{"id":203864,"date":"2025-03-21T02:22:41","date_gmt":"2025-03-21T02:22:41","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=203864"},"modified":"2025-03-21T02:22:43","modified_gmt":"2025-03-21T02:22:43","slug":"draw-the-lewis-structure-of-chooch","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/21\/draw-the-lewis-structure-of-chooch\/","title":{"rendered":"Draw the Lewis structure of (CHO)OCH"},"content":{"rendered":"\n

Draw the Lewis structure of (CHO)OCH<\/p>\n\n\n\n

3<\/p>\n\n\n\n

and then choose the appropriate set of molecular geometries of the three central atoms. Your answer choice is independent of the orientation of your drawn structure.<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

The molecular formula (CHO)OCH\u2083<\/strong> corresponds to methyl formate<\/strong>, an ester with the structural formula HCOOCH\u2083<\/strong>. Below is the Lewis structure<\/strong> and the correct molecular geometries<\/strong> of the three central atoms.<\/p>\n\n\n\n

Step 1: Draw the Lewis Structure<\/strong><\/h3>\n\n\n\n
    \n
  1. Count the valence electrons:<\/strong>\n
      \n
    • Carbon (C) (2 atoms)<\/strong> = 4 \u00d7 2 = 8<\/li>\n\n\n\n
    • Hydrogen (H) (4 atoms)<\/strong> = 1 \u00d7 4 = 4<\/li>\n\n\n\n
    • Oxygen (O) (2 atoms)<\/strong> = 6 \u00d7 2 = 12<\/li>\n\n\n\n
    • Total valence electrons<\/strong> = 24 electrons<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
    • Structure of Methyl Formate (HCOOCH\u2083):<\/strong>\n
        \n
      • The carbonyl (C=O) group<\/strong> is present in the formate part (HCOO\u207b<\/strong>).<\/li>\n\n\n\n
      • The methoxy group (OCH\u2083)<\/strong> is attached to the second oxygen.<\/li>\n\n\n\n
      • Carbonyl carbon (C=O<\/strong>) has a double bond<\/strong> with oxygen.<\/li>\n\n\n\n
      • The other oxygen forms a single bond<\/strong> with carbon and another single bond<\/strong> with the methyl group (CH\u2083<\/strong>).<\/li>\n<\/ul>\n<\/li>\n\n\n\n
      • Formal charge check and octet rule:<\/strong>\n
          \n
        • All atoms follow the octet rule.<\/li>\n\n\n\n
        • The Lewis structure is valid.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

          Step 2: Identify Molecular Geometries<\/strong><\/h3>\n\n\n\n

          The three central atoms are:<\/p>\n\n\n\n

            \n
          1. Carbonyl Carbon (C=O)<\/strong>\n
              \n
            • Geometry:<\/strong> Trigonal Planar (AX\u2083)<\/strong><\/li>\n\n\n\n
            • Bonding pairs:<\/strong> 3 (C-H, C-O, C=O)<\/li>\n\n\n\n
            • Lone pairs:<\/strong> 0<\/li>\n\n\n\n
            • Bond angle:<\/strong> ~120\u00b0<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Oxygen in OCH\u2083 group (O-CH\u2083)<\/strong>\n
                \n
              • Geometry:<\/strong> Bent (AX\u2082E\u2082)<\/strong><\/li>\n\n\n\n
              • Bonding pairs:<\/strong> 2 (O-C, O-CH\u2083)<\/li>\n\n\n\n
              • Lone pairs:<\/strong> 2<\/li>\n\n\n\n
              • Bond angle:<\/strong> <109.5\u00b0 (~104.5\u00b0)<\/strong><\/li>\n<\/ul>\n<\/li>\n\n\n\n
              • Methoxy Carbon (O-CH\u2083 group)<\/strong>\n
                  \n
                • Geometry:<\/strong> Tetrahedral (AX\u2084)<\/strong><\/li>\n\n\n\n
                • Bonding pairs:<\/strong> 4 (C-H, C-H, C-H, C-O)<\/li>\n\n\n\n
                • Lone pairs:<\/strong> 0<\/li>\n\n\n\n
                • Bond angle:<\/strong> ~109.5\u00b0<\/strong><\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

                  Final Answer:<\/strong><\/h3>\n\n\n\n

                  Trigonal planar \u2013 Bent \u2013 Tetrahedral<\/strong><\/p>\n\n\n\n

                  Now, I will generate the Lewis structure image.<\/p>\n\n\n\n

                  Here is the Lewis structure of methyl formate (HCOOCH\u2083), showing the molecular geometries:<\/p>\n\n\n\n

                    \n
                  • Trigonal planar (C=O carbonyl carbon)<\/strong><\/li>\n\n\n\n
                  • Bent (Oxygen in OCH\u2083 group)<\/strong><\/li>\n\n\n\n
                  • Tetrahedral (Methyl carbon, CH\u2083 group)<\/strong><\/li>\n<\/ul>\n\n\n\n

                    Let me know if you need further modifications!<\/p>\n\n\n\n

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

                    Draw the Lewis structure of (CHO)OCH 3 and then choose the appropriate set of molecular geometries of the three central atoms. Your answer choice is independent of the orientation of your drawn structure. The correct answer and explanation is: The molecular formula (CHO)OCH\u2083 corresponds to methyl formate, an ester with the structural formula HCOOCH\u2083. Below […]<\/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-203864","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203864","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=203864"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/203864\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=203864"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=203864"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=203864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}