{"id":193239,"date":"2025-02-19T08:37:07","date_gmt":"2025-02-19T08:37:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=193239"},"modified":"2025-02-19T08:37:09","modified_gmt":"2025-02-19T08:37:09","slug":"based-on-the-orbital-filling-configurations-predict-which-of-the-following-species-would-have-the-longest-bond","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/19\/based-on-the-orbital-filling-configurations-predict-which-of-the-following-species-would-have-the-longest-bond\/","title":{"rendered":"Based on the orbital filling configurations, predict which of the following species would have the longest bond"},"content":{"rendered":"\n<p>Based on the orbital filling configurations, predict which of the following species would have the longest bond.<br>A) O_{2} ^ 2-<br>B) O_{2}<br>C) O_{2}<br>D) O_{2} ^ +<br>E) O_{2} ^ 2+<\/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>The correct answer is <strong>B) O\u2082<\/strong> (neutral dioxygen).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The bond length in diatomic oxygen species can be predicted by examining the molecular orbital (MO) theory and the number of electrons in the bonding and antibonding molecular orbitals.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Molecular Orbitals for O\u2082:<\/h4>\n\n\n\n<p>For diatomic oxygen, the molecular orbitals are filled in the following order based on the energy levels:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c3(2s)<\/li>\n\n\n\n<li>\u03c3*(2s)<\/li>\n\n\n\n<li>\u03c3(2p_z)<\/li>\n\n\n\n<li>\u03c0(2p_x) = \u03c0(2p_y)<\/li>\n\n\n\n<li>\u03c0<em>(2p_x) = \u03c0<\/em>(2p_y)<\/li>\n\n\n\n<li>\u03c3*(2p_z)<\/li>\n<\/ul>\n\n\n\n<p>In the case of O\u2082, each oxygen atom contributes 8 electrons, and therefore the O\u2082 molecule has 16 electrons in total. These 16 electrons fill the molecular orbitals as follows:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 electrons in \u03c3(2s)<\/li>\n\n\n\n<li>2 electrons in \u03c3*(2s)<\/li>\n\n\n\n<li>4 electrons in \u03c3(2p_z)<\/li>\n\n\n\n<li>4 electrons in the degenerate \u03c0 orbitals (2 in each \u03c0 orbital)<\/li>\n\n\n\n<li>2 electrons in the antibonding \u03c0* orbitals (1 in each)<\/li>\n<\/ul>\n\n\n\n<p>This results in a bond order of <strong>2<\/strong> (calculated as: bond order = (number of bonding electrons &#8211; number of antibonding electrons) \/ 2).<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Effect of Ionization and Electron Addition:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>O\u2082\u00b2\u207b (superoxide)<\/strong>: The addition of 2 electrons to the antibonding \u03c0* orbitals reduces the bond order to 1, leading to a longer bond.<\/li>\n\n\n\n<li><strong>O\u2082\u207a (Oxygen ion)<\/strong>: The removal of 1 electron from the antibonding \u03c0* orbitals increases the bond order to 2.5, resulting in a shorter bond.<\/li>\n\n\n\n<li><strong>O\u2082\u00b2\u207a (Oxygen dication)<\/strong>: The removal of 2 electrons from the antibonding \u03c0* orbitals reduces the bond order to 3, leading to an even shorter bond.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the <strong>neutral O\u2082 molecule<\/strong> (B) has a bond order of 2, which results in a moderate bond length. In comparison, the ionized and charged species tend to have either shorter or longer bonds due to the change in bond order.<\/p>\n\n\n\n<p>I will generate an image for the orbital configuration of O\u2082 for further clarity.<\/p>\n\n\n\n<p>Here is a diagram showing the molecular orbital filling for the O\u2082 molecule. It illustrates the bonding and antibonding orbitals, and you can observe how the electrons fill these orbitals. The neutral O\u2082 molecule has a bond order of 2, which correlates to a bond length that is neither too long nor too short compared to its ionized forms.<\/p>\n\n\n\n<p>If you have any further questions or need more details, feel free to ask!<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-969.png\" alt=\"\" class=\"wp-image-193240\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Based on the orbital filling configurations, predict which of the following species would have the longest bond.A) O_{2} ^ 2-B) O_{2}C) O_{2}D) O_{2} ^ +E) O_{2} ^ 2+ The Correct Answer and Explanation is : The correct answer is B) O\u2082 (neutral dioxygen). Explanation: The bond length in diatomic oxygen species can be predicted by [&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-193239","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193239","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=193239"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193239\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193239"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193239"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193239"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}