{"id":201088,"date":"2025-03-15T02:34:10","date_gmt":"2025-03-15T02:34:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=201088"},"modified":"2025-03-15T02:34:13","modified_gmt":"2025-03-15T02:34:13","slug":"draw-the-lewis-structure-for-the-no2-ion-nitrogen-is-the-central-atom","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/15\/draw-the-lewis-structure-for-the-no2-ion-nitrogen-is-the-central-atom\/","title":{"rendered":"Draw the Lewis structure for the NO2+ ion (nitrogen is the central atom)"},"content":{"rendered":"\n<p>Draw the Lewis structure for the NO2+ ion (nitrogen is the central atom). The correct Lewis structure has<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>two single bonds and a total of 8 lone pairs<\/li>\n\n\n\n<li>two double bonds and a total of 4 lone pairs<\/li>\n\n\n\n<li>two single bonds and a total of 6 lone pairs<\/li>\n\n\n\n<li>one single bond, one double bond and a total of 6 lone pairs<\/li>\n<\/ul>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>The correct Lewis structure for the <strong>NO\u2082\u207a (Nitronium ion)<\/strong> has:<\/p>\n\n\n\n<p>\u2705 <strong>Two double bonds and a total of 4 lone pairs.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Determine the total valence electrons:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Nitrogen (N):<\/strong> 5 valence electrons<\/li>\n\n\n\n<li><strong>Oxygen (O):<\/strong> 6 valence electrons \u00d7 2 = 12<\/li>\n\n\n\n<li><strong>Total before charge adjustment:<\/strong> 5 + 12 = 17<\/li>\n\n\n\n<li>**Since NO\u2082\u207a has a <strong>+1 charge<\/strong>, we <strong>subtract<\/strong> one electron:\n<ul class=\"wp-block-list\">\n<li><strong>Final total = 16 valence electrons.<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Determine the skeletal structure:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Nitrogen (N) is the central atom<\/strong> since it is the least electronegative.<\/li>\n\n\n\n<li>The two <strong>oxygen (O) atoms<\/strong> will be bonded to nitrogen.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Form bonds between N and O:<\/strong>\n<ul class=\"wp-block-list\">\n<li>We start by connecting <strong>N to O with single bonds<\/strong>, which initially uses <strong>4 electrons<\/strong> (2 per bond).<\/li>\n\n\n\n<li>This leaves us with <strong>12 electrons<\/strong> to distribute as lone pairs.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Distribute remaining electrons:<\/strong>\n<ul class=\"wp-block-list\">\n<li>If we place lone pairs on <strong>oxygen atoms<\/strong> first, we realize that <strong>nitrogen has an incomplete octet<\/strong> (only 4 electrons around it).<\/li>\n\n\n\n<li>To complete nitrogen\u2019s octet, we form <strong>double bonds<\/strong> with each oxygen.<\/li>\n\n\n\n<li>Now, <strong>each oxygen has a double bond with nitrogen<\/strong> and <strong>two lone pairs<\/strong> (4 electrons each).<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Check formal charges:<\/strong>\n<ul class=\"wp-block-list\">\n<li><strong>Nitrogen:<\/strong> <strong>5 &#8211; (0 non-bonding + 4 bonding pairs) = 5 &#8211; 4 = +1<\/strong><\/li>\n\n\n\n<li><strong>Oxygen (each):<\/strong> <strong>6 &#8211; (4 non-bonding + 4 bonding) = 6 &#8211; 6 = 0<\/strong><\/li>\n\n\n\n<li>The overall charge is <strong>+1<\/strong>, matching the ion\u2019s charge.<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Final verification:<\/strong>\n<ul class=\"wp-block-list\">\n<li>All atoms have <strong>complete octets<\/strong>.<\/li>\n\n\n\n<li>The sum of formal charges equals <strong>+1<\/strong>, which is correct.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n<p>Thus, the best <strong>Lewis structure<\/strong> for NO\u2082\u207a has <strong>two double bonds and 4 lone pairs<\/strong> (2 per oxygen).<\/p>\n\n\n\n<p>Now, I&#8217;ll generate an image of the Lewis structure.<\/p>\n\n\n\n<p>Here is the Lewis structure diagram for the <strong>NO\u2082\u207a (Nitronium ion)<\/strong>. The nitrogen is the central atom, forming <strong>two double bonds<\/strong> with oxygen, and each oxygen has <strong>two lone pairs<\/strong>. The entire structure carries a <strong>+1 charge<\/strong>. Let me know if you need any modifications!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/03\/image-980.png\" alt=\"\" class=\"wp-image-201089\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis structure for the NO2+ ion (nitrogen is the central atom). The correct Lewis structure has The correct answer and explanation is: The correct Lewis structure for the NO\u2082\u207a (Nitronium ion) has: \u2705 Two double bonds and a total of 4 lone pairs. Explanation: Thus, the best Lewis structure for NO\u2082\u207a has two [&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-201088","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201088","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=201088"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/201088\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=201088"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=201088"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=201088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}