{"id":269180,"date":"2025-07-24T11:27:16","date_gmt":"2025-07-24T11:27:16","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=269180"},"modified":"2025-07-24T11:30:42","modified_gmt":"2025-07-24T11:30:42","slug":"draw-a-lewis-structure-for-the-molecule-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/24\/draw-a-lewis-structure-for-the-molecule-below\/","title":{"rendered":"Draw a Lewis structure for the molecule below."},"content":{"rendered":"\n

Draw a Lewis structure for the molecule below. Draw any resonance structures. Determine the formal charges on atoms in each of your resonance structures. Based on this, determine which resonance structure(s) are most important. Circle the most important contributor(s) to the resonance hybrid. If they are all equivalent, circle them all. COF2<\/p>\n\n\n\n

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

To draw the Lewis structure of COF2<\/strong> (carbonyl fluoride), we follow these steps:<\/p>\n\n\n\n

Step 1: Count the Total Number of Valence Electrons<\/h3>\n\n\n\n
    \n
  • Carbon (C):<\/strong> 4 valence electrons<\/li>\n\n\n\n
  • Oxygen (O):<\/strong> 6 valence electrons<\/li>\n\n\n\n
  • Fluorine (F):<\/strong> 7 valence electrons (per fluorine atom)<\/li>\n<\/ul>\n\n\n\n

    Total = 4 (C) + 6 (O) + 7(2) (F) = 24 valence electrons<\/strong>.<\/p>\n\n\n\n

    Step 2: Draw the Skeleton Structure<\/h3>\n\n\n\n

    Place the least electronegative atom in the center (Carbon), and attach the atoms around it. In this case:<\/p>\n\n\n\n

      \n
    • Carbon (C) will be in the center.<\/li>\n\n\n\n
    • Oxygen (O) and Fluorine (F) will surround it.<\/li>\n<\/ul>\n\n\n\n

      So, the basic skeleton looks like this: C \u2013 O \u2013 F<\/strong> with a fluorine atom attached to the carbon.<\/p>\n\n\n\n

      Step 3: Complete the Octets<\/h3>\n\n\n\n

      Start by completing the octet for the outer atoms (oxygen and fluorine). Fluorine will have a single bond with carbon, and oxygen will have a double bond with carbon.<\/p>\n\n\n\n

        \n
      • Oxygen<\/strong> will form a double bond with carbon<\/strong> (C=O).<\/li>\n\n\n\n
      • Each fluorine<\/strong> will form a single bond with carbon<\/strong> (C-F).<\/li>\n<\/ul>\n\n\n\n

        Step 4: Add Remaining Electrons<\/h3>\n\n\n\n

        After placing bonds, we can assign the remaining electrons to the atoms. The oxygen atom will have 4 additional electrons (in lone pairs), and each fluorine atom will have 6 lone pairs.<\/p>\n\n\n\n

        Step 5: Formal Charge Calculation<\/h3>\n\n\n\n

        The formal charge<\/strong> (FC) on an atom is calculated using the formula:FC=Valence electrons\u2212(Lone pair electrons+Bonding electrons2)FC = \\text{Valence electrons} – (\\text{Lone pair electrons} + \\frac{\\text{Bonding electrons}}{2})FC=Valence electrons\u2212(Lone pair electrons+2Bonding electrons\u200b)<\/p>\n\n\n\n

        Let’s calculate for each atom in the structure:<\/p>\n\n\n\n

          \n
        1. Carbon (C):<\/strong>\n
            \n
          • Valence electrons = 4<\/li>\n\n\n\n
          • Bonding electrons = 4 (two single bonds to F and one double bond to O)<\/li>\n\n\n\n
          • Lone pair electrons = 0<\/li>\n\n\n\n
          • Formal charge on C = 4 – (0 + 4\/2) = 0<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
          • Oxygen (O):<\/strong>\n
              \n
            • Valence electrons = 6<\/li>\n\n\n\n
            • Bonding electrons = 4 (from the double bond to C)<\/li>\n\n\n\n
            • Lone pair electrons = 4 (two lone pairs)<\/li>\n\n\n\n
            • Formal charge on O = 6 – (4 + 4\/2) = 0<\/strong>.<\/li>\n<\/ul>\n<\/li>\n\n\n\n
            • Fluorine (F):<\/strong>\n
                \n
              • Valence electrons = 7<\/li>\n\n\n\n
              • Bonding electrons = 2 (one single bond to C)<\/li>\n\n\n\n
              • Lone pair electrons = 6 (three lone pairs)<\/li>\n\n\n\n
              • Formal charge on F = 7 – (6 + 2\/2) = 0<\/strong>.<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n\n\n\n

                Step 6: Resonance Structures<\/h3>\n\n\n\n

                In the case of COF2<\/strong>, no significant resonance structures are possible, as moving electrons around would either violate octet rules or result in high formal charges on atoms. The given structure is already stable and does not have any alternative forms with lower energy.<\/p>\n\n\n\n

                Conclusion:<\/h3>\n\n\n\n
                  \n
                • The Lewis structure for COF2<\/strong> has no formal charges, with the carbon in the center double-bonded to oxygen and single-bonded to two fluorine atoms.<\/li>\n\n\n\n
                • There are no major resonance structures to consider here, as all formal charges are zero and the structure is stable.<\/li>\n<\/ul>\n\n\n\n

                  Thus, this is the most important resonance structure<\/strong> (which is also the only one).<\/p>\n\n\n\n

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

                  Draw a Lewis structure for the molecule below. Draw any resonance structures. Determine the formal charges on atoms in each of your resonance structures. Based on this, determine which resonance structure(s) are most important. Circle the most important contributor(s) to the resonance hybrid. If they are all equivalent, circle them all. COF2 The Correct Answer […]<\/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-269180","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/269180","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=269180"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/269180\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=269180"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=269180"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=269180"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}