{"id":183806,"date":"2025-01-17T06:16:55","date_gmt":"2025-01-17T06:16:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183806"},"modified":"2025-01-17T06:16:57","modified_gmt":"2025-01-17T06:16:57","slug":"use-molecular-orbital-theory-to-complete-the-ground-state-electron-configuration-for-each-of-the-molecules","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/17\/use-molecular-orbital-theory-to-complete-the-ground-state-electron-configuration-for-each-of-the-molecules\/","title":{"rendered":"Use molecular orbital theory to complete the ground state electron configuration for each of the molecules"},"content":{"rendered":"\n

Use molecular orbital theory to complete the ground state electron configuration for each of the molecules.<\/p>\n\n\n\n

Molecule<\/p>\n\n\n\n

Ground state electron configuration<\/p>\n\n\n\n

NF<\/p>\n\n\n\n

(\u03c31s)(\u03c31s)(\u03c32s)(\u03c32s<\/em>)(\u03c02p)(\u03c32p)(\u03c02p*)<\/p>\n\n\n\n

NF+<\/p>\n\n\n\n

(\u03c31s)(\u03c31s)(\u03c32s)(\u03c32s<\/em>)(\u03c02p)(\u03c32p)(\u03c02p*)<\/p>\n\n\n\n

NF\u2212<\/p>\n\n\n\n

(\u03c31s)(\u03c31s)(\u03c32s)(\u03c32s<\/em>)(\u03c02p)(\u03c32p)(\u03c02p*)<\/p>\n\n\n\n

Answer Bank<\/p>\n\n\n\n

0<\/p>\n\n\n\n

1<\/p>\n\n\n\n

2<\/p>\n\n\n\n

3<\/p>\n\n\n\n

4<\/p>\n\n\n\n

Determine the bond order of each of the molecules.<\/p>\n\n\n\n

NF bond order:<\/p>\n\n\n\n

NF+ bond order:<\/p>\n\n\n\n

NF\u2212 bond order:<\/p>\n\n\n\n

Classify each molecule according to its magnetic properties.<\/p>\n\n\n\n

Diamagnetic Paramagnetic<\/p>\n\n\n\n

Answer Bank<\/p>\n\n\n\n

NF<\/p>\n\n\n\n

NF+<\/p>\n\n\n\n

NF\u2212<\/p>\n\n\n\n

The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Molecular Orbital Theory Configurations for ( \\text{NF}, \\text{NF}^+, \\text{NF}^- ):<\/h3>\n\n\n\n

The ground-state electron configuration of these molecules can be completed based on the molecular orbital (MO) diagram for heteronuclear diatomic molecules, considering nitrogen (( Z = 7 )) and fluorine (( Z = 9 )). The key orbital energy ordering for NF is as follows:
[
(\\sigma_{1s})(\\sigma_{1s}^)(\\sigma_{2s})(\\sigma_{2s}^<\/em>)(\\pi_{2p})(\\sigma_{2p})(\\pi_{2p}^*)
]<\/p>\n\n\n\n

    \n
  1. NF (neutral):<\/strong>
    Total electrons: ( 7 + 9 = 16 ).
    Configuration: ( (\\sigma_{1s})^2(\\sigma_{1s}^)^2(\\sigma_{2s})^2(\\sigma_{2s}^<\/em>)^2(\\pi_{2p})^4(\\sigma_{2p})^2 ).<\/li>\n\n\n\n
  2. ( \\text{NF}^+ ):<\/strong>
    Total electrons: ( 16 – 1 = 15 ).
    Configuration: ( (\\sigma_{1s})^2(\\sigma_{1s}^)^2(\\sigma_{2s})^2(\\sigma_{2s}^<\/em>)^2(\\pi_{2p})^4(\\sigma_{2p})^1 ).<\/li>\n\n\n\n
  3. ( \\text{NF}^- ):<\/strong>
    Total electrons: ( 16 + 1 = 17 ).
    Configuration: ( (\\sigma_{1s})^2(\\sigma_{1s}^)^2(\\sigma_{2s})^2(\\sigma_{2s}^<\/em>)^2(\\pi_{2p})^4(\\sigma_{2p})^2(\\pi_{2p}^*)^1 ).<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Bond Order Calculation:<\/h3>\n\n\n\n

