{"id":225888,"date":"2025-06-04T17:57:17","date_gmt":"2025-06-04T17:57:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225888"},"modified":"2025-06-04T17:57:19","modified_gmt":"2025-06-04T17:57:19","slug":"considering-the-c2cl4-molecule","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/considering-the-c2cl4-molecule\/","title":{"rendered":"Considering the C2Cl4 molecule"},"content":{"rendered":"\n<p>Considering the C2Cl4 molecule: a) Draw the box and arrow configuration and the Lewis structure of C2Cl4. b) Draw the overlap of the orbitals (Hybridization diagram) of the C2Cl4 molecule. c) Indicate the quantity and type of bond (\u03c3 or \u03c0 ) that results. d) Indicate the approximate value of the angle Cl-C-Cl. e) Explain whether the C2Cl4 molecule will tend to orient or not under the influence of an electric field<\/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<p>Certainly! Let\u2019s walk through each part of the question for the <strong>C\u2082Cl\u2084 (tetrachloroethylene)<\/strong> molecule.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>a) Box and Arrow Configuration &amp; Lewis Structure<\/strong><\/h3>\n\n\n\n<p><strong>Box and Arrow Configuration:<\/strong><\/p>\n\n\n\n<p>Each <strong>carbon (C)<\/strong> has 4 valence electrons. Each <strong>chlorine (Cl)<\/strong> has 7 valence electrons.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>For <strong>C (1s\u00b2 2s\u00b2 2p\u00b2)<\/strong> \u2192 4 valence electrons: <code>C: 1s\u00b2 | 2s\u00b2 | 2p: \u2191 \u2191 (two unpaired in 2p)<\/code><\/li>\n<\/ul>\n\n\n\n<p>Carbon needs to form 4 bonds \u2192 it undergoes <strong>sp\u00b2 hybridization<\/strong>.<\/p>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>Cl     Cl\n \\     \/\n  C == C\n \/     \\\nCl     Cl\n<\/code><\/pre>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each C shares a <strong>double bond<\/strong> (C=C), and two single bonds with Cl atoms.<\/li>\n\n\n\n<li>Each Cl has 3 lone pairs.<\/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>b) Hybridization (Orbital Overlap) Diagram<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each <strong>C<\/strong> atom is <strong>sp\u00b2 hybridized<\/strong>:\n<ul class=\"wp-block-list\">\n<li>Three <strong>sp\u00b2 hybrid orbitals<\/strong> form \u03c3 bonds: two with Cl atoms and one with the other carbon.<\/li>\n\n\n\n<li>One <strong>unhybridized p orbital<\/strong> on each carbon forms the \u03c0 bond in the C=C double bond.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n<p><strong>Overlap:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u03c3 (C\u2013Cl): overlap of C(sp\u00b2) \u2013 Cl(p)<\/li>\n\n\n\n<li>\u03c3 (C\u2013C): overlap of C(sp\u00b2) \u2013 C(sp\u00b2)<\/li>\n\n\n\n<li>\u03c0 (C=C): lateral overlap of unhybridized p orbitals on each C<\/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>c) Bond Types and Quantity<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>C=C<\/strong>: 1 \u03c3 and 1 \u03c0 bond<\/li>\n\n\n\n<li><strong>4 C\u2013Cl bonds<\/strong>: 4 \u03c3 bonds<\/li>\n<\/ul>\n\n\n\n<p><strong>Total Bonds:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>\u03c3 bonds<\/strong>: 5 (4 C\u2013Cl + 1 C\u2013C)<\/li>\n\n\n\n<li><strong>\u03c0 bonds<\/strong>: 1 (from C=C)<\/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>d) Cl\u2013C\u2013Cl Bond Angle<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Since each carbon is <strong>sp\u00b2 hybridized<\/strong>, bond angles are approximately <strong>120\u00b0<\/strong>.<\/li>\n\n\n\n<li>So, \u2220Cl\u2013C\u2013Cl \u2248 <strong>120\u00b0<\/strong><\/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>e) Electric Field Orientation (Polarity)<\/strong><\/h3>\n\n\n\n<p><strong>C\u2082Cl\u2084 is a nonpolar molecule.<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Despite individual <strong>C\u2013Cl<\/strong> bonds being <strong>polar<\/strong>, the molecule is <strong>symmetric<\/strong>.<\/li>\n\n\n\n<li>The <strong>dipoles cancel<\/strong> due to the tetra-substituted symmetric structure.<\/li>\n<\/ul>\n\n\n\n<p><strong>Conclusion:<\/strong><br>C\u2082Cl\u2084 <strong>does not orient<\/strong> in an electric field because it has <strong>no net dipole moment<\/strong> (nonpolar molecule).<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary Explanation<\/strong><\/h3>\n\n\n\n<p>Tetrachloroethylene (C\u2082Cl\u2084) is a planar, symmetrical molecule featuring a central C=C double bond with each carbon atom also bonded to two chlorine atoms. In drawing the <strong>box and arrow configuration<\/strong>, each carbon has four valence electrons, forming bonds via <strong>sp\u00b2 hybridization<\/strong>. Each carbon uses three sp\u00b2 orbitals to form \u03c3 bonds\u2014two with chlorine atoms and one with the other carbon atom. The remaining unhybridized p orbital on each carbon overlaps side-to-side to form a \u03c0 bond, completing the double bond between the carbon atoms.<\/p>\n\n\n\n<p>The <strong>Lewis structure<\/strong> illustrates the connectivity, with a C=C double bond and four single C\u2013Cl bonds. The <strong>orbital hybridization diagram<\/strong> confirms the geometry and bonding: 4 \u03c3 bonds with chlorine atoms and 1 \u03c3 + 1 \u03c0 bond between carbons.<\/p>\n\n\n\n<p>In terms of bonding, the molecule has <strong>5 \u03c3 bonds<\/strong> and <strong>1 \u03c0 bond<\/strong>. Due to the sp\u00b2 hybridization, each carbon center forms bond angles of approximately <strong>120\u00b0<\/strong>, including the Cl\u2013C\u2013Cl angles.<\/p>\n\n\n\n<p>Although each C\u2013Cl bond is polar, C\u2082Cl\u2084 is a <strong>nonpolar molecule overall<\/strong>. Its symmetry ensures the dipole moments of the bonds cancel out. Thus, <strong>the molecule does not orient in an electric field<\/strong>, unlike polar molecules which align due to net dipole moments.<\/p>\n\n\n\n<p>This combination of bonding, hybridization, and symmetry makes tetrachloroethylene a chemically stable and nonpolar compound, with wide use in industrial applications as a solvent.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner7-187.jpeg\" alt=\"\" class=\"wp-image-225889\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Considering the C2Cl4 molecule: a) Draw the box and arrow configuration and the Lewis structure of C2Cl4. b) Draw the overlap of the orbitals (Hybridization diagram) of the C2Cl4 molecule. c) Indicate the quantity and type of bond (\u03c3 or \u03c0 ) that results. d) Indicate the approximate value of the angle Cl-C-Cl. e) Explain [&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 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