{"id":190382,"date":"2025-02-12T08:13:10","date_gmt":"2025-02-12T08:13:10","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=190382"},"modified":"2025-02-12T08:13:13","modified_gmt":"2025-02-12T08:13:13","slug":"plot-a-molecular-orbital-diagram-for-n2-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/02\/12\/plot-a-molecular-orbital-diagram-for-n2-molecule\/","title":{"rendered":"Plot a molecular orbital diagram for N2 molecule"},"content":{"rendered":"\n<ul class=\"wp-block-list\">\n<li>Plot a molecular orbital diagram for N2 molecule.<\/li>\n\n\n\n<li>Write down the ground-state electronic configuration for a following series of species: N + 2 , N2, N &#8211; 2 and N 2- 2 . Assign the bond order and determine the molecular term symbol.<\/li>\n<\/ul>\n\n\n\n<p>Sketch the molecular orbital energy diagram for for a radical OH molecule<\/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<h3 class=\"wp-block-heading\">Molecular Orbital Diagram for ( \\text{N}_2 ) Molecule:<\/h3>\n\n\n\n<p>The molecular orbital (MO) diagram for ( \\text{N}_2 ) is based on the fact that nitrogen has 7 electrons in each atom, so for a ( \\text{N}_2 ) molecule, there will be a total of 14 electrons. The arrangement of molecular orbitals (MOs) follows the Aufbau principle, and we fill them from lowest to highest energy, following Hund&#8217;s rule and the Pauli exclusion principle.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Step-by-Step Explanation:<\/h4>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Energy Levels<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For ( \\text{N}_2 ), molecular orbitals are derived from the atomic orbitals of nitrogen. The following are the MOs for diatomic molecules involving second-period elements:\n<ul class=\"wp-block-list\">\n<li>( \\sigma_{1s} ), ( \\sigma_{1s}^* ) (lowest)<\/li>\n\n\n\n<li>( \\sigma_{2s} ), ( \\sigma_{2s}^* )<\/li>\n\n\n\n<li>( \\sigma_{2p_z} ), ( \\pi_{2p_x} = \\pi_{2p_y} ), ( \\pi_{2p_x}^* = \\pi_{2p_y}^* ), ( \\sigma_{2p_z}^* ) (highest)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Electron Configuration<\/strong>:<br>The 14 electrons from both nitrogen atoms will fill the molecular orbitals as follows:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>2 electrons in ( \\sigma_{1s} )<\/li>\n\n\n\n<li>2 electrons in ( \\sigma_{1s}^* )<\/li>\n\n\n\n<li>2 electrons in ( \\sigma_{2s} )<\/li>\n\n\n\n<li>2 electrons in ( \\sigma_{2s}^* )<\/li>\n\n\n\n<li>4 electrons in ( \\pi_{2p_x} = \\pi_{2p_y} ) (these orbitals are degenerate)<\/li>\n\n\n\n<li>2 electrons in ( \\sigma_{2p_z} )<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Bond Order:<\/h4>\n\n\n\n<p>The bond order can be calculated using the formula:<\/p>\n\n\n\n<p>[<br>\\text{Bond Order} = \\frac{1}{2} \\left( \\text{Number of electrons in bonding MOs} &#8211; \\text{Number of electrons in anti-bonding MOs} \\right)<br>]<\/p>\n\n\n\n<p>For ( \\text{N}_2 ):<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Bonding MOs: 10 electrons (from ( \\sigma_{1s}, \\sigma_{2s}, \\pi_{2p_x}, \\pi_{2p_y}, \\sigma_{2p_z} ))<\/li>\n\n\n\n<li>Anti-bonding MOs: 4 electrons (from ( \\sigma_{1s}^<em>, \\sigma_{2s}^<\/em> ))<\/li>\n<\/ul>\n\n\n\n<p>Bond Order = ( \\frac{1}{2} \\times (10 &#8211; 4) = 3 )<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Molecular Term Symbol:<\/h4>\n\n\n\n<p>The molecular term symbol for ( \\text{N}_2 ) is ( ^1\\Sigma_g^+ ), where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( ^1 ) indicates singlet state (paired electrons),<\/li>\n\n\n\n<li>( \\Sigma ) indicates no orbital angular momentum (since ( \\pi )-orbitals are not involved in the ground state),<\/li>\n\n\n\n<li>( g ) indicates a gerade (symmetric) molecule, and<\/li>\n\n\n\n<li>( + ) indicates the symmetry of the wavefunction.<\/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\">Electronic Configuration for ( N^{+2} ), ( N_2 ), ( N^{-2} ), and ( N_2^{2-} ):<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>( N^{+2} )<\/strong> (Ion with 2 fewer electrons):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For ( N^{+2} ), the electronic configuration is based on ( N_2 ) but with 2 electrons removed, resulting in:\n<ul class=\"wp-block-list\">\n<li>( \\sigma_{1s}^2 ), ( \\sigma_{1s}^* ), ( \\sigma_{2s}^2 ), ( \\sigma_{2s}^* ), ( \\pi_{2p_x}^2, \\pi_{2p_y}^2 ), ( \\sigma_{2p_z}^2 ) (10 electrons)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Bond order = 3<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>( N_2 )<\/strong> (Neutral molecule):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( \\sigma_{1s}^2 ), ( \\sigma_{1s}^* ), ( \\sigma_{2s}^2 ), ( \\sigma_{2s}^* ), ( \\pi_{2p_x}^2, \\pi_{2p_y}^2 ), ( \\sigma_{2p_z}^2 ) (14 electrons)<\/li>\n\n\n\n<li>Bond order = 3<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>( N^{-2} )<\/strong> (Ion with 2 additional