{"id":225359,"date":"2025-06-04T07:27:00","date_gmt":"2025-06-04T07:27:00","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=225359"},"modified":"2025-06-04T07:27:02","modified_gmt":"2025-06-04T07:27:02","slug":"draw-the-lewis-structure-for-dichloroethane-c2h4cl2-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/04\/draw-the-lewis-structure-for-dichloroethane-c2h4cl2-2\/","title":{"rendered":"Draw the Lewis Structure for dichloroethane, C2H4Cl2"},"content":{"rendered":"\n

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

\"\"<\/figure>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Let’s analyze the molecule 1,2-dichloroethane (C\u2082H\u2084Cl\u2082)<\/strong> and answer the questions in detail, including the correct Lewis structure, geometry, and bonding.<\/p>\n\n\n\n

\n\n\n\n

Answers:<\/strong><\/h3>\n\n\n\n

a. Electron geometry around each carbon atom:<\/strong><\/p>\n\n\n\n

    \n
  • Tetrahedral<\/strong>, because each carbon is bonded to four atoms (single bonds only).<\/li>\n<\/ul>\n\n\n\n

    b. Molecular geometry around each carbon atom:<\/strong><\/p>\n\n\n\n

      \n
    • Tetrahedral<\/strong> for the same reason as above.<\/li>\n<\/ul>\n\n\n\n

      c. Bond angle around each carbon atom:<\/strong><\/p>\n\n\n\n

        \n
      • Approximately 109.5\u00b0<\/strong>, the ideal bond angle for tetrahedral geometry.<\/li>\n<\/ul>\n\n\n\n

        d. How many sigma bonds are in this molecule?<\/strong><\/p>\n\n\n\n

          \n
        • 7 sigma bonds<\/strong>:\n
            \n
          • 1 C\u2013C<\/li>\n\n\n\n
          • 2 C\u2013Cl<\/li>\n\n\n\n
          • 4 C\u2013H<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n

            e. How many pi bonds are in this molecule?<\/strong><\/p>\n\n\n\n

              \n
            • 0 pi bonds<\/strong>, since all bonds are single bonds (sigma only).<\/li>\n<\/ul>\n\n\n\n

              f. Redraw the Lewis structure (3D shape):<\/strong>
              Draw the two central carbon atoms connected by a single bond. Each carbon is bonded to two other atoms \u2014 one Cl and two H. The 3D structure should show tetrahedral geometry for each carbon.<\/p>\n\n\n\n

              g. What sort of hybrid orbitals are used by each carbon atom?<\/strong><\/p>\n\n\n\n

                \n
              • sp\u00b3 hybridization<\/strong>
                Each carbon forms four sigma bonds, consistent with sp\u00b3 hybrid orbitals.<\/li>\n<\/ul>\n\n\n\n

                h. Is C\u2082H\u2084Cl\u2082 polar?<\/strong><\/p>\n\n\n\n

                  \n
                • Yes<\/strong>, it is polar. The two chlorine atoms create a dipole moment because of their high electronegativity, and their positions prevent cancellation.<\/li>\n<\/ul>\n\n\n\n

                  i. What is the molarity of a solution made by dissolving 45.9 g of C\u2082H\u2084Cl\u2082 in 600.0 mL of solution?<\/strong><\/p>\n\n\n\n

                    \n
                  1. Molar mass of C\u2082H\u2084Cl\u2082 = (2\u00d712.01) + (4\u00d71.008) + (2\u00d735.45) = 98.96 g\/mol<\/strong><\/li>\n\n\n\n
                  2. Moles = 45.9 g \/ 98.96 g\/mol = 0.464 moles<\/strong><\/li>\n\n\n\n
                  3. Volume in liters = 600.0 mL = 0.600 L<\/strong><\/li>\n\n\n\n
                  4. Molarity = moles \/ volume = 0.464 mol \/ 0.600 L = 0.773 M<\/strong><\/li>\n<\/ol>\n\n\n\n
                    \n\n\n\n

                    Explanation<\/strong><\/h3>\n\n\n\n

                    1,2-Dichloroethane (C\u2082H\u2084Cl\u2082) is a haloalkane where two carbon atoms are bonded together and each is also bonded to hydrogen atoms and one chlorine atom. In terms of bonding and shape, each carbon forms four single (sigma) bonds, meaning it adopts sp\u00b3 hybridization<\/strong>, resulting in a tetrahedral geometry<\/strong> with bond angles close to 109.5\u00b0<\/strong>.<\/p>\n\n\n\n

                    Each C\u2013H, C\u2013Cl, and C\u2013C bond in the molecule is a sigma bond. There are no pi bonds in this structure because there are no double or triple bonds. Thus, the molecule has 7 sigma bonds<\/strong> in total.<\/p>\n\n\n\n

                    The Lewis structure shows each carbon atom in the center, bonded to two hydrogen atoms (for one carbon), one chlorine atom, and the other carbon atom. The three-dimensional geometry should reflect the tetrahedral shape around each carbon. Because chlorine is more electronegative than hydrogen and placed asymmetrically, the molecule is polar<\/strong>, with a net dipole moment pointing roughly from the hydrogen side toward the chlorine side.<\/p>\n\n\n\n

                    To calculate the molarity<\/strong> of a solution, we use the formula:
                    Molarity (M) = moles of solute \/ liters of solution<\/strong>.
                    Given 45.9 g of C\u2082H\u2084Cl\u2082 and its molar mass of 98.96 g\/mol, we calculate 0.464 moles. Dissolving this in 600.0 mL (0.600 L) gives a solution with a molarity of 0.773 M<\/strong>.<\/p>\n\n\n\n

                    Overall, this analysis demonstrates how molecular structure, polarity, hybridization, and stoichiometry interrelate in the study of simple organic molecules like dichloroethane.<\/p>\n\n\n\n

                    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

                    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 […]<\/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-225359","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225359","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=225359"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/225359\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=225359"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=225359"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=225359"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}