{"id":222921,"date":"2025-06-01T05:52:26","date_gmt":"2025-06-01T05:52:26","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=222921"},"modified":"2025-06-01T05:52:28","modified_gmt":"2025-06-01T05:52:28","slug":"draw-the-lewis-structure-for-the-polyatomic-hydroperoxyl-ho2-anion-3","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/01\/draw-the-lewis-structure-for-the-polyatomic-hydroperoxyl-ho2-anion-3\/","title":{"rendered":"Draw the lewis structure for the polyatomic hydroperoxyl ho2- anion"},"content":{"rendered":"\n<p>Draw the lewis structure for the polyatomic hydroperoxyl ho2- anion. be sure to include all resonance structures that satisfy the octet rule. c<br>Draw the lewis structure for the polyatomic hydroperoxyl ho2- anion. be sure to include all resonance structures that satisfy the octet rule. c<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Lewis Structure for Hydroperoxyl Anion (HO\u2082\u207b)<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Step-by-step:<\/h4>\n\n\n\n<p><strong>1. Count valence electrons:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>H:<\/strong> 1 valence electron<\/li>\n\n\n\n<li><strong>O (\u00d72):<\/strong> 6 valence electrons \u00d7 2 = 12<\/li>\n\n\n\n<li><strong>Charge (\u207b1):<\/strong> Add 1 extra electron<\/li>\n<\/ul>\n\n\n\n<p><strong>Total:<\/strong> 1 (H) + 12 (O) + 1 (charge) = <strong>14 valence electrons<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>2. Sketch the basic skeleton:<\/strong><\/p>\n\n\n\n<p>Hydroperoxyl has a linear structure: <strong>H\u2013O\u2013O\u207b<\/strong><\/p>\n\n\n\n<p>Attach H to one oxygen (because H can only form 1 bond), then bond that oxygen to the second oxygen.<\/p>\n\n\n\n<p><strong>Initial structure:<\/strong><br>H\u2013O\u2013O<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>3. Add bonding electrons:<\/strong><\/p>\n\n\n\n<p>Use single bonds between atoms:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H\u2013O (1 bond = 2 e\u207b)<\/li>\n\n\n\n<li>O\u2013O (1 bond = 2 e\u207b)<br>Total used: 4 electrons<br>Remaining: 14 \u2013 4 = 10 electrons<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>4. Distribute remaining electrons to satisfy octets:<\/strong><\/p>\n\n\n\n<p>Place remaining electrons as lone pairs on the two oxygens.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Place 6 electrons (3 lone pairs) on terminal O (the one not bonded to H)<\/li>\n\n\n\n<li>Place 4 electrons (2 lone pairs) on central O (it already has 2 bonds)<\/li>\n<\/ul>\n\n\n\n<p>That uses all 10 remaining electrons.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>5. Check formal charges:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Terminal O (rightmost): 6 valence \u2013 (6 nonbonding + 1 bond) = \u20131 \u2192 matches the anion charge<\/li>\n\n\n\n<li>Central O: 6 \u2013 (4 nonbonding + 2 bonds) = 0<\/li>\n\n\n\n<li>H: 1 \u2013 (0 nonbonding + 1 bond) = 0<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>6. Resonance Structures:<\/strong><\/p>\n\n\n\n<p>Hydroperoxyl anion has <strong>one significant resonance structure<\/strong> that satisfies the octet rule:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Main Resonance Structure:<\/h3>\n\n\n\n<pre class=\"wp-block-code\"><code>    ..\nH\u2013O\u2013O\u207b\n    ..\n<\/code><\/pre>\n\n\n\n<p>(With 3 lone pairs on the negatively charged O, and 2 lone pairs on the central O)<\/p>\n\n\n\n<p>Alternatively, you <strong>can draw a second resonance<\/strong> with a <strong>single bond and a negative charge localized on the oxygen attached to hydrogen<\/strong>, but this structure is <strong>less significant<\/strong> because it places a negative charge on the oxygen with a hydrogen \u2014 less favorable.<\/p>\n\n\n\n<p>So in total, the resonance forms are:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Major:<\/strong> Negative charge on terminal oxygen<\/li>\n\n\n\n<li><strong>Minor (less stable):<\/strong> Negative charge on the oxygen bonded to hydrogen<\/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\">Explanation<\/h3>\n\n\n\n<p>The hydroperoxyl anion (HO\u2082\u207b) consists of a hydrogen atom bonded to a pair of oxygen atoms with an overall negative charge. To draw its Lewis structure, we first determine the total number of valence electrons. Hydrogen contributes 1 electron, each oxygen provides 6, and the extra negative charge adds 1 more, giving a total of 14 valence electrons.<\/p>\n\n\n\n<p>The most likely bonding framework is H\u2013O\u2013O, since hydrogen can only form one bond. We begin by placing single bonds between H\u2013O and O\u2013O, using 4 electrons, leaving us with 10 to assign as lone pairs. These are distributed to satisfy the octet rule: the terminal (right) oxygen gets 6 electrons (3 lone pairs), and the central oxygen gets 4 electrons (2 lone pairs), for a total of 14.<\/p>\n\n\n\n<p>After assigning electrons, we check formal charges. The central oxygen has a formal charge of zero, while the terminal oxygen has a formal charge of \u20131, matching the charge of the anion. The hydrogen has a formal charge of zero.<\/p>\n\n\n\n<p>There is one major resonance structure where the negative charge is on the terminal oxygen, which is more stable due to minimized electron repulsion and full octets. A minor resonance structure can be drawn with the negative charge on the central oxygen (the one bonded to hydrogen), but this is less favorable due to the electronegativity and bond positioning.<\/p>\n\n\n\n<p>In summary, the hydroperoxyl anion is best represented with one dominant Lewis structure, possibly with a minor contributor, both obeying the octet rule. The molecule exhibits some resonance, but only one structure is significantly stable.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner4-11.jpeg\" alt=\"\" class=\"wp-image-222922\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the lewis structure for the polyatomic hydroperoxyl ho2- anion. be sure to include all resonance structures that satisfy the octet rule. cDraw the lewis structure for the polyatomic hydroperoxyl ho2- anion. be sure to include all resonance structures that satisfy the octet rule. c The Correct Answer and Explanation is: Lewis Structure for Hydroperoxyl [&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-222921","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222921","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=222921"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/222921\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=222921"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=222921"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=222921"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}