{"id":221341,"date":"2025-05-29T12:18:57","date_gmt":"2025-05-29T12:18:57","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221341"},"modified":"2025-05-29T12:18:59","modified_gmt":"2025-05-29T12:18:59","slug":"using-vb-theory-and-the-lewis-structure-model-determine-the-bond-order-in-a-h2-b-na2-c-s2-d-n2-and-e-cl2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/using-vb-theory-and-the-lewis-structure-model-determine-the-bond-order-in-a-h2-b-na2-c-s2-d-n2-and-e-cl2\/","title":{"rendered":"Using VB theory and the Lewis structure model, determine the bond order in (a) H2, (b) Na2, (c) S2, (d) N2 and (e) Cl2."},"content":{"rendered":"\n<p>Using VB theory and the Lewis structure model, determine the bond order in (a) H2, (b) Na2, (c) S2, (d) N2 and (e) Cl2. Is there any ambiguity with finding the bond orders by this method?<\/p>\n\n\n\n<ol start=\"2\" class=\"wp-block-list\">\n<li>Does VB theory indicate that the diatomic molecule He2 is a viable species? Rationalize your answer.<\/li>\n<\/ol>\n\n\n\n<p>(a) Use MO theory to determine the bond order in each of [He2] 2 and [He2] 2. (b) Does the MO picture of the bonding in these ions suggest that they are viable species?<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Bond Order Using Valence Bond (VB) Theory and Lewis Structures<\/strong><\/h3>\n\n\n\n<p>Using <strong>VB theory<\/strong> and <strong>Lewis structures<\/strong>, bond order is the number of bonds shared between atoms in a molecule. It&#8217;s calculated as:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p><strong>Bond Order = (Number of bonding electron pairs between atoms)<\/strong><\/p>\n<\/blockquote>\n\n\n\n<h4 class=\"wp-block-heading\">(a) <strong>H\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each H atom has 1 valence electron.<\/li>\n\n\n\n<li>H\u2082 shares a pair of electrons \u2192 1 single bond.<\/li>\n\n\n\n<li><strong>Bond Order = 1<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(b) <strong>Na\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each Na has 1 valence electron (3s\u00b9).<\/li>\n\n\n\n<li>Like H\u2082, they share a pair of electrons \u2192 1 bond.<\/li>\n\n\n\n<li><strong>Bond Order = 1<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(c) <strong>S\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Sulfur has 6 valence electrons (Group 16).<\/li>\n\n\n\n<li>S\u2082 forms a double bond (like O\u2082): S=S.<\/li>\n\n\n\n<li><strong>Bond Order = 2<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(d) <strong>N\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Nitrogen has 5 valence electrons.<\/li>\n\n\n\n<li>Triple bond in N\u2082: N\u2261N.<\/li>\n\n\n\n<li><strong>Bond Order = 3<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(e) <strong>Cl\u2082<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chlorine has 7 valence electrons.<\/li>\n\n\n\n<li>Shares one pair \u2192 single bond.<\/li>\n\n\n\n<li><strong>Bond Order = 1<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>Ambiguity in VB\/Lewis Bond Orders:<\/strong><\/h4>\n\n\n\n<p>Yes. <strong>VB theory and Lewis structures<\/strong> can be ambiguous, especially:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>When dealing with <strong>resonance structures<\/strong>, delocalized bonding (e.g., O\u2083).<\/li>\n\n\n\n<li><strong>Transition metals<\/strong> or molecules with <strong>unpaired electrons<\/strong>.<\/li>\n\n\n\n<li>Fails to explain <strong>bonding in excited states<\/strong>, <strong>magnetism<\/strong>, or <strong>bonding energies<\/strong> accurately.<\/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\"><strong>2. He\u2082 and VB Theory<\/strong><\/h3>\n\n\n\n<p>According to <strong>Valence Bond theory<\/strong>, <strong>He\u2082<\/strong> is <em>not a viable species<\/em>:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each He atom has a full 1s\u00b2 shell.<\/li>\n\n\n\n<li>No need to share electrons; forming a bond would <strong>increase repulsion<\/strong>.<\/li>\n\n\n\n<li>No unpaired electrons to overlap; so <strong>no covalent bond can form<\/strong>.<\/li>\n\n\n\n<li>Hence, <strong>He\u2082 is unstable<\/strong> and does not exist under normal conditions.