{"id":225355,"date":"2025-06-04T07:23:40","date_gmt":"2025-06-04T07:23:40","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225355"},"modified":"2025-06-04T07:23:43","modified_gmt":"2025-06-04T07:23:43","slug":"draw-the-lewis-structure-for-dichloroethane-c2h4cl2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/draw-the-lewis-structure-for-dichloroethane-c2h4cl2\/","title":{"rendered":"Draw the Lewis Structure for dichloroethane, C2H4Cl2"},"content":{"rendered":"\n<p>Draw the Lewis Structure for dichloroethane, C2H4Cl2: The two carbon atoms are in the center of this molecule. What is the electron geometry around each carbon atom? What is the molecular geometry around each carbon atom? What is the bond angle around each carbon atom? How many sigma bonds are in this molecule? How many pi bonds are in this molecule? Redraw your Lewis structure to show the three-dimensional shape that you described in parts above. What sort of hybrid orbitals are used by each carbon atom? Is CH2Cl2 polar? If so, add an arrow to your drawing above to show the dipole moment. What is the molarity of a solution made by dissolving 45.9 g of CH2Cl2 in enough water to make 600.0 mL of solution?<\/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>Let&#8217;s answer each part about <strong>dichloroethane (C\u2082H\u2084Cl\u2082)<\/strong> step by step:<\/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. Lewis Structure of C\u2082H\u2084Cl\u2082<\/strong><\/h3>\n\n\n\n<p>There are two isomers of dichloroethane, but the most common is <strong>1,2-dichloroethane<\/strong> (ClCH\u2082CH\u2082Cl).<\/p>\n\n\n\n<p><strong>Lewis Structure:<\/strong><\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>H   H       H   H\n \\ \/         \\ \/\n  C \u2014 C   or  C \u2014 C\n \/ \\         \/ \\\nCl  H       H  Cl\n<\/code><\/pre>\n\n\n\n<p>Each carbon has four single bonds: two to hydrogen atoms, one to the other carbon, and one to a chlorine atom.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Electron Geometry Around Each Carbon Atom<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Electron geometry:<\/strong> <strong>Tetrahedral<\/strong><\/li>\n\n\n\n<li>Because each carbon is bonded to four atoms (no lone pairs), it adopts a tetrahedral shape.<\/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. Molecular Geometry Around Each Carbon Atom<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molecular geometry:<\/strong> <strong>Tetrahedral<\/strong><\/li>\n\n\n\n<li>Same as electron geometry due to lack of 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>4. Bond Angles Around Each Carbon Atom<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Ideal tetrahedral bond angle: <strong>109.5\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>5. Sigma (\u03c3) and Pi (\u03c0) Bonds<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each single bond is a <strong>sigma bond<\/strong>.<\/li>\n\n\n\n<li>Total:\n<ul class=\"wp-block-list\">\n<li><strong>6 C-H \u03c3 bonds<\/strong><\/li>\n\n\n\n<li><strong>2 C-Cl \u03c3 bonds<\/strong><\/li>\n\n\n\n<li><strong>1 C-C \u03c3 bond<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n<li><strong>Total sigma bonds = 9<\/strong><\/li>\n\n\n\n<li><strong>Total pi bonds = 0<\/strong> (no double or triple bonds)<\/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>6. 3D Shape Drawing<\/strong><\/h3>\n\n\n\n<p>In 3D, each carbon atom forms a tetrahedron. Here&#8217;s a simplified 3D drawing for one carbon:<\/p>\n\n\n\n<pre class=\"wp-block-code\"><code>       Cl\n        |\nH \u2014 C \u2014 H\n      \/\n    H\n<\/code><\/pre>\n\n\n\n<p>(Use wedges\/dashes to show 3D bonds if drawing formally.)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>7. Hybrid Orbitals Used<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Each carbon atom uses <strong>sp\u00b3 hybrid orbitals<\/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>8. Is CH\u2082Cl\u2082 Polar?<\/strong><\/h3>\n\n\n\n<p>Yes, <strong>CH\u2082Cl\u2082 is polar<\/strong> due to:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Asymmetrical geometry.<\/li>\n\n\n\n<li>More electronegative chlorine atoms pulling electron density.<\/li>\n<\/ul>\n\n\n\n<p><strong>Dipole moment:<\/strong> Arrows point from C toward each Cl.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>9. Molarity Calculation<\/strong><\/h3>\n\n\n\n<p>Molar mass of CH\u2082Cl\u2082 (methylene chloride) =<br>= 12.01 + (2 \u00d7 1.008) + (2 \u00d7 35.45)<br>= <strong>84.93 g\/mol<\/strong><\/p>\n\n\n\n<p><strong>Moles =<\/strong> 45.9 g \u00f7 84.93 g\/mol \u2248 <strong>0.540 moles<\/strong><br><strong>Volume =<\/strong> 600.0 mL = 0.600 L<br><strong>Molarity =<\/strong> 0.540 mol \u00f7 0.600 L = <strong>0.900 M<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 <strong>Summary<\/strong><\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Geometry: Tetrahedral at both carbons.<\/li>\n\n\n\n<li>Bond angle: ~109.5\u00b0.<\/li>\n\n\n\n<li>Sigma bonds: 9.<\/li>\n\n\n\n<li>Pi bonds: 0.<\/li>\n\n\n\n<li>Hybridization: sp\u00b3.<\/li>\n\n\n\n<li>Polarity: Polar.<\/li>\n\n\n\n<li>Molarity: <strong>0.900 M<\/strong> for CH\u2082Cl\u2082 solution.<\/li>\n<\/ul>\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-banner4-302.jpeg\" alt=\"\" class=\"wp-image-225356\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw the Lewis Structure for dichloroethane, C2H4Cl2: The two carbon atoms are in the center of this molecule. What is the electron geometry around each carbon atom? What is the molecular geometry around each carbon atom? What is the bond angle around each carbon atom? How many sigma bonds are in this molecule? How many [&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-225355","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225355","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=225355"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225355\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225355"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225355"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225355"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}