    Bond order is calculated as:
    [
    \\text{Bond order} = \\frac{1}{2} \\left( N_b – N_a \\right)
    ]
    where ( N_b ) = bonding electrons, ( N_a ) = antibonding electrons.<\/p>\n\n\n\n

      \n
    1. NF:<\/strong>
      ( N_b = 10 ) (\u03c31s, \u03c32s, \u03c02p, \u03c32p), ( N_a = 6 ) (\u03c31s, \u03c32s<\/em>).
      Bond order = ( \\frac{1}{2}(10 – 6) = 2.0 ).<\/li>\n\n\n\n
    2. ( \\text{NF}^+ ):<\/strong>
      ( N_b = 9 ), ( N_a = 6 ).
      Bond order = ( \\frac{1}{2}(9 – 6) = 1.5 ).<\/li>\n\n\n\n
    3. ( \\text{NF}^- ):<\/strong>
      ( N_b = 10 ), ( N_a = 7 ).
      Bond order = ( \\frac{1}{2}(10 – 7) = 1.5 ).<\/li>\n<\/ol>\n\n\n\n
      \n\n\n\n

      Magnetic Properties:<\/h3>\n\n\n\n
        \n
      • Diamagnetic:<\/strong> Molecules with all electrons paired.<\/li>\n\n\n\n
      • Paramagnetic:<\/strong> Molecules with unpaired electrons.<\/li>\n<\/ul>\n\n\n\n
          \n
        1. NF:<\/strong> All electrons are paired \u2192 Diamagnetic<\/strong>.<\/li>\n\n\n\n
        2. ( \\text{NF}^+ ):<\/strong> One unpaired electron in ( \\sigma_{2p} ) \u2192 Paramagnetic<\/strong>.<\/li>\n\n\n\n
        3. ( \\text{NF}^- ):<\/strong> One unpaired electron in ( \\pi_{2p}^* ) \u2192 Paramagnetic<\/strong>.<\/li>\n<\/ol>\n\n\n\n
          \n\n\n\n

          Summary:<\/h3>\n\n\n\n
            \n
          1. Ground state configurations:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • NF: ( (\\sigma_{1s})^2(\\sigma_{1s}^)^2(\\sigma_{2s})^2(\\sigma_{2s}^<\/em>)^2(\\pi_{2p})^4(\\sigma_{2p})^2 )<\/li>\n\n\n\n
            • ( \\text{NF}^+ ): ( (\\sigma_{1s})^2(\\sigma_{1s}^)^2(\\sigma_{2s})^2(\\sigma_{2s}^<\/em>)^2(\\pi_{2p})^4(\\sigma_{2p})^1 )<\/li>\n\n\n\n
            • ( \\text{NF}^- ): ( (\\sigma_{1s})^2(\\sigma_{1s}^)^2(\\sigma_{2s})^2(\\sigma_{2s}^<\/em>)^2(\\pi_{2p})^4(\\sigma_{2p})^2(\\pi_{2p}^*)^1 )<\/li>\n<\/ul>\n\n\n\n
                \n
              1. Bond orders:<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • NF: 2.0<\/li>\n\n\n\n
                • ( \\text{NF}^+ ): 1.5<\/li>\n\n\n\n
                • ( \\text{NF}^- ): 1.5<\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. Magnetic properties:<\/strong><\/li>\n<\/ol>\n\n\n\n
                      \n
                    • NF: Diamagnetic<\/li>\n\n\n\n
                    • ( \\text{NF}^+ ): Paramagnetic<\/li>\n\n\n\n
                    • ( \\text{NF}^- ): Paramagnetic<\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"

                      Use molecular orbital theory to complete the ground state electron configuration for each of the molecules. Molecule Ground state electron configuration NF (\u03c31s)(\u03c31s)(\u03c32s)(\u03c32s)(\u03c02p)(\u03c32p)(\u03c02p*) NF+ (\u03c31s)(\u03c31s)(\u03c32s)(\u03c32s)(\u03c02p)(\u03c32p)(\u03c02p*) NF\u2212 (\u03c31s)(\u03c31s)(\u03c32s)(\u03c32s)(\u03c02p)(\u03c32p)(\u03c02p*) Answer Bank 0 1 2 3 4 Determine the bond order of each of the molecules. NF bond order: NF+ bond order: NF\u2212 bond order: Classify each molecule […]<\/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-183806","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183806","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=183806"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183806\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183806"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183806"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183806"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}