{"id":183421,"date":"2025-01-16T12:43:48","date_gmt":"2025-01-16T12:43:48","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183421"},"modified":"2025-01-16T12:43:50","modified_gmt":"2025-01-16T12:43:50","slug":"consider-the-ground-state-electron-configuration-of-titanium","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/consider-the-ground-state-electron-configuration-of-titanium\/","title":{"rendered":"Consider the ground state electron configuration of titanium"},"content":{"rendered":"\n<p>Consider the ground state electron configuration of titanium, Ti: [Ar] 4s2\u00c3\u201a\u00c2 3d2.\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>The total orbital angular momentum, L, for the ground state of Ti is <strong><em>_ and is represented by the letter <\/em><\/strong>?<\/p>\n\n\n\n<p>The total spin angular momentum, S, for the ground state of Ti is <em>__<\/em> ?<\/p>\n\n\n\n<p>The multiplicity, 2S+1, for the ground state of Ti is <strong><em>_ which is a <strong><em>_______<\/em><\/strong><\/em><\/strong>?<\/p>\n\n\n\n<p>The total angular momentum, J, of Ti can have the following possible values when L and S are determined from the ground state electron configuration.<\/p>\n\n\n\n<p>The actual ground state term symbol for Ti has a J value of <em>_<\/em>?<\/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>To determine the required properties for titanium (Ti\\text{Ti}), consider its ground state electron configuration:<\/p>\n\n\n\n<p><strong>Electron configuration:<\/strong><br>[Ar]\u20094s2\u20093d2[\\text{Ar}] \\, 4s^2 \\, 3d^2<\/p>\n\n\n\n<p>Titanium has two unpaired electrons in the 3d3d-orbital. Let\u2019s calculate each property step by step:<\/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>Total Orbital Angular Momentum (LL):<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each dd-electron has l=2l = 2 (orbital angular momentum quantum number). The possible mlm_l values are \u22122,\u22121,0,+1,+2-2, -1, 0, +1, +2.<\/li>\n\n\n\n<li>For two dd-electrons, we sum their mlm_l values:\n<ul class=\"wp-block-list\">\n<li>Maximum L=2+2=4L = 2 + 2 = 4, and we reduce LL using electron pairing rules (Hund&#8217;s rules). The two electrons are unpaired and occupy different mlm_l values to minimize repulsion. The resulting L=3L = 3 (corresponding to FF).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Answer:<\/strong> L=3L = 3, represented by FF.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">2. <strong>Total Spin Angular Momentum (SS):<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each electron has spin s=1\/2s = 1\/2. For two unpaired electrons:\n<ul class=\"wp-block-list\">\n<li>S=s1+s2=1\/2+1\/2=1S = s_1 + s_2 = 1\/2 + 1\/2 = 1 (maximum spin state due to parallel spins).<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Answer:<\/strong> S=1S = 1.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">3. <strong>Multiplicity (2S+12S+1):<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The multiplicity is given by 2S+12S + 1:\n<ul class=\"wp-block-list\">\n<li>2(1)+1=32(1) + 1 = 3, corresponding to a <strong>triplet state<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Answer:<\/strong> Multiplicity = 33, which is a <strong>triplet<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">4. <strong>Total Angular Momentum (JJ):<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The total angular momentum is J=\u2223L\u2212S\u2223J = |L &#8211; S| to L+SL + S:\n<ul class=\"wp-block-list\">\n<li>Possible JJ values: \u22233\u22121\u2223,3,3+1|3 &#8211; 1|, 3, 3 + 1 \u2192 J=2,3,4J = 2, 3, 4.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p>The actual ground state follows Hund\u2019s rules:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Maximize JJ for less-than-half-filled subshells.<\/li>\n\n\n\n<li><strong>Answer:<\/strong> J=2J = 2.<\/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\">Ground State Term Symbol:<\/h3>\n\n\n\n<p>The ground state term symbol combines L,S,L, S, and JJ:<br>\\text{Term Symbol: } \\, ^3F_2<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Consider the ground state electron configuration of titanium, Ti: [Ar] 4s2\u00c3\u201a\u00c2 3d2.\u00c3\u201a\u00c2 The total orbital angular momentum, L, for the ground state of Ti is _ and is represented by the letter ? The total spin angular momentum, S, for the ground state of Ti is __ ? The multiplicity, 2S+1, for the ground state [&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-183421","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183421","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=183421"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183421\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183421"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183421"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183421"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}