{"id":180905,"date":"2025-01-09T07:53:03","date_gmt":"2025-01-09T07:53:03","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=180905"},"modified":"2025-01-09T07:53:06","modified_gmt":"2025-01-09T07:53:06","slug":"determine-the-lewis-dot-structure-of-c2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/09\/determine-the-lewis-dot-structure-of-c2\/","title":{"rendered":"Determine the Lewis dot structure of C2"},"content":{"rendered":"\n<p>a. Determine the Lewis dot structure of C2? that satisfies the octet rule for carbon. b. Describe the bonding in terms of VBT for your Lewis structure, i.e. how are the bonds formed. c. Draw and label the MO diagram for C2?. Draw the orbital overlap next to the MO levels. The complete MO should be labeled appropriately, MOs should be labeled with the orbital type (s,p). d. Does your MO give a diamagnetic or paramagnetic molecule? e. How do the Lewis structure and MO diagram compare? f. What other Lewis structure(s) might be possible that do not obey the octet rule? g. Pick the Lewis structure you favor, and explain why it is most favorable.<\/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\">a. Lewis Dot Structure of C\u2082\u00b2\u207b (C\u2082 ion with a 2- charge):<\/h3>\n\n\n\n<p>To begin with, we calculate the total number of valence electrons for the C\u2082\u00b2\u207b ion:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon atom has 4 valence electrons, so for two carbon atoms: 4 * 2 = 8 electrons.<\/li>\n\n\n\n<li>The 2- charge means we add 2 more electrons, bringing the total to 8 + 2 = 10 valence electrons.<\/li>\n<\/ul>\n\n\n\n<p>For C\u2082\u00b2\u207b, to satisfy the octet rule for carbon (which requires 8 electrons), a plausible Lewis structure is as follows:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Both carbon atoms will share electrons to form bonds.<\/li>\n\n\n\n<li>To achieve the octet, a double bond is formed between the two carbon atoms. Since there are 10 electrons, two lone pairs will be placed on each carbon.<\/li>\n<\/ul>\n\n\n\n<p><strong>Lewis structure for C\u2082\u00b2\u207b:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>: C = C :\n<\/code><\/pre>\n\n\n\n<p>Here, each carbon atom has two lone pairs, and the two carbon atoms are bonded by a double bond. This structure satisfies the octet rule for carbon.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">b. Bonding in Terms of Valence Bond Theory (VBT):<\/h3>\n\n\n\n<p>In Valence Bond Theory, bonds are formed by the overlap of atomic orbitals. For C\u2082\u00b2\u207b:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon atom has 4 valence electrons in the 2s and 2p orbitals.<\/li>\n\n\n\n<li>The two carbon atoms will overlap their 2p orbitals to form a \u03c0-bond (side-to-side overlap) and overlap their 2s orbitals to form a \u03c3-bond (end-to-end overlap).<\/li>\n\n\n\n<li>The double bond consists of one \u03c3 bond and one \u03c0 bond.<\/li>\n\n\n\n<li>The lone pairs on each carbon atom occupy the remaining orbitals.<\/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\">c. Molecular Orbital (MO) Diagram:<\/h3>\n\n\n\n<p>The MO diagram for C\u2082\u00b2\u207b can be constructed by considering the combination of atomic orbitals of the two carbon atoms. The molecular orbitals will be labeled as follows:<\/p>\n\n\n\n<ol class=\"wp-block-list\">\n<li>The atomic 2s orbitals combine to form bonding and antibonding orbitals: \u03c3(2s) and \u03c3*(2s).<\/li>\n\n\n\n<li>The atomic 2p orbitals combine to form bonding (\u03c3(2p) and \u03c0(2p)) and antibonding orbitals (\u03c0*(2p) and \u03c3*(2p)).<\/li>\n<\/ol>\n\n\n\n<p><strong>MO Diagram for C\u2082\u00b2\u207b:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>         \u03c3*(2s)  \u03c0*(2p)  \u03c3*(2p)\n           |       |       |\n           |       |       |\n\u03c3(2s)  \u03c0(2p)  \u03c3(2p)  \n         ---------\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The bonding orbitals (\u03c3(2s), \u03c0(2p), \u03c3(2p)) are filled first, followed by the antibonding orbitals (\u03c3*(2s), \u03c0*(2p)).<\/li>\n\n\n\n<li>Since C\u2082\u00b2\u207b has 10 electrons, we place 8 electrons in the bonding orbitals and 2 electrons in the antibonding orbitals.<\/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\">d. Diamagnetic or Paramagnetic?<\/h3>\n\n\n\n<p>To determine whether C\u2082\u00b2\u207b is diamagnetic or paramagnetic, we look at the electron configuration in the molecular orbitals:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>If all electrons are paired, the molecule is diamagnetic.<\/li>\n\n\n\n<li>If there are unpaired electrons, the molecule is paramagnetic.<\/li>\n<\/ul>\n\n\n\n<p>For C\u2082\u00b2\u207b, the electrons in the antibonding orbitals (\u03c0*(2p)) are paired, meaning there are no unpaired electrons.<\/p>\n\n\n\n<p>Thus, C\u2082\u00b2\u207b is <strong>diamagnetic<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">e. Comparison Between Lewis Structure and MO Diagram:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Lewis Structure<\/strong> provides a simple representation, showing bonding as shared electron pairs (in this case, a double bond between two carbon atoms with lone pairs on each atom).<\/li>\n\n\n\n<li><strong>MO Diagram<\/strong> provides a more detailed quantum mechanical picture, considering molecular orbitals formed from the linear combination of atomic orbitals, and it shows how electrons occupy these orbitals based on energy levels.<\/li>\n<\/ul>\n\n\n\n<p>In both models, the bond order is 2 (due to the double bond), and both indicate that the molecule is diamagnetic.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">f. Other Possible Lewis Structures That Do Not Obey the Octet Rule:<\/h3>\n\n\n\n<p>In this case, another possible structure might be one where carbon atoms share more electrons, violating the octet rule. For example:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A structure with a triple bond and no lone pairs on either carbon atom could also be considered, though it does not satisfy the octet rule for carbon.<\/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\">g. Favored Lewis Structure:<\/h3>\n\n\n\n<p>The structure with a double bond and lone pairs on each carbon atom is the most favorable because:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>It satisfies the octet rule for both carbon atoms.<\/li>\n\n\n\n<li>It minimizes formal charges, with both carbon atoms having no formal charge.<\/li>\n\n\n\n<li>The structure is consistent with the observed experimental properties of C\u2082\u00b2\u207b and is the most stable configuration based on the principles of Lewis bonding.<\/li>\n<\/ul>\n\n\n\n<p>Thus, the structure with a double bond and lone pairs on both carbon atoms is the most favorable.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>a. Determine the Lewis dot structure of C2? that satisfies the octet rule for carbon. b. Describe the bonding in terms of VBT for your Lewis structure, i.e. how are the bonds formed. c. Draw and label the MO diagram for C2?. Draw the orbital overlap next to the MO levels. The complete MO should [&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-180905","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180905","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=180905"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/180905\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=180905"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=180905"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=180905"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}