{"id":199035,"date":"2025-03-10T17:07:36","date_gmt":"2025-03-10T17:07:36","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=199035"},"modified":"2025-03-10T17:07:38","modified_gmt":"2025-03-10T17:07:38","slug":"draw-the-lewis-structure-of-h2co-2","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/10\/draw-the-lewis-structure-of-h2co-2\/","title":{"rendered":"Draw the Lewis structure of H2CO"},"content":{"rendered":"\n

Draw the Lewis structure of H2<\/sub>CO, then answer the following questions.<\/p>\n\n\n\n

The electron group geometry of H2<\/sub>CO is –linear bent or angular\/ trigonal planar\/ tetrahedral\/ trigonal pyramid.<\/p>\n\n\n\n

The molecular shape of H2<\/sub>CO is —linear bent or angular\/ trigonal planar\/ tetrahedral\/ trigonal pyramid.<\/p>\n\n\n\n

The C-O bond is –polar\/ non polar.<\/p>\n\n\n\n

The H-C-H bond angle is approximately –90\u00b0 \/109\u00b0\/ 120\u00b0\/180\u00b0.<\/p>\n\n\n\n

A H2<\/sub>CO molecule is –polar\/ nonpolar.<\/p>\n\n\n\n

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

Lewis Structure of H\u2082CO (Formaldehyde)<\/strong><\/h3>\n\n\n\n

Formaldehyde (H\u2082CO) consists of one carbon (C), one oxygen (O), and two hydrogen (H) atoms. The Lewis structure<\/strong> of H\u2082CO is drawn as follows:<\/p>\n\n\n\n

    \n
  1. Determine the total number of valence electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
      \n
    • Carbon (C) has 4<\/strong> valence electrons.<\/li>\n\n\n\n
    • Oxygen (O) has 6<\/strong> valence electrons.<\/li>\n\n\n\n
    • Hydrogen (H) has 1<\/strong> valence electron (\u00d72 = 2<\/strong> total).<\/li>\n\n\n\n
    • Total: 4 + 6 + 2 = 12<\/strong> valence electrons.<\/li>\n<\/ul>\n\n\n\n
        \n
      1. Arrange the atoms:<\/strong><\/li>\n<\/ol>\n\n\n\n
          \n
        • Carbon is the central atom<\/strong> since it is the least electronegative (excluding hydrogen).<\/li>\n\n\n\n
        • Oxygen and hydrogen are attached to carbon.<\/li>\n<\/ul>\n\n\n\n
            \n
          1. Distribute electrons:<\/strong><\/li>\n<\/ol>\n\n\n\n
              \n
            • Carbon forms two single bonds<\/strong> with hydrogen.<\/li>\n\n\n\n
            • Carbon forms a double bond<\/strong> with oxygen.<\/li>\n\n\n\n
            • The oxygen has two lone pairs<\/strong>.<\/li>\n<\/ul>\n\n\n\n

              Thus, the Lewis structure<\/strong> is:<\/p>\n\n\n\n

                    O\n     \u2551\nH \u2014 C \u2014 H<\/code><\/pre>\n\n\n\n
              Oxygen has a double bond<\/strong> with carbon, and each hydrogen forms a single bond<\/strong> with carbon.<\/p>\n\n\n\n
              \n\n\n\n
              Answers to the Questions:<\/strong><\/h3>\n\n\n\n
                \n
              1. Electron Group Geometry:<\/strong> Trigonal Planar<\/strong><\/li>\n<\/ol>\n\n\n\n
                  \n
                • The electron geometry is determined by the number of electron domains (bonding and lone pairs) around the central atom. Carbon has three regions<\/strong> of electron density (two single bonds to hydrogen and one double bond to oxygen), which leads to a trigonal planar<\/strong> electron geometry.<\/li>\n<\/ul>\n\n\n\n
                    \n
                  1. Molecular Shape:<\/strong> Trigonal Planar<\/strong><\/li>\n<\/ol>\n\n\n\n
                      \n
                    • Since there are no lone pairs on carbon, the molecular shape remains trigonal planar<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                        \n
                      1. C-O Bond Polarity:<\/strong> Polar<\/strong><\/li>\n<\/ol>\n\n\n\n
                          \n
                        • The carbon-oxygen bond is polar because oxygen is significantly more electronegative than carbon, causing a partial negative charge on oxygen and a partial positive charge on carbon.<\/li>\n<\/ul>\n\n\n\n
                            \n
                          1. H-C-H Bond Angle:<\/strong> 120\u00b0<\/strong><\/li>\n<\/ol>\n\n\n\n
                              \n
                            • In a trigonal planar<\/strong> structure, the ideal bond angle is 120\u00b0<\/strong>.<\/li>\n<\/ul>\n\n\n\n
                                \n
                              1. Polarity of the H\u2082CO Molecule:<\/strong> Polar<\/strong><\/li>\n<\/ol>\n\n\n\n
                                  \n
                                • The molecule is polar because the C=O bond creates a dipole moment<\/strong>, which does not cancel out due to the asymmetrical distribution of charge.<\/li>\n<\/ul>\n\n\n\n
                                  \n\n\n\n
                                  Explanation (300 Words)<\/strong><\/h3>\n\n\n\n
                                  The structure of formaldehyde (H\u2082CO) consists of a central carbon atom bonded to two hydrogen atoms and one oxygen atom via a double bond. To determine its electron geometry<\/strong>, we count the electron domains around carbon, which include three regions<\/strong> (one double bond and two single bonds). Since three regions of electron density adopt a trigonal planar<\/strong> arrangement to minimize repulsion, the electron geometry<\/strong> of H\u2082CO is trigonal planar<\/strong>.<\/p>\n\n\n\n
                                  The molecular shape<\/strong> is also trigonal planar<\/strong>, since there are no lone pairs on the central carbon to distort the geometry. The C-O bond is polar<\/strong> due to the electronegativity difference between carbon (~2.5) and oxygen (~3.5), making oxygen partially negative and carbon partially positive. This dipole does not cancel out because the molecular shape is asymmetric.<\/p>\n\n\n\n
                                  Since the molecule follows a trigonal planar geometry, the H-C-H bond angle is approximately 120\u00b0<\/strong>, consistent with the expected angles in a planar structure.<\/p>\n\n\n\n
                                  Regarding molecular polarity, although the H-C bonds are relatively nonpolar, the strong C=O bond dipole<\/strong> dominates the molecule’s polarity, making H\u2082CO a polar molecule<\/strong>. This polarity allows H\u2082CO to dissolve in polar solvents like water and participate in hydrogen bonding interactions via its oxygen atom.<\/p>\n\n\n\n
                                  Overall, the shape, bond polarity, and molecular polarity of H\u2082CO significantly influence its physical and chemical properties, including its ability to act as a reactive electrophile in organic synthesis and biological systems.<\/p>\n","protected":false},"excerpt":{"rendered":"
                                  Draw the Lewis structure of H2CO, then answer the following questions. The electron group geometry of H2CO is –linear bent or angular\/ trigonal planar\/ tetrahedral\/ trigonal pyramid. The molecular shape of H2CO is —linear bent or angular\/ trigonal planar\/ tetrahedral\/ trigonal pyramid. The C-O bond is –polar\/ non polar. The H-C-H bond angle is approximately […]<\/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-199035","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199035","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=199035"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/199035\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=199035"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=199035"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=199035"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}