Molecular Orbital Diagram for F\u2082\u207b<\/h3>\n\n\n\n
Steps to draw the molecular orbital diagram:<\/strong><\/p>\n\n\n\n In summary, F\u2082\u207b has a bond order of 1\/2, and its molecular orbital diagram shows the placement of 19 electrons in bonding and antibonding orbitals. The additional electron in the antibonding \u03c3*(2pz) orbital reduces the bond strength<\/p>\n\n\n\n Draw the molecular orbital diagram for F2- The Correct Answer and Explanation is: Molecular Orbital Diagram for F\u2082\u207b Steps to draw the molecular orbital diagram: In summary, F\u2082\u207b has a bond order of 1\/2, and its molecular orbital diagram shows the placement of 19 electrons in bonding and antibonding orbitals. The additional electron in the […]<\/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-253789","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253789","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=253789"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253789\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253789"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253789"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253789"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}\n
Fluorine has an atomic number of 9, so the electron configuration is: 1s2 2s2 2p51s^2 \\ 2s^2 \\ 2p^51s2 2s2 2p5 When two fluorine atoms come together to form the F\u2082 molecule, they each contribute 9 electrons, giving a total of 18 electrons in the bonding molecular orbitals.<\/li>\n\n\n\n
The negative charge (F\u2082\u207b) adds one more electron, increasing the total electron count to 19. This electron will be placed in one of the molecular orbitals.<\/li>\n\n\n\n
The molecular orbitals for diatomic molecules like F\u2082 are filled in the following order based on the energy level:\n\n
The electrons are placed in the molecular orbitals, starting with the lowest energy levels:\n\n
Bond order is calculated as: Bond order=12(Number of bonding electrons\u2212Number of antibonding electrons)\\text{Bond order} = \\frac{1}{2} (\\text{Number of bonding electrons} – \\text{Number of antibonding electrons})Bond order=21\u200b(Number of bonding electrons\u2212Number of antibonding electrons) In F\u2082\u207b:
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<\/figure>\n","protected":false},"excerpt":{"rendered":"