<\/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\"><strong>3. Bond Order of [He\u2082]\u00b2\u207a and [He\u2082]\u00b2\u207b Using MO Theory<\/strong><\/h3>\n\n\n\n<p>In <strong>Molecular Orbital (MO) Theory<\/strong>, we calculate bond order as:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p><strong>Bond Order = (Bonding electrons \u2013 Antibonding electrons) \/ 2<\/strong><\/p>\n<\/blockquote>\n\n\n\n<h4 class=\"wp-block-heading\">(a) <strong>[He\u2082]\u00b2\u207a<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>He\u2082 has 4 electrons (2 per He).<\/li>\n\n\n\n<li>[He\u2082]\u00b2\u207a loses 2 electrons \u2192 2 electrons total.<\/li>\n\n\n\n<li>Fill \u03c3(1s): 2 electrons.<\/li>\n\n\n\n<li>No electrons in \u03c3*(1s).<\/li>\n\n\n\n<li><strong>Bond Order = (2 \u2013 0) \/ 2 = 1<\/strong><\/li>\n\n\n\n<li><strong>Viable species<\/strong>: Yes, stable in high-energy environments.<\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">(b) <strong>[He\u2082]\u00b2\u207b<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>He\u2082 has 4 electrons + 2 = 6 electrons.<\/li>\n\n\n\n<li>\u03c3(1s): 2, \u03c3*(1s): 2, then next 2 go into \u03c3*(1s) \u2192 total of 4 in antibonding.<\/li>\n\n\n\n<li><strong>Bond Order = (2 \u2013 4) \/ 2 = \u20131<\/strong><\/li>\n\n\n\n<li><strong>Viable species<\/strong>: No, <strong>negative bond order<\/strong> \u2192 repulsion > attraction.<\/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\"><strong>Explanation<\/strong><\/h3>\n\n\n\n<p>The stability and bonding of diatomic species like H\u2082, N\u2082, or He\u2082 can be understood using both Valence Bond (VB) theory and Molecular Orbital (MO) theory, though each has its strengths and limitations. VB theory, based on Lewis structures, predicts bond order by counting shared electron pairs. For instance, H\u2082 has one shared pair and thus a bond order of 1, while N\u2082 has a triple bond (bond order 3). However, this model oversimplifies bonding in molecules with delocalized electrons or resonance, such as O\u2082 or benzene, and cannot explain magnetic or energetic behavior accurately.<\/p>\n\n\n\n<p>He\u2082 provides a good example of the limitations of VB theory. Helium atoms have full 1s orbitals, and thus no tendency to form bonds; VB theory correctly predicts He\u2082 as non-viable. MO theory, however, gives a quantitative basis: He\u2082 has two electrons in bonding and two in antibonding orbitals, yielding a bond order of 0, confirming its instability.<\/p>\n\n\n\n<p>Charged helium species give more insight. [He\u2082]\u00b2\u207a, which has two electrons, fills only the bonding \u03c3(1s) orbital, with no electrons in the antibonding \u03c3*(1s) orbital. This gives a bond order of 1, indicating a stable bond. On the other hand, [He\u2082]\u00b2\u207b has six electrons: two in bonding and four in antibonding orbitals, resulting in a bond order of \u20131. This negative bond order suggests that the molecule is inherently unstable and will not exist under normal conditions.<\/p>\n\n\n\n<p>Thus, MO theory extends VB theory by explaining why certain molecules exist or not based on energy considerations and orbital overlap. It offers a more complete picture, especially for ions, excited states, and molecules with unusual electron configurations.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner4-102.jpeg\" alt=\"\" class=\"wp-image-221342\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Using VB theory and the Lewis structure model, determine the bond order in (a) H2, (b) Na2, (c) S2, (d) N2 and (e) Cl2. Is there any ambiguity with finding the bond orders by this method? (a) Use MO theory to determine the bond order in each of [He2] 2 and [He2] 2. (b) Does [&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-221341","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221341","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=221341"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221341\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221341"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221341"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221341"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}