{"id":256948,"date":"2025-07-16T14:35:02","date_gmt":"2025-07-16T14:35:02","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=256948"},"modified":"2025-07-16T14:35:05","modified_gmt":"2025-07-16T14:35:05","slug":"in-what-type-of-orbital-does-the-lone-pair-on-each-n-atom-reside","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/16\/in-what-type-of-orbital-does-the-lone-pair-on-each-n-atom-reside\/","title":{"rendered":"In what type of orbital does the lone pair on each N atom reside"},"content":{"rendered":"\n<p>In what type of orbital does the lone pair on each N atom reside<\/p>\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 lone pair on each nitrogen (N) atom in a molecule typically resides in an <strong>sp\u00b3 hybrid orbital<\/strong>. Nitrogen, which is in Group 15 of the periodic table, has five valence electrons. In molecules like ammonia (NH\u2083), nitrogen forms three sigma bonds with hydrogen atoms. The fourth electron remains in an orbital that is not involved in bonding, forming a lone pair. This lone pair occupies an <strong>sp\u00b3 hybrid orbital<\/strong>.<\/p>\n\n\n\n<p>To understand why nitrogen uses an sp\u00b3 orbital, it&#8217;s important to recognize its electronic configuration. The nitrogen atom in its ground state has an electron configuration of 1s\u00b2 2s\u00b2 2p\u00b3. When nitrogen forms a molecule like NH\u2083, one of the 2s electrons is promoted to the empty 2p orbital. This results in an excited state where nitrogen has three unpaired electrons available for bonding.<\/p>\n\n\n\n<p>In a molecule, nitrogen&#8217;s 2s and three 2p orbitals hybridize to form four equivalent sp\u00b3 hybrid orbitals. Three of these sp\u00b3 orbitals overlap with the hydrogen 1s orbitals to form bonds, while the fourth sp\u00b3 orbital holds the lone pair of electrons.<\/p>\n\n\n\n<p>The lone pair affects the geometry of the molecule. In ammonia, the molecule adopts a trigonal pyramidal shape because the lone pair occupies one of the sp\u00b3 hybrid orbitals, pushing the hydrogen atoms closer together. The lone pair itself is in an sp\u00b3 orbital, and it exerts a repulsive force on the bonding pairs, which slightly compresses the bond angles from the ideal 109.5 degrees of a perfect tetrahedron.<\/p>\n\n\n\n<p>Thus, in summary, the lone pair on each nitrogen atom typically resides in an <strong>sp\u00b3 hybrid orbital<\/strong>, and the type of orbital plays a crucial role in determining the molecule&#8217;s geometry and reactivity.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>In what type of orbital does the lone pair on each N atom reside The correct answer and explanation is: The lone pair on each nitrogen (N) atom in a molecule typically resides in an sp\u00b3 hybrid orbital. Nitrogen, which is in Group 15 of the periodic table, has five valence electrons. In molecules like [&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-256948","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256948","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=256948"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/256948\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=256948"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=256948"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=256948"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}