{"id":40175,"date":"2025-06-27T14:09:48","date_gmt":"2025-06-27T14:09:48","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=40175"},"modified":"2025-06-27T14:09:49","modified_gmt":"2025-06-27T14:09:49","slug":"draw-out-the-lewis-structure-of-h3po4-and-consider-its-major-resonance-forms","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/draw-out-the-lewis-structure-of-h3po4-and-consider-its-major-resonance-forms\/","title":{"rendered":"Draw out the Lewis Structure of H3PO4, and consider its major resonance forms"},"content":{"rendered":"\n<p>Draw out the Lewis Structure of H3PO4, and consider its major resonance forms. Box the major resonance contributor to the hybrid. List the intermolecular forces the compound would exhibit.<\/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 <strong>H\u2083PO\u2084 (Phosphoric Acid)<\/strong>:<\/h3>\n\n\n\n<p>To draw the Lewis structure of phosphoric acid, follow these steps:<\/p>\n\n\n\n<p><strong>Total Valence Electrons Calculation:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hydrogen (H): 3 \u00d7 1 = 3 electrons<\/li>\n\n\n\n<li>Phosphorus (P): 1 \u00d7 5 = 5 electrons<\/li>\n\n\n\n<li>Oxygen (O): 4 \u00d7 6 = 24 electrons<\/li>\n<\/ul>\n\n\n\n<p><strong>Total<\/strong> = 3 + 5 + 24 = <strong>32 valence electrons<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Structure Drawing:<\/strong><\/p>\n\n\n\n<p>Phosphoric acid has the following structure:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Phosphorus is the central atom.<\/li>\n\n\n\n<li>Three hydroxyl groups (OH) are bonded to phosphorus via oxygen atoms.<\/li>\n\n\n\n<li>One double-bonded oxygen connects directly to phosphorus.<\/li>\n<\/ul>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>       O  \n       ||  \nH - O - P - O - H  \n       |  \n      O - H  \n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The double-bonded oxygen satisfies octet requirements for phosphorus and oxygen.<\/li>\n\n\n\n<li>Each hydroxyl group (O-H) has two lone pairs on oxygen.<\/li>\n\n\n\n<li>Phosphorus has five bonds in total, but this is allowed for elements in period 3 or higher like phosphorus.<\/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\">Resonance Forms:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The double bond between phosphorus and oxygen can shift to different oxygen atoms.<\/li>\n\n\n\n<li>Major resonance contributors include structures where the double bond alternates between each of the three oxygen atoms bonded to phosphorus.<\/li>\n<\/ul>\n\n\n\n<p>However, the structure where the double bond is on the oxygen <strong>not bonded to hydrogen<\/strong> (the terminal oxygen) is the <strong>major resonance contributor<\/strong>. This is because the hydroxyl groups (O-H) maintain their single bonds, and formal charges are minimized.<\/p>\n\n\n\n<p><strong>Boxed Major Resonance Contributor:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-preformatted\">mathematicaCopyEdit<code>       O  \n       ||  \nH - O - P - O - H  \n       |  \n      O - H  \n<\/code><\/pre>\n\n\n\n<p>The double bond is with the terminal oxygen, and no formal charges appear on atoms in this form.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Intermolecular Forces in H\u2083PO\u2084:<\/h3>\n\n\n\n<p>Phosphoric acid exhibits several types of intermolecular forces:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Hydrogen Bonding<\/strong><br>Due to the presence of O-H bonds, strong hydrogen bonds form between H\u2083PO\u2084 molecules.<\/li>\n\n\n\n<li><strong>Dipole-Dipole Interactions<\/strong><br>The molecule is polar because of the asymmetrical arrangement of polar bonds, contributing to dipole-dipole interactions.<\/li>\n\n\n\n<li><strong>London Dispersion Forces<\/strong><br>These weak forces exist in all molecules, including H\u2083PO\u2084, but they are not the dominant force here.<\/li>\n<\/ol>\n\n\n\n<p>Thus, hydrogen bonding is the strongest and most significant intermolecular force in phosphoric acid.<\/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-banner8-1233.jpeg\" alt=\"\" class=\"wp-image-40176\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1233.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1233-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-1233-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw out the Lewis Structure of H3PO4, and consider its major resonance forms. Box the major resonance contributor to the hybrid. List the intermolecular forces the compound would exhibit. The Correct Answer and Explanation is: Lewis Structure of H\u2083PO\u2084 (Phosphoric Acid): To draw the Lewis structure of phosphoric acid, follow these steps: Total Valence Electrons [&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-40175","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40175","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=40175"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40175\/revisions"}],"predecessor-version":[{"id":40177,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/40175\/revisions\/40177"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=40175"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=40175"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=40175"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}