electrons):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The configuration for ( N^{-2} ) would be:\n<ul class=\"wp-block-list\">\n<li>( \\sigma_{1s}^2 ), ( \\sigma_{1s}^* ), ( \\sigma_{2s}^2 ), ( \\sigma_{2s}^* ), ( \\pi_{2p_x}^2, \\pi_{2p_y}^2 ), ( \\sigma_{2p_z}^2 ) (16 electrons)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Bond order = 4<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>( N_2^{2-} )<\/strong> (Ion with 2 additional electrons):<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The configuration is the same as ( N^{-2} ), resulting in:\n<ul class=\"wp-block-list\">\n<li>( \\sigma_{1s}^2 ), ( \\sigma_{1s}^* ), ( \\sigma_{2s}^2 ), ( \\sigma_{2s}^* ), ( \\pi_{2p_x}^2, \\pi_{2p_y}^2 ), ( \\sigma_{2p_z}^2 ) (16 electrons)<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Bond order = 4<\/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\">Molecular Orbital Energy Diagram for ( \\text{OH} ) Radical:<\/h3>\n\n\n\n<p>The OH radical has 9 valence electrons in total (8 from oxygen and 1 from hydrogen). The molecular orbital diagram for this species involves considering the bonding and anti-bonding orbitals derived from the atomic orbitals of oxygen and hydrogen.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>MO diagram<\/strong> involves ( \\sigma_{1s} ), ( \\sigma_{1s}^* ), ( \\sigma_{2s} ), ( \\sigma_{2s}^* ), ( \\sigma_{2p_z} ), ( \\pi_{2p_x} ), ( \\pi_{2p_y} ), and their anti-bonding counterparts. The energy levels are filled according to the number of electrons in these orbitals.<\/li>\n<\/ul>\n\n\n\n<p>I will now generate a plot for the molecular orbital diagram for the OH radical and provide a more detailed explanation.<\/p>\n\n\n\n<p>The molecular orbital diagram for the OH radical (which has 9 electrons) is shown above. Here&#8217;s a breakdown of the key components:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Energy Levels<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The molecular orbitals are arranged according to their energy levels, with lower energy levels at the bottom. The ( \\sigma )-orbitals are filled before the ( \\pi )-orbitals.<\/li>\n\n\n\n<li>Bonding MOs are at lower energy, and anti-bonding MOs are at higher energy.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Electron Distribution<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The 9 electrons are distributed across the available molecular orbitals. The bonding MOs (such as ( \\sigma_{1s}, \\sigma_{2s}, \\sigma_{2p_z}, \\pi_{2p_x}, \\pi_{2p_y} )) are filled first, and the remaining electrons occupy the anti-bonding orbitals, such as ( \\pi_{2p_x}^<em>, \\pi_{2p_y}^<\/em> ).<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Bonding and Anti-Bonding Orbitals<\/strong>:<\/li>\n<\/ol>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The OH radical has a bonding interaction between oxygen&#8217;s orbitals and hydrogen&#8217;s orbitals, contributing to its stability, but it also has some anti-bonding character due to the presence of unpaired electrons in higher-energy orbitals.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The OH radical has an odd number of electrons (9 total), meaning it has a single unpaired electron in one of the anti-bonding orbitals, which is typical for a radical species.<\/li>\n\n\n\n<li>This unpaired electron leads to the radical behavior of the molecule.<\/li>\n\n\n\n<li>The bond order is calculated based on the difference between the number of electrons in bonding and anti-bonding orbitals.<\/li>\n<\/ul>\n\n\n\n<p>The bond order for the OH radical is ( \\frac{1}{2} \\times (8 \\text{ bonding electrons} &#8211; 1 \\text{ anti-bonding electron}) = 3.5 ), indicating a strong bonding interaction with a small degree of destabilization from the radical electron.<\/p>\n\n\n\n<p>Let me know if you&#8217;d like further details on any part of this diagram!<\/p>\n\n\n\n<figure class=\"wp-block-image size-large\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/02\/image-376-768x1024.png\" alt=\"\" class=\"wp-image-190383\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Sketch the molecular orbital energy diagram for for a radical OH molecule The Correct Answer and Explanation is : Molecular Orbital Diagram for ( \\text{N}_2 ) Molecule: The molecular orbital (MO) diagram for ( \\text{N}_2 ) is based on the fact that nitrogen has 7 electrons in each atom, so for a ( \\text{N}_2 ) [&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-190382","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190382","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=190382"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/190382\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=190382"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=190382"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=190382"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}