{"id":208409,"date":"2025-04-26T16:39:46","date_gmt":"2025-04-26T16:39:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=208409"},"modified":"2025-04-26T16:39:49","modified_gmt":"2025-04-26T16:39:49","slug":"what-is-the-lewis-dot-structure-for-hofo","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/26\/what-is-the-lewis-dot-structure-for-hofo\/","title":{"rendered":"What is the Lewis dot structure for HOFO"},"content":{"rendered":"\n<p>What is the Lewis dot structure for HOFO?<\/p>\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>Sure! Let&#8217;s go through it carefully:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p><strong>Lewis Structure for HOFO<\/strong>:<\/p>\n\n\n\n<p>The molecule <strong>HOFO<\/strong> (hypofluorous acid) can be drawn as:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>H\u2013O\u2013F=O<\/code><\/pre>\n\n\n\n<p>Here\u2019s the breakdown:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Hydrogen (H) is bonded to an oxygen (O).<\/li>\n\n\n\n<li>That oxygen is bonded to a fluorine (F).<\/li>\n\n\n\n<li>The fluorine is double bonded to another oxygen (O).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-Step Explanation:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Count total valence electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H: 1 electron<\/li>\n\n\n\n<li>O: 6 electrons \u00d7 2 = 12 electrons<\/li>\n\n\n\n<li>F: 7 electrons<\/li>\n\n\n\n<li>Total = 1 + 12 + 7 = <strong>20 valence electrons<\/strong>.<\/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>H always forms one bond, so it must be connected to O.<\/li>\n\n\n\n<li>F usually forms one bond (very electronegative, prefers single bond).<\/li>\n\n\n\n<li>Remaining O can form double bonds. A logical starting skeleton: <strong>H\u2013O\u2013F\u2013O<\/strong>.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Connect atoms with single bonds first<\/strong>:<br>Each single bond uses <strong>2 electrons<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H\u2013O bond \u2192 2 electrons<\/li>\n\n\n\n<li>O\u2013F bond \u2192 2 electrons<\/li>\n\n\n\n<li>F\u2013O bond \u2192 2 electrons<br>\u2192 3 bonds \u00d7 2 = <strong>6 electrons used<\/strong>.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Distribute the remaining electrons<\/strong>:<br>20 &#8211; 6 = <strong>14 electrons left<\/strong>.<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Start filling outer atoms&#8217; octets (especially F and O).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Adjust to satisfy octet rule<\/strong>:<br>Fluorine prefers a single bond, so the bond between F and O must be a single bond.<br>The second oxygen (the one after F) would prefer to complete its octet via a <strong>double bond<\/strong> with F. Corrected structure:<\/li>\n<\/ol>\n\n\n\n<pre class=\"wp-block-code\"><code>   H\u2013O\u2013F=O<\/code><\/pre>\n\n\n\n<ol start=\"6\" class=\"wp-block-list\">\n<li><strong>Final check<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>H: 2 electrons (1 bond) \u2192 full for hydrogen.<\/li>\n\n\n\n<li>First O: 2 bonds (H and F) + 2 lone pairs \u2192 8 electrons.<\/li>\n\n\n\n<li>F: 1 bond (single) + 3 lone pairs \u2192 8 electrons.<\/li>\n\n\n\n<li>Second O: 1 double bond + 2 lone pairs \u2192 8 electrons.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Why this structure is correct:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Formal charges<\/strong> are minimized:<\/li>\n\n\n\n<li>Hydrogen: 0<\/li>\n\n\n\n<li>Oxygen: 0<\/li>\n\n\n\n<li>Fluorine: 0<\/li>\n\n\n\n<li>F, being highly electronegative, avoids extra bonds.<\/li>\n\n\n\n<li>O can form double bonds, satisfying octet.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the <strong>Lewis structure for HOFO<\/strong> is:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>H\u2013O\u2013F=O<\/code><\/pre>\n\n\n\n<p>where every atom satisfies its octet (or duet for hydrogen), and the molecule is neutral with formal charges close to zero.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n","protected":false},"excerpt":{"rendered":"<p>What is the Lewis dot structure for HOFO? The correct answer and explanation is : Sure! Let&#8217;s go through it carefully: Lewis Structure for HOFO: The molecule HOFO (hypofluorous acid) can be drawn as: Here\u2019s the breakdown: Step-by-Step Explanation: Why this structure is correct: Thus, the Lewis structure for HOFO is: where every atom satisfies [&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-208409","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208409","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=208409"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/208409\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=208409"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=208409"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=208409"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}