To determine the correct electron geometry, molecular geometry, and bond angles for NO\u2082 (nitrogen dioxide)<\/strong>, we must analyze its Lewis structure<\/strong> and apply VSEPR (Valence Shell Electron Pair Repulsion) theory<\/strong>.<\/p>\n\n\n\n The Lewis structure of NO\u2082<\/strong> shows that nitrogen has 2 bonding groups<\/strong> (the two oxygen atoms) and 1 nonbonding group<\/strong> (a lone pair). Therefore, it has trigonal planar electron geometry<\/strong> and bent molecular geometry<\/strong>. Nitrogen dioxide (NO\u2082) is a small molecule made up of one nitrogen atom and two oxygen atoms. To understand its geometry, we begin by drawing the Lewis structure<\/strong>. Nitrogen is the central atom, forming one double bond and one single bond with two oxygen atoms, with an odd number of total valence electrons<\/strong> (17). This makes NO\u2082 a radical<\/strong>, which is somewhat unusual and affects its stability and bonding.<\/p>\n\n\n\n In NO\u2082, nitrogen is surrounded by two bonding pairs<\/strong> and one lone pair<\/strong> of electrons. According to VSEPR theory<\/strong>, this gives nitrogen a total of three regions of electron density<\/strong>, which arrange themselves in a trigonal planar electron geometry<\/strong> to minimize repulsion.<\/p>\n\n\n\n However, because one of those regions is a lone pair, the molecular geometry<\/strong> becomes bent (or angular)<\/strong>. Lone pairs repel bonding pairs more strongly than bonding pairs repel each other, which compresses the bond angle<\/strong>. While the ideal bond angle for a trigonal planar arrangement is 120\u00b0<\/strong>, the actual bond angle in NO\u2082 is slightly less than 120\u00b0<\/strong>, typically around 115\u2013118\u00b0<\/strong>.<\/p>\n\n\n\n Some confusion arises when comparing this to tetrahedral<\/strong> geometries, which have bond angles of 109.5\u00b0<\/strong>. That value applies to molecules with four regions of electron density<\/strong> (e.g., ammonia, NH\u2083). NO\u2082, with only three regions, does not<\/strong> fall into this category.<\/p>\n\n\n\n Thus, the correct classification<\/strong> for NO\u2082 is:<\/p>\n\n\n\n This understanding helps predict molecular shape and polarity, both critical in chemistry and chemical reactivity.<\/p>\n\n\n\n Electrons Trigonal planar Trigonal pyramidal The Lewis structure of NO2 shows that nitrogen has bonding group(s) and tetrahedral nonbonding group(s). Therefore, it has electron geometry and molecular geometry. For this molecular geometry, the idealized bond angles are 109.5\u00c2\u00b0. The Lewis structure of NO2 shows that nitrogen has bonding group(s) and nonbonding group(s). Therefore, it has […]<\/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-227311","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227311","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=227311"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/227311\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=227311"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=227311"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=227311"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n\u2705 Corrected Answer:<\/strong><\/h3>\n\n\n\n
For this molecular geometry, the idealized bond angle<\/strong> is 120\u00b0<\/strong>, but the actual bond angle is slightly less than 120\u00b0<\/strong> because lone pairs exert greater repulsion than bonding pairs.<\/p>\n\n\n\n
\n\n\n\n\ud83d\udd0d Detailed Explanation<\/strong><\/h3>\n\n\n\n
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