{"id":37811,"date":"2025-06-26T06:19:42","date_gmt":"2025-06-26T06:19:42","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37811"},"modified":"2025-06-26T07:35:14","modified_gmt":"2025-06-26T07:35:14","slug":"draw-the-lewis-structure-including-important-resonance-structures-of-h2c2o4","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-the-lewis-structure-including-important-resonance-structures-of-h2c2o4\/","title":{"rendered":"Draw the Lewis structure (including important resonance structures) of H2C2O4"},"content":{"rendered":"\n<p>Draw the Lewis structure (including important resonance structures) of H2C2O4 (oxalic acid). Circle all important resonance structures and label all formal charges.<\/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<h3 class=\"wp-block-heading\">Lewis Structure of H\u2082C\u2082O\u2084 (Oxalic Acid)<\/h3>\n\n\n\n<p><strong>Molecular formula<\/strong>: H\u2082C\u2082O\u2084<br><strong>Structure<\/strong>: Oxalic acid contains two carboxylic acid groups: \u2013COOH.<br>Its structure can be written as:<\/p>\n\n\n\n<p><strong>HOOC\u2013COOH<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Count valence electrons<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H (2 atoms): 2 \u00d7 1 = 2 electrons<\/li>\n\n\n\n<li>C (2 atoms): 2 \u00d7 4 = 8 electrons<\/li>\n\n\n\n<li>O (4 atoms): 4 \u00d7 6 = 24 electrons<br><strong>Total valence electrons = 2 + 8 + 24 = 34 electrons<\/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\">Step 2: Draw the basic skeleton<\/h3>\n\n\n\n<p>Attach atoms in this way:<\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">scssCopyEdit<code>HO\u2013C(=O)\u2013C(=O)\u2013OH\n<\/code><\/pre>\n\n\n\n<p>Each carbon is bonded to one hydroxyl group (\u2013OH) and double bonded to another oxygen atom.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Add electrons to satisfy the octet rule<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Structure 1 (most stable):<\/h4>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>      O        O\n      ||       ||\nH\u2013O\u2013C\u2013\u2013\u2013C\u2013O\u2013H\n<\/code><\/pre>\n\n\n\n<p>This shows both carboxylic acid groups with C=O and O\u2013H bonds.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each O in C=O has 2 lone pairs<\/li>\n\n\n\n<li>Each hydroxyl O has 2 lone pairs<\/li>\n\n\n\n<li>No formal charges<\/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\">Step 4: Resonance structures<\/h3>\n\n\n\n<p>In each carboxyl group, the negative charge can be delocalized between the two O atoms after deprotonation (loss of H\u207a). So for the <strong>deprotonated form<\/strong>, resonance occurs:<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Structure 2 (after deprotonation of one \u2013OH):<\/h4>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>      O\u207b       O\n       |        ||\nH\u2013O\u2013C\u2013\u2013\u2013C\u2013O\u2013H   &lt;==&gt;   H\u2013O\u2013C=O  \u2194  H\u2013O\u2013C\u207b\u2013O\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The C\u2013O\u207b bond can form a double bond while the C=O bond becomes C\u2013O\u207b<\/li>\n\n\n\n<li>This gives two resonance forms for each carboxylate group<\/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\">Formal Charges (in resonance forms)<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>On O\u207b: \u20131 formal charge (6 valence e\u207b \u2013 [6 non-bonding + 1 bonding e\u207b]) = \u20131<\/li>\n\n\n\n<li>All other atoms in the neutral form have formal charges of 0<\/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\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The neutral H\u2082C\u2082O\u2084 has no formal charges in the fully protonated structure<\/li>\n\n\n\n<li>Resonance occurs in the deprotonated forms (common in acid-base reactions)<\/li>\n\n\n\n<li>Important resonance structures are those with delocalization of negative charge in carboxylate ions<\/li>\n<\/ul>\n\n\n\n<p><strong>Note<\/strong>: You should draw the two carboxyl groups with delocalized electrons when discussing the conjugate base. Circle those with delocalized negative charges as the <strong>important resonance structures<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-978.jpeg\" alt=\"\" class=\"wp-image-37812\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-978.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-978-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-978-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure (including important resonance structures) of H2C2O4 (oxalic acid). Circle all important resonance structures and label all formal charges. The Correct Answer and Explanation is: Lewis Structure of H\u2082C\u2082O\u2084 (Oxalic Acid) Molecular formula: H\u2082C\u2082O\u2084Structure: Oxalic acid contains two carboxylic acid groups: \u2013COOH.Its structure can be written as: HOOC\u2013COOH Step 1: Count valence [&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-37811","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37811","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=37811"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37811\/revisions"}],"predecessor-version":[{"id":37813,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37811\/revisions\/37813"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37811"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37811"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37811"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}