\/strong><\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine Total Valence Electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hydrogen (H): 1 electron \u00d7 2 = 2 electrons<\/li>\n\n\n\n<li>Carbon (C): 4 electrons<\/li>\n\n\n\n<li>Oxygen (O): 6 electrons \u00d7 3 = 18 electrons<\/li>\n\n\n\n<li>Total valence electrons = 2 + 4 + 18 = 24 electrons<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Identify the Central Atom:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carbon is less electronegative than oxygen and can form four bonds, making it the central atom.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Sketch the Skeleton Structure:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place carbon at the center, bonded to two hydroxyl groups and double-bonded to one oxygen atom.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Distribute Electrons as Bonds:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each single bond (C\u2013O) uses 2 electrons, and the double bond (C=O) uses 4 electrons.<\/li>\n\n\n\n<li>Total electrons used in bonds = 2 \u00d7 2 (for two single bonds) + 4 (for the double bond) = 8 electrons.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Assign Lone Pairs to Oxygen Atoms:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each oxygen atom needs 8 electrons to complete its octet.<\/li>\n\n\n\n<li>The oxygen double-bonded to carbon has two lone pairs.<\/li>\n\n\n\n<li>Each oxygen in the hydroxyl groups has two lone pairs.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Verify the Octet Rule:<\/strong><\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All atoms (carbon, oxygen, and hydrogen) satisfy the octet rule, with hydrogen achieving a duet.<\/li>\n<\/ul>\n\n\n\n<p><strong>Formal Charge Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Formal charge = Valence electrons \u2013 (Nonbonding electrons + \u00bd Bonding electrons)<\/li>\n<\/ul>\n\n\n\n<p>For each atom:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Hydrogen:<\/strong> 1 \u2013 (0 + \u00bd \u00d7 2) = 0<\/li>\n\n\n\n<li><strong>Carbon:<\/strong> 4 \u2013 (0 + \u00bd \u00d7 8) = 0<\/li>\n\n\n\n<li><strong>Oxygen (double-bonded to carbon):<\/strong> 6 \u2013 (4 + \u00bd \u00d7 4) = 0<\/li>\n\n\n\n<li><strong>Oxygen (single-bonded to carbon and hydrogen):<\/strong> 6 \u2013 (4 + \u00bd \u00d7 4) = 0<\/li>\n<\/ul>\n\n\n\n<p>All atoms have a formal charge of zero, indicating a stable structure.<\/p>\n\n\n\n<p><strong>Molecular Geometry:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The central carbon atom is sp\u00b2 hybridized, leading to a trigonal planar geometry with bond angles of approximately 120\u00b0.<\/li>\n<\/ul>\n\n\n\n<p><strong>Polarity:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecule is polar due to the presence of polar O\u2013H bonds and the asymmetrical arrangement of atoms.<\/li>\n<\/ul>\n\n\n\n<p>This Lewis structure accurately represents the bonding and electron distribution in carbonic acid, adhering to the octet rule and minimizing formal charges.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Which is a correct Lewis structure for carbonic acid, H2CO3? The Correct Answer and Explanation is : The correct Lewis structure for carbonic acid (H\u2082CO\u2083) features a central carbon atom double-bonded to one oxygen atom and single-bonded to two hydroxyl groups (\u2013OH). Each oxygen atom in the hydroxyl groups carries two lone pairs of electrons, [&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-182171","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182171","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=182171"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/182171\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=182171"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=182171"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=182171"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}