{"id":218586,"date":"2025-05-24T05:53:50","date_gmt":"2025-05-24T05:53:50","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=218586"},"modified":"2025-05-24T05:53:52","modified_gmt":"2025-05-24T05:53:52","slug":"determine-the-shape-and-bond-angle-of-these-oxynitrogen-ions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/24\/determine-the-shape-and-bond-angle-of-these-oxynitrogen-ions\/","title":{"rendered":"Determine the shape and bond angle of these oxynitrogen ions."},"content":{"rendered":"\n<p>Determine the shape and bond angle of these oxynitrogen ions. Linear Trigonal planar Tetrahedral Bent Pyramidal 180\u00b0 120\u00b0 109.5\u00b0 ~120\u00b0 ~109\u00b0 ~109\u00b0 Answer Bank NO- NO3- NO2- None<\/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<p>To determine the shape and bond angles of the oxynitrogen ions (NO\u207b, NO\u2083\u207b, and NO\u2082\u207b), we use <strong>Valence Shell Electron Pair Repulsion (VSEPR) theory<\/strong>, which predicts the geometry of a molecule based on the repulsion between electron pairs around a central atom.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">1. <strong>NO\u207b (Nitrosyl anion)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis structure<\/strong>: One nitrogen atom is bonded to one oxygen atom with a <strong>double bond<\/strong> and has a <strong>lone pair<\/strong> on nitrogen.<\/li>\n\n\n\n<li><strong>Electron domains<\/strong> around N: 2 (1 bonding region, 1 lone pair).<\/li>\n\n\n\n<li><strong>Electron geometry<\/strong>: Linear.<\/li>\n\n\n\n<li><strong>Molecular shape<\/strong>: <strong>Linear<\/strong>.<\/li>\n\n\n\n<li><strong>Bond angle<\/strong>: <strong>180\u00b0<\/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\">2. <strong>NO\u2083\u207b (Nitrate ion)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis structure<\/strong>: Nitrogen is bonded to three oxygens. It forms one double bond and two single bonds with resonance, making all N\u2013O bonds equivalent.<\/li>\n\n\n\n<li><strong>Electron domains<\/strong> around N: 3 bonding pairs, no lone pairs.<\/li>\n\n\n\n<li><strong>Electron geometry<\/strong>: Trigonal planar.<\/li>\n\n\n\n<li><strong>Molecular shape<\/strong>: <strong>Trigonal planar<\/strong>.<\/li>\n\n\n\n<li><strong>Bond angle<\/strong>: <strong>120\u00b0<\/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\">3. <strong>NO\u2082\u207b (Nitrite ion)<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis structure<\/strong>: Nitrogen is bonded to two oxygens with one double bond and one single bond (resonance between them). There is also a <strong>lone pair<\/strong> on nitrogen.<\/li>\n\n\n\n<li><strong>Electron domains<\/strong> around N: 3 (2 bonding pairs, 1 lone pair).<\/li>\n\n\n\n<li><strong>Electron geometry<\/strong>: Trigonal planar.<\/li>\n\n\n\n<li><strong>Molecular shape<\/strong>: <strong>Bent<\/strong>.<\/li>\n\n\n\n<li><strong>Bond angle<\/strong>: Slightly less than 120\u00b0, due to lone pair repulsion, typically <strong>~115\u2013118\u00b0<\/strong> (approximately <strong>~120\u00b0<\/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\">Final Answers:<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Ion<\/th><th>Shape<\/th><th>Bond Angle<\/th><\/tr><\/thead><tbody><tr><td><strong>NO\u207b<\/strong><\/td><td>Linear<\/td><td>180\u00b0<\/td><\/tr><tr><td><strong>NO\u2083\u207b<\/strong><\/td><td>Trigonal planar<\/td><td>120\u00b0<\/td><\/tr><tr><td><strong>NO\u2082\u207b<\/strong><\/td><td>Bent<\/td><td>~120\u00b0<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>These geometries arise from minimizing electron pair repulsions. Lone pairs repel more strongly than bonding pairs, distorting ideal angles slightly, as seen in NO\u2082\u207b. Resonance also plays a role, especially in NO\u2083\u207b, where the structure is delocalized.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/image-286.png\" alt=\"\" class=\"wp-image-218587\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the shape and bond angle of these oxynitrogen ions. Linear Trigonal planar Tetrahedral Bent Pyramidal 180\u00b0 120\u00b0 109.5\u00b0 ~120\u00b0 ~109\u00b0 ~109\u00b0 Answer Bank NO- NO3- NO2- None The Correct Answer and Explanation is: To determine the shape and bond angles of the oxynitrogen ions (NO\u207b, NO\u2083\u207b, and NO\u2082\u207b), we use Valence Shell Electron Pair [&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-218586","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218586","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=218586"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/218586\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=218586"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=218586"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=218586"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}